From c118ab699f82451062c81ec05a6c8f73971d1ee6 Mon Sep 17 00:00:00 2001
From: Zheyuan Wu <60459821+Trance-0@users.noreply.github.com>
Date: Fri, 28 Mar 2025 10:52:50 -0500
Subject: [PATCH] Update Math4121_L27.md

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-# Lecture 27
\ No newline at end of file
+# Math4121 Lecture 27
+
+## Lebesgue Measure
+
+### Outer Measure
+
+$$
+m_e(S)=\inf\left\{\sum_{n=1}^\infty \ell(I_n): S\subset \bigcup_{n=1}^\infty I_n\right\}
+$$
+
+where $I_j$ is an open interval
+
+**Properties:**
+
+1. $m_e(I)=\ell(I)$
+2. Countably sub-additive: $m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq \sum_{n=1}^\infty m_e(S_n)$ (Prove today)
+3. does not repect complementation (Build in to Borel measure)
+
+Why does Jordan content respect complementation?
+
+$(\text{Finite union of intervals })^C=\text{another finite union of intervals}$
+
+We know this failed for countable unions.
+
+Example:
+
+$$
+\bigcup_{n=1}^\infty \left(q_n-\frac{\epsilon}{2^n},q_n+\frac{\epsilon}{2^n}\right)
+$$
+
+Where $q_n$ is dense.
+
+### Inner Measure
+
+Say $S\subset I$
+
+$$
+m_i(S)=m(I)-m_e(I\setminus S)
+$$
+
+where $m(I)=\ell(I)$
+
+Say $S$ is (Lebesgue) measurable if $m_i(S)=m_e(S)$, call this value $m(S)=m_e(S)=m_i(S)$ the (Lebesgue) measure of $S$.
+
+#### Corollary of measurability of subsets
+
+If $S$ is measurable, and $S\subset T$, then
+
+$$
+m(S)=m_e(S)=m(I)-m_e(I\setminus S)
+$$
+
+$$
+m(I\setminus S)=m(I)-m(S)
+$$
+
+$I\setminus S$ is Lebesgue measurable and $m(I)=m(S)+m(I\setminus S)$
+
+#### Proposition 5.8 (Countable additivity over measurable sets)
+
+If $S_n$ are measurable, then
+
+$$
+m_e\left(\bigcup_{n=1}^\infty S_n\right)\leq\sum_{n=1}^\infty m(S_n)
+$$
+
+Proof:
+
+Let $\epsilon>0$ and for each $j$, let $\{I_{i,j}\}_{i=1}^\infty$ be a cover of $S_j$ s.t.
+
+$$
+\sum_{i=1}^\infty \ell(I_{i,j})<m(S_j)+\frac{\epsilon}{2^j}
+$$
+
+Then $\bigcup_{j=1}^\infty \bigcup_{i=1}^\infty I_{i,j}$ is a countable cover of $\bigcup_{j=1}^\infty S_j$ and 
+
+$$
+m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq \sum_{j=1}^\infty \sum_{i=1}^\infty \ell(I_{i,j})<\sum_{j=1}^\infty \left(m_e(S_j)+\frac{\epsilon}{2^j}\right)=\sum_{j=1}^\infty m_e(S_j)+\epsilon
+$$
+
+Since $\epsilon$ is arbitrary, we have
+
+$$
+m_e\left(\bigcup_{j=1}^\infty S_j\right)\leq\sum_{j=1}^\infty m_e(S_j)=\sum_{j=1}^\infty m(S_j)
+$$
+
+QED
+
+#### Corollary
+
+$$
+m_i(S)\leq m_e(S)
+$$
+
+Proof:
+
+$$
+m_i(S)=m(I)-m_e(I\setminus S)\leq m(I)-m_i(I\setminus S)=m_e(S)
+$$
+
+QED
+
+### Caratheodory's Criterion
+
+#### Lemma 5.7 (Local additivity)
+
+If $\{I_j\}_{j=1}^\infty$ are pairwise disjoint open intervals, then
+
+$$
+m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)=m_e\left(\bigcup_{j=1}^\infty (S\cap I_j)\right)=\sum_{j=1}^\infty m_e(S\cap I_j)
+$$
+
+Proof:
+
+For each $j$, let $\{J_i\}_{i=1}^\infty$ be a cover of $S\cap \left(\bigcup_{j=1}^\infty I_j\right)$ such that $\sum_{i=1}^\infty \ell(J_i)<c_e(S\cap \left(\bigcup_{j=1}^\infty I_j\right))+\epsilon$. Since $\{I_j\}_{j=1}^\infty$ are pairwise disjoint, so is $\{J_i\cap I_j\}_{j=1}^\infty$ for each $i$.
+
+$$
+\sum_{j=1}^\infty m_e(J_i\cap I_j)=m_e(J_i)
+$$
+
+$$
+\begin{aligned}
+m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)&\leq \sum_{j=1}^\infty m_e(S\cap I_j)\\
+&\leq \sum_{j=1}^\infty m_e(\bigcup_{i=1}^\infty J_i\cap I_j)\\
+&= \sum_{j=1}^\infty m_e(J_i)+\epsilon
+\end{aligned}
+$$
+
+Since $\epsilon$ is arbitrary, we have
+
+$$
+m_e\left(S\cap \left(\bigcup_{j=1}^\infty I_j\right)\right)\leq \sum_{j=1}^\infty m_e(S\cap I_j)
+$$
+
+QED
+
+#### Theorem 5.6 (Caratheodory's Criterion)
+
+A set $S$ is measurable if and only if for every set $X\in \mathbb{R}$ of finite outer measure,
+
+$$
+m_e(X)=m_e(X\cap S)+m_e(X\setminus S)
+$$
+
+Lebesgue: $X=I$ and $S\subset I$ we can cut any set by a measurable set to get a measurable set. (no matter how big the set is)
+