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}
+
+
+