Polished A-A and added new lines for broken enumerates.

src/measure/bochner-integral/bochner.tex

@@ -52,7 +52,8 @@
     \begin{enumerate}
         \item[(a)] $f_n \to f$ strongly pointwise.
         \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
     then $\int f d\mu = \limv{n}\int f_n d\mu$. 
 \end{theorem}
@@ -76,7 +77,8 @@
     \begin{enumerate}
         \item For any $x \in E$ and $A \in \cm$, $I_\lambda(x \cdot \one_A) = x \mu(A)$.
         \item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     
     For any $f \in L^1(X; E)$, $I_\lambda f = \int \lambda(f, d\mu)$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. 
 \end{definition}
@@ -96,7 +98,8 @@
     \begin{enumerate}
         \item[(a)] $f_n \to f$ strongly pointwise.
         \item[(b)] There exists $g \in L^1(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
     then $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. 
 \end{theorem}

src/measure/bochner-integral/strongly.tex

@@ -11,7 +11,8 @@
         \begin{enumerate}
             \item[(a)] For each $n \in \natp$, $\norm{f_n}_E \le \norm{f}_E$.
             \item[(b)] $\norm{f_n(x) - f(x)}_E \to 0$ pointwise as $n \to \infty$.
-        \end{enumerate}
+        \end\{enumerate\}
+
     \end{enumerate}
 
     If the above holds, then $f$ is a \textbf{strongly measurable} function. 
@@ -39,7 +40,8 @@
     \begin{enumerate}
         \item For any strongly measurable functions $f, g: X \to E$ and $\lambda \in K$, $\lambda f + g$ is strongly measurable. 
         \item For any strongly measurable functions $\bracs{f_n: X \to E|n \in \natp}$ and $f: X \to E$, if $f_n \to f$ strongly pointwise, then $f$ is strongly measurable. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{proposition}
 \begin{proof}

src/measure/lebesgue-integral/complex.tex

@@ -104,7 +104,8 @@
     \begin{enumerate}
         \item $f_n \to f$ pointwise.
         \item There exists $g \in \mathcal{L}^1(X)$ such that $\abs{f_n} \le \abs{g}$ for all $n \in \natp$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $\int fd\mu = \limv{n}\int f_n d\mu$.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 2.24]{Folland}}}. ]

src/measure/measurable-maps/metric.tex

@@ -76,7 +76,8 @@
         \item[(a)] For each $y \in Y$, $y \in \ol{N(y)^o}$.
         \item[(b)] $\bigcap_{y \in Y}N(y) \ne \emptyset$.
         \item[(c)] For any $y_0 \in Y$, $\bracs{y \in Y|y_0 \in N(y)} \in \cb_Y$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Then, for any $f: X \to Y$, the following are equivalent:
     \begin{enumerate}
         \item $f$ is $(\cm, \cb_Y)$-measurable.
@@ -84,7 +85,8 @@
         \begin{enumerate}
             \item[(i)] For each $x \in X$ and $n \in \natp$, $f_n(x) \in N(f(x))$.
             \item[(ii)] $f_n \to f$ pointwise.
-        \end{enumerate}
+        \end\{enumerate\}
+
         \item There exists a sequence $\seq{f_n}$ of $(\cm, \cb_Y)$-measurable simple functions such that $f_n \to f$ pointwise.
     \end{enumerate}
 \end{proposition}
@@ -146,6 +148,7 @@
         \bracs{y \in E|y_0 \in N(y)} = \bracs{y \in E|\norm{y_0}_E \le \norm{y}_E} \in \cb_E
         \]
 
-    \end{enumerate}
+    \end\{enumerate\}
+
     By \autoref{proposition:measurable-simple-separable}, (1) and (2) are equivalent.
 \end{proof}

src/measure/measurable-maps/real-valued.tex

@@ -24,7 +24,8 @@
         \item $G = \limsup_{n \to \infty}f_n$.
         \item $g = \limsup_{n \to \infty}f_n$.
         \item $\limv{n}f_n$ (if it exists).
-    \end{enumerate}
+    \end\{enumerate\}
+
     In addition, if the above functions are $\real$-valued, then they are $(\cm, \cb_{\real})$-measurable.
 \end{proposition}
 \begin{proof}

src/measure/measure/complete.tex

@@ -20,7 +20,8 @@
         \item There exists an extension $\ol{\mu}$ of $\mu$ as a measure on $\cm$.
         \item $(X, \ol{\cm}, \ol{\mu})$ is a complete measure space.
         \item[(U)] For any complete measure space $(X, \cf, \nu)$ where $\cf \supset \cm$ and $\nu|_\cm = \mu$, $\cf \supset \ol{\cm}$ and $\nu|_{\ol{\cm}} = \ol{\mu}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     and space $(X, \ol{\cm}, \mu)$ is the \textbf{completion} of $(X, \cm, \mu)$.
 \end{definition}
 \begin{proof}

src/measure/measure/kolmogorov.tex

@@ -16,13 +16,15 @@
         \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular.
         \item[(b)] $X_n$ is Hausdorff.
         \item[(c)] $X_n$ is separable.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Let $\bracs{\mu_{I}| I \subset \natp \text{ finite}}$ be consistent Borel probability measures, then for any $\seq{B_n}$ where:
     \begin{enumerate}
         \item[(d)] For each $n \in \nat$, $B_n \in \cb_{\prod_{j = 1}^n X_j}$.
         \item[(e)] For each $n \in \nat$, $B_{n+1} \subset B_n \times X_{n+1}$.
         \item[(f)] There exists $\eps > 0$ such that $\mu_{[n]}(B_n) > \eps$ for all $n \in \natp$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Then there exists $\seq{K_n}$ such that for every $n \in \natp$,
     \begin{enumerate}
         \item $K_n \subset \prod_{j = 1}^n X_j$ is compact.
@@ -52,7 +54,8 @@
         \item[(a)] Every finite measure on $\prod_{j \in J} X_j$ is regular.
         \item[(b)] $X_j$ is Hausdorff.
         \item[(c)] $X_j$ is separable.
-    \end{enumerate}
+    \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}[Proof {{\cite[Theorem 1.14]{Baudoin}}}. ]
@@ -71,7 +74,8 @@
         \item $K_n \subset B_n$.
         \item $K_{n+1} \subset K_n \times X_{n+1}$.
         \item $\mu(K_n) \ge \eps/2$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Let $N \in \natp$ and $x \in \prod_{j = 1}^N X_j$ such that $x \in \bigcap_{n \ge N}\pi_{[N]}(K_n)$. By compactness and (4), there exists $x_{N+1} \in X_{N+1}$ such that $(x_1, \cdots, x_N, x_{N+1}) \in \bigcap_{n > N}\pi_{[N+1]}(K_n)$. Thus there exists $x \in \prod_{i \in I}X_i$ such that $x \in \pi_{[n]}^{-1}(K_n)$ for all $n \in \natp$, and
     \[
     x \in \bigcap_{n \in \natp}\pi_{[n]}^{-1}(K_n) \subset \bigcap_{n \in \natp}\pi_{[n]}^{-1}(B_n) \ne \emptyset

src/measure/measure/lebesgue-stieltjes.tex

@@ -44,7 +44,8 @@
         \]
 
         \item[(U)] For any function $G: \real \to \real$ satisfying (1), $F - G$ is constant.
-    \end{enumerate}
+    \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}[Proof {{\cite[Theorem 1.16]{Folland}}}. ]

src/measure/measure/measure.tex

@@ -14,13 +14,15 @@
     \begin{enumerate}
         \item[(M1)] $\mu(\emptyset) = 0$.
         \item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     In which case, $(X, \cm, \mu)$ is a \textbf{measure space}.
 
     If $\mu: \cm \to [0, \infty]$ instead satisfies (M1) and
     \begin{enumerate}
         \item[(M2')] For any $\seqf{E_j} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{j = 1}^n E_j} = \sum_{j = 1}^n \mu(E_j)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $\mu$ is a \textbf{finitely-additive measure}.
 \end{definition}
 
@@ -100,7 +102,8 @@
         \item[(a)] $\sigma(\mathcal{P}) = \cm$.
         \item[(b)] $\mu(E) = \nu(E)$ for all $E \in \mathcal{P}$.
         \item[(c)] There exists $\seq{E_n} \subset \mathcal{P}$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \natp$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $\mu = \nu$.
 \end{theorem}
 \begin{proof}
@@ -122,7 +125,8 @@
         \]
 
         by continuity from below (\autoref{proposition:measure-properties}).
-    \end{enumerate}
+    \end\{enumerate\}
+
     so $\alg(F)$ is a $\lambda$-system. By (a) and Dynkin's $\pi$-$\lambda$ theorem (\autoref{theorem:pi-lambda}), $\alg(F) = \cm$.
 
     Let $\seq{E_n}$ as in assumption (c), then $\mu(E_n \cap F) = \mu(E_n \cap F)$ for all $n \in \natp$ and $F \in \cm$. Thus by continuity from below (\autoref{proposition:measure-properties}),
@@ -163,7 +167,8 @@
         \[
         \int g df_*\mu = \int g \circ f d\mu
         \]
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{definition}
 \begin{proof}

src/measure/measure/outer.tex

@@ -115,7 +115,8 @@
             \item[(a)] $\nu|_{\cm \cap \cn} \le \mu_{\cm \cap \cn}$.
             \item[(b)] For any $E \in \cm \cap \cn$ with $\mu(E) < \infty$, $\mu(E) = \nu(E)$.
             \item[(c)] If $\mu$ is $\sigma$-finite, then $\nu = \mu$.
-        \end{enumerate}
+        \end\{enumerate\}
+
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof {{\cite[Theorem 1.14]{Folland}}}. ]

src/measure/measure/product.tex

@@ -8,7 +8,8 @@
     \begin{enumerate}
         \item For each $E \in \cm$ and $F \in \cn$, $\mu \otimes \nu(E \times F) = \mu(E)\nu(F)$. 
         \item[(U)] For any measure $\lambda: \cm \otimes \cn \to [0, \infty]$, $\lambda \le \mu$. For any $A \in \cm \otimes \cn$ with $\mu(A) < \infty$, $\lambda(A) = \mu(A)$. In particular, if $\mu$ is $\sigma$-finite, then $\lambda = \mu$. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
     The measure $\mu \otimes \nu$ is the \textbf{product} of $\mu$ and $\nu$. 
 \end{definition}
@@ -57,7 +58,8 @@
     \begin{enumerate}
         \item For any $E \in \cm \otimes \cn$, $x \in X$, and $y \in Y$, $\bracs{z \in Y|(x, z) \in E} \in \cm$ and $\bracs{z \in X|(z, y) \in E} \in \cn$.
         \item For any measure space $(Z, \cf)$, $(\cm \otimes \cn, \cf)$-measurable function $f: X \times Y \to Z$, $x \in X$, and $y \in Y$, $f(x, \cdot)$ is $(\cn, \cf)$-measurable and $f(\cdot, y)$ is $(\cm, \cf)$-measurable. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{lemma}
 \begin{proof}
@@ -100,8 +102,10 @@
         \int_{X \times Y}f(z)\mu \otimes \nu(dz) &= \int_X\int_Y f(x, y)\nu(dy)\mu(dx) \\
         &= \int_{Y}\int_X f(x, y)\mu(dx)\nu(dy)
         \end{align*}
-        \end{enumerate}
-    \end{enumerate}
+        \end\{enumerate\}
+
+    \end\{enumerate\}
+
     
 \end{theorem}
 \begin{proof}

src/measure/measure/regular.tex

@@ -31,6 +31,7 @@
         \item[(a)] $X$ is a LCH space.
         \item[(b)] Every open set of $X$ is $\sigma$-compact.
         \item[(c)] For any $K \subset X$ compact, $\mu(K) < \infty$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $\mu$ is a regular measure.
 \end{theorem}

src/measure/measure/semifinite.tex

@@ -11,7 +11,8 @@
         \mu(E) = \sup\bracs{\mu(F)| F \in \cm, F \subset E, \mu(F) < \infty}
         \]
 
-    \end{enumerate}
+    \end\{enumerate\}
+
     If the above holds, then $\mu$ is a \textbf{semifinite measure}.
 \end{definition}
 \begin{proof}

src/measure/measure/sigma-finite.tex

@@ -7,6 +7,7 @@
     \begin{enumerate}
         \item There exists $\seq{E_n} \subset \cm$ pairwise disjoint such that $\bigsqcup_{n \in \nat}E_n = X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
         \item There exists $\seq{E_n} \subset \cm$ such that $E_n \upto X$ and $\mu(E_n) < \infty$ for all $n \in \nat$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     If the above holds, then $\mu$ is a \textbf{$\sigma$-finite measure}.
 \end{definition}

src/measure/radon/c0.tex

@@ -8,7 +8,8 @@
         \item $I = I^+ - I^-$.
         \item $I^+ \perp I^-$. 
         \item $\norm{I^+}_{C_0(X; \real)^*}, \norm{I^-}_{C_0(X; \real)^*} \le \norm{I}_{C_0(X; \real)^*}$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{lemma}
 \begin{proof}

src/measure/radon/radon.tex

@@ -115,7 +115,8 @@
     \begin{enumerate}
         \item[(a)] For any $U \subset X$ open, $U$ is $\sigma$-compact.
         \item[(b)] For any $K \subset X$ compact, $\mu(K) < \infty$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     then $\mu$ is a regular measure on $X$.
 \end{proposition}
 \begin{proof}

src/measure/radon/riesz.tex

@@ -71,7 +71,8 @@
         \]
 
         As this holds for all $\eps > 0$, $\mu^*(E) \le \sum_{n \in \natp}\mu^*(E_n)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     Therefore $\mu^*: 2^E \to [0, \infty]$ is an outer measure.
 
     To see that all Borel sets are $\mu^*$-measurable, let $E \subset X$ and $U \in \topo$. First suppose that $E$ is open. Let $f \prec E \cap U$, then for any $g \prec E \setminus \supp{f}$, $\supp{f} \cap \supp{g} = \emptyset$ and $f + g \prec E$, so

src/measure/sets/algebra.tex

@@ -38,7 +38,8 @@
     \begin{enumerate}
         \item For any $A, B \in \alg$, $A \cap B \in \alg$.
         \item For any $A, B \in \alg$, $A \setminus B \in \alg$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     If $\alg$ is a $\sigma$-algebra, then:
     \begin{enumerate}
         \item[(1')] For any $\seq{A_n} \in \alg$, $\bigcap_{n \in \natp}A_n \in \alg$.

src/measure/sets/borel.tex

@@ -98,7 +98,8 @@
         \item Open sets of $X$.
         \item $\bracs{B(x, r)|x \in X, r > 0}$.
         \item $\bracsn{\ol{B(x, r)}|x \in X, r > 0}$. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{proposition}
 \begin{proof}

src/measure/sets/elementary.tex

@@ -8,7 +8,8 @@
         \item[(P1)] $\emptyset \in \ce$.
         \item[(P2)] For any $A, B \in \ce$, $A \cap B \in \ce$.
         \item[(E)] For any $E, F \in \ce$ with $E \subset F$, there exists $\seqf{E_j} \subset \ce$ such that $E \setminus F = \bigsqcup_{j = 1}^n E_j$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     If $X \in \ce$, then (E) may be replaced with
     \begin{enumerate}
         \item[(E')] For any $E \in \ce$, there exists $\seqf{E_j} \subset \ce$ such that $E^c = \bigsqcup_{j = 1}^n E_j$.

src/measure/sets/lambda.tex

@@ -60,7 +60,8 @@
         \braks{\bigcup_{n \in \natp}E_n} \cap F = \bigcup_{n \in \natp}E_n \cap F \in \lambda(\mathcal{P})
         \]
 
-    \end{enumerate}
+    \end\{enumerate\}
+
     so $\cm(\ce)$ is a $\lambda$-system.
 
     Since $\mathcal{P}$ is a $\pi$-system, $\cm(\mathcal{P}) \supset \mathcal{P}$, so $\cm(\mathcal{P}) = \lambda(\mathcal{P})$. Thus for any $E \in \lambda(\mathcal{P})$ and $F \in \lambda(\mathcal{P})$, $E \cap F \in \lambda(\mathcal{P})$. Therefore $\cm(\lambda(\mathcal{P})) \supset \mathcal{P}$, $\cm(\lambda(\mathcal{P})) = \lambda(\mathcal{P})$, and $\lambda(\mathcal{P})$ satisfies (P2). By \autoref{lemma:pi-lambda}, $\lambda(\mathcal{P})$ is a $\sigma$-algebra.

src/measure/vector/complex.tex

@@ -7,7 +7,8 @@
     \begin{enumerate}
         \item[(M1)] $\mu(\emptyset) = 0$.
         \item[(M2)] For any $\seq{E_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}E_n} = \sum_{n \in \natp} \mu(E_n)$ where the sum converges absolutely. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     By \hyperref[Riemann's Rearrangement Theorem]{theorem:riemann-rearrangement}, (M2) implies that $\mu$ can only take at most one value in $\bracs{-\infty, \infty}$. 
 \end{definition}
 
@@ -101,7 +102,8 @@
     \begin{enumerate}
         \item[(i)] For each $1 \le k \le n$, $M_k = m(A_k) - \mu(A_k) > 0$.
         \item[(ii)] For each $1 \le k \le n - 1$, $A_{k+1} \subset A_k$ with $\mu(A_{k+1}) - \mu(A_k) > M_{k}/2$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     
     Let $A_n \in \cm$ with $A_{n+1} \subset A_n$ such that $\mu(A_{n+1}) - \mu(A_n) > M_n/2$. Since $A_{n+1} \cap P = \emptyset$ and $\mu(P) = M$, $A_{n+1}$ cannot be positive, so
     \[
@@ -157,7 +159,8 @@
         \item $\mu^+ \perp \mu^-$.
         \item $\mu = \mu^+ - \mu^-$.
         \item[(U)] For any other pair $(\nu^+, \nu^-)$ satisfying (1) and (2), $\mu^+ = \nu^+$ and $\mu^- = \nu^-$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{theorem}
 \begin{proof}

src/measure/vector/variation.tex

@@ -96,7 +96,8 @@
     \begin{enumerate}
         \item $M(X, \cm; E)$ equipped with $\norm{\cdot}_{\text{var}}$ is a normed vector space over $K$.
         \item If $E$ is a Banach space, then so is $M(X, \cm; E)$.
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{definition}
 \begin{proof}

src/measure/vector/vector.tex

@@ -7,7 +7,8 @@
         \begin{enumerate}
         \item[(M1)] $\mu(\emptyset) = 0$.
         \item[(M2)] For any $\seq{A_n} \subset \cm$ pairwise disjoint, $\mu\paren{\bigsqcup_{n \in \natp}A_n} = \sum_{n \in \natp} \mu(A_n)$ where the sum converges absolutely. 
-    \end{enumerate}
+    \end\{enumerate\}
+
     
 \end{definition}