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

src/dg/derivative/higher.tex

@@ -20,11 +20,11 @@
     is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
 \end{definition}
 
-\begin{theorem}[{{\cite[Theorem 5.1.1]{Cartan}}}]
+\begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric-frechet}
     Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x \in U$, then $D^nf(x) \in L^n(E; F)$ is symmetric.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 5.1.1]{Cartan}}}. ]
     First suppose that $n = 2$. Let $r > 0$ such that $B(x, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
     \[
     A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
@@ -85,11 +85,11 @@
     Since any element $\sigma \in S_n$ that does not fix $x_1$ is the composition of the transposition $(12)$ and an element that fixes $x_1$, $Df(x)$ is symmetric.
 \end{proof}
 
-\begin{theorem}[{{\cite[Proposition 4.5.14]{Bogachev}}}]
+\begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric}
     Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed system that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
     Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
     \[
     D_{B(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)

src/dg/derivative/sets.tex

@@ -46,7 +46,7 @@
 \end{definition}
 
 
-\begin{proposition}[{{\cite[Proposition 4.5.2]{Bogachev}}}]
+\begin{proposition}[Chain Rule]
 \label{proposition:chain-rule-sets}
     Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
     \begin{enumerate}
@@ -59,7 +59,7 @@
     \]
 
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 4.5.2]{Bogachev}}}. ]
     Since $g$ is $\tau$-differentiable at $f(x_0)$, there exists $s \in \mathcal{R}_\tau(F; G)$ such that
     \[
     g(f(x_0) + h) = g \circ f (x_0) + D_\tau g(f(x_0))h + s(h)

src/dg/derivative/taylor.tex

@@ -1,7 +1,7 @@
 \section{Taylor's Formula}
 \label{section:taylor}
 
-\begin{theorem}[Taylor's Formula, Lagrange Remainder {{\cite[Theorem 4.7.1]{Bogachev}}}]
+\begin{theorem}[Taylor's Formula, Lagrange Remainder]
 \label{theorem:taylor-lagrange}
     Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, $n \in \natp$, and $f \in C^{n}([a, b]; E)$ be $(n+1)$-fold differentiable on $[a, b] \setminus N$, then
     \[
@@ -14,7 +14,7 @@
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 4.7.1]{Bogachev}}}. ]
     If $n = 0$, then the theorem is the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
 
     Suppose inductively that the theorem holds for $n$. Let
@@ -60,7 +60,7 @@
 
 \end{proof}
 
-\begin{theorem}[Taylor's Formula, Peano Remainder {{\cite[Theorem 4.7.3]{Bogachev}}}]
+\begin{theorem}[Taylor's Formula, Peano Remainder]
 \label{theorem:taylor-peano}
     Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
     \[
@@ -68,7 +68,7 @@
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 4.7.3]{Bogachev}}}. ]
     Let
     \[
     r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})

src/fa/lc/hahn-banach.tex

@@ -35,7 +35,7 @@
     \end{align*}
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, Analytic Form {{\cite[Theorem 5.6, 5.7]{Folland}}}]
+\begin{theorem}[Hahn-Banach, Analytic Form]
 \label{theorem:hahn-banach}
     Let $E$ be a vector space over $K \in \RC$, $\rho: E \to \real$ be a sublinear functional, and $F \subsetneq E$ be a subspace, then
     \begin{enumerate}
@@ -43,7 +43,7 @@
         \item If $\rho$ is a seminorm, then for any $\phi \in \hom(F; \complex)$ with $\abs{\phi} \le \rho|_F$, there exists $\Phi \in \hom(E; \complex)$ such that $\abs{\Phi} \le \rho$ and $\Phi|_F = \phi$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 5.6, 5.7]{Folland}}}. ]
     (1): Let $x_0 \in E \setminus F$, then by \autoref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
     \[
     \phi_{x_0, \lambda}: F + \real x_0 \to \real \quad (x + tx_0) \mapsto x + \lambda t
@@ -96,21 +96,21 @@
     Finally, let $y \in A$, then $\dpb{y, \phi}{E} \le [y]_A < 1 = \dpb{x, \phi}{E}$.
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, First Geometric Form {{\cite[Theorem 1.6]{Brezis}}}]
+\begin{theorem}[Hahn-Banach, First Geometric Form]
 \label{theorem:hahn-banach-geometric-1}
     Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.6]{Brezis}}}. ]
     Let $C = A - B$, then $C \subset E$ is an open convex set by \autoref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
 
     By \autoref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
 \end{proof}
 
-\begin{theorem}[Hahn-Banach, Second Geometric Form {{\cite[Theorem 1.7]{Brezis}}}]
+\begin{theorem}[Hahn-Banach, Second Geometric Form]
 \label{theorem:hahn-banach-geometric-2}
     Let $E$ be a locally convex space over $\real$, $A, B \subset E$ be non-empty convex sets such that $A \cap B = \emptyset$, $A$ is closed, and $B$ is compact, then there exists $\phi \in E^* \setminus \bracs{0}$, $\alpha \in \real$, and $r > 0$ with $A \subset \bracs{\phi \le \alpha - r}$ and $B \subset \bracs{\phi \ge \alpha + r}$.
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.7]{Brezis}}}. ]
     Let $C = A - B$, then $C$ is closed by \autoref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
 
     By \autoref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,

src/fa/lc/tensor.tex

@@ -54,7 +54,7 @@
 
 
 
-\begin{definition}[Cross Seminorm, {{\cite[III.6.3]{SchaeferWolff}}}]
+\begin{definition}[Cross Seminorm]
 \label{definition:cross-seminorm}
     Let $E, F$ be locally convex spaces over $K \in \RC$. For any convex circled sets $U \in \cn_E(0)$ and $V \in \cn_F(0)$, let $p: E \to [0, \infty)$ and $q: F \to [0, \infty)$ be their \hyperref[gauges]{definition:gauge}. For any $z \in E \otimes_\pi F$, let
     \[
@@ -75,7 +75,7 @@
     \end{enumerate}
     
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[III.6.3]{SchaeferWolff}}}. ]
     (1): Let $\lambda \in K$, then for any $\seqf{(x_j,y_j)} \subset E \times F$, 
     \[
     |\lambda| \sum_{j = 1}^n p(x_j)q(y_j) = \sum_{j = 1}^n p(\lambda x_j)q(y_j)

src/fa/rs/rs.tex

@@ -27,7 +27,7 @@
     The set $RS([a, b], G)$ is the vector space of all \textbf{Riemann-Stieltjes integrable functions} with respect to $G$.
 \end{definition}
 
-\begin{lemma}[Summation by Parts, {{\cite[Proposition 1.4]{Lang}}}]
+\begin{lemma}[Summation by Parts]
 \label{lemma:sum-by-parts}
     Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $f: [a, b] \to E_1$, $G: [a, b] \to E_2$, and $(P, c) \in \scp_t([a, b])$, then
     \[
@@ -36,7 +36,7 @@
 
     where $P' = \seqfz[n+1]{y_j} = [a, c_1, \cdots, c_n, b]$ and $c' = \seqf[n+1]{d_j} = [x_0, \cdots, x_n]$.
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 1.4]{Lang}}}. ]
     Denote $c_0 = a$ and $c_{n+1} = b$, then
     \begin{align*}
         S(P, c, f, G) &= \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]

src/fa/tvs/complete-metric.tex

@@ -15,7 +15,7 @@
     Let $x = x_{n_1} + \limv{N}\sum_{k = 1}^N (x_{n_{k+1}} - x_{n_k})$, then $x = \limv{k}x_{n_k} \in E$. Since $\seq{x_n}$ is a Cauchy sequence that admits a convergent subsequence, it is convergent.
 \end{proof}
 
-\begin{theorem}[Successive Approximations {{\cite[Section III.2]{SchaeferWolff}}}]
+\begin{theorem}[Successive Approximations]
 \label{theorem:successive-approximations}
     Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorm $\rho$ and $\eta$, respectively. Let $T \in L(E; F)$, $r > 0$, $\gamma \in (0, 1)$, and $C \ge 0$. Suppose that for every $y \in B_F(0, r)$, there exists $x \in E$ such that:
     \begin{enumerate}
@@ -33,7 +33,7 @@
     \]
 
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Section III.2]{SchaeferWolff}}}. ]
     Let $y_0 = y$ and $x_0 = 0$. Let $N \in \natz$ and suppose inductively that $\seqf[N]{x_n} \subset E$ has been constructed such that:
     \begin{enumerate}
         \item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.

src/fa/tvs/dual.tex

@@ -1,7 +1,7 @@
 \section{The Dual Space}
 \label{section:tvs-dual}
 
-\begin{proposition}[Polarisation, {{\cite[Proposition 5.5]{Folland}}}]
+\begin{proposition}[Polarisation]
 \label{proposition:polarisation-linear}
     Let $E$ be a vector space over $\complex$, $\phi \in \hom(E; \complex)$, and $u = \re{\phi}$, then
     \begin{enumerate}
@@ -10,7 +10,7 @@
     \end{enumerate}
     Conversely, if $u \in \hom(E; \real)$ and $\phi \in \hom(E; \complex)$ is defined by $\dpb{x, \phi}{E} = \dpb{x, u}{E} - i \dpb{ix, u}{E}$ for all $x \in E$, then $f \in \hom(E; \complex)$.
 \end{proposition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 5.5]{Folland}}}. ]
     (1): Given that $\phi \in \hom(E; \real)$, $u \in \hom(E; \real)$. For any $x \in E$,
     \[
     \im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}

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 $\sup_{n \in \natp}\mu(P_n) \le \mu(P) \le M$.

src/topology/main/lch.tex

@@ -35,11 +35,11 @@
     so $V = \bigcup_{j = 1}^n V_{x_j} \in \cn^o(K)$ is precompact.
 \end{proof}
 
-\begin{lemma}[Urysohn's Lemma (LCH), {{\cite[Lemma 4.32]{Folland}}}]
+\begin{lemma}[Urysohn's Lemma (LCH)]
 \label{lemma:lch-urysohn}
     Let $X$ be a LCH space, $K \subset X$ be compact, and $U \in \cn(K)$, then there exists $f \in C_c(X; [0, 1])$ such that $\supp{f} \subset U$.
 \end{lemma}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Lemma 4.32]{Folland}}}. ]
     By \autoref{lemma:lch-compact-neighbour}, there exists $V, W \in \cn^o(K)$ precompact such that
     \[
     K \subset V \subset \ol{V} \subset W \subset \ol{W} \subset U

src/topology/main/regular.tex

@@ -1,7 +1,7 @@
 \section{Regular Spaces}
 \label{section:regularspaces}
 
-\begin{definition}[Regular Space, {{\cite[Proposition 1.4.11]{Bourbaki}}}]
+\begin{definition}[Regular Space]
 \label{definition:regular}
     Let $X$ be a topological space, then the following are equivalent:
     \begin{enumerate}
@@ -10,7 +10,7 @@
     \end{enumerate}
     If $X$ is a T1 space such that the above holds, then $X$ is \textbf{regular}.
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 1.4.11]{Bourbaki}}}. ]
     $(1) \Rightarrow (2)$: Let $U \in \cn^o(x)$, then $U^c$ is closed with $x \not\in U^c$ by (V1). Thus there exists $V \in \cn^o(U^c)$ such that $x \not\in V$, so $V^c \in \cn(x)$ is closed with $V^c \subset U$.
 
     $(2) \Rightarrow (1)$: Since $X$ is Hausdorff, $X$ is T1 as well. Let $x \in X$ and $A \subset X$ closed such that $x \not\in A$, then $A^c \in \cn(x)$. So there exists $K \in \cn(x)$ closed such that $x \in K \subset A^c$. Thus $K \in \cn(x)$ and $K^c \in \cn(A)$ are the desired neighbourhoods.

src/topology/uniform/complete.tex

@@ -48,7 +48,7 @@
 \end{proof}
 
 
-\begin{theorem}[Extension of Uniformly Continuous Functions, {{\cite[Theorem 1.3.2]{Bourbaki}}}]
+\begin{theorem}[Extension of Cauchy Continuous Functions]
 \label{theorem:uniform-continuous-extension}
     Let $(X, \fU)$ be a uniform space, $(Y, \fV)$ be a complete Hausdorff uniform space, $A \subset Y$ be a dense subset, and $f \in C(A; Y)$ be Cauchy continuous, then:
     \begin{enumerate}
@@ -56,7 +56,7 @@
         \item If $f \in UC(A; Y)$, then $F \in UC(X; Y)$.
     \end{enumerate}
 \end{theorem}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 1.3.2]{Bourbaki}}}. ]
     (1): By \autoref{proposition:imagecauchy}, $f(\bracs{U \cap A| U \in \cn(x)})$ is a Cauchy filter base for all $x \in X$. Thus the existence and uniqueness of the extension follows from \autoref{proposition:uniformextension}.
 
     (2): Let $V \in \fV$. Using \autoref{proposition:goodentourages}, assume without loss of generality that $V$ is symmetric and closed. Since $f \in UC(A; Y)$, there exists $U \in \fU$ such that $(f(x), f(y)) \in V$ whenever $(x, y) \in U$. Using \autoref{proposition:subspace-entourage}, assume without loss of generality that $U = \overline{U} = \ol{U \cap (A \times A)}$. In which case, since $F$ is continuous,

src/topology/uniform/completion.tex

@@ -2,7 +2,7 @@
 \label{section:completion}
 
 
-\begin{definition}[Hausdorff Completion, {{\cite[Theorem 2.3.3]{Bourbaki}}}]
+\begin{definition}[Hausdorff Completion]
 \label{definition:hausdorff-completion}
     Let $(X, \fU)$ be a uniform space, then there exists a $(\wh X, \iota)$ such that:
     \begin{enumerate}
@@ -35,7 +35,7 @@
 
     The pair $(\wh X, \iota)$ is the \textbf{Hausdorff completion} of $X$.
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Theorem 2.3.3]{Bourbaki}}}. ]
     (1, Uniform), (4): Let $\wh X$ be the set of all minimal Cauchy filters on $X$. For each $V \in \fU$, let
     \[
     \wh V = \bracsn{(\fF, \mathfrak{G}) \in \wh X| \exists U \in \fF \cap \mathfrak{G}: U \times U \subset V}
@@ -89,7 +89,7 @@
     Since the mapping $(\iota \times \iota)^{-1}$ is a bijection between two bases of uniformities of $X$ and $\wh X$, it is sufficient to show that $\iota$ is injective. To this end, observe that for any $x, y \in X$, $\cn(x) = \cn(y)$ if and only if $x = y$ by (4) of \autoref{definition:hausdorff}.
 \end{proof}
 
-\begin{definition}[Associated Hausdorff Uniform Space, {{\cite[Proposition 2.8.16]{Bourbaki}}}]
+\begin{definition}[Associated Hausdorff Uniform Space]
 \label{definition:associated-hausdorff}
     Let $(X, \fU)$ be a uniform space, then there exists $(X', i)$ such that:
     \begin{enumerate}
@@ -107,7 +107,7 @@
     \end{enumerate}
     known as the \textbf{Hausdorff uniform space associated with} $(X, \fU)$.
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 2.8.16]{Bourbaki}}}. ]
     Let $(\wh X, \iota)$ be the Hausdorff completion of $X$, $X' = \iota(X)$, and $i = \iota$, then $(X', i)$ satisfies (1) and (2).
 
     (U): Let $(\wh Y, \iota)$ be the Hausdorff completion of $Y$. Using \autoref{lemma:completion-of-hausdorff}, identify $Y$ as a subspace of $\wh Y$. By (U) of the Hausdorff completion, there exists a unique $\ol F \in UC(\wh X; \wh Y)$ such that the following diagram commutes:

src/topology/uniform/definition.tex

@@ -113,7 +113,7 @@
     so $\fb_S$ is a fundamental system of entourages.
 \end{proof}
 
-\begin{definition}[Topology of a Uniform Space, {{\cite[Proposition 2.1.2]{Bourbaki}}}]
+\begin{definition}[Topology of a Uniform Space]
 \label{definition:uniformtopology}
     Let $(X, \fU)$ be a uniform space and
     \[
@@ -122,7 +122,7 @@
 
     then there exists a unique topology $\topo \subset 2^X$ such that $\cn_\topo = \cn$, known as the \textbf{topology induced by the uniform structure $\fU$}.
 \end{definition}
-\begin{proof}
+\begin{proof}[Proof {{\cite[Proposition 2.1.2]{Bourbaki}}}. ]
 Using \autoref{proposition:neighbourhoodcharacteristic}, it is sufficient to show that $\cn(x)$ is non-empty for all $x \in X$, and that it satisfies (F1), (F2), (V1), and (V2). Firstly, since $\fU \ne \emptyset$, $\cn(x) \ne \emptyset$ for all $x \in X$.
 
 (F1): Let $U \in \fU$ and $V \supset U(x)$, then