Added monotone convergence for LSC functions.

src/fa/lc/hahn-banach.tex

@@ -132,17 +132,31 @@
     \begin{enumerate}
         \item For any subspace $M \subset E$, $x \in M \setminus E$, and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ such that
         \begin{enumerate}
-            \item $\phi \le \rho$.
+            \item $|\phi| \le \rho$.
             \item $\dpb{x, \phi}{E} = \inf_{y \in M}\rho(x + y)$.
         \end{enumerate}
-        \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $\phi \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
+        \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $|\phi| \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
         \item If $E$ is Hausdorff, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
+    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $|\phi| \le \rho_M \le \rho$.
 
     (2): By (1) applied to $M = \bracs{0}$.
 
     (3): By (2) applied to $x - y$.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:seminorm-lsc}
+    Let $E$ be a locally convex space and $\rho: E \to [0, \infty)$ be a continuous seminorm, then $\rho: E_w \to [0, \infty)$ is lower semicontinuous and Borel measurable. 
+\end{proposition}
+\begin{proof}
+    Let $x \in E$, then there exists $\phi_x \in E^*$ such that $\dpn{x, \phi_x}{E} = \rho(x)$ and $|\phi_x| \le \rho$. Thus
+    \[
+    \rho(x) = \sup_{y \in E}\dpn{x, \phi_y}{E}
+    \]
+
+    is lower semicontinuous and Borel measurable by \autoref{proposition:semicontinuous-properties}. 
+\end{proof}
+

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}
+
+
 

src/topology/main/index.tex

@@ -22,4 +22,5 @@
 \input{./support.tex}
 \input{./lch.tex}
 \input{./c0.tex}
+\input{./semicontinuity.tex}
 \input{./baire.tex}

src/topology/main/semicontinuity.tex

@@ -0,0 +1,64 @@
+\section{Semicontinuity}
+\label{section:semicontinuity}
+
+\begin{definition}[Semicontinuous]
+\label{definition:semicontinuous}
+    Let $X$ be a topological space, $f: X \to (-\infty, \infty]$, and $g: X \to [-\infty, \infty)$, then $f$ is \textbf{lower semicontinuous} if for each $a \in \real$, $\bracs{f > \alpha}$ is open, and $g$ is \textbf{upper semicontinuous} if for each $a \in \real$, $\bracs{f < \alpha}$ is open.
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 7.11]{Folland}}}]
+\label{proposition:semicontinuous-properties}
+    Let $X$ be a topological space, then
+    \begin{enumerate}
+        \item For any $U \subset X$ open, $\one_U$ is lower semicontinuous.
+        \item For any $f: X \to (-\infty, \infty]$ lower semicontinuous and $\alpha \ge 0$, $\alpha f$ is lower semicontinuous. 
+        \item For any $f, g: X \to (-\infty, \infty]$ lower semicontinuous, $f + g$ is lower semicontinuous.
+        \item For any collection $\mathcal{F} \subset (-\infty, \infty]^X$ of lower semicontinuous functions, $F = \sup_{f \in F}f$ is lower semicontinuous. 
+        \item For any $f: X \to (-\infty, \infty]$ lower semicontinuous, $f$ is Borel measurable. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For any $\alpha \in \real$, 
+    \[
+    \bracs{f > \alpha} = \begin{cases}
+        \emptyset &\alpha \ge 1 \\
+        U &\alpha \in [0, 1) \\
+        X &\alpha < 0
+    \end{cases}
+    \]
+
+    (2): If $\alpha = 0$, then $\alpha f = 0$ is continuous. If $\alpha > 0$, then for any $a \in \real$, $\bracs{\alpha f > a} = \bracs{f > a/\alpha}$ is open.
+
+    (3): Let $a \in \real$, $x_0 \in \bracs{f + g > a}$, and $\eps \in (0, ((f + g)(x_0) - a)/2)$, then
+    \[
+    \bracs{f + g > a} \supset \bracs{f > f(x_0) - \eps} \cap \bracs{g > g(x_0) - \eps} \in \cn^o(x_0)
+    \]
+
+    As this holds for all $x_0 \in \bracs{f + g > a}$, $\bracs{f + g > a}$ is open by \autoref{lemma:openneighbourhood}.
+
+    (4): For any $a \in \real$,
+    \[
+    \bracs{F > a} = \bigcup_{f \in \mathcal{F}} \bracs{f > a}
+    \]
+
+    is open. 
+
+    (5): By \autoref{proposition:borel-sigma-real-generators}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:semicontinuous-lch}
+    Let $X$ be a LCH space and $f: X \to [0, \infty]$ be lower semicontinuous, then
+    \[
+    f = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi
+    \]
+\end{proposition}
+\begin{proof}
+    Let $x \in X$ such that $f(x) > 0$ and $a \in (0, f(x))$, then $\bracs{f > a}$ is open. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\phi \in C_c(X; [0, a])$ such that $\phi(x) = a$ and $\supp{\phi} \subset \bracs{f > a}$. As this holds for all $a \in (0, f(x))$, 
+    \[
+    f(x) = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi(x)
+    \]
+\end{proof}
+
+
+