Added monotone convergence for LSC functions.

.vscode/project.code-snippets

@@ -60,6 +60,17 @@
       "$0"
     ]
   },
+  "Corollary Block": {
+  "scope": "latex",
+    "prefix": "cor",
+    "body": [
+      "\\begin{corollary}",
+      "\\label{corollary:$1}",
+      "    $2",
+      "\\end{corollary}",
+      "$0"
+    ]
+  },
   "Proof Block": {
   "scope": "latex",
     "prefix": "proof",

preamble.sty

@@ -8,12 +8,13 @@
 
 % ------------- Block Environmets ---------------
 
-\newtheorem{theorem}[subsection]{Theorem}
-\newtheorem{proposition}[subsection]{Proposition}
-\newtheorem{lemma}[subsection]{Lemma}
+\newtheorem{theorem}{Theorem}[section]
+\newtheorem{proposition}[theorem]{Proposition}
+\newtheorem{corollary}[theorem]{Corollary}
+\newtheorem{lemma}[theorem]{Lemma}
 
-\newtheorem{definition}[subsection]{Definition}
-\newtheorem{example}[subsection]{Example}
+\newtheorem{definition}[theorem]{Definition}
+\newtheorem{example}[theorem]{Example}
 % \newtheorem{exercise}[subsection]{Exercise}
 % \newtheorem{situation}[subsection]{Situation}
 

src/dg/derivative/higher.tex

@@ -35,26 +35,33 @@
         A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
         &- [f(x + k) - f(x) - Df(x)(k)] \\
-        &= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k)
+        &= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
         &- [f(x + k) - f(x) - Df(x)(k)] \\
     \end{align*}
     Let $B_h: B_E(0, r) \to F$ be defined by
-    \[
-    B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
-    \]
+    \begin{align*}
+    B_h(k) &= f(x + h + k) - f(x + k) \\
+    &- Df(x + h)(k) + Df(x)(k)
+    \end{align*}
+    
 
     then
-    \[
-    B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
-    \]
+    \begin{align*}
+    B_h(k) - B_h(0) &= f(x + h + k) - f(x + k) \\
+    &- Df(x + h)(k) + Df(x)(k) \\
+    &-f(x + h) + f(x)
+    \end{align*}
+    
 
     Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
     \begin{align*}
     DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
-    &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) - Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
+    &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) \\
+    &- Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
     &=r_2(k) + r_3(h)
     \end{align*}
+
     By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
     \[
     \norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
@@ -68,9 +75,12 @@
     so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
 
     Now suppose that the proposition holds for $n$. Identify $L^n(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j} \subset E$,
-    \[
-    Df(x)(x_1, \cdots, x_n) = Df(x)(x_1, x_2)(x_3, \cdots, x_n) = Df(x)(x_2, x_1)(x_3, \cdots, x_n) = Df(x)(x_2, x_1, x_3, \cdots, x_n)
-    \]
+    \begin{align*}
+    Df(x)(x_1, \cdots, x_n) &= Df(x)(x_1, x_2)(x_3, \cdots, x_n) \\
+    &= Df(x)(x_2, x_1)(x_3, \cdots, x_n) \\
+    &= Df(x)(x_2, x_1, x_3, \cdots, x_n)
+    \end{align*}
+    
 
     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}

src/fa/lc/bornologic.tex

@@ -0,0 +1,56 @@
+\section{Bornologic Spaces}
+\label{section:bornologic}
+
+\begin{definition}[Bornologic Space]
+\label{definition:bornologic-space}
+    Let $E$ be a locally convex space, then the following are equivalent:
+    \begin{enumerate}
+        \item For any $U \subset E$ convex and balanced, if $U$ absorbs every bounded set of $E$, then $U \in \cn_E(0)$.
+        \item For any seminorm $\rho: E \to [0, \infty)$ that is bounded on all bounded sets of $E$, $\rho$ is continuous. 
+    \end{enumerate}
+    
+    If the above holds, then $E$ is a \textbf{bornologic space}.
+\end{definition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): Let $B \subset E$ be bounded, then there exists $R > 0$ such that $\rho(B) \subset [0, R)$. In which case, 
+    \[
+    B \subset \bracs{\rho < R} = R\bracs{\rho < 1}
+    \]
+
+    By assumption, $\bracs{\rho < 1} \in \cn_E(0)$, so $\rho$ is continuous by \autoref{lemma:continuous-seminorm}.
+
+    (2) $\Rightarrow$ (1): Let $\rho$ be the \hyperref[gauge]{definition:gauge} of $U$, then for any $B \subset E$ bounded, there exists $R > 0$ such that $B \subset RU$. In which case, $\rho(B) \subset [0, R]$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:metrisable-bornologic}
+    Let $E$ be a metrisable locally convex space, then $E$ is bornologic.
+\end{proposition}
+\begin{proof}
+    Let $U \subset E$ be convex and balanced such that $U$ absorbs every bounded set of $E$. Let $\seq{U_n} \subset \cn^o(0)$ be a decreasing countable fundamental system of neighbourhoods at $0$. If $U_n \setminus nA \ne \emptyset$ for all $n \in \natp$, then there exists $\seq{x_n}$ such that $x_n \in U_n \setminus nA$ for all $n \in \natp$. In which case, $x_n \to 0$ as $n \to \infty$, so $\seq{x_n}$ is bounded. By assumption, there exists $n \in\natp$ such that $nA \supset \seq{x_n}$, which contradicts the fact that $\seq{x_n} \cap A = \emptyset$. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:bornologic-bounded}
+    Let $E$ be a bornologic space, $F$ be a locally convex space, and $T \in \hom(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $T$ is continuous.
+        \item $T$ is bounded.
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Rightarrow$ (2): By \autoref{proposition:continuous-bounded}. 
+
+    (2) $\Rightarrow$ (1): Let $\rho: F \to [0, \infty)$ be a continuous seminorm, then $\rho \circ T$ is a seminorm on $E$ that is bounded on bounded sets. Since $E$ is bornologic, $\rho \circ T$ is continuous. Therefore $T$ is continuous by \autoref{proposition:tvs-convex-morphism}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:bornologic-continuous-complete}
+    Let $E$ be a bornologic space and $F$ be a complete Hausdorff locally convex space, then $L_b(E; F)$ is complete. In particular, $E^*$ equipped with the topology of bounded convergence is complete.
+\end{proposition}
+\begin{proof}
+    By \autoref{proposition:bornologic-bounded}, $L_b(E; F) = B(E; F)$. By \autoref{proposition:operator-space-completeness}, $B(E; F)$ is complete, so $L_b(E; F)$ is complete as well.
+\end{proof}
+
+

src/fa/lc/convex.tex

@@ -99,10 +99,16 @@
         \item $[\cdot]$ is uniformly continuous.
         \item $[\cdot]$ is continuous.
         \item $[\cdot]$ is continuous at $0$.
+        \item $\bracs{x \in E| [x] < 1} \in \cn_E(0)$.
     \end{enumerate}
 \end{lemma}
 \begin{proof}
-    $(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \autoref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
+    $(4) \Rightarrow (1)$: Let $x, y \in E$ and $r > 0$. If
+    \[
+    x - y \in \bracs{x \in E|[x] < r} = r\bracs{x \in E|[x] < 1} \in \cn_E(0)
+    \]
+    
+    then $[x - y] < r$.
 \end{proof}
 
 

src/fa/lc/hahn-banach.tex

@@ -132,17 +132,31 @@
     \begin{enumerate}
         \item For any subspace $M \subset E$, $x \in M \setminus E$, and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ such that
         \begin{enumerate}
-            \item $\phi \le \rho$.
+            \item $|\phi| \le \rho$.
             \item $\dpb{x, \phi}{E} = \inf_{y \in M}\rho(x + y)$.
         \end{enumerate}
-        \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $\phi \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
+        \item For any $x \in E$ and continuous seminorm $\rho: E \to [0, \infty)$, there exists $\phi \in E^*$ with $|\phi| \le \rho$ and $\dpb{x, \phi}{E} = \rho(x)$.
         \item If $E$ is Hausdorff, then for any $x, y \in E$, there exists $\phi \in E^*$ with $\dpb{x, \phi}{E} \ne \dpb{y, \phi}{E}$.
     \end{enumerate}
 \end{proposition}
 \begin{proof}
-    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
+    (1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $|\phi| \le \rho_M \le \rho$.
 
     (2): By (1) applied to $M = \bracs{0}$.
 
     (3): By (2) applied to $x - y$.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:seminorm-lsc}
+    Let $E$ be a locally convex space and $\rho: E \to [0, \infty)$ be a continuous seminorm, then $\rho: E_w \to [0, \infty)$ is lower semicontinuous and Borel measurable. 
+\end{proposition}
+\begin{proof}
+    Let $x \in E$, then there exists $\phi_x \in E^*$ such that $\dpn{x, \phi_x}{E} = \rho(x)$ and $|\phi_x| \le \rho$. Thus
+    \[
+    \rho(x) = \sup_{y \in E}\dpn{x, \phi_y}{E}
+    \]
+
+    is lower semicontinuous and Borel measurable by \autoref{proposition:semicontinuous-properties}. 
+\end{proof}
+

src/fa/lc/index.tex

@@ -4,6 +4,7 @@
 
 \input{./convex.tex}
 \input{./continuous.tex}
+\input{./bornologic.tex}
 \input{./quotient.tex}
 \input{./projective.tex}
 \input{./inductive.tex}

src/fa/norm/absolute.tex

@@ -0,0 +1,125 @@
+\section{Conditional and Absolute Convergence}
+\label{section:conditional-absolute-convergence}
+
+\begin{definition}[Absolute Convergence]
+\label{definition:absolute-convergence}
+    Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges absolutely} if $\sum_{n \in \natp}\norm{x_n}_E < \infty$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:banach-absolute}
+    Let $E$ be a normed space, then the following are equivalent:
+    \begin{enumerate}
+        \item $E$ is a Banach space.
+        \item For any absolutely convergent series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$, there exists $x \in E$ such that $x = \sum_{n = 1}^\infty x_n$.
+    \end{enumerate}
+    
+\end{lemma}
+\begin{proof}
+    (2) $\Rightarrow$ (1): Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence ${n_k}$ such that $\norm{x_{n_{k+1}} - x_{n_k}} \le 2^{-k}$ for all $k \in \natp$. 
+    
+    For each $k \in \natp$, let $y_k = x_{n_{k+1}} - x_{n_k}$, then $\sum_{n = 1}^\infty y_k$ is absolutely convergent, and there exists $y \in E$ such that $y = \sum_{n = 1}^\infty y_n$. Let $x = x_{n_1} + y$, then $x_{n_k} \to x$ as $k \to \infty$. Since $\seq{x_n}$ is Cauchy, $x_n \to x$ as well by \autoref{lemma:cauchy-subnet}.
+\end{proof}
+
+
+\begin{definition}[Unconditional Convergence]
+\label{definition:unconditional-convergence}
+    Let $E$ be a normed space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$, 
+    \[
+    \sum_{n = 1}^\infty x_n = \sum_{n = 1}^\infty x_{\sigma(n)}
+    \]
+\end{definition}
+
+\begin{theorem}[Riemann's Rearrangement Theorem]
+\label{theorem:riemann-rearrangement}
+    Let $\seq{x_n} \subset \real$ and $N = P \sqcup N$ such that $x_n \ge 0$ for all $n \in P$ and $x_n \le 0$ for all $n \in N$, then
+    \begin{enumerate}
+        \item If $\sum_{n \in P}x_n = \infty$ and $\sum_{n \in N}x_n = -\infty$, then there exists bijections $\sigma, \tau: \natp \to \natp$ such that
+        
+        \[
+        \sum_{n = 1}^\infty x_{\sigma(n)} = \infty \quad \sum_{n = 1}^\infty x_{\tau(n)} = -\infty
+        \]
+
+        \item If $\sum_{n \in P}x_n = \infty$, $\sum_{n \in N}x_n = -\infty$, and $\sum_{n =1}^\infty x_n$ converges, then for any $x \in \ol{\real}$, there exists a bijection $\sigma: \natp \to \natp$ such that $\sum_{n = 1}^\infty x_{\sigma(n)} = x$.
+        \item If $\sum_{n \in P}x_n = \infty$ but $\sum_{n \in N}x_n > -\infty$, then $\sum_{n = 1}^\infty x_n$ converges to $\infty$ unconditionally.
+        \item If $\sum_{n \in N}x_n = -\infty$ but $\sum_{n \in P}x_n < \infty$, then $\sum_{n = 1}^\infty x_n$ converges to $-\infty$ unconditionally. 
+        \item If $\sum_{n \in \natp}|x_n| < \infty$, then $\sum_{n = 1}^\infty x_n$ converges unconditionally.
+    \end{enumerate}
+    In other words, a series in $\real$ converges unconditionally if and only if its positive parts or its negative parts are finite. 
+\end{theorem}
+\begin{proof}
+    By inserting zeroes, assume without loss of generality that $P$ and $N$ are both infinite. Let $\seq{p_k} = P$ be an enumeration of $P$ and $\seq{n_k} = N$ be an enumeration of $N$. 
+
+    (1): By taking $\sum_{n \in P}-x_n$, assume without loss of generality that $\sum_{n \in P}x_n = \infty$. 
+
+    Let $k_0 = 0$. Let $n \in \natp$ and suppose inductively that $k_n$ has been constructed such that
+    \[
+    \sum_{k = 1}^{k_n}x_{p_k} + \sum_{j = 1}^n x_{n_j} \ge n
+    \]
+
+    Since $\sum_{n \in \natp}p_n = \infty$, there exists $k_{n+1} \in \natp$ such that $\sum_{k = k_{n}+1}^{k_{n+1}}x_{p_k} \ge x_{n_{n+1}} + 1$. 
+    
+    Therefore there exists $\bracs{k_n}_1^K \subset \natp$ such that
+    \[
+    \sum_{k = 1}^{k_n}x_{p_k} + \sum_{j = 1}^n x_{n_j} \ge n
+    \]
+    
+    for all $1 \le n < K$. In which case, if $\sigma: \natp \to \natp$ is defined by
+    \[
+    (p_1, \cdots, p_{k_1}, n_1, p_1, \cdots, p_{k_2}, n_2, \cdots)
+    \]
+
+    Let $n \in \natp$, then for any $K \ge k_{n} + n$, $\sum_{k = 1}^K x_{\sigma(k)} \ge n$, so $\sum_{k = 1}^\infty x_{\sigma(k)} = \infty$.
+
+    (2): Assume without loss of generality that $x > 0$. Define $m_0 = 0$ and $n_0 = 0$. Let $n \in \natz$. If $n \ge 1$, suppose inductively that $m_n, n_n \in \natp$ has been constructed such that
+    \[
+    x < \sum_{k = 1}^{m_{n}}x_{p_k} + \sum_{k = 1}^{n_{-1}}x_{n_k} < x + x_{m_{n}}
+    \]
+    and
+    \[
+    x + x_{n_n} < \sum_{k = 1}^{m_n}x_{p_k} + \sum_{k = 1}^{n_n}x_{n_k} < x
+    \]
+
+    Since $\sum_{k \in \natp}x_{p_k} = \infty$, there exists $m_{n+1} > m_n$ such that
+    \[
+    x < \sum_{k = 1}^{m_{n+1}}x_{p_k} + \sum_{k = 1}^{n_n}x_{n_k} < x + x_{m_{n+1}}
+    \]
+
+    As $\sum_{k \in \natp}x_{n_k} = -\infty$, there exists $n_{n+1} > n_n$ such that
+    \[
+    x + x_{n_{n+1}} < \sum_{k = 1}^{m_{n+1}}x_{p_k} + \sum_{k = 1}^{n_{n+1}}x_{n_k} < x
+    \]
+
+    Let $\sigma: \natp \to \natp$ be defined by
+    \[
+    (p_1, \cdots, p_{m_1}, n_{1}, \cdots, n_{n_1}, p_{m_1 + 1}, \cdots, p_{m_2}, \cdots)
+    \]
+
+    then for any $K \in [m_n + n_n, m_{n+1} + n_{n+1}]$,
+    \[
+    x + x_{n_n} + x_{n_{n+1}} \le \sum_{k = 1}^K x_{\sigma(k)} \le x + x_{p_{n+1}}
+    \]
+
+    Since $x_n \to 0$ as $n \to \infty$, $\sum_{k = 1}^K x_{\sigma(k)} \to x$ as $K \to \infty$. 
+
+    (3): Let $\sigma: \natp \to \natp$ be a bijection and $\alpha > 0$, then there exists $K \in \natp$ such that $\sum_{k = 1}^K x_{p_k} > \alpha - \sum_{k \in \natp}x_{n_k}$. Since $\sigma$ is a bijection, there exists $K' \in \natp$ such that $\sigma([1, K']) \supset \bracs{p_k|1 \le k \le K}$. In which case, for any $k \ge K'$,
+    \[
+    \sum_{j = 1}^k x_{\sigma(j)} \ge \sum_{j = 1}^K x_{p_k} + \sum_{k \in \natp}x_{n_k} > \alpha
+    \]
+
+    (4): By applying (3) to $\sum_{n = 1}^\infty -x_n$. 
+
+    (5): Let $\sigma: \natp \to \natp$ be a bijection, then for any $N \in \natp$, there exists $K \in \natp$ such that $\sigma([1, K]) \supset [1, N]$. In which case,
+    \[
+        \abs{\sum_{n = 1}^N x_n - \sum_{n = 1}^K x_{\sigma(n)}} \le \sum_{n > N}|x_n|
+    \]
+
+    so
+    \[
+    \sum_{n = 1}^\infty x_n = \sum_{n = 1}^\infty x_{\sigma(n)}
+    \]
+\end{proof}
+
+
+
+

src/fa/norm/index.tex

@@ -2,5 +2,6 @@
 \label{chap:normed-spaces}
 
 \input{./normed.tex}
+\input{./absolute.tex}
 \input{./linear.tex}
 \input{./multilinear.tex}

src/fa/norm/normed.tex

@@ -31,6 +31,7 @@
     is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
 \end{proof}
 
+
 \begin{theorem}[Successive Approximation]
 \label{theorem:successive-approximation}
     Let $E, F$ be normed spaces, $T \in L(E; F)$, $C \ge 0$, and $\gamma \in (0, 1)$. If for all $y \in F$, there exists $x \in E$ such that:

src/fa/tvs/spaces-of-linear.tex

@@ -39,9 +39,53 @@
     for all $x \in S$.
 \end{proof}
 
+
+\begin{definition}[Space of Bounded Functions]
+\label{definition:bounded-function-space}
+    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+    \begin{enumerate}
+        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
+    
+    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+
+    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
+    \[
+    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+    \]
+    so $f \in B(T; E)$.
+
+    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+\end{proof}
+
+
+\begin{definition}[Space of Bounded Continuous Functions]
+\label{definition:bounded-continuous-function-space}
+    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
+    \begin{enumerate}
+        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
+
+    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
+    
+    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+\end{proof}
+
+
+
 \begin{definition}[Space of Bounded Linear Maps]
 \label{definition:bounded-linear-map-space}
     Let $E, F$ be TVSs over $K \in \RC$, $\mathfrak{S} \subset 2^E$ be an upward-directed system, and $k \in \nat$. The space $B_{\mathfrak{S}}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \mathfrak{S}$, equipped with the $\bracsn{S^k| S \in \mathfrak{S}}$-uniform topology.
+
+    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\mathfrak{S}}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
 \end{definition}
 
 \begin{proposition}
@@ -127,3 +171,35 @@
 
     The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
 \end{definition}
+
+\begin{proposition}
+\label{proposition:operator-space-completeness}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
+    \begin{enumerate}
+        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
+        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
+    \[
+    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
+    \]
+
+    and
+    \[
+    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - Tx
+    \]
+
+    are continuous with respect to the product topology. Since
+    \[
+    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
+    \]
+
+    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 
+
+    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
+
+\end{proof}
+
+

src/measure/index.tex

@@ -3,6 +3,8 @@
 
 \input{./sets/index.tex}
 \input{./measure/index.tex}
+\input{./vector/index.tex}
+\input{./radon/index.tex}
 \input{./measurable-maps/index.tex}
 \input{./lebesgue-integral/index.tex}
 \input{./bochner-integral/index.tex}

src/measure/measure/index.tex

@@ -6,8 +6,6 @@
 \input{./semifinite.tex}
 \input{./sigma-finite.tex}
 \input{./regular.tex}
-\input{./radon.tex}
 \input{./outer.tex}
 \input{./lebesgue-stieltjes.tex}
-\input{./riesz.tex}
 \input{./kolmogorov.tex}

src/measure/radon/index.tex

@@ -0,0 +1,6 @@
+\chapter{Radon Measures}
+\label{chap:radon}
+
+\input{./radon.tex}
+\input{./riesz.tex}
+

src/measure/radon/radon.tex

@@ -202,4 +202,51 @@
     (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}}}]
+\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}
+    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
+    \[
+    \int f d\mu \ge \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+
+    Let $\phi \in \Sigma^+(X, \cm)$ with $0 \le \phi < f$ and $\beta < \int \phi d\mu$. Let $\seqf{a_j} \subset (0, \infty)$ and $\seqf{E_j} \subset \cb_X$ such that $\phi = \sum_{j = 1}^n a_j \one_{E_j}$. Since $\int \phi d\mu < \infty$, for each $1 \le j \le n$, $\mu(E_j) < \infty$ by \hyperref[Markov's Inequality]{theorem:markov-inequality}. By \autoref{proposition:radon-regular-sigma-finite}, there exists compact sets $\seqf{K_j} \subset 2^X$ such that $K_j \subset E_j$ for each $1 \le j \le n$ and $\int \sum_{j = 1}^n a_j \one_{K_j}d\mu > \beta$. 
+
+    Let $\psi = \sum_{j = 1}^n a_j \one_{K_j}$ and $K = \bigcup_{j = 1}^n K_j$, then $K$ is compact by \autoref{proposition:compact-extensions} and $-\psi$ is lower semicontinuous by \autoref{proposition:semicontinuous-properties}.
+    
+    For any $x_0 \in K$, since $f_\alpha \upto f$ and $\psi \le \phi < f$, there exists $\alpha(x_0) \in A$ such that $f_{\alpha(x_0)}(x_0) > \psi(x_0)$. By \autoref{proposition:semicontinuous-properties}, $f_{\alpha(x_0)} - \psi$ is lower semicontinuous, so
+    \[
+    \bracsn{\bracsn{f_{\alpha(x_0)} > \psi}|x_0 \in K}
+    \]
+
+    is an open cover of $K$. Let $\bracs{x_j}_1^N \subset K$ such that $K \subset \bigcup_{j = 1}^N \bracsn{f_{\alpha(x_j)} > \psi}$. Since $f_\alpha \upto f$, there exists $\alpha_0 \in A$ such that
+    \[
+    f_{\alpha_0} \ge \max_{1 \le j \le N}f_{\alpha(x_j)} \ge \psi
+    \]
+    
+    so
+    \[
+    \int f_{\alpha_0} d\mu \ge \int \psi d\mu \ge \beta
+    \]
+
+    As such an $\alpha_0 \in A$ exists for all $\beta < \int \phi d\mu$, 
+    \[
+    \int \phi d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+
+    Since the above holds for all $\phi \in \Sigma^+(X, \cm)$ with $0 \le \phi < f$, 
+    \[
+    \int f d\mu \le \sup_{\alpha \in A}\int f_\alpha d\mu
+    \]
+
+    by \autoref{lemma:lebesgue-non-negative-strict}.
+
+\end{proof}
+
+
 

src/measure/vector/ac.tex

@@ -0,0 +1,10 @@
+\section{Absolutely Continuous}
+\label{section:absolutely-continuous-measure}
+
+\begin{definition}[Absolutely Continuous]
+\label{definition:absolutely-continuous}
+    Let $(X, \cm)$ be a measurable space and $\mu, \nu$ be signed/vector measures on $X$, then $\nu$ is \textbf{absolutely continuous} with respect to $\mu$, denoted $\nu \ll \mu$, if every $\mu$-null set is $\nu$-null.
+\end{definition}
+
+
+

src/measure/vector/complex.tex

@@ -0,0 +1,181 @@
+\section{Signed and Complex Measures}
+\label{section:signed-complex-measure}
+
+\begin{definition}[Signed Measure]
+\label{definition:signed-measure}
+    Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$, then $\mu$ is a \textbf{signed measure} if
+    \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}
+    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}
+
+
+\begin{definition}[Complex Measure]
+\label{definition:complex-measure}
+    Let $(X, \cm)$ be a measurable space and $\mu: \cm \to \complex$, then $\mu$ is a \textbf{complex measure} if
+    \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{definition}
+
+
+\begin{definition}[Positive/Negative/Null Sets]
+\label{definition:positive-negative-sets}
+    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, and $A \in \cm$, then $A$ is...
+    \begin{enumerate}
+        \item \textbf{positive} if $\mu(B) \ge 0$ for all $B \in \cm$ with $B \subset A$.
+        \item \textbf{negative} if $\mu(B) \le 0$ for all $B \in \cm$ with $B \subset A$.
+        \item \textbf{null} if $\mu(B) = 0$ for all $B \in \cm$ with $B \subset A$.
+    \end{enumerate}
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:signed-measure-properties}
+    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then:
+    \begin{enumerate}
+        \item For any $\seq{E_n} \subset \cm$ and $E \in \cm$ with $E_n \upto E$, $\mu\paren{\bigcup_{n \in \natp}E_n} = \limv{n}\mu(E_n)$.
+        \item For any $\seq{E_n} \subset \cm$ and $E \in \cm$ with $E_n \downto E$ and $|\mu(E_1)| < \infty$, $\mu\paren{\bigcap_{n \in \natp}E_n} = \limv{n}\mu(E_n)$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For each $N \in \natp$, let $F_N = E_N \setminus \bigcup_{n = 1}^{N-1}E_n$, then $E_N = \bigsqcup_{n = 1}^N F_n$. By (M2), 
+    \[
+    \mu(E) = \limv{N}\sum_{n = 1}^N \mu(F_N) = \limv{N}\mu(E_N)
+    \]
+
+    (2): By (1), 
+    \[
+    \mu(E) = \mu(E_1) - \mu(E_1 \setminus E) = \limv{n}[\mu(E_1) - \mu(E_1 \setminus E_n)] = \limv{n}\mu(E_n)
+    \]
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:positive-sets-properties}
+    Let $(X, \cm)$ be a measurable space, $\mu: \cm \to [-\infty, \infty]$ be a signed measure, then
+    \begin{enumerate}
+        \item For any $A \in \cm$ positive and $B \in \cm$ with $B \subset A$, $B$ is positive. 
+        \item For any $\seq{A_n} \subset \cm$ positive, $\bigcup_{n \in \natp}A_n$ is positive. 
+    \end{enumerate}
+\end{lemma}
+\begin{proof}
+    (1): For any $C \in \cm$ with $C \subset B$, $C \subset A$, so $\mu(C) \ge 0$. Hence $B$ is positive.
+
+    (2): For each $N \in \natp$, let $B_N = A_N \setminus \sum_{n = 1}^{N-1} A_n$. By (1), $B_N$ is positive with $\bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}B_n$. For any $C \subset \bigcup_{n \in \natp}B_n$, 
+    \[
+        \mu(C) = \sum_{n \in \natp}\mu(C \cap B_n) \ge 0
+    \]
+
+\end{proof}
+
+
+\begin{theorem}[Hahn Decomposition, {{\cite[Theorem 3.3]{Folland}}}]
+\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}
+        \item $P$ is positive.
+        \item $N$ is negative.
+        \item $X = P \sqcup N$.
+        \item[(U)] For any $P', N' \in \cm$ satisfying (1)-(3), $P \Delta P'$ and $N \Delta N'$ are null. 
+    \end{enumerate}
+
+    The disjoint union $X = P \sqcup N$ is the \textbf{Hahn decomposition} of $\mu$.
+\end{theorem}
+\begin{proof}
+    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 $\sup_{n \in \natp}\mu(P_n) \le \mu(P) \le M$.
+
+    (3): Let $N = X \setminus P$, then $P$ is positive and $X = P \sqcup N$. 
+    
+    (2): Suppose for contradiction that $N$ is not negative, then
+    \[
+    m(N) = \sup\bracs{\mu(A)|A \in \cm, A \subset N} > 0
+    \]
+
+    Let $A_1 \in \cm$ such that $A_1 \subset N$ and $\mu(A_1) > 0$. Since $A_{1} \cap P = \emptyset$ and $\mu(P) = M$, $A_{1}$ cannot be positive, so $M_1 = m(A_1) - \mu(A_1) > 0$. Let $n \in \natp$ and suppose inductively that $\bracs{A_n}_1^n$ has been constructed such that
+    \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}
+    
+    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
+    \[
+    M_{n+1} = m(A_{n+1}) - \mu(A_{n+1}) > 0
+    \]
+
+    Let $A = \bigcap_{n \in \natp}A_n$, then since $0 < \mu(A_2) < \infty$, $\mu(A) = \limv{n}\mu(A_n)$ by \autoref{proposition:signed-measure-properties}, so
+    \[
+    \frac{1}{2}\sum_{n = 1}^\infty M_n \le \sum_{n = 1}^{\infty}\mu(A_{n+1}) - \mu(A_n) = \mu(A) - \mu(A_1) < \infty
+    \]
+    and $M_n \to 0$ as $n \to \infty$. 
+    
+    Since $A \cap P = \emptyset$ and $\mu(P) = M$, $A$ cannot be positive. Thus there exists $B \in \cm$ with $B \subset A$ with $\mu(B) - \mu(A) > 0$. As $M_n \to 0$ as $n \to \infty$, there exists $n \in \natp$ such that
+    \[
+    \mu(B) - \mu(A_n) > \mu(B) - \mu(A) > M_n = m(A_n) - \mu(A_n)
+    \]
+
+    As $\mu(B) - \mu(A_n) > \mu(B) - \mu(A)$, this contradicts the definition of $m(A_n)$.
+
+    Therefore $N$ must be negative, and $X = P \sqcup N$ is the desired decomposition. 
+
+    (U): Since $X = P \sqcup N = P' \sqcup N'$, 
+    \[
+    P \Delta P' = P \setminus P' \sqcup P' \setminus P = P \cap N' \sqcup P' \cap N
+    \]
+
+    is a union of two null sets, which is null. Likewise, 
+    \[
+    N \Delta N' = N \cap P' \sqcup N' \cap P
+    \]
+
+    is also a null set. 
+\end{proof}
+
+\begin{corollary}
+\label{corollary:measure-bounded}
+    Let $(X, \cm)$ be a measurable space, then
+    \begin{enumerate}
+        \item For any signed measure $\mu: \cm \to (-\infty, \infty)$, $\mu$ is bounded.
+        \item For any complex measure $\mu: \cm \to \complex$, $\mu$ is bounded.
+    \end{enumerate}
+\end{corollary}
+\begin{proof}
+    (1): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$, then for any $E \in \cm$, $\mu(E) \in [\mu(N), \mu(P)]$.
+
+    (2): Since $\text{Re}(\mu)$ and $\text{Im}(\mu)$ are both signed measures taking values in $(-\infty, \infty)$, they are bounded by (1). Thus $\mu$ is also bounded. 
+\end{proof}
+
+\begin{theorem}[Jordan Decomposition]
+\label{theorem:jordan-decomposition}
+    Let $(X, \cm)$ be a measurable space and $\mu: \cm \to [-\infty, \infty]$ be a signed measure on $X$, then there exists positive measures $\mu^+, \mu^-: \cm \to [0, \infty]$ such that
+    \begin{enumerate}
+        \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{theorem}
+\begin{proof}
+    (1), (2): Let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$. For each $E \in \cm$, let
+    \[
+    \mu^+(E) = \mu(E \cap P) \quad \mu^-(E) = -\mu(E \cap N)
+    \]
+
+    then $(\mu^+, \mu^-)$ satisfies (1) and (2). 
+
+    (U): Let $X = P' \sqcup N'$ such that $P'$ is $\nu^-$-null and $N'$ is $\nu^+$-null, then $X = P' \sqcup N'$ is a Hahn decomposition of $\mu$. Since $P' \Delta P = 0$ and $N' \Delta N = 0$, for any $E \in \cm$,
+    \[
+    \mu^+(E) = \mu(E \cap P) = \mu(E \cap P') = \nu^+(E)
+    \]
+
+    Likewise,
+    \[
+    \mu^-(E) = -\mu(E \cap N) = -\mu(E \cap N') = \nu^-(E)
+    \]
+\end{proof}
+

src/measure/vector/index.tex

@@ -0,0 +1,9 @@
+\chapter{Signed, Complex, and Vector Measures}
+\label{chap:vector-measures}
+
+\input{./complex.tex}
+\input{./vector.tex}
+\input{./ac.tex}
+\input{./ms.tex}
+\input{./variation.tex}
+

src/measure/vector/ms.tex

@@ -0,0 +1,15 @@
+\section{Mutually Singular}
+\label{section:mutually-singular}
+
+
+\begin{definition}[Mutually Singular]
+\label{definition:mutually-singular}
+    Let $(X, \cm)$ be a measurable space and $\mu, \nu$ be signed/vector measures, then $\mu$ and $\nu$ are \textbf{mutually singular}, denoted $\mu \perp \nu$, if there exists $U, V \in \cm$ such that:
+    \begin{enumerate}
+        \item $U$ is $\nu$-null.
+        \item $V$ is $\mu$-null.
+        \item $X = U \sqcup V$.
+    \end{enumerate}
+\end{definition}
+
+

src/measure/vector/variation.tex

@@ -0,0 +1,124 @@
+\section{Total Variation}
+\label{section:total-variation-measure}
+
+\begin{definition}[Total Variation of Vector Measures]
+\label{definition:total-variation-vector}
+    Let $(X, \cm)$ be a measure space, $\mu$ be a vector measure taking values in a normed space $E$, and
+    \[
+    |\mu|(A) = \sup\bracs{\sum_{j = 1}^n \norm{\mu(A_j)}_E \bigg | \seqf{A_j} \subset \cm, A = \bigsqcup_{j = 1}^n A_j}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For any $A \in \cm$,
+        \[
+        |\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}\norm{\mu(A_i)} \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}
+        \]
+        \item $|\mu|$ is a measure on $(X, \cm)$.
+    \end{enumerate}
+
+    and the measure $|\mu|$ is the \textbf{total variation} of $\mu$.
+\end{definition}
+\begin{proof}
+    (1): Since all finite partitions are partitions, $|\mu| \le \nu$. On the other hand, let $\seqi{A} \subset \cm$ and $A = \bigsqcup_{i \in I}A_i$, then for any $J \subset I$ finite, 
+    \[
+    |\mu|(A) \ge \norm{\mu\paren{A \setminus \bigsqcup_{j \in J}A_j}}_E + \sum_{j \in J}\norm{\mu(A_j)}_E \ge \sum_{j \in J}\norm{\mu(A_j)}_E
+    \]
+
+    so
+    \[
+    |\mu|(A) \ge \sum_{i \in I}\norm{\mu(A_i)}_E
+    \]
+
+    As this holds for all $\seqi{A}$ such that $A = \bigsqcup_{i \in I}A_i$, $|\mu|(A) \ge \nu(A)$.
+
+    (2): Let $A \in \cm$ and $\seq{A_n}$ such that $A = \bigsqcup_{n \in \natp}A_n$. For each $n \in \natp$, let $\seq{B_{n, k}} \subset \cm$ such that $A_n = \bigsqcup_{k \in \natp}B_{n, k}$, then for any $N \in \natp$,
+    \[
+    |\mu|(A) \ge \sum_{n, k \in \natp}\norm{\mu(B_{n, k})}_E \ge \sum_{n = 1}^N \sum_{k \in \natp}\norm{\mu(B_{n, k})}_E
+    \]
+
+    As the above inequality is independent of the choice of $\bracsn{B_{n, k}|n, k \in \natp}$, $|\mu|(A) \ge \sum_{n = 1}^N |\mu|(A_n)$ for all $N \in \natp$. Thus $|\mu|(A) \ge \sum_{n \in \natp}|\mu|(A_n)$.
+
+    On the other hand, let $\bracs{B_k}_1^K \subset \cm$ such that $A = \bigsqcup_{k = 1}^K B_k$, then for each $N \in \natp$,
+    \[
+    \sum_{n = 1}^N |\mu|(A_n) \ge \sum_{k = 1}^K \sum_{n = 1}^N  \norm{\mu(A_n \cap B_k)}_E 
+    \]
+
+    so
+    \[
+    \sum_{n \in \natp}|\mu|(A_n) \ge \sum_{k = 1}^K \sum_{n \in \natp} \norm{\mu(A_n \cap B_k)}_E \ge \sum_{k = 1}^K \normn{\mu(B_k)}_E
+    \]
+
+    As the above holds for all choices of $\bracs{B_k}_1^K$, $\sum_{n \in \natp}|\mu|(A_n) \ge |\mu|(A)$.
+\end{proof}
+
+\begin{definition}[Total Variation of Signed Measures]
+\label{definition:total-variation-signed}
+    Let $(X, \cm)$ be a measure space, $\mu$ be a signed measure on $(X, \cm)$, and
+    \[
+    |\mu|(A) = \sup\bracs{\sum_{j = 1}^n |\mu(A_j)| \bigg | \seqf{A_j} \subset \cm, A = \bigsqcup_{j = 1}^n A_j}
+    \]
+
+    then:
+    \begin{enumerate}
+        \item For any $A \in \cm$,
+        \[
+        |\mu|(A) = \nu(A) = \sup\bracs{\sum_{i \in I}|\mu(A_i)| \bigg | \seqi{A} \subset \cm, A = \bigsqcup_{i \in I}A_i}
+        \]
+        \item $|\mu|$ is a measure on $(X, \cm)$.
+        \item Let $\mu = \mu^+ - \mu^-$ be the \hyperref[Jordan decomposition]{theorem:jordan-decomposition} of $\mu$, then $|\mu| = \mu^+ + \mu^-$.
+    \end{enumerate}
+
+    and the measure $|\mu|$ is the \textbf{total variation} of $\mu$.
+\end{definition}
+\begin{proof}
+    (1) and (2): Same as \autoref{definition:total-variation}.
+
+    (3): For any $A \in \cm$, $|\mu(A)| \le \mu^+(A) + \mu^-(A)$, so $(\mu^+ + \mu^-) \ge |\mu|$. On the other hand, let $X = P \sqcup N$ be a \hyperref[Hahn decomposition]{theorem:hahn-decomposition} of $\mu$, then
+    \[
+        |\mu|(A) \ge \mu(A \cap P) + \mu(A \cap N) = \mu^+(A) + \mu^-(A)
+    \]
+
+    so $(\mu^+ + \mu^-) \le |\mu|$.
+    
+
+\end{proof}
+
+
+\begin{definition}[Finite Measure]
+\label{definition:vector-measure-finite}
+    Let $(X, \cm)$ be a measurable space and $\mu$ be a signed/vector measure, then $\mu$ is \textbf{finite} if $|\mu|$ is finite. 
+\end{definition}
+
+\begin{definition}[Space of Finite Measures]
+\label{definition:vector-measure-finite-space}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $M(X, \cm; E)$ be the set of all finite $E$-valued vector measures on $(X, \cm)$. For each $\mu \in M(X, \cm; E)$, let $\norm{\mu}_{\text{var}} = |\mu|(X)$, then:
+    \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{definition}
+\begin{proof}
+    (1): For any $\mu, \nu \in M(X, \cm; E)$ and $\seqf{A_j} \subset \cm$ such that $X = \bigsqcup_{j = 1}^n A_j$, 
+    \[
+    |\mu|(X) + |\nu|(X) \ge \sum_{j = 1}^n \norm{\mu(A_j)}_E + \norm{\nu(A_j)}_E \ge \sum_{j = 1}^n \norm{(\mu + \nu)(A_j)}_E
+    \]
+
+    As this holds for all choices of $\seqf{A_j}$, $\norm{\mu + \nu}_{\text{var}} \le \norm{\mu}_{\text{var}} + \norm{\nu}_{\text{var}}$. 
+
+    (2): Let $\seq{\mu_n} \subset M(X, \cm; E)$ such that $\sum_{n \in \natp}\norm{\mu_n}_{\text{var}} < \infty$. For each $A \in \cm$, since $E$ is complete, let
+    \[
+    \mu(A) = \sum_{n = 1}^\infty \mu_n(A)
+    \]
+
+    then for any $\seq{A_k} \subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{k \in \natp}A_k$, 
+    \[
+        \mu(A) = \sum_{n = 1}^\infty \mu_n(A) = \sum_{n = 1}^\infty \sum_{k = 1}^\infty \mu_n(A_k) = \sum_{k = 1}^\infty \mu(A_k)
+    \]
+
+    by Fubini's theorem.
+    
+    % TODO: Actually link Fubini once it's there.
+\end{proof}
+

src/measure/vector/vector.tex

@@ -0,0 +1,45 @@
+\section{Vector Measures}
+\label{section:vector-measures}
+
+\begin{definition}[Vector Measure]
+\label{definition:vector-measure}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed space over $K \in \RC$, and $\mu: \cm \to E$, then $\mu$ is a \textbf{vector measure} if:
+        \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{definition}
+
+\begin{proposition}
+\label{proposition:vector-measure-bounded}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measure, then
+    \[
+    \sup_{A \in \cm}\norm{\mu(A)}_E < \infty
+    \]
+\end{proposition}
+\begin{proof}
+    For each $\phi \in E^*$, the mapping
+    \[
+    \mu_\phi: \cm \to \complex \quad A \mapsto \dpn{\mu(A), \phi}{E}
+    \]
+
+    is a complex measure. For each $A \in \cm$, let
+    \[
+    x_A: E^* \to \complex \quad \phi \mapsto \dpn{\mu(A), \phi}{E}
+    \]
+
+    then for each $\phi \in E^*$,
+    \[
+    \sup_{A \in \cm}|\dpn{\phi, x_A}{E^{**}}| = \sup_{A \in \cm}|\dpn{\mu(A),\phi}{E}| < \infty
+    \]
+
+    By \autoref{proposition:bornologic-continuous-complete} and \autoref{proposition:metrisable-bornologic}, $E^*$ is a Banach space. The \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness} implies that
+    \[
+    \sup_{A \in \cm}\norm{\mu(A)}_{E} = \sup_{A \in \cm}\norm{x_A}_{E^{**}} < \infty
+    \]
+
+\end{proof}
+
+
+

src/topology/functions/uniform.tex

@@ -17,29 +17,6 @@
 \end{remark}
 
 
-\begin{definition}[Space of Bounded Functions]
-\label{definition:bounded-function-space}
-    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
-    \begin{enumerate}
-        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
-    
-    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
-
-    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
-    \[
-    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
-    \]
-    so $f \in B(T; E)$.
-
-    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
-\end{proof}
-
-
 \begin{proposition}
 \label{proposition:uniform-limit-continuous}
     Let $X$ be a topological space and $Y$ be a uniform space, then:
@@ -72,20 +49,3 @@
     If $Y$ is complete, then $Y^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $C(T; X)$ and $UC(T; X)$ are both complete subspaces by \autoref{proposition:complete-closed}.
 \end{proof}
 
-
-\begin{definition}[Space of Bounded Continuous Functions]
-\label{definition:bounded-continuous-function-space}
-    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
-    \begin{enumerate}
-        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
-        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
-    \end{enumerate}
-\end{definition}
-\begin{proof}
-    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
-
-    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
-    
-    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
-\end{proof}
-

src/topology/main/c0.tex

@@ -5,7 +5,7 @@
 \label{definition:vanish-at-infinity}
     Let $X$ be a topological space, $E$ be a TVS over $K \in \RC$, and $f \in C(X; E)$, then $f$ \textbf{vanishes at infinity} if for every $U \in \cn_E^o(0)$, $\bracs{f \not\in U}$ is compact.
 
-    The set $C_0(X; \complex)$ is the space of all functions that vanish at infinity. 
+    The set $C_0(X; E)$ is the space of all functions that vanish at infinity. 
 \end{definition}
 
 \begin{proposition}

src/topology/main/index.tex

@@ -22,4 +22,5 @@
 \input{./support.tex}
 \input{./lch.tex}
 \input{./c0.tex}
+\input{./semicontinuity.tex}
 \input{./baire.tex}

src/topology/main/semicontinuity.tex

@@ -0,0 +1,64 @@
+\section{Semicontinuity}
+\label{section:semicontinuity}
+
+\begin{definition}[Semicontinuous]
+\label{definition:semicontinuous}
+    Let $X$ be a topological space, $f: X \to (-\infty, \infty]$, and $g: X \to [-\infty, \infty)$, then $f$ is \textbf{lower semicontinuous} if for each $a \in \real$, $\bracs{f > \alpha}$ is open, and $g$ is \textbf{upper semicontinuous} if for each $a \in \real$, $\bracs{f < \alpha}$ is open.
+\end{definition}
+
+\begin{proposition}[{{\cite[Proposition 7.11]{Folland}}}]
+\label{proposition:semicontinuous-properties}
+    Let $X$ be a topological space, then
+    \begin{enumerate}
+        \item For any $U \subset X$ open, $\one_U$ is lower semicontinuous.
+        \item For any $f: X \to (-\infty, \infty]$ lower semicontinuous and $\alpha \ge 0$, $\alpha f$ is lower semicontinuous. 
+        \item For any $f, g: X \to (-\infty, \infty]$ lower semicontinuous, $f + g$ is lower semicontinuous.
+        \item For any collection $\mathcal{F} \subset (-\infty, \infty]^X$ of lower semicontinuous functions, $F = \sup_{f \in F}f$ is lower semicontinuous. 
+        \item For any $f: X \to (-\infty, \infty]$ lower semicontinuous, $f$ is Borel measurable. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): For any $\alpha \in \real$, 
+    \[
+    \bracs{f > \alpha} = \begin{cases}
+        \emptyset &\alpha \ge 1 \\
+        U &\alpha \in [0, 1) \\
+        X &\alpha < 0
+    \end{cases}
+    \]
+
+    (2): If $\alpha = 0$, then $\alpha f = 0$ is continuous. If $\alpha > 0$, then for any $a \in \real$, $\bracs{\alpha f > a} = \bracs{f > a/\alpha}$ is open.
+
+    (3): Let $a \in \real$, $x_0 \in \bracs{f + g > a}$, and $\eps \in (0, ((f + g)(x_0) - a)/2)$, then
+    \[
+    \bracs{f + g > a} \supset \bracs{f > f(x_0) - \eps} \cap \bracs{g > g(x_0) - \eps} \in \cn^o(x_0)
+    \]
+
+    As this holds for all $x_0 \in \bracs{f + g > a}$, $\bracs{f + g > a}$ is open by \autoref{lemma:openneighbourhood}.
+
+    (4): For any $a \in \real$,
+    \[
+    \bracs{F > a} = \bigcup_{f \in \mathcal{F}} \bracs{f > a}
+    \]
+
+    is open. 
+
+    (5): By \autoref{proposition:borel-sigma-real-generators}.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:semicontinuous-lch}
+    Let $X$ be a LCH space and $f: X \to [0, \infty]$ be lower semicontinuous, then
+    \[
+    f = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi
+    \]
+\end{proposition}
+\begin{proof}
+    Let $x \in X$ such that $f(x) > 0$ and $a \in (0, f(x))$, then $\bracs{f > a}$ is open. By \hyperref[Urysohn's lemma]{lemma:lch-urysohn}, there exists $\phi \in C_c(X; [0, a])$ such that $\phi(x) = a$ and $\supp{\phi} \subset \bracs{f > a}$. As this holds for all $a \in (0, f(x))$, 
+    \[
+    f(x) = \sup_{\substack{\phi \in C_c(X) \\ 0 \le \phi \le f}}\phi(x)
+    \]
+\end{proof}
+
+
+

src/topology/uniform/cauchy.tex

@@ -24,6 +24,21 @@
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a filter on $X$, then $\fF$ is \textbf{Cauchy} if for every $V \in \fU$, there exists $E \in \fF$ such that $E$ is $V$-small.
 \end{definition}
 
+\begin{definition}[Cauchy Net]
+\label{definition:cauchy-net}
+    Let $(X, \fU)$ be a uniform space and $\net{x} \subset X$ be a net, then $\net{x}$ is \textbf{Cauchy} if for any $V \in \fU$, there exists $\alpha_0 \in A$ such that $(x_\alpha, x_\beta) \in V$ for all $\alpha, \beta \ge \alpha_0$.
+\end{definition}
+
+\begin{lemma}
+\label{lemma:cauchy-subnet}
+    Let $(X, \fU)$ be a uniform space and $\net{x} \subset X$ be a Cauchy net. If there exists a subnet $\angles{x_\beta}_B \subset \net{x}$ and $x \in X$ such that $x_\beta \to x$, then $x_\alpha \to x$.
+\end{lemma}
+\begin{proof}
+    Let $U \in \fU$ and $V \in \fU$ such that $V \circ V \subset U$, then there exists $\alpha_0 \in A$ such that $(x_\alpha, x_{\alpha'}) \in V$ for all $\alpha, \alpha' \ge \alpha_0$, and $\beta \in B$ with $\beta \ge \alpha_0$ such that $(x_\beta, x) \in V$. Thus, $(x_\alpha, x) \in V \circ V \subset U$ for all $\alpha \ge \alpha_0$, so $x_\alpha \to x$.
+\end{proof}
+
+
+
 \begin{proposition}[Cauchy Criterion]
 \label{proposition:cauchycriterion}
     Let $(X, \fU)$ be a uniform space and $\fF \subset 2^X$ be a convergent filter, then $\fF$ is Cauchy.

src/topology/uniform/metric.tex

@@ -70,14 +70,15 @@ The axioms of uniform spaces strongly resembles working in a metric space. In fa
     \begin{enumerate}
         \item[(FB1)] For any $J, J' \subset I$ finite and $r, r' > 0$,
         \[
-        \bigcap_{j \in J \cup J'}E(d_j, \min(r,r')E(d_j, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)} \cap \paren{\bigcap_{j \in J'}E(d_j, r')}
+        \bigcap_{j \in J \cup J'}E(d_j, r \wedge r') \subset \paren{\bigcap_{j \in J}E(d_j, r)} \cap \paren{\bigcap_{j \in J'}E(d_j, r')}
         \]
 
         \item[(UB1)] For each $i \in I$, $d(x, x) = 0$ for all $x \in X$. Thus for any $i \in I$ and $r > 0$, $E(d_i, r)$ contains the diagonal.
         \item[(UB2)] For each $J \subset I$ finite and $r > 0$,
-        \[
-        \paren{\bigcap_{j \in J}E(d_j, r/2)} \circ \paren{\bigcap_{j \in J}E(d_j, r)} \subset \bigcap_{j \in J}E(d_j, r/2) \circ E(d_j, r/2) \subset \bigcap_{j \in J}E(d_j, r)
-        \]
+        \begin{align*}
+        \paren{\bigcap_{j \in J}E(d_j, r/2)} \circ \paren{\bigcap_{j \in J}E(d_j, r)} &\subset \bigcap_{j \in J}E(d_j, r/2) \circ E(d_j, r/2) \\
+        &\subset \bigcap_{j \in J}E(d_j, r)
+        \end{align*}
         
         by the triangle inequality.
     \end{enumerate}