Me when I forget to commit.

src/measure/index.tex

@@ -3,3 +3,4 @@
 
 \input{./src/measure/sets/index.tex}
 \input{./src/measure/measure/index.tex}
+\input{./src/measure/measurable-maps/index.tex}

src/measure/measurable-maps/index.tex

@@ -0,0 +1,5 @@
+\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

@@ -0,0 +1,26 @@
+\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

@@ -0,0 +1,52 @@
+\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}

src/measure/measure/index.tex

@@ -8,3 +8,4 @@
 \input{./src/measure/measure/outer.tex}
 \input{./src/measure/measure/regular.tex}
 \input{./src/measure/measure/lebesgue-stieltjes.tex}
+\input{./src/measure/measure/kolmogorov.tex}

src/measure/measure/kolmogorov.tex

@@ -0,0 +1,76 @@
+\section{Kolmogorov's Extension Theorem}
+\label{section:kolmogorov-extension}
+
+
+\begin{definition}[Consistent]
+\label{definition:consistent-measures}
+    Let $\bracs{(\Omega_i, \cf_i)}_{i \in I}$ be a family of measurable spaces. For each $J \subset J' \subset I$, denote $\pi_J: \prod_{i \in I}X_i \to \prod_{j \in J}X_j$ and $\pi_{J', J}: \prod_{j \in J'}X_j \to \prod_{j \in J}X_j$ as the projection maps.
+
+    Let $\bracs{\mu_J|J \subset I \text{ finite}}$ such that each $\mu_J$ is a probability measure on $\prod_{j \in J}\Omega_j$, then $\bracs{\mu_J}$ is \textbf{consistent} if for all $J \subset J' \subset I$, $\mu_{J'} = \mu_J \circ \pi_{J', J}^{-1}$.
+\end{definition}
+
+\begin{lemma}[{{\cite[Lemma 1.17]{Baudoin}}}]
+\label{lemma:kolmogorov-compact-sequence}
+    Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$,
+    \begin{enumerate}
+        \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular\footnote{A potential sufficient condition for this is that each $X_n$ is LCH where every open set is $\sigma$-compact. However, I have yet to verify if this condition persists over products.}.
+        \item[(b)] $X_n$ is Hausdorff.
+        \item[(c)] $X_n$ is separable.
+    \end{enumerate}
+    Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where:
+    \begin{enumerate}
+        \item[(d)] For each $n \in \nat$, $B_n \in \cb(\prod_{j = 1}^n X_j)$.
+        \item[(e)] For each $n \in \nat$, $B_{n+1} \subset B_n \times X_{n+1}$.
+        \item[(f)] There exists $\eps > 0$ such that $\mu_{[n]}(B_n) > \eps$ for all $n \in \natp$.
+    \end{enumerate}
+    Then there exists $\seq{K_n}$ such that for every $n \in \natp$,
+    \begin{enumerate}
+        \item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
+        \item $K_n \subset B_n$.
+        \item $K_{n+1} \subset K_n \times X_{n+1}$.
+        \item $\mu(K_n) \ge \eps/2$.
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    Let $n \in \natp$. By (c), $\mu_{[n]}$ is a Borel probability measure. Thus $\mu_{[n]}$ is regular by (a). Thus there exists $C_n \subset \prod_{j = 1}^n X_j$ compact such that $C_n \subset B_n$ and $\mu_{[n]}(B_n \setminus C_n) < \eps/2^{n+1}$. Let
+    \[
+    K_n = \bigcap_{j = 1}^n \pi_{[n], [j]}^{-1}(C_n)
+    \]
+    Since each $X_j$ is Hausdorff, $K_n \subset \prod_{j = 1}^n X_j$ is compact with $K_n \subset B_n$ and $K_{n+1} \subset K_n \times X_{n+1}$. Moreover,
+    \[
+    \mu_{[n]}(B_n \setminus K_n) \le \sum_{j = 1}^n\mu_{[n]}\braks{\pi_{[n], [j]}^{-1}(B_j \setminus C_j)} \le \sum_{j = 1}^n\mu_{[j]}(B_j \setminus C_j) \le \eps/2
+    \]
+    Thus $\mu_{[n]}(K_n) \ge \eps/2$.
+\end{proof}
+
+\begin{theorem}[Kolmogorov's Extension Theorem, {{\cite[Theorem 1.14]{Baudoin}}}]
+\label{theorem:kolmogorov-extension}
+    Let $\seqi{X}$ be topological spaces where for each $J \subset I$ finite,
+    \begin{enumerate}
+        \item[(a)] Every finite measure on $\prod_{j \in J} X_j$ is regular.
+        \item[(b)] $X_j$ is Hausdorff.
+        \item[(c)] $X_j$ is separable.
+    \end{enumerate}
+    Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then there exists a unique probability measure $\mu: \bigotimes_{i \in I}\cb_{X_i} \to [0, 1]$ such that for any $J \subset I$ finite, $\mu = \mu_J \circ \pi_J^{-1}$.
+\end{theorem}
+\begin{proof}
+    Let
+    \[
+    \alg = \bracs{\pi_J^{-1}(B)| B \in \bigotimes_{j \in J}\cb_{X_j}, J \subset I \text{ finite}}
+    \]
+    then $\alg$ is an algebra. For any $B \in \bigotimes_{j \in J}\cb_{X_j}$, define $\mu_0(\pi_J^{-1}(B)) = \mu_J(B)$, then $\mu_0: \alg \to [0, 1]$ is well-defined and finitely additive by the consistency of $\bracs{\mu_J}$.
+
+    To show that $\mu_0$ is a premeasure, it is sufficient to show that for any $\{\pi_{J_n}^{-1}(B_n)\}_1^\infty$ with $\pi_{J_n}^{-1}(B_n) \downto \emptyset$, $\mu_0(\pi_{J_n}^{-1}(B_n)) \downto 0$. To this end, suppose for contradiction that $\limv{n}\mu_0(\pi_{J_n}^{-1}(B_n)) = \eps > 0$.
+
+    By inserting additional elements into the sequence and relabeling the indices, assume without loss of generality that $J_n = [n]$ for all $n \in \natp$. By \ref{lemma:kolmogorov-compact-sequence}, there exists $\seq{K_n}$ such that:
+    \begin{enumerate}
+        \item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
+        \item $K_n \subset B_n$.
+        \item $K_{n+1} \subset K_n \times X_{n+1}$.
+        \item $\mu(K_n) \ge \eps/2$.
+    \end{enumerate}
+    Let $N \in \natp$ and $x \in \prod_{j = 1}^N X_j$ such that $x \in \bigcap_{n \ge N}\pi_{[N]}(K_n)$. By compactness and (4), there exists $x_{N+1} \in X_{N+1}$ such that $(x_1, \cdots, x_N, x_{N+1}) \in \bigcap_{n > N}\pi_{[N+1]}(K_n)$. Thus there exists $x \in \prod_{i \in I}X_i$ such that $x \in \pi_{[n]}^{-1}(K_n)$ for all $n \in \natp$, and
+    \[
+    x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset
+    \]
+\end{proof}

src/measure/measure/regular.tex

@@ -21,3 +21,14 @@
 \label{definition:regular-measure}
     Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a measure, then $\mu$ is \textbf{regular} if it is inner regular and outer regular.
 \end{definition}
+
+\begin{theorem}[{{\cite[Theorem 7.8]{Folland}}}]
+\label{theorem:sigma-finite-regular-measure}
+    Let $X$ be a topological space and $\mu: \cb_X \to [0, \infty]$ be a Borel measure. If:
+    \begin{enumerate}
+        \item[(a)] $X$ is a LCH space.
+        \item[(b)] Every open set of $X$ is $\sigma$-compact.
+        \item[(c)] For any $K \subset X$ compact, $\mu(K) < \infty$.
+    \end{enumerate}
+    then $\mu$ is a regular measure.
+\end{theorem}