Adjusted citation formats. Moved citation off of named theorems if possible.

src/measure/radon/c0.tex

@@ -89,7 +89,7 @@
 \end{proof}
 
 
-\begin{theorem}[Singer's Representation Theorem, {{\cite{HensgenSinger}}}]
+\begin{theorem}[Singer's Representation Theorem]
 \label{theorem:singer-representation}
     Let $X$ be an LCH space and $E$ be a normed space over $K \in \RC$. For each $\mu \in M_R(X; E^*)$, let
     \[
@@ -103,7 +103,7 @@
 
     is an isometric isomorphism. 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite{HensgenSinger}}}. ]
     (Isometric): Let $\mu \in M_R(X; E^*)$, then for any $f \in C_0(X; E)$, 
     \[
     |\dpn{f, I_\mu}{C_0(X; E)}| \le \int \norm{f}_E d|\mu| \le \norm{f}_u \cdot \norm{\mu}_{\text{var}}

src/measure/radon/radon.tex

@@ -182,7 +182,7 @@
     \]
 \end{proof}
 
-\begin{theorem}[Lusin, {{\cite[Theorem 7.10]{Folland}}}]
+\begin{theorem}[Lusin]
 \label{theorem:lusin}
     Let $X$ be a LCH space, $\mu$ be a Radon measure on $X$, $E$ be a normed vector space, and $f: X \to E$ be a measurable function with $\mu\bracs{f \ne 0} < \infty$, then for any $\eps > 0$,
     \begin{enumerate}
@@ -191,7 +191,7 @@
         \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}
+\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. 
@@ -223,14 +223,14 @@
     (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, {{\cite[Proposition 7.12]{Folland}}}]
+\begin{proposition}[Monotone Convergence Theorem for Lower Semicontinuous Functions]
 \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}
+\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

src/measure/radon/riesz.tex

@@ -26,7 +26,7 @@
 \end{proof}
 
 
-\begin{theorem}[Riesz Representation Theorem, {{\cite[Theorem 7.2]{Folland}}}]
+\begin{theorem}[Riesz Representation Theorem]
 \label{theorem:riesz-radon}
     Let $(X, \topo)$ be a LCH space and $I \in \hom(C_c(X; \real); \real)$ be a positive linear functional, then there exists a Borel measure $\mu: \cb_X \to [0, \infty]$ such that:
     \begin{enumerate}
@@ -37,7 +37,7 @@
         \item[(U)] If $\nu: \cb_X \to [0, \infty]$ is a Borel measure that satisfies (3) and (4), then $\mu = \nu$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 7.2]{Folland}}}. ]
     (1): For any $U \in \topo$ and $f \in C_c(X; \real)$, denote $f \prec U$ if $f \in C_c(X; [0, 1])$ and $\supp{f} \subset U$. Let
     \[
     \mu_0: \topo \to [0, \infty] \quad U \mapsto \sup_{f \prec U}\dpb{f, I}{C_c(X; \real)}