Added Egoroff's theorem.

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

@@ -40,3 +40,40 @@
 
     is the smallest $\sigma$-algebra on $X$ such that each $\seqi{f}$ is measurable.
 \end{definition}
+
+
+\begin{theorem}[Egoroff]
+\label{theorem:egoroff}
+    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 for any $\eps > 0$, there exists $E \in \cm$ such that:
+    \begin{enumerate}
+        \item $f_n \to f$ uniformly on $E$.
+        \item $\mu(X \setminus E) < \eps$
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Since $f_n \to f$ almost everywhere, for any $\eps > 0$, 
+    \[
+    \mu\paren{\limsup_{n \to \infty}\bracs{d(f_n, f) > \eps}} = 0
+    \]
+    
+    By continuity from above (\autoref{proposition:measure-properties}), 
+    \[
+    \lim_{N \to \infty}\paren{\bigcup_{n \ge N}\bracs{d(f_n, f) > \eps}} = 0
+    \]
+
+    For each $k \in \natp$, let $N_k \in \natp$ such that
+    \[
+    \paren{\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}} < 2^{-k}\eps
+    \]
+    then by subadditivity, 
+    \[
+    \mu\paren{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}} \le \eps
+    \]
+    and $f_n \to f$ uniformly on
+    \[
+    \braks{\bigcup_{k \in \natp}\bigcup_{n \ge N_k}\bracs{d(f_n, f) > 1/k}}^c
+    \]
+\end{proof}
+
+
+