Added monotone convergence for LSC functions.

src/measure/radon/radon.tex

@@ -202,4 +202,51 @@
     (2, unbounded): By (1) applied to $g$, there exists $\phi \in C_c(X; \complex)$ such that $\mu\bracs{\phi \ne g} < \eps/2$, and $\mu(\bracs{\phi \ne f}) < \eps$.
 \end{proof}
 
+\begin{proposition}[Monotone Convergence Theorem for Lower Semicontinuous Functions, {{\cite[Proposition 7.12]{Folland}}}]
+\label{proposition:mct-radon}
+    Let $X$ be a LCH space, $\net{f}$ and $f: X \to [0, \infty]$ be non-negative lower semicontinuous functions such that $f_\alpha \upto f$, then for any Radon measure $\mu$ on $X$, 
+    \[
+    \int f d\mu = \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+\end{proposition}
+\begin{proof}
+    Assume without loss of generality that $\int f d\mu < \infty$. By \autoref{proposition:semicontinuous-properties}, $f$ is Borel measurable, so $f \ge f_\alpha$ for all $\alpha \in A$ implies that
+    \[
+    \int f d\mu \ge \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+
+    Let $\phi \in \Sigma^+(X, \cm)$ with $0 \le \phi < f$ and $\beta < \int \phi d\mu$. Let $\seqf{a_j} \subset (0, \infty)$ and $\seqf{E_j} \subset \cb_X$ such that $\phi = \sum_{j = 1}^n a_j \one_{E_j}$. Since $\int \phi d\mu < \infty$, for each $1 \le j \le n$, $\mu(E_j) < \infty$ by \hyperref[Markov's Inequality]{theorem:markov-inequality}. By \autoref{proposition:radon-regular-sigma-finite}, there exists compact sets $\seqf{K_j} \subset 2^X$ such that $K_j \subset E_j$ for each $1 \le j \le n$ and $\int \sum_{j = 1}^n a_j \one_{K_j}d\mu > \beta$. 
+
+    Let $\psi = \sum_{j = 1}^n a_j \one_{K_j}$ and $K = \bigcup_{j = 1}^n K_j$, then $K$ is compact by \autoref{proposition:compact-extensions} and $-\psi$ is lower semicontinuous by \autoref{proposition:semicontinuous-properties}.
+    
+    For any $x_0 \in K$, since $f_\alpha \upto f$ and $\psi \le \phi < f$, there exists $\alpha(x_0) \in A$ such that $f_{\alpha(x_0)}(x_0) > \psi(x_0)$. By \autoref{proposition:semicontinuous-properties}, $f_{\alpha(x_0)} - \psi$ is lower semicontinuous, so
+    \[
+    \bracsn{\bracsn{f_{\alpha(x_0)} > \psi}|x_0 \in K}
+    \]
+
+    is an open cover of $K$. Let $\bracs{x_j}_1^N \subset K$ such that $K \subset \bigcup_{j = 1}^N \bracsn{f_{\alpha(x_j)} > \psi}$. Since $f_\alpha \upto f$, there exists $\alpha_0 \in A$ such that
+    \[
+    f_{\alpha_0} \ge \max_{1 \le j \le N}f_{\alpha(x_j)} \ge \psi
+    \]
+    
+    so
+    \[
+    \int f_{\alpha_0} d\mu \ge \int \psi d\mu \ge \beta
+    \]
+
+    As such an $\alpha_0 \in A$ exists for all $\beta < \int \phi d\mu$, 
+    \[
+    \int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+
+    Since the above holds for all $\phi \in \Sigma^+(X, \cm)$ with $0 \le \phi < f$, 
+    \[
+    \int f d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+
+    by \autoref{lemma:lebesgue-non-negative-strict}.
+
+\end{proof}
+
+