Added convergence in measure.

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

@@ -0,0 +1,83 @@
+\section{Convergence in Measure}
+\label{section:convergence-in-measure}
+
+\begin{definition}[Convergence in Measure]
+\label{definition:convergence-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{in measure} if for every $\eps > 0$,
+    \[
+    \lim_{n \to \infty}\mu(\bracs{d(f_n, f) > \eps}) = 0
+    \]
+\end{definition}
+
+\begin{definition}[Cauchy in Measure]
+\label{definition:cauchy-in-measure}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ be Borel measurable functions from $X$ to $Y$, then $\seq{f_n}$ is \textbf{Cauchy in measure} if for every $\eps > 0$,
+    \[
+    \mu(\bracs{d(f_m, f_n) > \eps}) \to 0 \quad \text{as}\ m, n \to \infty
+    \]
+\end{definition}
+
+\begin{lemma}
+\label{lemma:ae-in-measure}
+    Let $(X, \cm, \mu)$ be a finite measure space, $(Y, d)$ be a metric space, and $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$ such that $f_n \to f$ almost everywhere, then $f_n \to f$ in measure.
+\end{lemma}
+\begin{proof}
+    Let $\eps > 0$, then for almost every $x \in X$, there exists $N \in \natp$ such that $d(f_n(x), f(x)) < \eps$ for all $n \ge N$, so
+    \[
+    \mu\paren{\bigcup_{N \in \natp}\bigcap_{n \ge N}\bracs{d(f_n, f) < \eps}} = \mu(X)
+    \]
+
+    By continuity from above (\autoref{proposition:measure-properties}), 
+    \[
+    \limv{N}\mu{\bracs{d(f_N, f) \ge \eps}} \le \limv{N}\mu\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) \ge \eps}} = 0
+    \]
+\end{proof}
+
+
+\begin{proposition}[{{\cite[Theorem 2.30]{Folland}}}]
+\label{proposition:cauchy-in-measure-limit}
+    Let $(X, \cm, \mu)$ be a measure space, $(Y, d)$ be a complete metric space, and $\seq{f_n}$ be Borel measurable functions from $X \to Y$, then:
+    \begin{enumerate}
+        \item There exists a Borel measurable function $f: X \to Y$ such that $f_n \to f$ in measure.
+        \item There exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.
+        \item For any Borel measurable function $g: X \to Y$ such that $f_n \to g$ in measure, $f = g$ almost everywhere.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (2): Since $\seq{f_n}$ is Cauchy in measure, there exists a subsequence $\seq{n_k}$ such that for each $k \in \natp$, $\mu(\bracsn{d(f_{n_k}, f_{n_{k+1}}) > 2^{-k}}) \le 2^{-k}$.
+    
+    In this case, for any $K \in \natp$ and $j \ge k \ge K$, 
+    \[
+    \bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}} \subset \bigcup_{\ell = k}^{j - 1}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
+    \subset \bigcup_{\ell = K}^{\infty}\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
+    \]
+
+    By monotonicity and subadditivity (\autoref{proposition:measure-properties}), 
+    \[
+    \mu\paren{\bigcup_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) > 2^{-K+1}}} 
+    \le \sum_{\ell \ge K}\mu\bracsn{d(f_{n_\ell}, f_{n_{\ell + 1}}) > 2^{-\ell}}
+    \le 2^{-K+1}
+    \]
+
+    so
+    \[
+    \mu\braks{\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} \le 2^{-K+1}
+    \]
+
+    By the \hyperref[First Borel-Cantelli Lemma]{lemma:borel-cantelli-1},
+    \[
+    \mu\braks{\limsup_{K \to \infty}\paren{\bigcap_{j, k \ge K}\bracsn{d(f_{n_j}, f_{n_k}) \le 2^{-K+1}}}^c} = 0
+    \]
+
+    Thus, for almost every $x \in X$, there exists $K \in \natp$ such that $d(f_{n_j}(x), f_{n_k}(x)) < 2^{-K+1}$ for all $j, k \ge K$. Therefore $\seq{f_n(x)}$ is Cauchy for almost every $x$, and converges to a Borel measurable function $f: X \to Y$.
+
+    (1): Let $\eps, \delta > 0$, then there exists $N \in \natp$ such that $\mu(\bracs{d(f_m, f_n) > \delta/2}) < \eps/2$ for all $m, n \ge N$. Let $k \in \natp$ such that $n_k \ge N$ and $\mu(\bracs{d(f, f_{n_k}) > \delta/2}) < \eps/2$, then for any $m \ge N$,
+    \[
+    \mu\bracs{d(f_m, f) > \delta} \le \mu\bracs{d(f_m, f_{n_k}) > \delta/2} + \mu\bracs{d(f, f_{n_k}) > \delta/2} < \eps
+    \]
+
+    (3): By (2), there exists a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ and $f_{n_k} \to g$ almost everywhere, so $f = g$ almost everywhere. 
+\end{proof}
+
+
+

src/measure/measurable-maps/index.tex

@@ -6,3 +6,4 @@
 \input{./real-valued.tex}
 \input{./simple.tex}
 \input{./metric.tex}
+\input{./in-measure.tex}

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

@@ -6,11 +6,17 @@
     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{definition}[Convergence Almost Everywhere]
+\label{definition:convergence-ae}
+    Let $(X, \cm, \mu)$ be a measure space, $Y$ be a topological space, $\seq{f_n}$ and $f$ be Borel measurable functions from $X$ to $Y$, then $f_n \to f$ \textbf{almost everywhere/a.e.} if there exists a $\mu$-null set $N \in \cm$ such that $f_n \to f$ pointwise on $X \setminus N$.
+\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.

src/measure/measure/measure.tex

@@ -64,15 +64,19 @@
 
 
     (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,
-    \[
-    \mu\paren{\bigcup_{n \in \natp}E_n} = \mu\paren{\bigsqcup_{n \in \natp}F_n} = \limv{n}\sum_{k = 1}^n \mu(F_n) = \limv{n}\mu(E_n)
-    \]
+    \begin{align*}
+    \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)
+    \end{align*}
+    
 
 
     (4): Since $\mu(E_1) < \infty$,
-    \[
-    \limv{n}\mu(E_n) = \mu(E_1) - \limv{n}\mu(E_1 \setminus E_n) = \mu(E_1) - \mu\paren{E_1 \setminus \bigcap_{n \in \natp}E_n} = \mu\paren{\bigcap_{n \in \natp}E_n}
-    \]
+    \begin{align*}
+    \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}
+    \end{align*}
 
     by (3).
 
@@ -125,5 +129,23 @@
     \[
     \mu(F) = \limv{n}\mu(E_n \cap F) = \limv{n}\nu(E_n \cap F) = \nu(F)
     \]
-
 \end{proof}
+
+\begin{lemma}[First Borel-Cantelli Lemma]
+\label{lemma:borel-cantelli-1}
+    Let $(X, \cm, \mu)$ be a measure space and $\seq{E_n} \subset \cm$. If $\sum_{n \in \natp}\mu(E_n) < \infty$, then
+    \[
+    \mu\paren{\limsup_{n \to \infty}E_n} = 0
+    \]
+\end{lemma}
+\begin{proof}
+    For any $n \in \natp$, by monotonicity and subadditivity (\autoref{proposition:measure-properties}), 
+    \[
+    \mu\paren{\limsup_{k \to \infty}E_k} \le \mu\paren{\bigcup_{k \ge n}E_k} \le \sum_{k \ge n}\mu(E_k)
+    \]
+    As $\sum_{k \in \natp}\mu(E_k) < \infty$,
+    \[
+    \mu\paren{\limsup_{k \to \infty}E_k} \le \inf_{n \in \natp}\sum_{k \ge n}\mu(E_k) = 0
+    \]
+\end{proof}
+

src/measure/sets/index.tex

@@ -4,3 +4,4 @@
 \input{./algebra.tex}
 \input{./lambda.tex}
 \input{./elementary.tex}
+\input{./limits.tex}

src/measure/sets/limits.tex

@@ -0,0 +1,35 @@
+\section{Limits of Sets}
+\label{section:set-limits}
+
+\begin{definition}[Limit of Sets]
+\label{definition:set-limits}
+    Let $X$ be a set and $\seq{E_n} \subset 2^X$, then the \textbf{limit superior} of $\seq{E_n}$ is
+    \[
+    \limsup_{n \to \infty} E_n = \bigcap_{n \in \natp}\bigcup_{k \ge n}E_k
+    \]
+    and the \textbf{limit inferior} of $\seq{E_n}$ is
+    \[
+    \liminf_{n \to \infty}E_n = \bigcup_{n \in \natp}\bigcap_{k \ge n}E_k
+    \]
+\end{definition}
+
+
+\begin{lemma}
+\label{lemma:set-limits-characterisation}
+    Let $X$ be a set and $\seq{X_n} \subset 2^X$, then
+    \[
+    \limsup_{n \to \infty} E_n = \bracs{x \in X|x \in E_n \text{for infinitely many }n}
+    \]
+    and
+    \[
+    \liminf_{n \to \infty}E_n = \bracs{x \in X|x \in E_n \text{for all but finitely many }n}
+    \]
+\end{lemma}
+\begin{proof}
+    For any $x \in X$, $x \in \limsup_{n \to \infty}E_n$ if and only if for any $n \in \natp$, there exists $k \ge n$ such that $x \in E_k$, if and only if $x \in E_n$ for infinitely many $n \in \natp$.
+
+    For any $x \in X$, $x \in \liminf_{n \to \infty}$ if and only if there exists $n \in \natp$ such that $x \in E_k$ for all $k \ge n$, if and only if $x \in E_n$ for all but finitely many $n \in \natp$.
+\end{proof}
+
+
+