Progress over the past week.

src/measure/sets/algebra.tex

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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}

src/measure/sets/elementary.tex

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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}

src/measure/sets/index.tex

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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}

src/measure/sets/lambda.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}