Quartz sync: Mar 6, 2025, 2:14 PM

content/Definitions/Measure Theory/Sigma-Algebra/Borel Sets.md

@@ -1,2 +1,2 @@
 # Definition
-The $\sigma$-algebra generated by the [[Open Sets|open sets]] in a [[Topological Space|topological space]] $X$ is the $\sigma$-algebra of **Borel Sets** of $X$.
+The $\sigma$[[Sigma-Algebra|-algebra]] generated by the [[Open Sets|open sets]] in a [[Topological Space|topological space]] $X$ is the $\sigma$[[Sigma-Algebra|-algebra]] of **Borel Sets** of $X$.

content/Definitions/Measure Theory/Sigma-Algebra/Borel Sigma-Algebra.md

@@ -1 +1,2 @@
-**Borel $\sigma$-algebra** on a [[Topological Space|topological space]] $X$ is the one generated by $\tau$.
+# Definition
+**Borel $\sigma$-algebra** on a [[Topological Space|topological space]] $X$ is the one generated by the [[Open Sets|open sets]] $\tau$.

content/Lectures/Lecture 12 - Induced Topologies.md

@@ -24,7 +24,7 @@ By assumption $f_{n}(x_{i}) \to f_{n}(x)$, so $f_{n}(x_{i}) \in A_{n} \; \forall
 Then $x_{i} \in \cap f^{-1}_{n}(A_{n}) \subset A$ for all $i \geq j$, so $x_i \to x$.
 
 QED.
-# COR
+# Corollary
 $X$ has [[Initial Topology|initial topology]] induced by $F$. Say $Z$ is a [[Topological Space|topological space]], then:
 
 $g : Z \to X$ is [[Continuous|continuous]] $f \circ g$ is [[Continuous|continuous]] $\forall f \in F$.
@@ -67,7 +67,7 @@ Then $f^{-1}(U)$ and $f^{-1}(V)$ will be disjoint neighbourhoods of $x$ and $y$.
 > Which is not possible
 
 QED.
-# COR
+# Corollary
 A product of [[Hausdorff]] spaces is [[Hausdorff]] in the [[Product Topology|product topology]].
 
 $\Pi X_{\lambda}$

content/Lectures/Lecture 15.md

@@ -110,3 +110,67 @@ Set $s_{n} = h_{n} \circ f$.
 ![[Drawing 2025-03-06 12.25.26.excalidraw.dark.svg]]
 %%[[Drawing 2025-03-06 12.25.26.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 12.25.26.excalidraw.light.svg|light exported image]]%%
 
+---
+Exercise part of the session.
+
+These exercises are from Exercise 8.
+# Question 3
+Prove $\mu(A) \leq \mu(B)$ when $A \subset B$
+## Proof
+Have $B = A \cup (\underbrace{A^{\complement} \cap B}_{B \setminus A})$, so
+> [!example]- What this set looks like
+> ![[Drawing 2025-03-06 13.17.23.excalidraw.dark.svg]]
+%%[[Drawing 2025-03-06 13.17.23.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 13.17.23.excalidraw.light.svg|light exported image]]%%
+
+$\mu(B) = \mu(A) + \underbrace{\mu(B \setminus A)}_{\geq 0}$
+$\implies \mu(B) >+ \mu(A)$.
+# Question 4
+Show that if $X$ has a $\sigma$[[Sigma-Algebra|-algebra]] and $f : X \to Y$ set. Then the collection $N$ of subsets $A \subset Y$ such that $f^{-1}(A)$ [[Measurable|measurable]], is a $\sigma$[[Sigma-Algebra|-algebra]].
+
+$N \equiv \{ A \subset Y \, | \, f^{-1}(A) \in M \}$
+prove that $N$ is a $\sigma$[[Sigma-Algebra|-algebra]].
+## Proof
+1. Have $\emptyset \in N$ since $f^{-1}(\emptyset) = \emptyset \in M$.
+2. If $A \in N$, then $f^{-1}(A) \in M$, so $f^{-1}(A)^{\complement} \in M = f^{-1}(A^{\complement}) \implies A^{\complement} \in N$.
+3. If $A_{n} \in N$, then $f^{-1}(A_{n} \in M)$, so $f^{-1}(\cup A_{n}) = \cup f^{-1}(A_{n}) \in M$, so $\cup A_{n} \in N$.
+
+# A question I don't know the number of
+Say $f : X \to Y$ and they are [[Topological Space|topological spaces]].
+Show that if $f$ is [[Measurable|measurable]], then $f^{-1}(A)$ is [[Borel Measurable|Borel measurable]] for any [[Borel Sets|Borel set]] $A \subset Y$.
+## Proof
+Consider $N = \{ A \subset Y \, | \, f^{-1}(A)\ \text{Borel} \}$. This is a $\sigma$[[Sigma-Algebra|-algebra]].
+> [!example]-
+> ![[Drawing 2025-03-06 13.41.17.excalidraw.dark.svg]]
+%%[[Drawing 2025-03-06 13.41.17.excalidraw.md|🖋 Edit in Excalidraw]], and the [[Drawing 2025-03-06 13.41.17.excalidraw.light.svg|light exported image]]%%
+
+It contains all the [[Open Sets|open sets]] in $Y$ since $f$ is [[Measurable|measurable]] and then $f^{-1}(V)$ is [[Borel Measurable|Borel measurable]] for $V$ [[Open Sets|open]].
+
+Hence $N$ contains all the [[Borel Sets|Borel sets]] in $Y$. If $A$ is [[Borel Sets|Borel]], then $A \in N$, so $f^{-1}(A)$ is [[Borel Sets|Borel]]. (**Note**: not sure if on this if the "Borel"s are about them being Borel sets or Borel measurable)
+# Question 5
+$X$ with $\sigma$[[Sigma-Algebra|-algebra]] $M$.
+$f: X \to [0, \infty]$ is [[Measurable|measurable]] $\iff f^{-1}(\langle a, \infty]) \in M,\ \forall a \gt 0$.
+## Proof
+### $\Rightarrow$
+"Obvious".
+### $\Leftarrow$
+In $[0, \infty]$ [[Ball|balls]] are $[0, a\rangle$, $\langle a, \infty ]$, $[0, \infty]$, or $\langle a, b \rangle$ for $a, b \in \mathbb{R}$.
+
+Any open set in $[0, \infty]$ is a [[Countable|countable]] union of these "building blocks".
+
+Checking they belong to $M$:
+- $[0, a]$
+- $[0, b\rangle \cap \langle a, \infty = \langle a, b \rangle$
+- $[0, a \rangle = \cup_{n=1}^{\infty}\left[ 0, a-\frac{1}{n} \right]$
+QED.
+
+$f_{n} : X \to [0, \infty]$ is [[Measurable|measurable]]
+$f \equiv \sup f_{n}$ is [[Measurable|measurable]]
+> [!note] Can also use [[Infimum|infimum]] instead
+> $\inf f_{n} = -\sup(-f_{n})$
+
+Now need to check if $f^{-1}(\langle a, \infty ])$ is [[Measurable|measurable]] and is in $M$
+
+$f^{-1}(\langle a, \infty ]) = \{ x \in X \, | \, f(x) \gt a \}$
+$= \cup_{n}f_{n}^{-1}(\langle a, \infty ]) \in M$
+Why is the set and the union both the same?
+$\cup_{n} \{ x \in X \, | \, f_{n}(x) \gt a \} = \{ x \in X \, | \, f_{n}(x) \gt a\ \forall n \}$