Adjusted some things in Radon meaasures.

src/measure/radon/radon.tex

@@ -51,7 +51,7 @@
 
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 7.5]{Folland}}}]
+\begin{proposition}
 \label{proposition:radon-regular-sigma-finite}
     Let $X$ be a LCH space and $\mu: \cb_X \to [0, \infty]$ be a Radon measure, then
     \begin{enumerate}
@@ -59,7 +59,7 @@
         \item If $X$ is $\sigma$-compact or $\mu$ is $\sigma$-finite, then $\mu$ is regular.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.5]{Folland}}}. ]
     (1): Let $E \in \cb_X$ with $\mu(E) < \infty$ and $\eps > 0$, then there exists
     \begin{itemize}
     \item $U \in \cn^o(E)$ with $\mu(U \setminus E) < \eps/2$.
@@ -85,7 +85,7 @@
 
 \end{proof}
 
-\begin{proposition}[{{\cite[Proposition 7.7]{Folland}}}]
+\begin{proposition}
 \label{proposition:radon-measurable-description}
     Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a $\sigma$-finite Radon measure, and $E \in \cb_X$, then
     \begin{enumerate}
@@ -93,7 +93,7 @@
         \item There exists a $F_\sigma$ set $A$ and a $G_\delta$ set $B$ such that $A \subset E \subset B$ and $\mu(B \setminus A) = 0$.
     \end{enumerate}
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof, {{\cite[Proposition 7.7]{Folland}}}. ]
     (1): Let $\seq{E_n} \subset \cb_X$ such that $\mu(E_n) < \infty$. By outer regularity, there exists $\seq{U_n}$ open such that $U_n \in \cn^o(E_n)$ and $\mu(U_n) < \mu(E_n) + \eps/2^n$ for all $n \in \natp$. In which case,
     \[
     \mu\paren{\bigcup_{n \in \natp}U_n \setminus E} \le \sum_{n \in \natp}\mu(U_n \setminus E_n) < \eps
@@ -194,7 +194,7 @@
         \item If $E = \complex$ and $f$ is bounded, then $\phi$ can be taken such that $\norm{\phi}_u \le \norm{f}_u$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}[Proof {{\cite[Theorem 7.10]{Folland}}}. ]
+\begin{proof}[Proof, {{\cite[Theorem 7.10]{Folland}}}. ]
     First assume that $f$ is bounded.
 
     (1, bounded): If $f$ is bounded, then $f \in L^1(X; E)$. By \autoref{proposition:radon-cc-dense}, there exists $\seq{\phi_n} \subset C_c(X)$ such that $\phi_n \to f$ in $L^1(\mu)$. Since $\phi_n \to f$ in $L^1(\mu)$, $\phi_n \to f$ in measure by \autoref{proposition:lp-in-measure}. By taking a subsequence using \ref{proposition:cauchy-in-measure-limit}, assume without loss of generality that $\phi_n \to f$ almost everywhere. 
@@ -226,14 +226,14 @@
     (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]
+\begin{proposition}[Monotone Convergence Theorem (LSC)]
 \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}[Proof {{\cite[Proposition 7.12]{Folland}}}. ]
+\begin{proof}[Proof, {{\cite[Proposition 7.12]{Folland}}}. ]
     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