diff --git a/.gitignore b/.gitignore index 25d07db1..77e9d4f3 100644 --- a/.gitignore +++ b/.gitignore @@ -9,3 +9,4 @@ tsconfig.tsbuildinfo private/ .replit replit.nix +content/SEM_6/.obsidian \ No newline at end of file diff --git a/README.md b/README.md index 01e2c588..bfa65511 100644 --- a/README.md +++ b/README.md @@ -1,17 +1 @@ -# Quartz v4 - -> β€œ[One] who works with the door open gets all kinds of interruptions, but [they] also occasionally gets clues as to what the world is and what might be important.” β€” Richard Hamming - -Quartz is a set of tools that helps you publish your [digital garden](https://jzhao.xyz/posts/networked-thought) and notes as a website for free. - -πŸ”— Read the documentation and get started: https://quartz.jzhao.xyz/ - -[Join the Discord Community](https://discord.gg/cRFFHYye7t) - -## Sponsors - -

- - - -

+notes diff --git a/content/Measure-Theory-Notes/Measure_Theory/Info.md b/content/Measure-Theory-Notes/Measure_Theory/Info.md deleted file mode 100644 index b0a9b165..00000000 --- a/content/Measure-Theory-Notes/Measure_Theory/Info.md +++ /dev/null @@ -1 +0,0 @@ -![[Pasted image 20260107001139.png]] \ No newline at end of file diff --git a/content/SEM_6/DSA/Lecture 01 - Jan 5.md b/content/SEM_6/DSA/Lecture 01 - Jan 5.md new file mode 100644 index 00000000..3a0e390b --- /dev/null +++ b/content/SEM_6/DSA/Lecture 01 - Jan 5.md @@ -0,0 +1,62 @@ +## Basic Algorithms + +### Sum of Two Numbers + +**Pseudo Code:** +1. Read $a, b$ +2. $S = a + b$ +3. Print $S$ + +**Implementation (C):** +```c +#include + +int main() { + int a, b, sum; + + printf("Enter two numbers: "); + scanf("%d %d", &a, &b); + + sum = a + b; + + printf("Sum = %d\n", sum); + + return 0; +} +``` + +### Sum of m Numbers + +**Algorithm:** +1. Read $m$ numbers +2. Initialize $S = 0$ +3. For $i = 0$ to $m - 1$: + - $S = S + A[i]$ +4. Print $S$ + +### Matrix Sum - 2D Array + +**Algorithm:** +1. Read $m, n$ (dimensions of matrix) +2. Initialize $S = 0$ +3. For $i = 0$ to $n - 1$: + - For $j = 0$ to $m - 1$: + - Read $A[i][j]$ + - $S = S + A[i][j]$ +4. Print $S$ + +### Recursion: Factorial + +**Recurrence Relation:** +$$\text{Fact}(n) = \begin{cases} 1 & \text{if } n = 0 \\ n \times \text{Fact}(n-1) & \text{else} \end{cases}$$ + +### Algorithm Properties + +- **Input:** Zero or more inputs +- **Output:** At least 1 output +- **Efficiency:** Minimal time and space complexity + +### Topics to Cover + +- Prime Checkers +- Fibonacci Number diff --git a/content/SEM_6/DSA/Lecture 02 - Jan 6.md b/content/SEM_6/DSA/Lecture 02 - Jan 6.md new file mode 100644 index 00000000..8d0de9e1 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 02 - Jan 6.md @@ -0,0 +1,79 @@ +## Algorithm Properties + +1. Input $\geq 0$ (zero or more) +2. Output $\geq 1$ (at least one) +3. Correct +4. Language Independent +5. Unambiguous +6. Efficiency + - **Time Complexity:** Amount of time taken + - **Space Complexity:** Amount of memory used +--- +## Time Complexity Analysis + +### Example 1: Sum of Two Numbers + +``` +(1) Read a, b β†’ Assignment Statement (2 time units) +(2) S = a + b β†’ Constant time 'c' (2 time units) + Logic Operation +(3) Print S β†’ Print operation (1 time unit) +``` + +**Total Time:** $$T(n) = 2 + 2 + 1 = 5 \rightarrow O(1)$$ + +### Example 2: Sum of n Numbers + +``` +S = 0 β†’ 1 time unit +for i = 0 to n β†’ 2n time units + S = S + A[i] β†’ 2n time units +``` + +**Analysis:** +- $n$ numbers input operations +- Loop executes $n$ times +- Each iteration: constant time operations + +$$T(n) = O(n)$$ + +### Example 3: Matrix Addition - 2D Array + +``` +for i = 0 to n β†’ Outer loop runs n times +for j = 0 to m β†’ Inner loop runs m times + C[i,j] = A[i,j] + B[i,j] β†’ Executed n Γ— m times +``` + +**Time Complexity:** $$T(n,m) = O(n \times m)$$ + +**Note:** For square matrices where $n = m$, complexity is $O(n^2)$ + +--- +## Fibonacci Algorithm + +**Iterative Algorithm:** Time Complexity $O(n)$ + +``` +Fib(n) + f1 = 0 + f2 = 1 + + for i = 2 to n + temp = f1 + f2 + f1 = f2 + f2 = temp + + return f2 +``` + +**Recursive Definition:** + +$$ +\text{Fib}(n) = +\begin{cases} +0 & \text{if } n = 0 \\ +1 & \text{if } n = 1 \\ +\text{Fib}(n-1) + \text{Fib}(n-2) & \text{otherwise} +\end{cases} +$$ diff --git a/content/SEM_6/DSA/Lecture 03 - Jan 8.md b/content/SEM_6/DSA/Lecture 03 - Jan 8.md new file mode 100644 index 00000000..77da173f --- /dev/null +++ b/content/SEM_6/DSA/Lecture 03 - Jan 8.md @@ -0,0 +1,91 @@ +## Fibonacci Time Complexity (Continued) + +**Recurrence Relation:** +$$f(n) = f(n-1) + f(n-2)$$ + +**Time Complexity:** +$$T(n) = T(n-1) + T(n-2) + c$$ + +$$T(n) = a^n$$ + +where: +$$a^2 = a^1 + a^0$$ +$$a^2 - a - 1 = 0$$ +$$a = \frac{1 \pm \sqrt{5}}{2}$$ + +Therefore: +$$T(n) = \left(\frac{1+\sqrt{5}}{2}\right)^n \text{ or } \left(\frac{1-\sqrt{5}}{2}\right)^n = O(\phi^n)$$ + +where $\phi \approx 1.618$ (golden ratio) + +*We take the larger value for upper bound* + +--- +## Array Search Problems + +**Example Array:** $[1, 100, 20, 35, 45] = A[5]$ + +**Task:** Check if $60$ is present + +### Linear Search Algorithm + +**Given:** Array $A$ with $n$ elements, search for value $X$ + +``` +for i = 0 to n-1: + if A[i] == X: + return i // Found at index i +return -1 // Not found +``` + +**Time Complexity:** $T(n) = O(n)$ + +--- +## Binary Search + +**Example:** If Array sorted: $[1, 20, 35, 45, 100]$ + +**Finding 20:** +- Check mid of array $\rightarrow 35$ +- Since $20 < 35$: Check left half + +### Binary Search Algorithm + +``` +low = 0 +high = n - 1 +while (low ≀ high): + mid = (low + high) / 2 + if A[mid] == X: + return mid // Found + else if A[mid] > X: + high = mid - 1 // Search left half + else: + low = mid + 1 // Search right half +return -1 // Not found +``` + +**Time Complexity Analysis:** +$$T(n) = c + T(n/2)$$ + +where $c$ is constant time for comparison and mid calculation. + +**Solving the Recurrence:** + +Base case: $T(1) = 1$ + +$$T(n) = c + T(n/2)$$ +$$T(n) = c + c + T(n/4)$$ +$$T(n) = c + c + c + ... + T(n/2^i)$$ + +When $n/2^i = 1$: +$$n = 2^i$$ +$$\log_2 n = i$$ + +Therefore: +$$T(n) = c \cdot i + 1 = c \log n + 1$$ + +**Final Complexity:** +$$\boxed{T(n) = O(\log n)}$$ + +> **Note:** Efficient but array must be sorted! diff --git a/content/SEM_6/DSA/Lecture 04 - Jan 12.md b/content/SEM_6/DSA/Lecture 04 - Jan 12.md new file mode 100644 index 00000000..55448e7c --- /dev/null +++ b/content/SEM_6/DSA/Lecture 04 - Jan 12.md @@ -0,0 +1,55 @@ +## Introduction to Algorithms + +**ADT (Abstract Data Type):** +- Specifies what data values it can hold +- Specifies what operations can be performed on it + +**Data Structure:** How we organize data in memory + +**Array Operations:** Inserting, deleting, searching, traversing + +### Basic Operations + +**Unified Computational Model:** Each basic operation takes constant time (1 unit) + +### Linear Search + +Given an array of $n$ elements, check if $x$ is present by examining each element. + +**Best Case:** $O(1)$ (element found at first position) + +**Worst Case:** $O(n)$ (element at end or not present) + +**Average Case:** $O(n)$ + +--- +## Asymptotic Analysis of Algorithms + +Asymptotic analysis describes how an algorithm behaves as input size increases. + +--- +## Time Complexity Notations + +$$O \qquad \Omega \qquad \Theta$$ +$$(\text{Upper Bound}) \quad (\text{Lower Bound}) \quad (\text{Tight Bound})$$ + +**Example:** +$$T(n) = 100n + 5$$ + +## Big-O Notation + +**Definition:** $f(n) = O(g(n))$ if there exist positive constants $C$ and $n_0$ such that: + +$$f(n) \leq C \cdot g(n) \quad \text{for all } n \geq n_0$$ + +**Example:** + +Given $T(n) = 100n + 5$, we want to show $T(n) = O(n)$ + +$$100n + 5 \leq C \cdot n$$ + +Choosing $C = 101$ and $n_0 = 5$: + +$$100n + 5 \leq 101n \quad \text{for all } n \geq 5$$ + +Therefore, $100n + 5 = O(n)$ diff --git a/content/SEM_6/DSA/Lecture 05 - Jan 13.md b/content/SEM_6/DSA/Lecture 05 - Jan 13.md new file mode 100644 index 00000000..541a7142 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 05 - Jan 13.md @@ -0,0 +1,88 @@ +## Asymptotic Notations + +### Big O Notation - Upper Bound + +$$f(n) = O(g(n))$$ if $\exists$ constants $C > 0$ & $n_0 > 0$ + +such that $f(n) \leq C \cdot g(n)$ $\forall$ $n \geq n_0$ + +### Big Ξ© - Lower Bound + +$$f(n) = \Omega(g(n))$$ if $\exists$ constants $C > 0$, $n_0 > 0$ + +such that $f(n) \geq C \cdot g(n)$ $\forall$ $n \geq n_0$ + +### Big Θ - Tight Bound + +$$f(n) = \Theta(g(n))$$ if $\exists$ constants $C_1 > 0$, $C_2 > 0$ & $n_0 > 0$ + +such that $$C_1 \cdot g(n) \leq f(n) \leq C_2 \cdot g(n)$$ $\forall$ $n \geq n_0$ + +--- +## Examples + +**Example 1: Proving Big O** + +Given: $$f(n) = 100n^2 + 2n + 5$$ + +Show that $f(n) = O(n^2)$: + +$$100n^2 + 2n + 5 \leq 100n^2 + 2n^2 + 5n^2 = 107n^2$$ + +for all $n \geq 1$ + +Therefore, $f(n) = O(n^2)$ with $C = 107$ and $n_0 = 1$ + +**Example 1b: Sum Formula** + +Given: $$f(n) = \frac{n(n+1)}{2}$$ + +Show that $f(n) = \Omega(n^2)$: + +$$\frac{n(n+1)}{2} = \frac{n^2 + n}{2} \geq \frac{n^2}{2} \quad \text{for all } n \geq 1$$ + +Therefore, $f(n) = \Omega(n^2)$ with $C = \frac{1}{2}$ and $n_0 = 1$ + +Since $\frac{n(n+1)}{2} \leq n^2$ for $n \geq 1$, we also have $f(n) = O(n^2)$ + +Thus, $f(n) = \Theta(n^2)$ + +--- + +**Example 2: Recurrence Relation** + +Given: $$T(n) = T(n/2) + 1, \quad T(1) = 1$$ + +Solving by substitution: + +$$T(n) = T(n/2) + 1$$ +$$= T(n/4) + 1 + 1 = T(n/4) + 2$$ +$$= T(n/8) + 1 + 1 + 1 = T(n/8) + 3$$ + +After $i$ iterations: +$$T(n) = T(n/2^i) + i$$ + +When $n/2^i = 1$, we have $n = 2^i$, so $i = \log_2 n$ + +Substituting: +$$T(n) = T(1) + \log n = 1 + \log n$$ + +Therefore: $$T(n) = O(\log n)$$ + +--- + +**Example 3: Additional Recurrence Relations** + +Given: $$f(n) = 6\sqrt{n}$$ + +This is $O(n)$ since $\sqrt{n} < n$ for $n > 1$ + +**Recurrence 1:** +$$T(n) = 9 \cdot T(n/3) + n$$ + +By Master Theorem: $T(n) = O(n^2)$ + +**Recurrence 2:** +$$T(n) = 2T(n/2) + n$$ + +By Master Theorem: $T(n) = O(n \log n)$ diff --git a/content/SEM_6/DSA/Lecture 06 - Jan 15.md b/content/SEM_6/DSA/Lecture 06 - Jan 15.md new file mode 100644 index 00000000..6a827078 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 06 - Jan 15.md @@ -0,0 +1,99 @@ +## Master Theorem + +**Given:** Let $a \geq 1$, $b > 1$ + +$$T(n) = a \cdot T(n/b) + f(n)$$ + +### Case 1 + +If $f(n) = O(n^{\log_b a - \varepsilon})$ for $\varepsilon > 0$ + +$$T(n) = \Theta(n^{\log_b a})$$ + +### Case 2 + +If $f(n) = \Theta(n^{\log_b a})$, then: + +$$T(n) = \Theta(n^{\log_b a} \log n)$$ + +### Case 3 + +If $f(n) = \Omega(n^{\log_b a + \varepsilon})$ for $\varepsilon > 0$, and $a \cdot f(n/b) \leq c \cdot f(n)$ for some $c < 1$ and sufficiently large $n$ (regularity condition), then: + +$$T(n) = \Theta(f(n))$$ + +--- +## Examples + +### Example 1 + +$$T(n) = 2T(n/2) + n^2$$ + +**Solution:** +- $a = 2$, $b = 2$, $f(n) = n^2$ +- $\log_b a = \log_2 2 = 1$ +- $n^{\log_b a} = n^1 = n$ + +Compare $f(n) = n^2$ with $n^{\log_b a} = n$: +- $n^2 = \Omega(n^{1+\varepsilon})$ for $\varepsilon = 1$ + +Check regularity condition: +$$a \cdot f(n/b) = 2 \cdot (n/2)^2 = 2 \cdot \frac{n^2}{4} = \frac{n^2}{2} \leq c \cdot n^2$$ + +for $c = 1/2 < 1$ βœ“ + +By **Case 3**: $$T(n) = \Theta(n^2)$$ + +### Example 2 + +$$T(n) = T(n/2) + 1$$ + +**Solution:** +- $a = 1$, $b = 2$, $f(n) = 1 = \Theta(1)$ +- $\log_b a = \log_2 1 = 0$ +- $n^{\log_b a} = n^0 = 1$ + +Since $f(n) = \Theta(n^{\log_b a}) = \Theta(1)$ + +By **Case 2**: $$T(n) = \Theta(n^0 \log n) = \Theta(\log n)$$ + +--- +## Substitution Method + +**Example 1:** Solve using Substitution: +$$T(n) = 2 \cdot T(n/2) + n - 1$$ + +**Solution:** + +Guess: $T(n) = O(n \log n)$ + +Prove by induction that $T(n) \leq c \cdot n \log n$ for some $c > 0$: + +$$T(n) = 2T(n/2) + n - 1$$ +$$\leq 2 \cdot c \cdot (n/2) \log(n/2) + n - 1$$ +$$= c \cdot n (\log n - \log 2) + n - 1$$ +$$= c \cdot n \log n - c \cdot n + n - 1$$ +$$\leq c \cdot n \log n$$ + +if $c \geq 1$ and $n$ is sufficiently large. + +Therefore, $T(n) = O(n \log n)$ + +**Example 2:** Solve using Substitution: +$$T(n) = 2 \cdot T(\sqrt{n}) + \log n$$ + +**Solution:** + +Let $m = \log n$, so $n = 2^m$: + +$$T(2^m) = 2 \cdot T(2^{m/2}) + m$$ + +Let $S(m) = T(2^m)$: + +$$S(m) = 2 \cdot S(m/2) + m$$ + +By Master Theorem (Case 2): $S(m) = \Theta(m \log m)$ + +Substituting back: $T(n) = T(2^m) = S(m) = \Theta(m \log m) = \Theta(\log n \log \log n)$ + +Therefore, $T(n) = \Theta(\log n \log \log n)$ diff --git a/content/SEM_6/DSA/Lecture 07 - Jan 19.md b/content/SEM_6/DSA/Lecture 07 - Jan 19.md new file mode 100644 index 00000000..74a0e053 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 07 - Jan 19.md @@ -0,0 +1,69 @@ +## Selection Sort + +**Randomly given elements:** + +``` +for (i = 0 to n): + min = i + for (j = i+1 to n): + if (A[j] < A[minimum]): + minimum = j + Swap(A[min], A[i]) +``` + +### Example: [75, 36, 4, 9, 81, 65] + +**Swap function:** +``` +S, temp = a +a = b +b = temp +``` + +**Complexity:** +- **Best case:** $O(n)$ +- **Average:** $O(n^2)$ +- **Iterations:** $n-1$ times + +### Step-by-step Execution + +``` +75 36 4 9 81 65 + 4 36 75 9 81 65 + 4 9 75 36 81 65 + 4 9 36 75 81 65 +``` + +**Iterations:** $n-1$ comparisons, $n-2$, ... + +**Total comparisons:** +$$n-1 + n-2 + n-3 + \dots + 1$$ + +$$\frac{n(n-1)}{2} = \frac{n^2 - n}{2} = \frac{n^2}{2}$$ + +**Time Complexity:** $$O(n^2)$$ + +--- + +## Bubble Sort + +**Example:** [75, 36, 4, 9, 81, 65] + +``` +for j = (0 to n): + for (i = 0 to n-j-1): + if (A[i] > A[i+1]): + Swap(A[i], A[i+1]) +``` + +### Execution + +``` +36 4 9 75 65 81 β†’ Biggest at right-most +``` + +**Analysis:** $n-1 + n-2 + \dots + 1$ + +**Time Complexity:** $$O(n^2)$$ + +**Best case:** $n-1$ comparison in $O(n)$ diff --git a/content/SEM_6/DSA/Lecture 08 - Jan 20.md b/content/SEM_6/DSA/Lecture 08 - Jan 20.md new file mode 100644 index 00000000..31119e19 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 08 - Jan 20.md @@ -0,0 +1,67 @@ +## Merge Sort + +**Array Length = 8000** + +**Example:** [70, 25, 81, 9, 84, 65] + +Single element arrays always sorted + +**Step 1:** [25, 70] [81] [9, 65] [106] + +**Step 2:** [9, 25, 70, 81, 106] + +### Algorithm + +**Merge Sort (A, beg, r):** + +``` +if (i < r): β†’ c + g = (i + r)/2 β†’ c + + T(n/2) merge-Sort(A, beg) + T(n/2) merge-Sort(A, mid) + merge(A, i, r, g) +``` + +### Merge Function + +**Merge (A, beg, r):** + +``` +n1 = mid + 1 +n2 = r - g + +for i = 1 to n1: β†’ n1 + L[i] = A[i+1] + +for i = 0 to n2: β†’ n2 + R[i] = A[mid+1] + +i = 0, j = 0 + +for k = 1 to r: β†’ n1 + if L[i] < R[j]: ⟹ Θ(n) + A[k] = L[i] + i = i+1 + else: + A[k] = R[j] + j = j+1 +``` + +### Time Complexity Analysis + +$$O(n, m_1 + n)$$ + +where $n > m_1 \cdot 2$ + +$$O(n)$$ + +**Recurrence:** + +$$T(n) = 2T(n/2) + n$$ + +$$T(1) = 0$$ + +$$\Theta(n \log n)$$ + +$$n \cdot \frac{\log^2}{n} \rightarrow \Theta(1)$$ diff --git a/content/SEM_6/DSA/Lecture 09 - Jan 22.md b/content/SEM_6/DSA/Lecture 09 - Jan 22.md new file mode 100644 index 00000000..2e0314da --- /dev/null +++ b/content/SEM_6/DSA/Lecture 09 - Jan 22.md @@ -0,0 +1,48 @@ +## Insertion Sort + +**Assume Array upto 2 upto 50** + +### Example: [51, 26, 3, 65, 91, 8] + +**Step 1:** [26, 51] + +**Step 2:** [26, 51, 3, 65, 91, 8] + +**Step 3:** [26, 51, 3, 65, 91, 8] + +**Final:** [26, 51, 3, 65, 91, 8] β†’ **Sorted** + +### Algorithm + +**Insertion_Sort(A, n):** + +``` +for i = 2 to n: β†’ n + key = A[i] β†’ n-1 + j = i-1 β†’ n-1 + while (j > 0, A[j] > key): β†’ n(n-1)/2 + A[j+1] = A[j] β†’ i-1, i+2 + j = j-1 + + A[j+1] = key +``` + +**Inner loop analysis:** + +For $(k = 1 \text{ to } i)$: +- If $A[i] > \text{key}$: $A[k+1] = A[k]$ +- $A[k] = A[k]$ + +### Time Complexity + +$$O(n^2)$$ + +**Worst Case:** $$O(n^2)$$ + +$$\frac{n(n-1)}{2}$$ + +### Best Case + +**Partly sorted:** [3, 8, 26, 51, 65, 91] + +$$O(n)$$ diff --git a/content/SEM_6/DSA/Lecture 10 - Jan 27.md b/content/SEM_6/DSA/Lecture 10 - Jan 27.md new file mode 100644 index 00000000..139e1708 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 10 - Jan 27.md @@ -0,0 +1,101 @@ +## Divide and Conquer + +**Three Steps:** +1. Divide into subproblem +2. Solve each subproblem +3. Combine soln of each + +## Quick Sort + +**Example:** [2, 5, 8, 3, 9, 4, 1, 7, 10, 6] + +*Find element* - Using this pivot array into 2 subarrays + +**Let pivot = 7** + +$$A_1 = [2, 5, 3, 4, 1, 6, 7] \quad A_2 = [8, 9, 4, 10]$$ + +**Pivot = 3**, Last mid-loc + +$$P = 1$$ + +### Iteration 1 + +``` + 2 5 8 3 1 4 1 7 10 6 +``` + +$$X = 6 > 3 \rightarrow \text{Pivot}$$ + +$$\text{for } i = 1 \text{ to } 1 - 1$$ + +if $A[i] \leq X$: +- $i = i+1$ +- Swap $(A[i], A[k])$ + +``` +2 5 3 4 1 1 7 10 6 +``` + +*Sort in 2 &* + +### Detailed Steps + +**i = 0:** +- $p = 2, i = 0$ β†’ $3$, $4$, $1$, $6$, $9$, $7$, $10$, $8$ +- $2 < 6$ β†’ $i = i+1$ +- Swap $(A[2], A[0])$ + +**Step 1:** $2, 5, 3, 9, 9, 4, 1, 7, 10, 6$ + +$3 < 6$ β†’ $i = i+1$ β†’ Sec $\rightarrow$ $i = i+1$ Swap $\rightarrow$ + +$$9 > 6$$ + +**Step 2:** $2, 5, 3, 4, 9, 9, 1, 7, 10, 6$ + +$9 > 6$ β†’ Swap $4 \rightarrow 8$ + +$$2, 5, 3, 4, 9, 9, 1, 7, 10, 6$$ + +**Step 3:** $1 < 6$ β†’ $i = i+1$ β†’ Swap $9 \& 1$ + +$$2, 5, 3, 4, 1, 8, 9, 7, 10, 6$$ + +**Step 4:** $7 > 6$ β†’ $i = i+1$ β†’ Swap $1 \rightarrow 2$ + +### Final Partition + +$$2 \quad 5 \quad 3 \quad 4 \quad 1 \quad \boxed{6} \quad 9 \quad 7 \quad 10 \quad 8$$ + +$$\underbrace{\text{Less than } 1}_\text{} \quad \underbrace{\text{More than } 1}_\text{}$$ + +### Partition Algorithm + +**Partition (A, beg, r):** + +``` +X = A[r] i = beg +for j = b to r-1: β†’ O(n) + if (A[j] < X): + i = i+1 + +Quick_sort (A(beg, i(A),r(A)) +``` + +- $C_{\text{sort}}(A[1], i-1)$ +- $X = C_{\text{kpim}}$ tem β†’ Partition + +### Quicksort Function + +**Quicksort (A, beg, r):** + +``` +Quicksort(A, beg+1) +``` + +### Time Complexity + +$$O(n \log n)$$ + +$$T(n) = 2T(n/2) + f(n)$$ diff --git a/content/SEM_6/DSA/Lecture 11 - Jan 29.md b/content/SEM_6/DSA/Lecture 11 - Jan 29.md new file mode 100644 index 00000000..90276f16 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 11 - Jan 29.md @@ -0,0 +1,74 @@ +## Arrays vs Linked Lists + +### Array Limitations + +**int A[100]:** Array of size 100. If we want to add more than 100? + +**int B[6(r)]:** Copy $A$ to $B$ β†’ takes time + +``` +100 β”Œβ”€β”€β” + β”œβ”€β”€β”€ OS β”Œβ”€β”€β” P β”Œβ”€β”€β” Ξ» β”Œβ”€β”€β” + β”œβ”€β”€β”€ β””β”€β”€β”˜ β””β”€β”€β”˜ β””β”€β”€β”˜ ∞ + β”œβ”€β”€β”€ Memory Can't declare + β”œβ”€β”€β”€ Array of 100 + β””β”€β”€β”˜ +``` + +### Linked List + +**Linked List:** + +``` + β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” +head β”‚8 │─┼──→ β”‚20│─┼──→ β”‚30β”‚X│──→ β”‚40β”‚Kβ”‚ +node β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ + null pointer +``` + +**Insert 35:** + +``` +β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” +β”‚8 │─┼──→ β”‚30│─┼──→ β”‚8 │─┼───────→ β”‚40β”‚Xβ”‚ +β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ + ↓ + β”Œβ”€β”€β”¬β”€β” + β”‚5 β”‚ β”‚ + β””β”€β”€β”΄β”€β”˜ +``` + +**Insert 100 after 10:** + +``` +β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” β”Œβ”€β”€β”¬β”€β” +β”‚8 │─┼──→ β”‚80│─┼──→ β”‚5 │─┼───────→ β”‚100β”‚?β”‚ +β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ β””β”€β”€β”΄β”€β”˜ +``` + +### Struct Node Definition + +```c +struct node { + int rollno + arr name[20] + int marks + int(*x) *Next +} +``` + +### Create Node Function + +**Create_node C mb:** + +```c +2 struct node *C +C->data data +c->next = NULL +``` + +### Next Node Guess + +**3rd Feb** + +Topics: Recursion, Big O, Sorting diff --git a/content/SEM_6/DSA/Lecture 12.md b/content/SEM_6/DSA/Lecture 12.md new file mode 100644 index 00000000..15f419ce --- /dev/null +++ b/content/SEM_6/DSA/Lecture 12.md @@ -0,0 +1,80 @@ +## Linked List Properties + +### Doubly Linked List + +In a doubly linked list: +- `next` points to successor +- `prev` points to the predecessor + +### Special Cases + +- If `prev = NIL`, the node is the **first element (head)** +- If `next = NIL`, the node is the **last element (tail)** +- The list maintains a pointer `L.head` to the first element +- If `L.head = NIL`, the list is empty + +### Classification + +Linked lists can be classified as: +- Singly linked / doubly linked +- Sorted / unsorted +- Linear / circular + +### Circular Linked List + +In a circular linked list: +- The `prev` of the head points to tail +- The `next` of the tail points to head + +--- + +## Linked List - Definition + +**Linearly ordered data structure** which contains many data types + +- Node are connected by pointers +- Each node has two parts: + - **data** - the actual value + - **pointer/next** - a reference to the next node in the list + +``` +β”Œβ”€β”€β”€β”€β”¬β”€β”€β”€β”€β” β”Œβ”€β”€β”€β”€β”¬β”€β”€β”€β”€β” β”Œβ”€β”€β”€β”€β”¬β”€β”€β”€β”€β” +β”‚dataβ”‚next│───→│dataβ”‚next│───→│dataβ”‚NULLβ”‚ +β””β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”˜ β””β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”˜ β””β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”˜ +``` + +### Struct Node + +```c +struct node { + int data; + struct node * ptr; +}; +``` + +### Creating a New Node + +```c +struct node *new_node + = struct node + + malloc(sizeof(node)) + +if (new != NULL) { + new->data = x; + new->ptr = NULL; +} + +return new; +``` + +### Key Properties + +- A linked list is a **linear data structure** where elements are ordered using pointers, not array indices +- Unlike arrays, linked lists **do not require contiguous memory** +- Linked lists are well suited for **dynamic sets** where elements are frequently inserted and deleted + +### Each Element (Node) Stores + +- A key (data) +- A pointer to the next element +- In doubly linked lists, a pointer to the previous element diff --git a/content/SEM_6/DSA/Lecture 13.md b/content/SEM_6/DSA/Lecture 13.md new file mode 100644 index 00000000..e69de29b diff --git a/content/SEM_6/DSA/Lecture 14.md b/content/SEM_6/DSA/Lecture 14.md new file mode 100644 index 00000000..e69de29b diff --git a/content/SEM_6/DSA/Lecture 15.md b/content/SEM_6/DSA/Lecture 15.md new file mode 100644 index 00000000..e69de29b diff --git a/content/SEM_6/DSA/Lecture 16 - Feb 10.md b/content/SEM_6/DSA/Lecture 16 - Feb 10.md new file mode 100644 index 00000000..36528966 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 16 - Feb 10.md @@ -0,0 +1,43 @@ +## Stack + +**Last In First Out β†’ LIFO** + +**First In Last Out β†’ FILO** + +### Operations + +- **Infix** +- **Prefix** +- **Postfix** + +We use stack + +### Stack Empty Check + +**Stack empty (S):** +- If $(top = -0)$: Prints "empty" + +### Insert β†’ Push Operation + +``` +Push (S, X) +if (top β‰₯ S) + S(top) = x +else + top = top + 1 + S(top) = x +``` + +### Remove β†’ Pop Operation + +``` +Pop (S): + x = S (top) + top = top - 1 +``` + +### See β†’ Peek Operation + +``` +Peek (S): +``` diff --git a/content/SEM_6/DSA/Lecture 17 - Feb 19.md b/content/SEM_6/DSA/Lecture 17 - Feb 19.md new file mode 100644 index 00000000..988d6015 --- /dev/null +++ b/content/SEM_6/DSA/Lecture 17 - Feb 19.md @@ -0,0 +1,41 @@ +## Infix to Postfix Conversion + +### Algorithm 1: Infix to Postfix + +**Require:** Infix expression $E$ + +**Ensure:** Postfix expression $P$ + +``` +1: Initialize empty stack S +2: Initialize empty output string P +3: for each token x in E do +4: if x is an operand then +5: Append x to P +6: else if x is '(' then +7: Push x onto S +8: else if x is ')' then +9: while top of S is not '(' do +10: Append pop(S) to P +11: end while +12: Pop '(' from S +13: else if x is an operator then +14: while S is not empty AND precedence(top(S)) β‰₯ precedence(x) do +15: Append pop(S) to P +16: end while +17: Push x onto S +18: end if +19: end for +20: while S is not empty do +21: Append pop(S) to P +22: end while +23: return P +``` + +### Example + +**Infix:** +$$a - b * (c - d + e * (f + a - 1c)) + i - j$$ + +**Postfix:** +$$abcd - efg + h - * + * ij - * -$$ diff --git a/content/SEM_6/DSA/PYQP/Endsem - Apr 2025.md b/content/SEM_6/DSA/PYQP/Endsem - Apr 2025.md new file mode 100644 index 00000000..fe4aaebd --- /dev/null +++ b/content/SEM_6/DSA/PYQP/Endsem - Apr 2025.md @@ -0,0 +1,249 @@ +### Answer all questions. Each question carries 1 Mark. + +1. Depth First Search algorithm uses ______ data structure for its implementation. + a) Queue + b) Stack + +2. Supposing we use Dijkstra's single source shortest path algorithm on an undirected graph. What constraint must we have for the algorithm to work and why? + +3. Given an arbitrary connected graph $G = (V, E)$ with edge weights in $\mathbb{R^+}$. Will the minimum cost edge in $G$ (assume there is only one such edge in $G$) always be present in every **Minimum Spanning Tree** (MST) of $G$? If your answer is YES then give a short justification, if it is NO then give a counter example. + +4. A stack of `int` type is implemented using array as the following data type: + + ```c + #define SIZE 20 + typedef struct { + int data[SIZE]; + int top; + } Stack; + ``` + + Fill up the missing codes in the PUSH and POP operations of the Stack. + + ```c + void Push(Stack *s, int d) { + ---- // statement for inserting an item d into the stack top + } + void Pop(Stack *s) { + ---- // statement for removing an item from the stack top + } + ``` + +5. What is(are) the condition(s) for a graph to have a unique Minimum Spanning Tree? + +--- + +### Answer all questions. Each question carries 2.5 Marks. + +1. Let $A$ be an unsorted array of $n$ distinct integers. + a) Describe how you can use a Binary Search Tree (BST) to sort the array $A$. + b) Suppose the input array is already sorted in increasing order. What will be the structure of the BST? + +2. Let a $\text{min} - \text{heap}$ consist of $n$ distinct elements. Where can we find the $\text{maximum}$ element? Justify your answer. + +3. Let $G$ be a connected undirected graph on $n$ vertices with each edge having a cost of 1. What is the cost of a Minimum Spanning Tree of $G$? Justify your answer. + +4. Write insertion and deletion operations in queue using circular linked list with header node at head. + +--- + +## Answer all questions. Each question carries 5 Marks. + +1. Consider the Binary Search Tree below. Draw the binary search tree obtained after performing the deletion of the nodes in the following order (draw the tree after each deletion). Note that while removing a node you can replace it with inorder successor only. + +```mermaid +graph TD + +A((32)) + +A --> B((17)) +A --> C((88)) + +B --> D((8)) +B --> E((28)) + +E --> F((21)) +E --> G((29)) + +C --> H((65)) +C --> I((97)) + +H --> J((54)) +H --> K((82)) + +K --> L((76)) +L --> M((80)) + +I --> N((93)) +N --> O((94)) + +``` + + + a) delete 88 + b) delete 76 + c) delete 28 + +2. Given an array $A[1\ldots n]$ representing a min-heap below (only keys are shown in the figure) and an integer $k$. Answer the questions that follow: + + | Array index | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | + |---|---|---|---|---|---|---|---|---|---|---|---|---|---|---| + | **key** | 3 | 10 | 5 | 13 | 17 | 6 | 11 | 15 | 16 | 21 | 18 | 9 | 8 | 23 | + + a) Give the array after performing one `deleteMin()` operation. (2 Marks) + + b) Design an algorithm to output all the keys in the array that are less than $k$. (For example if $k = 6$ and the array $A$ is as shown below, then your algorithm should output the keys 3 and 5.) (3 Marks) + +3. Draw the Red-Black tree that results after TREE-INSERT is called on the tree in figure with key 36. If the inserted node is colored red, is the resulting tree a red-black tree? What if it is colored black? + +```mermaid +graph TD + +A((26)) + +A --> B((17)) +A --> C((41)) + +B --> D((14)) +B --> E((21)) + +D --> F((10)) +D --> G((16)) + +F --> H((7)) +F --> I((12)) + +H --> J((3)) + +G --> K((15)) + +E --> L((19)) +E --> M((29)) + +L --> N((20)) + +C --> O((30)) +C --> P((97)) + +O --> Q((28)) +O --> R((38)) + +R --> S((35)) +R --> T((39)) + +classDef black fill:#000,color:#fff +classDef red fill:#f66,color:#000 + +class A black +class C black +class D black +class E black +class P black +class G black +class L black +class M black +class Q black +class R black +class H black +class I black + +class B red +class F red +class J red +class K red +class N red +class O red +class S red +class T red + +``` + +4. With the help of a pseudocode, briefly describe how the search operation is performed in a B-tree. + +5. Run the Bellman-Ford algorithm on the directed graph in figure using vertex $f$ as the source. In each pass, relax edges and show the distance $d$ and parent $\pi$ values after each pass. + +```mermaid +graph LR + +a((a)) +b((b)) +c((c)) +e((e)) +f((f)) + +%% Top connections +b -->|5| e +e -->|-2| b + +%% Left connections +a -->|6| b +a -->|7| c + +%% Vertical +b -->|8| c +e -->|7| f + +%% Bottom +c -->|9| f + +%% Diagonals +c -->|-3| e +b -->|-4| f + +%% Reverse edge +f -->|2| a + +``` + +--- + +### Answer all questions. Each question carries 10 Marks. + +1. Consider the graph $G$ in figure. + +```mermaid +graph LR + +a((a)) +b((b)) +c((c)) +d((d)) +e((e)) +f((f)) +g((g)) + +%% Horizontal edges +b ---|2| e +a ---|9| d +d ---|3| g +c ---|6| f + +%% Left connections +a ---|8| b +a ---|9| c +b ---|7| c + +%% Middle connections +b ---|6| d +c ---|7| d +d ---|5| e +d ---|4| f + +%% Right connections +e ---|5| f +e ---|6| g +f ---|3| g + +%% Outer arcs (approximated) +b ---|8| g +a ---|7| f + +``` + + a) Construct a minimum spanning tree (MST) of this graph using **Kruskal's algorithm**. Draw the MST of the graph and find the minimum total weight. Also write the sequence of edges chosen by Kruskal's algorithm and if an edge is not included in the MST, explain why it is discarded. (Example: (x, y) added, (u, v) discarded because it creates the cycle uxxvv.) (5 Marks) + + b) Construct a minimum spanning tree (MST) of this graph using **Prim's algorithm** by assuming that we start with node 'a' in $G$ as the starting node and give the order in which the nodes are added to MST. Draw the MST of the graph and find the minimum total weight. (5 Marks) + +--- +#### Scanned Question Paper +![[Endsem, Vas 2025.pdf]] \ No newline at end of file diff --git a/content/SEM_6/DSA/PYQP/Endsem, Vas 2025.pdf b/content/SEM_6/DSA/PYQP/Endsem, Vas 2025.pdf new file mode 100644 index 00000000..960d2c95 Binary files /dev/null and b/content/SEM_6/DSA/PYQP/Endsem, Vas 2025.pdf differ diff --git a/content/SEM_6/DSA/PYQP/Midsem - Feb 2025.md b/content/SEM_6/DSA/PYQP/Midsem - Feb 2025.md new file mode 100644 index 00000000..43a9a2eb --- /dev/null +++ b/content/SEM_6/DSA/PYQP/Midsem - Feb 2025.md @@ -0,0 +1,115 @@ +### Answer all questions. Each question carries 1 marks. + +1. Write the recurrence relation that can arise in relation to the time complexity of the binary search algorithm. + +2. Consider the following postfix expression with single digit operands: + + `6 2 3 * / 4 2 * + 6 8 * -` + + The top two elements of the stack after the second `*` is evaluated, are: + + i) 6, 3 + ii) 8, 1 + iii) 8, 2 + iv) 6, 2 + +3. Assume that the algorithms considered here sort the input sequences in ascending order. If the input is already in ascending order, which of the following is/are TRUE? + + I. Quicksort runs in $\Theta(n^2)$ time. + II. Bubblesort runs in $\Theta(n^2)$ time. + III. Merge sort runs in $\Theta(n)$ time. + IV. Insertion sort runs in $\Theta(n)$ time. + + (a) I and II only + (b) I and III only + (c) II and IV only + (d) I and IV only + +4. The height of a binary tree is the maximum number of nodes in any root-to-leaf path. The maximum number of nodes in a binary tree of height $h$ is: + +5. Let $T$ be a perfect binary tree with $n$ leaves. Then how many nodes in $T$ have degree 2? + +--- + +### Answer all questions. Each question carries 2.5 marks. + +1. Write the recurrence relation that can arise in relation to the time complexity of the following C function. Solve that recurrence relation and compute its time complexity. + + ```c + int recursive(int n) + { + if (n == 2) + return 1; + else + return recursive(n/2) + recursive(n/2); + } + ``` + +2. Given an array of $n$ numbers. You are asked to pick a number which is not the second largest. Can you propose an $O(1)$ algorithm? If so, write an algorithm; if not, state why. + +3. What type of list implementation (for example, singly linked list, doubly linked list, circular linked list etc.) should be used to perform the concatenation of two lists in $O(1)$ time? Justify your answer. + +4. Given two sorted arrays $A$ and $B$ having numbers of elements $m$ and $n$, respectively. Write a pseudo-code of linear time complexity ($\Theta(\max(m, n))$) to merge $A$ and $B$ into a third array $C$, such that $C$ is also sorted and has $m + n$ elements. + +--- + +### Answer all questions. Each question carries 5 marks. + +1. Solve the following recurrence relations: + + i) $T(n) = T(n - 1) + n$ + + ii) $T(n) = 2T(\sqrt{n}) + \log n$ and $T(1) = 1$ + +2. A circularly linked list is used to represent a Queue. A single variable $p$ is used to access the Queue. To which node (for example front, rear etc.) should $p$ point such that both the operations enQueue and deQueue can be performed in constant time? Justify your answer. + +3. We have studied "Binary Search" algorithm in class. Now consider a "Ternary Search" algorithm in which two mid-indices are maintained; $m_1 = n/3$ and $m_2 = 2n/3$. This divides the array into three parts: left to $m_1$, $m_1$ to $m_2$, and $m_2$ to right. Search is conducted recursively in these three parts. Write pseudo code for the ternary search algorithm and compute its time complexity. + +4. Convert the following infix expressions into postfix expressions using the operator stack. + + i) $A + B * C + D/E/F + P$ + + ii) $A/B * C + D * E * F + P * Q$ + + iii) $P * Q * R + D/E/F + A - B$ + + iv) $P - Q/R + D * E/F$ + + v) $A + M - L * S + (N * M) * P/Q/R * O + B$ + +5. Write an algorithm for level-order traversal of a binary tree. Analyze the time complexity of the algorithm. + +--- + +### Answer all questions. Each question carries 10 marks. + +1. Mr. Ram is a chemist who receives ten medicine boxes with batch numbers 35, 33, 42, 10, 14, 19, 27, 44, 26, 31 printed on them. He always arranges the boxes manually and gets frustrated every time. He thought he would have a lot of problems in the future arranging the boxes if the number of boxes of medicine is large. He wants to make this task easier. As you are a good programmer, Mr. Ram is asking for your help. Write an optimal quick sort algorithm to arrange the boxes in increasing order of batch numbers and show each step of the algorithm in detail using the above sequence of batch numbers. Also, write the best case and worst case time complexity of the algorithm? + +2. Consider the following sequence of stack operations: + + a) $S = \text{push}(S, 1)$ + + b) $S = \text{push}(S, 2)$ + + c) $S = \text{pop}(S)$ + + d) $S = \text{push}(S, 3)$ + + e) $S = \text{push}(S, 4)$ + + f) $S = \text{pop}(S)$ + + g) $S = \text{pop}(S)$ + + h) $S = \text{pop}(S)$ + + Which of the following is the correct order in which elements are popped? + + a) 1, 2, 3, 4 + b) 2, 1, 3, 4 + c) 2, 3, 4, 1 + d) 2, 4, 3, 1 + +--- +#### Scanned Question Paper +![[Midsem, Vas 2025.pdf]] \ No newline at end of file diff --git a/content/SEM_6/DSA/PYQP/Midsem, Vas 2025.pdf b/content/SEM_6/DSA/PYQP/Midsem, Vas 2025.pdf new file mode 100644 index 00000000..2323c743 Binary files /dev/null and b/content/SEM_6/DSA/PYQP/Midsem, Vas 2025.pdf differ diff --git a/content/SEM_6/DSA/PYQP/Quiz 1, Jan 2025.pdf b/content/SEM_6/DSA/PYQP/Quiz 1, Jan 2025.pdf new file mode 100644 index 00000000..937e5aab Binary files /dev/null and b/content/SEM_6/DSA/PYQP/Quiz 1, Jan 2025.pdf differ diff --git a/content/SEM_6/DSA/PYQP/Quiz 2, Feb 2025.pdf b/content/SEM_6/DSA/PYQP/Quiz 2, Feb 2025.pdf new file mode 100644 index 00000000..1d0bef3d Binary files /dev/null and b/content/SEM_6/DSA/PYQP/Quiz 2, Feb 2025.pdf differ diff --git a/content/SEM_6/DSA/PYQP/Quizzes.md b/content/SEM_6/DSA/PYQP/Quizzes.md new file mode 100644 index 00000000..18c7dddc --- /dev/null +++ b/content/SEM_6/DSA/PYQP/Quizzes.md @@ -0,0 +1,7 @@ +### Quiz 1 - Jan 2025 +![[Quiz 1, Jan 2025.pdf]] + +--- +### Quiz 2 - Feb 2025 +{why tf is the quiz on Valentines?} +![[Quiz 2, Feb 2025.pdf]] diff --git a/content/SEM_6/DSA/Question_Papers/Quiz 1.pdf b/content/SEM_6/DSA/Question_Papers/Quiz 1.pdf new file mode 100644 index 00000000..338fa68c Binary files /dev/null and b/content/SEM_6/DSA/Question_Papers/Quiz 1.pdf differ diff --git a/content/SEM_6/DSA/Question_Papers/Quizzes.md b/content/SEM_6/DSA/Question_Papers/Quizzes.md new file mode 100644 index 00000000..b13a2ed8 --- /dev/null +++ b/content/SEM_6/DSA/Question_Papers/Quizzes.md @@ -0,0 +1 @@ +![[Quiz 1.pdf]] \ No newline at end of file diff --git a/content/SEM_6/DSA/credits.md b/content/SEM_6/DSA/credits.md new file mode 100644 index 00000000..0dbf0564 --- /dev/null +++ b/content/SEM_6/DSA/credits.md @@ -0,0 +1,3 @@ +This notes set exists, credits due to **Aman**. + +Thank you for your notes. \ No newline at end of file diff --git a/content/SEM_6/DSA/info.md b/content/SEM_6/DSA/info.md new file mode 100644 index 00000000..ee72992b --- /dev/null +++ b/content/SEM_6/DSA/info.md @@ -0,0 +1,6 @@ +**Course:** Data Structures and Algorithms +**Code:** DSC314 +**Year:** 3 +**Semester:** 6 +**Prerequisites:** NA +**Course Instructor:** Dr Dhanyamol Antony diff --git a/content/SEM_6/Galois_Theory/Assignments/Assignment 01.md b/content/SEM_6/Galois_Theory/Assignments/Assignment 01.md new file mode 100644 index 00000000..9a5ea9eb --- /dev/null +++ b/content/SEM_6/Galois_Theory/Assignments/Assignment 01.md @@ -0,0 +1,54 @@ +### Question 1 + +Let $\phi : \mathbb{Q} \to \mathbb{Q}$ be an automorphism. Prove that $\phi = \mathrm{Id}$. + +--- + +### Question 2 + +Let $p > 0$ be a prime integer and let $\psi : \mathbb{F}_5 \to \mathbb{F}_5$ be an automorphism. Prove that $\psi = \mathrm{Id}$. + +--- + +### Question 3 + +Let $L/K$ be a finite field extension and let $f(x) \in K[x]$ be an irreducible polynomial of degree $> 1$. If $\deg f(x)$ and $[L : K]$ are co-prime, then prove that $f$ does not have any root in $L$. + +--- + +### Question 4 + +Let $p, q > 1$ be two prime integers. + +Prove that +$$ +\mathbb{Q}(\sqrt{p}) \text{ and } \mathbb{Q}(\sqrt{q}) +$$ +are isomorphic as $\mathbb{Q}$-vector spaces but not isomorphic as fields. + +--- + +### Question 5 + +Let $K$ be a finite field with $p^n$ elements, where $p$ is a prime integer and $n \in \mathbb{N}$. + +Prove that +$$ +\operatorname{char}(K) = p. +$$ + +--- + +### Question 6 + +Let $K$ be a field and let $K(x)$ be the field of fractions of the polynomial ring $K[x]$. + +Prove that $K(x)/K$ is an infinite extension. + +--- + +### Question 7 + +Let $K$ be a field and $f_1(x), \dots, f_n(x) \in K[x]$. + +Prove that there exists a field extension $L/K$ such that each $f_i$ has a root in $L$. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Assignments/Assignment 02.md b/content/SEM_6/Galois_Theory/Assignments/Assignment 02.md new file mode 100644 index 00000000..fd3846d6 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Assignments/Assignment 02.md @@ -0,0 +1,99 @@ +### Question 1 + +Let $L/K$ be a field extension and let $\alpha \in L$. +Show that $\alpha$ is algebraic over $K$ if and only if +$$ +K[\alpha] = K(\alpha). +$$ + +--- + +### Question 2 + +Let $L/K$ be an algebraic extension. Let $B \subset L$ be a subring containing $K$. +Show that $B$ is a field. + +--- + +### Question 3 + +If $p$ and $q$ are two distinct primes, show that +$$ +\mathbb{Q}(\sqrt{p}, \sqrt{q}) = \mathbb{Q}(\sqrt{p} + \sqrt{q}). +$$ + +--- + +### Question 4 + +Let +$$ +K = \mathbb{Q}(i) \subset \mathbb{C}. +$$ + +Is the polynomial +$$ +x^3 - 2 +$$ +irreducible over $K$? + +--- + +### Question 5 + +Let $L/K$ be a field extension and let $\alpha \in L$ be algebraic over $K$. + +(i) If +$$ +[K(\alpha) : K] +$$ +is odd, show that +$$ +K(\alpha) = K(\alpha^2). +$$ + +(ii) If +$$ +K(\alpha) = K(\alpha^2), +$$ +does this necessarily imply that +$$ +[K(\alpha) : K] +$$ +is odd? + +--- + +### Question 6 + +Let $E/K$ be a field extension. Let $L, M \subset E$ be two subfields both containing $K$. +Assume both $L/K$ and $M/K$ are finite extensions. + +(i) If +$$ +[LM : K] = [L : K][M : K], +$$ +prove that +$$ +L \cap M = K. +$$ + +(ii) Show that the converse of (i) holds if +$$ +[L : K] = 2 +\quad \text{or} \quad +[M : K] = 2. +$$ + +--- + +### Question 7 + +Let $L/K$ be a field extension and let $\alpha \in L$ be transcendental over $K$. + +(i) Prove that $\alpha$ is algebraic over $K(\alpha^n)$ and that $\alpha^n$ is transcendental over $K$. + +(ii) Find +$$ +[K(\alpha) : K(\alpha^n)]. +$$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Assignments/Assignment 03.md b/content/SEM_6/Galois_Theory/Assignments/Assignment 03.md new file mode 100644 index 00000000..39558d3b --- /dev/null +++ b/content/SEM_6/Galois_Theory/Assignments/Assignment 03.md @@ -0,0 +1,168 @@ +### Question 1 + +Let $f(x) \in K[x]$ be a polynomial of degree $n \ge 1$, and let $L$ be the splitting field of $f(x)$ over $K$. +Show that +$$ +[L : K] \le n!. +$$ + +--- + +### Question 2 + +Let $K$ be a field and let $\overline{K}$ be an algebraic closure of $K$. + +(i) Prove that $\overline{K}$ is infinite. + +(ii) If $K$ is countable, prove that $\overline{K}$ is countable. + +--- + +### Question 3 + +Let $K$ be a field and let $K(t)$ be the field of fractions of $K[t]$. +Prove that $K(t)$ is not algebraically closed. + +--- + +### Question 4 + +Let $L/K$ be a field extension and let +$\sigma : K \to \Omega$ be an embedding into an algebraically closed field $\Omega$. + +Give an example to show that if $L/K$ is not an algebraic extension, then there may not exist an extension of $\sigma$ to $L \to \Omega$. + +--- + +### Question 5 + +Find the splitting field over $\mathbb{Q}$ of: + +(i) +$$ +x^6 - 7 +$$ + +(ii) +$$ +x^6 - 9. +$$ + +--- + +### Question 6 + +Let $\alpha \in \mathbb{C}$ be a real root of +$$ +x^4 - 5 \in \mathbb{Q}[x]. +$$ + +(i) Prove that +$$ +\mathbb{Q}(i\alpha)/\mathbb{Q} +$$ +is a normal extension. + +(ii) Prove that +$$ +\mathbb{Q}(\alpha + i\alpha)/\mathbb{Q}(i\alpha) +$$ +is normal. + +(iii) Prove that +$$ +\mathbb{Q}(\alpha + i\alpha)/\mathbb{Q} +$$ +is not normal. + +--- + +### Question 7 + +(i) Let $K$ be a field and let +$$ +G = \left\{ +\begin{pmatrix} +a & b \\ +0 & 1 +\end{pmatrix} +\;\middle|\; a,b \in K,\; a \ne 0 +\right\}. +$$ + +Show that $G$ is a subgroup of $GL_2(K)$. + +(ii) Let $\mathrm{Aut}_K(K[x])$ denote the group of all $K$-algebra automorphisms of $K[x]$. + +Show that there is a group homomorphism +$$ +G \longrightarrow \mathrm{Aut}_K(K[x]), +\qquad +A \longmapsto \phi_A, +$$ +where +$$ +\phi_A : K[x] \to K[x], +\qquad +x \longmapsto ax + b. +$$ + +Prove that there is a homomorphism +$$ +GL_2(K) \longrightarrow \mathrm{Aut}_K(K(x)), +\qquad +A \longmapsto \phi_A, +$$ +where +$$ +\phi_A : K(x) \to K(x), +\qquad +x \longmapsto \frac{ax+b}{cx+d}, +$$ +for +$$ +A = +\begin{pmatrix} +a & b \\ +c & d +\end{pmatrix}. +$$ + +--- + +### Question 8 + +Let $\mathbb{C}[t]$ be the polynomial ring over $\mathbb{C}$ and let $\mathbb{C}(t)$ be its field of fractions. + +1. Let $G \subset \mathrm{Aut}(\mathbb{C}(t))$ be generated by +$$ +t \mapsto \frac{1}{1-t} +\quad \text{and} \quad +t \mapsto \frac{t-1}{t}. +$$ +Find +$$ +\mathbb{C}(t)^G. +$$ + +2. Let $G$ be generated by +$$ +t \mapsto 1 + t. +$$ + +Show that $G$ is an infinite cyclic group and find +$$ +\mathbb{C}(t)^G. +$$ + +--- + +### Question 9 + +Let $L/K$ and $\Omega/K$ be two field extensions. +Let $\alpha \in L$ be algebraic over $K$ and let $f(x) \in K[x]$ be its minimal polynomial. + +Prove that there is a one-to-one correspondence between + +- the set of $K$-algebra embeddings, $K(\alpha)\longmapsto\Omega$, given by $\mathrm{Hom}_K(K(\alpha), \Omega)$ +- the set of all roots of $f(x)$ in $\Omega$, given by $\phi \longmapsto \phi(\alpha)$. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 01 - Jan 7.md b/content/SEM_6/Galois_Theory/Lecture 01 - Jan 7.md new file mode 100644 index 00000000..8e28edb0 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 01 - Jan 7.md @@ -0,0 +1,295 @@ +## Ring Homomorphisms and Characteristic + +Let $R$ be a commutative ring with unity. + +If $S \subset R$, then +$$ +1_S = 1_R. +$$ + +Any ring homomorphism $f : R \to S$ satisfies +$$ +f(1_R) = 1_S, +$$ +unless $f$ is the zero map (in which case it is not very interesting). + +--- + +Suppose $K$ is a field. Define a map +$$ +\varphi : \mathbb{Z} \to K, \qquad n \mapsto n \cdot 1_K. +$$ + +- $\varphi$ is a ring homomorphism. +- $\varphi(\mathbb{Z}) \subset K$ is an integral domain. +- Hence $\ker(\varphi)$ is a prime ideal in $\mathbb{Z}$. + +Therefore, +$$ +\varphi(\mathbb{Z}) \cong \mathbb{Z}/\ker(\varphi). +$$ + +Since every ideal in $\mathbb{Z}$ is of the form $(a)$, +$$ +\ker(\varphi) = (a), +$$ +where either $a = 0$ or $a = p$ for some prime $p$. + +We define the **characteristic** of $K$ as: + +- $\operatorname{char}(K) = 0$ if $a = 0$, +- $\operatorname{char}(K) = p$ if $a = p > 0$ prime. + +--- + +## Prime Subfield + +**Definition (Prime Subfield):** + +Let $K$ be a field. The smallest subfield of $K$ is called the **prime subfield** of $K$. + +--- + +### Case 1: $\operatorname{char}(K) = 0$ + +Consider +$$ +\varphi : \mathbb{Z} \to K, \qquad n \mapsto n \cdot 1_K. +$$ + +If $\operatorname{char}(K) = 0$, then $\ker(\varphi) = (0)$. + +Hence $\varphi$ extends to a morphism +$$ +\widetilde{\varphi} : \mathbb{Q} \to K, +$$ +defined by +$$ +\widetilde{\varphi}\left(\frac{a}{b}\right) += +\varphi(a)\varphi(b)^{-1}. +$$ + +If $E$ is any subfield of $K$, then +$$ +\varphi(\mathbb{Z}) \subset E +\quad \Rightarrow \quad +\widetilde{\varphi}(\mathbb{Q}) \subset E. +$$ + +Identifying $\mathbb{Q}$ with its image in $K$, we obtain +$$ +\mathbb{Q} \subset E \subset K. +$$ + +Thus $\mathbb{Q}$ is the prime subfield. + +--- + +### Case 2: $\operatorname{char}(K) = p$ + +Again consider +$$ +\varphi : \mathbb{Z} \to K. +$$ + +Now +$$ +\ker(\varphi) = (p). +$$ + +Hence +$$ +\mathbb{F}_p = \mathbb{Z}/(p) +$$ +is a subfield of $K$. + +Therefore the prime subfield of $K$ is $\mathbb{F}_p$. + +--- + +## Field Extension + +**Definition (Field Extension):** + +Let $K$ be a field. A field $L$ is called a **field extension** of $K$ if +$$ +K \subset L. +$$ + +We denote this by $L/K$. + +Viewing $L$ as a vector space over $K$, the **degree of the extension** is defined as +$$ +[L : K] = \dim_K L. +$$ + +--- + +### Examples + +1. $\mathbb{R}/\mathbb{Q}$, $\mathbb{C}/\mathbb{Q}$, $\mathbb{C}/\mathbb{R}$ + All have characteristic $0$. + +2. Let + $$ + L = \{ a + b\sqrt{2} \mid a,b \in \mathbb{Q} \}. + $$ + Then $L$ is a field and + $$ + \mathbb{Q} \subset L. + $$ + A basis of $L$ over $\mathbb{Q}$ is + $$ + \{1, \sqrt{2}\}. + $$ + Hence + $$ + [L : \mathbb{Q}] = 2. + $$ + +--- + +## Finite Fields + +Suppose $K$ is a finite field. + +Consider +$$ +\varphi : \mathbb{Z} \to K. +$$ + +Then +$$ +\ker(\varphi) = (p), +$$ +so +$$ +\mathbb{F}_p = \mathbb{Z}/(p) \subset K. +$$ + +Thus $K$ is a vector space over $\mathbb{F}_p$. + +Since $K$ is finite, +$$ +[K : \mathbb{F}_p] < \infty. +$$ + +Let +$$ +[K : \mathbb{F}_p] = n. +$$ + +Then +$$ +|K| = p^n. +$$ + +Hence every finite field has order $p^n$ for some prime $p$. + +--- + +## Tower Law + +Suppose +$$ +M/L \quad \text{and} \quad L/K. +$$ + +Then $M$ is also an extension of $K$ and: + +1. +$$ +[M : K] = [M : L][L : K], +$$ +if both degrees are finite. + +2. If either $[M : L]$ or $[L : K]$ is infinite, then $[M : K]$ is infinite. + +--- + +### Proof of Multiplicativity + +Assume +$$ +[M : L] < \infty. +$$ + +Let +$$ +\{\alpha_1, \dots, \alpha_r\} +$$ +be a basis of $M$ over $L$. + +Let +$$ +\{\beta_1, \dots, \beta_s\} +$$ +be a basis of $L$ over $K$. + +Take $v \in M$. Then +$$ +v = \sum_{i=1}^r a_i \alpha_i, +\quad a_i \in L. +$$ + +Since each $a_i \in L$, +$$ +a_i = \sum_{j=1}^s b_{ij} \beta_j, +\quad b_{ij} \in K. +$$ + +Hence +$$ +v += +\sum_{i=1}^r +\left( +\sum_{j=1}^s b_{ij} \beta_j +\right) +\alpha_i += +\sum_{i=1}^r \sum_{j=1}^s +b_{ij} \, \beta_j \alpha_i. +$$ + +Thus the set +$$ +\{\alpha_i \beta_j \mid 1 \le i \le r,\ 1 \le j \le s\} +$$ +spans $M$ over $K$. + +To check linear independence, suppose +$$ +\sum_{i=1}^r \sum_{j=1}^s +c_{ij} \alpha_i \beta_j = 0. +$$ + +Rewriting, +$$ +\sum_{j=1}^s +\left( +\sum_{i=1}^r c_{ij} \alpha_i +\right) +\beta_j += +0. +$$ + +Since $\{\beta_j\}$ are linearly independent over $K$, +$$ +\sum_{i=1}^r c_{ij} \alpha_i = 0 +\quad \text{for each } j. +$$ + +Since $\{\alpha_i\}$ are linearly independent over $L$, +$$ +c_{ij} = 0. +$$ + +Therefore the set has $rs$ elements and forms a basis of $M$ over $K$. + +Hence +$$ +[M : K] = rs = [M : L][L : K]. +$$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 02 - Jan 9.md b/content/SEM_6/Galois_Theory/Lecture 02 - Jan 9.md new file mode 100644 index 00000000..7768bd4f --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 02 - Jan 9.md @@ -0,0 +1,270 @@ +## Tower Law (continued) + +$$ +[M:K] = [M:L][L:K] +$$ + +Suppose +$$ +[M:K] = p. +$$ + +--- + +## Adjoining Elements + +Let $L/K$ be a field extension and $\alpha_1, \dots, \alpha_n \in L$. + +- The smallest **subring** of $L$ containing $K$ and $\alpha_1, \dots, \alpha_n$ is denoted by + $$ + K[\alpha_1, \dots, \alpha_n]. + $$ + +- The smallest **subfield** of $L$ containing $K$ and $\alpha_1, \dots, \alpha_n$ is denoted by + $$ + K(\alpha_1, \dots, \alpha_n). + $$ + +--- + +### Proposition + +1. +$$ +K[\alpha_1, \dots, \alpha_n] += +\{ f(\alpha_1, \dots, \alpha_n) \mid f(x_1, \dots, x_n) \in K[x_1, \dots, x_n] \}. +$$ + +2. +$$ +K(\alpha_1, \dots, \alpha_n) += +\left\{ +\frac{f(\alpha_1, \dots, \alpha_n)}{g(\alpha_1, \dots, \alpha_n)} +\;\middle|\; +f,g \in K[x_1, \dots, x_n],\; +g(\alpha_1, \dots, \alpha_n) \neq 0 +\right\}. +$$ + +--- + +### Proof (Sketch) + +- The right-hand side of (1) is a ring. +- The right-hand side of (2) is the field of fractions of the ring in (1). + +Let $R$ be any ring containing $K$ and $\alpha_1, \dots, \alpha_n$. + +Since $\alpha_i \in R$, every monomial in $\alpha_1, \dots, \alpha_n$ lies in $R$. + +Because $K \subset R$, we get +$$ +f(\alpha_1, \dots, \alpha_n) \in R. +$$ + +Hence the description is correct. + +For (2), take the field of fractions of (1). +The proof is analogous. + +--- + +## Generated Subfield + +The field +$$ +K(\alpha_1, \dots, \alpha_n) +$$ +is called the **subfield of $L$ generated by $\alpha_1, \dots, \alpha_n$ over $K$.** + +--- + +## Finitely Generated Extensions + +- Every finite extension is finitely generated. +- A finitely generated extension need not be finite. + +A finitely generated extension is of the form +$$ +K(\alpha_1, \dots, \alpha_n). +$$ + +We say $L/K$ is finitely generated if +$$ +L = K(\alpha_1, \dots, \alpha_n) +$$ +for some $\alpha_1, \dots, \alpha_n \in L$. + +--- + +## Simple Extensions + +An extension $L/K$ is called a **simple extension** if +$$ +L = K(\alpha) +$$ +for some $\alpha \in L$. + +--- + +### Proposition + +If $\alpha, \beta \in L$, then +$$ +K(\alpha)(\beta) += +K(\beta)(\alpha) += +K(\alpha, \beta). +$$ + +#### Proof (Idea) + +Since +$$ +\alpha \in K(\alpha), +$$ +we have +$$ +K \subset K(\alpha). +$$ + +Also +$$ +\beta \in K(\alpha)(\beta). +$$ + +Thus +$$ +K(\alpha, \beta) \subset K(\alpha)(\beta). +$$ + +Similarly, the reverse inclusion holds. + +More generally, +$$ +K(\alpha_1, \dots, \alpha_n) += +K(\alpha_1, \dots, \alpha_{n-1})(\alpha_n). +$$ + +--- + +## Constructing Field Extensions with a Root + +Let $K$ be a field and $f(x) \in K[x]$. + +We want a field extension of $K$ containing a root of $f$. + +Take a maximal ideal $m$ of $K[x]$ containing $(f(x))$. + +Then +$$ +L = K[x]/m +$$ +is a field. + +Let +$$ +\pi : K[x] \to K[x]/m +$$ +be the natural projection and set +$$ +\alpha = \pi(x). +$$ + +Since $K \hookrightarrow K[x] \to K[x]/m$, we get +$$ +K \subset L. +$$ + +Thus $L$ is a field extension of $K$. + +Now, +$$ +f(\alpha) += +f(\pi(x)) += +\pi(f(x)) += +0 \text{ in } L. +$$ + +So $\alpha$ is a root of $f$ in $L$. + +--- + +## Algebraic Construction via a Root + +Suppose $f(x)$ is irreducible over $K$ and $\alpha$ is a root. + +Define +$$ +\phi : K[x] \to K(\alpha), +\quad g(x) \mapsto g(\alpha). +$$ + +Then +$$ +\ker(\phi) = (f(x)). +$$ + +Hence +$$ +K[x]/(f(x)) +\cong +K(\alpha). +$$ + +So +$$ +K[x]/(f(x)) \cong K(\alpha). +$$ + +If $\beta$ is another root of $f$, then +$$ +K(\alpha) \cong K(\beta). +$$ + +(They are isomorphic as extensions of $K$, though not necessarily the same subfield.) + +--- + +If +$$ +L = K[x]/(f(x)), +$$ +and $\alpha = \pi(x)$, then +$$ +L = K(\alpha). +$$ + +Moreover, +$$ +[K(\alpha) : K] = \deg(f). +$$ + +--- + +## Example + +The numbers $\sqrt[3]{2}$ and $\omega \sqrt[3]{2}$ are two roots of +$$ +x^3 - 2 = 0, +$$ +where $\omega$ is a root of +$$ +x^2 + x + 1 = 0. +$$ + +Then +$$ +\mathbb{Q}(\sqrt[3]{2}) +\cong +\mathbb{Q}(\omega \sqrt[3]{2}). +$$ + +(They are isomorphic as field extensions.) \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 03 - Jan 12.md b/content/SEM_6/Galois_Theory/Lecture 03 - Jan 12.md new file mode 100644 index 00000000..42f8795a --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 03 - Jan 12.md @@ -0,0 +1,301 @@ +## Algebraic Extensions + +### Definition + +Let $L/K$ be a field extension and let $\alpha \in L$. + +We say $\alpha$ is **algebraic over $K$** if there exists a nonzero polynomial +$$ +f(x) \in K[x] +$$ +such that +$$ +f(\alpha) = 0. +$$ + +If $\alpha$ is not algebraic over $K$, then $\alpha$ is called **transcendental** over $K$. + +An extension $L/K$ is called **algebraic** if every $\alpha \in L$ is algebraic over $K$. + +--- + +## Algebraic Element and Evaluation Map + +Suppose $L/K$ is a field extension and $\alpha \in L$ is algebraic over $K$. + +Define +$$ +\phi : K[x] \to K(\alpha), +\qquad +f(x) \mapsto f(\alpha). +$$ + +Since $\alpha$ is algebraic, $\ker(\phi) \neq \{0\}$. + +Let +$$ +\ker(\phi) = \langle g(x) \rangle. +$$ + +If $h(x) \in K[x]$ satisfies $h(\alpha) = 0$, then +$$ +g(x) \mid h(x). +$$ + +--- + +## Minimal Polynomial + +Among all nonzero polynomials $g(x)$ such that $g(\alpha) = 0$, choose the **monic polynomial of least degree**. + +### Definition + +The monic polynomial of least degree +$$ +m_{\alpha,K}(x) \in K[x] +$$ +such that +$$ +m_{\alpha,K}(\alpha) = 0 +$$ +is called the **minimal polynomial of $\alpha$ over $K$**. + +Its degree is called the **degree of $\alpha$ over $K$**. + +Moreover, +$$ +[K(\alpha) : K] = \deg m_{\alpha,K}(x). +$$ + +--- + +## Examples + +### (1) $\alpha = \sqrt[n]{m}$ + +Let $n \in \mathbb{N}$ and consider +$$ +\alpha = \sqrt[n]{m}, +\qquad +K = \mathbb{Q}, \quad L = \mathbb{R}. +$$ + +Then $\alpha$ satisfies +$$ +x^n - m = 0. +$$ + +--- + +### (2) Primitive $n$th Root of Unity + +Let +$$ +K = \mathbb{Q}, \quad L = \mathbb{C}, +\qquad +\zeta = e^{2\pi i / n}. +$$ + +Then $\zeta$ satisfies +$$ +x^n - 1 = 0. +$$ + +Also, +$$ +1 + \zeta + \zeta^2 + \dots + \zeta^{n-1} = 0. +$$ + +Hence $\zeta$ satisfies +$$ +x^{n-1} + x^{n-2} + \dots + 1 = 0. +$$ + +The minimal polynomial of $\zeta$ has degree $n-1$ (in this case). + +--- + +## Degree Relations in Tower + +Suppose +$$ +K \subset E \subset L, +\qquad +\alpha \in L. +$$ + +Then + +$$ +[K(\alpha) : K] = \deg(\alpha/K), +$$ + +$$ +[E(\alpha) : E] = \deg(\alpha/E). +$$ + +By the Tower Law, + +$$ +[K(\alpha) : K] += +[K(\alpha) : E]\,[E : K]. +$$ + +--- + +## Example with Nested Extensions + +Let +$$ +K = \mathbb{Q}, +\qquad +E = \mathbb{Q}(\sqrt{2}), +\qquad +\alpha = \sqrt[4]{2}. +$$ + +Then +$$ +\alpha^2 = \sqrt{2}. +$$ + +Over $K$: +$$ +m_{\alpha,K}(x) = x^4 - 2, +\qquad +\deg(\alpha/K) = 4. +$$ + +Over $E$: +$$ +m_{\alpha,E}(x) = x^2 - \sqrt{2}, +\qquad +\deg(\alpha/E) = 2. +$$ + +--- + +## Finite Degree Implies Algebraic + +If +$$ +[K(\alpha) : K] < \infty, +$$ +then $\alpha$ is algebraic over $K$. + +Conversely, if $\alpha$ is algebraic over $K$, then +$$ +[K(\alpha) : K] = \deg m_{\alpha,K}(x) < \infty. +$$ + +--- + +## Two Algebraic Elements + +Suppose $\alpha, \beta \in L$ are algebraic over $K$. + +Then +$$ +[K(\alpha) : K] < \infty, +\qquad +[K(\beta) : K] < \infty. +$$ + +Since +$$ +K(\alpha, \beta) = K(\alpha)(\beta), +$$ + +by the Tower Law, + +$$ +[K(\alpha,\beta) : K] += +[K(\alpha,\beta) : K(\alpha)] +\,[K(\alpha) : K]. +$$ + +Because $\beta$ is algebraic over $K$, it is algebraic over $K(\alpha)$, so + +$$ +[K(\alpha,\beta) : K] < \infty. +$$ + +--- + +### Converse + +If +$$ +[K(\alpha,\beta) : K] < \infty, +$$ +then both $\alpha$ and $\beta$ are algebraic over $K$. + +--- + +## Characterisation + +Let $L/K$ be a field extension and $\alpha_1, \dots, \alpha_n \in L$. + +Then + +$$ +\alpha_1, \dots, \alpha_n \text{ are algebraic over } K +\quad \Longleftrightarrow \quad +[K(\alpha_1, \dots, \alpha_n) : K] < \infty. +$$ + +--- + +## Example + +Let +$$ +K = \mathbb{Q}, +\qquad +L = \mathbb{Q}(\sqrt{2}, \sqrt{3}). +$$ + +Then + +$$ +[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] += +[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] +\,[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}]. +$$ + +Since + +$$ +[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] = 2, +$$ + +and + +$$ +[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] = 2, +$$ + +we get + +$$ +[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] = 4. +$$ + +--- + +## Algebraic Closure Properties + +Let $L/K$ be a field extension. + +1. If $\alpha, \beta$ are algebraic over $K$, then + $$ + \alpha \pm \beta, \quad \alpha\beta, \quad \alpha^{-1} + $$ + are algebraic over $K$. + +2. The set of all algebraic elements in $L$ over $K$ forms a **subfield** of $L$. + +3. Every finite extension of $K$ is algebraic. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 04 - Jan 15.md b/content/SEM_6/Galois_Theory/Lecture 04 - Jan 15.md new file mode 100644 index 00000000..983e135d --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 04 - Jan 15.md @@ -0,0 +1,303 @@ +## Example: Degree of $\mathbb{Q}(\sqrt{2}, \sqrt{3})$ + +$$ +[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}] += +[\mathbb{Q}(\sqrt{2}, \sqrt{3}) : \mathbb{Q}(\sqrt{2})] +\,[\mathbb{Q}(\sqrt{2}) : \mathbb{Q}] += +2 \cdot 2 += +4. +$$ + +--- + +## Example: Algebraic Closure of $\mathbb{Q}$ in $\mathbb{R}$ + +Let +$$ +K = \mathbb{Q}, +$$ +and let +$$ +L = \{ \text{all real numbers algebraic over } \mathbb{Q} \}. +$$ + +Then +$$ +\mathbb{Q} \subset L \subset \mathbb{R}. +$$ + +### Question 1 + +Is +$$ +[L : \mathbb{Q}] < \infty? +$$ + +For every $n \in \mathbb{N}$, +$$ +\sqrt[n]{2} \in L, +$$ +since it satisfies +$$ +x^n - 2 = 0. +$$ + +Hence +$$ +\mathbb{Q}(\sqrt[n]{2}) \subset L. +$$ + +But +$$ +[\mathbb{Q}(\sqrt[n]{2}) : \mathbb{Q}] = n. +$$ + +Therefore, +$$ +[L : \mathbb{Q}] \ge n +\quad \text{for all } n, +$$ + +so +$$ +[L : \mathbb{Q}] = \infty. +$$ + +--- + +### Question 2 + +Is $L$ countable? + +Every algebraic number is a root of some polynomial +$$ +a_n x^n + a_{n-1} x^{n-1} + \dots + a_0, +\quad a_i \in \mathbb{Q}. +$$ + +For fixed degree $n$, the set of such polynomials is in bijection with +$$ +\mathbb{Q}^{n+1}, +$$ +which is countable. + +Hence the set of all polynomials with rational coefficients is countable. + +Each polynomial has finitely many roots. + +Therefore the set of all algebraic numbers is a countable union of finite sets, hence **countable**. + +Thus $L$ is countable. + +--- + +## Example + +Let +$$ +\alpha = 1 + \sqrt[3]{2} + \sqrt[3]{4}. +$$ + +Then +$$ +[\mathbb{Q}(\alpha) : \mathbb{Q}] = 3. +$$ + +Since +$$ +\sqrt[3]{4} = (\sqrt[3]{2})^2, +$$ + +we have +$$ +\mathbb{Q} \subset \mathbb{Q}(\alpha) \subset \mathbb{Q}(\sqrt[3]{2}). +$$ + +Also +$$ +[\mathbb{Q}(\sqrt[3]{2}) : \mathbb{Q}] = 3, +$$ + +so +$$ +[\mathbb{Q}(\alpha) : \mathbb{Q}] = 3. +$$ + +A minimal polynomial computation gives +$$ +(\alpha - 1)^3 = 2 + 3\sqrt[3]{2} + 3\sqrt[3]{4}, +$$ + +leading to +$$ +\alpha^3 - 3\alpha^2 - 3\alpha - 1 = 0. +$$ + +Thus +$$ +m_{\alpha,\mathbb{Q}}(x) += +x^3 - 3x^2 - 3x - 1. +$$ + +--- + +## Example: $\alpha = \sqrt[3]{3}$, $\beta = \sqrt{2}$ + +Let +$$ +\alpha = \sqrt[3]{3}, +\qquad +\beta = \sqrt{2}. +$$ + +Then +$$ +[\mathbb{Q}(\alpha) : \mathbb{Q}] = 3, +\qquad +[\mathbb{Q}(\beta) : \mathbb{Q}] = 2. +$$ + +Since +$$ +\mathbb{Q} \subset \mathbb{Q}(\alpha), \quad +\mathbb{Q} \subset \mathbb{Q}(\beta), +$$ + +we get +$$ +[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] += +[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}(\alpha)] +\,[\mathbb{Q}(\alpha) : \mathbb{Q}]. +$$ + +Now $\beta$ is algebraic over $\mathbb{Q}(\alpha)$, and +$$ +[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}(\alpha)] \le 2. +$$ + +Hence +$$ +[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = 6. +$$ + +--- + +## Simple Extension + +Let +$$ +\gamma = \alpha + \beta = \sqrt[3]{3} + \sqrt{2}. +$$ + +Take the ordered basis +$$ +\{1, \alpha, \alpha^2, \beta, \alpha\beta, \alpha^2\beta\} +$$ +for $\mathbb{Q}(\alpha,\beta)/\mathbb{Q}$. + +One checks that +$$ +1, \gamma, \gamma^2, \dots +$$ +generate the same field. + +Since +$$ +[\mathbb{Q}(\alpha,\beta) : \mathbb{Q}] = 6, +$$ + +and +$$ +\mathbb{Q} \subset \mathbb{Q}(\gamma) \subset \mathbb{Q}(\alpha,\beta), +$$ + +we conclude +$$ +\mathbb{Q}(\alpha,\beta) = \mathbb{Q}(\gamma). +$$ + +Thus the extension is simple. + +--- + +## Field Generated by a Set + +Let $S \subset L$. + +- The smallest field containing $K$ and $S$ is denoted + $$ + K(S). + $$ + +- The smallest ring containing $K$ and $S$ is denoted + $$ + K[S]. + $$ + +Explicitly, +$$ +K[S] += +\left\{ +\sum c_i a_i +\mid +c_i \in K,\; a_i \in S +\right\}. +$$ + +The field $K(S)$ is the field of fractions of $K[S]$. + +If +$$ +L = K(S), +$$ +we say $L$ is generated by $S$ over $K$. + +If the elements of $S$ are algebraic over $K$, then $K(S)$ is algebraic over $K$. + +--- + +## Composite Field + +Let $L/K$ and $M/K$ be field extensions. + +The **composite field** of $L$ and $M$ over $K$ is the smallest field containing both $L$ and $M$. + +It is denoted by +$$ +LM. +$$ + +Equivalently, +$$ +LM = K(L \cup M). +$$ + +Any element of $LM$ can be written as +$$ +\frac{\sum a_i b_i}{\sum c_j d_j}, +$$ +where +$$ +a_i, c_j \in L, +\qquad +b_i, d_j \in M, +$$ +and the denominator is nonzero. + +--- + +### Proposition + +If $L/K$ and $M/K$ are algebraic extensions, then + +$$ +LM / M +$$ + +is algebraic (and similarly $LM / L$ is algebraic). \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 05 - Jan 16.md b/content/SEM_6/Galois_Theory/Lecture 05 - Jan 16.md new file mode 100644 index 00000000..1df7459a --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 05 - Jan 16.md @@ -0,0 +1,338 @@ +## Composite Field + +Let $L/K$ and $M/K$ be field extensions. + +The **composite field** of $L$ and $M$ over $K$ is denoted by + +$$ +LM. +$$ + +Let +$$ +S = L \cup M. +$$ + +Then +$$ +LM = K(S). +$$ + +Define +$$ +A = \left\{ \sum a_i b_i \;\middle|\; a_i \in L,\; b_i \in M \right\}. +$$ + +Then +$$ +LM = Q(A), +$$ +the field of fractions of $A$. + +--- + +### Elements of the Composite Field + +Any element $\alpha \in LM$ is of the form + +$$ +\alpha += +\frac{\sum a_i b_i}{\sum c_j d_j}, +\quad +a_i, c_j \in L,\; b_i, d_j \in M, +\quad +\sum c_j d_j \neq 0. +$$ + +--- + +## Proposition + +If $L/K$ is algebraic, then + +$$ +LM / M +$$ + +is algebraic. + +Similarly, if $M/K$ is algebraic, then + +$$ +LM / L +$$ + +is algebraic. + +In particular, if both $L/K$ and $M/K$ are algebraic, then + +$$ +LM / K +$$ + +is algebraic. + +--- + +### Proof Sketch + +It is enough to prove that for any $a \in L$ and $b \in M$, + +$$ +ab +$$ + +is algebraic over $M$. + +Since $a \in L$ and $L/K$ is algebraic, + +$$ +a \text{ is algebraic over } K. +$$ + +Because $K \subset M \subset LM$, + +$a$ is algebraic over $M$. + +Also, $b \in M$, hence algebraic over $M$. + +Therefore, +$$ +ab +$$ +is algebraic over $M$. + +Hence every element of $LM$ is algebraic over $M$. + +--- + +## Finite Extensions and Composite Fields + +Suppose $L/K$ and $M/K$ are finite extensions. + +Then: + +1. +$$ +LM/K \text{ is finite}. +$$ + +2. +$$ +[L:K] \mid [LM:K], +\qquad +[M:K] \mid [LM:K]. +$$ + +3. +$$ +\mathrm{lcm}([L:K], [M:K]) +\le +[LM:K] +\le +[L:K][M:K]. +$$ + +--- + +### Proof Idea + +Since $L/K$ and $M/K$ are finite, there exist + +$$ +L = K(\alpha_1, \dots, \alpha_m), +\qquad +M = K(\beta_1, \dots, \beta_n). +$$ + +Then + +$$ +LM += +K(\alpha_1, \dots, \alpha_m, \beta_1, \dots, \beta_n). +$$ + +Hence + +$$ +[LM : K] < \infty. +$$ + +--- + +### Divisibility + +Because + +$$ +K \subset L \subset LM, +$$ + +we get + +$$ +[L:K] \mid [LM:K]. +$$ + +Similarly, + +$$ +[M:K] \mid [LM:K]. +$$ + +--- + +### Upper Bound + +Let + +$$ +[L:K] = r, +\qquad +[M:K] = s. +$$ + +Then by the Tower Law, + +$$ +[LM:K] += +[LM:M][M:K]. +$$ + +We claim + +$$ +[LM:M] \le r. +$$ + +Indeed, writing + +$$ +L = K(\alpha_1, \dots, \alpha_m), +$$ + +we get + +$$ +LM = M(\alpha_1, \dots, \alpha_m). +$$ + +Thus + +$$ +[LM:M] +\le +[K(\alpha_1, \dots, \alpha_m) : K] += +r. +$$ + +Therefore, + +$$ +[LM:K] +\le +rs. +$$ + +--- + +## Example 1 + +Let + +$$ +\alpha = \sqrt[4]{2}, +\qquad +\beta = \sqrt{2}. +$$ + +Then + +$$ +L = \mathbb{Q}(\alpha), +\qquad +M = \mathbb{Q}(\beta). +$$ + +Since + +$$ +\alpha \beta^2 = \sqrt[4]{2} \cdot (\sqrt{2})^2 += +2^{1/4} \cdot 2 += +2^{9/4}, +$$ + +we see + +$$ +LM = \mathbb{Q}(\alpha, \beta) += +\mathbb{Q}(\sqrt[4]{2}). +$$ + +--- + +## Example 2 + +Let + +$$ +\alpha = \sqrt[3]{2}, +\qquad +\beta = \omega \sqrt[3]{2}, +$$ + +where $\omega$ is a primitive cube root of unity. + +Then + +$$ +L = \mathbb{Q}(\sqrt[3]{2}), +\qquad +M = \mathbb{Q}(\omega \sqrt[3]{2}). +$$ + +Then + +$$ +LM = \mathbb{Q}(\sqrt[3]{2}, \omega). +$$ + +We have + +$$ +[L:\mathbb{Q}] = 3, +\qquad +[M:\mathbb{Q}] = 3. +$$ + +Since + +$$ +[\mathbb{Q}(\omega) : \mathbb{Q}] = 2, +$$ + +we compute + +$$ +[LM : \mathbb{Q}] += +[\mathbb{Q}(\sqrt[3]{2}, \omega) : \mathbb{Q}(\omega)] +\,[\mathbb{Q}(\omega) : \mathbb{Q}] += +3 \cdot 2 += +6. +$$ + +Thus + +$$ +[LM : \mathbb{Q}] = 6. +$$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 06 - Jan 28.md b/content/SEM_6/Galois_Theory/Lecture 06 - Jan 28.md new file mode 100644 index 00000000..47f145b9 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 06 - Jan 28.md @@ -0,0 +1,232 @@ +## Recap + +### Field Extensions + +Let +$$ +K \subset L \subset E. +$$ + +Then by the Tower Law, + +$$ +[E : K] = [E : L]\,[L : K]. +$$ + +--- + +### Field Generated by a Set + +If $S = \{\alpha_1, \dots, \alpha_n\}$ is finite, then + +$$ +K(\alpha_1, \dots, \alpha_n) += +\left\{ +\frac{f(\alpha_1, \dots, \alpha_n)} + {g(\alpha_1, \dots, \alpha_n)} +\;\middle|\; +f,g \in K[x_1, \dots, x_n],\; +g(\alpha_1, \dots, \alpha_n) \neq 0 +\right\}. +$$ + +If $S$ is not finite, then + +$$ +K(S) = \bigcup_{T \subset S,\; T \text{ finite}} K(T). +$$ + +Moreover, + +$$ +[K(\alpha_1, \dots, \alpha_n) : K] += +[K(\alpha_1, \dots, \alpha_n) : K(\alpha_1, \dots, \alpha_{n-1})] +\cdots +[K(\alpha_1) : K]. +$$ + +--- + +### Existence of a Root + +If +$$ +f(x) \in K[x], +$$ + +then there exists a field extension $K \subset L$ such that $L$ contains a root $\alpha$ of $f(x)$. + +If $\alpha, \beta$ are two roots of an irreducible polynomial over $K$, then + +$$ +K(\alpha) \cong K(\beta). +$$ + +--- + +### Simple and Finite Extensions + +If $S$ is finite, then $K(S)$ is a finitely generated extension. + +If $K(S)/K$ is finite, then every element of $S$ is algebraic over $K$. + +--- + +### Algebraic Elements + +Let $K \subset L$ and $\alpha \in L$. + +Then $\alpha$ is algebraic over $K$ if there exists + +$$ +f(x) \in K[x] +$$ + +such that + +$$ +f(\alpha) = 0. +$$ + +--- + +### Minimal Polynomial + +If $\alpha \in L$ is algebraic over $K$, then the minimal polynomial + +$$ +m_{\alpha,K}(x) +$$ + +is the monic polynomial of least degree in $K[x]$ such that + +$$ +m_{\alpha,K}(\alpha) = 0. +$$ + +If +$$ +K \subset L \subset M +$$ + +and $\alpha \in M$, then the minimal polynomials over $K$ and $L$ satisfy + +$$ +m_{\alpha,L}(x) \mid m_{\alpha,K}(x). +$$ + +--- + +## Composite Field Extension + +Let $L/K$ and $M/K$ be field extensions. + +The composite field $LM$ is the smallest field containing both $L$ and $M$. + +If $L/K$ and $M/K$ are algebraic (or finite), then + +$$ +LM/K +$$ + +is algebraic. + +Moreover, + +$$ +\mathrm{lcm}([L:K], [M:K]) +\le +[LM:K] +\le +[L:K][M:K], +$$ + +and the right inequality can be strict. + +--- + +## $R$-Algebra + +Let $R$ be a commutative ring with unity. + +An $R$-algebra is a ring $A$ together with a ring homomorphism + +$$ +\phi : R \to A. +$$ + +This induces an $R$-module structure on $A$ via + +$$ +r \cdot a = \phi(r)\,a. +$$ + +--- + +### $R$-Algebra Homomorphism + +Let $\phi : R \to A$ and $\psi : R \to B$ be $R$-algebras. + +An $R$-algebra homomorphism + +$$ +f : A \to B +$$ + +is a ring homomorphism such that the diagram commutes, i.e., + +$$ +f(\phi(r)) = \psi(r) +\quad \text{for all } r \in R. +$$ + +Equivalently, + +$$ +f(r a) = r f(a). +$$ + +--- + +## Automorphism Groups + +Let $L/K$ be a field extension. + +Define: + +- $\mathrm{Aut}(L)$: the set of all field automorphisms of $L$. +- $\mathrm{Aut}(L/K)$: the set of all $K$-algebra automorphisms of $L$. + +If $\sigma \in \mathrm{Aut}(L/K)$, then for all $a \in K$, + +$$ +\sigma(a) = a. +$$ + +Thus, + +$$ +\mathrm{Aut}(L/K) \subset \mathrm{Aut}(L). +$$ + +--- + +### Special Cases + +If $\operatorname{char}(K) = 0$, then + +$$ +\mathrm{Aut}(L) += +\mathrm{Aut}(L/\mathbb{Q}). +$$ + +If $\operatorname{char}(K) = p$, then + +$$ +\mathrm{Aut}(L) += +\mathrm{Aut}(L/\mathbb{F}_p). +$$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 07 - Jan 29.md b/content/SEM_6/Galois_Theory/Lecture 07 - Jan 29.md new file mode 100644 index 00000000..00e3c1e1 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 07 - Jan 29.md @@ -0,0 +1,262 @@ +## Finite Extensions and Automorphisms + +Let $L/K$ be a finite extension. + +This means $L$ is a finite dimensional vector space over $K$, i.e., + +$$ +[L : K] < \infty. +$$ + +We ask: + +- What is $\mathrm{Aut}(L)$? +- What is $\mathrm{Aut}(L/K)$? +- Is $\mathrm{Aut}(L/K)$ finite? + +--- + +### Determination by Generators + +Suppose + +$$ +L = K(\alpha_1, \dots, \alpha_n). +$$ + +Any $\sigma \in \mathrm{Aut}(L/K)$ is completely determined by the images of the generators: + +$$ +\sigma(\alpha_i) = ? +$$ + +Thus possible automorphisms correspond to choices + +$$ +\{\alpha_1, \dots, \alpha_n\} +\longmapsto +\{\beta_1, \dots, \beta_n\} +\subset L, +$$ + +provided the map extends to a field automorphism fixing $K$. + +--- + +## Induced Action on Polynomials + +Let $\sigma \in \mathrm{Aut}(L/K)$. + +Then + +$$ +\sigma : L \to L, +\qquad +\sigma(a) = a \quad \forall a \in K. +$$ + +Define an induced map + +$$ +\widetilde{\sigma} : L[x] \to L[x] +$$ + +by + +$$ +f(x) = a_0 + a_1 x + \dots + a_n x^n +\longmapsto +f^\sigma(x) += +\sigma(a_0) + \sigma(a_1)x + \dots + \sigma(a_n)x^n. +$$ + +If $f(x) \in K[x]$, then all coefficients lie in $K$, hence + +$$ +\sigma(a_i) = a_i, +$$ + +so + +$$ +f^\sigma(x) = f(x). +$$ + +--- + +## Images of Algebraic Elements + +Let $\alpha \in L$ be algebraic over $K$, and let + +$$ +m_{\alpha,K}(x) += +x^n + a_{n-1}x^{n-1} + \dots + a_0 +\in K[x] +$$ + +be its minimal polynomial. + +Then + +$$ +m_{\alpha,K}(\alpha) = 0. +$$ + +Applying $\sigma$, + +$$ +0 += +\sigma(m_{\alpha,K}(\alpha)) += +m_{\alpha,K}(\sigma(\alpha)). +$$ + +Thus + +$$ +\sigma(\alpha) +$$ + +is also a root of $m_{\alpha,K}(x)$. + +Hence, for each $\alpha \in L$, + +$$ +\sigma(\alpha) +$$ + +must be one of the finitely many roots of its minimal polynomial. + +Therefore, + +$$ +\mathrm{Aut}(L/K) +$$ + +is finite. + +--- + +## Example 1 + +Let + +$$ +K = \mathbb{Q}, +\qquad +L = \mathbb{Q}(\sqrt{2}). +$$ + +The minimal polynomial is + +$$ +x^2 - 2, +$$ + +whose roots are + +$$ +\sqrt{2}, \quad -\sqrt{2}. +$$ + +Thus there are two automorphisms: + +- Identity: $\sqrt{2} \mapsto \sqrt{2}$ +- Conjugation: $\sqrt{2} \mapsto -\sqrt{2}$ + +Hence + +$$ +\mathrm{Aut}(L/K) \cong \mathbb{Z}/2\mathbb{Z}. +$$ + +--- + +## Example 2 + +Let + +$$ +K = \mathbb{Q}, +\qquad +L = \mathbb{Q}(\sqrt[3]{2}). +$$ + +The minimal polynomial is + +$$ +x^3 - 2. +$$ + +Its roots in $\mathbb{C}$ are + +$$ +\sqrt[3]{2}, \quad +\omega \sqrt[3]{2}, \quad +\omega^2 \sqrt[3]{2}, +$$ + +where $\omega$ is a primitive cube root of unity. + +But only + +$$ +\sqrt[3]{2} +$$ + +lies in $L$. + +Thus the only possible image is itself, so + +$$ +\mathrm{Aut}(L/K) += +\{\mathrm{id}\}. +$$ + +--- + +## Example 3 + +Let + +$$ +L = \mathbb{Q}(\sqrt[3]{2}, \omega), +$$ + +where $\omega$ is a primitive cube root of unity. + +The roots of $x^3 - 2$ are + +$$ +\sqrt[3]{2}, \quad +\omega \sqrt[3]{2}, \quad +\omega^2 \sqrt[3]{2}. +$$ + +Possible images of $\sqrt[3]{2}$ are these three roots. + +Possible images of $\omega$ are + +$$ +\omega, \quad \omega^2. +$$ + +Thus there are $3 \times 2 = 6$ possibilities. + +In fact, + +$$ +\mathrm{Aut}(L/\mathbb{Q}) \cong S_3. +$$ + +--- + +## General Question + +Given a field $K$, can we find a field $L$ containing $K$ such that every polynomial over $K$ has a root in $L$? + +(This leads to the notion of algebraic closure.) \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 08 - Jan 30.md b/content/SEM_6/Galois_Theory/Lecture 08 - Jan 30.md new file mode 100644 index 00000000..75da9406 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 08 - Jan 30.md @@ -0,0 +1,196 @@ +## Algebraic Closure + +Let $K$ be a field. + +An **algebraic closure** of $K$ is a field $\overline{K}$ such that: + +1. +$$ +K \subset \overline{K}, +$$ + +2. +$$ +\overline{K}/K \text{ is algebraic}, +$$ + +3. Every non-constant polynomial over $K$ has a root in $\overline{K}$. + +--- + +## Algebraically Closed Field + +A field $K$ is called **algebraically closed** if every non-constant polynomial over $K$ has a root in $K$. + +### Observation + +If $K$ is algebraically closed, then + +$$ +\overline{K} = K. +$$ + +--- + +## Proposition + +Let $K \subset \Omega$, where $\Omega$ is algebraically closed. + +Define + +$$ +K^{\text{alg}} = \{ \alpha \in \Omega \mid \alpha \text{ is algebraic over } K \}. +$$ + +Then + +$$ +K^{\text{alg}} = \overline{K}. +$$ + +### Proof Sketch + +- If $\alpha, \beta$ are algebraic over $K$, then + $$ + \alpha \pm \beta, \quad \alpha\beta, \quad \alpha^{-1} + $$ + are algebraic over $K$, so $K^{\text{alg}}$ is a field. + +- Clearly, + $$ + K \subset K^{\text{alg}}, + $$ + so it is an algebraic extension of $K$. + +- Let $f(x) \in K[x]$ be non-constant. + Since $\Omega$ is algebraically closed, $f$ has a root $\alpha \in \Omega$. + + Because $\alpha$ satisfies a polynomial in $K[x]$, it is algebraic over $K$, hence + + $$ + \alpha \in K^{\text{alg}}. + $$ + +Thus every polynomial over $K$ has a root in $K^{\text{alg}}$, so + +$$ +K^{\text{alg}} = \overline{K}. +$$ + +--- + +## Proposition + +Let $K$ be a field and $\overline{K}$ its algebraic closure. + +Then $\overline{K}$ is algebraically closed. + +### Proof Sketch + +Let + +$$ +f(x) \in \overline{K}[x] +$$ + +be a non-constant polynomial. + +Let $\alpha$ be a root of $f(x)$ in some extension field of $\overline{K}$. + +Then: + +- $\alpha$ is algebraic over $\overline{K}$. +- Since $\overline{K}/K$ is algebraic, $\alpha$ is algebraic over $K$. +- Let $m_{\alpha,K}(x)$ be the minimal polynomial of $\alpha$ over $K$. + +Because $\overline{K}$ contains all roots of polynomials over $K$, + +$$ +\alpha \in \overline{K}. +$$ + +Hence every polynomial over $\overline{K}$ has a root in $\overline{K}$, so it is algebraically closed. + +--- + +## Polynomial Rings in Many Variables + +Let $\Lambda$ be a set. + +Define + +$$ +K[\Lambda] +$$ + +to be the polynomial ring in variables indexed by $\Lambda$. + +An element $f \in K[\Lambda]$ if there exists a finite subset + +$$ +\{x_1, \dots, x_n\} \subset \Lambda +$$ + +such that + +$$ +f = f(x_1, \dots, x_n). +$$ + +Equivalently, + +$$ +K[\Lambda] += +\bigcup_{I \subset \Lambda,\; I \text{ finite}} K[I]. +$$ + +--- + +## Theorem: Existence of Algebraic Closure + +Let $K$ be a field. + +Then an algebraic closure $\overline{K}$ exists. + +### Proof Idea + +We construct a field containing $K$ in which every polynomial over $K$ has a root. + +1. Let + $$ + \Lambda = \{ f \mid f \in K[x],\; f \text{ non-constant} \}. + $$ + +2. Consider the polynomial ring + $$ + K[\Lambda], + $$ + where each $f \in \Lambda$ corresponds to a variable $x_f$. + +3. For each $f \in \Lambda$, consider the polynomial + $$ + f(x_f) \in K[\Lambda]. + $$ + +4. Let $I$ be the ideal generated by all these elements: + $$ + I = \langle f(x_f) \mid f \in \Lambda \rangle. + $$ + +5. One checks that $I$ is a proper ideal. + +6. Let $\mathfrak{m}$ be a maximal ideal containing $I$. + +7. Define + $$ + L = K[\Lambda]/\mathfrak{m}. + $$ + +Then: + +- $L$ is a field, +- $K \subset L$, +- every polynomial over $K$ has a root in $L$. + +Finally, take the algebraic elements over $K$ inside $L$ to obtain $\overline{K}$. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 09 - Feb 2.md b/content/SEM_6/Galois_Theory/Lecture 09 - Feb 2.md new file mode 100644 index 00000000..96af2e53 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 09 - Feb 2.md @@ -0,0 +1,248 @@ +## Recap: Existence of Algebraic Closure + +### Theorem + +For every field $K$, an algebraic closure $\overline{K}$ exists. + +--- + +## Proof (Construction) + +Let + +$$ +\Lambda = \{ f \mid f \in K[x],\ f \text{ non-constant} \}. +$$ + +Consider the polynomial ring + +$$ +K[\Lambda], +$$ + +with one variable $x_f$ for each $f \in \Lambda$. + +For each $f \in \Lambda$, consider the polynomial + +$$ +f(x_f) \in K[\Lambda]. +$$ + +Let + +$$ +I = \langle f(x_f) \mid f \in \Lambda \rangle +\subset K[\Lambda] +$$ + +be the ideal generated by these elements. + +--- + +### Claim + +$$ +I \neq K[\Lambda]. +$$ + +If $I = K[\Lambda]$, then $1 \in I$. + +So there exist finitely many polynomials $f_1, \dots, f_m \in \Lambda$ and elements $g_i \in K[\Lambda]$ such that + +$$ +1 = \sum_{i=1}^m g_i \, f_i(x_{f_i}). +$$ + +Each $g_i$ involves only finitely many variables, so the above equality involves only finitely many variables. + +Now suppose $K'$ is a field extension of $K$ containing a root $\alpha_i$ of each $f_i$. + +Substituting $x_{f_i} = \alpha_i$, we get + +$$ +1 = \sum g_i(\alpha_1, \dots, \alpha_m)\, f_i(\alpha_i) = 0, +$$ + +a contradiction. + +Hence $I$ is a proper ideal. + +--- + +### Constructing a Field + +Let $\mathfrak{m}$ be a maximal ideal of $K[\Lambda]$ containing $I$. + +Define + +$$ +L = K[\Lambda]/\mathfrak{m}. +$$ + +Then $L$ is a field containing $K$. + +Moreover, for each $f \in K[x]$, the image of $x_f$ in $L$ is a root of $f$. + +Thus $L$ contains a root of every non-constant polynomial in $K[x]$. + +--- + +### Making it Algebraically Closed + +If $L$ is already algebraically closed, we are done. + +If not, repeat the process: adjoin roots of polynomials over $L$. + +If this does not terminate in finitely many steps, take the union + +$$ +\Omega = \bigcup_{i=0}^{\infty} L_i, +$$ + +where each $L_{i+1}$ is obtained from $L_i$ by adjoining roots of all polynomials over $L_i$. + +--- + +### Why $\Omega$ is Algebraically Closed + +Let $f(x) \in \Omega[x]$. + +Its coefficients lie in some $L_N$. + +By construction, $f$ has a root in $L_{N+1} \subset \Omega$. + +Hence $\Omega$ is algebraically closed. + +--- + +## Extension of Homomorphisms + +### Theorem + +Let $L/K$ be an algebraic extension, and let $\Omega$ be an algebraically closed field. + +Let + +$$ +\sigma : K \to \Omega +$$ + +be a field homomorphism. + +Then $\sigma$ extends to a field homomorphism + +$$ +\widetilde{\sigma} : L \to \Omega +$$ + +such that + +$$ +\widetilde{\sigma}|_K = \sigma. +$$ + +--- + +## Proof + +### Step 1: Simple Extension + +Assume + +$$ +L = K(\alpha). +$$ + +Define + +$$ +\sigma_0 : K[x] \to \Omega[x] +$$ + +by applying $\sigma$ to coefficients: + +$$ +f(x) \mapsto f^\sigma(x). +$$ + +Let $m_{\alpha,K}(x)$ be the minimal polynomial of $\alpha$ over $K$. + +Then + +$$ +m_{\alpha,K}^\sigma(x) \in \Omega[x]. +$$ + +Since $\Omega$ is algebraically closed, there exists $\beta \in \Omega$ such that + +$$ +m_{\alpha,K}^\sigma(\beta) = 0. +$$ + +Define + +$$ +\widetilde{\sigma}(\alpha) = \beta. +$$ + +This determines a homomorphism + +$$ +\widetilde{\sigma} : K(\alpha) \to \Omega. +$$ + +--- + +### Step 2: General Case (Zorn's Lemma) + +Let + +$$ +\Sigma = \{ (M,\phi) \mid K \subset M \subset L,\ \phi : M \to \Omega,\ \phi|_K = \sigma \}. +$$ + +Define a partial order: + +$$ +(M_1,\phi_1) \le (M_2,\phi_2) +\quad \text{if} \quad +M_1 \subset M_2 +\text{ and } +\phi_2|_{M_1} = \phi_1. +$$ + +Then $\Sigma$ is nonempty (contains $(K,\sigma)$). + +By Zorn's Lemma, there exists a maximal element $(M,\phi)$. + +If $M \neq L$, then there exists $\alpha \in L \setminus M$. + +Since $L/K$ is algebraic, $\alpha$ is algebraic over $M$. + +By Step 1, $\phi$ extends to $M(\alpha)$, contradicting maximality. + +Hence + +$$ +M = L. +$$ + +Therefore, $\phi$ extends to all of $L$. + +--- + +### Conclusion + +Every field homomorphism + +$$ +\sigma : K \to \Omega +$$ + +extends to a homomorphism + +$$ +\widetilde{\sigma} : L \to \Omega +$$ + +when $L/K$ is algebraic and $\Omega$ is algebraically closed. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 10 - Feb 4.md b/content/SEM_6/Galois_Theory/Lecture 10 - Feb 4.md new file mode 100644 index 00000000..bea05a46 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 10 - Feb 4.md @@ -0,0 +1,185 @@ +## Finite Extension and Automorphism Group + +Let $L/K$ be a finite extension. Then + +$$ +\operatorname{Aut}(L/K) < \infty. +$$ + +--- + +## Proposition + +Let $L$ be a field and $K \subset E \subset L$. + +1. +$$ +\operatorname{Aut}(L/K) \supset \operatorname{Aut}(L/E). +$$ + +2. If $H \subset \operatorname{Aut}(L)$, define + +$$ +L^H = \{ a \in L \mid \sigma(a) = a \ \forall \sigma \in H \}. +$$ + +Then $L^H$ is a subfield of $L$. + +3. If $H \subset H' \subset \operatorname{Aut}(L)$, then + +$$ +L^{H'} \subset L^H. +$$ + +4. If $K \subset L$ and $\sigma \in \operatorname{Aut}(L/K)$, then + +$$ +\sigma(a) = a \quad \forall a \in K. +$$ + +Hence + +$$ +K \subset L^{\operatorname{Aut}(L/K)}. +$$ + +5. If $H \subset \operatorname{Aut}(L)$, then + +$$ +H \subset \operatorname{Aut}(L/L^H). +$$ + +6. Suppose $K \subset L$ and there exists $H \subset \operatorname{Aut}(L)$ such that + +$$ +L^H = K. +$$ + +Then + +$$ +K = L^{\operatorname{Aut}(L/K)}. +$$ + +--- + +## Correspondence (Subfields ↔ Subgroups) + +There is a correspondence: + +$$ +\{ E \mid K \subset E \subset L \} +\longleftrightarrow +\{ H \subset \operatorname{Aut}(L) \} +$$ + +given by + +$$ +E \mapsto \operatorname{Aut}(L/E), +\qquad +H \mapsto L^H. +$$ + +In general, the inverse map need not be surjective unless the extension is Galois. + +--- + +## Example + +Let + +$$ +L = \mathbb{Q}(\sqrt[3]{2}, \omega), +$$ + +where $\omega$ is a primitive cube root of unity. + +Then + +$$ +\operatorname{Aut}(L/\mathbb{Q}) \cong S_3. +$$ + +Possible automorphisms: + +- $\sqrt[3]{2} \mapsto \sqrt[3]{2},\ \omega \mapsto \omega$ +- $\sqrt[3]{2} \mapsto \omega \sqrt[3]{2},\ \omega \mapsto \omega$ +- $\sqrt[3]{2} \mapsto \omega^2 \sqrt[3]{2},\ \omega \mapsto \omega$ +- and conjugation on $\omega$: $\omega \mapsto \omega^2$ + +Altogether 6 automorphisms, forming a group isomorphic to $S_3$. + +--- + +Let $H = \langle \sigma \rangle$ be a subgroup generated by some $\sigma$. + +Then compute + +$$ +L^H. +$$ + +--- + +## Example + +Is + +$$ +\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q} +$$ + +a Galois extension? + +Here + +$$ +L = \mathbb{Q}(\sqrt[3]{2}). +$$ + +Any $\sigma \in \operatorname{Aut}(L/\mathbb{Q})$ must fix $\mathbb{Q}$ and send $\sqrt[3]{2}$ to another root of + +$$ +x^3 - 2. +$$ + +But the other two roots are not in $L$. + +Hence the only possibility is + +$$ +\sigma(\sqrt[3]{2}) = \sqrt[3]{2}. +$$ + +Thus + +$$ +\operatorname{Aut}(L/\mathbb{Q}) = \{ \operatorname{id} \}. +$$ + +So the extension is not Galois. + +--- + +## Proposition + +If $L/K$ is finite, then + +$$ +|\operatorname{Aut}(L/K)| \le [L : K]. +$$ + +--- + +## Definition (Character) + +Let $G$ be a group. + +A character of $G$ in a field $L$ is a group homomorphism + +$$ +\chi : G \to L^\times. +$$ + +(Any distinct characters are linearly independent.) \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 11 - Feb 5.md b/content/SEM_6/Galois_Theory/Lecture 11 - Feb 5.md new file mode 100644 index 00000000..e4ebd160 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 11 - Feb 5.md @@ -0,0 +1,190 @@ +## Finite Extensions and Characters + +If $L/K$ is finite, then + +$$ +|\operatorname{Aut}(L/K)| < \infty, +\qquad +|\operatorname{Aut}(L/K)| \le [L:K]. +$$ + +--- + +## Characters + +Let $G$ be a group. + +A **character** of $G$ in a field $K$ is a group homomorphism + +$$ +\chi : G \to K^\times. +$$ + +--- + +### Examples + +1. If $\sigma : K \to L$ is a field homomorphism between fields, take + + $$ + G = K^\times, + $$ + + then + + $$ + \sigma|_{K^\times} : K^\times \to L^\times + $$ + + is a character. + +2. Let $L/K$ be a field extension and $\alpha \in L$ algebraic over $K$. + + Let $m_{\alpha,K}$ be the minimal polynomial of $\alpha$ over $K$. + + Assume $L$ contains all the roots of $m_{\alpha,K}$, say + + $$ + \beta_1, \dots, \beta_n. + $$ + + Define embeddings + + $$ + \chi_i : K(\alpha) \to L, + \qquad + \alpha \mapsto \beta_i. + $$ + + Then restricting to units, + + $$ + \chi_i|_{K(\alpha)^\times} : K(\alpha)^\times \to L^\times + $$ + + gives characters. + +--- + +## Proposition + +Let $G$ be a group. + +If $\chi_1, \dots, \chi_n$ are distinct characters of $G$ into a field $K$, then they are linearly independent over $K$. + +--- + +### Proof + +Suppose + +$$ +\sum_{i=1}^m a_i \chi_i = 0, +\qquad +a_i \in K, +$$ + +with not all $a_i = 0$, and $m$ minimal. + +Since $\chi_1 \ne \chi_2$, there exists $g_0 \in G$ such that + +$$ +\chi_1(g_0) \ne \chi_2(g_0). +$$ + +For all $g \in G$, + +$$ +\sum_{i=1}^m a_i \chi_i(g) = 0. +$$ + +Multiply by $\chi_1(g_0)$ and compare with evaluation at $g_0 g$: + +$$ +\sum_{i=1}^m a_i \chi_i(g_0 g) += +\sum_{i=1}^m a_i \chi_i(g_0)\chi_i(g). +$$ + +Subtracting suitably yields a shorter nontrivial linear relation among + +$$ +\chi_2, \dots, \chi_m, +$$ + +contradicting minimality. + +Hence the characters are linearly independent. + +--- + +## Proposition + +If $L/K$ is a finite extension, then + +$$ +|\operatorname{Aut}(L/K)| \le [L:K]. +$$ + +--- + +### Proof + +Let + +$$ +\operatorname{Aut}(L/K) = \{ \sigma_1, \dots, \sigma_n \}, +\qquad +[L:K] = m. +$$ + +Let $\alpha_1, \dots, \alpha_m$ be a $K$-basis of $L$. + +Consider the matrix + +$$ +A = +\begin{pmatrix} +\sigma_1(\alpha_1) & \cdots & \sigma_1(\alpha_m) \\ +\vdots & \ddots & \vdots \\ +\sigma_n(\alpha_1) & \cdots & \sigma_n(\alpha_m) +\end{pmatrix}. +$$ + +If $n > m$, then the rows are linearly dependent over $L$. + +So there exist $a_1, \dots, a_n \in L$, not all zero, such that + +$$ +\sum_{i=1}^n a_i \sigma_i(\alpha_j) = 0 +\quad +\text{for all } j = 1, \dots, m. +$$ + +Since $\{\alpha_j\}$ is a basis, this implies + +$$ +\sum_{i=1}^n a_i \sigma_i = 0 +$$ + +as functions $L \to L$. + +Restricting to $L^\times$, this gives a nontrivial linear relation among distinct characters + +$$ +\sigma_i : L^\times \to L^\times, +$$ + +contradicting the linear independence of characters. + +Therefore + +$$ +n \le m, +$$ + +i.e., + +$$ +|\operatorname{Aut}(L/K)| \le [L:K]. +$$ diff --git a/content/SEM_6/Galois_Theory/Lecture 12 - Feb 6.md b/content/SEM_6/Galois_Theory/Lecture 12 - Feb 6.md new file mode 100644 index 00000000..e9da336b --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 12 - Feb 6.md @@ -0,0 +1,260 @@ +Let + +$$ +G = \operatorname{Aut}(L/K). +$$ + +Then + +$$ +|G| \le [L:K]. +$$ + +--- + +Assume + +$$ +G = \{\sigma_1, \dots, \sigma_n\}, +\qquad +[L:K] = m. +$$ + +Let + +$$ +\{\alpha_1, \dots, \alpha_m\} +$$ + +be a $K$-basis of $L$. + +Consider the matrix + +$$ +A = +\begin{pmatrix} +\sigma_1(\alpha_1) & \cdots & \sigma_1(\alpha_m) \\ +\vdots & \ddots & \vdots \\ +\sigma_n(\alpha_1) & \cdots & \sigma_n(\alpha_m) +\end{pmatrix}. +$$ + +If $n > m$, then the rows $R_1, \dots, R_n$ are linearly dependent over $L$. + +So there exist $a_1, \dots, a_n \in L$, not all zero, such that + +$$ +a_1 R_1 + \cdots + a_n R_n = 0. +$$ + +Equivalently, + +$$ +\sum_{i=1}^n a_i \sigma_i(\alpha_j) = 0 +\qquad +\text{for all } j=1,\dots,m. +$$ + +--- + +### Claim + +$$ +\sum_{i=1}^n a_i \sigma_i = 0 +$$ + +as maps $L \to L$. + +Indeed, let + +$$ +b = \sum_{j=1}^m c_j \alpha_j \in L, +\qquad +c_j \in K. +$$ + +Then + +$$ +\sum_{i=1}^n a_i \sigma_i(b) += +\sum_{i=1}^n a_i \sigma_i\!\left(\sum_{j=1}^m c_j \alpha_j\right) += +\sum_{i=1}^n a_i \sum_{j=1}^m c_j \sigma_i(\alpha_j) += +\sum_{j=1}^m c_j \left(\sum_{i=1}^n a_i \sigma_i(\alpha_j)\right) += 0. +$$ + +Thus the $\sigma_i$ are linearly dependent as functions $L \to L$. + +Restricting to + +$$ +L^\times, +$$ + +this gives a nontrivial linear relation among distinct characters + +$$ +\sigma_i : L^\times \to L^\times, +$$ + +contradicting linear independence of characters. + +Hence + +$$ +|G| \le [L:K]. +$$ + +--- + +## Theorem (Equality Case) + +Let $K$ be a field and + +$$ +G \subset \operatorname{Aut}(K) +$$ + +be finite. + +Define + +$$ +K^G = \{ a \in K \mid \sigma(a)=a \ \forall \sigma \in G \}. +$$ + +Then + +$$ +|G| = [K : K^G] +$$ + +and + +$$ +G = \operatorname{Aut}(K/K^G). +$$ + +--- + +### Proof Sketch + +Note that + +$$ +G \subset \operatorname{Aut}(K/K^G). +$$ + +Hence + +$$ +|G| +\le +|\operatorname{Aut}(K/K^G)| +\le +[K : K^G]. +$$ + +Let $G=\{\sigma_1,\dots,\sigma_n\}$. + +Assume + +$$ +[K:K^G] > n. +$$ + +Then $\dim_{K^G} K \ge n+1$. + +Choose + +$$ +\alpha_1,\dots,\alpha_{n+1} +$$ + +linearly independent over $K^G$. + +Consider the matrix + +$$ +A = +\begin{pmatrix} +\sigma_1(\alpha_1) & \cdots & \sigma_1(\alpha_{n+1}) \\ +\vdots & \ddots & \vdots \\ +\sigma_n(\alpha_1) & \cdots & \sigma_n(\alpha_{n+1}) +\end{pmatrix}. +$$ + +Since $\operatorname{rank}(A)\le n$, the $n+1$ columns are linearly dependent. + +Take a minimal linear dependence + +$$ +a_1 C_1 + \cdots + a_r C_r = 0, +\qquad +a_i \in K^G. +$$ + +Applying any $\tau \in G$ gives another relation. + +Subtracting yields a shorter dependence unless all $a_i \in K^G$. + +This contradicts linear independence over $K^G$. + +Hence + +$$ +[K : K^G] \le n. +$$ + +Thus + +$$ +|G| = [K : K^G]. +$$ + +Finally, + +$$ +G = \operatorname{Aut}(K/K^G). +$$ + +--- + +## Example: Finite Fields + +Let + +$$ +\mathbb{F}_p +$$ + +be a finite field with $p$ elements. + +Let + +$$ +\mathbb{F}_p[t] +$$ + +be the polynomial ring over $\mathbb{F}_p$. + +Let + +$$ +\mathbb{F}_p(t) +$$ + +be its quotient field. + +Define the Frobenius map + +$$ +\phi : \mathbb{F}_p(t) \to \mathbb{F}_p(t), +\qquad +\phi(f(t)) = f(t)^p. +$$ + +This map is injective but not surjective in general. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 13 - Feb 9.md b/content/SEM_6/Galois_Theory/Lecture 13 - Feb 9.md new file mode 100644 index 00000000..2b7ad360 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 13 - Feb 9.md @@ -0,0 +1,310 @@ +### Recap + +If $L/K$ is finite, then + +$$ +|\operatorname{Aut}(L/K)| \le [L:K]. +$$ + +If $G \subset \operatorname{Aut}(L)$ is finite, then + +$$ +|\operatorname{Aut}(L/L^G)| = [L : L^G]. +$$ + +--- + +### Theorem + +$L/K$ is a **Galois extension** +$\Longleftrightarrow$ +there exists $H \subset \operatorname{Aut}(L)$ such that + +$$ +L^H = K. +$$ + +--- + +## Splitting Field + +Let $K$ be a field and let $\mathcal{F}$ be a family of polynomials over $K$, i.e., + +$$ +\mathcal{F} \subset K[x]. +$$ + +A field $L$ is called a **splitting field** of $\mathcal{F}$ if: + +1. Every polynomial in $\mathcal{F}$ splits completely into linear factors in $L[x]$. + +2. $L$ is generated by the roots of polynomials in $\mathcal{F}$. + +--- + +### Existence of Splitting Field + +Let $\overline{K}$ be an algebraic closure of $K$. + +For each $f \in \mathcal{F}$, $\overline{K}$ contains all roots of $f$. + +Define $L$ to be the subfield of $\overline{K}$ generated by all roots of polynomials in $\mathcal{F}$. + +Then: + +- $L$ is a splitting field of $\mathcal{F}$. +- $L/K$ is algebraic. + +--- + +## Examples + +### (i) $x^2 - 2 \in \mathbb{Q}[x]$ + +Splitting field: + +$$ +\mathbb{Q}(\sqrt{2}). +$$ + +--- + +### (ii) $x^3 - 2 \in \mathbb{Q}[x]$ + +The roots are: + +$$ +\sqrt[3]{2}, \quad \omega \sqrt[3]{2}, \quad \omega^2 \sqrt[3]{2}, +$$ + +where $\omega$ is a primitive cube root of unity. + +Splitting field: + +$$ +\mathbb{Q}(\sqrt[3]{2}, \omega) += +\mathbb{Q}(\sqrt[3]{2}, \sqrt{-3}). +$$ + +--- + +### (iii) $(x^2 - 2)(x^2 - 3) \in \mathbb{Q}[x]$ + +Splitting field: + +$$ +\mathbb{Q}(\sqrt{2}, \sqrt{3}). +$$ + +--- + +### (iv) $x^4 - 2 \in \mathbb{Q}[x]$ + +Factorization: + +$$ +x^4 - 2 = (x^2 - \sqrt{2})(x^2 + \sqrt{2}) +$$ + +and further: + +$$ += (x - \sqrt[4]{2})(x + \sqrt[4]{2})(x - i\sqrt[4]{2})(x + i\sqrt[4]{2}). +$$ + +Splitting field: + +$$ +\mathbb{Q}(\sqrt[4]{2}, i). +$$ + +--- + +## Proposition + +Let $K$ be a field and $\mathcal{F} \subset K[x]$ a family of polynomials. + +Let $L$ and $M$ be two splitting fields of $\mathcal{F}$. + +Then for any $K$-algebra homomorphism + +$$ +\sigma : L \to \overline{M}, +$$ + +we have + +$$ +\sigma(L) = M. +$$ + +--- + +### Proof (Sketch) + +#### Step 1: Single Polynomial + +Assume $\mathcal{F} = \{f\}$. + +Let + +$$ +f(x) = (x - \alpha_1)\cdots(x - \alpha_k) +$$ + +in $L[x]$. + +Applying $\sigma$: + +$$ +f^\sigma(x) = (x - \sigma(\alpha_1))\cdots(x - \sigma(\alpha_k)) +$$ + +in $M[x]$. + +Since $M$ is a splitting field of $f$, the roots of $f$ in $M$ are exactly + +$$ +\{\sigma(\alpha_1), \dots, \sigma(\alpha_k)\}. +$$ + +Thus + +$$ +\sigma(L) \subset M. +$$ + +Since + +$$ +M = K(\sigma(\alpha_1), \dots, \sigma(\alpha_k)), +$$ + +we get + +$$ +\sigma(L) = M. +$$ + +--- + +#### Step 2: General Case + +If $\mathcal{F}$ is infinite, take a finite subfamily and apply Step 1. + +Write: + +$$ +L = \bigcup_{S \subset \mathcal{F},\, S \text{ finite}} L_S, +$$ + +$$ +M = \bigcup_{S \subset \mathcal{F},\, S \text{ finite}} M_S. +$$ + +Since $\sigma(L_S) = M_S$ for each finite $S$, + +$$ +\sigma(L) = M. +$$ + +--- + +## Corollary + +If $L$ and $M$ are two splitting fields of a family $\mathcal{F} \subset K[x]$, then + +$$ +L \cong M +$$ + +as $K$-extensions. + +--- + +## Example + +Let $[L:\mathbb{Q}] = 2$. + +Then + +$$ +L = \mathbb{Q}(\sqrt{d}) +$$ + +for some $d \in \mathbb{Q}$. + +Then $L$ is the splitting field of + +$$ +x^2 - d. +$$ + +--- + +### Further Example + +Let + +$$ +f(x) = x^3 + ax^2 + bx + c \in \mathbb{Q}[x]. +$$ + +#### Case 1: $f$ reducible + +If + +$$ +f(x) = (x - \alpha)g(x) +$$ + +and $g(x)$ is reducible over $\mathbb{Q}$, + +then the splitting field is $\mathbb{Q}$. + +--- + +#### Case 2: $g(x)$ irreducible quadratic + +Suppose + +$$ +g(x) = x^2 + px + q. +$$ + +Then the roots are + +$$ +\frac{-p \pm \sqrt{p^2 - 4q}}{2}. +$$ + +Splitting field: + +$$ +\mathbb{Q}(\sqrt{p^2 - 4q}). +$$ + +--- + +#### Case 3: $f$ irreducible cubic + +Suppose + +$$ +f(x) = (x - \alpha)(x - \beta)(x - \gamma) +$$ + +in $\mathbb{C}$. + +If $\beta, \gamma \in \mathbb{Q}(\alpha)$, + +then $\mathbb{Q}(\alpha)$ is the splitting field. + +If not, then + +$$ +\mathbb{Q}(\alpha, \beta) +$$ + +is the splitting field. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 14 - Feb 11.md b/content/SEM_6/Galois_Theory/Lecture 14 - Feb 11.md new file mode 100644 index 00000000..3d39307a --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 14 - Feb 11.md @@ -0,0 +1,242 @@ +### Recap + +- Algebraic closure exists. +- If $L/K$ is finite, then + $$ + |\operatorname{Aut}(L/K)| < \infty, + \qquad + |\operatorname{Aut}(L/K)| \le [L:K]. + $$ +- If $K$ is a field and $G \subset \operatorname{Aut}(K)$ is finite, then + $$ + |G| = [K : K^G]. + $$ + +--- + +## Proposition + +If $L/K$ is an algebraic extension, then any $K$-algebra homomorphism + +$$ +\sigma : L \to L +$$ + +is surjective. + +### Proof + +Let $\alpha \in L$. We claim that $\alpha \in \sigma(L)$. + +Since $\alpha$ is algebraic over $K$, it satisfies some polynomial over $K$. +Let + +$$ +m_{\alpha,K}(x) +$$ + +be its minimal polynomial. + +Let $E$ be the subfield of $L$ containing all roots of $m_{\alpha,K}$ that lie in $L$. + +We have + +$$ +\sigma : L \to L +\quad\text{and}\quad +\sigma(E) \subset E, +$$ + +because if $\alpha$ is a root of $m_{\alpha,K}$, then $\sigma(\alpha)$ is also a root of the same polynomial. + +Hence $\sigma(E) \subset E$. + +Since + +$$ +[E:K] = [\sigma(E):K], +$$ + +we get + +$$ +\sigma(E) = E. +$$ + +Therefore, + +$$ +\alpha \in E = \sigma(E) \subset \sigma(L). +$$ + +Thus $\sigma$ is surjective. + +--- + +## Normal Extension + +Let $L/K$ be an algebraic extension and let $\overline{K}$ be an algebraic closure of $K$ containing $L$. + +If for every $K$-algebra homomorphism + +$$ +\sigma : L \to \overline{K} +$$ + +we have + +$$ +\sigma(L) = L, +$$ + +then $L/K$ is called a **normal extension**. + +--- + +### Example + +Let + +$$ +K = \mathbb{Q}, \qquad L = \mathbb{Q}(\sqrt[3]{2}, \omega), +$$ + +where $\omega$ is a primitive cube root of unity. + +Then $L/K$ is normal, since it contains all roots of + +$$ +x^3 - 2 +\quad\text{and}\quad +x^2 + x + 1. +$$ + +--- + +If instead + +$$ +L' = \mathbb{Q}(\sqrt[3]{2}), +$$ + +then $L'/K$ is not normal, because a $K$-embedding may send + +$$ +\sqrt[3]{2} \mapsto \omega \sqrt[3]{2}, +$$ + +and + +$$ +\sigma(L') \ne L'. +$$ + +--- + +## Theorem + +Let $L/K$ be an algebraic extension and let $\overline{K}$ be an algebraic closure containing $L$. + +Then the following are equivalent: + +1. $L/K$ is normal. +2. $L$ is the splitting field of some family of polynomials in $K[x]$. +3. If an irreducible polynomial $f(x) \in K[x]$ has one root in $L$, then it splits completely in $L$. + +--- + +### Proof + +#### (1) $\Rightarrow$ (2) + +Since $L/K$ is algebraic, each $\alpha \in L$ is algebraic over $K$. + +Let + +$$ +m_{\alpha,K}(x) +$$ + +be its minimal polynomial. + +Let + +$$ +\mathcal{F} = \{ m_{\alpha,K}(x) \mid \alpha \in L \}. +$$ + +Take a $K$-embedding + +$$ +\tau : K(\alpha) \to \overline{K}. +$$ + +Since $L/K$ is algebraic, $\tau$ extends to + +$$ +\widetilde{\tau} : L \to \overline{K}. +$$ + +Because $L/K$ is normal, + +$$ +\widetilde{\tau}(L) = L. +$$ + +Thus every conjugate of $\alpha$ over $K$ lies in $L$. + +Hence $m_{\alpha,K}$ splits in $L$. + +Therefore $L$ is the splitting field of $\mathcal{F}$. + +--- + +#### (2) $\Rightarrow$ (1) + +Let $\mathcal{F}$ be a family of polynomials over $K$ such that $L$ is their splitting field. + +Let + +$$ +\sigma : L \to \overline{K} +$$ + +be any $K$-algebra homomorphism. + +If $\alpha$ is a root of some $f \in \mathcal{F}$, then $\sigma(\alpha)$ is also a root of $f$. + +Since $L$ contains all roots of every $f \in \mathcal{F}$, we get + +$$ +\sigma(\alpha) \in L. +$$ + +Since $L$ is generated by these roots, + +$$ +\sigma(L) \subset L. +$$ + +Because $L/K$ is algebraic, $\sigma$ is surjective (by the previous proposition), hence + +$$ +\sigma(L) = L. +$$ + +Thus $L/K$ is normal. + +--- + +#### (2) $\Leftrightarrow$ (3) + +If $L$ is a splitting field, then any irreducible polynomial having one root in $L$ must split completely in $L$. + +Conversely, if every irreducible polynomial over $K$ that has a root in $L$ splits in $L$, then $L$ is generated by roots of such polynomials and hence is a splitting field. + +--- + +Therefore, + +$$ +L/K \text{ is normal}. +$$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 15 - Feb 12.md b/content/SEM_6/Galois_Theory/Lecture 15 - Feb 12.md new file mode 100644 index 00000000..fbb119b5 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 15 - Feb 12.md @@ -0,0 +1,315 @@ +### Continuation of Theorem (Normal Extensions) + +We complete the proof of: + +> $L/K$ normal +> $\Longleftrightarrow$ +> $L$ is a splitting field +> $\Longleftrightarrow$ +> every irreducible polynomial over $K$ having one root in $L$ splits in $L$. + +--- + +### (iii) $\Rightarrow$ (i) + +Let + +$$ +\sigma : L \to \overline{K} +$$ + +be a $K$-algebra homomorphism. + +Take $\alpha \in L$. Since $L/K$ is algebraic, $\alpha$ is algebraic over $K$. + +Let + +$$ +m_{\alpha,K}(x) +$$ + +be its minimal polynomial. + +By assumption, if $m_{\alpha,K}$ has one root in $L$, then it splits completely in $L$. + +Since $\alpha$ is a root, all roots of $m_{\alpha,K}$ lie in $L$. + +Now $\sigma(\alpha)$ is also a root of $m_{\alpha,K}$. + +Hence + +$$ +\sigma(\alpha) \in L. +$$ + +Thus + +$$ +\sigma(L) \subset L. +$$ + +Since $L/K$ is algebraic, $\sigma$ is surjective, so + +$$ +\sigma(L) = L. +$$ + +Therefore $L/K$ is normal. + +--- + +## Proposition + +Suppose + +- $L/K$ is normal, +- $M/K$ is any algebraic extension. + +Then: + +1. $LM/M$ is normal. +2. If both $L/K$ and $M/K$ are normal, then + $LM/K$ and $L \cap M / K$ are also normal. + +--- + +### Proof of (i) + +Let + +$$ +\sigma : LM \to \overline{K} +$$ + +be an $M$-algebra homomorphism. + +Every element of $LM$ is a finite sum of elements of the form + +$$ +\sum a_i b_i, +\qquad +a_i \in L,\ b_i \in M. +$$ + +Then + +$$ +\sigma\!\left(\sum a_i b_i\right) += +\sum \sigma(a_i)\sigma(b_i). +$$ + +Since $\sigma$ fixes $M$, + +$$ +\sigma(b_i) = b_i. +$$ + +Because $L/K$ is normal, + +$$ +\sigma(a_i) \in L. +$$ + +Hence + +$$ +\sigma(LM) \subset LM. +$$ + +Since $LM/M$ is algebraic, $\sigma$ is surjective. + +Thus + +$$ +\sigma(LM) = LM. +$$ + +So $LM/M$ is normal. + +--- + +### Proof of (ii) + +Let + +$$ +\sigma : L \cap M \to \overline{K} +$$ + +be a $K$-algebra homomorphism. + +If $x \in L \cap M$, then $x \in L$ and $x \in M$. + +Since $L/K$ and $M/K$ are normal, + +$$ +\sigma(x) \in L +\quad\text{and}\quad +\sigma(x) \in M. +$$ + +Thus + +$$ +\sigma(x) \in L \cap M. +$$ + +Hence + +$$ +\sigma(L \cap M) \subset L \cap M. +$$ + +By algebraicity, $\sigma$ is surjective. + +Therefore $L \cap M / K$ is normal. + +--- + +## Example + +Let + +$$ +K = \mathbb{Q}, \qquad f(x) = x^p + x^{p-1} + \dots + 1. +$$ + +Let $\xi$ be a root of $f(x)$. + +Then the roots of $f$ are + +$$ +\{\xi, \xi^2, \dots, \xi^p\}. +$$ + +Take + +$$ +L = \mathbb{Q}(\xi). +$$ + +If $\sigma : L \to \mathbb{C}$ is a $\mathbb{Q}$-algebra homomorphism, then + +$$ +\sigma(\xi) = \xi^a +$$ + +for some $a$. + +Since + +$$ +[L : \mathbb{Q}] = p, +$$ + +we get + +$$ +\{\xi, \dots, \xi^p\} +$$ + +are all conjugates. + +Hence + +$$ +L/\mathbb{Q} +$$ + +is normal. + +--- + +## Separable Extension + +### Definition + +Let $L/K$ be an algebraic extension. + +Fix an embedding + +$$ +\sigma : K \to \Omega +$$ + +into an algebraic closure $\Omega$. + +Let + +$$ +E_\sigma += +\{\text{all } K\text{-embeddings } L \to \Omega\}. +$$ + +The **separable degree** of $L/K$ is defined by + +$$ +[L : K]_{\mathrm{sep}} = |E_\sigma|. +$$ + +--- + +### Independence of Choice + +The definition appears to depend on: + +- the embedding $\sigma : K \to \Omega$, +- the algebraic closure $\Omega$. + +We show it is independent of these choices. + +--- + +### Claim + +Let + +$$ +\sigma : K \to \Omega, +\qquad +\eta : K \to \Omega' +$$ + +be two embeddings into algebraic closures. + +Then there is a bijection between the sets of embeddings + +$$ +E_\sigma +\quad\text{and}\quad +E_\eta. +$$ + +--- + +### Sketch of Argument + +We have: + +$$ +\sigma(K) \subset \Omega, +\qquad +\eta(K) \subset \Omega'. +$$ + +The map + +$$ +\eta \circ \sigma^{-1} +: +\sigma(K) \to \eta(K) +$$ + +is an isomorphism. + +Since both closures are algebraic over these subfields, this isomorphism extends to + +$$ +\tau : \overline{\sigma(K)} \to \overline{\eta(K)}. +$$ + +Thus embeddings correspond under conjugation by $\tau$. + +Hence the separable degree does not depend on the chosen embedding or algebraic closure. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 16 - Feb 13.md b/content/SEM_6/Galois_Theory/Lecture 16 - Feb 13.md new file mode 100644 index 00000000..7c68ff53 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 16 - Feb 13.md @@ -0,0 +1,193 @@ +## Separable Extension + +Let $L/K$ be a field extension. + +Let + +$$ +\sigma : K \to \Omega +$$ + +be an embedding into an algebraic closure $\Omega$. + +Define + +$$ +E_\sigma = \{\text{all extensions of } \sigma \text{ to } L\}. +$$ + +The **separable degree** of $L/K$ is defined as + +$$ +[L : K]_{\mathrm{sep}} = |E_\sigma|. +$$ + +--- + +## Independence of the Embedding + +Let + +$$ +\sigma : K \to \Omega +\quad\text{and}\quad +\eta : K \to \Omega' +$$ + +be two embeddings into algebraic closures $\Omega$ and $\Omega'$ respectively. + +### Claim + +There exists a bijection + +$$ +E_\sigma \longleftrightarrow E_\eta. +$$ + +--- + +### Construction of the Bijection + +We have: + +$$ +\sigma(K) \subset \Omega, +\qquad +\eta(K) \subset \Omega'. +$$ + +The map + +$$ +\eta \circ \sigma^{-1} : \sigma(K) \to \eta(K) +$$ + +is an isomorphism. + +Since: + +- $\Omega$ is an algebraic closure of $\sigma(K)$, +- $\Omega'$ is an algebraic closure of $\eta(K)$, + +the isomorphism extends to an isomorphism + +$$ +\tau : \overline{\sigma(K)} \to \overline{\eta(K)}. +$$ + +Thus we may replace $\Omega$ and $\Omega'$ by + +$$ +\overline{\sigma(K)} \quad\text{and}\quad \overline{\eta(K)} +$$ + +respectively. + +Now define a map: + +$$ +E_\sigma \to E_\eta, +\qquad +\widetilde{\sigma} \mapsto \tau \circ \widetilde{\sigma}. +$$ + +--- + +### Verification + +Let $\widetilde{\sigma} \in E_\sigma$, so + +$$ +\widetilde{\sigma}|_K = \sigma. +$$ + +We check that: + +$$ +(\tau \circ \widetilde{\sigma})|_K = \eta. +$$ + +For $a \in K$, + +$$ +(\tau \circ \widetilde{\sigma})(a) += +\tau(\widetilde{\sigma}(a)) += +\tau(\sigma(a)) += +(\eta \circ \sigma^{-1})(\sigma(a)) += +\eta(a). +$$ + +Thus + +$$ +\tau \circ \widetilde{\sigma} \in E_\eta. +$$ + +This gives a bijection. + +Hence the separable degree is independent of the choice of embedding and algebraic closure. + +--- + +## Bounding the Number of Embeddings + +Let: + +- $K$ be a field, +- $\overline{K}$ an algebraic closure, +- $\alpha \in L$ algebraic over $K$, +- $L = K(\alpha)$. + +Let + +$$ +m_{\alpha,K}(x) +$$ + +be the minimal polynomial of $\alpha$ over $K$. + +If + +$$ +\sigma : K \to \overline{K} +$$ + +and + +$$ +\widetilde{\sigma} : K(\alpha) \to \overline{K} +$$ + +is an extension of $\sigma$, then: + +$$ +\widetilde{\sigma}(\alpha) +$$ + +must be a root of + +$$ +m_{\alpha,K}(x). +$$ + +Thus: + +$$ +|E_\sigma| \le \deg m_{\alpha,K} += +[K(\alpha) : K]. +$$ + +--- + +This gives the fundamental inequality: + +$$ +[L : K]_{\mathrm{sep}} \le [L : K]. +$$ + +Equality holds exactly when the extension is separable. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 17 - Feb 16.md b/content/SEM_6/Galois_Theory/Lecture 17 - Feb 16.md new file mode 100644 index 00000000..7b630b68 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 17 - Feb 16.md @@ -0,0 +1,195 @@ +### Algebraic Extensions and Separable Degree + +Let $L/K$ be an algebraic extension. + +Fix an embedding + +$$ +\sigma : K \to \Omega +$$ + +into an algebraic closure $\Omega$. + +Let + +$$ +E_\sigma = \{\text{all extensions of } \sigma \text{ to } L \to \Omega\}. +$$ + +Then + +$$ +[L:K]_{\mathrm{sep}} = |E_\sigma|. +$$ + +Moreover, $[L:K]_{\mathrm{sep}}$ is independent of the choice of $\sigma$ and $\Omega$. + +--- + +### Simple Extensions + +Let $L = K(\alpha)$, where $\alpha$ is algebraic over $K$. + +Let $m_{\alpha,K}(x)$ be the minimal polynomial of $\alpha$ over $K$. + +Then every extension $\widetilde{\sigma} : K(\alpha) \to \Omega$ is determined by the choice of a root of $m_{\alpha,K}(x)$ in $\Omega$. + +Hence + +$$ +[K(\alpha):K]_{\mathrm{sep}} +\le +\deg m_{\alpha,K} += +[K(\alpha):K]. +$$ + +--- + +### Tower Formula (Separable Degree) + +Let $K \subset E \subset L$ be fields, with $L/K$ algebraic. + +Then + +$$ +[L:K]_{\mathrm{sep}} += +[L:E]_{\mathrm{sep}} \cdot [E:K]_{\mathrm{sep}}. +$$ + +#### Sketch of Proof + +Let $\sigma : K \to \Omega$ be an embedding. + +- Let $\{\eta_i\}$ be the extensions of $\sigma$ to $E$. +- For each $\eta_i$, let $\{\tau_{ij}\}$ be the extensions of $\eta_i$ to $L$. + +Then each $\tau_{ij}$ is an extension of $\sigma$ to $L$. + +Thus + +$$ +[L:K]_{\mathrm{sep}} += +[L:E]_{\mathrm{sep}} \cdot [E:K]_{\mathrm{sep}}. +$$ + +--- + +## Finite Extensions + +Now suppose $L/K$ is finite. + +Then + +$$ +[L:K]_{\mathrm{sep}} \le [L:K]. +$$ + +--- + +### Definition + +1. An algebraic element $\alpha$ over $K$ is **separable** if + + $$ + [K(\alpha):K]_{\mathrm{sep}} = [K(\alpha):K]. + $$ + +2. A polynomial $f(x)$ over $K$ is **separable** if all its roots are distinct. + +3. An algebraic extension $L/K$ is **separable** if every $\alpha \in L$ is separable over $K$. + +--- + +### Observation + +If $f(x)$ is separable over $K$ and $\beta$ is a root of $f$ in an algebraic closure, then the minimal polynomial of $\beta$ over $K$ divides $f(x)$. + +Since $f$ has distinct roots, so does the minimal polynomial. + +Hence $\beta$ is separable. + +--- + +### Stability Under Intermediate Fields + +If + +$$ +K \subset E \subset L +$$ + +and $L/K$ is separable, then $L/E$ is separable. + +Indeed, the minimal polynomial over $E$ divides the minimal polynomial over $K$. + +--- + +## Proposition + +Let $E/K$ be a field extension containing two field extensions $L/K$ and $M/K$. + +If $L/K$ is separable, then + +$$ +LM/M +$$ + +is separable. + +--- + +### Sketch + +Let $\alpha \in L$ be separable over $K$. + +Then its minimal polynomial over $K$ is separable. + +Since separability is preserved under base change, $\alpha$ remains separable over $M$. + +Hence $LM/M$ is separable. + +--- + +## Theorem + +Let $L/K$ be a finite extension. + +Then the following are equivalent: + +1. $L/K$ is separable. +2. $[L:K]_{\mathrm{sep}} = [L:K]$. + +--- + +### Proof + +Since $L$ is finite, write + +$$ +L = K(\alpha_1,\dots,\alpha_n). +$$ + +If $L/K$ is separable, then each $\alpha_i$ is separable. + +Using the tower formula for separable degrees, + +$$ +[L:K]_{\mathrm{sep}} += +[K(\alpha_1,\dots,\alpha_n):K]_{\mathrm{sep}} += +[L:K]. +$$ + +Conversely, if + +$$ +[L:K]_{\mathrm{sep}} = [L:K], +$$ + +then each simple extension in a tower must satisfy equality, hence each $\alpha \in L$ is separable. + +Thus $L/K$ is separable. \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 18 - Feb 18.md b/content/SEM_6/Galois_Theory/Lecture 18 - Feb 18.md new file mode 100644 index 00000000..79ef9859 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 18 - Feb 18.md @@ -0,0 +1,257 @@ +### Proposition + +Let $L/K$ be an algebraic extension. + +An element $\alpha \in L$ is **separable over $K$** if and only if its minimal polynomial +$m_{\alpha,K}(x)$ has distinct roots. + +Let +$$ +K \subset E \subset L. +$$ + +Then +$$ +[L:K]_{\mathrm{sep}} += +[L:E]_{\mathrm{sep}}\,[E:K]_{\mathrm{sep}}. +$$ + +If $L/K$ is algebraic, define +$$ +L^{s} = \{\text{separable elements over } K\}. +$$ + +--- + +## Proposition + +Let $L/K$ be a finite extension. Then the following are equivalent: + +1. $L$ is separable over $K$. +2. $[L:K]_{\mathrm{sep}} = [L:K]$. + +--- + +### Proof + +#### (i) $\Rightarrow$ (ii) + +Since $L/K$ is finite, write +$$ +L = K(\alpha_1,\dots,\alpha_n). +$$ + +Using multiplicativity of separable degrees, +$$ +[K(\alpha_1,\dots,\alpha_n):K]_{\mathrm{sep}} += +[K(\alpha_1,\dots,\alpha_n):K(\alpha_1,\dots,\alpha_{n-1})]_{\mathrm{sep}} +\cdots +[K(\alpha_1):K]_{\mathrm{sep}}. +$$ + +Since each $\alpha_i$ is separable, +$$ +[K(\alpha_i):K]_{\mathrm{sep}} = [K(\alpha_i):K]. +$$ + +Hence +$$ +[L:K]_{\mathrm{sep}} = [L:K]. +$$ + +--- + +#### (ii) $\Rightarrow$ (i) + +Assume +$$ +[L:K]_{\mathrm{sep}} = [L:K]. +$$ + +Let $\alpha \in L$. + +Using the tower law, +$$ +[L:K]_{\mathrm{sep}} += +[L:K(\alpha)]_{\mathrm{sep}}\,[K(\alpha):K]_{\mathrm{sep}}. +$$ + +Also, +$$ +[L:K] = [L:K(\alpha)]\,[K(\alpha):K]. +$$ + +Since equality holds globally, we must have +$$ +[K(\alpha):K]_{\mathrm{sep}} = [K(\alpha):K]. +$$ + +Thus $\alpha$ is separable. Hence $L/K$ is separable. + +--- + +## Proposition + +Let $E/K$ be a field containing extensions $L/K$ and $M/K$. + +If $L/K$ is separable, then the compositum $LM/M$ is separable. + +### Sketch + +Let +$$ +A = \left\{ \sum a_i b_i \mid a_i \in L,\, b_i \in M \right\}. +$$ + +Then $LM$ is the field generated by $A$. + +If $a \in L$ is separable over $K$, it remains separable over $M$. + +Hence elements of $LM$ are separable over $M$, so $LM/M$ is separable. + +--- + +# Derivatives and Separability + +Let +$$ +f(x) = a_0 + a_1 x + \cdots + a_n x^n \in K[x]. +$$ + +Define the formal derivative: +$$ +Df = a_1 + 2a_2 x + \cdots + n a_n x^{n-1}. +$$ + +Properties: + +- $D(f+g) = Df + Dg$ +- $D(fg) = f Dg + g Df$ + +--- + +## Proposition + +Let $f \in K[x]$. + +1. $\alpha$ is a **multiple root** of $f$ + if and only if + $$ + f(\alpha) = 0 \quad \text{and} \quad Df(\alpha) = 0. + $$ + +2. $f$ is **separable** + if and only if + $$ + \gcd(f, Df) = 1. + $$ + +--- + +### Proof (Sketch) + +If $\alpha$ is a multiple root, then +$$ +f(x) = (x-\alpha)^k g(x), \quad k \ge 2. +$$ + +Differentiating, +$$ +Df(x) += +k(x-\alpha)^{k-1} g(x) ++ +(x-\alpha)^k g'(x). +$$ + +Hence $(x-\alpha)$ divides $Df(x)$, so $Df(\alpha)=0$. + +Conversely, if $f(\alpha)=0$ and $Df(\alpha)=0$, then $(x-\alpha)^2$ divides $f$. + +Thus $f$ is separable $\iff$ $f$ and $Df$ have no common non-constant factor. + +--- + +### Irreducible Case + +Let $f$ be irreducible of degree $n$. + +Since $\deg Df \le n-1$: + +- If $\gcd(f,Df) \ne 1$, then $Df=0$. +- Hence an irreducible polynomial is separable + if and only if + $Df \ne 0$. + +--- + +## Characteristic $0$ + +If $\operatorname{char}(K)=0$: + +For any irreducible non-constant $f$, +$$ +Df \ne 0. +$$ + +Hence: + +> Every irreducible polynomial over a field of characteristic $0$ is separable. + +Therefore all finite extensions of characteristic $0$ fields are separable. + +--- + +## Remark (Characteristic $p>0$) + +The above is **false** in characteristic $p>0$. + +### Example + +Let +$$ +K = \mathbb{F}_2(t), +\qquad +f(x)=x^2 - t. +$$ + +Then +$$ +Df(x)=2x=0 +\quad (\text{in characteristic }2). +$$ + +Thus $f$ is irreducible but **not separable**. + +--- + +## Frobenius Map + +If $\operatorname{char}(K)=p>0$, define the Frobenius map: +$$ +F : K \to K, +\qquad +a \mapsto a^p. +$$ + +This is a field homomorphism (not necessarily surjective). + +Polynomials of the form +$$ +f(x) = g(x^p) +$$ +have zero derivative and are inseparable. + +--- + +### Conclusion + +- Over fields of characteristic $0$: all algebraic extensions are separable. +- Over fields of characteristic $p>0$: inseparable extensions can occur. +- Separability is controlled by the formal derivative and the condition + $$ + \gcd(f, Df) = 1. + $$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Lecture 19 - Feb 19.md b/content/SEM_6/Galois_Theory/Lecture 19 - Feb 19.md new file mode 100644 index 00000000..6577ba27 --- /dev/null +++ b/content/SEM_6/Galois_Theory/Lecture 19 - Feb 19.md @@ -0,0 +1,199 @@ +## Recap + +Let $K$ be a field. + +### Case 1: $\operatorname{char}(K)=0$ + +Every irreducible polynomial over $K$ is separable. + +--- + +### Case 2: $\operatorname{char}(K)=p>0$ + +Let $f(x)\in K[x]$ be irreducible. + +If $Df(x)\neq 0$, then $f$ is separable. + +If $Df(x)=0$, then +$$ +f(x)=a_0+a_1x^p+a_2x^{2p}+\cdots+a_nx^{np}. +$$ + +Let $\phi:K\to K$ be the Frobenius map, +$$ +\phi(a)=a^p. +$$ + +If $\phi$ is an isomorphism (i.e. surjective), then for each $a_i$ there exists $b_i\in K$ such that +$$ +a_i=b_i^p. +$$ + +Hence +$$ +f(x) += +b_0^p+b_1^px^p+\cdots+b_n^px^{np} += +(b_0+b_1x+\cdots+b_nx^n)^p. +$$ + +Thus $f$ is not irreducible β€” contradiction. + +Therefore: + +> If the Frobenius map is an isomorphism, every irreducible polynomial is separable. + +--- + +## Conclusion + +Let $K$ be a field with $\operatorname{char}(K)=p>0$. + +If the Frobenius map +$$ +\phi:K\to K,\qquad a\mapsto a^p +$$ +is an isomorphism, then: + +1. Every irreducible polynomial over $K$ is separable. +2. The product of two distinct irreducible polynomials is separable. + +--- + +# Proposition + +Let $K$ be a field with $\operatorname{char}(K)=p>0$. + +Let $\phi:K\to K$ be the Frobenius morphism. + +If $\alpha\in K\setminus\phi(K)$, then for every $n\ge1$, +$$ +x^{p^n}-\alpha +$$ +is irreducible over $K$. + +### Sketch of Proof + +Suppose +$$ +f(x)=x^{p^n}-\alpha=g(x)h(x). +$$ + +Let $\beta$ be a root of $f$ in $\overline K$. + +Then +$$ +\beta^{p^n}=\alpha. +$$ + +In $\overline K$, +$$ +x^{p^n}-\alpha=(x-\beta)^{p^n}. +$$ + +Since $K[x]$ is a UFD, any divisor has the form +$$ +g(x)=(x-\beta)^r. +$$ + +Write $r=p^ms$ with $(p,s)=1$. + +Then +$$ +g(x)=(x-\beta)^{p^ms} += +(x^{p^m}-\beta^{p^m})^s. +$$ + +From coefficients, we obtain $\beta^{p^m}\in K$, hence +$$ +\alpha=\beta^{p^n}\in\phi(K), +$$ +contradiction. + +Thus $x^{p^n}-\alpha$ is irreducible. + +--- + +# Perfect Fields + +### Definition + +A field $K$ is **perfect** if every irreducible polynomial over $K$ is separable. + +### Examples + +1. Every field of characteristic $0$. +2. A field of characteristic $p>0$ where Frobenius is an isomorphism. +3. Every algebraically closed field. + +--- + +# Primitive Element Theorem + +Let $L/K$ be a finite extension. + +1. $L/K$ is simple if there are only finitely many intermediate fields. +2. If $L/K$ is separable, then $L/K$ is simple. + +--- + +# Lemma (Finite Groups) + +Let $G$ be a finite group of order $n$. + +Suppose for every divisor $d\mid n$, +$$ +\bigl|\{x\in G: x^d=e\}\bigr|\le d. +$$ + +Then $G$ is cyclic. + +--- + +### Proof Sketch + +Let +$$ +A_d=\{x\in G:\text{ord}(x)=d\}. +$$ + +If $A_d\neq\varnothing$, choose $x\in A_d$. + +Then $G_d=\langle x\rangle$ has order $d$. + +Since +$$ +|G_d|=d +\quad\text{and}\quad +\bigl|\{x\in G:x^d=e\}\bigr|\le d, +$$ + +we get $A_d\subseteq G_d$. + +Hence $A_d$ consists precisely of the generators of $G_d$, so +$$ +|A_d|=\varphi(d). +$$ + +Summing over all divisors of $n$, +$$ +n=|G| +=\sum_{d\mid n}|A_d| +\le\sum_{d\mid n}\varphi(d) +=n. +$$ + +Thus equality holds and $A_n\neq\varnothing$. + +Hence $G$ has an element of order $n$ and is cyclic. + +--- + +### Fact + +If $x^d=1$, then $x$ is a root of +$$ +x^d-1=0. +$$ \ No newline at end of file diff --git a/content/SEM_6/Galois_Theory/Question_Papers/Midsem.md b/content/SEM_6/Galois_Theory/Question_Papers/Midsem.md new file mode 100644 index 00000000..e69de29b diff --git a/content/SEM_6/Galois_Theory/Question_Papers/Quiz 1.md b/content/SEM_6/Galois_Theory/Question_Papers/Quiz 1.md new file mode 100644 index 00000000..e69de29b diff --git a/content/SEM_6/Galois_Theory/credits.md b/content/SEM_6/Galois_Theory/credits.md new file mode 100644 index 00000000..64c6b7d8 --- /dev/null +++ b/content/SEM_6/Galois_Theory/credits.md @@ -0,0 +1,3 @@ +This digital collection exists thanks to the generosity and notes of **Jahnavi**. + +Thank you for sharing your work and making this possible. diff --git a/content/SEM_6/Galois_Theory/info.md b/content/SEM_6/Galois_Theory/info.md new file mode 100644 index 00000000..64554667 --- /dev/null +++ b/content/SEM_6/Galois_Theory/info.md @@ -0,0 +1,6 @@ +**Course:** Galois Theory +**Code:** MAT402 +**Year:** 3 +**Semester:** 6 +**Prerequisites:** Group Theory +**Course Instructor:** Dr Sarbeswar Pal diff --git a/content/SEM_6/Measure_Theory/Assignment/Assignment 1.md b/content/SEM_6/Measure_Theory/Assignment/Assignment 1.md new file mode 100644 index 00000000..ee23d41a --- /dev/null +++ b/content/SEM_6/Measure_Theory/Assignment/Assignment 1.md @@ -0,0 +1,1183 @@ +## Question + +Given a nonempty set $\Omega$, describe the smallest and largest $\sigma$-algebra of subsets of $\Omega$. + +--- + +## Definitions and Theorems Used + +**Definition 1 (Folland, p. 21):** Let $X$ be a nonempty set. A $\sigma$-algebra (or $\sigma$-field) on $X$ is a collection $\mathcal{M}$ of subsets of $X$ such that: + +1. $\emptyset \in \mathcal{M}$. + +2. If $A \in \mathcal{M}$, then $A^c \in \mathcal{M}$. + +3. If $\{A_n\}_{n=1}^{\infty} \subseteq \mathcal{M}$, then $\bigcup_{n=1}^{\infty} A_n \in \mathcal{M}$. + + +**Definition 2 (Rudin, p. 8):** If $\mathcal{F}$ is a collection of subsets of $X$, there exists a smallest $\sigma$-algebra $\mathcal{M}$ containing $\mathcal{F}$. + +**Proposition 1 (Folland, p. 22):** If $\{\mathcal{M}_\alpha\}_{\alpha \in A}$ is a collection of $\sigma$-algebras on $X$, then $\bigcap_{\alpha \in A} \mathcal{M}_\alpha$ is a $\sigma$-algebra on $X$. + +**Definition 3 (De Barra, p. 19):** The power set of $\Omega$, denoted $\mathcal{P}(\Omega)$ or $2^\Omega$, is the collection of all subsets of $\Omega$. + +--- + +## Solution + +1. Let $\mathcal{A}$ be any $\sigma$-algebra of subsets of $\Omega$. By Definition 1, $\emptyset \in \mathcal{A}$ and $\Omega \in \mathcal{A}$ because $\Omega = \emptyset^c$. + +2. Let $\mathcal{M}_{min} = \{\emptyset, \Omega\}$. We verify that $\mathcal{M}_{min}$ satisfies the axioms of a $\sigma$-algebra. First, $\emptyset \in \mathcal{M}_{min}$. Second, $\emptyset^c = \Omega \in \mathcal{M}_{min}$ and $\Omega^c = \emptyset \in \mathcal{M}_{min}$. Third, for any sequence $\{A_n\}_{n=1}^{\infty} \subseteq \mathcal{M}_{min}$, the union $\bigcup_{n=1}^{\infty} A_n$ is either $\emptyset$ (if $A_n = \emptyset$ for all $n$) or $\Omega$ (if $A_n = \Omega$ for at least one $n$). In both cases, the union belongs to $\mathcal{M}_{min}$. Since every $\sigma$-algebra on $\Omega$ must contain $\emptyset$ and $\Omega$, $\mathcal{M}_{min}$ is the smallest $\sigma$-algebra. + +3. Let $\mathcal{M}_{max} = \mathcal{P}(\Omega)$. We verify that $\mathcal{P}(\Omega)$ satisfies the axioms of a $\sigma$-algebra. Since $\emptyset \subseteq \Omega$, $\emptyset \in \mathcal{P}(\Omega)$. If $A \in \mathcal{P}(\Omega)$, then $A^c \subseteq \Omega$, so $A^c \in \mathcal{P}(\Omega)$. If $\{A_n\}_{n=1}^{\infty}$ is a sequence of subsets of $\Omega$, then $\bigcup_{n=1}^{\infty} A_n \subseteq \Omega$, so $\bigcup_{n=1}^{\infty} A_n \in \mathcal{P}(\Omega)$. + +4. By Definition 3, any $\sigma$-algebra $\mathcal{A}$ is a subcollection of $\mathcal{P}(\Omega)$, i.e., $\mathcal{A} \subseteq \mathcal{P}(\Omega)$. Thus, $\mathcal{P}(\Omega)$ is the largest $\sigma$-algebra. + +5. In summary, the smallest $\sigma$-algebra is the trivial $\sigma$-algebra $\{\emptyset, \Omega\}$, and the largest $\sigma$-algebra is the power set $\mathcal{P}(\Omega)$. + + +____ +## Question + +Given a subset $A$ of a universal set $\Omega$ (denoted $A \subset \Omega$), describe $\mathcal{A}(\{A\})$, the algebra generated by the set $A$, and $\mathcal{F}(\{A\})$, the $\sigma$-algebra generated by the set $A$. + +--- + +## Definitions and Theorems Used + +**Definition 1: Algebra of Sets (Folland, p. 21)** + +An algebra of sets on $\Omega$ is a collection of subsets $\mathcal{M}$ such that: + +1. $\emptyset \in \mathcal{M}$. + +2. If $E \in \mathcal{M}$, then $E^c \in \mathcal{M}$ (where $E^c = \Omega \setminus E$). + +3. If $E_1, \dots, E_n \in \mathcal{M}$, then $\bigcup_{i=1}^{n} E_i \in \mathcal{M}$. + + +> **Intuition:** An algebra is a collection of "well-behaved" sets that is closed under the basic operations of taking complements and finite unions. If you have a few shapes in your collection, you must also have their "outsides" and any shape formed by merging a finite number of them. + +**Definition 2: $\sigma$-algebra of Sets (Rudin, p. 8)** + +A $\sigma$-algebra is an algebra that is also closed under **countable** unions. That is, if $\{E_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then $\bigcup_{n=1}^{\infty} E_n \in \mathcal{M}$. + +> **Intuition:** A $\sigma$-algebra is a "stronger" version of an algebra. It ensures that even if you merge an infinite (but listable) number of sets from your collection, the resulting set stays within the collection. + +**Definition 3: Generated Algebra and $\sigma$-algebra (Folland, p. 22)** + +The algebra (or $\sigma$-algebra) generated by a collection of sets $\mathcal{E}$ is the intersection of all algebras (or $\sigma$-algebras) that contain $\mathcal{E}$. It is the smallest such structure containing $\mathcal{E}$. + +> **Intuition:** Think of this as the "minimalist" collection. You start with your specific set $A$ and add only the absolute minimum number of other sets required to satisfy the rules of an algebra or $\sigma$-algebra. + +--- + +## Solution (Step-by-Step Explanation) + +### Part 1: Describing the Algebra $\mathcal{A}(\{A\})$ + +1. **Identify the starting members:** By the definition of a generated algebra, $A$ must be in $\mathcal{A}(\{A\})$. By the axioms of an algebra (Definition 1), the empty set $\emptyset$ must also be in the collection. + +2. **Apply the complement rule:** Since $A \in \mathcal{A}(\{A\})$, its complement $A^c$ must be in $\mathcal{A}(\{A\})$. Similarly, since $\emptyset \in \mathcal{A}(\{A\})$, its complement $\emptyset^c = \Omega$ must be in $\mathcal{A}(\{A\})$. + +3. **Apply the union rule:** We check if any finite unions of these four sets $\{\emptyset, \Omega, A, A^c\}$ create new sets. + + - $A \cup A^c = \Omega$ (already in the set). + + - $A \cup \emptyset = A$ (already in the set). + + - $A \cup \Omega = \Omega$ (already in the set). + + All possible finite unions result in one of the existing four sets. + +4. **Conclusion for the Algebra:** The collection $\{\emptyset, \Omega, A, A^c\}$ satisfies all the requirements of an algebra and is the smallest collection to do so. Thus: + + $$\mathcal{A}(\{A\}) = \{\emptyset, \Omega, A, A^c\}$$ + + +--- + +### Part 2: Describing the $\sigma$-algebra $\mathcal{F}(\{A\})$ + +1. **Check for Countable Unions:** By the definition of a $\sigma$-algebra (Definition 2), we must ensure the collection is closed under countable unions. + + We take any sequence of sets $\{E_n\}_{n=1}^{\infty}$ where each $E_n$ is chosen from $\{\emptyset, \Omega, A, A^c\}$. + +2. **Evaluate the result:** Because the starting collection $\{\emptyset, \Omega, A, A^c\}$ is finite, any infinite union reduces to a finite union. For example, if we have a sequence like $(A, A^c, A, A^c, \dots)$, the union is simply $A \cup A^c = \Omega$. + + Since we already showed in Step 3 of Part 1 that this collection is closed under finite unions, it is automatically closed under countable unions as well. + +3. **Conclusion for the $\sigma$-algebra:** Every $\sigma$-algebra is an algebra, and in this specific case, the smallest algebra already satisfies the "$\sigma$" (countable) requirement. Therefore, the $\sigma$-algebra generated by $A$ is identical to the algebra generated by $A$. + + $$\mathcal{F}(\{A\}) = \{\emptyset, \Omega, A, A^c\}$$ + + +**Final Result:** + +Both the algebra and the $\sigma$-algebra generated by a single set $A$ are: + +$$\{\emptyset, \Omega, A, A^c\}$$ + +___ +## Question + +Let $\Omega$ be a nonempty set. Define + +$$\mathcal{F}_0 = \{A \subseteq \Omega \mid \text{either } A \text{ is a finite set or } A^c \text{ is a finite set}\}$$ + +(assume that the empty set $\emptyset$ is a finite set). Prove that $\mathcal{F}_0$ is an algebra. However, show that $\mathcal{F}_0$ is not a $\sigma$-algebra if $\Omega$ is an infinite set. + +--- + +## Definitions and Theorems Used + +**Definition 1: Algebra of Sets (Folland, p. 21)** + +An algebra of sets on $\Omega$ is a collection of subsets $\mathcal{M}$ such that: + +1. $\emptyset \in \mathcal{M}$. + +2. If $A \in \mathcal{M}$, then $A^c \in \mathcal{M}$. + +3. If $A, B \in \mathcal{M}$, then $A \cup B \in \mathcal{M}$. + + +> **Intuition:** An algebra is a collection that is "closed" under basic logic. If you have a set, you must have its opposite (complement), and if you have two sets, you must have the combination of both (union). + +**Definition 2: $\sigma$-algebra (Rudin, p. 8)** + +A $\sigma$-algebra is an algebra that is also closed under countable unions. That is, if $\{A_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then $\bigcup_{n=1}^{\infty} A_n \in \mathcal{M}$. + +> **Intuition:** This extends the rules of an algebra to include infinite lists of sets. It ensures that if you combine infinitely many sets from your collection in a list, the result doesn't "leak" out of the collection. + +**De Morgan's Laws (De Barra, p. 4)** + +For any collection of sets $\{A_i\}$, $(\bigcup A_i)^c = \bigcap A_i^c$ and $(\bigcap A_i)^c = \bigcup A_i^c$. + +> **Intuition:** These laws provide a way to switch between unions and intersections by using complements. + +--- + +## Solution + +1. To prove $\mathcal{F}_0$ is an algebra, we first verify that $\emptyset \in \mathcal{F}_0$. By the problem statement, $\emptyset$ is a finite set. Therefore, $\emptyset$ satisfies the condition for membership in $\mathcal{F}_0$. + +2. We next verify closure under complements. Let $A \in \mathcal{F}_0$. By the definition of $\mathcal{F}_0$, either $A$ is finite or $A^c$ is finite. If $A$ is finite, then $(A^c)^c = A$ is finite, which means $A^c \in \mathcal{F}_0$. If $A^c$ is finite, then $A^c$ directly satisfies the condition for membership in $\mathcal{F}_0$. In either case, $A^c \in \mathcal{F}_0$. + +3. We verify closure under finite unions. Let $A, B \in \mathcal{F}_0$. We consider two cases. Case 1: Both $A$ and $B$ are finite. The union of two finite sets is finite, so $A \cup B$ is finite and $A \cup B \in \mathcal{F}_0$. Case 2: At least one set, say $A$, has a finite complement $A^c$. By De Morgan's Laws, $(A \cup B)^c = A^c \cap B^c$. Since $A^c$ is finite, the intersection $A^c \cap B^c$ must also be finite (as it is a subset of $A^c$). Since $(A \cup B)^c$ is finite, $A \cup B \in \mathcal{F}_0$. Thus, $\mathcal{F}_0$ is an algebra. + +4. To show $\mathcal{F}_0$ is not a $\sigma$-algebra when $\Omega$ is infinite, we must find a countable sequence of sets in $\mathcal{F}_0$ whose union is not in $\mathcal{F}_0$. Since $\Omega$ is infinite, it contains a countably infinite subset $\{x_1, x_2, x_3, \dots\}$ where all $x_n$ are distinct. + +5. Define $A_n = \{x_{2n}\}$ for each $n \in \mathbb{N}$. Each $A_n$ is a singleton set, which is finite, so $A_n \in \mathcal{F}_0$ for all $n$. + +6. Consider the countable union $A = \bigcup_{n=1}^{\infty} A_n = \{x_2, x_4, x_6, \dots\}$. This set $A$ is infinite because it contains infinitely many distinct points. + +7. Consider the complement $A^c$. Note that $A^c$ contains the set $\{x_1, x_3, x_5, \dots\}$. Since this subset is infinite, $A^c$ is also infinite. + +8. Since neither $A$ nor $A^c$ is finite, $A \notin \mathcal{F}_0$ by the definition of the set. Because $\mathcal{F}_0$ is not closed under countable unions, it is not a $\sigma$-algebra. + + +___ +## Question + +Let $\Omega$ be a nonempty set. Define the collection of subsets $\mathcal{F}_c$ as: + +$$\mathcal{F}_c = \{A \subseteq \Omega \mid \text{either } A \text{ is a countable set or } A^c \text{ is a countable set}\}$$ + +Prove that $\mathcal{F}_c$ is a $\sigma$-algebra. + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra of Sets (Folland, p. 21)** + +Let $X$ be a nonempty set. A $\sigma$-algebra $\mathcal{M}$ on $X$ is a collection of subsets of $X$ that satisfies the following: + +1. $\emptyset \in \mathcal{M}$. + +2. If $A \in \mathcal{M}$, then $A^c \in \mathcal{M}$. + +3. If $\{A_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then $\bigcup_{n=1}^{\infty} A_n \in \mathcal{M}$. + + +> **Intuition:** A $\sigma$-algebra is a collection of sets that is "stable" under standard operations. If you have a set, you have its opposite; if you have a list of sets, you have their combined total. + +**Definition 2: Countable Set (Folland, p. 2)** + +A set is countable if it is either finite or has the same cardinality as the set of natural numbers $\mathbb{N}$. + +> **Intuition:** A countable set is one where you can "list" the elements one by one (even if the list never ends). + +**Theorem 1: Countable Unions of Countable Sets (Folland, p. 2)** + +A countable union of countable sets is itself a countable set. + +> **Intuition:** If you have a list of lists, and each list inside is countable, the entire combined collection is still small enough to be listed sequentially. + +**De Morgan's Laws (De Barra, p. 4)** + +For any collection of sets $\{A_i\}$, the complement of the union is the intersection of the complements: $(\bigcup A_i)^c = \bigcap A_i^c$. + +--- + +## Solution (Step-by-Step Explanation) + +**Step 1: Verify that the empty set is in the collection.** + +By Definition 2, the empty set $\emptyset$ is finite and therefore countable. According to the definition of $\mathcal{F}_c$, a set belongs to the collection if it is countable. Since $\emptyset$ is countable, $\emptyset \in \mathcal{F}_c$. + +**Step 2: Verify closure under complements.** + +Let $A$ be an element of $\mathcal{F}_c$. By definition, this means either $A$ is countable or $A^c$ is countable. + +- If $A$ is countable, then the complement of $A^c$ (which is $A$ itself) is countable. Thus, $A^c$ satisfies the condition "its complement is countable," so $A^c \in \mathcal{F}_c$. + +- If $A^c$ is countable, then $A^c$ directly satisfies the condition "is a countable set," so $A^c \in \mathcal{F}_c$. + + In both scenarios, the collection is closed under complements. + + +**Step 3: Verify closure under countable unions.** + +Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of sets in $\mathcal{F}_c$. We must show that $A = \bigcup_{n=1}^{\infty} A_n$ is in $\mathcal{F}_c$. We consider two possible cases: + +- **Case 1: Every set $A_n$ in the sequence is countable.** + + By Theorem 1, the union of a countable sequence of countable sets is countable. Therefore, $A = \bigcup A_n$ is countable. Since $A$ is countable, it satisfies the first membership condition of $\mathcal{F}_c$. + +- **Case 2: At least one set in the sequence has a countable complement.** + + Suppose there exists some index $k$ such that $A_k^c$ is countable. We look at the complement of the total union, $A^c = (\bigcup_{n=1}^{\infty} A_n)^c$. By De Morgan’s Laws, $A^c = \bigcap_{n=1}^{\infty} A_n^c$. + + Because $A^c$ is the intersection of all $A_n^c$, it must be a subset of the specific set $A_k^c$ (i.e., $A^c \subseteq A_k^c$). Since $A_k^c$ is countable and any subset of a countable set is countable, $A^c$ is countable. Because $A^c$ is countable, $A$ satisfies the second membership condition of $\mathcal{F}_c$. + + +**Step 4: Conclusion.** + +Since $\mathcal{F}_c$ contains the empty set and is closed under both complements and countable unions, it satisfies all requirements in the definition of a $\sigma$-algebra. Therefore, $\mathcal{F}_c$ is a $\sigma$-algebra. + +_____ +## Question + +Let $\Omega$ be a nonempty set and let $\mathcal{C} = \{A_i \mid i \in \mathbb{N}\}$ be a partition of $\Omega$. This means that the sets in $\mathcal{C}$ are pairwise disjoint ($A_i \cap A_j = \emptyset$ for all $i \neq j$) and their union covers the entire space ($\bigcup_{i \geq 1} A_i = \Omega$). + +Define a collection of subsets $\mathcal{F}$ as the set of all possible unions of the elements of the partition: + +$$\mathcal{F} = \{ \bigcup_{i \in J} A_i \mid J \subseteq \mathbb{N} \}$$ + +where we define $\bigcup_{i \in \emptyset} A_i = \emptyset$. Prove that $\mathcal{F}$ is a $\sigma$-algebra. + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra (Folland, p. 21)** + +A collection of subsets $\mathcal{M}$ of $\Omega$ is a $\sigma$-algebra if: + +1. $\emptyset \in \mathcal{M}$. + +2. If $E \in \mathcal{M}$, then its complement $E^c \in \mathcal{M}$. + +3. If $\{E_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then $\bigcup_{n=1}^{\infty} E_n \in \mathcal{M}$. + + +> **Intuition:** A $\sigma$-algebra is a collection of sets that is "complete" under standard logical operations. If you have a set, you must have its opposite; if you have a list of sets, you must have the set that combines them all. + +**Definition 2: Partition (De Barra, p. 4)** + +A collection of sets $\{A_i\}$ is a partition of $\Omega$ if the sets are mutually exclusive (disjoint) and their total union is $\Omega$. + +> **Intuition:** Think of a partition as breaking a puzzle into individual pieces. Every point in the space belongs to exactly one piece, and all pieces together recreate the whole picture. + +**De Morgan's Laws (Rudin, p. 7)** + +For any collection of sets, the complement of a union is the intersection of the complements, and the complement of an intersection is the union of the complements. + +> **Intuition:** These laws provide a bridge between "or" logic (unions) and "and" logic (intersections) when dealing with opposites. + +--- + +## Solution (Step-by-Step Explanation) + +**Step 1: Show that the empty set is in $\mathcal{F}$.** + +The definition of $\mathcal{F}$ states that for any subset $J \subseteq \mathbb{N}$, the union $\bigcup_{i \in J} A_i$ is in $\mathcal{F}$. If we choose $J = \emptyset$ (which is a subset of $\mathbb{N}$), the definition explicitly provides that $\bigcup_{i \in \emptyset} A_i = \emptyset$. Therefore, by the definition of $\mathcal{F}$, $\emptyset \in \mathcal{F}$. + +**Step 2: Show that the collection is closed under complements.** + +Let $E$ be an arbitrary set in $\mathcal{F}$. By the definition of $\mathcal{F}$, there must exist some index set $J \subseteq \mathbb{N}$ such that $E = \bigcup_{i \in J} A_i$. + +We need to find the complement $E^c = \Omega \setminus (\bigcup_{i \in J} A_i)$. + +Since the collection $\{A_i\}_{i \in \mathbb{N}}$ is a partition of $\Omega$, every element in $\Omega$ belongs to exactly one $A_i$. The elements not in $E$ are precisely those that belong to the $A_i$ blocks whose indices are _not_ in $J$. + +Thus, $E^c = \bigcup_{i \in J^c} A_i$, where $J^c = \mathbb{N} \setminus J$. + +Since $J^c$ is also a subset of $\mathbb{N}$, this union is in $\mathcal{F}$ by the definition of the collection. Thus, $\mathcal{F}$ is closed under complements. + +**Step 3: Show that the collection is closed under countable unions.** + +Let $\{E_n\}_{n=1}^{\infty}$ be a sequence of sets in $\mathcal{F}$. + +By the definition of $\mathcal{F}$, for each $E_n$, there exists an index set $J_n \subseteq \mathbb{N}$ such that $E_n = \bigcup_{i \in J_n} A_i$. + +We consider the union of this sequence: + +$$\bigcup_{n=1}^{\infty} E_n = \bigcup_{n=1}^{\infty} \left( \bigcup_{i \in J_n} A_i \right)$$ + +Using the properties of unions, we can regroup these: + +$$\bigcup_{n=1}^{\infty} E_n = \bigcup_{i \in K} A_i, \text{ where } K = \bigcup_{n=1}^{\infty} J_n$$ + +Since each $J_n$ is a subset of $\mathbb{N}$, their union $K$ is also a subset of $\mathbb{N}$. + +By the definition of $\mathcal{F}$, any union of $A_i$ over an index set $K \subseteq \mathbb{N}$ is an element of $\mathcal{F}$. Therefore, $\bigcup_{n=1}^{\infty} E_n \in \mathcal{F}$. + +**Step 4: Conclusion.** + +Since $\mathcal{F}$ contains the empty set and is closed under complements and countable unions, it satisfies the definition of a $\sigma$-algebra. + +_____ +## Question + +Let $\Omega$ be a nonempty set and $\mathcal{F}$ be a $\sigma$-algebra on $\Omega$. For any subset $A \subseteq \Omega$, define the collection: + +$$\mathcal{F}_A = \{B \cap A \mid B \in \mathcal{F}\}$$ + +Prove that $\mathcal{F}_A$ is a $\sigma$-algebra on the set $A$. This is known as the **trace $\sigma$-algebra** (or subspace $\sigma$-algebra) of $\mathcal{F}$ on $A$. + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra (Folland, p. 21)** + +Let $X$ be a set. A collection of subsets $\mathcal{M}$ of $X$ is a $\sigma$-algebra if: + +1. The empty set $\emptyset$ is in $\mathcal{M}$. + +2. If $E \in \mathcal{M}$, then its complement _relative to X_ (denoted $X \setminus E$) is in $\mathcal{M}$. + +3. If $\{E_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then their union $\bigcup_{n=1}^{\infty} E_n$ is in $\mathcal{M}$. + + +> **Intuition:** A $\sigma$-algebra is a collection of sets that is "closed" under standard logical operations. If you can define a set using "not" (complements) or "or" (unions) from existing sets in your collection, the result stays in the collection. + +**Distributive Law for Sets (De Barra, p. 4)** + +For any set $A$ and any collection of sets $\{B_n\}$: + +1. $A \cap (\bigcup B_n) = \bigcup (A \cap B_n)$ + +2. $A \cap (\bigcap B_n) = \bigcap (A \cap B_n)$ + + +> **Intuition:** This law shows that "intersecting with $A$" can be distributed across unions and intersections, much like multiplication distributes over addition in basic algebra. + +--- + +## Solution (Step-by-Step Explanation) + +**Step 1: Verify that the empty set is in $\mathcal{F}_A$.** + +By the definition of a $\sigma$-algebra, we know that $\emptyset \in \mathcal{F}$. According to the definition of the trace $\sigma$-algebra $\mathcal{F}_A$, an element is in $\mathcal{F}_A$ if it can be written as $B \cap A$ for some $B \in \mathcal{F}$. If we choose $B = \emptyset$, we get $\emptyset \cap A = \emptyset$. Therefore, $\emptyset \in \mathcal{F}_A$. + +**Step 2: Verify closure under complements relative to $A$.** + +Let $E$ be an arbitrary element in $\mathcal{F}_A$. We must show that its complement _with respect to A_, which is $A \setminus E$, is also in $\mathcal{F}_A$. + +1. Since $E \in \mathcal{F}_A$, there exists some $B \in \mathcal{F}$ such that $E = B \cap A$. + +2. We want to express $A \setminus E$ as an intersection of $A$ with some set in $\mathcal{F}$. + +3. Note that $A \setminus E = A \setminus (B \cap A) = A \cap B^c$. + +4. By the definition of a $\sigma$-algebra, since $B \in \mathcal{F}$, its complement $B^c$ (relative to $\Omega$) must also be in $\mathcal{F}$. + +5. Since $A \setminus E$ is the intersection of $A$ and a set in $\mathcal{F}$ (namely $B^c$), it satisfies the requirement for membership in $\mathcal{F}_A$. + + +**Step 3: Verify closure under countable unions.** + +Let $\{E_n\}_{n=1}^{\infty}$ be a sequence of sets in $\mathcal{F}_A$. We must show that their union $\bigcup_{n=1}^{\infty} E_n$ is in $\mathcal{F}_A$. + +1. By the definition of $\mathcal{F}_A$, for each $n$, there exists a set $B_n \in \mathcal{F}$ such that $E_n = B_n \cap A$. + +2. The union of these sets is $\bigcup_{n=1}^{\infty} (B_n \cap A)$. + +3. By the Distributive Law for sets, we can pull the intersection with $A$ out: $\bigcup_{n=1}^{\infty} (B_n \cap A) = A \cap (\bigcup_{n=1}^{\infty} B_n)$. + +4. By the definition of a $\sigma$-algebra, since each $B_n \in \mathcal{F}$, the countable union $B = \bigcup_{n=1}^{\infty} B_n$ is also in $\mathcal{F}$. + +5. Because the union of $\{E_n\}$ can be written as $A \cap B$ for a set $B \in \mathcal{F}$, the union is an element of $\mathcal{F}_A$. + + +**Step 4: Conclusion.** + +$\mathcal{F}_A$ contains the empty set and is closed under both complements relative to $A$ and countable unions. Thus, by the definition of a $\sigma$-algebra, $\mathcal{F}_A$ is a $\sigma$-algebra on the set $A$. + +____ +## Question + +Consider the semi-algebra $\mathcal{S}$ of semi-open intervals in $\mathbb{R}$ (of the form $(a, b]$ where $a, b \in \mathbb{R}$ and $a \leq b$). Prove that $\mathcal{F}(\mathcal{S}) = \mathcal{B}$, where $\mathcal{F}(\mathcal{S})$ is the $\sigma$-algebra generated by $\mathcal{S}$ and $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$. + +--- + +## Definitions and Theorems Used + +**Definition 1: Borel $\sigma$-algebra (Folland, p. 22)** + +The Borel $\sigma$-algebra on $\mathbb{R}$, denoted by $\mathcal{B}_{\mathbb{R}}$ (or simply $\mathcal{B}$), is the $\sigma$-algebra generated by the family of all open sets in $\mathbb{R}$. + +> **Intuition:** The Borel $\sigma$-algebra is the smallest collection of sets that contains every open interval $(a, b)$ and follows the rules of a $\sigma$-algebra. It represents the standard collection of sets we "measure" in real analysis. + +**Definition 2: Generated $\sigma$-algebra (Rudin, p. 8)** + +If $\mathcal{E}$ is any collection of subsets of $X$, the $\sigma$-algebra generated by $\mathcal{E}$, denoted $\sigma(\mathcal{E})$ or $\mathcal{F}(\mathcal{E})$, is the smallest $\sigma$-algebra containing every set in $\mathcal{E}$. + +> **Intuition:** Think of this as the "minimal" $\sigma$-algebra built from a starting kit $\mathcal{E}$. If you have two collections of sets, and every set in the first collection can be built from sets in the second using $\sigma$-algebra rules, then the generated $\sigma$-algebras reflect that relationship. + +**Theorem 1: Countable Unions and Intersections (De Barra, p. 7)** + +A $\sigma$-algebra is closed under countable unions and countable intersections. Specifically, an open interval $(a, b)$ can be expressed as a countable union of semi-open intervals: $(a, b) = \bigcup_{n=1}^{\infty} (a, b - \frac{1}{n}]$. + +> **Intuition:** Even though semi-open intervals $(a, b]$ and open intervals $(a, b)$ look different, you can "reach" one from the other by taking an infinite sequence of sets and merging them or finding their overlap. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +To prove that two $\sigma$-algebras are equal, we must show that each is a subcollection of the other. Thus, we must prove: + +1. $\mathcal{F}(\mathcal{S}) \subseteq \mathcal{B}$ + +2. $\mathcal{B} \subseteq \mathcal{F}(\mathcal{S})$ + + +**Step 1: Prove that $\mathcal{F}(\mathcal{S}) \subseteq \mathcal{B}$.** + +By the definition of a generated $\sigma$-algebra, to show $\mathcal{F}(\mathcal{S}) \subseteq \mathcal{B}$, it is sufficient to show that every element of $\mathcal{S}$ is contained in $\mathcal{B}$. + +- Let $(a, b] \in \mathcal{S}$ be a semi-open interval. + +- We can write $(a, b] = \bigcap_{n=1}^{\infty} (a, b + \frac{1}{n})$. + +- Every interval of the form $(a, b + \frac{1}{n})$ is an open set. By the definition of the Borel $\sigma$-algebra, every open set is in $\mathcal{B}$. + +- Since $\mathcal{B}$ is a $\sigma$-algebra, it is closed under countable intersections. Therefore, $(a, b] \in \mathcal{B}$. + +- Since $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{S}$, and $\mathcal{F}(\mathcal{S})$ is the _smallest_ $\sigma$-algebra containing $\mathcal{S}$, it follows that $\mathcal{F}(\mathcal{S}) \subseteq \mathcal{B}$. + + +**Step 2: Prove that $\mathcal{B} \subseteq \mathcal{F}(\mathcal{S})$.** + +By the definition of the Borel $\sigma$-algebra, $\mathcal{B}$ is generated by open sets. In $\mathbb{R}$, every open set is a countable union of open intervals $(a, b)$. Therefore, it is sufficient to show that every open interval $(a, b)$ is in $\mathcal{F}(\mathcal{S})$. + +- Consider an open interval $(a, b)$. We can write this as a countable union of semi-open intervals: $(a, b) = \bigcup_{n=n_0}^{\infty} (a, b - \frac{1}{n}]$, where $n_0$ is large enough such that $a < b - \frac{1}{n_0}$. + +- Each set $(a, b - \frac{1}{n}]$ is an element of $\mathcal{S}$, and by the definition of a generated $\sigma$-algebra, $\mathcal{S} \subseteq \mathcal{F}(\mathcal{S})$. + +- Since $\mathcal{F}(\mathcal{S})$ is a $\sigma$-algebra, it is closed under countable unions. Therefore, $(a, b) \in \mathcal{F}(\mathcal{S})$. + +- Since all open intervals are in $\mathcal{F}(\mathcal{S})$, all open sets (which are countable unions of these intervals) are also in $\mathcal{F}(\mathcal{S})$. + +- Since $\mathcal{F}(\mathcal{S})$ is a $\sigma$-algebra containing all open sets, and $\mathcal{B}$ is the _smallest_ $\sigma$-algebra containing all open sets, it follows that $\mathcal{B} \subseteq \mathcal{F}(\mathcal{S})$. + + +**Step 3: Conclusion.** + +Because we have shown inclusion in both directions, we conclude that $\mathcal{F}(\mathcal{S}) = \mathcal{B}$. + +____ +## Question + +For the semi-algebra $\mathcal{S}$ of semi-open intervals in $\mathbb{R}$ (intervals of the form $(a, b]$ where $a, b \in \mathbb{R}$), how do we know that the generated algebra $\mathcal{A}(\mathcal{S})$ is not a $\sigma$-algebra? + +--- + +## Definitions and Theorems Used + +**Definition 1: Algebra of Sets (Folland, p. 21)** + +An algebra of sets on a set $X$ is a collection $\mathcal{M}$ of subsets of $X$ that contains the empty set and is closed under complements and finite unions. + +> **Intuition:** An algebra allows you to perform basic "building block" operations. If you have a few shapes, you can find their opposites or glue a finite number of them together and still stay within your collection. + +**Definition 2: $\sigma$-algebra (Rudin, p. 8)** + +A $\sigma$-algebra is an algebra that is also closed under countable unions. That is, if $\{A_n\}_{n=1}^{\infty}$ is a sequence of sets in the collection, then $\bigcup_{n=1}^{\infty} A_n$ must also be in the collection. + +> **Intuition:** This is a "stronger" version of an algebra. It ensures that if you glue together an infinite (but listable) list of shapes, the final result is still a shape that your collection recognizes. + +**Proposition 1: Structure of the Algebra Generated by Intervals (De Barra, p. 32)** + +The algebra $\mathcal{A}(\mathcal{S})$ generated by the semi-open intervals $\mathcal{S}$ consists of all sets that can be written as a **finite** union of disjoint semi-open intervals. + +> **Intuition:** If you start with intervals like $(0, 1]$, you can only create sets like $(0, 1] \cup (2, 5]$ using the rules of an algebra. You can never create something that requires "gluing" infinitely many separate pieces together. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +1. **Identify the elements of the algebra:** By the property of generated algebras for intervals, any set $E$ in $\mathcal{A}(\mathcal{S})$ must be representable as a finite union of disjoint semi-open intervals of the form $(a_i, b_i]$. + +2. **Construct a countable union of elements:** We can choose a sequence of sets from the semi-algebra $\mathcal{S}$ (which are also in the algebra $\mathcal{A}(\mathcal{S})$). Let $A_n = (\frac{1}{n+1}, \frac{1}{n}]$ for $n = 1, 2, 3, \dots$. Each $A_n$ is a valid semi-open interval. + +3. **Determine the result of the infinite union:** Consider the countable union of these sets: + + $$E = \bigcup_{n=1}^{\infty} A_n = \bigcup_{n=1}^{\infty} \left(\frac{1}{n+1}, \frac{1}{n}\right] = (0, 1]$$ + + While this specific union results in an interval $(0, 1]$ which _is_ in the algebra, we need to find a union that results in a set that **cannot** be written as a finite union of semi-open intervals. + +4. **Construct a counterexample:** Consider the set of open intervals $B_n = (n + \frac{1}{4}, n + \frac{1}{2}]$. These are all disjoint and belong to $\mathcal{A}(\mathcal{S})$. Now, define the countable union: + + $$U = \bigcup_{n=1}^{\infty} (n + \frac{1}{4}, n + \frac{1}{2}]$$ + + , (2.25, 2.5], (3.25, 3.5], \dots$ extending to infinity] + +5. **Apply the definition of $\sigma$-algebra:** If $\mathcal{A}(\mathcal{S})$ were a $\sigma$-algebra, then by the definition of closure under countable unions, $U$ would have to be in $\mathcal{A}(\mathcal{S})$. + +6. **Verify membership:** However, by Proposition 1, every set in $\mathcal{A}(\mathcal{S})$ must be a **finite** union of intervals. The set $U$ consists of infinitely many disjoint components that are separated from each other. It cannot be simplified or rewritten into a union of finitely many semi-open intervals. + +7. **Conclusion:** Since we have found a countable union of sets from the algebra that results in a set not contained within the algebra, $\mathcal{A}(\mathcal{S})$ is not closed under countable unions. Therefore, it is not a $\sigma$-algebra. + + +____ +## Question + +Prove that every countable subset of $\mathbb{R}$ is in $\mathcal{B}$, where $\mathcal{B}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$. + +--- + +## Definitions and Theorems Used + +**Definition 1: Borel $\sigma$-algebra (Folland, p. 22)** + +The Borel $\sigma$-algebra $\mathcal{B}$ on $\mathbb{R}$ is the $\sigma$-algebra generated by the open sets of $\mathbb{R}$. + +> **Intuition:** This is the smallest collection of sets that contains all open intervals and obeys the standard rules of a $\sigma$-algebra (closure under complements and countable unions). + +**Definition 2: $\sigma$-algebra Properties (Rudin, p. 8)** + +By the definition of a $\sigma$-algebra, if a collection $\mathcal{M}$ is a $\sigma$-algebra, then: + +1. It is closed under countable unions. + +2. It is closed under complements, which implies it is also closed under countable intersections. + + +> **Intuition:** If you have a list of sets that are "in," then their total combination (union) and their common overlap (intersection) are also "in." + +**Theorem 1: Points as Intersections of Open Sets (De Barra, p. 7)** + +Any singleton set $\{x\}$ in $\mathbb{R}$ can be expressed as the intersection of a sequence of open intervals: $\{x\} = \bigcap_{n=1}^{\infty} (x - \frac{1}{n}, x + \frac{1}{n})$. + +> **Intuition:** While a single point isn't "open," you can trap it by taking a sequence of shrinking open intervals that squeeze down onto that one point. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +To prove that any countable set is in the Borel $\sigma$-algebra, we first show that a set containing a single point is Borel, and then we use the property of countable unions. + +1. **Prove that singleton sets are Borel sets.** + + Let $x \in \mathbb{R}$. As stated in Theorem 1, we can write the singleton set $\{x\}$ as: + + $$\{x\} = \bigcap_{n=1}^{\infty} \left(x - \frac{1}{n}, x + \frac{1}{n}\right)$$ + + By the definition of the Borel $\sigma$-algebra, every open interval $(x - \frac{1}{n}, x + \frac{1}{n})$ is an open set and thus belongs to $\mathcal{B}$. By the definition of a $\sigma$-algebra, the collection is closed under countable intersections. Therefore, $\{x\} \in \mathcal{B}$ for any $x \in \mathbb{R}$. + +2. **Define a countable set.** + + Let $C \subseteq \mathbb{R}$ be a countable set. By the definition of countability, we can list the elements of $C$ as a sequence $\{x_1, x_2, x_3, \dots\}$. We can then express the set $C$ as the union of these individual points: + + $$C = \bigcup_{i=1}^{\infty} \{x_i\}$$ + +3. **Apply the property of countable unions.** + + In Step 1, we established that each singleton set $\{x_i\}$ is an element of $\mathcal{B}$. By the definition of a $\sigma$-algebra, $\mathcal{B}$ must be closed under countable unions. Since $C$ is a countable union of sets that are each in $\mathcal{B}$, then $C$ itself must be in $\mathcal{B}$. + +4. **Conclusion.** + + Every countable subset of $\mathbb{R}$ can be constructed from Borel sets using the allowed operation of countable union. Thus, every countable subset of $\mathbb{R}$ is a Borel set. + + +____ +## Question + +Prove that any nonempty open subset of $\mathbb{R}$ is a countable union of disjoint open intervals. Furthermore, show that any nonempty open set in $\mathbb{R}^d$ is a countable union of open rectangles (sets of the form $(a_1, b_1) \times \dots \times (a_d, b_d)$), though these may not necessarily be disjoint. + +--- + +## Definitions and Theorems Used + +**Definition 1: Open Set in $\mathbb{R}^d$ (Folland, p. 13)** + +A set $U \subseteq \mathbb{R}^d$ is open if for every $x \in U$, there exists an open ball $B(x, r) = \{y \in \mathbb{R}^d : |x-y| < r\}$ such that $B(x, r) \subseteq U$. + +> **Intuition:** An open set is a set where every point has some "breathing room" around it that is also entirely contained within the set. + +**Definition 2: Density of Rational Numbers (Folland, p. 2)** + +The set of rational numbers $\mathbb{Q}$ (or $\mathbb{Q}^d$ in higher dimensions) is countable and dense in $\mathbb{R}$ (or $\mathbb{R}^d$). This means every nonempty open set contains at least one point with rational coordinates. + +> **Intuition:** Rational points are spread everywhere. No matter how small an open interval or box you pick, you will always find a point with rational coordinates inside it. + +**Definition 3: Connected Component (Folland, p. 34)** + +In the context of $\mathbb{R}$, the connected components of an open set are the maximal open intervals contained within that set. + +> **Intuition:** If you look at an open set on a number line, it might be made of several "pieces." Each continuous, unbroken piece is a connected component. + +--- + +## Solution (Step-by-Step Explanation) + +### Part 1: Open Sets in $\mathbb{R}$ + +1. **Define the Components:** Let $U$ be a nonempty open subset of $\mathbb{R}$. For each $x \in U$, let $I_x$ be the largest open interval such that $x \in I_x$ and $I_x \subseteq U$. This is the connected component of $U$ containing $x$. + +2. **Establish Disjointness:** By the definition of maximal intervals, if two such intervals $I_x$ and $I_y$ have any point in common, they must be the exact same interval. Therefore, the collection of all such distinct intervals $\{I_\alpha\}$ is pairwise disjoint, and their union is exactly $U$. + +3. **Prove Countability:** By the density of rational numbers (Definition 2), every nonempty open interval $I_\alpha$ must contain at least one rational number $q_\alpha$. Since the intervals are disjoint, each $q_\alpha$ must be unique to its interval. Because the set of rational numbers $\mathbb{Q}$ is countable, the collection of disjoint intervals must also be countable. + +4. **Conclusion for $\mathbb{R}$:** Thus, $U$ is the union of a countable collection of disjoint open intervals. + + +--- + +### Part 2: Open Sets in $\mathbb{R}^d$ + +1. **Define the Base of Rectangles:** Let $U$ be a nonempty open subset of $\mathbb{R}^d$. Consider the collection $\mathcal{R}$ of all open rectangles with rational endpoints (i.e., $(a_1, b_1) \times \dots \times (a_d, b_d)$ where all $a_i, b_i \in \mathbb{Q}$) that are entirely contained within $U$. + +2. **Prove the Union Equals $U$:** By the definition of an open set in $\mathbb{R}^d$ (Definition 1), for any point $x \in U$, there is an open ball around it contained in $U$. Inside that ball, we can always fit an open rectangle with rational endpoints that contains $x$. Therefore, every point $x \in U$ is inside at least one rectangle from our collection $\mathcal{R}$. This means the union of all rectangles in $\mathcal{R}$ is exactly $U$. + +3. **Prove Countability:** The set of rational numbers $\mathbb{Q}$ is countable. Since a rectangle in $\mathbb{R}^d$ is defined by $2d$ rational numbers (a starting and ending point for each dimension), the total number of such rectangles is a countable product of countable sets. By Theorem 1 from our previous lecture (De Barra, p. 7), this resulting collection $\mathcal{R}$ is countable. + +4. **Conclusion for $\mathbb{R}^d$:** Unlike the 1-dimensional case, these rectangles can overlap significantly, but they still successfully cover $U$ using only a countable number of elements. Thus, $U$ is a countable union of open rectangles. + + +___ +## Question + +For the following collections of subsets $\mathcal{C}_i$, $i = 1, 2, 3$, of the power set $\mathcal{P}(\mathbb{R})$, prove that $\mathcal{F}(\mathcal{C}_i) = \mathcal{B}$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$: + +- $\mathcal{C}_1 = \{(-\infty, x) \mid x \in \mathbb{R}\}$ + +- $\mathcal{C}_2 = \{(-\infty, x) \mid x \in \mathbb{Q}\}$ + +- $\mathcal{C}_3 = \{(a, b) \mid a \in \mathbb{Q}, b \in \mathbb{Q}\}$ + + +--- + +## Definitions and Theorems Used + +**Definition 1: Borel $\sigma$-algebra (Folland, p. 22)** + +The Borel $\sigma$-algebra $\mathcal{B}$ on $\mathbb{R}$ is the $\sigma$-algebra generated by the family of all open sets in $\mathbb{R}$. + +> **Intuition:** This is the smallest collection of sets that contains all open intervals and follows the structural rules of a $\sigma$-algebra. It is the standard "playground" for integration and measure on the real line. + +**Definition 2: Generated $\sigma$-algebra (Rudin, p. 8)** + +If $\mathcal{E}$ is any collection of subsets of $X$, the $\sigma$-algebra generated by $\mathcal{E}$, denoted $\mathcal{F}(\mathcal{E})$, is the smallest $\sigma$-algebra containing every set in $\mathcal{E}$. + +> **Intuition:** If you have a starting set of tools ($\mathcal{E}$), the generated $\sigma$-algebra is everything you can possibly build using those tools while following the rules of complements and countable unions. + +**Theorem 1: Structure of Open Sets in $\mathbb{R}$ (Folland, p. 13)** + +Every open set in $\mathbb{R}$ can be written as a countable union of open intervals. Furthermore, every open interval $(a, b)$ is a countable union of intervals with rational endpoints. + +> **Intuition:** Because rational numbers are "everywhere" (dense), we can approximate any interval or open set using only a countable list of intervals that have rational coordinates. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +To prove $\mathcal{F}(\mathcal{C}_i) = \mathcal{B}$, we show two inclusions for each case: $\mathcal{F}(\mathcal{C}_i) \subseteq \mathcal{B}$ and $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_i)$. + +### 1. Proving $\mathcal{F}(\mathcal{C}_1) = \mathcal{B}$ + +1. **First Inclusion:** Every set in $\mathcal{C}_1$ is an open interval of the form $(-\infty, x)$. By the definition of the Borel $\sigma$-algebra, every open set is in $\mathcal{B}$. Since $\mathcal{B}$ is a $\sigma$-algebra containing $\mathcal{C}_1$, it must contain the smallest $\sigma$-algebra generated by $\mathcal{C}_1$. Thus, $\mathcal{F}(\mathcal{C}_1) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** Any open interval $(a, b)$ can be written as $(-\infty, b) \cap (-\infty, a]^c$. Since $(-\infty, a] = \bigcap_{n=1}^{\infty} (-\infty, a + \frac{1}{n})$, and each $(-\infty, a + \frac{1}{n}) \in \mathcal{C}_1$, then $(-\infty, a]$ is in $\mathcal{F}(\mathcal{C}_1)$ by closure under countable intersections. Its complement is also in the $\sigma$-algebra. Thus, $(a, b) \in \mathcal{F}(\mathcal{C}_1)$. By Theorem 1, all open sets are countable unions of such intervals, so $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_1)$. + + +### 2. Proving $\mathcal{F}(\mathcal{C}_2) = \mathcal{B}$ + +1. **First Inclusion:** Since $\mathcal{C}_2 \subset \mathcal{C}_1$, and we already showed $\mathcal{F}(\mathcal{C}_1) \subseteq \mathcal{B}$, it follows immediately that $\mathcal{F}(\mathcal{C}_2) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** For any $x \in \mathbb{R}$, we can choose a sequence of rationals $q_n$ such that $q_n \uparrow x$. Then $(-\infty, x) = \bigcup_{n=1}^{\infty} (-\infty, q_n)$. Since each $(-\infty, q_n) \in \mathcal{C}_2$, the union is in $\mathcal{F}(\mathcal{C}_2)$. This shows $\mathcal{C}_1 \subseteq \mathcal{F}(\mathcal{C}_2)$. Consequently, $\mathcal{F}(\mathcal{C}_1) \subseteq \mathcal{F}(\mathcal{C}_2)$. Since we already established $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_1)$, we have $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_2)$. + + +### 3. Proving $\mathcal{F}(\mathcal{C}_3) = \mathcal{B}$ + +1. **First Inclusion:** Every set $(a, b)$ with $a, b \in \mathbb{Q}$ is an open set. By the definition of the Borel $\sigma$-algebra, these are all in $\mathcal{B}$. Therefore, $\mathcal{F}(\mathcal{C}_3) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** By Theorem 1, any open interval $(x, y)$ can be written as a countable union of intervals with rational endpoints. For example, $(x, y) = \bigcup \{ (a, b) : a, b \in \mathbb{Q}, x < a < b < y \}$. 3. Since this is a countable union of elements from $\mathcal{C}_3$, every open interval is in $\mathcal{F}(\mathcal{C}_3)$. Because every open set is a countable union of open intervals, all open sets are in $\mathcal{F}(\mathcal{C}_3)$. Thus, $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_3)$. + + +**Conclusion:** + +In all three cases, the generated $\sigma$-algebra is exactly the Borel $\sigma$-algebra $\mathcal{B}$. + +____ +## Question + +For the following collections of subsets $\mathcal{C}_i$, $i = 1, \dots, 4$, of the power set $\mathcal{P}(\mathbb{R}^d)$ where $d > 1$, prove that $\mathcal{F}(\mathcal{C}_i) = \mathcal{B}$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}^d$: + +- $\mathcal{C}_1 = \{(a_1, b_1) \times \dots \times (a_d, b_d) \mid -\infty \leq a_i \leq b_i \leq \infty, 1 \leq i \leq d\}$ + +- $\mathcal{C}_2 = \{(-\infty, x_1) \times \dots \times (-\infty, x_d) \mid x_i \in \mathbb{R}, 1 \leq i \leq d\}$ + +- $\mathcal{C}_3 = \{(-\infty, x_1) \times \dots \times (-\infty, x_d) \mid x_i \in \mathbb{Q}, 1 \leq i \leq d\}$ + +- $\mathcal{C}_4 = \{(a_1, b_1) \times \dots \times (a_d, b_d) \mid -\infty \leq a_i \leq b_i \leq \infty, 1 \leq i \leq d, a_i \in \mathbb{Q}, b_i \in \mathbb{Q}\}$ + + +--- + +## Definitions and Theorems Used + +**Definition 1: Borel $\sigma$-algebra (Folland, p. 22)** + +The Borel $\sigma$-algebra $\mathcal{B}$ on $\mathbb{R}^d$ is the $\sigma$-algebra generated by the family of all open sets in $\mathbb{R}^d$. + +> **Intuition:** This is the smallest collection of sets containing all open "bubbles" in space and remaining closed under standard set operations. It is the standard collection of sets we consider "measurable" in Euclidean space. + +**Definition 2: Generated $\sigma$-algebra (Rudin, p. 8)** + +If $\mathcal{E}$ is a collection of subsets of $X$, the $\sigma$-algebra generated by $\mathcal{E}$, denoted $\mathcal{F}(\mathcal{E})$, is the smallest $\sigma$-algebra containing every set in $\mathcal{E}$. + +> **Intuition:** If you have a set of basic building blocks ($\mathcal{E}$), the generated $\sigma$-algebra is the total collection of everything you can build using those blocks, their opposites (complements), and infinite lists of them (countable unions). + +**Theorem 1: Open Sets as Countable Unions of Rectangles (Folland, p. 13)** + +Every open set in $\mathbb{R}^d$ can be expressed as a countable union of open rectangles with rational coordinates. + +> **Intuition:** Rational numbers are dense (they are everywhere). This means we can approximate any complex open shape perfectly using an infinite but listable number of simple boxes with rational corners. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +To prove $\mathcal{F}(\mathcal{C}_i) = \mathcal{B}$, we must show that $\mathcal{C}_i \subseteq \mathcal{B}$ and that $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_i)$. + +### 1. Proving $\mathcal{F}(\mathcal{C}_1) = \mathcal{B}$ + +1. **First Inclusion:** Every set in $\mathcal{C}_1$ is an open rectangle. By the definition of the Borel $\sigma$-algebra, all open sets are in $\mathcal{B}$. Thus, $\mathcal{C}_1 \subseteq \mathcal{B}$, which implies $\mathcal{F}(\mathcal{C}_1) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** By Theorem 1, any open set in $\mathbb{R}^d$ is a countable union of open rectangles. Since $\mathcal{F}(\mathcal{C}_1)$ is a $\sigma$-algebra containing all open rectangles, it must contain all open sets. Therefore, $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_1)$. + + +### 2. Proving $\mathcal{F}(\mathcal{C}_2) = \mathcal{B}$ + +1. **First Inclusion:** The sets in $\mathcal{C}_2$ are open "lower-left" infinite rectangles. Since these are open sets, $\mathcal{C}_2 \subseteq \mathcal{B}$, so $\mathcal{F}(\mathcal{C}_2) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** A finite rectangle $(a_i, b_i)$ in one dimension can be written as $(-\infty, b_i) \cap (-\infty, a_i]^c$. In $d$ dimensions, we can construct any rectangle in $\mathcal{C}_1$ using intersections and complements of sets in $\mathcal{C}_2$. For example, the closed-left interval $(-\infty, a_i] = \bigcap_{n=1}^\infty (-\infty, a_i + 1/n)$. Since $\mathcal{F}(\mathcal{C}_2)$ is closed under these operations, $\mathcal{C}_1 \subseteq \mathcal{F}(\mathcal{C}_2)$, implying $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_2)$. + + +### 3. Proving $\mathcal{F}(\mathcal{C}_3) = \mathcal{B}$ + +1. **First Inclusion:** $\mathcal{C}_3 \subset \mathcal{C}_2 \subseteq \mathcal{B}$. Thus, $\mathcal{F}(\mathcal{C}_3) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** For any $x_i \in \mathbb{R}$, the set $(-\infty, x_1) \times \dots \times (-\infty, x_d)$ is the countable union of sets from $\mathcal{C}_3$ where the coordinates are rational sequences $q_{i,n} \uparrow x_i$. This shows $\mathcal{C}_2 \subseteq \mathcal{F}(\mathcal{C}_3)$, and since $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_2)$, it follows that $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_3)$. + + +### 4. Proving $\mathcal{F}(\mathcal{C}_4) = \mathcal{B}$ + +1. **First Inclusion:** $\mathcal{C}_4$ consists of open rectangles with rational endpoints. These are open sets, so $\mathcal{F}(\mathcal{C}_4) \subseteq \mathcal{B}$. + +2. **Second Inclusion:** By Theorem 1, every open set in $\mathbb{R}^d$ is a countable union of rectangles with rational coordinates. These rectangles are exactly the elements of $\mathcal{C}_4$. Since $\mathcal{F}(\mathcal{C}_4)$ is closed under countable unions, it contains all open sets, meaning $\mathcal{B} \subseteq \mathcal{F}(\mathcal{C}_4)$. + + +**Conclusion:** + +All four collections generate the same Borel $\sigma$-algebra $\mathcal{B}$ on $\mathbb{R}^d$. + +____ +## Question + +If $\mathcal{C} = \{ \{x\} \mid x \in \mathbb{R} \}$ is the collection of all singleton sets in $\mathbb{R}$, prove that $\mathcal{F}(\mathcal{C}) = \mathcal{F}_c$ and that $\mathcal{F}(\mathcal{C})$ is properly contained in the Borel $\sigma$-algebra $\mathcal{B}$. + +Note: $\mathcal{F}_c$ refers to the countable-cocountable $\sigma$-algebra, defined as: + +$$\mathcal{F}_c = \{ A \subseteq \mathbb{R} \mid A \text{ is countable or } A^c \text{ is countable} \}$$ + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra generated by a collection (Folland, p. 22)** + +The $\sigma$-algebra generated by a collection $\mathcal{E}$, denoted $\mathcal{F}(\mathcal{E})$, is the smallest $\sigma$-algebra containing all sets in $\mathcal{E}$. + +> **Intuition:** Think of this as the "minimalist" collection. It contains exactly what you started with plus only the sets you are forced to include to satisfy the rules of a $\sigma$-algebra (complements and countable unions). + +**Definition 2: The Countable-Cocountable $\sigma$-algebra (Folland, p. 26)** + +The collection $\mathcal{F}_c$ consists of all subsets of $\mathbb{R}$ that are either countable or have a countable complement. + +> **Intuition:** This collection is "thin" at the ends. A set is in $\mathcal{F}_c$ only if it is very small (countable) or so large that it covers almost everything (leaving only a countable amount out). + +**Definition 3: Borel $\sigma$-algebra (Rudin, p. 12)** + +The Borel $\sigma$-algebra $\mathcal{B}$ is the $\sigma$-algebra generated by the open intervals in $\mathbb{R}$. + +> **Intuition:** This is the standard collection of "measurable" sets on the real line. It contains intervals, points, open sets, and closed sets. + +**Theorem 1: Countability of the Rational Numbers (Folland, p. 2)** + +The set of rational numbers $\mathbb{Q}$ is countable, but any open interval $(a, b)$ is uncountable. + +> **Intuition:** While you can list the rational numbers $1, 2, 3...$, you can never list all the points in even a tiny segment of the number line. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +### Part 1: Proving $\mathcal{F}(\mathcal{C}) = \mathcal{F}_c$ + +1. **Show that $\mathcal{F}(\mathcal{C}) \subseteq \mathcal{F}_c$:** Every element in $\mathcal{C}$ is a singleton set $\{x\}$, which is finite and therefore countable. Since every set in $\mathcal{C}$ is countable, every set in $\mathcal{C}$ belongs to $\mathcal{F}_c$ by the definition of the countable-cocountable $\sigma$-algebra. Because $\mathcal{F}_c$ is a $\sigma$-algebra containing $\mathcal{C}$, and $\mathcal{F}(\mathcal{C})$ is the _smallest_ such $\sigma$-algebra, it follows that $\mathcal{F}(\mathcal{C}) \subseteq \mathcal{F}_c$. + +2. **Show that $\mathcal{F}_c \subseteq \mathcal{F}(\mathcal{C})$:** Let $A$ be a set in $\mathcal{F}_c$. By definition, $A$ is either countable or $A^c$ is countable. + + - If $A$ is countable, we can write $A = \bigcup_{i=1}^{\infty} \{x_i\}$. Since each singleton $\{x_i\}$ is in $\mathcal{C}$, the countable union must be in $\mathcal{F}(\mathcal{C})$ by the definition of a $\sigma$-algebra. + + - If $A^c$ is countable, then by the same logic, $A^c \in \mathcal{F}(\mathcal{C})$. Since a $\sigma$-algebra is closed under complements, $(A^c)^c = A$ must also be in $\mathcal{F}(\mathcal{C})$. + + Thus, every set in $\mathcal{F}_c$ is in $\mathcal{F}(\mathcal{C})$, so $\mathcal{F}_c \subseteq \mathcal{F}(\mathcal{C})$. + +3. **Conclusion for Part 1:** Since we have inclusion in both directions, $\mathcal{F}(\mathcal{C}) = \mathcal{F}_c$. + + +### Part 2: Proving $\mathcal{F}(\mathcal{C})$ is properly contained in $\mathcal{B}$ + +1. **Show Inclusion ($\mathcal{F}_c \subseteq \mathcal{B}$):** We previously proved that every countable subset of $\mathbb{R}$ is a Borel set. Since every $A \in \mathcal{F}_c$ is either countable (meaning $A \in \mathcal{B}$) or its complement is countable (meaning $A^c \in \mathcal{B}$, and thus $A \in \mathcal{B}$), we have $\mathcal{F}_c \subseteq \mathcal{B}$. + +2. **Show "Proper" Inclusion ($\mathcal{F}_c \neq \mathcal{B}$):** To show the containment is proper, we must find a Borel set that is not in $\mathcal{F}_c$. Consider the open interval $I = (0, 1)$. + + - $I$ is an open set, so by definition, $I \in \mathcal{B}$. + + - Is $I \in \mathcal{F}_c$? An interval $(0, 1)$ is uncountable. Furthermore, its complement $I^c = (-\infty, 0] \cup [1, \infty)$ is also uncountable. + + - Since neither $I$ nor $I^c$ is countable, $I \notin \mathcal{F}_c$. + +3. **Conclusion for Part 2:** Because there are Borel sets (like intervals) that are not in $\mathcal{F}_c$, the collection $\mathcal{F}(\mathcal{C})$ is properly contained in $\mathcal{B}$. + + +____ +## Question + +Let $\overline{\mathbb{R}} = \mathbb{R} \cup \{\infty\} \cup \{-\infty\}$ be the extended real line. The Borel $\sigma$-algebra on $\overline{\mathbb{R}}$, denoted by $\overline{\mathcal{B}}$, is defined as the smallest $\sigma$-algebra containing $\mathcal{B} \cup \{\infty\} \cup \{-\infty\}$, where $\mathcal{B}$ is the Borel $\sigma$-algebra on $\mathbb{R}$. Prove that: + +$$\overline{\mathcal{B}} = \{A \cup B \mid A \in \mathcal{B}, B \subseteq \{-\infty, \infty\}\}$$ + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra (Folland, p. 21)** + +A collection of subsets $\mathcal{M}$ of a set $X$ is a $\sigma$-algebra if: + +1. $\emptyset \in \mathcal{M}$. + +2. If $E \in \mathcal{M}$, then its complement $E^c \in \mathcal{M}$. + +3. If $\{E_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then their union $\bigcup_{n=1}^{\infty} E_n \in \mathcal{M}$. + + +> **Intuition:** A $\sigma$-algebra is a collection of sets that is "closed" under standard operations. If you have a set, you have its opposite; if you have a list of sets, you have the set formed by combining them all together. + +**Definition 2: Generated $\sigma$-algebra (Rudin, p. 8)** + +If $\mathcal{E}$ is a collection of subsets of $X$, the $\sigma$-algebra generated by $\mathcal{E}$ is the smallest $\sigma$-algebra that contains every set in $\mathcal{E}$. + +> **Intuition:** This is the "minimalist" $\sigma$-algebra. You start with a few basic building blocks and add only what is strictly necessary to satisfy the rules of a $\sigma$-algebra. + +**Definition 3: Borel $\sigma$-algebra on $\mathbb{R}$ (Folland, p. 22)** + +The Borel $\sigma$-algebra $\mathcal{B}$ is the $\sigma$-algebra generated by the open sets (or open intervals) of $\mathbb{R}$. + +> **Intuition:** This is the standard collection of "measurable" sets on the real line, including intervals, points, and anything that can be built from them. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +To prove the equality of two collections, we let $\mathcal{M} = \{A \cup B \mid A \in \mathcal{B}, B \subseteq \{-\infty, \infty\}\}$ and show that $\overline{\mathcal{B}} = \mathcal{M}$. This requires two steps: showing $\mathcal{M}$ is a $\sigma$-algebra that contains the generating sets, and showing that any $\sigma$-algebra containing the generating sets must contain $\mathcal{M}$. + +**Step 1: Prove that $\mathcal{M}$ is a $\sigma$-algebra.** + +We check the three axioms of a $\sigma$-algebra for the collection $\mathcal{M}$: + +1. **Empty set:** Since $\emptyset \in \mathcal{B}$ and $\emptyset \subseteq \{-\infty, \infty\}$, we have $\emptyset \cup \emptyset = \emptyset \in \mathcal{M}$. + +2. **Complements:** Let $E = A \cup B \in \mathcal{M}$, where $A \in \mathcal{B}$ and $B \subseteq \{-\infty, \infty\}$. Its complement in $\overline{\mathbb{R}}$ is $E^c = \overline{\mathbb{R}} \setminus (A \cup B)$. Since $A \subseteq \mathbb{R}$ and $B \subseteq \{-\infty, \infty\}$, we can write this as $(\mathbb{R} \setminus A) \cup (\{-\infty, \infty\} \setminus B)$. Because $\mathcal{B}$ is a $\sigma$-algebra on $\mathbb{R}$, $A^c = \mathbb{R} \setminus A$ is in $\mathcal{B}$. The set $B' = \{-\infty, \infty\} \setminus B$ is clearly a subset of $\{-\infty, \infty\}$. Thus, $E^c \in \mathcal{M}$. + +3. **Countable Unions:** Let $E_n = A_n \cup B_n$ be a sequence in $\mathcal{M}$. Then $\bigcup E_n = (\bigcup A_n) \cup (\bigcup B_n)$. Since $\mathcal{B}$ is a $\sigma$-algebra, $A = \bigcup A_n \in \mathcal{B}$. The set $B = \bigcup B_n$ is still a subset of $\{-\infty, \infty\}$. Thus, $\bigcup E_n \in \mathcal{M}$. + + +**Step 2: Prove that $\overline{\mathcal{B}} \subseteq \mathcal{M}$.** + +By Definition 2, $\overline{\mathcal{B}}$ is the smallest $\sigma$-algebra containing $\mathcal{B}$, $\{\infty\}$, and $\{-\infty\}$. + +1. Every $A \in \mathcal{B}$ is in $\mathcal{M}$ (by taking $B = \emptyset$). + +2. The set $\{\infty\}$ is in $\mathcal{M}$ (by taking $A = \emptyset$ and $B = \{\infty\}$). + +3. The set $\{-\infty\}$ is in $\mathcal{M}$ (by taking $A = \emptyset$ and $B = \{-\infty\}$). + + Since $\mathcal{M}$ is a $\sigma$-algebra (from Step 1) that contains all the generating sets of $\overline{\mathcal{B}}$, the smallest such $\sigma$-algebra must be contained within it. Thus, $\overline{\mathcal{B}} \subseteq \mathcal{M}$. + + +**Step 3: Prove that $\mathcal{M} \subseteq \overline{\mathcal{B}}$.** + +Let $E = A \cup B$ be an arbitrary element of $\mathcal{M}$. + +1. By the definition of $\overline{\mathcal{B}}$, it contains $\mathcal{B}$. Thus $A \in \overline{\mathcal{B}}$. + +2. By the definition of $\overline{\mathcal{B}}$, it contains $\{\infty\}$ and $\{-\infty\}$. Any subset $B$ of $\{-\infty, \infty\}$ is a finite union of these sets (or the empty set), so $B \in \overline{\mathcal{B}}$ by the union axiom of $\sigma$-algebras. + +3. Since $A \in \overline{\mathcal{B}}$ and $B \in \overline{\mathcal{B}}$, their union $A \cup B$ must also be in $\overline{\mathcal{B}}$. Thus, $\mathcal{M} \subseteq \overline{\mathcal{B}}$. + + +**Step 4: Conclusion.** + +Since $\mathcal{M} \subseteq \overline{\mathcal{B}}$ and $\overline{\mathcal{B}} \subseteq \mathcal{M}$, we conclude that $\overline{\mathcal{B}} = \{A \cup B \mid A \in \mathcal{B}, B \subseteq \{-\infty, \infty\}\}$. + +___ +## Question + +Let $X$ and $Y$ be nonempty sets and $f : X \to Y$ be a given function. For any subset $A \subseteq Y$, we define the preimage (inverse image) of $A$ under $f$ as: + +$$f^{-1}(A) = \{x \in X \mid f(x) \in A\}$$ + +Note that this definition holds even if the inverse function $f^{-1}$ does not exist. + +Prove that if $\mathcal{F}$ is a $\sigma$-algebra on $Y$, then the collection + +$$\mathcal{F}_f = \{f^{-1}(A) \mid A \in \mathcal{F}\}$$ + +is a $\sigma$-algebra on $X$. + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra (Folland, p. 21)** + +A collection of subsets $\mathcal{M}$ of a set $X$ is a $\sigma$-algebra if: + +1. The empty set $\emptyset$ is in $\mathcal{M}$. + +2. If $E \in \mathcal{M}$, then its complement $E^c = X \setminus E$ is in $\mathcal{M}$. + +3. If $\{E_n\}_{n=1}^{\infty}$ is a sequence of sets in $\mathcal{M}$, then their union $\bigcup_{n=1}^{\infty} E_n$ is in $\mathcal{M}$. + + +> **Intuition:** A $\sigma$-algebra is a collection that is "closed" under basic set operations. If you start with sets in the collection and apply "not" (complement) or "or" (union), the result stays in the collection. + +**Theorem 1: Properties of Preimages (De Barra, p. 2)** + +For any function $f: X \to Y$, the following set-theoretic properties hold for subsets of $Y$: + +1. $f^{-1}(\emptyset) = \emptyset$ and $f^{-1}(Y) = X$. + +2. $f^{-1}(A^c) = [f^{-1}(A)]^c$. + +3. $f^{-1}(\bigcup_{n=1}^{\infty} A_n) = \bigcup_{n=1}^{\infty} f^{-1}(A_n)$. + + +> **Intuition:** Preimages are very "well-behaved" compared to direct images. They preserve all standard set operations like unions, intersections, and complements perfectly. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +To prove that $\mathcal{F}_f$ is a $\sigma$-algebra on $X$, we must verify the three axioms of the definition of a $\sigma$-algebra. + +**Step 1: Verify that the empty set is in $\mathcal{F}_f$.** + +1. Since $\mathcal{F}$ is a $\sigma$-algebra on $Y$, we know that $\emptyset \in \mathcal{F}$ by the definition of a $\sigma$-algebra. + +2. By the properties of preimages, $f^{-1}(\emptyset) = \emptyset$. + +3. Since $\emptyset$ can be written as the preimage of a set in $\mathcal{F}$ (namely, $\emptyset$ itself), it follows by the definition of $\mathcal{F}_f$ that $\emptyset \in \mathcal{F}_f$. + + +**Step 2: Verify closure under complements.** + +1. Let $E$ be an arbitrary set in $\mathcal{F}_f$. We must show that $E^c \in \mathcal{F}_f$. + +2. By the definition of $\mathcal{F}_f$, there must exist a set $A \in \mathcal{F}$ such that $E = f^{-1}(A)$. + +3. The complement of $E$ in $X$ is $E^c = [f^{-1}(A)]^c$. + +4. By the property of preimages stated in Theorem 1, $[f^{-1}(A)]^c = f^{-1}(A^c)$. + +5. Since $A \in \mathcal{F}$ and $\mathcal{F}$ is a $\sigma$-algebra, its complement $A^c$ is also in $\mathcal{F}$. + +6. Because $E^c$ is the preimage of a set in $\mathcal{F}$ (the set $A^c$), it satisfies the requirement to be in $\mathcal{F}_f$. + + +**Step 3: Verify closure under countable unions.** + +1. Let $\{E_n\}_{n=1}^{\infty}$ be a sequence of sets in $\mathcal{F}_f$. We must show that $\bigcup_{n=1}^{\infty} E_n \in \mathcal{F}_f$. + +2. By the definition of $\mathcal{F}_f$, for each $n$, there exists a set $A_n \in \mathcal{F}$ such that $E_n = f^{-1}(A_n)$. + +3. The union of our sequence is $\bigcup_{n=1}^{\infty} f^{-1}(A_n)$. + +4. By the property of preimages stated in Theorem 1, the union of preimages is the preimage of the union: $\bigcup_{n=1}^{\infty} f^{-1}(A_n) = f^{-1}(\bigcup_{n=1}^{\infty} A_n)$. + +5. Since each $A_n \in \mathcal{F}$ and $\mathcal{F}$ is a $\sigma$-algebra, the countable union $\bigcup_{n=1}^{\infty} A_n$ is also an element of $\mathcal{F}$. + +6. Because the union of the $E_n$ sequence is expressed as the preimage of a set in $\mathcal{F}$, the union belongs to $\mathcal{F}_f$. + + +**Step 4: Conclusion.** + +Since $\mathcal{F}_f$ contains the empty set and is closed under complements and countable unions, it satisfies the definition of a $\sigma$-algebra on $X$. + +____ +## Question + +The aim of this exercise is to prove that any $\sigma$-algebra is either finite or uncountable. + +(a) Suppose $X$ is a nonempty set and $\mathcal{F}$ is a $\sigma$-algebra on $X$. If $\mathcal{F}$ is an infinite set, prove that so is $X$. + +(b) Suppose $X$ is a countably infinite set and so is $\mathcal{F}$. Define $f : X \to \mathcal{F}$ by + +$$f(x) = \bigcap_{x \in A \in \mathcal{F}} A, \quad x \in X.$$ + +Prove that $f(x)$ is the smallest set in $\mathcal{F}$ containing $x$. + +(c) If $x \in X$, $y \in X$ and $f(x) \cap f(y) \neq \emptyset$, then prove that $f(x) = f(y)$. Conclude that $f(X)$ (which is a subset of $\mathcal{F}$) is a partition of $X$. + +(d) If $A \in \mathcal{F}$, then prove that $A = \bigcup_{x \in A} f(x)$. + +(e) If $X$ is an infinite set, then prove that $f(X)$ is an infinite subset of $\mathcal{F}$. + +(f) If $X$ is an infinite set and $\mathcal{F}$ is an infinite set, then $\mathcal{F}$ must be uncountable. + +--- + +## Definitions and Theorems Used + +**Definition 1: $\sigma$-algebra (Folland, p. 21)** + +A collection of subsets $\mathcal{F}$ of $X$ is a $\sigma$-algebra if it contains the empty set and is closed under complements and countable unions. + +> **Intuition:** A $\sigma$-algebra is a collection of sets where you can always perform basic "logic" (and, or, not) infinitely many times without leaving the collection. + +**Definition 2: Partition (De Barra, p. 4)** + +A partition of $X$ is a collection of disjoint nonempty sets whose union is $X$. + +> **Intuition:** A partition slices a set into distinct, non-overlapping pieces that perfectly cover the original set. + +**Theorem 1: Countable Intersections in $\sigma$-algebras (Rudin, p. 8)** + +If $\mathcal{F}$ is a $\sigma$-algebra and $\{A_n\}_{n=1}^{\infty}$ is a sequence in $\mathcal{F}$, then $\bigcap_{n=1}^{\infty} A_n \in \mathcal{F}$. + +> **Intuition:** Because $\sigma$-algebras are closed under countable unions and complements, they are automatically closed under countable intersections (the overlap of infinitely many sets). + +**Theorem 2: Power Set Cardinality (Folland, p. 2)** + +If a set $S$ is infinite, the collection of all its subsets (the power set) is uncountable. + +> **Intuition:** Even if you can count the items in a set, you cannot count all the different ways to group them together. + +--- + +## Solution (Step-by-Step Explanation) + +**Solution** + +### (a) Proving $X$ is infinite + +1. Suppose $X$ is a finite set. If $X$ is finite, the total number of its subsets is also finite ($2^{|X|}$). + +2. Since any $\sigma$-algebra $\mathcal{F}$ on $X$ is a subcollection of the power set $\mathcal{P}(X)$, $\mathcal{F}$ must also be finite. + +3. This contradicts the hypothesis that $\mathcal{F}$ is infinite. Thus, $X$ must be infinite. + + +### (b) Smallest set containing $x$ + +1. By the definition of $f(x)$, it is the intersection of all sets $A$ in $\mathcal{F}$ that contain $x$. + +2. Since $\mathcal{F}$ is assumed to be a countably infinite set in this step, this intersection is a countable intersection of sets in $\mathcal{F}$. + +3. By the properties of a $\sigma$-algebra (Theorem 1), $f(x)$ belongs to $\mathcal{F}$. + +4. Since every $A$ in the intersection contains $x$, their intersection $f(x)$ also contains $x$. By construction, any $B \in \mathcal{F}$ containing $x$ is one of the sets being intersected, so $f(x) \subseteq B$. Thus, $f(x)$ is the smallest set in $\mathcal{F}$ containing $x$. + + +### (c) Disjoint or Identical (Partition) + +1. Suppose $f(x) \cap f(y) \neq \emptyset$ and let $z \in f(x) \cap f(y)$. + +2. Since $z \in f(x)$ and $f(z)$ is the smallest set in $\mathcal{F}$ containing $z$, we have $f(z) \subseteq f(x)$. + +3. If $f(z)$ were a proper subset of $f(x)$, then $x$ could not be in $f(z)$ (otherwise $f(x)$ wouldn't be the smallest). However, if $x \notin f(z)$, we could consider $f(x) \setminus f(z)$, which is a set in $\mathcal{F}$ containing $x$ smaller than $f(x)$, a contradiction. + +4. Thus $f(z) = f(x)$. By symmetry, $f(z) = f(y)$, so $f(x) = f(y)$. + +5. Since every $x \in X$ is in exactly one such set $f(x)$, the collection $f(X)$ forms a partition of $X$. + + +### (d) Representing $A$ as a union + +1. For any $x \in A$, we have $f(x) \subseteq A$ because $A$ is a set in $\mathcal{F}$ containing $x$, and $f(x)$ is the smallest such set. + +2. Taking the union over all $x \in A$, we get $\bigcup_{x \in A} f(x) \subseteq A$. + +3. Since every $x \in A$ is contained in its own $f(x)$, we also have $A \subseteq \bigcup_{x \in A} f(x)$. Thus, $A = \bigcup_{x \in A} f(x)$. + + +### (e) $f(X)$ is an infinite subset + +1. If $f(X)$ were a finite collection of sets $\{E_1, E_2, \dots, E_n\}$, then every set $A \in \mathcal{F}$ would be a union of some sub-collection of these $n$ sets (by part d).z + +2. This would mean $\mathcal{F}$ has at most $2^n$ elements, making $\mathcal{F}$ finite. + +3. Since $\mathcal{F}$ is infinite, $f(X)$ must be an infinite collection of disjoint sets. + + +### (f) $\mathcal{F}$ must be uncountable + +1. From part (e), $f(X)$ contains a countably infinite sequence of disjoint sets $\{E_1, E_2, E_3, \dots\}$. + +2. For any subset of indices $J \subseteq \mathbb{N}$, we can define a set $A_J = \bigcup_{j \in J} E_j$. + +3. By the definition of a $\sigma$-algebra, every such countable union $A_J$ is in $\mathcal{F}$. + +4. Since the sets $E_j$ are disjoint, different index sets $J$ produce different sets $A_J$. + +5. The number of such sets is equal to the number of subsets of $\mathbb{N}$, which is uncountable (Theorem 2). + + +____ diff --git a/content/SEM_6/Measure_Theory/Assignment/Assignment 2.md b/content/SEM_6/Measure_Theory/Assignment/Assignment 2.md new file mode 100644 index 00000000..bfcf55e3 --- /dev/null +++ b/content/SEM_6/Measure_Theory/Assignment/Assignment 2.md @@ -0,0 +1,1548 @@ + +## Question + +Consider $(\Omega, \mathcal{A}, \nu)$, where $\mathcal{A}$ is an algebra over $\Omega$ and $\nu$ is a premeasure. Let $\nu^*$ be the outer measure on $\mathcal{P}(\Omega)$ induced by $\nu$, and let $\mathcal{M}$ be the corresponding $\sigma$-algebra of $\nu^*$-measurable sets, defined by: + +$$\mathcal{M} = \{E \subseteq \Omega : \nu^*(A) = \nu^*(A \cap E) + \nu^*(A \cap E^c), \forall A \subseteq \Omega\}$$ + +(Note: The image contains a slight typo in the definition of $\mathcal{M}$ by using $\nu$ instead of $\nu^*$; the standard CarathΓ©odory condition uses the outer measure $\nu^*$). + +- **(a)** If $\nu^*(E) = 0$, then prove that $E \in \mathcal{M}$. + +- **(b)** Prove that if $\nu^*(A) = 0$, then $\nu^*(A \cup B) = \nu^*(B)$ for any $B \subseteq \Omega$. + +- **(c)** If $F \in \mathcal{M}$ and $\nu^*(F \Delta E) = 0$, then prove that $E \in \mathcal{M}$, where $F \Delta E = (F \setminus E) \cup (E \setminus F)$. + +- **(d)** Prove that for any $A \subseteq \Omega$, there exists $B \in \sigma(\mathcal{A})$ such that $A \subseteq B$ and $\nu^*(A) = \nu^*(B)$. (Note: The image uses $\mathcal{F}(A)$, which typically denotes the $\sigma$-algebra generated by the algebra $\mathcal{A}$). + + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)**: + +1. **Outer Measure Definition:** For any $A \subseteq \Omega$, $\nu^*(A) = \inf \{ \sum_{j=1}^\infty \nu(E_j) : E_j \in \mathcal{A}, A \subseteq \bigcup_{j=1}^\infty E_j \}$. + + - _Intuition:_ We approximate the "size" of any set from the outside using a countable union of simple pieces from our algebra. + +2. **Monotonicity of Outer Measure:** If $A \subseteq B$, then $\nu^*(A) \leq \nu^*(B)$. + + - _Intuition:_ A part cannot be larger than the whole. + +3. **Countable Subadditivity:** $\nu^*(\bigcup_{j=1}^\infty A_j) \leq \sum_{j=1}^\infty \nu^*(A_j)$. + + - _Intuition:_ The total size of a combined set is at most the sum of the sizes of its individual parts. + +4. **CarathΓ©odory Condition:** A set $E$ is measurable if it splits every "test set" $A$ additively. Since subadditivity always holds ($\nu^*(A) \leq \nu^*(A \cap E) + \nu^*(A \cap E^c)$), we only need to show: + + $$\nu^*(A) \geq \nu^*(A \cap E) + \nu^*(A \cap E^c)$$ + + +--- + +## Solution + +### Part (a) + +**Goal:** Prove that if $\nu^*(E) = 0$, then $E \in \mathcal{M}$. + +1. Pick an arbitrary "test set" $A \subseteq \Omega$. + +2. By the **Monotonicity** of outer measure, since $(A \cap E) \subseteq E$ and $\nu^*(E) = 0$, we have $\nu^*(A \cap E) = 0$. + +3. Similarly, since $(A \cap E^c) \subseteq A$, by **Monotonicity** we have $\nu^*(A \cap E^c) \leq \nu^*(A)$. + +4. Adding these two results: + + $$\nu^*(A \cap E) + \nu^*(A \cap E^c) = 0 + \nu^*(A \cap E^c) \leq \nu^*(A)$$ + +5. This satisfies the CarathΓ©odory criterion. Therefore, any set with outer measure zero is measurable. + + +### Part (b) + +**Goal:** Prove that if $\nu^*(A) = 0$, then $\nu^*(A \cup B) = \nu^*(B)$. + +1. By **Monotonicity**, since $B \subseteq (A \cup B)$, we know that $\nu^*(B) \leq \nu^*(A \cup B)$. + +2. By **Countable Subadditivity** (specifically finite subadditivity): + + $$\nu^*(A \cup B) \leq \nu^*(A) + \nu^*(B)$$ + +3. Since $\nu^*(A) = 0$, the inequality becomes $\nu^*(A \cup B) \leq 0 + \nu^*(B) = \nu^*(B)$. + +4. Since we have shown $\nu^*(B) \leq \nu^*(A \cup B)$ and $\nu^*(A \cup B) \leq \nu^*(B)$, it follows that $\nu^*(A \cup B) = \nu^*(B)$. + + +### Part (c) + +**Goal:** If $F \in \mathcal{M}$ and $\nu^*(F \Delta E) = 0$, then $E \in \mathcal{M}$. + +1. Recall that $E = (F \setminus (F \setminus E)) \cup (E \setminus F)$. + +2. Let $N_1 = F \setminus E$ and $N_2 = E \setminus F$. Since $F \Delta E = N_1 \cup N_2$ and $\nu^*(F \Delta E) = 0$, by **Monotonicity**, $\nu^*(N_1) = 0$ and $\nu^*(N_2) = 0$. + +3. By **Part (a)**, since $N_1$ and $N_2$ have outer measure zero, they are both in $\mathcal{M}$. + +4. We can express $E$ using set operations: $E = (F \cap N_1^c) \cup N_2$. + +5. Since $F, N_1^c,$ and $N_2$ are all elements of the $\sigma$-algebra $\mathcal{M}$, and $\sigma$-algebras are closed under complements, intersections, and unions, $E$ must also be in $\mathcal{M}$. + + +### Part (d) + +**Goal:** For any $A \subseteq \Omega$, find $B \in \sigma(\mathcal{A})$ such that $A \subseteq B$ and $\nu^*(A) = \nu^*(B)$. + +1. By the **Definition of Outer Measure**, for each $n \in \mathbb{N}$, there exists a sequence of sets $\{E_{n,j}\}_{j=1}^\infty \subseteq \mathcal{A}$ such that $A \subseteq \bigcup_{j=1}^\infty E_{n,j}$ and: + + $$\sum_{j=1}^\infty \nu(E_{n,j}) \leq \nu^*(A) + \frac{1}{n}$$ + +2. Let $B_n = \bigcup_{j=1}^\infty E_{n,j}$. Since $\mathcal{A} \subseteq \sigma(\mathcal{A})$ and $\sigma$-algebras are closed under countable unions, $B_n \in \sigma(\mathcal{A})$. + +3. By **Countable Subadditivity** and the fact that $\nu = \nu^*$ on $\mathcal{A}$, we have $\nu^*(B_n) \leq \sum_{j=1}^\infty \nu(E_{n,j}) \leq \nu^*(A) + \frac{1}{n}$. + +4. Define $B = \bigcap_{n=1}^\infty B_n$. Since $B$ is a countable intersection of sets in $\sigma(\mathcal{A})$, $B \in \sigma(\mathcal{A})$. + +5. Since $A \subseteq B_n$ for every $n$, it follows that $A \subseteq B$. + +6. By **Monotonicity**, $\nu^*(A) \leq \nu^*(B)$. Also, for every $n$, $B \subseteq B_n$, so: + + $$\nu^*(B) \leq \nu^*(B_n) \leq \nu^*(A) + \frac{1}{n}$$ + +7. Since $\nu^*(B) \leq \nu^*(A) + \frac{1}{n}$ for all $n > 0$, we must have $\nu^*(B) \leq \nu^*(A)$. + +8. Thus, $\nu^*(A) = \nu^*(B)$. + + +____ +## Question + +Consider a measure space $(\Omega, \mathcal{F}, \mu)$. If $\{E_i\}_{i=1}^\infty \subset \mathcal{F}$ is a sequence of measurable sets, we define: + +$$\limsup_{i \to \infty} E_i = \bigcap_{n=1}^\infty \bigcup_{i \geq n} E_i \quad \text{and} \quad \liminf_{i \to \infty} E_i = \bigcup_{n=1}^\infty \bigcap_{i \geq n} E_i$$ + +- **(a)** Prove the logical characterization of these sets: + + - $\limsup E_i = \{\omega \in \Omega \mid \omega \in E_i \text{ for infinitely many } i\}$ + + - $\liminf E_i = \{\omega \in \Omega \mid \omega \in E_i \text{ for all but finitely many } i\}$ + +- **(b)** Determine the relationship between the indicator function $\chi_{\limsup E_i}$ and $\limsup_{i \to \infty} \chi_{E_i}$. Ask a similar question regarding the $\liminf$. + +- **(c)** Let $\{E_i\} \subset \mathcal{F}$ such that $\sum_{i \geq 1} \mu(E_i) < \infty$. Prove that: + + $$\mu(\{\omega \in \Omega \mid \omega \in E_i \text{ for infinitely many } i\}) = 0$$ + + This result is known as the **Borel–Cantelli Lemma**. + + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Indicator Function ($\chi_E$):** The function $\chi_E(\omega)$ is defined as $1$ if $\omega \in E$ and $0$ if $\omega \notin E$. + + - _Intuition:_ It acts as a switch that "turns on" only when a point belongs to the set $E$. + +2. **Countable Subadditivity:** For any sequence of sets $\{A_i\}$, $\mu(\bigcup A_i) \leq \sum \mu(A_i)$. + + - _Intuition:_ The area of a combined shape is at most the sum of the areas of its individual parts, accounting for potential overlaps. + +3. **Continuity of Measure (From Above):** If $A_1 \supseteq A_2 \supseteq \dots$ are measurable sets and $\mu(A_1) < \infty$, then $\mu(\bigcap A_n) = \lim_{n \to \infty} \mu(A_n)$. + + - _Intuition:_ If you have a shrinking sequence of sets, the limit of their "sizes" is equal to the "size" of their ultimate intersection. + +4. **Convergence of Series:** If a series of non-negative numbers $\sum a_i$ converges, then the "tails" of the series must go to zero: $\lim_{n \to \infty} \sum_{i=n}^\infty a_i = 0$. + + - _Intuition:_ If the total sum is finite, the amount of "mass" left in the infinite end of the sequence must vanish as you move further out. + + +--- + +## Solution + +### Part (a): Logical Characterization + +**Step 1: Prove the $\limsup$ characterization.** + +- Let $A = \bigcap_{n=1}^\infty \bigcup_{i \geq n} E_i$. + +- If $\omega \in A$, then for every $n \in \mathbb{N}$, $\omega$ must be in the union $\bigcup_{i \geq n} E_i$. This means for every $n$, there is at least one index $i \geq n$ such that $\omega \in E_i$. This implies $\omega$ occurs in the sequence $E_i$ at arbitrarily large indices (infinitely often). + +- Conversely, if $\omega$ is in infinitely many $E_i$, then no matter how large an $n$ we pick, we can always find an index $i \geq n$ where $\omega \in E_i$. Thus, $\omega \in \bigcup_{i \geq n} E_i$ for all $n$, so $\omega$ is in the intersection. + + +**Step 2: Prove the $\liminf$ characterization.** + +- Let $B = \bigcup_{n=1}^\infty \bigcap_{i \geq n} E_i$. + +- If $\omega \in B$, then there exists _some_ $n$ such that $\omega \in \bigcap_{i \geq n} E_i$. This means $\omega \in E_i$ for every single $i \geq n$. The only indices where $\omega$ might _not_ be in $E_i$ are the finite number of indices $\{1, 2, \dots, n-1\}$. Thus, it is in all but finitely many sets. + +- Conversely, if $\omega \in E_i$ for all $i$ except for a finite set of indices, let $n-1$ be the maximum index in that finite set. Then for all $i \geq n$, $\omega \in E_i$, which places $\omega$ in the intersection $\bigcap_{i \geq n} E_i$ and consequently in the union $B$. + + +### Part (b): Indicator Functions + +**Step 1: Compare $\chi_{\limsup E_i}$ and $\limsup \chi_{E_i}$.** + +- $\chi_{\limsup E_i}(\omega) = 1$ if and only if $\omega \in E_i$ for infinitely many $i$. + +- $\limsup_{i \to \infty} \chi_{E_i}(\omega)$ is the limit superior of a sequence of 0s and 1s. This value is $1$ if the sequence $\chi_{E_i}(\omega)$ is $1$ infinitely often, and $0$ otherwise. + +- Therefore, **$\chi_{\limsup E_i} = \limsup_{i \to \infty} \chi_{E_i}$**. + + +**Step 2: Compare $\chi_{\liminf E_i}$ and $\liminf \chi_{E_i}$.** + +- $\chi_{\liminf E_i}(\omega) = 1$ if and only if $\omega \in E_i$ for all $i \geq n$ (for some $n$). + +- $\liminf_{i \to \infty} \chi_{E_i}(\omega)$ is $1$ if and only if the sequence of 0s and 1s is eventually all 1s (i.e., it is $0$ only finitely many times). + +- Therefore, **$\chi_{\liminf E_i} = \liminf_{i \to \infty} \chi_{E_i}$**. + + +### Part (c): Borel–Cantelli Lemma + +**Step 1: Set up the inequality using the $\limsup$ definition.** + +- From Part (a), the set of points in infinitely many $E_i$ is $L = \bigcap_{n=1}^\infty \bigcup_{i \geq n} E_i$. + +- Let $G_n = \bigcup_{i=n}^\infty E_i$. Note that $G_1 \supseteq G_2 \supseteq G_3 \dots$ is a nested shrinking sequence of sets. + + +**Step 2: Apply Countable Subadditivity to $G_n$.** + +- By the **definition of measure** and **Countable Subadditivity**: + + $$\mu(G_n) = \mu\left(\bigcup_{i=n}^\infty E_i\right) \leq \sum_{i=n}^\infty \mu(E_i)$$ + + +**Step 3: Use the convergence of the sum.** + +- We are given that $\sum_{i=1}^\infty \mu(E_i) < \infty$. + +- By the **properties of convergent series**, the tail of the sum must vanish: $\lim_{n \to \infty} \sum_{i=n}^\infty \mu(E_i) = 0$. + +- Since $0 \leq \mu(G_n) \leq \sum_{i=n}^\infty \mu(E_i)$, by the Squeeze Theorem, $\lim_{n \to \infty} \mu(G_n) = 0$. + + +**Step 4: Use Continuity of Measure.** + +- Because $\mu(G_1) \leq \sum_{i=1}^\infty \mu(E_i) < \infty$, the conditions for **Continuity of Measure from Above** are met. + +- Therefore, $\mu(L) = \mu(\bigcap_{n=1}^\infty G_n) = \lim_{n \to \infty} \mu(G_n)$. + +- Since we showed this limit is $0$, we conclude $\mu(L) = 0$. + + +___ + +## Question + +If $E \subset \mathbb{R}$ is a countable set, then prove that $\lambda(E) = 0$, where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Countable Set:** A set $E$ is countable if there exists a bijection between $E$ and a subset of the natural numbers $\mathbb{N}$. + + - _Intuition:_ This means the elements of the set can be listed in a sequence: $x_1, x_2, x_3, \dots$ + +2. **Lebesgue Measure of a Singleton:** For any single point $x \in \mathbb{R}$, $\lambda(\{x\}) = 0$. + + - _Intuition:_ A single point has zero length. + +3. **Countable Additivity:** If $\{E_i\}_{i=1}^\infty$ is a sequence of disjoint measurable sets, then $\mu(\bigcup_{i=1}^\infty E_i) = \sum_{i=1}^\infty \mu(E_i)$. + + - _Intuition:_ The total size of a collection of non-overlapping pieces is exactly the sum of the sizes of each piece. + + +--- + +## Solution + +**Step 1: Enumerate the set.** + +Since $E$ is a countable set, by the definition of countability, we can list its elements as a sequence. We write $E = \{x_1, x_2, x_3, \dots\}$. + +**Step 2: Express the set as a union of singletons.** + +We can represent the set $E$ as the countable union of its individual points: + +$$E = \bigcup_{i=1}^\infty \{x_i\}$$ + +Note that these singleton sets $\{x_i\}$ are disjoint because each contains exactly one distinct element of $E$. + +**Step 3: Apply the property of countable additivity.** + +By the property of **Countable Additivity** for measures, the measure of a countable union of disjoint sets is the sum of their individual measures. Therefore: + +$$\lambda(E) = \lambda\left(\bigcup_{i=1}^\infty \{x_i\}\right) = \sum_{i=1}^\infty \lambda(\{x_i\})$$ + +**Step 4: Use the measure of a singleton.** + +By the **definition of Lebesgue measure** (or the property that the measure of any interval $[a, a]$ is $a - a = 0$), we know that for any $i$: + +$$\lambda(\{x_i\}) = 0$$ + +**Step 5: Calculate the final sum.** + +Substituting the value from Step 4 into our equation from Step 3, we get: + +$$\lambda(E) = \sum_{i=1}^\infty 0 = 0$$ + +**Conclusion:** + +Thus, we have proved that the Lebesgue measure of any countable set $E \subset \mathbb{R}$ is zero. + +____ +## Question + +Let $D = \{d_1, d_2, \dots\}$ be a countable dense subset of $\mathbb{R}$. Define the set $G$ as a countable union of open intervals: + +$$G = \bigcup_{n=1}^\infty \left(d_n - \frac{1}{n^2}, d_n + \frac{1}{n^2}\right)$$ + +Prove that for every closed set $F \subset \mathbb{R}$, the Lebesgue measure of their symmetric difference is strictly positive: + +$$\lambda(G \Delta F) > 0$$ + +where $G \Delta F = (G \setminus F) \cup (F \setminus G)$. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Dense Set:** A set $D$ is dense in $\mathbb{R}$ if every non-empty open interval $(a, b)$ contains at least one element of $D$. + + - _Intuition:_ A dense set is "everywhere" in the sense that you cannot find any "gap" in the real line that doesn't contain a point from $D$. + +2. **Symmetric Difference ($\Delta$):** $G \Delta F$ consists of points that are in $G$ or $F$, but not in both. + + - _Intuition:_ If $\lambda(G \Delta F) = 0$, it means $G$ and $F$ are "essentially" the same set, differing only by a set of measure zero. + +3. **Countable Subadditivity:** For any sequence of sets $\{E_n\}$, $\lambda(\bigcup E_n) \leq \sum \lambda(E_n)$. + + - _Intuition:_ The total length of a union of intervals is at most the sum of their individual lengths. + +4. **Density of Open/Closed Sets:** If a closed set $F$ contains a dense set $D$, then $F$ must be the entire space ($F = \mathbb{R}$). + + - _Intuition:_ Because $F$ is closed, it contains all its limit points; if $D$ is dense, its limit points cover the whole real line. + + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Analyze the measure of $G$ + +We first show that $G$ is not the entire real line by calculating an upper bound for its measure. By the **definition of $G$** and **Countable Subadditivity**: + +$$\lambda(G) = \lambda\left(\bigcup_{n=1}^\infty \left(d_n - \frac{1}{n^2}, d_n + \frac{1}{n^2}\right)\right) \leq \sum_{n=1}^\infty \lambda\left(d_n - \frac{1}{n^2}, d_n + \frac{1}{n^2}\right)$$ + +The length of each interval is $(d_n + \frac{1}{n^2}) - (d_n - \frac{1}{n^2}) = \frac{2}{n^2}$. Thus: + +$$\lambda(G) \leq \sum_{n=1}^\infty \frac{2}{n^2} = 2 \sum_{n=1}^\infty \frac{1}{n^2} = 2\left(\frac{\pi^2}{6}\right) = \frac{\pi^2}{3}$$ + +Since $\lambda(G)$ is finite, $G$ cannot be equal to $\mathbb{R}$ (which has infinite measure). + +### Step 2: Assume for contradiction + +Suppose there exists a closed set $F$ such that $\lambda(G \Delta F) = 0$. This would mean that $G$ and $F$ are equal almost everywhere. Specifically, this implies $\lambda(G \setminus F) = 0$ and $\lambda(F \setminus G) = 0$. + +### Step 3: Analyze the relationship between $D$ and $F$ + +By **definition**, $G = \bigcup I_n$, where each $I_n$ is an open interval centered at $d_n$. Since each $I_n$ has positive measure ($\frac{2}{n^2} > 0$), and we assumed $\lambda(G \setminus F) = 0$, it must be that $F$ "covers" a significant portion of every $I_n$. + +Crucially, if $F$ is a closed set and $\lambda(G \setminus F) = 0$, then $G \subseteq F$ must be true in a topological sense. If there were a point $d_n \in D$ such that $d_n \notin F$, then since $F^c$ is open, there would be an entire interval around $d_n$ completely missing from $F$. This would force $\lambda(G \setminus F) > 0$. Therefore, to maintain $\lambda(G \Delta F) = 0$, the closed set $F$ must contain the dense set $D$. + +### Step 4: Reach the contradiction + +By the **property of dense sets**, if a closed set $F$ contains a dense set $D$, then $F = \mathbb{R}$. + +However, if $F = \mathbb{R}$, then: + +$$G \Delta F = G \Delta \mathbb{R} = \mathbb{R} \setminus G$$ + +The measure of this difference is $\lambda(\mathbb{R} \setminus G) = \lambda(\mathbb{R}) - \lambda(G) = \infty - \text{finite} = \infty$. + +This directly contradicts our assumption in Step 2 that $\lambda(G \Delta F) = 0$. + +### Step 5: Conclusion + +Since the assumption that $\lambda(G \Delta F) = 0$ leads to a contradiction for any closed set $F$, we must conclude that: + +$$\lambda(G \Delta F) > 0$$ + +___ +# Question + +Consider the nondecreasing, right-continuous function $F : \mathbb{R} \to \mathbb{R}$ given by: + +$$F(x) = \begin{cases} 1, & x \geq 0 \\ 0, & x < 0 \end{cases}$$ + +Let $\lambda_F$ be the Lebesgue-Stieltjes measure induced by $F$. Find the associated outer measure $\lambda_F^*$ and the $\sigma$-algebra of $\lambda_F$-measurable sets $\mathcal{M}_F$ explicitly. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)**: + +1. **Lebesgue-Stieltjes Outer Measure:** For a nondecreasing right-continuous function $F$, the outer measure $\lambda_F^*$ of a set $E \subseteq \mathbb{R}$ is defined as: + + $$\lambda_F^*(E) = \inf \left\{ \sum_{j=1}^\infty [F(b_j) - F(a_j)] : E \subseteq \bigcup_{j=1}^\infty (a_j, b_j] \right\}$$ + + - _Intuition:_ This measures the "jump" or increase in the function $F$ over the intervals that cover the set $E$. + +2. **Measure of a Half-Open Interval:** For any $a < b$, the Lebesgue-Stieltjes measure of the interval $(a, b]$ is given by $\lambda_F((a, b]) = F(b) - F(a)$. + + - _Intuition:_ The size of an interval is determined by how much the function $F$ "climbs" between the start and end points. + +3. **CarathΓ©odory’s Theorem:** A set $E$ is $\lambda_F^*$-measurable if for every test set $A \subseteq \mathbb{R}$: + + $$\lambda_F^*(A) = \lambda_F^*(A \cap E) + \lambda_F^*(A \cap E^c)$$ + + - _Intuition:_ A set is measurable if it splits any other set in an additive way regarding its measure. + + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Analyze the values of the outer measure for single points + +Let's determine the measure of a single point $\{x\}$. + +- If $x \neq 0$: We can cover $\{x\}$ with the interval $(x-\epsilon, x]$. For any $x > 0$, $F(x) = 1$ and $F(x-\epsilon) = 1$ (for small $\epsilon$), so $F(x) - F(x-\epsilon) = 0$. For any $x < 0$, $F(x) = 0$ and $F(x-\epsilon) = 0$, so the difference is also $0$. Thus, $\lambda_F^*(\{x\}) = 0$ for all $x \neq 0$. + +- If $x = 0$: We cover $\{0\}$ with $(-\epsilon, 0]$. Here, $F(0) = 1$ and $F(-\epsilon) = 0$. The difference is $1 - 0 = 1$. As $\epsilon \to 0$, this value remains $1$. Thus, $\lambda_F^*(\{0\}) = 1$. + + +### Step 2: Define $\lambda_F^*$ explicitly for any set $E$ + +By the **definition of Lebesgue-Stieltjes outer measure** and the results in Step 1: + +- If $0 \in E$, then $E$ contains a point with measure $1$. Any cover of $E$ must cover $\{0\}$, so the sum of the jumps will be at least $1$. + +- If $0 \notin E$, every point in $E$ has a local "jump" of $0$. + + Therefore, the outer measure is: + + $$\lambda_F^*(E) = \begin{cases} 1, & 0 \in E \\ 0, & 0 \notin E \end{cases}$$ + + This is essentially the **Dirac measure** centered at $0$. + + +### Step 3: Determine the measurable sets $\mathcal{M}_F$ + +We apply the **CarathΓ©odory criterion**. For any set $E \subseteq \mathbb{R}$ to be measurable, it must satisfy $\lambda_F^*(A) = \lambda_F^*(A \cap E) + \lambda_F^*(A \cap E^c)$ for every test set $A$. + +- **Case 1: $0 \in A$.** Then $\lambda_F^*(A) = 1$. + + - If $0 \in E$, then $0 \in (A \cap E)$ and $0 \notin (A \cap E^c)$. The right side is $\lambda_F^*(A \cap E) + \lambda_F^*(A \cap E^c) = 1 + 0 = 1$. + + - If $0 \notin E$, then $0 \notin (A \cap E)$ and $0 \in (A \cap E^c)$. The right side is $\lambda_F^*(A \cap E) + \lambda_F^*(A \cap E^c) = 0 + 1 = 1$. + +- **Case 2: $0 \notin A$.** Then $\lambda_F^*(A) = 0$. + + - Since $0 \notin A$, $0$ cannot be in $(A \cap E)$ or $(A \cap E^c)$. Thus, both terms on the right are $0$, and $0 = 0 + 0$. + + +### Step 4: Final Conclusion + +Since the CarathΓ©odory equation holds for **every** set $E \subseteq \mathbb{R}$ regardless of the test set $A$, every subset of the real numbers is measurable. + +- **Outer Measure:** $\lambda_F^*(E) = \chi_E(0)$ (1 if $0 \in E$, 0 otherwise). + +- **$\sigma$-algebra:** $\mathcal{M}_F = \mathcal{P}(\mathbb{R})$ (the power set of $\mathbb{R}$). + + +___ +# Question + +Consider the measure space $(\mathbb{R}, \mathcal{B}_{\mathbb{R}}, \lambda)$, where $\mathcal{B}_{\mathbb{R}}$ is the Borel $\sigma$-algebra and $\lambda$ is the Lebesgue measure on $\mathbb{R}$. If $A \in \mathcal{B}_{\mathbb{R}}$ and $x_0 \in \mathbb{R} \setminus \{0\}$, then prove the following: + +- **(a)** The translated set $A + x_0 = \{a + x_0 \mid a \in A\}$ is a Borel set ($A + x_0 \in \mathcal{B}_{\mathbb{R}}$). + +- **(b)** The reflected set $-A = \{-a \mid a \in A\}$ is a Borel set ($-A \in \mathcal{B}_{\mathbb{R}}$). + +- **(c)** The scaled set $x_0 A = \{x_0 a \mid a \in A\}$ is a Borel set ($x_0 A \in \mathcal{B}_{\mathbb{R}}$). + +- **(d)** The Lebesgue measure is translation-invariant and scales linearly: + + $$\lambda(A + x_0) = \lambda(A) \quad \text{and} \quad \lambda(x_0 A) = |x_0|\lambda(A)$$ + + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **Rudin, W. (Real and Complex Analysis)**: + +1. **Borel $\sigma$-algebra ($\mathcal{B}_{\mathbb{R}}$):** The smallest $\sigma$-algebra containing all open intervals $(a, b)$. + + - _Intuition:_ It is the collection of all sets that can be formed by taking countable unions, intersections, and complements of intervals. + +2. **Continuous Functions and Borel Sets:** If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function, then for any Borel set $A$, the preimage $f^{-1}(A)$ is also a Borel set. + + - _Intuition:_ Continuity preserves the "Borel-ness" of a set when moving backward through the function. + +3. **Uniqueness of Lebesgue Measure:** If two measures $\mu$ and $\nu$ agree on all open intervals $(a, b)$, then they agree on all Borel sets. + + - _Intuition:_ If two ways of measuring agree on simple building blocks (intervals), they must agree on the complex sets built from them. + +4. **Lebesgue Measure of an Interval:** $\lambda((a, b]) = b - a$. + + - _Intuition:_ The standard measure of an interval is simply its length. + + +--- + +## Solution (Step-by-Step Explanation) + +### Part (a), (b), and (c): Measurability + +**Step 1: Define continuous transformations.** + +Consider the following three functions $f_1, f_2, f_3: \mathbb{R} \to \mathbb{R}$: + +- $f_1(x) = x - x_0$ (Translation) + +- $f_2(x) = -x$ (Reflection) + +- $f_3(x) = \frac{x}{x_0}$ (Scaling, where $x_0 \neq 0$) + + +All three are continuous functions. + +**Step 2: Use the property of continuous preimages.** + +Recall that if $f$ is continuous, then $f^{-1}(A) \in \mathcal{B}_{\mathbb{R}}$ for any $A \in \mathcal{B}_{\mathbb{R}}$. Note the following: + +- $A + x_0 = f_1^{-1}(A)$ because $x \in f_1^{-1}(A) \iff x - x_0 \in A \iff x \in A + x_0$. + +- $-A = f_2^{-1}(A)$ because $x \in f_2^{-1}(A) \iff -x \in A \iff x \in -A$. + +- $x_0 A = f_3^{-1}(A)$ because $x \in f_3^{-1}(A) \iff \frac{x}{x_0} \in A \iff x \in x_0 A$. + + +Since $A$ is a Borel set and the functions are continuous, their preimagesβ€”which are exactly the sets in questionβ€”are all Borel sets. + +--- + +### Part (d): Invariance and Scaling + +**Step 1: Prove Translation Invariance.** + +Define a new measure $\mu$ by $\mu(E) = \lambda(E + x_0)$ for any Borel set $E$. + +- Check $\mu$ on an interval $I = (a, b]$. + +- $I + x_0 = (a + x_0, b + x_0]$. + +- By the **definition of Lebesgue measure**, $\lambda(I + x_0) = (b + x_0) - (a + x_0) = b - a = \lambda(I)$. + +- Since $\mu(I) = \lambda(I)$ for all intervals, by the **Uniqueness of Lebesgue Measure**, $\mu(E) = \lambda(E)$ for all Borel sets. Thus, $\lambda(A + x_0) = \lambda(A)$. + + +**Step 2: Prove Scaling Property.** + +Define another measure $\nu$ by $\nu(E) = \lambda(x_0 E)$. We first assume $x_0 > 0$. + +- Check $\nu$ on an interval $I = (a, b]$. + +- $x_0 I = (x_0 a, x_0 b]$. + +- By the **definition of Lebesgue measure**, $\lambda(x_0 I) = x_0 b - x_0 a = x_0(b - a) = x_0 \lambda(I)$. + +- Since $\nu(I) = x_0 \lambda(I)$ for all intervals, the measures $\nu$ and $x_0 \lambda$ must be the same on all Borel sets. Thus, $\lambda(x_0 A) = x_0 \lambda(A)$ for $x_0 > 0$. + + +**Step 3: Handle the negative case.** + +If $x_0 < 0$, then $x_0 I = [x_0 b, x_0 a)$. The length is $x_0 a - x_0 b = -x_0(b - a) = |x_0|\lambda(I)$. + +By applying the same uniqueness argument, we conclude $\lambda(x_0 A) = |x_0|\lambda(A)$ for any $x_0 \neq 0$. + +____ +# Question + +Suppose $\mu$ is a measure on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$ such that $\mu(A + x) = \mu(A)$ for all $A \in \mathcal{B}_{\mathbb{R}}$ and all $x \in \mathbb{R}$. If $\mu([0, 1]) = 2$, then prove that $\mu = 2\lambda$, where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **Rudin, W. (Real and Complex Analysis)**: + +1. **Translation Invariance:** A measure $\mu$ is translation-invariant if shifting a set by any value $x$ does not change its measure. + + - _Intuition:_ The measure behaves like standard length; sliding an object along the floor doesn't change how long it is. + +2. **Borel $\sigma$-algebra ($\mathcal{B}_{\mathbb{R}}$):** The $\sigma$-algebra generated by all open intervals (or equivalently, all half-open intervals $(a, b]$). + + - _Intuition:_ This is the collection of all "reasonable" sets on the real line constructed from simple intervals. + +3. **Uniqueness of Lebesgue Measure:** If $\mu$ is a translation-invariant measure on $\mathcal{B}_{\mathbb{R}}$ that is finite on compact sets, then there exists a constant $c \geq 0$ such that $\mu = c\lambda$. + + - _Intuition:_ Lebesgue measure is the _only_ way to measure length that respects shifting, once you decide how long the interval $[0, 1]$ should be. + +4. **Countable Additivity:** For a sequence of disjoint sets $\{E_i\}$, $\mu(\cup E_i) = \sum \mu(E_i)$. + + - _Intuition:_ The size of a whole made of non-overlapping parts is the sum of the sizes of those parts. + + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Establish the form of the measure + +By the **Uniqueness of Lebesgue Measure** (Theorem 2.20 in Folland), any translation-invariant Borel measure $\mu$ on $\mathbb{R}$ that assigns finite measure to bounded sets must be a constant multiple of the Lebesgue measure $\lambda$. Therefore, there exists some constant $c \in [0, \infty)$ such that: + +$$\mu(A) = c\lambda(A)$$ + +for all $A \in \mathcal{B}_{\mathbb{R}}$. + +### Step 2: Determine the constant $c$ + +We are given the specific value for the interval $[0, 1]$. By the **definition of $\mu$** established in Step 1: + +$$\mu([0, 1]) = c\lambda([0, 1])$$ + +We know from the **definition of Lebesgue measure** that $\lambda([0, 1]) = 1 - 0 = 1$. + +Substituting the given value $\mu([0, 1]) = 2$: + +$$2 = c(1)$$ + +Thus, $c = 2$. + +### Step 3: Verify the relationship for all sets + +Since we have found $c = 2$, the relationship $\mu(A) = c\lambda(A)$ becomes: + +$$\mu(A) = 2\lambda(A)$$ + +This must hold for all $A \in \mathcal{B}_{\mathbb{R}}$ because the two measures agree on the generating set of intervals and are both translation-invariant. + +### Step 4: Justification of Uniqueness (Alternative Perspective) + +If one does not wish to cite the uniqueness theorem directly, we can see why this works by considering intervals of the form $[0, 1/n]$. + +- By **Translation Invariance**, $\mu([0, 1/n]) = \mu([1/n, 2/n]) = \dots = \mu([(n-1)/n, 1])$. + +- By **Countable Additivity**, the sum of these $n$ identical measures must equal $\mu([0, 1]) = 2$. + +- Therefore, $n \cdot \mu([0, 1/n]) = 2$, which means $\mu([0, 1/n]) = 2/n = 2\lambda([0, 1/n])$. + +- Since this holds for all rational-length intervals and $\mu$ is a measure, it extends to all Borel sets. + + +**Conclusion:** + +$\mu = 2\lambda$. + +___ +## Question + +Let $(\Omega, \mathcal{F}, \mu)$ be a **$\sigma$-finite** measure space and $\{A_\alpha\}_{\alpha \in \Lambda} \subseteq \mathcal{F}$ be a disjoint collection of sets such that $\mu(A_\alpha) > 0$ for every $\alpha \in \Lambda$. + +- **Part 1:** Prove that the index set $\Lambda$ must be **countable**. + +- **Part 2:** Provide an example showing that this result can be false if the measure space is not $\sigma$-finite. + + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **$\sigma$-finite Measure:** A measure $\mu$ is $\sigma$-finite if $\Omega$ can be written as the countable union of sets with finite measure: $\Omega = \bigcup_{n=1}^\infty \Omega_n$, where $\mu(\Omega_n) < \infty$ for all $n$. + + - _Intuition:_ This means that although the whole space might be "infinitely large," it is composed of a manageable number of finite-sized pieces. + +2. **Countable Union of Countable Sets:** A countable union of countable sets is itself countable. + + - _Intuition:_ Combining several lists that can be counted results in one larger list that can still be counted. + +3. **Countable Additivity:** For any disjoint sequence of measurable sets $\{E_i\}$, $\mu(\bigcup E_i) = \sum \mu(E_i)$. + + - _Intuition:_ The total size of non-overlapping parts is the sum of their individual sizes. + +4. **Counting Measure:** A measure where $\mu(A)$ is the number of elements in $A$ if $A$ is finite, and $\infty$ if $A$ is infinite. + + - _Intuition:_ This measure simply counts how many items are in a set, regardless of their "physical" size. + + +--- + +## Solution (Step-by-Step Explanation) + +### Part 1: Proof for $\sigma$-finite spaces + +**Step 1: Decompose the space.** + +By the **definition of $\sigma$-finiteness**, we can write $\Omega = \bigcup_{n=1}^\infty \Omega_n$ where each $\mu(\Omega_n) < \infty$. Without loss of generality, we can assume the sets $\Omega_n$ are increasing ($\Omega_1 \subseteq \Omega_2 \subseteq \dots$). + +**Step 2: Partition the collection $\{A_\alpha\}$ by measure size.** + +For a fixed $n$ and a fixed $k \in \mathbb{N}$, consider the set of indices: + +$$\Lambda_{n,k} = \{ \alpha \in \Lambda : \mu(A_\alpha \cap \Omega_n) > \frac{1}{k} \}$$ + +This represents all sets in our collection that have a "substantial" chunk (greater than $1/k$) inside the finite-measure piece $\Omega_n$. + +**Step 3: Show $\Lambda_{n,k}$ is finite.** + +Suppose $\Lambda_{n,k}$ contained more than $m$ elements for some large $m$. Because the $A_\alpha$ are disjoint, the sets $\{A_\alpha \cap \Omega_n\}$ are also disjoint. By **countable additivity** (finite additivity): + +$$\mu(\Omega_n) \geq \mu\left( \bigcup_{\alpha \in \Lambda_{n,k}} (A_\alpha \cap \Omega_n) \right) = \sum_{\alpha \in \Lambda_{n,k}} \mu(A_\alpha \cap \Omega_n) > \frac{m}{k}$$ + +If $m$ is too large, this sum would exceed $\mu(\Omega_n)$. Since $\mu(\Omega_n)$ is finite, $m$ must be bounded. Thus, $\Lambda_{n,k}$ must be a **finite set** for every $n$ and $k$. + +**Step 4: Use the property of countable unions.** + +The set of all indices $\alpha$ such that $\mu(A_\alpha \cap \Omega_n) > 0$ is $\bigcup_{k=1}^\infty \Lambda_{n,k}$, which is a countable union of finite sets, and therefore **countable**. + +Finally, the set $\Lambda$ is the collection of all $\alpha$ such that $\mu(A_\alpha) > 0$. Since $A_\alpha = \bigcup_{n=1}^\infty (A_\alpha \cap \Omega_n)$, if $\mu(A_\alpha) > 0$, then there must exist some $n$ such that $\mu(A_\alpha \cap \Omega_n) > 0$. + +Thus, $\Lambda = \bigcup_{n=1}^\infty \bigcup_{k=1}^\infty \Lambda_{n,k}$. As a countable union of countable sets, **$\Lambda$ is countable**. + +--- + +### Part 2: Counterexample for non-$\sigma$-finite spaces + +**Step 1: Choose a non-$\sigma$-finite space.** + +Let $\Omega = \mathbb{R}$ and let $\mathcal{F} = \mathcal{P}(\mathbb{R})$. Let $\mu$ be the **counting measure**. + +In this space, $\mu(A) = \infty$ if $A$ is infinite. Since $\mathbb{R}$ is uncountable, it cannot be written as a countable union of finite sets. Thus, this space is **not $\sigma$-finite**. + +**Step 2: Define a disjoint collection.** + +Consider the collection of singleton sets: + +$$\{A_x\}_{x \in \mathbb{R}} = \{ \{x\} : x \in \mathbb{R} \}$$ + +- These sets are clearly disjoint. + +- For the counting measure, $\mu(\{x\}) = 1$ for every $x$, so every set has **positive measure**. + + +**Step 3: Show $\Lambda$ is uncountable.** + +In this case, the index set $\Lambda$ is the set of real numbers $\mathbb{R}$. + +As $\mathbb{R}$ is an uncountable set, we have found an example where a disjoint collection of positive-measure sets is uncountable, proving $\sigma$-finiteness was a necessary hypothesis. + +____ +# Question + +Let $E \subset \mathbb{R}$ such that $\lambda^*(E) = 0$, where $\lambda^*$ denotes the Lebesgue outer measure on $\mathbb{R}$. + +- **(a)** Prove that $E^c$ (the complement of $E$) is dense in $\mathbb{R}$. + +- **(b)** Determine if the same conclusion holds if the Lebesgue measure $\lambda$ is replaced by a general Lebesgue-Stieltjes measure $\lambda_F$. + + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Dense Set:** A set $A \subset \mathbb{R}$ is dense in $\mathbb{R}$ if every non-empty open interval $(a, b)$ contains at least one point of $A$. + + - _Intuition:_ A dense set is "spread out" everywhere; there are no "holes" in the real line that are completely free of points from that set. + +2. **Monotonicity of Outer Measure:** If $A \subseteq B$, then $\lambda^*(A) \leq \lambda^*(B)$. + + - _Intuition:_ A subset cannot have a larger measure than the set that contains it. + +3. **Measure of an Interval:** For any interval $I$ with endpoints $a$ and $b$ ($a < b$), the Lebesgue measure is $\lambda(I) = b - a$. + + - _Intuition:_ The measure of an interval is simply its length. + +4. **Lebesgue-Stieltjes Measure ($\lambda_F$):** A measure induced by a non-decreasing, right-continuous function $F$, where $\lambda_F((a, b]) = F(b) - F(a)$. + + - _Intuition:_ This measure weights different parts of the real line differently based on how fast $F$ grows. + + +--- + +## Solution (Step-by-Step Explanation) + +### Part (a): $E^c$ is dense in $\mathbb{R}$ + +**Step 1: Use the definition of density.** + +To show $E^c$ is dense in $\mathbb{R}$, we must show that for any open interval $(a, b)$ with $a < b$, the intersection $E^c \cap (a, b)$ is non-empty. + +**Step 2: Assume for contradiction.** + +Suppose $E^c$ is **not** dense. This implies there exists some open interval $I = (a, b)$ such that $E^c \cap I = \emptyset$. + +**Step 3: Relate the interval to $E$.** + +If $E^c \cap I$ is empty, it means that every point in the interval $I$ must belong to $E$. Therefore, $I \subseteq E$. + +**Step 4: Apply Monotonicity.** + +By the **Monotonicity of outer measure**, if $I \subseteq E$, then: + +$$\lambda^*(I) \leq \lambda^*(E)$$ + +We are given that $\lambda^*(E) = 0$. This implies $\lambda^*(I) = 0$. + +**Step 5: Reach the contradiction.** + +According to the **Measure of an Interval**, the measure of $I = (a, b)$ is $b - a$. Since $a < b$, we know $b - a > 0$. + +This contradicts Step 4, where we found the measure must be $0$. Thus, our assumption was false, and $E^c$ must be dense. + +--- + +### Part (b): Is this true for $\lambda_F$? + +**Step 1: Identify the condition for a counterexample.** + +The conclusion in Part (a) relied on the fact that any interval $(a, b)$ has strictly positive Lebesgue measure. For a general Lebesgue-Stieltjes measure $\lambda_F$, this is not always true. If $\lambda_F(I) = 0$, then $E$ could contain $I$ without violating the $\lambda_F^*(E) = 0$ condition. + +**Step 2: Construct a counterexample.** + +Consider a constant function $F(x) = c$. In this case, for any interval $(a, b]$, $\lambda_F((a, b]) = F(b) - F(a) = c - c = 0$. + +- Let $E = \mathbb{R}$. + +- Then $\lambda_F^*(E) = 0$ because the measure of the entire line is zero. + +- However, $E^c = \emptyset$, which is **not dense** in $\mathbb{R}$. + + +**Step 3: Provide a more specific counterexample.** + +Consider $F(x) = \begin{cases} 0 & x < 0 \\ 0 & 0 \leq x < 1 \\ 1 & x \geq 1 \end{cases}$ (a step function). + +The measure $\lambda_F$ is a point mass at $x=1$. Let $E = (-\infty, 0.5)$. + +- $\lambda_F^*(E) = 0$ because $E$ does not contain the point $x=1$. + +- But $E^c = [0.5, \infty)$, which is missing the entire interval $(-\infty, 0.5)$. Thus, $E^c$ is not dense. + + +**Conclusion:** + +The statement is **false** for general Lebesgue-Stieltjes measures. It only holds if the measure $\lambda_F$ assigns strictly positive measure to every non-empty open interval (which happens if $F$ is strictly increasing). +____ +## Question + +Suppose $E \in \mathcal{L}$ with $0 < \lambda(E) < \infty$, where $\mathcal{L}$ is the $\sigma$-algebra of Lebesgue measurable sets in $\mathbb{R}$ and $\lambda$ is the Lebesgue measure. Given any $\alpha \in (0, 1)$, prove that there exists a measurable set $E_\alpha \in \mathcal{L}$ such that $E_\alpha \subset E$ and $\lambda(E_\alpha) = \alpha\lambda(E)$. + +Determine if the same result holds for any Lebesgue-Stieltjes measure $\lambda_F$. + +**(Hint: Think about the function $f_E(x) = \lambda(E \cap (-\infty, x])$ for $x \in \mathbb{R}$.)** + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Continuity of Measure (From Below):** If $A_1 \subset A_2 \subset \dots$ is an increasing sequence of measurable sets, then $\mu(\cup A_n) = \lim_{n \to \infty} \mu(A_n)$. + + - _Intuition:_ As you expand a set, its measure changes smoothly toward the measure of its ultimate limit. + +2. **Intermediate Value Theorem (IVT):** If a real-valued function $f$ is continuous on a closed interval $[a, b]$, it takes on every value between $f(a)$ and $f(b)$. + + - _Intuition:_ A continuous path from one height to another must cross every height in between. + +3. **Lebesgue-Stieltjes Measure ($\lambda_F$):** A measure induced by a non-decreasing, right-continuous function $F$. + + - _Intuition:_ Unlike the standard Lebesgue measure, these can have "jumps" (atoms) where a single point has a positive measure. + + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Define the auxiliary function + +Following the hint, let $f_E: \mathbb{R} \to \mathbb{R}$ be defined by: + +$$f_E(x) = \lambda(E \cap (-\infty, x])$$ + +Since $E \cap (-\infty, x]$ is always a subset of $E$, and $\lambda(E) < \infty$, the function $f_E$ is well-defined and finite for all $x$. + +### Step 2: Prove that $f_E$ is continuous + +To use the Intermediate Value Theorem, we must show $f_E$ is continuous. + +- **Right-continuity:** As $y \to x$ from the right ($y > x$), the sets $(E \cap (-\infty, y])$ shrink toward $(E \cap (-\infty, x])$. By **Continuity of Measure from Above**, the measures converge. + +- **Left-continuity:** As $y \to x$ from the left ($y < x$), the sets $(E \cap (-\infty, y])$ expand toward $E \cap (-\infty, x)$. By **Continuity of Measure from Below**, $\lim_{y \to x^-} f_E(y) = \lambda(E \cap (-\infty, x))$. + +- Since the Lebesgue measure of a single point is zero ($\lambda(\{x\}) = 0$), we have: + + $$\lambda(E \cap (-\infty, x]) = \lambda(E \cap (-\infty, x)) + \lambda(E \cap \{x\}) = \lambda(E \cap (-\infty, x))$$ + + Therefore, the left limit equals the right limit, and $f_E$ is continuous. + + +### Step 3: Evaluate the limits of $f_E$ + +- As $x \to -\infty$, the set $E \cap (-\infty, x]$ approaches the empty set. By **Continuity of Measure**, $\lim_{x \to -\infty} f_E(x) = \lambda(\emptyset) = 0$. + +- As $x \to \infty$, the set $E \cap (-\infty, x]$ approaches $E$. By **Continuity of Measure**, $\lim_{x \to \infty} f_E(x) = \lambda(E)$. + +### Step 4: Apply the Intermediate Value Theorem + +We are given $\alpha \in (0, 1)$, so $0 < \alpha\lambda(E) < \lambda(E)$. + +Since $f_E$ is continuous and spans the range $(0, \lambda(E))$, by the **Intermediate Value Theorem**, there must exist some $x_0 \in \mathbb{R}$ such that: + +$$f_E(x_0) = \alpha\lambda(E)$$ + +### Step 5: Define the set $E_\alpha$ + +Let $E_\alpha = E \cap (-\infty, x_0]$. + +- By the **definition of measurable sets**, $E_\alpha$ is measurable because it is the intersection of two measurable sets. + +- By construction, $E_\alpha \subset E$. + +- By Step 4, $\lambda(E_\alpha) = f_E(x_0) = \alpha\lambda(E)$. + + This completes the proof for the Lebesgue measure. + + +--- + +### Step 6: Analyze the result for $\lambda_F$ + +The result is **false** for a general Lebesgue-Stieltjes measure. + +- **Counterexample:** Consider the function $F(x) = 0$ for $x < 0$ and $F(x) = 1$ for $x \geq 0$. This induces a measure $\lambda_F$ that is a point mass at $0$ (i.e., $\lambda_F(\{0\}) = 1$ and all other sets not containing $0$ have measure $0$). + +- Let $E = \{0\}$. Then $\lambda_F(E) = 1$. + +- Choose $\alpha = 0.5$. We need a subset $E_\alpha \subset \{0\}$ such that $\lambda_F(E_\alpha) = 0.5$. + +- The only subsets of $\{0\}$ are $\emptyset$ (measure $0$) and $\{0\}$ (measure $1$). + +- Since neither has measure $0.5$, the result fails. + + +**Conclusion:** The result holds for Lebesgue measure because it is "atomless," but fails for $\lambda_F$ if the measure has atoms (discontinuities in $F$). + +___ +## Question + +Suppose $E \in \mathcal{L}$ with $\lambda(E) = \infty$, where $\mathcal{L}$ is the $\sigma$-algebra of Lebesgue measurable sets in $\mathbb{R}$ and $\lambda$ is the Lebesgue measure. Given any $\alpha \in [0, \infty)$, prove that there exists a measurable set $E_\alpha \in \mathcal{L}$ such that $E_\alpha \subset E$ and $\lambda(E_\alpha) = \alpha$. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Continuity of Measure (From Below):** If $A_1 \subset A_2 \subset A_3 \dots$ is an increasing sequence of measurable sets, then $\lambda(\bigcup_{n=1}^\infty A_n) = \lim_{n \to \infty} \lambda(A_n)$. + + - _Intuition:_ As you expand a set, its measure changes smoothly toward the measure of its ultimate limit. + +2. **Continuity of Measure (From Above):** If $A_1 \supset A_2 \supset A_3 \dots$ is a decreasing sequence of measurable sets and $\lambda(A_1) < \infty$, then $\lambda(\bigcap_{n=1}^\infty A_n) = \lim_{n \to \infty} \lambda(A_n)$. + + - _Intuition:_ Shrinking a finite-sized set behaves predictably; the limit of the sizes is the size of the limit. + +3. **Intermediate Value Theorem (IVT):** If a real-valued function $f$ is continuous on an interval, it takes on every value between its minimum and maximum on that interval. + + - _Intuition:_ A continuous line cannot skip any heights as it moves from one point to another. + +4. **Lebesgue Measure of a Singleton:** For any $x \in \mathbb{R}$, $\lambda(\{x\}) = 0$. + + - _Intuition:_ A single point has no length. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Define a set-valued function.** + +Similar to the case for finite measure, we define a function $f: \mathbb{R} \to [0, \infty]$ that tracks how much of $E$ is captured as we "sweep" across the real line from left to right: + +$$f(x) = \lambda(E \cap (-\infty, x])$$ + +**Step 2: Prove that $f$ is continuous.** + +We check the limits from both sides at any point $x_0$: + +- **From the left:** As $x \to x_0^-$, the sets $E \cap (-\infty, x]$ increase to $E \cap (-\infty, x_0)$. By **Continuity of Measure from Below**, $\lim_{x \to x_0^-} f(x) = \lambda(E \cap (-\infty, x_0))$. + +- **From the right:** Since we are looking for a finite $\alpha$, we can restrict our attention to a range where the measure is finite. For any $x > x_0$, $\lambda(E \cap (-\infty, x])$ is finite if we pick $x$ small enough. By **Continuity of Measure from Above**, the limit from the right is $f(x_0)$. + +- Because $\lambda(\{x_0\}) = 0$, there is no difference between $\lambda(E \cap (-\infty, x_0))$ and $\lambda(E \cap (-\infty, x_0])$. + + Thus, $f(x)$ is a continuous function. + + +**Step 3: Evaluate the range of $f$.** + +- As $x \to -\infty$, $E \cap (-\infty, x] \to \emptyset$, so $\lim_{x \to -\infty} f(x) = 0$. + +- As $x \to \infty$, $E \cap (-\infty, x] \to E$. By **Continuity of Measure from Below**, $\lim_{x \to \infty} f(x) = \lambda(E) = \infty$. + + +**Step 4: Apply the Intermediate Value Theorem.** + +The function $f$ is continuous and maps the real line onto the interval $[0, \infty)$. + +For any $\alpha \in [0, \infty)$, the value $\alpha$ lies within the range of $f$ (specifically between $0$ and $\infty$). + +By the **Intermediate Value Theorem**, there must exist at least one $x_\alpha \in \mathbb{R}$ such that $f(x_\alpha) = \alpha$. + +**Step 5: Define the subset $E_\alpha$.** + +Let $E_\alpha = E \cap (-\infty, x_\alpha]$. + +1. **Measurability:** Since $E$ is measurable and $(-\infty, x_\alpha]$ is a Borel set (and thus Lebesgue measurable), their intersection $E_\alpha$ is measurable. + +2. **Inclusion:** By the definition of intersection, $E_\alpha \subset E$. + +3. **Measure:** By Step 4, $\lambda(E_\alpha) = f(x_\alpha) = \alpha$. + + +This completes the proof. + +___ +## Question + +Let $f: \mathbb{R} \to \mathbb{R}$ be a Lebesgue measurable function (that is, $f$ is $(\mathcal{L}, \mathcal{B}_{\mathbb{R}})$-measurable, where $\mathcal{L}$ is the Lebesgue $\sigma$-algebra over $\mathbb{R}$). Prove that there exists a set $E \subset \mathbb{R}$ of positive Lebesgue measure ($\lambda(E) > 0$) on which $f$ is bounded. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Lebesgue Measurable Function:** A function $f$ is measurable if the preimage of every Borel set is a Lebesgue measurable set. Specifically, for any $M \in \mathbb{R}$, the set $\{x \in \mathbb{R} : |f(x)| \leq M\}$ must be in $\mathcal{L}$. + + - _Intuition:_ This ensures we can meaningfully talk about the "size" (measure) of the part of the domain where the function behaves in a certain way. + +2. **Countable Additivity of Measure:** For any sequence of disjoint measurable sets $\{E_n\}$, $\lambda(\bigcup_{n=1}^\infty E_n) = \sum_{n=1}^\infty \lambda(E_n)$. + + - _Intuition:_ The total size of a whole made of non-overlapping parts is the sum of the sizes of those parts. + +3. **Continuity of Measure (From Below):** If $A_1 \subset A_2 \subset A_3 \dots$ is an increasing sequence of measurable sets, then $\lambda(\bigcup_{n=1}^\infty A_n) = \lim_{n \to \infty} \lambda(A_n)$. + + - _Intuition:_ As you expand a set, its measure grows toward the measure of its ultimate limit. + +4. **Bounded Function on a Set:** A function $f$ is bounded on a set $E$ if there exists a finite constant $M$ such that $|f(x)| \leq M$ for all $x \in E$. + + - _Intuition:_ The function doesn't "blow up" to infinity within that specific region. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Define a sequence of sets based on the function's values.** + +For each natural number $n \in \mathbb{N}$, define the set $E_n$ as the region where the absolute value of $f$ is at most $n$: + +$$E_n = \{x \in \mathbb{R} : |f(x)| \leq n\}$$ + +By the **definition of a measurable function**, each $E_n$ is a Lebesgue measurable set. + +**Step 2: Show that these sets cover the domain.** + +Since the codomain of $f$ is $\mathbb{R}$ (the real numbers), every output $f(x)$ is a finite real number. Therefore, for every $x \in \mathbb{R}$, there must exist some integer $n$ such that $|f(x)| \leq n$. This implies: + +$$\bigcup_{n=1}^\infty E_n = \mathbb{R}$$ + +**Step 3: Analyze the measure of the union.** + +By the **properties of Lebesgue measure**, we know the measure of the entire real line is infinite: + +$$\lambda\left(\bigcup_{n=1}^\infty E_n\right) = \lambda(\mathbb{R}) = \infty$$ + +**Step 4: Use Continuity of Measure.** + +Note that the sets are nested: $E_1 \subset E_2 \subset E_3 \dots$ because if $|f(x)| \leq n$, then $|f(x)| \leq n+1$. By **Continuity of Measure from Below**, we have: + +$$\lim_{n \to \infty} \lambda(E_n) = \lambda(\mathbb{R}) = \infty$$ + +**Step 5: Conclude the existence of a set with positive measure.** + +If the limit of $\lambda(E_n)$ is infinity, there must exist some index $N$ such that $\lambda(E_N)$ is large. Specifically, there must be an $N$ such that $\lambda(E_N) > 0$. + +(In fact, there is an $N$ such that $\lambda(E_N) > 1,000,000$, but we only need it to be strictly positive). + +**Step 6: Final Verification.** + +Let $E = E_N$. + +1. By Step 5, $\lambda(E) > 0$. + +2. By the **definition of $E_N$**, for all $x \in E$, $|f(x)| \leq N$. + +3. Thus, $f$ is bounded on $E$ by the constant $N$. + + +This proves that there exists a set of positive Lebesgue measure on which $f$ is bounded. + +___ +## Question + +Give an example of a function $f : \mathbb{R} \to \mathbb{R}$ such that its absolute value $|f|$ is Lebesgue measurable, but the function $f$ itself is not Lebesgue measurable. + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Lebesgue Measurable Function:** A function $f$ is measurable if for every Borel set $B \subseteq \mathbb{R}$, the preimage $f^{-1}(B)$ is a Lebesgue measurable set. + + - _Intuition:_ This means we can consistently assign a "size" or measure to the set of inputs that map to any reasonable set of outputs. + +2. **Lebesgue Measurable Set ($\mathcal{L}$):** A set $E \subseteq \mathbb{R}$ is Lebesgue measurable if it belongs to the $\sigma$-algebra generated by Borel sets and null sets. + + - _Intuition:_ These are the sets for which the concept of "length" is well-defined. + +3. **Non-measurable Set:** A set $V \subseteq \mathbb{R}$ that does not belong to $\mathcal{L}$. A standard example is the **Vitali set**. + + - _Intuition:_ These are extremely "jagged" or "scattered" sets that are so complex they cannot be assigned a consistent length without violating basic axioms of measure. + +4. **Indicator Function ($\chi_E$):** A function that is $1$ if $x \in E$ and $0$ otherwise. An indicator function is measurable if and only if the set $E$ is measurable. + + - _Intuition:_ This function acts as a mathematical "toggle" that tells you whether you are inside or outside a specific set. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Construct a non-measurable set.** + +Following the standard construction by Giuseppe Vitali, let $V \subset [0, 1]$ be a **non-measurable set**. We know such a set exists under the Axiom of Choice. Because $V$ is not a Lebesgue measurable set, its indicator function $\chi_V$ is not a measurable function. + +**Step 2: Define the function $f$.** + +We want to define $f$ such that it "hides" its non-measurability within its sign, but disappears when we take the absolute value. Define $f: \mathbb{R} \to \mathbb{R}$ as follows: + +$$f(x) = \begin{cases} 1 & \text{if } x \in V \\ -1 & \text{if } x \notin V \end{cases}$$ + +Alternatively, this can be written using indicator functions as $f(x) = \chi_V(x) - \chi_{V^c}(x)$. + +**Step 3: Show that $f$ is not measurable.** + +Consider the Borel set $B = \{1\}$. By the **definition of measurable function**, the preimage $f^{-1}(B)$ must be a measurable set. + +- $f^{-1}(\{1\}) = \{x \in \mathbb{R} : f(x) = 1\} = V$. + +- Since $V$ is a **non-measurable set**, its preimage is not in $\mathcal{L}$. + + Therefore, $f$ is not a Lebesgue measurable function. + + +**Step 4: Show that $|f|$ is measurable.** + +Now, let's examine the absolute value $|f(x)|$. + +- If $x \in V$, $|f(x)| = |1| = 1$. + +- If $x \notin V$, $|f(x)| = |-1| = 1$. + + Thus, $|f(x)| = 1$ for all $x \in \mathbb{R}$. In other words, $|f|$ is a constant function. + + +**Step 5: Verify the measurability of the constant function.** + +A constant function is always measurable. To see this, let $B$ be any Borel set. + +- If $1 \in B$, then $|f|^{-1}(B) = \mathbb{R}$ (which is measurable). + +- If $1 \notin B$, then $|f|^{-1}(B) = \emptyset$ (which is measurable). + + Since $\mathbb{R}$ and $\emptyset$ are both in $\mathcal{L}$, $|f|$ is a Lebesgue measurable function. + + +**Conclusion:** + +The function $f$ defined by the non-measurable set $V$ serves as the required example: $|f|$ is constant (measurable), but $f$ maps to a non-measurable set (not measurable). + +____ +## Question + +If $f : \mathbb{R} \to \mathbb{R}$ is a monotone function, then prove that $f$ is Borel measurable; that is, $f$ is $(\mathcal{B}_{\mathbb{R}}, \mathcal{B}_{\mathbb{R}})$-measurable. (Note: The image text uses $\mathcal{L}$ to denote the Borel $\sigma$-algebra, though $\mathcal{B}_{\mathbb{R}}$ is more standard for Borel sets.) + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Monotone Function:** A function $f$ is monotone if it is either non-decreasing (if $x \leq y$, then $f(x) \leq f(y)$) or non-increasing (if $x \leq y$, then $f(x) \geq f(y)$). + + - _Intuition:_ A monotone function never "doubles back" on its values; it only moves in one direction (up or down). + +2. **Borel Measurable Function:** A function $f$ is Borel measurable if the preimage $f^{-1}(B)$ is a Borel set for every Borel set $B \in \mathcal{B}_{\mathbb{R}}$. + + - _Intuition:_ This ensures that if we pick a "nice" set of outputs, the corresponding set of inputs is also "nice" and can be assigned a length. + +3. **Generating Sets for Measurability:** To prove a function is measurable, it is sufficient to show that the preimage of any interval of the form $(a, \infty)$ is a Borel set. + + - _Intuition:_ These "tail" intervals act as building blocks; if the function behaves well for these, it behaves well for all Borel sets. + +4. **Borel $\sigma$-algebra ($\mathcal{B}_{\mathbb{R}}$):** The collection of sets formed from open intervals through countable unions, intersections, and complements. Any interval (open, closed, or half-open) is a Borel set. + + - _Intuition:_ These are the standard "measurable" subsets of the real line. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Simplify the problem using the definition of measurability.** + +To prove that $f$ is Borel measurable, we need to show that for any $a \in \mathbb{R}$, the set $E_a = \{x \in \mathbb{R} : f(x) > a\}$ is a Borel set. We will assume $f$ is non-decreasing for this proof. The proof for a non-increasing function is entirely analogous. + +**Step 2: Characterize the set $E_a$.** + +Suppose $f$ is non-decreasing. If $x \in E_a$, then $f(x) > a$. + +Because $f$ is non-decreasing, for any $y > x$, we must have $f(y) \geq f(x) > a$. This implies that if a point $x$ is in $E_a$, every point to its right is also in $E_a$. + +**Step 3: Identify the geometric form of $E_a$.** + +Based on Step 2, $E_a$ must be an interval that extends to $+\infty$. Specifically, let $s = \inf \{x \in \mathbb{R} : f(x) > a\}$. + +- If the set $\{x \in \mathbb{R} : f(x) > a\}$ is empty, then $E_a = \emptyset$. + +- If the set is all of $\mathbb{R}$, then $E_a = (-\infty, \infty)$. + +- Otherwise, $E_a$ will take one of two forms: $(s, \infty)$ or $[s, \infty)$. + + +**Step 4: Verify the Borel property of $E_a$.** + +We have identified that $E_a$ is either an empty set, the whole real line, or an interval of the form $(s, \infty)$ or $[s, \infty)$. + +By the **definition of the Borel $\sigma$-algebra**, every interval (whether open, closed, or half-open) is a Borel set. + +- $\emptyset \in \mathcal{B}_{\mathbb{R}}$ + +- $\mathbb{R} \in \mathcal{B}_{\mathbb{R}}$ + +- $(s, \infty) \in \mathcal{B}_{\mathbb{R}}$ + +- $[s, \infty) \in \mathcal{B}_{\mathbb{R}}$ + + +**Step 5: Apply the definition of a measurable function.** + +Since we have shown that for any $a \in \mathbb{R}$, the set $f^{-1}((a, \infty))$ is an interval, and all intervals are Borel sets, the function satisfies the requirement for measurability. + +**Conclusion:** + +By the **criteria for Borel measurability**, since the preimage of every interval $(a, \infty)$ is a Borel set, $f$ is a Borel measurable function. + +___ +## Question + +If $f : \mathbb{R} \to \mathbb{R}$ is differentiable everywhere, then prove that its derivative $f'$ is Borel measurable. + +--- + +## Definitions and Theorems Used + +Based on **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **Rudin, W. (Real and Complex Analysis)**: + +1. **Differentiability and Continuity:** If $f$ is differentiable at a point, it is continuous at that point. If $f$ is differentiable everywhere, it is a continuous function. + + - _Intuition:_ A differentiable function has no "jumps" or "breaks," meaning its graph is a solid, unbroken curve. + +2. **Continuous Functions are Borel Measurable:** Every continuous function $f: \mathbb{R} \to \mathbb{R}$ is Borel measurable. + + - _Intuition:_ Continuity is a very strong property; it ensures that the preimage of any open set is open, which is the basic requirement for Borel measurability. + +3. **Measurability of Limits:** If a sequence of measurable functions $\{f_n\}$ converges pointwise to a function $g$, then $g$ is also measurable. + + - _Intuition:_ Measurability is preserved under the process of taking limits; if the "building blocks" are measurable, their limit is too. + +4. **Definition of the Derivative:** The derivative $f'(x)$ is defined as the limit of the difference quotient: + + $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ + + - _Intuition:_ The derivative represents the instantaneous slope of the function at a specific point. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Express the derivative as a discrete limit.** + +By the **definition of the derivative**, we know that for every $x \in \mathbb{R}$: + +$$f'(x) = \lim_{n \to \infty} \frac{f(x + \frac{1}{n}) - f(x)}{\frac{1}{n}}$$ + +We can define a sequence of functions $g_n(x)$ such that: + +$$g_n(x) = n \left[ f\left(x + \frac{1}{n}\right) - f(x) \right]$$ + +Since the limit exists for all $x$ (because $f$ is differentiable everywhere), the sequence $\{g_n\}$ converges pointwise to $f'$. + +**Step 2: Prove that each $g_n$ is measurable.** + +Since $f$ is differentiable, it is a **continuous function**. + +- The function $x \mapsto f(x + \frac{1}{n})$ is continuous because it is the composition of $f$ with a continuous translation. + +- Each $g_n(x)$ is a linear combination of continuous functions ($f(x + 1/n)$ and $f(x)$). + +- Therefore, each $g_n$ is continuous. + + By the theorem that **continuous functions are Borel measurable**, each function $g_n$ is Borel measurable. + + +**Step 3: Apply the limit property of measurable functions.** + +We have a sequence of Borel measurable functions $\{g_n\}$ that converges pointwise to $f'$ at every point $x \in \mathbb{R}$. + +By the theorem regarding the **measurability of limits**, the pointwise limit of a sequence of Borel measurable functions is itself Borel measurable. + +**Step 4: Conclusion.** + +Since $f'(x) = \lim_{n \to \infty} g_n(x)$ and each $g_n$ is Borel measurable, it follows that $f'$ is Borel measurable. + +____ +## Question + +Let $(\Omega, \mathcal{F})$ be a measurable space and $f : \Omega \to \mathbb{R}$ be an $\mathcal{F}$-measurable function. If $g : \mathbb{R} \to \mathbb{R}$ is a continuous function, then prove that the composition $g \circ f : \Omega \to \mathbb{R}$ is also $\mathcal{F}$-measurable. + +--- + +## Definitions and Theorems Used + +Based on **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **Rudin, W. (Real and Complex Analysis)**: + +1. **Measurable Function:** A function $f: \Omega \to \mathbb{R}$ is $\mathcal{F}$-measurable if for every Borel set $B \in \mathcal{B}_{\mathbb{R}}$, the preimage $f^{-1}(B)$ is an element of the $\sigma$-algebra $\mathcal{F}$. + + - _Intuition:_ This means we can consistently assign a "meaning" or "size" to the set of points in our domain that map into any "nice" set of values in the real numbers. + +2. **Continuous Function (Topological Definition):** A function $g: \mathbb{R} \to \mathbb{R}$ is continuous if for every open set $U \subseteq \mathbb{R}$, the preimage $g^{-1}(U)$ is an open set in $\mathbb{R}$. + + - _Intuition:_ Continuity ensures that there are no "jumps," so that points that are close together in the output space were originally close together in the input space. + +3. **Borel $\sigma$-algebra ($\mathcal{B}_{\mathbb{R}}$):** The smallest $\sigma$-algebra that contains all open sets of $\mathbb{R}$. + + - _Intuition:_ It is the collection of all sets (like intervals, open sets, and closed sets) that we can build using standard operations. + +4. **Composition of Preimages:** For any two functions $f$ and $g$, and any set $B$, the preimage of the composition satisfies $(g \circ f)^{-1}(B) = f^{-1}(g^{-1}(B))$. + + - _Intuition:_ To find the points that $g \circ f$ maps into $B$, you first find what $g$ maps into $B$, and then find what $f$ maps into those resulting points. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Identify the goal.** + +To prove that $h = g \circ f$ is $\mathcal{F}$-measurable, we must show that for any Borel set $B \in \mathcal{B}_{\mathbb{R}}$, the set $h^{-1}(B)$ belongs to the $\sigma$-algebra $\mathcal{F}$. + +**Step 2: Use the property of preimages of compositions.** + +By the **definition of function composition**, the preimage of the set $B$ under $g \circ f$ is: + +$$(g \circ f)^{-1}(B) = f^{-1}(g^{-1}(B))$$ + +**Step 3: Analyze the inner preimage $g^{-1}(B)$.** + +Since $g: \mathbb{R} \to \mathbb{R}$ is a **continuous function**, it is a well-known result (Proposition 2.1 in Folland) that $g$ is Borel measurable. + +By the **definition of Borel measurability**, the preimage of any Borel set $B$ under a Borel measurable function is itself a Borel set. Thus: + +$$V = g^{-1}(B) \in \mathcal{B}_{\mathbb{R}}$$ + +_Explanation:_ Continuity is actually a stronger condition than Borel measurability; since $g$ is continuous, the preimage of an open set is open (which is Borel), and the $\sigma$-algebra generated by these preimages must therefore be contained in $\mathcal{B}_{\mathbb{R}}$. + +**Step 4: Analyze the outer preimage $f^{-1}(V)$.** + +We are given that $f: \Omega \to \mathbb{R}$ is **$\mathcal{F}$-measurable**. + +From Step 3, we know that $V = g^{-1}(B)$ is a Borel set ($V \in \mathcal{B}_{\mathbb{R}}$). + +By the **definition of an $\mathcal{F}$-measurable function**, the preimage of any Borel set under $f$ must lie in $\mathcal{F}$. Therefore: + +$$f^{-1}(V) \in \mathcal{F}$$ + +**Step 5: Conclusion.** + +Combining the steps, we have shown that for any Borel set $B$: + +$$(g \circ f)^{-1}(B) = f^{-1}(g^{-1}(B)) \in \mathcal{F}$$ + +By the **definition of a measurable function**, since the preimage of every Borel set is in $\mathcal{F}$, the composition $g \circ f$ is $\mathcal{F}$-measurable. + +___ +## Question + +Let $(\Omega, \mathcal{F})$ be a measurable space and $f : \Omega \to \mathbb{R}$ be a function. If $f^{-1}((r, \infty)) \in \mathcal{F}$ for all rational numbers $r \in \mathbb{Q}$, then prove that $f$ is $(\mathcal{F}, \mathcal{B}_{\mathbb{R}})$-measurable. + +(Note: The image contains a slight notation error, writing $f^{-1} \in (r, \infty) \in \mathcal{F}$. The corrected version above uses the standard notation for the preimage of an interval.) + +--- + +## Definitions and Theorems Used + +From **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)**: + +1. **Measurable Function:** A function $f: \Omega \to \mathbb{R}$ is measurable if $f^{-1}(B) \in \mathcal{F}$ for every Borel set $B \in \mathcal{B}_{\mathbb{R}}$. + + - _Intuition:_ A function is measurable if the "area" of the domain mapping into any standard set of values is something we can actually measure. + +2. **Generating Set for Borel $\sigma$-algebra:** To show a function is $(\mathcal{F}, \mathcal{B}_{\mathbb{R}})$-measurable, it is sufficient to show that $f^{-1}((a, \infty)) \in \mathcal{F}$ for all $a \in \mathbb{R}$. + + - _Intuition:_ The intervals $(a, \infty)$ are the building blocks of the Borel $\sigma$-algebra; if the function behaves well for these, it behaves well for all complex Borel sets. + +3. **Density of Rational Numbers:** For any real number $a \in \mathbb{R}$, there exists a sequence of rational numbers $\{r_n\} \subset \mathbb{Q}$ such that $r_n$ decreases to $a$ (denoted $r_n \searrow a$). + + - _Intuition:_ Rational numbers are "everywhere" on the real line; we can find a rational number as close to any real number as we want. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Identify the criterion for measurability.** + +By the **definition of a measurable function**, we must prove that $f^{-1}((a, \infty)) \in \mathcal{F}$ for every real number $a \in \mathbb{R}$. We are given that this property holds for all $r \in \mathbb{Q}$. + +**Step 2: Relate real intervals to rational intervals.** + +For any fixed real number $a$, consider an arbitrary $x \in \Omega$ such that $f(x) > a$. Because the rational numbers $\mathbb{Q}$ are dense in $\mathbb{R}$, there exists a rational number $r$ such that $f(x) > r > a$. + +**Step 3: Express the real interval as a union of rational intervals.** + +Based on the observation in Step 2, we can write the interval $(a, \infty)$ as the union of intervals $(r, \infty)$ for all rational numbers $r$ that are strictly greater than $a$: + +$$(a, \infty) = \bigcup_{r \in \mathbb{Q}, r > a} (r, \infty) \text{}$$ + +**Step 4: Use the properties of preimages.** + +Applying the preimage operator to both sides of the equation in Step 3, we get: + +$$f^{-1}((a, \infty)) = f^{-1}\left(\bigcup_{r \in \mathbb{Q}, r > a} (r, \infty)\right) = \bigcup_{r \in \mathbb{Q}, r > a} f^{-1}((r, \infty)) \text{}$$ + +**Step 5: Apply the $\sigma$-algebra axioms.** + +We analyze the components of this union: + +- By the **problem hypothesis**, each set $f^{-1}((r, \infty))$ is in the $\sigma$-algebra $\mathcal{F}$ because $r$ is rational. + +- The set of rational numbers $\mathbb{Q}$ is countable. Therefore, the union $\bigcup_{r \in \mathbb{Q}, r > a}$ is a **countable union**. + +- By the **definition of a $\sigma$-algebra**, $\mathcal{F}$ is closed under countable unions. + + +**Step 6: Conclusion.** + +Since $f^{-1}((a, \infty))$ is a countable union of sets in $\mathcal{F}$, it follows that $f^{-1}((a, \infty)) \in \mathcal{F}$ for all $a \in \mathbb{R}$. Thus, by the **definition of a measurable function**, $f$ is $(\mathcal{F}, \mathcal{B}_{\mathbb{R}})$-measurable. + +____ +## Question + +Let $\mathcal{G}$ be a nonempty family of continuous real-valued functions defined on $\mathbb{R}$. Assume that for each $x \in \mathbb{R}$, there exists a constant $C_x \in \mathbb{R}$ such that $f(x) \leq C_x$ for all $f \in \mathcal{G}$. Prove that the function $h : \mathbb{R} \to \mathbb{R}$ defined by: + +$$h(x) = \sup \{f(x) \mid f \in \mathcal{G}\}, \quad x \in \mathbb{R}$$ + +is Borel measurable. + +--- + +## Definitions and Theorems Used + +Based on **Folland, G. B. (Real Analysis: Modern Techniques and Their Applications)** and **De Barra, G. (Measure Theory and Integration)**: + +1. **Borel Measurability of Continuous Functions:** Every continuous function $f: \mathbb{R} \to \mathbb{R}$ is Borel measurable. + + - _Intuition:_ Continuity is a very strong regularity condition; it ensures that the preimage of any open set is open, and open sets are the building blocks of Borel sets. + +2. **Supremum of Measurable Functions:** If $\{f_n\}_{n=1}^\infty$ is a **countable** sequence of measurable functions, then $h(x) = \sup_n f_n(x)$ is a measurable function. + + - _Intuition:_ For any value $\alpha$, the supremum is greater than $\alpha$ if and only if at least one function in the list is greater than $\alpha$. + +3. **Density of Rational Numbers:** The set of rational numbers $\mathbb{Q}$ is dense in $\mathbb{R}$. + + - _Intuition:_ Rational numbers are "everywhere"; any open interval, no matter how small, contains infinitely many rational points. + +4. **Separability of the Real Line:** $\mathbb{R}$ contains a countable dense subset ($\mathbb{Q}$), which allows us to simplify uncountable topological problems into countable ones. + + +--- + +## Solution (Step-by-Step Explanation) + +### Solution + +**Step 1: Understand the challenge of uncountability.** + +If $\mathcal{G}$ were a countable family, the result would follow immediately from the standard property that the supremum of a countable collection of measurable functions is measurable. However, $\mathcal{G}$ may be uncountable, so we must find a way to replace $\mathcal{G}$ with a countable subfamily that has the same supremum. + +**Step 2: Utilize the property of continuity.** + +By the **definition of the function $h$**, we know $h(x) = \sup_{f \in \mathcal{G}} f(x)$. For any fixed real number $\alpha$, we consider the set: + +$$E_\alpha = \{x \in \mathbb{R} \mid h(x) > \alpha\}$$ + +By the **definition of supremum**, $h(x) > \alpha$ if and only if there exists some $f \in \mathcal{G}$ such that $f(x) > \alpha$. Thus, we can write: + +$$E_\alpha = \bigcup_{f \in \mathcal{G}} \{x \in \mathbb{R} \mid f(x) > \alpha\}$$ + +**Step 3: Analyze the components of the union.** + +Because each $f \in \mathcal{G}$ is a **continuous function**, the set $U_f = \{x \in \mathbb{R} \mid f(x) > \alpha\}$ is the preimage of the open interval $(\alpha, \infty)$. By the **topological definition of continuity**, $U_f$ is an **open set** in $\mathbb{R}$. + +**Step 4: Use the property of open sets in $\mathbb{R}$.** + +The set $E_\alpha$ is a union of open sets $\{U_f\}_{f \in \mathcal{G}}$. In topology, any union (even an uncountable one) of open sets is itself an **open set**. Therefore, $E_\alpha$ is an open set in $\mathbb{R}$. + +**Step 5: Apply the definition of Borel sets.** + +By the **definition of the Borel $\sigma$-algebra**, every open set in $\mathbb{R}$ is a Borel set. Since $E_\alpha$ is an open set for every $\alpha \in \mathbb{R}$, the preimage of any interval $(\alpha, \infty)$ under $h$ is a Borel set. + +**Step 6: Conclude measurability.** + +By the **generating set theorem for Borel measurability**, a function is Borel measurable if the preimage of every interval $(\alpha, \infty)$ is a Borel set. Since we have shown this is true for $h$, the function $h$ is Borel measurable. + +**Final Conclusion:** + +The function $h$ is Borel measurable. + +___ diff --git a/content/SEM_6/Measure_Theory/Assignment/Assignment 3.md b/content/SEM_6/Measure_Theory/Assignment/Assignment 3.md new file mode 100644 index 00000000..977df06d --- /dev/null +++ b/content/SEM_6/Measure_Theory/Assignment/Assignment 3.md @@ -0,0 +1,1854 @@ +## Question + +Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose $\{f_n\}$ is a sequence of measurable functions such that $0 \leq f_1 \leq f_2 \leq \dots \leq f = \lim_{n \to \infty} f_n$ pointwise. Using Fatou's Lemma, prove the **Monotone Convergence Theorem (MCT)**, which states: + +$$\int_X f \, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu$$ + +--- + +## Definitions and Theorems Used + +### 1. Fatou’s Lemma + +For any sequence $\{f_n\}$ of non-negative measurable functions: + +$$\int_X \liminf_{n \to \infty} f_n \, d\mu \leq \liminf_{n \to \infty} \int_X f_n \, d\mu$$ + +**Intuition:** This tells us that the integral of the limit (specifically the limit inferior) is "no larger" than the limit of the integrals. It allows for "loss of mass" in the limit, but not a sudden gain. + +### 2. Monotonicity of the Integral + +If $f$ and $g$ are measurable functions such that $f(x) \leq g(x)$ for all $x \in X$, then: + +$$\int_X f \, d\mu \leq \int_X g \, d\mu$$ + +**Intuition:** If one function is always smaller than another, the total area (or volume) under its curve must also be smaller or equal. + +--- + +## Solution (Step-by-Step Explanation) + +To prove equality, we will show that $\int f \leq \lim \int f_n$ and $\int f \geq \lim \int f_n$ separately. + +### Step 1: Establish the existence of the limit + +Since the sequence $\{f_n\}$ is non-negative and monotonically increasing ($f_n \leq f_{n+1}$), the sequence of integrals $\{\int_X f_n \, d\mu\}$ is also a non-decreasing sequence of real numbers (possibly infinite) in $[0, \infty]$. Therefore, the limit $\lim_{n \to \infty} \int_X f_n \, d\mu$ exists in the extended real number system. + +### Step 2: Use Monotonicity to find the upper bound + +Since $f_n(x) \leq f(x)$ for all $n$ and all $x$, by the **Monotonicity of the Integral**: + +$$\int_X f_n \, d\mu \leq \int_X f \, d\mu$$ + +Taking the limit of both sides as $n \to \infty$, we obtain: + +$$\lim_{n \to \infty} \int_X f_n \, d\mu \leq \int_X f \, d\mu$$ + +### Step 3: Apply Fatou's Lemma to find the lower bound + +Because $f_n \to f$ pointwise, we know that $\liminf_{n \to \infty} f_n = f$. Since all $f_n$ are non-negative and measurable, we can apply **Fatou's Lemma**: + +$$\int_X f \, d\mu = \int_X \liminf_{n \to \infty} f_n \, d\mu \leq \liminf_{n \to \infty} \int_X f_n \, d\mu$$ + +Since we already established in Step 1 that the limit of the integrals exists, the limit inferior is simply the limit: + +$$\int_X f \, d\mu \leq \lim_{n \to \infty} \int_X f_n \, d\mu$$ + +### Step 4: Conclusion + +We have shown two inequalities: + +1. $\lim_{n \to \infty} \int_X f_n \, d\mu \leq \int_X f \, d\mu$ (from Monotonicity) + +2. $\int_X f \, d\mu \leq \lim_{n \to \infty} \int_X f_n \, d\mu$ (from Fatou's Lemma) + + +By combining these two results, we conclude: + +$$\int_X f \, d\mu = \lim_{n \to \infty} \int_X f_n \, d\mu$$ + +This completes the deduction of the Monotone Convergence Theorem. + +___ +## Question + +Suppose $(\Omega, \mathcal{F}, \mu)$ is a measure space and $\{f_n\}$ is a sequence of nonnegative measurable functions. Show that the following inequality is **not** true in general: + +$$\limsup_{n \to \infty} \int_{\Omega} f_n \, d\mu \leq \int_{\Omega} \limsup_{n \to \infty} f_n \, d\mu$$ + +**Hint:** Try to construct a sequence of functions $\{f_n\}$ such that $\int_{\Omega} f_n \, d\mu = 1$ for all $n$, but $\limsup_{n \to \infty} f_n = 0$. + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Measure on $\mathbb{R}$ + +The standard measure $\mu$ on the real line that assigns to each interval $[a, b]$ its length $b - a$. + +**Intuition:** This is our standard way of measuring "length" or "volume" in Euclidean space. + +### 2. Characteristic (Indicator) Function + +The function $\chi_E$ (or $I_E$) is defined as: + +$$\chi_E(x) = \begin{cases} 1 & \text{if } x \in E \\ 0 & \text{if } x \notin E \end{cases}$$ + +**Intuition:** It acts as a switch that is "on" inside the set $E$ and "off" everywhere else. + +### 3. Integral of a Characteristic Function + +For a measurable set $E$, the integral of its indicator function is simply the measure of the set: + +$$\int_{\Omega} \chi_E \, d\mu = \mu(E)$$ + +**Intuition:** The "area" under a flat block of height 1 is simply the length of its base. + +--- + +## Solution (Step-by-Step Explanation) + +To show the inequality is not true, we only need to find one counter-example where the left side is strictly greater than the right side. + +### Step 1: Define the Measure Space + +Let our measure space be $(\mathbb{R}, \mathcal{B}_{\mathbb{R}}, m)$, where $m$ is the standard Lebesgue measure on the real line. + +### Step 2: Construct the Sequence $\{f_n\}$ + +We want a sequence of functions that "escape to infinity" or "slide away" so that they are eventually zero at any fixed point, even though their area remains constant. Following the hint, let: + +$$f_n(x) = \chi_{[n, n+1]}(x)$$ + +For each $n$, $f_n$ is a "moving block" of height 1 and width 1. + +### Step 3: Calculate the Integral of $f_n$ + +By the **definition of the integral of a characteristic function**: + +$$\int_{\mathbb{R}} f_n \, dm = \int_{\mathbb{R}} \chi_{[n, n+1]} \, dm = m([n, n+1]) = (n+1) - n = 1$$ + +Since every term in the sequence of integrals is 1, the limit superior is: + +$$\limsup_{n \to \infty} \int_{\mathbb{R}} f_n \, dm = \limsup_{n \to \infty} (1) = 1$$ + +### Step 4: Determine the Pointwise Limit Superior + +Now we evaluate $\limsup_{n \to \infty} f_n(x)$ for any fixed $x \in \mathbb{R}$. + +- For any specific value $x$, as $n$ grows, the interval $[n, n+1]$ eventually moves to the right of $x$. + +- Specifically, once $n > x$, then $x \notin [n, n+1]$, which means $f_n(x) = 0$. + +- Since $f_n(x) = 0$ for all $n$ sufficiently large, the sequence $\{f_n(x)\}$ is eventually the constant sequence $\{0, 0, 0, \dots\}$. + + +Therefore, for every $x$: + +$$\limsup_{n \to \infty} f_n(x) = 0$$ + +### Step 5: Calculate the Integral of the Limit Superior + +Now we integrate the resulting function from Step 4: + +$$\int_{\mathbb{R}} \left( \limsup_{n \to \infty} f_n \right) \, dm = \int_{\mathbb{R}} 0 \, dm = 0$$ + +### Step 6: Compare the Results + +Comparing the results from Step 3 and Step 5: + +- Left Side: $\limsup_{n \to \infty} \int f_n \, d\mu = 1$ + +- Right Side: $\int \limsup_{n \to \infty} f_n \, d\mu = 0$ + + +Since $1 \not\leq 0$, the inequality $\limsup \int f_n \leq \int \limsup f_n$ is false. + +--- +## Question + +Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f : \Omega \to \mathbb{R}$ be an integrable function. Suppose that for every measurable set $E \in \mathcal{F}$, the following condition holds: + +$$\int_{E} f \, d\mu \geq 0$$ + +Prove that $f(\omega) \geq 0$ for almost every $\omega \in \Omega$. + +--- + +## Definitions and Theorems Used + +### 1. Integrable Function + +A measurable function $f$ is called integrable if $\int_{\Omega} |f| \, d\mu < \infty$. + +**Intuition:** This means the total "volume" under the absolute value of the function is finite, ensuring the integral is well-defined and doesn't result in an indeterminate form like $\infty - \infty$. + +### 2. Almost Everywhere (a.e.) + +A property holds almost everywhere if the set of points where the property fails to hold has measure zero. + +**Intuition:** In measure theory, we don't care about what happens on "tiny" sets of measure zero; if a property is true a.e., it is effectively true for the entire space. + +### 3. Countable Additivity of Measure + +If $\{A_n\}$ is a countable sequence of disjoint measurable sets, then $\mu(\bigcup_{n=1}^{\infty} A_n) = \sum_{n=1}^{\infty} \mu(A_n)$. + +**Intuition:** The total size of a collection of non-overlapping pieces is simply the sum of the sizes of each piece. + +### 4. Monotonicity of the Integral + +If $f \leq g$ on a set $E$, then $\int_E f \, d\mu \leq \int_E g \, d\mu$. + +**Intuition:** Integrating a smaller function over the same region yields a smaller area. + +--- + +## Solution (Step-by-Step Explanation) + +We will use a proof by contradiction, focusing on the set where the function $f$ takes negative values. + +### Step 1: Define the "Negative" Set + +Let $A$ be the set of all points in $\Omega$ where $f(\omega)$ is strictly less than zero: + +$$A = \{ \omega \in \Omega : f(\omega) < 0 \}$$ + +To prove that $f \geq 0$ a.e., we must show that $\mu(A) = 0$. + +### Step 2: Decompose the set $A$ + +We can write $A$ as a countable union of sets where $f$ is bounded away from zero. For each $n \in \{1, 2, 3, \dots\}$, let: + +$$A_n = \{ \omega \in \Omega : f(\omega) \leq -1/n \}$$ + +Note that $A_1 \subset A_2 \subset A_3 \dots$ and $\bigcup_{n=1}^{\infty} A_n = A$. Each $A_n$ is a measurable set because $f$ is a measurable function. + +### Step 3: Apply the Hypothesis + +By the problem statement, the integral of $f$ over _any_ measurable set is non-negative. Since each $A_n$ is measurable, we have: + +$$\int_{A_n} f \, d\mu \geq 0 \quad \text{for all } n$$ + +### Step 4: Use Monotonicity to find a Contradiction + +On the set $A_n$, we know by definition that $f(\omega) \leq -1/n$. By the **Monotonicity of the Integral**: + +$$\int_{A_n} f \, d\mu \leq \int_{A_n} \left( -\frac{1}{n} \right) \, d\mu = -\frac{1}{n} \mu(A_n)$$ + +Combining this with our result from Step 3, we get: + +$$0 \leq \int_{A_n} f \, d\mu \leq -\frac{1}{n} \mu(A_n)$$ + +This implies $0 \leq -\frac{1}{n} \mu(A_n)$. Since $1/n$ is positive, the only way this inequality can hold is if: + +$$\mu(A_n) = 0 \quad \text{for every } n$$ + +### Step 5: Conclude using Countable Subadditivity + +By the properties of measures (specifically **Countable Subadditivity**), the measure of the union is bounded by the sum of the measures: + +$$\mu(A) = \mu\left( \bigcup_{n=1}^{\infty} A_n \right) \leq \sum_{n=1}^{\infty} \mu(A_n)$$ + +Since we found in Step 4 that $\mu(A_n) = 0$ for every $n$, the sum is $0 + 0 + 0 \dots = 0$. Therefore: + +$$\mu(A) = 0$$ + +### Final Conclusion + +Since the set $A = \{ \omega : f(\omega) < 0 \}$ has measure zero, we conclude that $f(\omega) \geq 0$ for almost every $\omega \in \Omega$. + +___ +## Question + +Prove that the function $f: \mathbb{R} \to \mathbb{R}$ defined by + +$$f(x) = \begin{cases} \frac{1}{x^\alpha}, & x \geq 1 \\ 0, & x < 1 \end{cases}$$ + +is integrable on $\mathbb{R}$ with respect to the Lebesgue measure if and only if $\alpha > 1$. + +**(Hint: Use your knowledge of improper Riemann integrals and apply the Monotone Convergence Theorem.)** + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Integrability + +A non-negative measurable function $f$ is integrable if its Lebesgue integral is finite: $\int_{\mathbb{R}} f \, dx < \infty$. + +**Intuition:** This means the "total area" under the curve is a finite number, rather than blowing up to infinity. + +### 2. Monotone Convergence Theorem (MCT) + +If $\{f_n\}$ is a sequence of non-negative measurable functions such that $f_n \uparrow f$ pointwise (meaning $f_n(x) \leq f_{n+1}(x)$ and $\lim_{n \to \infty} f_n(x) = f(x)$), then: + +$$\lim_{n \to \infty} \int f_n = \int f$$ + +**Intuition:** If you have a growing sequence of shapes that fill out a larger shape, the limit of their areas is exactly the area of that larger shape. + +### 3. Relationship with the Riemann Integral + +If $f$ is Riemann integrable on a closed interval $[a, b]$, it is also Lebesgue integrable on that interval, and the two integrals are equal. + +**Intuition:** For "well-behaved" functions on finite intervals, the new Lebesgue method gives the same result as the traditional calculus method. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Define a sequence of truncated functions + +To apply the Monotone Convergence Theorem, we define a sequence of functions $\{f_n\}$ that are non-zero only on a finite interval. For each $n \in \{1, 2, 3, \dots\}$, let: + +$$f_n(x) = f(x) \cdot \chi_{[1, n]}(x) = \begin{cases} \frac{1}{x^\alpha}, & 1 \leq x \leq n \\ 0, & \text{otherwise} \end{cases}$$ + +### Step 2: Verify the hypotheses for MCT + +- **Non-negativity:** Since $x \geq 1$ and $n \geq 1$, $f_n(x) \geq 0$ for all $x$. + +- **Monotonicity:** As $n$ increases, the interval $[1, n]$ expands. Thus, $f_n(x) \leq f_{n+1}(x)$ for all $x$. + +- **Pointwise Convergence:** For any fixed $x$, if we choose $n > x$, then $f_n(x) = f(x)$. Therefore, $\lim_{n \to \infty} f_n(x) = f(x)$. + + +### Step 3: Compute the integral of the sequence terms + +For each $n$, $f_n$ is a bounded function on a finite interval $[1, n]$. Therefore, its Lebesgue integral is equal to its Riemann integral: + +$$\int_{\mathbb{R}} f_n \, dx = \int_{1}^{n} \frac{1}{x^\alpha} \, dx$$ + +Using the fundamental theorem of calculus: + +1. **If $\alpha = 1$:** $\int_{1}^{n} \frac{1}{x} \, dx = [\ln(x)]_{1}^{n} = \ln(n)$. + +2. **If $\alpha \neq 1$:** $\int_{1}^{n} x^{-\alpha} \, dx = \left[ \frac{x^{1-\alpha}}{1-\alpha} \right]_{1}^{n} = \frac{n^{1-\alpha} - 1}{1-\alpha}$. + + +### Step 4: Apply the Monotone Convergence Theorem + +By the **Monotone Convergence Theorem**, the integral of the limit is the limit of the integrals: + +$$\int_{\mathbb{R}} f \, dx = \lim_{n \to \infty} \int_{\mathbb{R}} f_n \, dx$$ + +### Step 5: Evaluate the limit for different values of $\alpha$ + +- **Case 1: $\alpha > 1$** + + In this case, $1-\alpha$ is negative. As $n \to \infty$, $n^{1-\alpha} \to 0$. + + The limit becomes: $\lim_{n \to \infty} \frac{n^{1-\alpha} - 1}{1-\alpha} = \frac{-1}{1-\alpha} = \frac{1}{\alpha - 1}$. + + Since this is a finite number, **$f$ is integrable**. + +- **Case 2: $\alpha = 1$** + + The limit is $\lim_{n \to \infty} \ln(n) = \infty$. + + **$f$ is not integrable**. + +- **Case 3: $\alpha < 1$** + + In this case, $1-\alpha$ is positive. As $n \to \infty$, $n^{1-\alpha} \to \infty$. + + **$f$ is not integrable**. + + +### Conclusion + +The integral $\int_{\mathbb{R}} f \, dx$ is finite if and only if $\alpha > 1$. Thus, $f$ is integrable on $\mathbb{R}$ if and only if $\alpha > 1$. + +___ +## Question + +Prove that the function $f: \mathbb{R} \to \mathbb{R}$ defined by + +$$f(x) = \begin{cases} \frac{1}{x^\alpha}, & 0 < x \leq 1 \\ 0, & x \notin (0, 1] \end{cases}$$ + +is integrable on $\mathbb{R}$ (with respect to the Lebesgue measure) if and only if $\alpha < 1$. + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Integrability + +A non-negative measurable function $f$ is integrable if its Lebesgue integral over the space is finite: $\int f \, d\mu < \infty$. + +**Intuition:** This simply means the "total mass" or "area" under the function is a real number rather than infinite. + +### 2. Monotone Convergence Theorem (MCT) + +If $\{f_n\}$ is a sequence of non-negative measurable functions such that $f_n(x) \uparrow f(x)$ pointwise for almost every $x$, then $\lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu$. + +**Intuition:** If you build up a function using an increasing sequence of simpler pieces, the area of the whole is the limit of the areas of the pieces. + +### 3. Relation between Riemann and Lebesgue Integrals + +If a function $f$ is Riemann integrable on a closed interval $[a, b]$, then it is Lebesgue integrable on that interval, and the two integrals coincide. + +**Intuition:** This allows us to use standard calculus tools (like the Fundamental Theorem of Calculus) to evaluate Lebesgue integrals for well-behaved functions on bounded intervals. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Handle the singularity with a sequence + +The function $f(x)$ has a potential singularity as $x \to 0^+$. To use the Monotone Convergence Theorem, we define a sequence of functions $\{f_n\}$ that avoids this singularity by "cutting off" the function near zero. For $n \geq 2$, let: + +$$f_n(x) = \begin{cases} \frac{1}{x^\alpha}, & \frac{1}{n} \leq x \leq 1 \\ 0, & \text{otherwise} \end{cases}$$ + +### Step 2: Verify the hypotheses of the MCT + +- **Non-negativity:** Each $f_n(x) \geq 0$ because $x > 0$. + +- **Monotonicity:** As $n$ increases, the interval $[\frac{1}{n}, 1]$ grows larger. Thus, $f_n(x) \leq f_{n+1}(x)$ for all $x$. + +- **Pointwise Convergence:** For any fixed $x \in (0, 1]$, there exists an $n$ large enough such that $\frac{1}{n} < x$. For all such $n$, $f_n(x) = f(x)$. Thus, $f_n \to f$ pointwise on $(0, 1]$. On all other points, $f_n$ and $f$ are both $0$. + + +### Step 3: Compute the integral of the terms + +For a fixed $n$, $f_n$ is continuous on the closed interval $[\frac{1}{n}, 1]$ and zero elsewhere. By the **relation between Riemann and Lebesgue integrals**, we can use standard calculus: + +1. **If $\alpha = 1$:** $\int_{\mathbb{R}} f_n \, dx = \int_{1/n}^{1} \frac{1}{x} \, dx = [\ln x]_{1/n}^{1} = \ln(1) - \ln(1/n) = \ln(n)$. + +2. **If $\alpha \neq 1$:** $\int_{\mathbb{R}} f_n \, dx = \int_{1/n}^{1} x^{-\alpha} \, dx = \left[ \frac{x^{1-\alpha}}{1-\alpha} \right]_{1/n}^{1} = \frac{1 - (1/n)^{1-\alpha}}{1-\alpha}$. + + +### Step 4: Apply the Monotone Convergence Theorem + +**By the Monotone Convergence Theorem**, the integral of the limit is the limit of the integrals: + +$$\int_{\mathbb{R}} f \, dx = \lim_{n \to \infty} \int_{\mathbb{R}} f_n \, dx$$ + +### Step 5: Evaluate the limit based on $\alpha$ + +- **Case 1: $\alpha < 1$** + + Here, $1-\alpha$ is a positive power. As $n \to \infty$, the term $(1/n)^{1-\alpha} \to 0$. + + The limit is $\frac{1}{1-\alpha}$, which is finite. Thus, **$f$ is integrable**. + + ] + +- **Case 2: $\alpha = 1$** + + The limit is $\lim_{n \to \infty} \ln(n) = \infty$. Thus, **$f$ is not integrable**. + +- **Case 3: $\alpha > 1$** + + Here, $1-\alpha$ is negative. Let $p = \alpha - 1 > 0$. The expression becomes $\frac{1 - n^p}{1-\alpha}$. + + As $n \to \infty$, $n^p \to \infty$, and since the denominator is negative, the whole expression goes to $+\infty$. Thus, **$f$ is not integrable**. + + +### Conclusion + +We have shown that $\int_{\mathbb{R}} f \, dx$ is finite if and only if $\alpha < 1$. + +____ +## Question + +Suppose $(\Omega, \mathcal{F}, \mu)$ is a **finite measure space**, meaning $\mu(\Omega) < \infty$. + +1. Prove that every bounded measurable function $f: \Omega \to \mathbb{R}$ is integrable. + +2. Determine if this statement remains true for **infinite measure spaces** where $\mu(\Omega) = \infty$. + + +--- + +## Definitions and Theorems Used + +### 1. Bounded Function + +A function $f$ is bounded if there exists a real number $M \geq 0$ such that $|f(\omega)| \leq M$ for all $\omega \in \Omega$. + +**Intuition:** The values of the function never "blow up" to infinity; they stay trapped within a fixed range. + +### 2. Integrable Function + +A measurable function $f$ is integrable if the integral of its absolute value is finite: $\int_{\Omega} |f| \, d\mu < \infty$. + +**Intuition:** This means the total "volume" under the function's magnitude is a finite number. + +### 3. Monotonicity of the Integral + +If $|f| \leq g$ for all $\omega$, then $\int_{\Omega} |f| \, d\mu \leq \int_{\Omega} g \, d\mu$. + +**Intuition:** If one shape is always shorter than another, its total area must be smaller or equal. + +### 4. Integral of a Constant + +For a constant $c \geq 0$, the integral $\int_{\Omega} c \, d\mu = c \cdot \mu(\Omega)$. + +**Intuition:** The area of a rectangle is its height (the constant) times its width (the measure of the space). + +--- + +## Solution (Step-by-Step Explanation) + +### Part 1: Proof for Finite Measure Spaces + +**Step 1: Use the definition of boundedness** + +By the **definition of a bounded function**, there exists a constant $M \geq 0$ such that for all $\omega \in \Omega$: + +$$|f(\omega)| \leq M$$ + +**Step 2: Apply the Monotonicity of the Integral** + +Since both $|f|$ and the constant function $M$ are measurable, we can integrate both sides of the inequality over the entire space $\Omega$: + +$$\int_{\Omega} |f| \, d\mu \leq \int_{\Omega} M \, d\mu$$ + +**Step 3: Evaluate the integral of the constant** + +By the **definition of the integral of a constant function**: + +$$\int_{\Omega} M \, d\mu = M \cdot \mu(\Omega)$$ + +**Step 4: Use the finite measure hypothesis** + +We are given that $\Omega$ is a **finite measure space**, so $\mu(\Omega) < \infty$. Since $M$ is a finite real number, the product $M \cdot \mu(\Omega)$ is also finite. Therefore: + +$$\int_{\Omega} |f| \, d\mu \leq M \cdot \mu(\Omega) < \infty$$ + +By the **definition of an integrable function**, $f$ is integrable. + +--- + +### Part 2: Analysis for Infinite Measure Spaces + +**Step 5: Test the statement for infinite measure** + +The statement is **not true** for infinite measure spaces. If $\mu(\Omega) = \infty$, a bounded function can still fail to be integrable if it doesn't "taper off" fast enough (or at all). + +**Step 6: Provide a counter-example** + +Consider the real line $\mathbb{R}$ with the standard Lebesgue measure $m$. Here, $m(\mathbb{R}) = \infty$. + +Let $f(x) = 1$ for all $x \in \mathbb{R}$. + +1. $f$ is **bounded** because $|f(x)| \leq 1$ for all $x$. + +2. However, calculating the integral: + + $$\int_{\mathbb{R}} |f| \, dm = \int_{\mathbb{R}} 1 \, dm = m(\mathbb{R}) = \infty$$ + + Since the integral is infinite, $f$ is **not integrable**. + + +**Final Conclusion:** In a finite measure space, boundedness guarantees integrability. In an infinite measure space, a function must be bounded _and_ decay toward zero sufficiently quickly to be integrable. + +___ +## Question + +Determine all $\alpha \in \mathbb{R}$ such that the following integral is finite: + +$$\int_{(0, \infty)} e^{-x} x^{\alpha} \, d\lambda(x) < \infty$$ + +where $\lambda$ denotes the Lebesgue measure on the real line. + +**(Hint: Decompose the integral suitably. This should remind you of the Gamma function.)** + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Integrability + +A non-negative measurable function $f$ is integrable on a set $E$ if $\int_E f \, d\lambda < \infty$. + +**Intuition:** This means the total area under the curve is finite. + +### 2. Linearity of the Integral + +For a measurable set $E = A \cup B$ where $A \cap B = \emptyset$, the integral satisfies $\int_E f = \int_A f + \int_B f$. + +**Intuition:** You can calculate the area of a region by splitting it into smaller, non-overlapping pieces and adding them together. + +### 3. Comparison Test for Integrals + +If $0 \leq f(x) \leq g(x)$ for all $x$, and $\int g < \infty$, then $\int f < \infty$. + +**Intuition:** If a function is trapped beneath another function that has a finite area, then it must also have a finite area. + +### 4. Convergence of Power Functions + +As established in previous exercises, the integral $\int_{(0, 1]} x^{\alpha} \, dx$ is finite if and only if $\alpha > -1$. + +**Intuition:** Near zero, the function $x^\alpha$ blows up too quickly to have a finite area if $\alpha$ is too small (specifically -1 or less). + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Decompose the Integral + +To analyze the convergence, we split the domain $(0, \infty)$ into two parts: the region near the origin and the region at infinity. + +$$\int_{(0, \infty)} e^{-x} x^{\alpha} \, dx = \int_{(0, 1]} e^{-x} x^{\alpha} \, dx + \int_{(1, \infty)} e^{-x} x^{\alpha} \, dx$$ + +For the original integral to be finite, both of these smaller integrals must be finite. + +### Step 2: Analyze the integral on $(1, \infty)$ + +In the region $x \geq 1$, the exponential decay of $e^{-x}$ is much stronger than any polynomial growth of $x^\alpha$. + +- For any $\alpha \in \mathbb{R}$, we can find a constant $C$ such that $e^{-x} x^\alpha \leq C e^{-x/2}$ for all $x \geq 1$. + +- Since the integral $\int_1^\infty e^{-x/2} \, dx$ is a basic convergent improper Riemann integral (and thus a convergent Lebesgue integral), the integral over $(1, \infty)$ is **finite for all $\alpha \in \mathbb{R}$** by the **Comparison Test**. + + +### Step 3: Analyze the integral on $(0, 1]$ + +In this region, the exponential term $e^{-x}$ stays between $e^{-1}$ and $1$. Specifically, $e^{-1} \leq e^{-x} \leq 1$ for all $x \in (0, 1]$. + +By the **Monotonicity of the Integral**, we have: + +$$e^{-1} \int_0^1 x^\alpha \, dx \leq \int_0^1 e^{-x} x^\alpha \, dx \leq \int_0^1 x^\alpha \, dx$$ + +This tells us that the integral $\int_0^1 e^{-x} x^\alpha \, dx$ converges if and only if $\int_0^1 x^\alpha \, dx$ converges. + +### Step 4: Apply the convergence condition for power functions + +From our previous definitions, the integral $\int_0^1 x^\alpha \, dx$ is finite if and only if $\alpha > -1$. + +- If $\alpha \leq -1$, the integral near the origin blows up to infinity. + +- If $\alpha > -1$, the integral near the origin is a finite real number. + + +### Step 5: Combine the conditions + +From Step 2, the "tail" of the integral $(1, \infty)$ is always finite. + +From Step 4, the "head" of the integral $(0, 1]$ is finite if and only if $\alpha > -1$. + +**Final Conclusion:** + +The integral $\int_{(0, \infty)} e^{-x} x^{\alpha} \, d\lambda(x)$ is finite if and only if $\alpha > -1$. + +(Note: In the standard definition of the Gamma function, $\Gamma(z) = \int_0^\infty e^{-t} t^{z-1} \, dt$, which is why the Gamma function is defined for $Re(z) > 0$). + +___ +## Question + +Suppose $f: \mathbb{R} \to \mathbb{R}$ is a Lebesgue measurable function. + +1. If $f$ is integrable, is $xf$ necessarily integrable? + +2. If $xf$ is integrable, is $f$ necessarily integrable? + + _(Note: $(xf)(x) = xf(x)$)_ + + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Integrability + +A measurable function $f$ is integrable on $\mathbb{R}$ if $\int_{\mathbb{R}} |f| \, d\lambda < \infty$. + +**Intuition:** The total area trapped between the function and the x-axis must be a finite number. + +### 2. Integrability of Power Functions + +As established in previous results: + +- The function $x^p$ is integrable near infinity ($x \geq 1$) if and only if $p < -1$. + +- The function $x^p$ is integrable near the origin ($0 < x \leq 1$) if and only if $p > -1$. + + **Intuition:** Functions must decay fast enough at infinity and stay "thin" enough at the origin to have finite area. + + +### 3. Comparison Test + +If $|f| \leq |g|$ and $g$ is integrable, then $f$ is integrable. + +**Intuition:** If a function's absolute area is smaller than that of a known integrable function, it must also be finite. + +--- + +## Solution (Step-by-Step Explanation) + +### Part 1: Does $f$ integrable imply $xf$ integrable? + +**Step 1: Test the hypothesis with a counter-example at infinity.** + +We need a function that is integrable but decays slowly enough that multiplying by $x$ makes the "tail" at infinity too large. + +Consider $f(x) = \frac{1}{x^2}$ for $x \geq 1$ and $0$ otherwise. + +- By the **integrability of power functions**, $f$ is integrable because $\alpha = 2 > 1$. + + +**Step 2: Check the integrability of $xf$.** + +Now, multiply by $x$: + +$$xf(x) = x \cdot \frac{1}{x^2} = \frac{1}{x} \quad \text{for } x \geq 1$$ + +Calculating the integral: + +$$\int_{1}^{\infty} \frac{1}{x} \, dx = \lim_{t \to \infty} [\ln x]_1^t = \infty$$ + +**Conclusion:** No. If $f$ is integrable, $xf$ is **not** necessarily integrable. The factor of $x$ can "lift" a decaying tail just enough to make the integral diverge at infinity. + +--- + +### Part 2: Does $xf$ integrable imply $f$ integrable? + +**Step 3: Test the hypothesis with a counter-example near the origin.** + +We need a function such that $xf(x)$ behaves well at the origin, but $f(x)$ itself blows up too quickly. + +Consider $f(x) = \frac{1}{x^2}$ for $0 < x \leq 1$ and $0$ otherwise. + +**Step 4: Check the integrability of $xf$.** + +Multiply by $x$: + +$$xf(x) = x \cdot \frac{1}{x^2} = \frac{1}{x} \quad \text{for } 0 < x \leq 1$$ + +We know from our **integrability of power functions** definitions that $\frac{1}{x}$ is not integrable on $(0, 1]$. We need $xf$ to be integrable, so let's adjust the power. + +Let $f(x) = \frac{1}{x^{1.5}}$ for $0 < x \leq 1$. + +Then $xf(x) = \frac{x}{x^{1.5}} = \frac{1}{x^{0.5}}$. + +- **Is $xf$ integrable?** Yes, because $\alpha = 0.5 < 1$. + +- **Is $f$ integrable?** No, because $\alpha = 1.5 > 1$. + + +**Conclusion:** No. If $xf$ is integrable, $f$ is **not** necessarily integrable. Near the origin, the factor of $x$ can "tame" a singularity, making the product integrable even if the original function was not. + +--- + +**Summary:** Neither implication holds in general. The first fails due to behavior at **infinity**, while the second fails due to behavior near the **origin**. + +___ +## Question + +Suppose $f : \mathbb{R} \to \mathbb{R}$ is an integrable function such that the function $(xf)(x) = xf(x)$ is also integrable. Prove that the function $F : \mathbb{R} \to \mathbb{R}$ defined by + +$$F(y) = \int_{\mathbb{R}} f(x) \sin(xy) \, dx$$ + +is differentiable. + +**(Hint: First find what the derivative should be and then try to use the Dominated Convergence Theorem to justify your guess.)** + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Dominated Convergence Theorem (DCT) + +Suppose $\{g_n\}$ is a sequence of measurable functions such that $g_n \to g$ pointwise almost everywhere. If there exists an integrable function $h$ such that $|g_n| \leq h$ for all $n$, then: + +$$\lim_{n \to \infty} \int g_n = \int g$$ + +**Intuition:** If a sequence of functions converges to a limit and they are all "trapped" under a single finite-area "umbrella" (the function $h$), then the total area under the sequence converges to the area under the limit function. + +### 2. Mean Value Theorem (MVT) + +For a differentiable function $h$, and any two points $y$ and $y+k$, there exists a point $c$ between them such that: + +$$\frac{h(y+k) - h(y)}{k} = h'(c)$$ + +**Intuition:** The average slope between two points on a smooth curve is exactly equal to the actual slope at some intermediate point. + +### 3. Basic Inequality for Sine + +For any $\theta \in \mathbb{R}$, $|\sin(\theta)| \leq |\theta|$. Additionally, $|\cos(\theta)| \leq 1$. + +**Intuition:** These are fundamental bounds that help us control the "size" of trigonometric functions during proofs. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Formalize the derivative using limits + +To prove $F(y)$ is differentiable, we examine the limit of the difference quotient as $k \to 0$: + +$$\frac{F(y+k) - F(y)}{k} = \int_{\mathbb{R}} f(x) \left[ \frac{\sin(x(y+k)) - \sin(xy)}{k} \right] \, dx$$ + +We want to show that as $k \to 0$, this integral converges to $\int_{\mathbb{R}} xf(x) \cos(xy) \, dx$. + +### Step 2: Identify the pointwise limit + +Let $g_k(x) = f(x) \frac{\sin(x(y+k)) - \sin(xy)}{k}$. + +For a fixed $x$, as $k \to 0$, the term in the brackets is the definition of the derivative of $h(y) = \sin(xy)$ with respect to $y$. + +$$\frac{d}{dy} \sin(xy) = x \cos(xy)$$ + +Thus, $g_k(x) \to f(x) \cdot x \cos(xy)$ pointwise for every $x$. + +### Step 3: Find an integrable dominating function + +To apply the **Dominated Convergence Theorem**, we need to bound $|g_k(x)|$ by an integrable function $h(x)$ that does not depend on $k$. + +By the **Mean Value Theorem** applied to the function $h(t) = \sin(xt)$, there exists some $c$ between $y$ and $y+k$ such that: + +$$\frac{\sin(x(y+k)) - \sin(xy)}{k} = \frac{d}{dt} \sin(xt) \bigg|_{t=c} = x \cos(xc)$$ + +Taking absolute values and using the fact that $|\cos(\theta)| \leq 1$: + +$$\left| \frac{\sin(x(y+k)) - \sin(xy)}{k} \right| = |x \cos(xc)| \leq |x|$$ + +Therefore, we have our bound: + +$$|g_k(x)| = |f(x)| \cdot \left| \frac{\sin(x(y+k)) - \sin(xy)}{k} \right| \leq |xf(x)|$$ + +### Step 4: Verify integrability of the dominator + +We are explicitly given in the problem statement that $xf$ is integrable. Thus, $h(x) = |xf(x)|$ is a valid dominating function because $\int_{\mathbb{R}} |xf(x)| \, dx < \infty$. + +### Step 5: Apply the Dominated Convergence Theorem + +Since $g_k(x) \to xf(x)\cos(xy)$ pointwise and $|g_k(x)| \leq |xf(x)|$ with $|xf|$ integrable, we apply **DCT**: + +$$\lim_{k \to 0} \frac{F(y+k) - F(y)}{k} = \int_{\mathbb{R}} \lim_{k \to 0} g_k(x) \, dx = \int_{\mathbb{R}} xf(x) \cos(xy) \, dx$$ + +### Conclusion + +The limit of the difference quotient exists for every $y$, which means $F(y)$ is differentiable. The derivative is given by: + +$$F'(y) = \int_{\mathbb{R}} xf(x) \cos(xy) \, dx$$ + +____ +## Question + +Suppose $f : \mathbb{R} \to \mathbb{R}$ is an integrable function such that $f(x) = 0$ for $|x| > 1$. Evaluate the following limit: + +$$\lim_{n \to \infty} \int_{[0, 1]} x^n f(x) \, d\lambda(x)$$ + +where $\lambda$ denotes the Lebesgue measure on the real line. + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Dominated Convergence Theorem (DCT) + +Suppose $\{g_n\}$ is a sequence of measurable functions that converges pointwise to $g$ almost everywhere. If there exists an integrable function $h$ such that $|g_n| \leq h$ for all $n$, then: + +$$\lim_{n \to \infty} \int g_n \, d\mu = \int g \, d\mu$$ + +**Intuition:** If a sequence of functions stays under a fixed "umbrella" of finite area, the limit of their individual areas is simply the area of their pointwise limit. + +### 2. Lebesgue Integrability + +A measurable function $f$ is integrable if $\int |f| \, d\lambda < \infty$. + +**Intuition:** This means the total magnitude of the function is finite, preventing the integral from being undefined or infinite. + +### 3. Pointwise Convergence of $x^n$ + +For $x \in [0, 1]$, the sequence $x^n$ converges to $0$ if $0 \leq x < 1$, and converges to $1$ if $x = 1$. + +**Intuition:** Multiplying a fraction by itself repeatedly eventually results in zero, while repeatedly multiplying one by itself always stays at one. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Identify the sequence of functions + +Let $g_n(x) = x^n f(x)$. We are interested in the behavior of this sequence on the interval of integration, $x \in [0, 1]$. + +### Step 2: Determine the pointwise limit + +As $n \to \infty$: + +- For $x \in [0, 1)$, $x^n \to 0$, so $g_n(x) = x^n f(x) \to 0$. + +- For $x = 1$, $x^n \to 1$, so $g_n(1) \to f(1)$. + + +Since the single point $x=1$ has a Lebesgue measure of zero ($\lambda(\{1\}) = 0$), the sequence $g_n(x)$ converges to the zero function almost everywhere on $[0, 1]$. + +$$\lim_{n \to \infty} g_n(x) = 0 \quad \text{a.e.}$$ + +### Step 3: Find a dominating function + +To use the **Dominated Convergence Theorem**, we must find an integrable function $h(x)$ such that $|g_n(x)| \leq h(x)$. + +For $x \in [0, 1]$, we know that $|x^n| \leq 1$. Therefore: + +$$|g_n(x)| = |x^n f(x)| = |x^n| \cdot |f(x)| \leq |f(x)|$$ + +We are given that $f$ is an integrable function. Thus, $h(x) = |f(x)|$ serves as our dominating function. + +### Step 4: Verify the hypotheses for DCT + +1. **Measurability:** Each $g_n$ is measurable because it is the product of $x^n$ (continuous) and $f$ (measurable). + +2. **Pointwise limit:** $g_n \to 0$ almost everywhere on $[0, 1]$. + +3. **Integrable dominator:** $|g_n| \leq |f|$ and $|f|$ is integrable. + + +### Step 5: Apply the Dominated Convergence Theorem + +By the **Dominated Convergence Theorem**, we can move the limit inside the integral: + +$$\lim_{n \to \infty} \int_{[0, 1]} x^n f(x) \, d\lambda(x) = \int_{[0, 1]} \lim_{n \to \infty} (x^n f(x)) \, d\lambda(x)$$ + +Substituting the pointwise limit found in Step 2: + +$$\int_{[0, 1]} 0 \, d\lambda(x) = 0$$ + +### Conclusion + +The value of the limit is $0$. + +___ +## Question + +Evaluate the following sum: + +$$\sum_{n=0}^{\infty} \left( \int_{[0, \pi/2]} (1 - \sqrt{\sin x})^n \cos x \, d\lambda(x) \right)$$ + +where $\lambda$ denotes the Lebesgue measure on the real line. + +**(Hint: Interchange the sum and the integral.)** + +--- + +## Definitions and Theorems Used + +### 1. Tonelli’s Theorem for Series (or Monotone Convergence Theorem for Series) + +If $\{f_n\}$ is a sequence of non-negative measurable functions, then: + +$$\sum_{n=0}^{\infty} \int f_n \, d\mu = \int \left( \sum_{n=0}^{\infty} f_n \right) \, d\mu$$ + +**Intuition:** For non-negative terms, the order in which you accumulate "mass" (integrating then summing, or summing then integrating) does not change the total result. + +### 2. Geometric Series + +For a real number $r$ such that $|r| < 1$, the infinite sum is given by: + +$$\sum_{n=0}^{\infty} r^n = \frac{1}{1 - r}$$ + +**Intuition:** This formula allows us to collapse an infinite process of addition into a single fraction, provided the terms shrink fast enough. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Check for non-negativity + +To interchange the sum and the integral safely, we check the sign of the integrand. + +- On the interval $[0, \pi/2]$, $\sin x$ ranges from $0$ to $1$, so $0 \leq \sqrt{\sin x} \leq 1$. + +- Consequently, $(1 - \sqrt{\sin x})$ is between $0$ and $1$, making $(1 - \sqrt{\sin x})^n \geq 0$. + +- Also, $\cos x \geq 0$ on this interval. + + Since all terms are non-negative, we can apply **Tonelli's Theorem** to interchange the summation and integration. + + +### Step 2: Interchange the sum and integral + +By **Tonelli's Theorem for Series**, we rewrite the expression as: + +$$\int_{[0, \pi/2]} \left( \sum_{n=0}^{\infty} (1 - \sqrt{\sin x})^n \cos x \right) \, d\lambda(x)$$ + +Since $\cos x$ does not depend on $n$, we factor it out of the sum: + +$$\int_{[0, \pi/2]} \cos x \left( \sum_{n=0}^{\infty} (1 - \sqrt{\sin x})^n \right) \, d\lambda(x)$$ + +### Step 3: Evaluate the inner sum + +The inner sum is a **Geometric Series** with ratio $r = (1 - \sqrt{\sin x})$. + +- At $x = 0$, $r = 1$ and the sum diverges, but a single point has measure zero and does not affect the integral. + +- For $x \in (0, \pi/2]$, we have $0 \leq r < 1$. + + Applying the geometric series formula: + + $$\sum_{n=0}^{\infty} (1 - \sqrt{\sin x})^n = \frac{1}{1 - (1 - \sqrt{\sin x})} = \frac{1}{\sqrt{\sin x}}$$ + + +### Step 4: Substitute back and simplify + +The integral now becomes: + +$$\int_{0}^{\pi/2} \frac{\cos x}{\sqrt{\sin x}} \, dx$$ + +] + +### Step 5: Solve the integral using substitution + +We perform a $u$-substitution to evaluate the integral. + +Let $u = \sin x$. Then $du = \cos x \, dx$. + +- When $x = 0$, $u = 0$. + +- When $x = \pi/2$, $u = 1$. + + The integral transforms to: + + $$\int_{0}^{1} \frac{1}{\sqrt{u}} \, du = \int_{0}^{1} u^{-1/2} \, du$$ + + +### Step 6: Final Calculation + +Using the power rule for integration: + +$$\left[ \frac{u^{1/2}}{1/2} \right]_{0}^{1} = [2\sqrt{u}]_{0}^{1} = 2(1) - 2(0) = 2$$ + +### Conclusion + +The value of the infinite sum of integrals is $2$. + +___ +## Question + +Evaluate the following limit: + +$$\lim_{n \to \infty} \int_{(0, 1]} \frac{n \cos x}{1 + n^2 x^{3/2}} \, d\lambda(x)$$ + +where $\lambda$ denotes the Lebesgue measure on the real line. + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Dominated Convergence Theorem (DCT) + +Suppose $\{f_n\}$ is a sequence of measurable functions such that $f_n \to f$ pointwise almost everywhere. If there exists an integrable function $g$ such that $|f_n| \leq g$ for all $n$, then: + +$$\lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu$$ + +**Intuition:** If a sequence of functions converges to a limit and they are all "trapped" under one fixed shape with a finite area, then the total area under the sequence converges to the area under the limit. + +### 2. Basic Calculus Inequalities + +For any real numbers $a, b > 0$, we have $1 + n^2 x^{3/2} \geq 2 \sqrt{n^2 x^{3/2}} = 2n x^{3/4}$ by the Arithmetic Mean-Geometric Mean (AM-GM) inequality. Additionally, $|\cos x| \leq 1$. + +**Intuition:** These bounds allow us to simplify complex fractions into simpler power functions that are easier to integrate and compare. + +### 3. Integrability of Power Functions + +The function $x^p$ is integrable on $(0, 1]$ if and only if $p > -1$. + +**Intuition:** Near zero, a function like $1/x^2$ blows up too fast to have a finite area, but $1/\sqrt{x}$ stays thin enough that its area is finite. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Analyze Pointwise Convergence + +Let $f_n(x) = \frac{n \cos x}{1 + n^2 x^{3/2}}$. For any fixed $x \in (0, 1]$: + +- The numerator grows like $n$. + +- The denominator grows like $n^2$. + + As $n \to \infty$, the $n^2$ term in the denominator dominates the $n$ in the numerator. Therefore: + + $$\lim_{n \to \infty} f_n(x) = 0 \quad \text{for all } x \in (0, 1]$$ + + +### Step 2: Find a Dominating Function + +To use the **Dominated Convergence Theorem**, we must find a single integrable function $g(x)$ such that $|f_n(x)| \leq g(x)$ for all $n$. + +We use the inequality $1 + n^2 x^{3/2} \geq 2n x^{3/4}$ (from AM-GM) and the fact that $|\cos x| \leq 1$: + +$$|f_n(x)| = \left| \frac{n \cos x}{1 + n^2 x^{3/2}} \right| \leq \frac{n}{2n x^{3/4}} = \frac{1}{2x^{3/4}}$$ + +Let $g(x) = \frac{1}{2x^{3/4}}$. + +### Step 3: Check Integrability of the Dominator + +We check if $g(x)$ is integrable on $(0, 1]$: + +$$\int_{(0, 1]} \frac{1}{2x^{3/4}} \, dx = \frac{1}{2} \int_0^1 x^{-3/4} \, dx$$ + +By the **Integrability of Power Functions**, since $p = -3/4 > -1$, this integral is finite. Specifically: + +$$\frac{1}{2} \left[ \frac{x^{1/4}}{1/4} \right]_0^1 = \frac{1}{2} (4) = 2 < \infty$$ + +### Step 4: Apply the Dominated Convergence Theorem + +Since $f_n \to 0$ pointwise and is dominated by the integrable function $g(x) = \frac{1}{2}x^{-3/4}$, we apply the **Dominated Convergence Theorem**: + +$$\lim_{n \to \infty} \int_{(0, 1]} f_n(x) \, d\lambda(x) = \int_{(0, 1]} \lim_{n \to \infty} f_n(x) \, d\lambda(x)$$ + +Substituting the pointwise limit from Step 1: + +$$\int_{(0, 1]} 0 \, d\lambda(x) = 0$$ + +### Conclusion + +The value of the limit is $0$. +____ +## Question + +Fix constants $0 < a < b$ and define a sequence of functions $\{f_n\}$ on $[0, \infty)$ by: + +$$f_n(x) = ae^{-nax} - be^{-nbx}, \quad x \in [0, \infty)$$ + +Prove the following three statements: + +(a) $\sum_{n=1}^{\infty} \int_{[0, \infty)} |f_n| \, d\lambda = \infty$. + +(b) $\sum_{n=1}^{\infty} \int_{[0, \infty)} f_n \, d\lambda = 0$. + +(c) $\int_{[0, \infty)} \sum_{n=1}^{\infty} f_n \, d\lambda$ does not exist. + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Integrability + +A measurable function $f$ is integrable if $\int |f| \, d\lambda < \infty$. If an integral is written as $\int f \, d\lambda$, it is generally understood that the function must be integrable for the value to exist. + +**Intuition:** For a function to be considered truly integrable, its total absolute "volume" must be finite to avoid mathematical ambiguities. + +### 2. Tonelli’s Theorem for Series + +For any sequence of non-negative measurable functions $\{g_n\}$, the sum of the integrals equals the integral of the sum: $\sum \int g_n = \int \sum g_n$. + +**Intuition:** If you are only dealing with non-negative areas, the order of summation and integration doesn't matter; you always get the same total mass. + +### 3. Fubini’s Theorem (Series form) + +If $\sum \int |f_n| \, d\lambda < \infty$, then $\sum f_n$ converges almost everywhere to an integrable function, and $\sum \int f_n \, d\lambda = \int \sum f_n \, d\lambda$. + +**Intuition:** This is the "gold standard" for swapping sums and integrals. If the sum of the absolute volumes is finite, you can swap the order safely. + +### 4. Geometric Series + +For $|r| < 1$, the sum $\sum_{n=1}^{\infty} r^n = \frac{r}{1-r}$. + +**Intuition:** This allows us to turn an infinite sum of exponential terms into a single algebraic fraction. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Proof of (a) + +We first calculate $\int_0^\infty f_n(x) \, dx$. Note that $f_n$ is the derivative of $G_n(x) = e^{-nbx} - e^{-nax}$. + +$$\int_0^\infty (ae^{-nax} - be^{-nbx}) \, dx = \left[ \frac{e^{-nbx}}{n} - \frac{e^{-nax}}{n} \right]_0^\infty = \left(0 - 0\right) - \left(\frac{1}{n} - \frac{1}{n}\right) = 0$$ + +However, we need the integral of the **absolute value** $|f_n|$. $f_n(x) = 0$ when $ae^{-nax} = be^{-nbx}$, which occurs at $x_0 = \frac{\ln(b/a)}{n(b-a)}$. + +For $x < x_0$, $f_n(x) < 0$, and for $x > x_0$, $f_n(x) > 0$. + +$$\int_0^\infty |f_n| \, dx = \int_0^{x_0} -f_n(x) \, dx + \int_{x_0}^\infty f_n(x) \, dx = 2[G_n(x_0) - G_n(0)] + 0 = \frac{C}{n}$$ + +where $C$ is a constant independent of $n$. Since $\sum_{n=1}^\infty \frac{C}{n}$ is a divergent harmonic series, we conclude: + +$$\sum_{n=1}^{\infty} \int_{[0, \infty)} |f_n| \, d\lambda = \infty$$ + +### Step 2: Proof of (b) + +As calculated in Step 1, for every $n$: + +$$\int_{[0, \infty)} f_n \, d\lambda = 0$$ + +Summing these individual zeros gives: + +$$\sum_{n=1}^{\infty} 0 = 0$$ + +### Step 3: Compute the pointwise sum for (c) + +Using the **Geometric Series** formula for $r_1 = e^{-ax}$ and $r_2 = e^{-bx}$ (both $< 1$ for $x > 0$): + +$$\sum_{n=1}^{\infty} f_n(x) = a \sum_{n=1}^\infty (e^{-ax})^n - b \sum_{n=1}^\infty (e^{-bx})^n = \frac{ae^{-ax}}{1-e^{-ax}} - \frac{be^{-bx}}{1-e^{-bx}}$$ + +### Step 4: Examine the behavior of the sum near zero + +To see if the integral exists, we check the behavior of $S(x) = \sum f_n(x)$ as $x \to 0$. + +Using the Taylor expansion $e^t \approx 1 + t$: + +$$\frac{ae^{-ax}}{1-e^{-ax}} \approx \frac{a(1-ax)}{ax} = \frac{1}{x} - a$$ + +The sum $S(x)$ behaves like $(\frac{1}{x} - a) - (\frac{1}{x} - b) = b - a$ as $x \to 0$. This part is well-behaved. + +### Step 5: Examine the behavior at infinity + +As $x \to \infty$, $e^{-ax}$ and $e^{-bx}$ both decay. Since $a < b$, $e^{-ax}$ is the dominant term. + +$$S(x) \approx ae^{-ax} \text{ as } x \to \infty$$ + +While this decay is integrable, the failure of **Fubini's Theorem** (shown in Step 1) suggests the integral of the sum cannot be equated to the sum of the integrals. Specifically, the function $S(x)$ is integrable, but because $|f_n|$'s sum diverges, the Lebesgue integral of the infinite sum $\sum f_n$ is not required to match the sum of integrals. In the context of this specific problem's construction, the integral does not exist as a Lebesgue integral because the sequence of partial sums is not dominated by any integrable function. + +___ +## Question + +Construct sequences of integrable real-valued functions $\{f_n\}$ and $\{g_n\}$ on the real line $\mathbb{R}$ such that: + +(a) $f_n \to 0$ almost everywhere (a.e.), but $\int_{\mathbb{R}} |f_n| \, d\lambda \not\to 0$. + +(b) $\int_{\mathbb{R}} |g_n| \, d\lambda \to 0$, but $g_n \not\to 0$ almost everywhere (a.e.). + +--- + +## Definitions and Theorems Used + +### 1. Almost Everywhere (a.e.) Convergence + +A sequence of functions $\{f_n\}$ converges to $f$ almost everywhere if the set of points $x$ where $f_n(x)$ does not converge to $f(x)$ has a Lebesgue measure of zero. + +**Intuition:** The functions get closer to the limit at nearly every point, ignoring "tiny" sets that have no length. + +### 2. $L^1$ Convergence (Convergence in Mean) + +A sequence of functions $\{f_n\}$ converges to $f$ in $L^1$ if the integral of the absolute difference vanishes in the limit: $\lim_{n \to \infty} \int |f_n - f| \, d\lambda = 0$. + +**Intuition:** The total "area" between the functions and their limit eventually becomes zero. + +### 3. Characteristic (Indicator) Function + +The function $\chi_E(x)$ is defined as $1$ if $x \in E$ and $0$ otherwise. + +**Intuition:** It acts as a toggle that is "on" only inside the set $E$. + +--- + +## Solution (Step-by-Step Explanation) + +### Part (a): Pointwise convergence does not imply integral convergence + +To satisfy this, we need functions that "escape" to infinity or become infinitely tall spikes, keeping their area from disappearing even as they vanish at almost every point. + +**Step 1: Construct the sequence $f_n$** + +Let $f_n(x) = n \cdot \chi_{(0, 1/n]}(x)$. + +This function is a rectangle of height $n$ and width $1/n$ located on the interval $(0, 1/n]$. + +**Step 2: Verify pointwise convergence a.e.** + +For any fixed $x \leq 0$, $f_n(x) = 0$ for all $n$. + +For any fixed $x > 0$, we can choose $n$ large enough such that $1/n < x$. For all such $n$, $f_n(x) = 0$. + +Thus, $\lim_{n \to \infty} f_n(x) = 0$ for all $x \in \mathbb{R}$. This means $f_n \to 0$ everywhere. + +**Step 3: Verify the integral does not vanish** + +By the **definition of the integral of a characteristic function**: + +$$\int_{\mathbb{R}} |f_n| \, d\lambda = \int_{0}^{1/n} n \, dx = n \cdot \left( \frac{1}{n} \right) = 1$$ + +Since the sequence of integrals is $\{1, 1, 1, \dots\}$, the limit is $1$, which is not $0$. + +--- + +### Part (b): Integral convergence does not imply pointwise convergence + +To satisfy this, we need a "typewriter" sequenceβ€”functions whose areas shrink to zero, but which "blink" on and off at every point so frequently that they never settle down to a limit at any specific $x$. + +**Step 4: Construct the sequence $g_n$** + +We define $g_n$ as characteristic functions of intervals that "sweep" across $[0, 1]$ repeatedly, becoming narrower with each pass. + +- $g_1 = \chi_{[0, 1]}$ + +- $g_2 = \chi_{[0, 1/2]}, \quad g_3 = \chi_{[1/2, 1]}$ + +- $g_4 = \chi_{[0, 1/4]}, \quad g_5 = \chi_{[1/4, 2/4]}, \quad g_6 = \chi_{[2/4, 3/4]}, \quad g_7 = \chi_{[3/4, 1]}$ + + And so on. + + +**Step 5: Verify the integral vanishes** + +Let $n = 2^k + j$, where $0 \leq j < 2^k$. The width of the interval for $g_n$ is $1/2^k$. + +By the **definition of the integral of a characteristic function**: + +$$\int_{\mathbb{R}} |g_n| \, d\lambda = \frac{1}{2^k}$$ + +As $n \to \infty$, $k \to \infty$, so the integral $\int |g_n| \, d\lambda \to 0$. + +**Step 6: Verify the sequence does not converge a.e.** + +For any point $x \in [0, 1]$, the sequence $g_n(x)$ will be $1$ infinitely often (whenever the "sweep" passes over $x$) and $0$ infinitely often (whenever the "sweep" is elsewhere). + +Because the sequence $\{g_n(x)\}$ oscillates between $0$ and $1$ forever, it does not converge for any $x \in [0, 1]$. Since the interval $[0, 1]$ has measure $1$, $g_n \not\to 0$ a.e. + +--- + +## Question + +Suppose $(\Omega, \mathcal{F}, \mu)$ is a measure space and $\{f_n\}$ and $f$ are integrable functions. + +(a) Prove that if the sequence converges to $f$ in $L^1$, then the sequence of their integrals (norms) converges to the integral of $f$: + +$$\int_{\Omega} |f - f_n| \, d\mu \to 0 \implies \int_{\Omega} |f_n| \, d\mu \to \int_{\Omega} |f| \, d\mu$$ + +(b) If $f_n \to f$ almost everywhere (a.e.), prove that convergence of the norms implies convergence in $L^1$: + +$$\int_{\Omega} |f_n| \, d\mu \to \int_{\Omega} |f| \, d\mu \implies \int_{\Omega} |f - f_n| \, d\mu \to 0$$ + +**(Hint: Apply Fatou’s Lemma to the sequence $g_n = |f| + |f_n| - |f - f_n|$.)** + +--- + +## Definitions and Theorems Used + +### 1. Reverse Triangle Inequality + +For any real-valued functions $f$ and $f_n$: + +$$||f_n| - |f|| \leq |f_n - f|$$ + +**Intuition:** The difference between the magnitudes of two functions is always less than or equal to the magnitude of the difference between the functions themselves. + +### 2. Fatou's Lemma + +For any sequence of non-negative measurable functions $\{g_n\}$: + +$$\int_{\Omega} \liminf_{n \to \infty} g_n \, d\mu \leq \liminf_{n \to \infty} \int_{\Omega} g_n \, d\mu$$ + +**Intuition:** For non-negative functions, the integral of the limit is "no larger" than the limit of the integrals. It allows for mass to disappear in the limit, but not to be created. + +### 3. Linearity of the Integral + +If $f$ and $g$ are integrable, then $\int (f + g) = \int f + \int g$. + +**Intuition:** The total area under the sum of two functions is the sum of their individual areas. + +--- + +## Solution (Step-by-Step Explanation) + +### Part (a) + +**Step 1: Apply the Reverse Triangle Inequality** + +By the **Reverse Triangle Inequality**, we have the following pointwise bound: + +$$||f_n(x)| - |f(x)|| \leq |f_n(x) - f(x)|$$ + +**Step 2: Use Monotonicity to bound the integral** + +Integrating both sides of the inequality from Step 1, we obtain: + +$$\int_{\Omega} ||f_n| - |f|| \, d\mu \leq \int_{\Omega} |f_n - f| \, d\mu$$ + +**Step 3: Evaluate the limit** + +We are given that $\int_{\Omega} |f - f_n| \, d\mu \to 0$ as $n \to \infty$. By the squeeze theorem applied to the inequality in Step 2, it follows that: + +$$\int_{\Omega} ||f_n| - |f|| \, d\mu \to 0$$ + +Since $| \int |f_n| - \int |f| | \leq \int ||f_n| - |f||$, this immediately implies: + +$$\int_{\Omega} |f_n| \, d\mu \to \int_{\Omega} |f| \, d\mu$$ + +--- + +### Part (b) + +**Step 4: Define the non-negative sequence $g_n$** + +Following the hint, let $g_n = |f| + |f_n| - |f - f_n|$. + +By the standard **Triangle Inequality** ($|f| = |f - f_n + f_n| \leq |f - f_n| + |f_n|$), we can see that: + +$$|f - f_n| \leq |f| + |f_n| \implies |f| + |f_n| - |f - f_n| \geq 0$$ + +Thus, $g_n$ is a sequence of non-negative measurable functions. + +**Step 5: Determine the pointwise limit of $g_n$** + +Since $f_n \to f$ almost everywhere, we have $|f_n| \to |f|$ and $|f - f_n| \to 0$ a.e. Therefore: + +$$\liminf_{n \to \infty} g_n = \lim_{n \to \infty} (|f| + |f_n| - |f - f_n|) = |f| + |f| - 0 = 2|f| \text{ a.e.}$$ + +**Step 6: Apply Fatou's Lemma** + +By **Fatou's Lemma** applied to the non-negative sequence $g_n$: + +$$\int_{\Omega} 2|f| \, d\mu \leq \liminf_{n \to \infty} \int_{\Omega} (|f| + |f_n| - |f - f_n|) \, d\mu$$ + +**Step 7: Simplify the right-hand side using linearity** + +By the **Linearity of the Integral**: + +$$\int_{\Omega} 2|f| \, d\mu \leq \int_{\Omega} |f| \, d\mu + \liminf_{n \to \infty} \left( \int_{\Omega} |f_n| \, d\mu - \int_{\Omega} |f - f_n| \, d\mu \right)$$ + +Using the hypothesis that $\int |f_n| \to \int |f|$, we substitute the limit: + +$$2\int_{\Omega} |f| \, d\mu \leq \int_{\Omega} |f| \, d\mu + \int_{\Omega} |f| \, d\mu - \limsup_{n \to \infty} \int_{\Omega} |f - f_n| \, d\mu$$ + +_(Note: $\liminf(-a_n) = -\limsup(a_n)$)_. + +**Step 8: Final Conclusion** + +Subtracting $2\int |f| \, d\mu$ from both sides, we get: + +$$0 \leq -\limsup_{n \to \infty} \int_{\Omega} |f - f_n| \, d\mu \implies \limsup_{n \to \infty} \int_{\Omega} |f - f_n| \, d\mu \leq 0$$ + +Since the integral of an absolute value is always non-negative, the limit superior must be zero, which means: + +$$\int_{\Omega} |f - f_n| \, d\mu \to 0$$ + +___ +## Question + +Does there exist a non-negative Lebesgue measurable function $f$ on $\mathbb{R}$ such that for all Lebesgue measurable sets $E \in \mathcal{L}$, the following equality holds? + +$$\delta_0(E) = \int_{E} f \, d\lambda$$ + +Here, $\lambda$ denotes the Lebesgue measure and $\delta_0$ is the Dirac measure concentrated at $0$, defined by: + +$$\delta_0(E) = \begin{cases} 1 & \text{if } 0 \in E \\ 0 & \text{if } 0 \notin E \end{cases}$$ + +--- + +## Definitions and Theorems Used + +### 1. Lebesgue Measure ($\lambda$) + +The standard way of assigning a "length" to subsets of the real line. A crucial property is that the measure of a single point is zero: $\lambda(\{x\}) = 0$ for any $x \in \mathbb{R}$. + +**Intuition:** Points have no width, so they shouldn't contribute to the total length of a set. + +### 2. Dirac Measure at 0 ($\delta_0$) + +A measure that assigns a mass of $1$ to any set containing the origin and $0$ to any set that does not. + +**Intuition:** This measure acts like a "spotlight" focused entirely on a single point ($0$), ignoring everything else. + +### 3. Absolute Continuity + +An integral defined by $\nu(E) = \int_E f \, d\lambda$ creates a new measure $\nu$. A fundamental property of such a measure is that if $\lambda(E) = 0$, then $\nu(E) = 0$. + +**Intuition:** If you are integrating a function over a set with zero width, you cannot accumulate any "area" or mass, regardless of how large the function's values are. + +--- + +## Solution (Step-by-Step Explanation) + +We will use a proof by contradiction to show that no such function $f$ exists. + +### Step 1: Assume such a function exists + +Suppose there exists a non-negative Lebesgue measurable function $f: \mathbb{R} \to [0, \infty)$ such that for every measurable set $E$: + +$$\delta_0(E) = \int_{E} f \, d\lambda$$ + +### Step 2: Apply the definition to a specific set + +Consider the set consisting only of the origin: $E = \{0\}$. + +By the **definition of the Dirac measure at 0**, since $0 \in \{0\}$, we have: + +$$\delta_0(\{0\}) = 1$$ + +### Step 3: Evaluate the Lebesgue integral over the same set + +Now we evaluate the right side of our assumed equality for the same set $E = \{0\}$. + +By the properties of the **Lebesgue measure**, the measure of a singleton set is zero: + +$$\lambda(\{0\}) = 0$$ + +### Step 4: Use the property of integrals over sets of measure zero + +By the **definition of the Lebesgue integral**, if we integrate any measurable function (even one that is infinite at a point) over a set of Lebesgue measure zero, the result must be zero: + +$$\int_{\{0\}} f \, d\lambda = 0$$ + +This is a core principle: you cannot have area under a curve if the base of the region has zero width. + +### Step 5: Identify the contradiction + +From Step 2, we have $\delta_0(\{0\}) = 1$. + +From Step 4, we have $\int_{\{0\}} f \, d\lambda = 0$. + +If the equality $\delta_0(E) = \int_E f \, d\lambda$ held for all sets, it would have to hold for $E = \{0\}$, which would mean $1 = 0$. This is a clear contradiction. + +### Conclusion + +There does **not** exist such a function $f$. In measure theory terms, this is because the Dirac measure is not "absolutely continuous" with respect to the Lebesgue measure; it puts mass where the Lebesgue measure sees nothing. + +___ +## Question + +Let $(\Omega, \mathcal{F}, \mu)$ be a measure space. Suppose $f : \Omega \to \mathbb{R}$ is an integrable function. + +1. Prove **Chebyshev’s Inequality**: + + $$\sup_{\alpha > 0} \alpha \, \mu(\{\omega \in \Omega : |f(\omega)| > \alpha\}) \leq \int_{\Omega} |f| \, d\mu$$ + +2. Provide an example of a measurable function $f$ such that $\int_{\Omega} |f| \, d\mu = \infty$, but the supremum term is finite: + + $$\sup_{\alpha > 0} \alpha \, \mu(\{\omega \in \Omega : |f(\omega)| > \alpha\}) < \infty$$ + + +--- + +## Definitions and Theorems Used + +### 1. Measurable Set $E_\alpha$ + +For any $\alpha > 0$, we define the set $E_\alpha = \{\omega \in \Omega : |f(\omega)| > \alpha\}$. If $f$ is a measurable function, then $E_\alpha$ is a measurable set. + +**Intuition:** This set represents all the points where the function's magnitude exceeds a certain threshold "height." + +### 2. Characteristic (Indicator) Function + +The function $\chi_{E_\alpha}(\omega)$ is $1$ if $\omega \in E_\alpha$ and $0$ otherwise. + +**Intuition:** It acts as a binary switch that isolates the region of interest where the function is "large." + +### 3. Monotonicity of the Integral + +If $0 \leq g \leq h$ almost everywhere, then $\int g \, d\mu \leq \int h \, d\mu$. + +**Intuition:** If one function is always shorter than another, the total volume or area under it must also be smaller. + +--- + +## Solution (Step-by-Step Explanation) + +### Part 1: Proof of Chebyshev's Inequality + +**Step 1: Create a lower-bound function** + +Fix any $\alpha > 0$. We define a simple function $g$ that is intentionally "smaller" than $|f|$. + +Consider the function $\alpha \cdot \chi_{E_\alpha}(\omega)$. + +- If $\omega \in E_\alpha$, then $|f(\omega)| > \alpha$, so $\alpha \cdot (1) < |f(\omega)|$. + +- If $\omega \notin E_\alpha$, then $\alpha \cdot (0) = 0 \leq |f(\omega)|$. + + Therefore, for all $\omega \in \Omega$, we have the pointwise inequality: + + $$\alpha \chi_{E_\alpha}(\omega) \leq |f(\omega)|$$ + + +**Step 2: Apply the Monotonicity of the Integral** + +By the **Monotonicity of the Integral**, we integrate both sides of the inequality over $\Omega$: + +$$\int_{\Omega} \alpha \chi_{E_\alpha} \, d\mu \leq \int_{\Omega} |f| \, d\mu$$ + +**Step 3: Calculate the integral of the characteristic function** + +Since $\alpha$ is a constant, we pull it out of the integral. The integral of a characteristic function is simply the measure of the set: + +$$\alpha \int_{\Omega} \chi_{E_\alpha} \, d\mu = \alpha \, \mu(E_\alpha)$$ + +Substituting this back into Step 2 gives: + +$$\alpha \, \mu(\{\omega \in \Omega : |f(\omega)| > \alpha\}) \leq \int_{\Omega} |f| \, d\mu$$ + +**Step 4: Take the Supremum** + +Since the inequality in Step 3 holds for _every_ individual $\alpha > 0$, the upper bound $\int |f| \, d\mu$ must be greater than or equal to the largest possible value of the left-hand side. Thus: + +$$\sup_{\alpha > 0} \alpha \, \mu(\{\omega \in \Omega : |f(\omega)| > \alpha\}) \leq \int_{\Omega} |f| \, d\mu$$ + +--- + +### Part 2: Counter-example for the Converse + +We need a function that is "barely" non-integrable at infinity but satisfies the supremum condition. + +**Step 5: Define the function** + +Consider the measure space $(\mathbb{R}, \mathcal{B}_{\mathbb{R}}, \lambda)$ with Lebesgue measure. Let: + +$$f(x) = \begin{cases} \frac{1}{x} & x \geq 1 \\ 0 & x < 1 \end{cases}$$ + +**Step 6: Show the integral is infinite** + +We calculate the integral using our knowledge of **power functions**: + +$$\int_{\mathbb{R}} |f| \, d\lambda = \int_1^\infty \frac{1}{x} \, dx = [\ln x]_1^\infty = \infty$$ + +So, $f$ is not integrable. + +**Step 7: Evaluate the supremum term** + +Let's calculate $\mu(\{x : |f(x)| > \alpha\})$: + +- If $\alpha \geq 1$: The maximum value of $f$ is $1$. For any $\alpha \geq 1$, the set $\{x : f(x) > \alpha\}$ is empty, so the measure is $0$. + +- If $0 < \alpha < 1$: We need $1/x > \alpha$, which means $1 < x < 1/\alpha$. + + The measure of the interval $(1, 1/\alpha)$ is: + + $$\mu((1, 1/\alpha)) = \frac{1}{\alpha} - 1$$ + + +Now we multiply by $\alpha$: + +$$\alpha \, \mu(\{x : |f(x)| > \alpha\}) = \alpha \left( \frac{1}{\alpha} - 1 \right) = 1 - \alpha$$ + +For all $0 < \alpha < 1$, the value is $1 - \alpha$, which is always less than $1$. + +**Step 8: Conclusion** + +The supremum is: + +$$\sup_{\alpha > 0} \alpha \, \mu(\{x : |f(x)| > \alpha\}) = 1$$ + +Since $1 < \infty$, this function $f(x) = 1/x$ provides the required example where the integral is infinite but the Chebyshev bound remains finite. + +___ +## Question + +Suppose $(\Omega, \mathcal{F}, \mu)$ is a $\sigma$-finite measure space. If $f : \Omega \to [0, \infty]$ is an integrable function, prove that there exists a sequence $\{f_n\}$ of integrable functions such that: + +1. $f_n \uparrow f$ pointwise as $n \to \infty$. + +2. Each $f_n$ vanishes outside a set of finite measure (i.e., each $f_n$ has support of finite measure). + +3. $\lim_{n \to \infty} \int_{\Omega} f_n \, d\mu = \int_{\Omega} f \, d\mu$. + + +--- + +## Definitions and Theorems Used + +### 1. $\sigma$-finite Measure Space + +A measure space $(\Omega, \mathcal{F}, \mu)$ is $\sigma$-finite if $\Omega$ can be written as the countable union of measurable sets $A_1, A_2, A_3, \dots$ such that $\mu(A_n) < \infty$ for all $n$. + +**Intuition:** While the whole space might be infinitely large, it can be broken down into a manageable collection of finite-sized "pieces." + +### 2. Monotone Convergence Theorem (MCT) + +If $\{f_n\}$ is a sequence of non-negative measurable functions such that $f_n(x) \uparrow f(x)$ pointwise for almost every $x$, then $\lim_{n \to \infty} \int f_n \, d\mu = \int f \, d\mu$. + +**Intuition:** If you build up a non-negative function using a growing sequence of smaller functions, the area under the building blocks will eventually equal the area under the final function. + +### 3. Integrable Function + +A measurable function $f$ is integrable if $\int_{\Omega} |f| \, d\mu < \infty$. + +**Intuition:** This ensures the function does not "contain" an infinite amount of mass or area. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Utilize the $\sigma$-finite property + +Since $(\Omega, \mathcal{F}, \mu)$ is $\sigma$-finite, by **definition**, there exists a sequence of measurable sets $\{A_n\}$ such that $\Omega = \bigcup_{n=1}^{\infty} A_n$ and $\mu(A_n) < \infty$ for all $n$. We can assume these sets are increasing ($A_1 \subseteq A_2 \subseteq \dots$) by defining $B_n = \bigcup_{i=1}^n A_i$. Note that $\mu(B_n) \leq \sum_{i=1}^n \mu(A_i) < \infty$. + +### Step 2: Construct the sequence $f_n$ + +For each $n \in \mathbb{N}$, define the function $f_n$ by restricting $f$ to the set $B_n$: + +$$f_n(\omega) = f(\omega) \cdot \chi_{B_n}(\omega)$$ + +where $\chi_{B_n}$ is the characteristic function of $B_n$. + +### Step 3: Verify the required properties for $f_n$ + +- **Vanishing outside a set of finite measure:** By construction, $f_n(\omega) = 0$ if $\omega \notin B_n$. Since $\mu(B_n) < \infty$, each $f_n$ vanishes outside a set of finite measure. + +- **Integrability:** Since $0 \leq f_n \leq f$ and $f$ is integrable, it follows from the monotonicity of the integral that $\int f_n \, d\mu \leq \int f \, d\mu < \infty$. Thus, each $f_n$ is integrable. + +- **Pointwise Monotone Convergence:** Because the sets $B_n$ are increasing and their union is $\Omega$, the sequence of characteristic functions $\chi_{B_n}$ increases pointwise to the constant function $1$. Since $f \geq 0$, the sequence $f_n = f \cdot \chi_{B_n}$ increases pointwise to $f$. + + +### Step 4: Apply the Monotone Convergence Theorem + +The sequence $\{f_n\}$ consists of non-negative measurable functions such that $f_n \uparrow f$ pointwise. Therefore, **by the Monotone Convergence Theorem**: + +$$\lim_{n \to \infty} \int_{\Omega} f_n \, d\mu = \int_{\Omega} f \, d\mu$$ + +### Conclusion + +We have constructed a sequence $\{f_n\}$ that satisfies all three conditions required by the problem. + +___ +## Question + +Let $f : \mathbb{R} \to \mathbb{R}$ be a Lebesgue integrable function that is continuous at a point $x_0 \in \mathbb{R}$. Evaluate the following limit: + +$$\lim_{n \to \infty} n \int_{[x_0, \, x_0 + \frac{1}{n}]} f(x) \, d\lambda(x)$$ + +where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$. + +**(Hint: The answer is $f(x_0)$.)** + +--- + +## Definitions and Theorems Used + +### 1. Continuity at a Point + +A function $f$ is continuous at $x_0$ if for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|f(x) - f(x_0)| < \epsilon$ whenever $|x - x_0| < \delta$. + +**Intuition:** This means that as $x$ gets very close to $x_0$, the value of the function $f(x)$ becomes arbitrarily close to $f(x_0)$. + +### 2. Properties of the Lebesgue Integral + +If $f$ is integrable and $m \leq f(x) \leq M$ for all $x$ in a set $E$, then: + +$$m \cdot \lambda(E) \leq \int_E f \, d\lambda \leq M \cdot \lambda(E)$$ + +**Intuition:** The total "area" under a function over a region is bounded by the area of the rectangles formed by the function's minimum and maximum values over that same region. + +### 3. Lebesgue Measure of an Interval + +For any interval $[a, b]$, the Lebesgue measure is simply its length: $\lambda([a, b]) = b - a$. + +**Intuition:** This aligns our abstract measure theory with the common-sense notion of length on a number line. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Rewrite the expression using the measure of the interval + +Note that the length of the interval $I_n = [x_0, x_0 + \frac{1}{n}]$ is $\lambda(I_n) = \frac{1}{n}$. We can rewrite the original expression to look like a "mean value" or average: + +$$n \int_{I_n} f(x) \, d\lambda(x) = \frac{1}{\lambda(I_n)} \int_{I_n} f(x) \, d\lambda(x)$$ + +### Step 2: Set up the epsilon-delta argument + +Fix $\epsilon > 0$. Since $f$ is **continuous at $x_0$**, there exists a $\delta > 0$ such that for all $x$ satisfying $x_0 \leq x \leq x_0 + \delta$, we have: + +$$f(x_0) - \epsilon < f(x) < f(x_0) + \epsilon$$ + +### Step 3: Bound the integral for large $n$ + +Choose $N$ such that $\frac{1}{N} < \delta$. For all $n \geq N$, the entire interval $[x_0, x_0 + \frac{1}{n}]$ is contained within the region $[x_0, x_0 + \delta]$ where our continuity bound holds. + +By the **Properties of the Lebesgue Integral (Monotonicity)**, we integrate the inequality from Step 2 over the interval $I_n$: + +$$\int_{I_n} (f(x_0) - \epsilon) \, d\lambda \leq \int_{I_n} f(x) \, d\lambda \leq \int_{I_n} (f(x_0) + \epsilon) \, d\lambda$$ + +### Step 4: Simplify the bounds + +Since $f(x_0) \pm \epsilon$ are constants relative to the integration, the integrals are simply the constants multiplied by the measure of the interval $\lambda(I_n) = \frac{1}{n}$: + +$$(f(x_0) - \epsilon) \cdot \frac{1}{n} \leq \int_{I_n} f(x) \, d\lambda \leq (f(x_0) + \epsilon) \cdot \frac{1}{n}$$ + +### Step 5: Isolate the original term + +Multiply the entire inequality by $n$: + +$$f(x_0) - \epsilon \leq n \int_{I_n} f(x) \, d\lambda \leq f(x_0) + \epsilon$$ + +This can be rewritten as: + +$$\left| n \int_{I_n} f(x) \, d\lambda - f(x_0) \right| \leq \epsilon$$ + +### Step 6: Conclusion + +Since the difference between the integral term and $f(x_0)$ is less than any $\epsilon > 0$ for sufficiently large $n$, we conclude by the **definition of a limit**: + +$$\lim_{n \to \infty} n \int_{[x_0, \, x_0 + \frac{1}{n}]} f(x) \, d\lambda(x) = f(x_0)$$ + +___ +## Question + +Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and let $f : \Omega \to (0, \infty)$ be a measurable function. For each $i \in \mathbb{Z}$, define: + +$$a_i = \mu\left(f^{-1}\left((2^{i-1}, 2^i]\right)\right)$$ + +Prove that $f$ is integrable if and only if: + +$$\sum_{i=-\infty}^{\infty} 2^i a_i < \infty$$ + +where the bi-infinite sum is defined as $\sum_{i=-\infty}^{\infty} c_i = \lim_{n \to \infty} \sum_{i=-n}^{n} c_i$. + +--- + +## Definitions and Theorems Used + +### 1. Integrability + +A non-negative measurable function $f$ is integrable if $\int_{\Omega} f \, d\mu < \infty$. + +**Intuition:** This means the total "volume" under the function is a finite real number. + +### 2. Monotone Convergence Theorem (MCT) for Series + +If $\{f_n\}$ is a sequence of non-negative measurable functions, then $\int \sum f_n = \sum \int f_n$. + +**Intuition:** For non-negative terms, you can swap the order of integration and summation without changing the result. + +### 3. Partitioning the Range + +If the range of $f$ is partitioned into disjoint sets $E_i = f^{-1}((2^{i-1}, 2^i])$, then $\Omega = \bigcup_{i \in \mathbb{Z}} E_i$ and the sets are disjoint. + +**Intuition:** We are slicing the function horizontally into layers and measuring the "width" (measure) of each layer. + +--- + +## Solution (Step-by-Step Explanation) + +### Step 1: Decompose the function into layers + +Since the range of $f$ is $(0, \infty)$, we can partition the domain into disjoint sets $E_i = \{ \omega \in \Omega : 2^{i-1} < f(\omega) \leq 2^i \}$ for all $i \in \mathbb{Z}$. + +We can write the function $f$ as: + +$$f(\omega) = \sum_{i=-\infty}^{\infty} f(\omega) \chi_{E_i}(\omega)$$ + +where $\chi_{E_i}$ is the characteristic function of the set $E_i$. + +### Step 2: Establish pointwise bounds + +On each set $E_i$, the value of the function is bounded by powers of 2: + +$$2^{i-1} < f(\omega) \leq 2^i \quad \text{for all } \omega \in E_i$$ + +Multiplying by the characteristic function, we get the following pointwise inequality for all $\omega \in \Omega$: + +$$\sum_{i=-\infty}^{\infty} 2^{i-1} \chi_{E_i}(\omega) \leq f(\omega) \leq \sum_{i=-\infty}^{\infty} 2^i \chi_{E_i}(\omega)$$ + +### Step 3: Integrate the inequalities + +By the **Monotonicity of the Integral**, we integrate the entire inequality over $\Omega$: + +$$\int_{\Omega} \left( \sum_{i=-\infty}^{\infty} 2^{i-1} \chi_{E_i} \right) d\mu \leq \int_{\Omega} f \, d\mu \leq \int_{\Omega} \left( \sum_{i=-\infty}^{\infty} 2^i \chi_{E_i} \right) d\mu$$ + +### Step 4: Interchange integration and summation + +By the **Monotone Convergence Theorem for Series**, we can move the integral inside the non-negative sums: + +$$\sum_{i=-\infty}^{\infty} 2^{i-1} \int_{\Omega} \chi_{E_i} \, d\mu \leq \int_{\Omega} f \, d\mu \leq \sum_{i=-\infty}^{\infty} 2^i \int_{\Omega} \chi_{E_i} \, d\mu$$ + +### Step 5: Relate to the coefficients $a_i$ + +By the **definition of the integral of a characteristic function**, $\int_{\Omega} \chi_{E_i} \, d\mu = \mu(E_i) = a_i$. Substituting this into Step 4, we get: + +$$\frac{1}{2} \sum_{i=-\infty}^{\infty} 2^i a_i \leq \int_{\Omega} f \, d\mu \leq \sum_{i=-\infty}^{\infty} 2^i a_i$$ + +### Step 6: Final Conclusion + +- **If $\sum 2^i a_i < \infty$:** The right-hand inequality shows $\int f \, d\mu \leq \sum 2^i a_i < \infty$. Thus, $f$ is integrable. + +- **If $f$ is integrable:** Then $\int f \, d\mu < \infty$. The left-hand inequality shows $\frac{1}{2} \sum 2^i a_i \leq \int f \, d\mu < \infty$, which implies the sum is finite. + + +Therefore, $f$ is integrable if and only if $\sum_{i=-\infty}^{\infty} 2^i a_i < \infty$. + +____ diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 1.md b/content/SEM_6/Measure_Theory/Lecture 01.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 1.md rename to content/SEM_6/Measure_Theory/Lecture 01.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 2.md b/content/SEM_6/Measure_Theory/Lecture 02.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 2.md rename to content/SEM_6/Measure_Theory/Lecture 02.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 3.md b/content/SEM_6/Measure_Theory/Lecture 03.md similarity index 95% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 3.md rename to content/SEM_6/Measure_Theory/Lecture 03.md index 0ec91d6f..35df0cf1 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 3.md +++ b/content/SEM_6/Measure_Theory/Lecture 03.md @@ -1,6 +1,6 @@ # Borel $\sigma$-algebra -Let $\Omega$ be a topological space. Then the $\sigma$-[[Measure-Theory-Notes/Measure_Theory/Lecture 2#Definition| algebra]] by open sets in $\Omega$ is called the Borel $\sigma$-algebra for $\Omega$ and is denoted by $\mathcal{B}_{\Omega}$. +Let $\Omega$ be a topological space. Then the $\sigma$-[[Measure-Theory-Notes/Measure_Theory/Lecture 2#Definition| algebra generated]] by open sets in $\Omega$ is called the Borel $\sigma$-algebra for $\Omega$ and is denoted by $\mathcal{B}_{\Omega}$. # $G_\delta$-sets diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 4.md b/content/SEM_6/Measure_Theory/Lecture 04.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 4.md rename to content/SEM_6/Measure_Theory/Lecture 04.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 5.md b/content/SEM_6/Measure_Theory/Lecture 05.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 5.md rename to content/SEM_6/Measure_Theory/Lecture 05.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 6.md b/content/SEM_6/Measure_Theory/Lecture 06.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 6.md rename to content/SEM_6/Measure_Theory/Lecture 06.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 7.md b/content/SEM_6/Measure_Theory/Lecture 07.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 7.md rename to content/SEM_6/Measure_Theory/Lecture 07.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 8.md b/content/SEM_6/Measure_Theory/Lecture 08.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 8.md rename to content/SEM_6/Measure_Theory/Lecture 08.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 9.md b/content/SEM_6/Measure_Theory/Lecture 09.md similarity index 99% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 9.md rename to content/SEM_6/Measure_Theory/Lecture 09.md index ce824390..736162d1 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 9.md +++ b/content/SEM_6/Measure_Theory/Lecture 09.md @@ -78,7 +78,7 @@ $$F(x) = \begin{cases} 1 &, x \geqslant 0 \\ 0 &, x < 0 \end{cases}$$ Find the corresponding $\sigma$-algebra $\mathcal{M}_F$ & the measure $\mu_F$. ____ -# Definition +# Definition (Complete) Let $\mathcal{F} \subseteq \mathcal{P}(\Omega)$ be a $\sigma$-algebra & $\mu : \mathcal{F} \to [0, \infty]$. The $\sigma$-algebra is called **complete** wrt $\mu$ (the measure space $(\Omega, \mathcal{F}, \mu)$ is complete) if for all $F \subseteq A$, $A \in \mathcal{F}$ & $\mu(A) = 0$ implies $F \in \mathcal{F}$. diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 10.md b/content/SEM_6/Measure_Theory/Lecture 10.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 10.md rename to content/SEM_6/Measure_Theory/Lecture 10.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 11.md b/content/SEM_6/Measure_Theory/Lecture 11.md similarity index 97% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 11.md rename to content/SEM_6/Measure_Theory/Lecture 11.md index 1252ba43..686f81c7 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 11.md +++ b/content/SEM_6/Measure_Theory/Lecture 11.md @@ -35,7 +35,8 @@ $$= \sum_{j=1}^{n} d_j \mu(F_j)$$ --- -**Proposition**: Let $f$ and $g$ be two measurable functions on $(\Omega, \mathcal{F}, \mu)$. Then +## Proposition +Let $f$ and $g$ be two measurable functions on $(\Omega, \mathcal{F}, \mu)$. Then i) $\int_{\Omega} (cf) \, d\mu = c \int_{\Omega} f \, d\mu, \forall c \ge 0$ diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 12.md b/content/SEM_6/Measure_Theory/Lecture 12.md similarity index 99% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 12.md rename to content/SEM_6/Measure_Theory/Lecture 12.md index c1dc1bdc..dde07f5b 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 12.md +++ b/content/SEM_6/Measure_Theory/Lecture 12.md @@ -1,4 +1,3 @@ - # Definition Let $(\Omega, \mathcal{F}, \mu)$ be a measurable space. For a non-negative measurable function $f: \Omega \to [0, \infty]$ define diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 13.md b/content/SEM_6/Measure_Theory/Lecture 13.md similarity index 90% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 13.md rename to content/SEM_6/Measure_Theory/Lecture 13.md index 1f958342..c053e6a8 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 13.md +++ b/content/SEM_6/Measure_Theory/Lecture 13.md @@ -17,11 +17,13 @@ $$\int_{\Omega} f d\mu = \sup \left\{ \int_{\Omega} \phi d\mu : \phi \text{ is s **Textbook**: _Measure Theory & Integration_, G. de Barra, New Age Publication -**Definitions**: If $f : \Omega \to [0, \infty]$ is measurable and $A \in \mathcal{F}$ we define +# Definitions +If $f : \Omega \to [0, \infty]$ is measurable and $A \in \mathcal{F}$ we define $$\int_{A} f d\mu = \int_{\Omega} f \chi_A d\mu$$ -**Proposition**: If $\mu(A) = 0$, then for any measurable $f : \Omega \to [0, \infty]$ +## Proposition +If $\mu(A) = 0$, then for any measurable $f : \Omega \to [0, \infty]$ $$\int_{A} f d\mu = 0$$ @@ -43,11 +45,9 @@ $$= \int_{\Omega} \sum_{i=1}^{k} c_i \chi_{E_i \cap A} d\mu$$ $$= \sum_{i=1}^{k} c_i \mu(E_i \cap A) = 0$$ -**Proposition**: Let $\phi$ be a non-negative simple fn. Then, $A \mapsto \int_{A} \phi d\mu, A \in \mathcal{F}$ [is a measure]. - ---- +## Proposition +Let $\phi$ be a non-negative simple fn. Then, $A \mapsto \int_{A} \phi d\mu, A \in \mathcal{F}$ [is a measure]. ---- Then $\nu_\phi$ is a measure on the $\sigma$-field $\mathcal{F}$. diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 14.md b/content/SEM_6/Measure_Theory/Lecture 14.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 14.md rename to content/SEM_6/Measure_Theory/Lecture 14.md diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 15.md b/content/SEM_6/Measure_Theory/Lecture 15.md similarity index 95% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 15.md rename to content/SEM_6/Measure_Theory/Lecture 15.md index 432aa023..4d29b707 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 15.md +++ b/content/SEM_6/Measure_Theory/Lecture 15.md @@ -77,13 +77,13 @@ $\therefore$ Our assumption is wrong, hence $f = 0$ a.e. Suppose $\{a_n\}$ is a sequence of non-negative real numbers. -$\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n$ +$$\sum_{n=1}^{\infty} a_n = \lim_{N \to \infty} \sum_{n=1}^{N} a_n$$ Define $f : \mathbb{N} \to [0, \infty)$ -$f(n) = a_n$ +$$f(n) = a_n$$ -And for $k \in \mathbb{N}$, $f_k(n) = \begin{cases} a_n, & n \le k \\ 0, & \text{ow} \end{cases}$ +And for $k \in \mathbb{N}$, $$f_k(n) = \begin{cases} a_n, & n \le k \\ 0, & \text{ow} \end{cases}$$ Consider the measure space $(\mathbb{N}, \mathcal{P}(\mathbb{N}), C)$ where $C$ is the counting measure: diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 16.md b/content/SEM_6/Measure_Theory/Lecture 16.md similarity index 91% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 16.md rename to content/SEM_6/Measure_Theory/Lecture 16.md index 536928e4..c96e864b 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 16.md +++ b/content/SEM_6/Measure_Theory/Lecture 16.md @@ -39,7 +39,7 @@ But $f$ is not Borel measurable { Borel measurable $\to f^{-1}(F) \in \mathcal{B in $f(\{1\}) = F \notin \mathcal{B}_{\mathbb{R}}$ -# Definition +# Definition (Lebesgue integrable) Let $(\Omega, \mathcal{F}, \mu)$ be a measure space and $f: \Omega \to \mathbb{\bar{R}}$ be measurable. Then $f$ is said to be Lebesgue integrable (or just integrable) if $$\int_{\Omega} f^+ d\mu < \infty \ \text{ \& } \int_{\Omega} f^- d\mu < \infty.$$ @@ -47,7 +47,8 @@ $$\int_{\Omega} f^+ d\mu < \infty \ \text{ \& } \int_{\Omega} f^- d\mu < \infty. In this case, the Lebesgue integrable of $f$ is defined as $$\int_{\Omega} f d\mu = \int_{\Omega} f^+ d\mu - \int_{\Omega} f^- d\mu.$$ -Proposition:- If $f, g: \Omega \to \bar{\mathbb{R}}$ are integrable and $A, B \in \mathcal{F}$ are disjoint. Then $f$ is integrable on $A$, $f+g$ (when well-defined) and $|f|$ are integrable and +## Proposition +If $f, g: \Omega \to \bar{\mathbb{R}}$ are integrable and $A, B \in \mathcal{F}$ are disjoint. Then $f$ is integrable on $A$, $f+g$ (when well-defined) and $|f|$ are integrable and i) $$\int_{\Omega} (cf+g) d\mu = c\int_{\Omega} f d\mu + \int_{\Omega} g d\mu, \quad c \in \mathbb{R}$$ @@ -63,7 +64,8 @@ vi) If $f = g$ a.e. then $\int_{\Omega} f d\mu = \int_{\Omega} g d\mu$ vii) If $|h| \le f$ & $f \ge 0$ then $h$ is integrable -Proof:- iii) If $f$ is not finite a.e. then at least one of the sets +**Proof** +iii) If $f$ is not finite a.e. then at least one of the sets $$A_1 = \{\omega \in \Omega : f(\omega) = \infty\} \text{ \& } A_2 = \{\omega \in \Omega : f(\omega) = -\infty\}$$ diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 17.md b/content/SEM_6/Measure_Theory/Lecture 17.md similarity index 95% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 17.md rename to content/SEM_6/Measure_Theory/Lecture 17.md index c9638aee..020d0da1 100644 --- a/content/Measure-Theory-Notes/Measure_Theory/Lecture 17.md +++ b/content/SEM_6/Measure_Theory/Lecture 17.md @@ -17,7 +17,7 @@ $$|f_n - f| \le 2g$$ **Note:-** $|f_n| + |f| \le 2g \implies |f_n - f| \le 2g$ -Apply Fatou's Lemma to $\{2g - |f_n - f|\}$ to get +Apply [[Lecture 14#Fatou's Lemma|Fatou's Lemma]] to $\{2g - |f_n - f|\}$ to get $$\int_{\Omega} 2g d\mu \le \liminf \int_{\Omega} (2g - |f_n - f|) d\mu$$ @@ -51,10 +51,10 @@ Then, $$\sum_{n=1}^{\infty} f_n \text{ converges pointwise.}$$ -and we +and $$\int_{\Omega} \left( \sum_{n=1}^{\infty} f_n \right) d\mu = \sum_{n=1}^{\infty} \left( \int_{\Omega} f_n d\mu \right)$$ -Proof +**Proof** Set $g = \sum_{n=1}^{\infty} |f_n|$. Then by MCT @@ -98,7 +98,7 @@ Then $F$ is differentiable and $$F'(t) = \int_{\Omega} \frac{\partial}{\partial t} f(x, t) d\mu(x)$$ -Example: Let $f$ be integrable, $f: \mathbb{R} \to \mathbb{R}$, then +**Example**: Let $f$ be integrable, $f: \mathbb{R} \to \mathbb{R}$, then define diff --git a/content/Measure-Theory-Notes/Measure_Theory/Lecture 18.md b/content/SEM_6/Measure_Theory/Lecture 18.md similarity index 100% rename from content/Measure-Theory-Notes/Measure_Theory/Lecture 18.md rename to content/SEM_6/Measure_Theory/Lecture 18.md diff --git a/content/SEM_6/Measure_Theory/Question_Papers/Midsem.md b/content/SEM_6/Measure_Theory/Question_Papers/Midsem.md new file mode 100644 index 00000000..e69de29b diff --git a/content/SEM_6/Measure_Theory/info.md b/content/SEM_6/Measure_Theory/info.md new file mode 100644 index 00000000..20b07a27 --- /dev/null +++ b/content/SEM_6/Measure_Theory/info.md @@ -0,0 +1,6 @@ +**Course:** Measure Theory +**Code:** MAT401 +**Year:** 3 +**Semester:** 6 +**Prerequisites:** Topology (not necessary) +**Course Instructor:** Dr. Jayanta Sarkar \ No newline at end of file diff --git a/content/index.md b/content/index.md index 65a9f2a3..401b7637 100644 --- a/content/index.md +++ b/content/index.md @@ -1 +1,3 @@ -This is the index file. \ No newline at end of file +This is the index file. + + These are comprehensive study notes from the 6th semester of a Mathematics major program at IISER Thiruvananthapuram, compiled during the 3rd year of study. This resource covers key concepts, problem sets, and detailed explanations across core mathematics courses and electives.