Me when I forget to commit.

src/measure/measurable-maps/index.tex

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+\chapter{Measurable Functions}
+\label{chap:measurable-maps}
+
+\input{./src/measure/measurable-maps/measurable-maps.tex}
+\input{./src/measure/measurable-maps/product.tex}

src/measure/measurable-maps/measurable-maps.tex

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+\section{Measurable Functions}
+\label{section:measurable-maps}
+
+\begin{definition}[Measurable Function]
+\label{definition:measurable-function}
+    Let $(X, \cm)$ and $(Y, \cn)$ be measurable spaces and $f: X \to Y$ be a mapping, then $f$ is \textbf{$(\cm, \cn)$-measurable} if $f^{-1}(E) \in \cm$ for all $E \in \cn$.
+\end{definition}
+
+\begin{definition}[Borel Measurable]
+\label{definition:borel-measurable-function}
+    Let $(X, \cm)$ be a measurable space, $Y$ be a topological space, and $f: X \to Y$ be a mapping, then $f$ is \textbf{Borel measurable} if $f$ is $(\cm, \cb_Y)$-measurable.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:continuous-borel-measurable}
+    Let $X, Y$ be topological spaces and $f: X \to Y$ be continuous, then $f$ is Borel measurable.
+\end{lemma}
+
+\begin{definition}[Generated $\sigma$-Algebra]
+\label{definition:generated-sigma-algebra-function}
+    Let $X$ be a set, $\bracs{(Y_i, \cn_i)}_{i \in I}$ be measurable spaces, and $\seqi{f}$ with $f_i: X \to Y_i$ for each $i \in I$. The \textbf{$\sigma$-algebra generated by} $\seqi{f}$,
+    \[
+    \sigma(\bracs{f_i| i \in I}) = \sigma\paren{f_i^{-1}(\cn_i)|i \in I}
+    \]
+    is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable.
+\end{definition}

src/measure/measurable-maps/product.tex

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+\section{Product $\sigma$-Algebras}
+\label{section:product-sigma-algebra}
+
+\begin{definition}[Product $\sigma$-Algebra]
+\label{definition:product-sigma-algebra}
+    Let $\bracs{(X_i, \cm_i)}_{i \in I}$ be measurable spaces, then the \textbf{product} $\sigma$-algebra $\bigotimes_{i \in I}\cm_i$ on $\prod_{i \in I}X_i$ is the $\sigma$-algebra generated by $\seqi{\pi_i}$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:rectangles}
+    Let $\bracs{(X_i, \cm_i)}_{i \in I}$, let
+    \[
+    \ce = \bracs{\bigcap_{j \in J}\pi_j^{-1}(E_j) \bigg | J \subset I \text{ finite}, E_j \in \cm_j}
+    \]
+    then $\ce$ is an elementary family, and $\sigma(\ce) = \bigotimes_{i \in I}\cm_i$.
+\end{lemma}
+\begin{proof}
+\begin{enumerate}
+    \item[(P1)] For any $j \in I$, $\emptyset = \pi_j^{-1}(\emptyset) \in \ce$.
+    \item[(P2)] Let
+    \[
+    \bigcap_{j \in J}\pi_j^{-1}(E_j), \bigcap_{j \in J'}\pi_j^{-1}(F_j) \in \ce
+    \]
+    Assume without loss of generality that $J = J'$, then
+    \[
+    \bigcap_{j \in J}\pi_j^{-1}(E_j) \cap \bigcap_{j \in J'}\pi_j^{-1}(F_j) = \bigcap_{j \in J}\pi_j^{-1}(E_j \cap F_j) \in \ce
+    \]
+    \item[(E)] For any $j \in I$, $\prod_{i \in I}X_i = \pi_j^{-1}(X_j) \in \ce$. Thus it is sufficient to show that $\ce$ is closed under complements. Let $\bigcap_{j \in J}\pi_j^{-1}(E_j) \in \ce$, then
+    \[
+    \braks{\bigcap_{j \in J}\pi_j^{-1}(E_j)}^c = \bigcup_{j \in J}\pi_j^{-1}(E_j^c) = \bigsqcup_{\emptyset \subsetneq K \subset J}
+    \underbrace{\paren{\bigcap_{k \in K}\pi_j^{-1}(E_k^c)} \cap \paren{\bigcap_{j \in J \setminus K}\pi_j^{-1}(E_k)}}_{\in \ce}
+    \]
+\end{enumerate}
+\end{proof}
+
+\begin{proposition}[{{\cite[Proposition 1.5]{Folland}}}]
+\label{proposition:product-sigma-algebra-metric}
+    Let $\seqf{X_j}$ be topological spaces, $X = \prod_{j = 1}^n X_j$, then:
+    \begin{enumerate}
+        \item $\bigotimes_{j = 1}^n \cb_{X_j} \subset \cb_{X}$.
+        \item If $\seq{X_j}$ are separable, then $\bigotimes_{j = 1}^n \cb_{X_j} = \cb_{X}$
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): By \ref{proposition:separable-product}, $X$ is also separable. Let $U \subset X$, then there exists $\seq{x_j} \subset X$ such that $\overline{\bracs{x_j|j \in \natp}} \supset U$. For each $j \in \nat$, let $\bracs{U_{j, k}}$ such that $U_{j, k} \in \cn_{X_k}(\pi_k(x_j))$ for each $1 \le k \le n$ and $x_j \in \bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \subset U$.
+
+    Since $\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k}) \in \bigotimes_{j = 1}^n \cb_{X_j}$ and
+    \[
+    U = \bigcup_{j \in \natp}\bigcap_{k = 1}^n \pi_k^{-1}(U_{j, k})
+    \]
+    The open set $U$ is in $\bigotimes_{j = 1}^n \cb_{X_j}$.
+\end{proof}