Cleaned up convergence in measure.

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

@@ -153,7 +153,7 @@
 \label{theorem:cauchy-in-measure-limit}
     Let $(X, \cm, \mu)$ be a measure space and $(Y, d)$ be a complete metric space, then:
     \begin{enumerate}
-        \item For any $\seq{f_n} \subset L^0(X; Y)$ that is Cauchy in measure, there exists $f \in L^0(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere.
+        \item For any $\seq{f_n} \subset L^0(X; Y)$ that is Cauchy in measure, there exists $f \in L^0(X; Y)$ and a subsequence $\seq{n_k}$ such that $f_{n_k} \to f$ almost everywhere and in measure.
         \item $L^0(X; Y)$ equipped with the uniform structure of convergence in measure is complete. 
     \end{enumerate}
 \end{theorem}
@@ -183,7 +183,15 @@
     \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 $f: X \to Y$.
+    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 almost everywhere to a Borel measurable function $f \in L^0(X; Y)$.
+
+    Finally, for each $K \in \natp$,  
+    \[
+        \mu\bracs{d(f_{n_K}, f) > 2^{-K+1}} \le \sum_{k \ge K}\mu\bracs{d(f_{n_k}, f_{n_K}) > 2^{-k}} \le \sum_{k \ge K}2^{-k}
+    \]
+
+    so $f_{n_k} \to f$ in measure as well. 
+    
 
     (2): Since the uniform structure of convergence in measure on $L^0(X; Y)$ is defined by the \hyperref[Ky Fan metric]{definition:ky-fan}, completeness follows from (1) and \autoref{proposition:complete-metric-space}. 
 \end{proof}