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\part{Measure Theory and Integration}
\label{part:measure}
\input{./src/measure/sets/index.tex}
\input{./src/measure/measure/index.tex}

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\section{Complete Measures}
\label{section:complete-measure}
\begin{definition}[Complete Measure]
\label{definition:complete-measure}
Let $(X, \cm, \mu)$ be a measure space, then $\mu$ is \textbf{complete} if for any null set $E \in \cm$ and $F \subset E$, $F \in \cm$.
\end{definition}
\begin{definition}[Completion]
\label{definition:measure-completion}
Let $(X, \cm, \mu)$ be a measure space, $\cn = \bracs{N \in \cm| \mu(N) = 0}$, and
\[
\ol{\cm} = \bracs{E \cup F| E \in \cm, F \subset N, N \in \cn}
\]
then
\begin{enumerate}
\item For any $A \in \ol{\cm}$, there exists $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $E \cap N = \emptyset$ and $A = E \sqcup F$.
\item $\ol{\cm} \supset \cm$ is a $\sigma$-algebra.
\item There exists an extension $\ol{\mu}$ of $\mu$ as a measure on $\cm$.
\item $(X, \ol{\cm}, \ol{\mu})$ is a complete measure space.
\item[(U)] For any complete measure space $(X, \cf, \nu)$ where $\cf \supset \cm$ and $\nu|_\cm = \mu$, $\cf \supset \ol{\cm}$ and $\nu|_{\ol{\cm}} = \ol{\mu}$.
\end{enumerate}
and space $(X, \ol{\cm}, \mu)$ is the \textbf{completion} of $(X, \cm, \mu)$.
\end{definition}
\begin{proof}
(1): Let $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Define $E' = E \setminus N$ and $F' = F \cup (E \cap N)$, then $E' \cap N = \emptyset$ and $A = E \cup F = E' \sqcup F'$.
(2): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \sqcup F$, then
\begin{align*}
A^c &= (A^c \cap N^c) \cup (A^c \cap N) = (N^c \setminus A) \cup (N \setminus A) \\
&= \underbrace{(N^c \setminus E)}_{\in \cm} \cup \underbrace{(N \setminus F)}_{\subset N} \in \cm
\end{align*}
Let $\seq{A_n} \subset \cm$. For each $n \in \natp$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N$ such that $A_n = E_n \cup F_n$, then
\[
\bigcup_{n \in \natp}A_n = \underbrace{\braks{\bigcup_{n \in \natp}E_n}}_{\in \cm} \cup \underbrace{\braks{\bigcup_{n \in \natp}F_n}}_{\subset \bigcup_{n \in \natp}N_n} \in \ol{\cm}
\]
(3): Let $A \in \ol{\cm}$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$. Since $A \setminus N = E \setminus N \in \cm$ and $A \cup N = E \cup N \in \cm$,
\[
\mu(A \setminus N) \le \mu(E) \le \mu(A \cup N)
\]
Let $E' \in \cm$, $N' \in \cn$, and $F' \subset N'$ such that $A = E' \cup F'$, then
\[
\mu(A \setminus (N \cup N')) \le \mu(A \setminus N') \le \mu(E') \le \mu(A \cup N') \le \mu(A \cup (N \cup N'))
\]
Now, since $N$ and $N'$ are null,
\[
\mu(A \cup (N \cup N')) = \mu(A \setminus (N \cup N')) + \mu(N \cup N') = \mu(A \setminus (N \cup N'))
\]
so
\[
\mu(A \setminus (N \cup N')) = \mu(E) = \mu(E') = \mu(A \cup (N \cup N'))
\]
Thus for any decomposition $A = E \cup F$ where $E \in \cm$ and $F$ is contained in a null set, $\ol{\mu}(A) = \mu(E)$ is well-defined.
To see that $\ol{\mu}$ is a measure, let $\seq{A_n} \subset \ol{\cm}$ be pairwise disjoint. For each $n \in \nat$, let $E_n \in \cm$, $N_n \in \cn$, and $F_n \subset N_n$ such that $A_n = E_n \cup F_n$, then
\[
\ol{\mu}\paren{\bigsqcup_{n \in \natp}A_n} = \mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu(E_n) = \sum_{n \in \natp}\mu(A_n)
\]
(4): Let $A \in \ol{\cm}$ with $\ol{\mu}(A) = 0$, $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup F$, then $\mu(E) = \mu(E \cup N) = 0$. By definition of $\ol{\cm}$, any subset of $E \cup N$ is also in $\ol{\cm}$. Thus $\ol{\mu}$ is complete.
(U): Let $N \in \cn \subset \cm \subset \cf$, then for any $F \subset N$, $\nu(F) \le \nu(N) = \mu(N)$, and $F \in \cf$ by completeness of $(X, \cf, \nu)$. Thus $\cf \supset \ol{\cm}$.
For any $A \in \ol{\cm}$, let $E \in \cm$, $N \in \cn$, and $F \subset N$ such that $A = E \cup N$, then
\[
\nu(A) = \nu(E \cup F) = \nu(E) = \mu(E) = \ol{\mu}(A)
\]
\end{proof}

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\chapter{Positive Measures}
\label{chap:measures}
\input{./src/measure/measure/measure.tex}
\input{./src/measure/measure/complete.tex}
\input{./src/measure/measure/semifinite.tex}
\input{./src/measure/measure/sigma-finite.tex}
\input{./src/measure/measure/outer.tex}

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\section{Measures}
\label{section:measures}
\begin{definition}[Measurable Space]
\label{definition:measurable-space}
Let $X$ be a set and $\cm \subset 2^X$ be a $\sigma$-algebra, then the pair $(X, \cm)$ is a \textbf{measurable space}.
\end{definition}
\begin{definition}[Measure]
\label{definition:measure}
Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [0, \infty]$, then $\mu$ is a \textbf{(countably-additive) measure} if
\begin{enumerate}
\item[(M1)] $\mu(\emptyset) = 0$.
\item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$.
\end{enumerate}
In which case, $(X, \cm, \mu)$ is a \textbf{measure space}.
If $\mu: \cm \to [0, \infty]$ instead satisfies (M1) and
\begin{enumerate}
\item[(M2')] For any $\seqf{E_j} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{j = 1}^n E_j} = \sum_{j = 1}^n \mu(E_j)$.
\end{enumerate}
then $\mu$ is a \textbf{finitely-additive measure}.
\end{definition}
\begin{definition}[Null Set]
\label{definition:null}
Let $(X, \cm, \mu)$ be a measure space, then $E \in \cm$ is \textbf{$\mu$-null/null} if $\mu(E) = 0$.
\end{definition}
\begin{definition}[Almost Everywhere]
\label{definition:almost-everywhere}
Let $(X, \cm, \mu)$ be a measure space. A statement holds \textbf{$\mu$-almost everywhere/$\mu$-a.e./a.e.} if it is false on a subset of a $\mu$-null set.
\end{definition}
\begin{proposition}[{{\cite[Theorem 1.8]{Folland}}}]
\label{proposition:measure-properties}
Let $(X, \cm, \mu)$ be a measure space, then:
\begin{enumerate}
\item For any $E, F \in \cm$ with $E \subset F$, $\mu(E) \le \mu(F)$.
\item For any $\seq{E_n} \subset \cm$, $\mu\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu(E_n)$.
\item For any $\seq{E_n} \subset \cm$ with $E_n \subset E_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}E_n} = \limv{n}\mu(E_n)$.
\item For any $\seq{E_n} \subset \cm$ with $E_n \supset E_{n+1}$ for all $n \in \nat$ and $\mu(E_1) < \infty$, $\mu(\bigcap_{n \in \natp}E_n) = \limv{n}\mu(E_n)$.
\item For any $\seq{E_n} \subset \cm$,
\[
\mu\paren{\liminf_{n \to \infty}E_n} \le \liminf_{n \to \infty}\mu(E_n)
\]
\item For any $\seq{E_n} \subset \cm$ with $\mu\paren{\bigcup_{n \in \nat}E_n} < \infty$,
\[
\mu\paren{\limsup_{n \to \infty}E_n} \ge \limsup_{n \to \infty}\mu(E_n)
\]
\end{enumerate}
\end{proposition}
\begin{proof}
(1): $\mu(F) = \mu(E) + \mu(F \setminus E) \ge \mu(E)$.
(2): For each $n \in \natp$, let $F_n = E_n \setminus \bigcup_{k = 1}^{n-1}E_k$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
\[
\mu\paren{\bigcup_{n \in \nat}E_n} = \mu\paren{\bigsqcup_{n \in \nat}F_n} = \sum_{n \in \natp}\mu(F_n) \le \sum_{n \in \natp}\mu(E_n)
\]
(3): Denote $E_0 = \emptyset$. For each $n \in \natp$, let $F_n = E_n \setminus E_{n - 1}$, then $\seq{F_n}$ is pairwise disjoint with $\bigsqcup_{n \in \natp}F_n = \bigcup_{n \in \natp}E_n$. In which case,
\[
\mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
\]
(4): Since $\mu(E_1) < \infty$,
\[
\limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n}
\]
by (3).
(5): Using (1) and (3),
\begin{align*}
\mu\paren{\liminf_{n \to \infty}E_n} &= \mu\paren{\bigcup_{n \in \natp}\bigcap_{m \ge n}E_m} = \limv{n}\mu\paren{\bigcap_{m \ge n}E_m} \\
&\le \limv{n} \inf_{m \ge n}\mu(E_m) = \liminf_{n \to \infty}\mu(E_n)
\end{align*}
(6): Using (1) and (4),
\begin{align*}
\mu\paren{\limsup_{n \to \infty}E_n} &= \mu\paren{\bigcap_{n \in \natp}\bigcup_{m \ge n}E_m} = \limv{n}\mu\paren{\bigcup_{m \ge n}E_m} \\
&\ge \limv{n}\sup_{m \ge n}\mu(E_m) = \limsup_{n \to \infty}\mu(E_n)
\end{align*}
\end{proof}
\begin{theorem}[Dynkin's Uniqueness Theorem]
\label{theorem:dynkin-uniqueness}
Let $(X, \cm)$ be a measurable space, $\mathcal{P} \subset \cm$ be a $\pi$-syste, and $\mu, \nu: \cm \to [0, \infty]$ be measures. If
\begin{enumerate}
\item[(a)] $\sigma(\mathcal{P}) = \cm$.
\item[(b)] $\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
\item[(c)] There exists $\seq{E_n} \subset \mathcal{P}$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$.
\end{enumerate}
then $\mu = \nu$.
\end{theorem}
\begin{proof}
Let $F \in \mathcal{P}$ with $\mu(F) < \infty$ and
\[
\alg(F) = \bracs{E \in \cm|\mu(E \cap F) = \nu(E \cap F)}
\]
then $\alg(F) \supset \mathcal{P}$ by (b), and
\begin{enumerate}
\item[(L2)] For any $E, E' \in \alg(F)$ with $E \subset E'$ and $F \in \mathcal{P}$,
\begin{align*}
\mu((E' \setminus E) \cap F) &= \mu(E' \cap F) - \mu(E \cap F) \\
&= \nu(E' \cap F) - \nu(E \cap F) = \nu((E' \setminus E) \cap F)
\end{align*}
\item[(L3)] For any $\seq{E_n} \subset \alg$ with $E_n \subset E_{n+1}$ for all $n \in \natp$ and $F \in \mathcal{P}$,
\[
\mu\paren{\bigcup_{n \in \nat}E_n \cap F} = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu\paren{\bigcup_{n \in \nat}E_n \cap F}
\]
by continuity from below (\ref{proposition:measure-properties}).
\end{enumerate}
so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\ref{theorem:pi-lambda}), $\alg(F) = \cm$.
Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\ref{proposition:measure-properties}),
\[
\mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F)
\]
\end{proof}

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\section{Carathéodory's Extension Theorem}
\label{section:caratheodory}
\begin{definition}[Outer Measure]
\label{definition:outer-measure}
Let $X$ be a set and $\mu^*: 2^X \to [0, \infty]$, then $\mu^*$ is an \textbf{outer measure} if:
\begin{enumerate}
\item[(M1)] $\mu^*(\emptyset) = 0$.
\item[(OM1)] For any $E, F \subset X$ with $E \subset F$, $\mu^*(E) \le \mu^*(F)$.
\item[(OM2)] For any $\seq{E_n} \subset X$,
\[
\mu^*\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu^*(E_n)
\]
\end{enumerate}
\end{definition}
\begin{proposition}[{{\cite[Proposition 1.10]{Folland}}}]
\label{proposition:outer-measure-inf}
Let $X$ be a set, $\ce \subset 2^X$, and $\rho: \ce \to [0, \infty]$ such that $\emptyset, X \in \ce$ and $\rho(\emptyset) = 0$. Define
\[
\mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\rho(E_n) \bigg | \seq{E_n} \subset \ce, \bigcup_{n \in \natp}E_n \supset E}
\]
then $\mu^*$ is an outer measure.
\end{proposition}
\begin{proof}
(OM2): Let $\seq{E_n} \subset 2^X$ and assume without loss of generality that $\sum_{n \in \natp}\mu^*(E_n) < \infty$. Let $\eps > 0$. For each $n \in \nat$, let $\bracs{F_{n, k}}_{k = 1}^\infty \subset \ce$ such that $\bigcup_{k \in \natp}F_{n, k} \supset E_n$ and $\sum_{n \in \natp}\rho(F_{n, k}) < \mu^*(E_n) + 2^{-n}\eps$, then
\begin{align*}
\mu^*\paren{\bigcup_{n \in \natp}E_n} &\le \sum_{(n, k) \in \natp \times \natp}\rho(F_{n, k}) = \sum_{n \in \natp}\sum_{k \in \natp}\rho(F_{n, k}) \\
&\le \sum_{n \in \natp}(\mu^*(E_n) + 2^{-n}\eps) \le \eps + \sum_{n \in \natp}\mu^*(E_n)
\end{align*}
Since $\eps$ is arbitrary, $\mu^*\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu^*(E_n)$.
\end{proof}
\begin{definition}[Outer Measurable]
\label{definition:outer-measurable}
Let $X$ be a set, $\mu^*: 2^X \to [0, \infty]$ be an outer measure, and $E \subset X$, then $E$ is \textbf{$\mu^*$-measurable} if for any $F \subset X$,
\[
\mu^*(F) = \mu^*(E \cap F) + \mu^*(E \setminus F)
\]
\end{definition}
\begin{theorem}[Carathéodory, {{\cite[Theorem 1.11]{Folland}}}]
\label{theorem:caratheodory}
Let $X$ be a set, $\mu^*: 2^X \to [0, \infty]$ be an outer measure, and $\cm \subset 2^X$ be the collection of all $\mu^*$-measurable sets, then:
\begin{enumerate}
\item For any $F \subset X$ and $\seq{E_n} \subset \cm$ pairwise disjoint,
\[
\mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp}\mu^*(F \cap E_n)
\]
\item $\cm$ is a $\sigma$-algebra.
\item $(X, \cm, \mu^*|_\cm)$ is a complete measure space.
\end{enumerate}
\end{theorem}
\begin{proof}
(1): Let $N \in \nat$, then
\[
\mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \mu^*\paren{F \cap \bigsqcup_{n =1}^N E_n} = \sum_{n = 1}^N \mu^*(F \cap E_n)
\]
As this holds for all $N \in \nat$,
\[
\mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \sum_{n \in \natp}\mu^*(F \cap E_n)
\]
(2): Firstly, $\emptyset, X \in \cm$ since $\mu^*(\emptyset) = 0$. For any $A \in \cm$, $A^c \in \cm$ as well because the definition of $\mu^*$-measurability is symmetric.
Let $A, B \in \cm$ and $F \subset X$, then
\begin{align*}
\mu^*(F) &= \mu^*(F \cap A) + \mu^*(F \setminus A) \\
&= \mu^*(F \cap A \cap B) + \mu^*(F \cap A \setminus B) + \mu^*(F \cap B \setminus A) + \mu^*(F \setminus (A \cup B)) \\
&= \mu^*(F \cap B) + \mu^*(F \cap A \setminus B) + \mu^*(F \setminus (A \cup B)) \\
&= \mu^*(F \cap (A \cup B)) + \mu^*(F \setminus (A \cup B))
\end{align*}
so $\cm$ is an algebra. Since (1) implies that $\cm$ is closed under countable disjoint unions, $\cm$ is a $\sigma$-algebra by \ref{lemma:sigma-algebra-condition}.
(3): By (1), $\mu|_\cm$ is a measure. Let $N \in \cm$ with $\mu^*(N) = 0$, then for any $E \subset N$ and $F \subset X$,
\begin{align*}
\mu^*(E) &= \underbrace{\mu^*(E \cap N)}_0 + \mu^*(E \setminus N) \\
&= \underbrace{\mu^*(E \cap N \setminus F)}_0 + \underbrace{\mu^*(E \cap F)}_0 + \mu^*(E \setminus N) \\
&= \mu^*((E \setminus F) \cap N) + \mu^*(E \cap F) + \mu^*((E \setminus F) \setminus N) \\
&= \mu^*(E \cap F) + \mu^*(E \setminus F)
\end{align*}
\end{proof}
\begin{definition}[Premeasure]
\label{definition:premeasure}
Let $X$ be a set, $\alg \subset 2^X$ be an algebra, and $\mu_0: \alg \to [0, \infty]$, then $\mu_0$ is a \textbf{premeasure} if:
\begin{enumerate}
\item[(M1)] $\mu_0(\emptyset) = 0$.
\item[(PM)] For any $\seq{A_n} \subset \alg$ pairwise disjoint with $\bigsqcup_{n \in \natp}A_n \in \alg$,
\[
\mu_0\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in\natp}\mu_0(A_n)
\]
\end{enumerate}
\end{definition}
\begin{theorem}[Carathéodory's Extension Theorem, {{\cite[Theorem 1.14]{Folland}}}]
\label{theorem:caratheodory-extension}
Let $X$ be a set, $\alg \subset 2^X$ be an algebra, and $\mu_0: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that
\begin{enumerate}
\item $\cm \supset \alg$.
\item $\mu|_\alg = \mu_0$.
\item $\mu$ is complete.
\item[(U)] For any measure space $(X, \cn, \nu)$ satisfying (1) and (2),
\begin{enumerate}
\item[(a)] $\nu|_{\cm \cap \cn} \le \mu_{\cm \cap \cn}$.
\item[(b)] For any $E \in \cm \cap \cn$ with $\mu(E) < \infty$, $\mu(E) = \nu(E)$.
\item[(c)] If $\mu$ is $\sigma$-finite, then $\nu = \mu$.
\end{enumerate}
\end{enumerate}
\end{theorem}
\begin{proof}
Let
\[
\mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}
\]
then $\mu^*$ is an outer measure by \ref{proposition:outer-measure-inf}. By Carathéodory's theorem (\ref{theorem:caratheodory}), there exists a measure space $(X, \cm, \mu)$ such that $\mu = \mu^*|_{\cm}$.
(1): Let $E \in \alg$ and $F \subset X$. For any $\seq{F_n} \subset \alg$ with $\bigcup_{n \in \natp}F_n \supset F$,
\[
\sum_{n \in \natp}\mu_0(F_n) = \sum_{n \in \natp}\mu_0(F_n \cap E) + \sum_{n \in \natp}\mu_0(F_n \setminus E) \ge \mu^*(F \cap E) + \mu^*(F \setminus E)
\]
As this holds for all such $\seq{F_n}$, $\mu_0^*(F) = \mu^*(F \cap E) + \mu^*(F \setminus E)$.
(2): Let $E \in \alg$ and $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$, then
\[
\sum_{n \in \natp}\mu_0(E_n) = \sum_{n \in \natp}\mu_0(E_n \cap E) + \sum_{n \in \natp}\mu_0(E_n \setminus E) \ge \sum_{n \in \natp}\mu_0(E_n \cap E) = \mu_0(E)
\]
As this holds for all such $\seq{E_n}$, $\mu^*(E) = \mu_0(E)$.
(U): Let $\seq{E_n} \subset \alg$, then by continuity from below (\ref{proposition:measure-properties}),
\[
\nu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\nu\paren{\bigcup_{k = 1}^nE_k} = \limv{n}\mu\paren{\bigcup_{k = 1}^n E_k} = \mu\paren{\bigcup_{n \in \natp}E_n}
\]
Thus if $E \in \cm \cap \cn$ with $\bigcup_{n \in \natp}E_n \supset E$, then
\[
\nu(E) \le \nu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)
\]
As this holds for all such $\seq{E_n}$, $\nu(E) \le \mu^*(E) = \mu(E)$.
Suppose that $\mu(E) < \infty$, then there exists $\seq{E_n} \subset \alg$ such that $\bigcup_{n \in \natp}E_n \supset E$ and $\sum_{n \in \natp}\mu_0(E_n) < \infty$. In which case,
\begin{align*}
\mu\paren{\bigcup_{n \in \natp}E_n} &= \nu\paren{\bigcup_{n \in \natp}E_n} = \nu(E) + \nu\paren{\bigcup_{n \in \natp}E_n \setminus E} \\
&\le \nu(E) + \mu\paren{\bigcup_{n \in \natp}E_n \setminus E} \\
\mu(E) &\le \nu(E)
\end{align*}
so $\nu(E) = \mu(E)$ whenever $\mu(E) < \infty$.
Finally, if $\mu$ is $\sigma$-finite, then there exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$. For each $n \in \natp$, there exists $\bracsn{E_{n, k}}_{k = 1}^\infty \subset \alg$ such that $E_n \subset \bigcup_{k \in \natp}E_{n, k}$ and $\mu\paren{\bigcup_{k \in \natp}E_{n, k}} < \infty$. Let $F_n = \bigcup_{j = 1}^n \bigcup_{k \in \natp}E_{n, k}$, then $\seq{F_n} \subset \sigma(\alg) \subset \cm \cap \cn$ with $F_n \upto X$ and $\mu(F_n) < \infty$ for all $n \in \nat$. For any $E \in \cm \cap \cn$, by continuity from below (\ref{proposition:measure-properties}),
\[
\nu(E) = \limv{n}\nu(E \cap F_n) = \limv{n}\mu(E \cap F_n) = \mu(E)
\]
so $\nu_{\cm \cap \cn} = \mu_{\cm \cap \cn}$.
\end{proof}

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\section{Semifinite Measures}
\label{section:semifinite}
\begin{definition}[Semifinite]
\label{definition:semifinite-measure}
Let $(X, \cm, \mu)$ be a measure space, then the following are equivalent:
\begin{enumerate}
\item For any $E \in \cm$ with $\mu(E) = \infty$, there exists $F \subset E$ with $0 < \mu(F) < \infty$.
\item For any $E \in \cm$,
\[
\mu(E) = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
\]
\end{enumerate}
If the above holds, then $\mu$ is a \textbf{semifinite measure}.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: If $\mu(E) < \infty$, then the result holds directly. Assume that $\mu(E) = \infty$, and suppose for contradiction that
\[
M = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty} < \infty
\]
Let $\seq{F_n} \subset \cm$ such that $F \subset E$, $\mu(F) < \infty$, and $\mu(F_n) \upto M$. Let $F = \bigcup_{n \in \nat}F_n$, then $\mu(F) = M$, and $\mu(E \setminus F) = \infty$. By (1), there exists $G \in \cm$ with $G \subset E \setminus F$ and $0 < \mu(G) < \infty$. Thus $F \cup G \subset E$ with $M < \mu(F) + \mu(G) = \mu(F \cup G) < \infty$. This contradicts the fact that $M$ is the supremum.
\end{proof}
\begin{definition}[Semifinite Part]
\label{definition:semifinite-part}
Let $(X, \cm, \mu)$ be a measure space and
\[
\mu_0: \cm \to [0, \infty] \quad E \mapsto \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
\]
then $\mu_0$ is a semifinite measure, and the \textbf{semifinite part} of $\mu$.
\end{definition}
\begin{proof}
Firstly, for any $F \in \cm$ with $\mu(F) < \infty$, $\mu(F) = \mu_0(F)$.
Let $\seq{E_n} \subset \cm$ be pairwise disjoint. For any $F \in \cm$ with $F \subset \bigsqcup_{n \in \natp}E_n$ and $\mu(F) < \infty$,
\[
\mu(F) = \sum_{n \in \natp}\mu(F \cap E_n) \le \sum_{n \in \natp}\mu_0(E_n)
\]
Thus $\mu_0\paren{\bigcup_{n \in \natp}E_n} \le \sum_{n \in \natp}\mu_0(E_n)$.
On the other hand, let $n \in \natp$ and $\seqf{F_k} \subset \cm$ such that $F_k \subset E_k$ for each $1 \le k \le n$ and $\mu(F_k) < \infty$, then $F = \bigsqcup_{k = 1}^n F_k \subset \bigsqcup_{k \in \natp}E_k$. Thus
\[
\sum_{k = 1}^n \mu(F_k) = \mu(F) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
\]
As this holds for all such $\seqf{F_k}$,
\[
\sum_{k = 1}^n \mu_0(E_k) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
\]
Since this holds for all $n \in \natp$,
\[
\sum_{n \in \natp}\mu_0(E_n) \le \mu_0\paren{\bigsqcup_{k \in \natp}E_k}
\]
\end{proof}

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\section{$\sigma$-Finite Measures}
\label{section:sigma-finite}
\begin{definition}[$\sigma$-Finite Measure]
\label{definition:sigma-finite}
Let $(X, \cm, \mu)$ be a measure space, then the following are equivalent:
\begin{enumerate}
\item There exists $\seq{E_n} \subset \cm$ pairwise disjoint such that $\bigsqcup_{n \in \nat}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
\item There exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
\end{enumerate}
\end{definition}

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\section{Algebras}
\label{section:set-algebra}
\begin{definition}[Algebra]
\label{definition:algebra}
Let $X$ be a set and $\alg \subset 2^X$, then $\alg$ is an \textbf{algebra} if:
\begin{enumerate}
\item[(A1)] $\emptyset, X \in \alg$.
\item[(A2)] For any $A \in \alg$, $A^c \in \alg$.
\item[(A3)] For any $A, B \in \alg$, $A \cup B \in \alg$.
\end{enumerate}
\end{definition}
\begin{definition}[$\sigma$-Algebra]
\label{definition:sigma-algebra}
Let $X$ be a set and $\cm \subset 2^X$, then $\cm$ is a \textbf{$\sigma$-algebra} if:
\begin{enumerate}
\item[(A1)] $\emptyset, X \in \cm$.
\item[(A2)] For any $A \in \cm$, $A^c \in \cm$.
\item[(A3')] For any $\seq{A_n} \in \cm$, $\bigcup_{n \in \nat^+}A_n \in \cm$.
\end{enumerate}
\end{definition}
\begin{proposition}
\label{proposition:algebra-properties}
Let $X$ be a set and $\alg \subset 2^X$ be an algebra, then:
\begin{enumerate}
\item For any $A, B \in \alg$, $A \cap B \in \alg$.
\item For any $A, B \in \alg$, $A \setminus B \in \alg$.
\end{enumerate}
If $\alg$ is a $\sigma$-algebra, then:
\begin{enumerate}
\item[(1')] For any $\seq{A_n} \in \alg$, $\bigcap_{n \in \natp}A_n \in \alg$.
\end{enumerate}
\end{proposition}
\begin{proof}
(1, 1'): $\bigcap_{i \in I}A_i = \braks{\bigcup_{i \in I}A_i^c}^c$.
(2): $A \setminus B = A \cap B^c$.
\end{proof}
\begin{lemma}
\label{lemma:sigma-algebra-condition}
Let $X$ be a set and $\alg \subset 2^X$ be an algebra, then the following are equivalent:
\begin{enumerate}
\item For any $\seq{A_n} \subset \alg$, $\bigcup_{n \in \nat^+}A_n \in \alg$.
\item For any $\seq{A_n} \subset \alg$ with $A_n \subset A_{n+1}$ for all $n \in \natp$, $\bigcup_{n \in \natp}A_n \in \alg$.
\item For any $\seq{A_n} \subset \alg$ pairwise disjoint, $\bigsqcup_{n \in \natp}A_n \in \alg$.
\end{enumerate}
\end{lemma}
\begin{proof}
(2) $\Rightarrow$ (1): For each $n \in \nat$, let $B_n = \bigcup_{k = 1}^n A_n \in \alg$, then $\bigcup_{n \in \natp}A_n = \bigcup_{n \in \nat}B_n \in \alg$.
(3) $\Rightarrow$ (2): Denote $A_0 = \emptyset$ and $B_n = A_n \setminus A_{n-1}$, then $\seq{B_n} \subset \alg$ is pairwise disjoint and $\bigcup_{n \in \nat}A_n = \bigsqcup_{n \in \nat}B_n \in \alg$.
\end{proof}
\begin{definition}[Generated $\sigma$-Algebra]
\label{definition:generated-sigma-algebra}
Let $X$ be a set and $\ce \subset 2^X$, then the $\sigma$-algebra $\sigma(\ce)$ \textbf{generated by} $\ce$ is the smallest $\sigma$-algebra on $X$ containing $\ce$.
\end{definition}
\begin{definition}[Borel $\sigma$-Algebra]
\label{definition:borel-sigma-algebra}
Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$.
\end{definition}

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\section{Elementary Families}
\label{section:elementary-families}
\begin{definition}[Elementary Family]
\label{definition:elementary-family}
Let $X$ be a set and $\ce \subset 2^X$, then $\ce$ is an \textbf{elementary family} if:
\begin{enumerate}
\item[(P1)] $\emptyset \in \mathcal{P}$.
\item[(P2)] For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
\item[(E1)] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.
\end{enumerate}
\end{definition}
\begin{proposition}[{{\cite[Proposition 1.7]{Folland}}}]
\label{proposition:elementary-family-algebra}
Let $X$ be a set and $\ce \subset 2^X$ be an elementary family, then
\[
\alg = \bracs{\bigsqcup_{i = 1}^n E_j \bigg | \seqf{E_j} \subset \text{ pairwise disjoint}}
\]
is an algebra.
\end{proposition}
\begin{proof}
(A1): By (P1) $\emptyset \in \ce$. By (E1), there exists $\seqf{E_j} \subset \ce$ such that $X = \emptyset^c = \bigsqcup_{j = 1}^nE_j \in \alg$.
(A2): Let $A = \bigsqcup_{j = 1}^n A_j \in \alg$. By including additional empty sets, assume without loss of generality that there exists $\bracs{E_{i, j}| 1 \le i \le n, 1 \le j \le m}$ such that $A_i^c = \bigsqcup_{j = 1}^m E_{i, j}$ for each $1 \le j \le n$. In which case,
\[
A^c = \bigcap_{i = 1}^n A_i^c = \bigcap_{i = 1}^n \bigsqcup_{j = 1}^m E_{i, j} = \bigsqcup_{\alpha \in [1, m]^n} \underbrace{\bigcap_{i = 1}^n E_{i, \alpha_i}}_{\in \ce} \in \ce
\]
(A3): Let $A = \bigsqcup_{j = 1}^n A_j, B = \bigsqcup_{j = 1}^m B_j \in \alg$, then
\[
A \cap B = \braks{\bigsqcup_{j = 1}^n A_j} \cap \braks{\bigsqcup_{j = 1}^m B_j} = \bigsqcup_{i = 1}^n \bigcup_{j = 1}^m \underbrace{A_i \cap B_j}_{\in \ce} \in \alg
\]
so $\alg$ is closed under intersections. Thus using (A2), $A \cup B = (A^c \cap B^c)^c \in \alg$.
\end{proof}

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\chapter{Set Systems}
\label{chap:set-system}
\input{./src/measure/sets/algebra.tex}
\input{./src/measure/sets/lambda.tex}
\input{./src/measure/sets/elementary.tex}

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\section{Lambda Systems}
\label{section:lambda-system}
\begin{definition}[$\pi$-System]
\label{definition:pi-system}
Let $X$ be a set and $\mathcal{P} \subset 2^X$, then $\mathcal{P}$ is a \textbf{$\pi$-system} if:
\begin{enumerate}
\item[(P1)] $\emptyset \in \mathcal{P}$.
\item[(P2)] For any $A, B \in \mathcal{P}$, $A \cap B \in \mathcal{P}$.
\end{enumerate}
\end{definition}
\begin{definition}[$\lambda$-System]
\label{definition:lambda-system}
Let $X$ be a set and $\alg \subset 2^X$, then $\alg$ is a \textbf{$\lambda$-system/$d$-system} if:
\begin{enumerate}
\item[(L1)] $\emptyset, X \in \alg$.
\item[(L2)] For any $E, F \in \alg$ with $E \subset F$, $F \setminus E \in \alg$.
\item[(L3)] For any $\seq{A_n} \subset \alg$ with $A_n \subset A_{n+1}$ for all $n \in \nat^+$, $\bigcup_{n \in \nat^+}A_n \in \alg$.
\end{enumerate}
\end{definition}
\begin{definition}[Generated $\lambda$-System]
\label{definition:generated-lambda-system}
Let $X$ be a set and $\ce \subset 2^X$, then the smallest $\lambda$-system $\lambda(\ce)$ over $X$ containing $\ce$ is the \textbf{$\lambda$-system generated by $\ce$}.
\end{definition}
\begin{lemma}
\label{lemma:pi-lambda}
Let $X$ be a set and $\alg \subset 2^X$, then the following are equivalent:
\begin{enumerate}
\item $\alg$ is a $\sigma$-algebra.
\item $\alg$ is a $\pi$-system and a $\lambda$-system.
\end{enumerate}
\end{lemma}
\begin{proof}
$(2) \Rightarrow (1)$: Let $A, B \in \alg$, then $A \cup B = (A^c \cap B^c)^c \in \alg$. Thus $\alg$ is an algebra. By \ref{lemma:sigma-algebra-condition}, $\alg$ is a $\sigma$-algebra.
\end{proof}
\begin{theorem}[Dynkin's $\pi$-$\lambda$ Theorem]
\label{theorem:pi-lambda}
Let $X$ be a set and $\mathcal{P} \subset 2^X$ be a $\pi$-system, then $\sigma(\mathcal{P}) = \lambda(\mathcal{P})$.
\end{theorem}
\begin{proof}
Let $\ce \subset \lambda(\mathcal{P})$ and
\[
\cm(\ce) = \bracs{E \in \lambda(\mathcal{P})| E \cap F \in \lambda(\mathcal{P}) \forall F \in \ce}
\]
then
\begin{enumerate}
\item[(L2)] Let $E, F \in \cm$ with $E \subset F$, then for any $G \in \ce$,
\[
(F \setminus E) \cap G = (F \cap G) \setminus (E \cap G) \in \lambda(\mathcal{P})
\]
\item[(L3)] Let $\seq{E_n} \in \cm$ with $E_n \subset E_{n+1}$ for all $n \in \natp$, and $F \in \ce$, then
\[
\braks{\bigcup_{n \in \natp}E_n} \cap F = \bigcup_{n \in \natp}E_n \cap F \in \lambda(\mathcal{P})
\]
\end{enumerate}
so $\cm(\ce)$ is a $\lambda$-system.
Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \ref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.
\end{proof}