Added MCT for convergence in measure.

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

@@ -7,11 +7,22 @@
     \[
     \lim_{g, \fF}\mu(\bracs{d(f, g) > \eps}) = 0
     \]
+
+    and $\fF \to f$ \textbf{locally in measure} if $\fF \to f$ in measure on every set of finite measure. 
+
+    Alternatively, if $\net{f}$ is a net of $(\cm, \cb_Y)$-measurable functions, then $f_\alpha \to f$ \textbf{in measure} if for each $\eps > 0$, 
+    \[
+    \lim_{\alpha \in A}\mu(\bracs{d(f, f_\alpha) > \eps}) = 0
+    \]
+
+    and $f_\alpha \to f$ \textbf{locally in measure} if $f_\alpha \to f$ in measure on every set of finite measure. 
 \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 $\fF$ be a filter of $(\cm, \cb_Y)$-measurable functions, then $\fF$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $A \in \fF$ such that $\mu(\bracs{d(f, g) > \delta}) < \eps$ for all $f, g \in A$. 
+
+    Alternatively, if $\net{f}$ is a net of $(\cm, \cb_Y)$-measurable functions, then $\net{f}$ is \textbf{Cauchy in measure} if for every $\eps, \delta > 0$, there exists $\alpha_0 \in A$ such that for each $\alpha, \beta \in A$ with $\alpha, \beta \ge \alpha_0$, $\mu(\bracs{d(f_\alpha, f_\beta) > \delta}) < \eps$. 
 \end{definition}
 
 \begin{lemma}
@@ -76,5 +87,34 @@
     (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}
 
+\begin{theorem}[Monotone Convergence Theorem (In Measure)]
+\label{theorem:mct-measure}
+    Let $(X, \cm, \mu)$ be a semifinite measure space, $\net{f} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that
+    \begin{enumerate}[label=(\alph*)]
+        \item For each $x \in X$, $f_\alpha(x) \upto f(x)$. 
+        \item $f_\alpha \to f$ locally in measure. 
+    \end{enumerate}
 
+    then
+    \[
+    \lim_{\alpha \in A}\int f_\alpha d\mu = \int f d\mu
+    \]
+\end{theorem}
+\begin{proof}
+    By \autoref{definition:lebesgue-non-negative}, $\int f_\alpha d\mu \le \int f d\mu$ for each $\alpha \in A$.
 
+    Assume without loss of generality that $f \ne 0$. Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $\mu$ is semifinite, assume further without loss of generality that $\phi \in L^1(X, \cm; \real)$ as well. 
+    
+    Let $\eps > 0$ and $\lambda \in (0, 1)$, then $\delta = \min_{y \in \phi(X) \setminus \bracs{0}}(1 - \lambda)y > 0$. By (b), there exists $\alpha \in A$ such that
+    \[
+    \mu\bracs{f_\alpha + \delta < f} < \frac{\epsilon}{\norm{\phi}_u}
+    \]
+
+    In which case, 
+    \begin{align*}
+        \int f_\alpha d\mu &\ge \int_{\bracs{f_\alpha + \delta \ge f}} f_\alpha d\mu \ge \int_{\bracs{f_\alpha + \delta \ge f}}\lambda\phi d\mu \\
+        &\ge \lambda\int \phi d\mu - \frac{\eps \norm{\phi}_u}{\norm{\phi}_u} = \lambda \int \phi d\mu - \eps
+    \end{align*}
+
+    As the above holds for all $\eps > 0$, $\lambda \in (0, 1)$, $\sup_{\alpha \in A}\int f_\alpha d\mu \ge \int \phi d\mu$. Therefore $\sup_{\alpha \in A}\int f_\alpha d\mu \ge \int f d\mu$ by \autoref{lemma:lebesgue-non-negative-strict}. 
+\end{proof}