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

src/measure/lebesgue-integral/complex.tex

@@ -98,7 +98,7 @@
 	The construction of the Lebesgue integral using positive/negative and real/imaginary parts, and the deduction of properties from this definition both appear cumbersome. This comes from the lack of usage of its linearity on integrable simple functions. A more functional analytic approach will use this linearity and the density of simple functions to construct the integral. This is the route through which the Bochner integral will be presented.
 \end{remark}
 
-\begin{theorem}[Dominated Convergence Theorem {{\cite[Theorem 2.24]{Folland}}}]
+\begin{theorem}[Dominated Convergence Theorem]
 \label{theorem:dct}
     Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^1(X)$, and $f:X \to \complex$ such that
     \begin{enumerate}
@@ -107,7 +107,7 @@
     \end{enumerate}
     then $\int fd\mu = \limv{n}\int f_n d\mu$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 2.24]{Folland}}}. ]
     By (1) and (2), $f$ is measurable with $\int \abs{f}d\mu \le \int \abs{g}d\mu < \infty$, so $f \in \mathcal{L}^1(X)$. Now, since $g + f, g - f \ge 0$, by \hyperref[Fatou's lemma]{lemma:fatou} and \autoref{proposition:lebesgue-integral-properties},
     \begin{align*}
         \int g d\mu + \int f d\mu &\le \liminf_{n \to \infty}\int g + f_n d\mu = \int g d\mu + \liminf_{n \to \infty}f_n d\mu \\

src/measure/lebesgue-integral/non-negative.tex

@@ -38,7 +38,7 @@
     the two sides are equal.
 \end{proof}
 
-\begin{theorem}[Monotone Convergence Theorem, {{\cite[Theorem 2.14]{Folland}}}]
+\begin{theorem}[Monotone Convergence Theorem]
 \label{theorem:mct}
     Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, and $f \in \mathcal{L}^+(X, \cm)$ such that $f_n \upto f$ pointwise, then
     \[
@@ -46,7 +46,7 @@
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 2.14]{Folland}}}. ]
     By \autoref{definition:lebesgue-non-negative}, $\int f_n d\mu \le \int f d\mu$ for each $n \in \natp$.
 
     Let $\phi \in \Sigma^+(X, \cm)$ with $\phi|_{\bracs{f > 0}} < f|_{\bracs{f > 0}}$ and $\phi \le f$. Since $f_n \upto f$, $\bracs{f_n \ge \phi} \upto X$. In which case, since $A \mapsto \int \one_A \phi d\mu$ is a measure (\autoref{proposition:lebesgue-simple-properties}),
@@ -57,7 +57,7 @@
     by continuity from below (\autoref{proposition:measure-properties}). Therefore $\limv{n}\int f_n d\mu \ge \int f$ by \autoref{lemma:lebesgue-non-negative-strict}.
 \end{proof}
 
-\begin{lemma}[Fatou, {{\cite[Lemma 2.18]{Folland}}}]
+\begin{lemma}[Fatou]
 \label{lemma:fatou}
     Let $(X, \cm, \mu)$ be a measure space, $\seq{f_n} \subset \mathcal{L}^+(X, \cm)$, then
     \[
@@ -65,7 +65,7 @@
     \]
 
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Lemma 2.18]{Folland}}}. ]
     For each $n \in \natp$, $\inf_{k \ge n}f_k \le f_n$. By the monotone convergence theorem,
     \[
     \int \liminf_{n \to \infty}f_n d\mu = \limv{n}\int \inf_{k \ge n}f_k d\mu

src/measure/measure/kolmogorov.tex

@@ -45,7 +45,7 @@
     Thus $\mu_{[n]}(K_n) \ge \eps/2$.
 \end{proof}
 
-\begin{theorem}[Kolmogorov's Extension Theorem, {{\cite[Theorem 1.14]{Baudoin}}}]
+\begin{theorem}[Kolmogorov's Extension Theorem]
 \label{theorem:kolmogorov-extension}
     Let $\seqi{X}$ be topological spaces where for each $J \subset I$ finite,
     \begin{enumerate}
@@ -55,7 +55,7 @@
     \end{enumerate}
     Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then there exists a unique probability measure $\mu: \bigotimes_{i \in I}\cb_{X_i} \to [0, 1]$ such that for any $J \subset I$ finite, $\mu = \mu_J \circ \pi_J^{-1}$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.14]{Baudoin}}}. ]
     Let
     \[
     \alg = \bracs{\pi_J^{-1}(B)| B \in \bigotimes_{j \in J}\cb_{X_j}, J \subset I \text{ finite}}

src/measure/measure/lebesgue-stieltjes.tex

@@ -34,7 +34,7 @@
     (2): By \autoref{proposition:elementary-family-algebra}, $\alg$ is a ring.
 \end{proof}
 
-\begin{definition}[Lebesgue-Stieltjes Measure, {{\cite[Theorem 1.16]{Folland}}}]
+\begin{definition}[Lebesgue-Stieltjes Measure]
 \label{definition:lebesgue-stieltjes-measure}
     Let $\mu: \cb_\real \to [0, \infty]$ be a Borel measure on $\real$ such that for any $K \subset \real$ compact, $\mu(K) < \infty$, then
     \begin{enumerate}
@@ -47,7 +47,7 @@
     \end{enumerate}
     Conversely, if $F: \real \to \real$ is a Stieltjes function, then there exists a unique Borel measure $\mu_F: \cb_\real \to [0, \infty]$ satisfying (1), and $\mu_F$ is the \textbf{Lebesgue-Stieltjes measure} associated with $F$.
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]
     (1): Let
     \[
     F: \real \to \real \quad x \mapsto \begin{cases}

src/measure/measure/outer.tex

@@ -44,7 +44,7 @@
 
 \end{definition}
 
-\begin{theorem}[Carathéodory, {{\cite[Theorem 1.11]{Folland}}}]
+\begin{theorem}[Carathéodory]
 \label{theorem:caratheodory}
     Let $X$ be a set, $\mu^*: 2^X \to [0, \infty]$ be an outer measure, and $\cm \subset 2^X$ be the collection of all $\mu^*$-measurable sets, then:
     \begin{enumerate}
@@ -57,7 +57,7 @@
         \item $(X, \cm, \mu^*|_\cm)$ is a complete measure space.
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.11]{Folland}}}. ]
     (1): Let $N \in \nat$, then
     \[
     \mu^*\paren{F \cap \bigsqcup_{n \in \natp}E_n} \ge \mu^*\paren{F \cap \bigsqcup_{n =1}^N E_n} = \sum_{n = 1}^N \mu^*(F \cap E_n)
@@ -103,7 +103,7 @@
     \end{enumerate}
 \end{definition}
 
-\begin{theorem}[Carathéodory's Extension Theorem, {{\cite[Theorem 1.14]{Folland}}}]
+\begin{theorem}[Carathéodory's Extension Theorem]
 \label{theorem:caratheodory-extension}
     Let $X$ be a set, $\alg \subset 2^X$ be a ring, and $\mu_0: \alg \to [0, \infty]$ be a premeasure, then there exists a measure space $(X, \cm, \mu)$ such that
     \begin{enumerate}
@@ -118,7 +118,7 @@
         \end{enumerate}
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.14]{Folland}}}. ]
     Let
     \[
     \mu^*: 2^X \to [0, \infty] \quad E \mapsto \inf\bracs{\sum_{n \in \natp}\mu_0(E_n) \bigg | \seq{E_n} \subset \alg, \bigcup_{n \in \natp}E_n \supset E}

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)}

src/measure/vector/complex.tex

@@ -73,7 +73,7 @@
 \end{proof}
 
 
-\begin{theorem}[Hahn Decomposition, {{\cite[Theorem 3.3]{Folland}}}]
+\begin{theorem}[Hahn Decomposition]
 \label{theorem:hahn-decomposition}
     Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then there exists $P, N \in \cm$ such that:
     \begin{enumerate}
@@ -85,7 +85,7 @@
 
     The disjoint union $X = P \sqcup N$ is the \textbf{Hahn decomposition} of $\mu$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 3.3]{Folland}}}. ]
     By flipping the sign of $\mu$, assume without loss of generality that $\mu < \infty$. 
 
     (1): Let $M = \sup\bracs{\mu(P)|P \in \cm \text{ positive}}$, then there exists positive sets $\seq{P_n} \subset \cm$ such that $\mu(P_n) \upto M$. Let $P = \bigcup_{n \in \natp}P_n$, then $P$ is positive by \autoref{lemma:positive-sets-properties}, and $\sup_{n \in \natp}\mu(P_n) \le \mu(P) \le M$.