Replaced mentions of normed spaces to normed vector spaces.

src/dg/derivative/taylor.tex

@@ -8,7 +8,7 @@
     f(b) - f(a) - \sum_{k = 1}^n \frac{1}{k!}D^kf(a)(b - a)^k
     \]
 
-    is contained in the closed convex hull\footnote{It may be possible to sharpen the below claim to include the $1/(n+1)!$ factor. However, I was not able to follow the proof for this.} of
+    is contained in the closed convex hull of
     \[
     \bracs{D^{n+1}f(s)(t - a)^{n+1}  | s \in (a, b) \setminus N, t \in [a, b]}
     \]

src/fa/lc/continuous.tex

@@ -35,3 +35,4 @@
 \begin{proof}
     $(1) \Rightarrow (2)$: By continuity of $T$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$ such that for any $x \in \prod_{j = 1}^n E_j$, $\max_{1 \le j \le n}[x_j]_{E_j} < 1$ implies that $[Tx]_F < 1$. In which case, the inequality follows from linearity.
 \end{proof}
+

src/fa/lp/definition.tex

@@ -63,7 +63,7 @@
 
 \begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
 \label{theorem:minkowski}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^p(X; E)$, then
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, and $f, g \in \mathcal{L}^p(X; E)$, then
     \[
     \norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
     \]
@@ -89,7 +89,7 @@
 
 \begin{definition}[$L^p$ Space]
 \label{definition:lp}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^p(X; E)$. The quotient
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $p \in [1, \infty]$, then $\norm{\cdot}_{L^p(X; E)}$ is a seminorm on $\mathcal{L}^p(X; E)$. The quotient
     \[
     L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
     \]
@@ -100,24 +100,36 @@
     By \hyperref[Minkowski's Inequality]{theorem:minkowski}.
 \end{proof}
 
+
 \begin{proposition}
-\label{proposition:lp-simple-dense}
+\label{proposition:dct-lp}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a Banach space over $K \in \RC$, $p \in [1, \infty)$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$. If
+    \begin{enumerate}
+        \item[(a)] $f_n \to f$ strongly pointwise.
+        \item[(b)] There exists $g \in L^p(X) \cap L^+(X)$ such that $\norm{f_n}_E \le g$ for all $n \in \natp$.
+    \end{enumerate}
+    
+    then $f_n \to f$ in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
-    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
+    By assumptions $a$ and $b$, $\norm{f_n - f}_E \le 2g$ for all $n \in \natp$. Since $\seq{f_n} \subset L^p(X; E)$, $f \in L^p(X; E)$, and $g \in L^p(X)$, $\norm{f_n - f}_E^p, g^p \in L^1(X)$ for all $n \in \natp$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct}, 
-
-    For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
     \[
-    \limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0
+    \limv{n}\norm{f_n - f}_{L^p(X; E)}^p = \limv{n}\int \norm{f_n - f}_E d\mu  = 0
     \]
+\end{proof}
 
-    Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
+
+\begin{proposition}
+\label{proposition:lp-simple-dense}
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
+\end{proposition}
+\begin{proof}
+    Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ strongly pointwise as $n \to \infty$. By the \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^p(X; E)$.
 \end{proof}
 
 \begin{theorem}[Markov's Inequality]
 \label{theorem:markov-inequality}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $f: X \to E$ be a Borel measurable function, then
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, and $f: X \to E$ be a Borel measurable function, then
     \begin{enumerate}
         \item For any $\alpha > 0$, 
         \[
@@ -146,7 +158,7 @@
 
 \begin{proposition}
 \label{proposition:lp-in-measure}
-    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
+    Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed vector space, $p \in [1, \infty]$, $\seq{f_n} \subset L^p(X; E)$, and $f \in L^p(X; E)$ such that $f_n \to f$ in $L^p$, then $f_n \to f$ in measure.
 \end{proposition}
 \begin{proof}
     Let $\eps > 0$. If $p < \infty$, then by \hyperref[Markov's Inequality]{theorem:markov-inequality},

src/fa/norm/absolute.tex

@@ -3,12 +3,12 @@
 
 \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 $\sum_{n \in \natp}\norm{x_n}_E < \infty$.
+    Let $E$ be a normed vector 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:
+    Let $E$ be a normed vector 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 $x = \sum_{n = 1}^\infty x_n$.
@@ -24,7 +24,7 @@
 
 \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 $\sigma: \natp \to \natp$, 
+    Let $E$ be a normed vector space, then a series $\sum_{n = 1}^\infty x_n$ with $\seq{x_n} \subset E$ \textbf{converges unconditionally} if for any bijection $\sigma: \natp \to \natp$, 
     \[
     \sum_{n = 1}^\infty x_n = \sum_{n = 1}^\infty x_{\sigma(n)}
     \]

src/fa/norm/index.tex

@@ -1,7 +1,8 @@
-\chapter{Normed Spaces}
+\chapter{Normed Vector Spaces}
 \label{chap:normed-spaces}
 
 \input{./normed.tex}
 \input{./absolute.tex}
 \input{./linear.tex}
+\input{./separable.tex}
 \input{./multilinear.tex}

src/fa/norm/linear.tex

@@ -12,3 +12,25 @@
 \begin{proof}
     By \autoref{proposition:lc-spaces-linear-map}.
 \end{proof}
+
+
+\begin{theorem}[Linear Extension Theorem (Normed)]
+\label{theorem:linear-extension-theorem-normed}
+    Let $E$ be a normed vector space over $K \in \RC$, $F$ be a Banach space over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
+    \begin{enumerate}
+        \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
+        \item $\normn{\ol T}_{L(E; F)} = \normn{T}_{L(D; F)}$.
+        \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
+    \end{enumerate}
+
+\end{theorem}
+\begin{proof}
+    (1), (U): By \autoref{theorem:linear-extension-theorem-tvs}.
+
+    (2): Since $\ol{T}$ is continuous, the function
+    \[
+    N: E \to \real \quad x \mapsto \normn{\ol T}_{F} - \norm{T}_{L(D; F)}\cdot \norm{x}_E
+    \]
+
+    is continuous, so $\bracs{N \le 0} \supset D$ is closed. By density of $D$, $\bracs{N \le 0} = E$. Therefore $\normn{\ol T}_{L(E; F)} = \normn{T}_{L(D; F)}$.
+\end{proof}

src/fa/norm/multilinear.tex

@@ -3,7 +3,7 @@
 
 \begin{proposition}
 \label{proposition:bilinear-separate}
-    Let $E, F, G$ be normed spaces and $T: E \times F \to G$ be a bilinear map. If:
+    Let $E, F, G$ be normed vector spaces and $T: E \times F \to G$ be a bilinear map. If:
     \begin{enumerate}
         \item For each $x \in E$, $y \mapsto T(x, y)$ is a continuous linear map from $F$ to $G$.
         \item For each $y \in F$, $x \mapsto T(x, y)$ is a continuous linear map from $E$ to $G$.

src/fa/norm/normed.tex

@@ -34,7 +34,7 @@
 
 \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:
+    Let $E, F$ be normed vector 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:
     \begin{enumerate}
         \item[(a)] $\norm{x}_E \le C\norm{y}_F$.
         \item[(b)] $\norm{y - Tx}_F \le \gamma \norm{y}_F$.
@@ -69,7 +69,7 @@
 
 \begin{theorem}[Uniform Boundedness Principle]
 \label{theorem:uniform-boundedness}
-    Let $E, F$ be normed spaces and $\mathcal{T} \subset L(E; F)$. If
+    Let $E, F$ be normed vector spaces and $\mathcal{T} \subset L(E; F)$. If
     \begin{enumerate}
         \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
         \item $E$ is a Banach space.
@@ -86,3 +86,17 @@
 
     so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
 \end{proof}
+
+
+\begin{proposition}
+\label{proposition:dual-norm}
+    Let $E$ be a normed vector space, then for any $x \in E$, 
+    \[
+    \norm{x}_E = \sup_{\substack{\phi \in E^* \\ \norm{\phi}_{E^*} = 1}}\dpn{x, \phi}{E}
+    \]
+\end{proposition}
+\begin{proof}
+    For any $\phi \in E^*$ with $\norm{\phi}_{E^*} = 1$, $\dpn{x, \phi}{E} \le \norm{x}_E \cdot \norm{\phi}_{E^*} = \norm{x}_E$. On the other hand, by the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, there exists $x \in E^*$ such that $\dpn{x, \phi}{E} = \norm{x}_E$ and $\norm{\phi}_{E^*} \le 1$. 
+\end{proof}
+
+

src/fa/norm/separable.tex

@@ -0,0 +1,48 @@
+\section{Separable Normed Vector Spaces}
+\label{section:separable-banach-space}
+
+\begin{proposition}
+\label{proposition:separable-dual}
+    Let $E$ be a separable normed vector space, then $E^*$ is separable with respect to the weak*-topology. 
+\end{proposition}
+\begin{proof}
+    Let $\seq{x_n} \subset E$ be a dense subset and $S = \bracsn{\phi \in E^*| \norm{\phi}_{E^*} \le 1}$. For each $N \in \natp$, let
+    \[
+    T_N: S \to \real^N \quad \phi \mapsto (\dpn{x_1, \phi}{E}, \cdots, \dpn{x_N, \phi}{E})
+    \]
+
+    then since $\real^N$ is separable, there exists $\bracs{\phi_{N, k}}_{k = 1}^\infty \subset S$ such that $\bracs{T_N\phi_{N, k}}_{k = 1}^\infty$ is dense in $T_N(S)$.
+
+    Let $\phi \in S$, then for each $N \in \natp$, there exists $k_N \in \natp$ such that for each $1 \le n \le N$, 
+    \[
+    |\dpn{x_n, \phi_{N, k_N}}{E} - \dpn{x_n, \phi}{E}| \le \frac{1}{n}
+    \] 
+
+    Thus for each $N \in \natp$, $\dpn{x_n, \phi_{N, k_N}}{E} \to \dpn{x_n, \phi}{E}$ as $N \to \infty$. Since $\phi_{N, k_N} \to \phi$ pointwise on a dense subset of $E$, and $\bracsn{\phi_{N, k_N}|N \in \natp} \subset S$ is uniformly equicontinuous, $\phi_{N, k_N} \to \phi$ in the weak*-topology by \autoref{proposition:strong-operator-dense}. 
+\end{proof}
+
+\begin{proposition}
+\label{proposition:separable-banach-borel-sigma-algebra}
+    Let $E$ be a separable normed vector space, then the Borel $\sigma$-algebra on $E$ is generated by the following families of sets:
+    \begin{enumerate}
+        \item Open sets in $E$ with respect to the strong topology. 
+        \item $\bracs{B(x, r)|x \in E, r > 0}$.
+        \item $\bracsn{\ol{B(x, r)}|x \in E, r > 0}$. 
+        \item Open sets in $E$ with respect to the weak topology. 
+    \end{enumerate}
+    
+\end{proposition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2) $\Leftrightarrow$ (3): By \autoref{proposition:separable-metric-borel-sigma-algebra}.
+
+    (4) $\subset$ (1): Every weakly open set is strongly open. 
+
+    (2) $\subset$ (4): By \autoref{proposition:seminorm-lsc}, $\norm{\cdot}_E: E \to [0, \infty)$ is Borel measurable with respect to the weak topology. For any $x \in E$, let
+    \[
+    \phi_x: E \to [0, \infty) \quad y \mapsto \norm{x - y}_E
+    \]
+
+    then $\phi_x$ is Borel measurable with respect to the weak topology, so $B(x, r) = \bracs{\phi_x < r}$ is a Borel set with respect to the weak topology. 
+\end{proof}
+
+

src/fa/rs/bv.tex

@@ -16,7 +16,7 @@
 
     is the \textbf{total variation} of $f$ on $[a, b]$ with respect to $\rho$.
 
-    If $E$ is a normed space, then the variation and total variation of $f$ is taken with respect to its norm.
+    If $E$ is a normed vector space, then the variation and total variation of $f$ is taken with respect to its norm.
 \end{definition}
 
 \begin{definition}[Bounded Variation, {{\cite[Proposition X.1.1]{Lang}}}]
@@ -35,7 +35,7 @@
         then $f \in BV([a, b]; E)$ with $[f]_{\text{var}, \rho} \le M_\rho$.
         \item For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\text{var}, \rho}$.
     \end{enumerate}
-    If $(E, \norm{\cdot}_E)$ is a normed space, then
+    If $(E, \norm{\cdot}_E)$ is a normed vector space, then
     \begin{enumerate}
         \item[(5)] $f$ has at most countably many discontinuities.
     \end{enumerate}

src/fa/tvs/continuous.tex

@@ -51,3 +51,30 @@
     \end{enumerate}
     The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
 \end{definition}
+
+\begin{theorem}[Linear Extension Theorem (TVS)]
+\label{theorem:linear-extension-theorem-tvs}
+    Let $E$ be a TVS over $K \in \RC$, $F$ be a complete Hausdorff TVS over $K$, $D \subset E$ be a dense subspace, and $T \in L(D; F)$, then:
+    \begin{enumerate}
+        \item There exists an extension $\ol T \in L(E; F)$ such that $\ol T|_D = T$.
+        \item[(U)] For any $S \in C(E; F)$ satisfying (1), $S = \ol T$.
+    \end{enumerate}
+
+\end{theorem}
+\begin{proof}
+    By (3) of \autoref{definition:continuous-linear}, $T \in UC(D; F)$. By \autoref{theorem:uniform-continuous-extension}, there exists a unique $\ol T \in C(E; F)$ such that $\ol T|_D = T$. 
+
+    It remains to show that $\ol T$ is linear. Since $\ol T$ is continuous, the maps
+    \[
+    A: E \times E \to E \quad (x, y) \mapsto \ol Tx + \ol Ty - \ol T(x + y)
+    \]
+
+    and
+    \[
+    M: K \times E \to E \quad (\lambda x) \mapsto \lambda \ol Tx - Tx
+    \]
+
+    are continuous. Thus $\bracs{A = 0} \supset D \times D$ and $\bracs{M = 0} \supset K \times D$ are both closed. By density of $D$, $\bracs{A = 0} = E \times E$ and $\bracs{M = 0} = K \times E$. Therefore $T$ is linear. 
+\end{proof}
+
+

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

@@ -153,14 +153,34 @@
 
 \begin{definition}[Strong Operator Topology]
 \label{definition:strong-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F)$ is the \textbf{strong operator topology}.
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
 
     The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
 \end{definition}
 
+
+\begin{proposition}
+\label{proposition:strong-operator-dense}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
+    \begin{enumerate}
+        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
+        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
+    \end{enumerate}
+    
+    then $T_\alpha \to T$ in $L_s(E; F)$.
+\end{proposition}
+\begin{proof}
+    Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
+    \[
+    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
+    \]
+\end{proof}
+
+
+
 \begin{definition}[Weak Operator Topology]
 \label{definition:weak-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $L(E; F_w)$ is the \textbf{weak operator topology}.
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_w^E$ is the \textbf{weak operator topology}.
 
     The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
 \end{definition}

src/measure/bochner-integral/bochner.tex

@@ -0,0 +1,60 @@
+\section{The Bochner Integral}
+\label{section:bochner-integral}
+
+\begin{definition}[Bochner Integral]
+\label{definition:bochner-integral}
+    Let $(X, \cm, \mu)$ be a measure space and $E$ be a Banach space over $K \in \RC$, then there exists a unique $I \in L(L^1(X; E); E)$ such that:
+    \begin{enumerate}
+        \item For any $x \in E$ and $A \in \cm$ with $\mu(A) < \infty$, $I(x \cdot \one_A) = x \cdot \mu(A)$. 
+        \item For all $f \in L^1(X; E)$, $\norm{If}_E \le \int \norm{f}_E d\mu$. 
+    \end{enumerate}
+
+    For any $f \in L^1(X; E)$, $If = \int f d\mu$ is the \textbf{Bochner integral} of $f$.
+\end{definition}
+\begin{proof}
+    (1): For any $\phi \in \Sigma(X; E) \cap L^1(X; E)$, let
+    \[
+    I\phi = \sum_{y \in \phi(X)}^n y \cdot \mu\bracs{\phi = y}
+    \]
+
+    For any $\lambda \in K$, if $\lambda \ne 0$, then the mapping $x \mapsto \lambda x$ is a bijection, so
+    \[
+    I(\lambda \phi) = \sum_{\lambda y \in \lambda \phi(X)}^n (\lambda y) \cdot \mu\bracs{\lambda \phi = \lambda y} 
+    = \lambda \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} = \lambda I\phi
+    \]
+
+    If $\lambda = 0$, then $I(\lambda \phi) = I(0) = 0$. 
+
+    Let $\phi, \psi \in \Sigma(X; E) \cap L^1(X; E)$, then
+    \begin{align*}
+        I\phi + I\psi &= \sum_{y \in \phi(X)}y \cdot \mu\bracs{\phi = y} + \sum_{z \in \psi(X)}z \cdot \mu\bracs{\psi = z} \\
+        &= \sum_{y \in \phi(X)} \sum_{z \in \psi(X)} (y + z) \cdot \mu\bracs{\phi = y, \psi = z} \\
+        &= \sum_{y \in (\phi + \psi)(X)}\sum_{{z \in \phi(X) \atop {z' \in \psi(X) \atop z + z' = y}}}(z + z') \cdot \mu\bracsn{\phi = g, \psi = z'} \\
+        &= \sum_{y \in (\phi + \psi)(X)}y \cdot \mu(\bracs{\phi + \psi = y}) = I\phi + I\psi
+    \end{align*}
+
+    so $I$ is a linear operator on $\Sigma(X; E) \cap L^1(X; E)$ that satisfies (1).
+
+    (2): For any $\phi \in \Sigma(X; E) \cap L^1(X; E)$, 
+    \[
+    \norm{I\phi}_E \le \sum_{y \in \phi(X)}\norm{y}_E \cdot \mu\bracs{\phi = y} = \int \norm{\phi}_E d\mu = \norm{\phi}_{L^1(X; E)}
+    \]
+    
+    By \autoref{proposition:lp-simple-dense}, $\Sigma(X; E) \cap L^1(X; E)$ is dense in $L^1(X; E)$. Therefore by the \hyperref[Linear Extension Theorem]{theorem:linear-extension-theorem-normed}, $I$ admits a unique norm-preserving extension to $L^1(X; E)$. 
+    
+\end{proof}
+
+\begin{theorem}[Dominated Convergence Theorem]
+\label{theorem:dct-bochner}
+    Let $(X, \cm, \mu)$ be a measure spacs, $E$ be a Banach space over $K \in \RC$, $\seq{f_n} \subset L^1(X; E)$, and $f \in L^1(X; E)$. If
+    \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}
+    
+    then $\int f d\mu = \limv{n}\int f_n d\mu$. 
+\end{theorem}
+\begin{proof}
+    By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X; E)$. Since $h \mapsto \int h d\mu$ is a bounded linear operator, $\int f d\mu = \limv{n}\int f_n d\mu$. 
+\end{proof}
+

src/measure/bochner-integral/index.tex

@@ -2,3 +2,4 @@
 \label{chap:bochner-integral}
 
 \input{./strongly.tex}
+\input{./bochner.tex}

src/measure/bochner-integral/strongly.tex

@@ -6,16 +6,22 @@
     Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $f: X \to E$, then the following are equivalent:
     \begin{enumerate}
         \item For each $\phi \in E^*$, $\phi \circ f$ is $(\cm, \cb_K)$-measurable and $f(X) \subset E$ is separable.
-        \item $f$ is $(\cm, \cb_E)$ measurable and $f(X) \subset E$ is separable.
+        \item $f$ is $(\cm, \cb_E)$-measurable and $f(X) \subset E$ is separable.
         \item There exists a sequence $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that
         \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}
+
+    If the above holds, then $f$ is a \textbf{strongly measurable} function. 
 \end{definition}
 \begin{proof}
-    (1) $\Rightarrow$ (2): TODO
+    (1) $\Rightarrow$ (2): First suppose that $E$ is separable. By \autoref{proposition:separable-banach-borel-sigma-algebra}, the Borel $\sigma$-algebra on $E$ coincides with the $\sigma$-algebra on $E$ generated by the weak topology. Thus if $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$, then $f$ is $(\cm, \cb_E)$-measurable. 
+
+    Now suppose that $E$ is arbitrary. Let $F \subset E$ be the closure of the linear span of $f(X)$, then $F$ is a separable closed subspace of $E$. For any $\phi \in F^*$, by the \hyperref[Hahn-Banch Theorem]{theorem:hahn-banach}, there exists an extension $\Phi \in E^*$ of $\phi$. In which case, since $f(X) \subset F$, for any Borel set $B \in \cb_{K}$, $\bracs{\phi \circ f \in B} = \bracs{\Phi \circ f \in B} \in \cm$. Thus $\phi \circ f$ is $(\cm, \cb_K)$-measurable for all $\phi \in E^*$. 
+
+    By the separable case, $f$ is $(\cm, \cb_{F})$-measurable. Let $B \in \cb_E$, then $B \cap F \in \cb_F$ by \autoref{lemma:borel-induced}. Therefore $\bracs{f \in B} = \bracs{f \in B \cap F} \in \cm$, and $f$ is $(\cm, \cb_E)$-measurable. 
 
     (2) $\Rightarrow$ (3): By \autoref{proposition:measurable-simple-separable-norm}.
 
@@ -26,3 +32,21 @@
 
     and each $f_n$ is finitely-valued, $f(X)$ is separable.
 \end{proof}
+
+\begin{proposition}
+\label{proposition:strongly-measurable-properties}
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, then:
+    \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{proposition}
+\begin{proof}
+    (1): Since $x \mapsto \lambda$ is continuous, $\lambda f$ is strongly measurable by (2) of \autoref{definition:strongly-measurable}. 
+
+    By (3) of \autoref{definition:strongly-measurable}, there exists $\seq{f_n}, \seq{g_n} \subset \Sigma(X, \cm; E)$ such that $f_n \to f$ and $g_n \to g$ strongly pointwise. In which case, $\seq{f_n + g_n} \subset \Sigma(X, \cm; E)$ and $f_n + g_n \to f + g$ strongly pointwise. Therefore $f + g$ is also strongly measurable. 
+
+    (2): By \autoref{proposition:metric-measurable-limit}, $f$ is $(\cm, \cb_E)$-measurable. Since $f(X) \subset \overline{\bigcup_{n \in \natp}}f_n(X)$, $f(X)$ is also separable, so $f$ is strongly measurable by (1) of \autoref{definition:strongly-measurable}.
+\end{proof}
+

src/measure/lebesgue-integral/simple.tex

@@ -8,7 +8,7 @@
 
 \begin{definition}[Integral of Non-Negative Simple Functions]
 \label{definition:lebesgue-simple}
-    Let $(X, \cm, \mu)$ be a measure space and $f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}} \in \Sigma^+(X, \cm)$ be a non-negative simple function in standard form, then\footnote{With the convention that $0 \cdot \infty = 0$.}
+    Let $(X, \cm, \mu)$ be a measure space and $f = \sum_{y \in f(X)}y \cdot \one_{\bracs{f = y}} \in \Sigma^+(X, \cm)$ be a non-negative simple function in standard form, then (with the convention that $0 \cdot \infty = 0$)
     \[
     \int f d\mu = \int f(x) \mu(dx) = \sum_{y \in f(X)}y \cdot \mu(\bracs{f = y})
     \]

src/measure/measurable-maps/metric.tex

@@ -71,7 +71,7 @@
 
 \begin{proposition}
 \label{proposition:measurable-simple-separable}
-    Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^Y$\footnote{This mapping is typically obtained as slices of the level sets of a continuous function $Y \times Y \to \real$.} such that
+    Let $(X, \cm)$ be a measurable space, $Y$ be a separable metric space, and $N: Y \to 2^Y$ such that
     \begin{enumerate}
         \item[(a)] For each $y \in Y$, $y \in \ol{N(y)^o}$.
         \item[(b)] $\bigcap_{y \in Y}N(y) \ne \emptyset$.
@@ -125,7 +125,7 @@
 
 \begin{proposition}
 \label{proposition:measurable-simple-separable-norm}
-    Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_E)$ be a separable normed space, and $f: X \to E$, then the following are equivalent:
+    Let $(X, \cm)$ be a measurable space, $(E, \norm{\cdot}_E)$ be a separable normed vector space, and $f: X \to E$, then the following are equivalent:
     \begin{enumerate}
         \item $f$ is $(\cm, \cb_E)$-measurable.
         \item There exists simple functions $\seq{f_n}$ such that $\abs{f_n} \le \abs{f}$ for all $n \in \natp$, and $f_n \to f$ pointwise.

src/measure/measure/kolmogorov.tex

@@ -13,7 +13,7 @@
 \label{lemma:kolmogorov-compact-sequence}
     Let $\seq{X_n}$ be topological spaces where for each $n \in \natp$,
     \begin{enumerate}
-        \item[(a)] Every finite measure on $\prod_{j = 1}^n X_j$ is regular\footnote{A potential sufficient condition for this is that each $X_n$ is LCH where every open set is $\sigma$-compact. However, I have yet to verify if this condition persists over products.}.
+        \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}

src/measure/radon/c0.tex

@@ -26,7 +26,7 @@
 
 \begin{definition}[Space of Finite Radon Measures]
 \label{definition:space-radon-measures}
-    Let $X$ be a LCH space and $E$ be a normed space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
+    Let $X$ be a LCH space and $E$ be a normed vector space over $K \in \RC$, then $M_R(X; E)$ is the \textbf{space of finite Radon measures} on $X$, which forms a vector space over $K$. 
 \end{definition}
 \begin{proof}
     Let $\mu, \nu \in M_R(X; E)$, then for any $A \in \cb_X$, $|\mu + \nu|(A) \le |\mu|(A) + |\nu|(A)$. Let $\eps > 0$, then by outer regularity and \autoref{proposition:radon-measurable-description}, there exists $K \subset A$ compact and $U \in \cn^o(A)$ such that $(|\mu| + |\nu|)(A \setminus K), (|\mu| + |\nu|)(U \setminus A) < \eps$. Therefore $|\mu + \nu|$ is regular on all Borel sets, and hence Radon. 

src/measure/radon/radon.tex

@@ -150,7 +150,7 @@
 
 \begin{proposition}
 \label{proposition:radon-cc-dense}
-    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
+    Let $X$ be a LCH space, $\mu: \cb_X \to [0, \infty]$ be a Radon measure, $E$ be a normed vector space, and $p \in [1, \infty)$, then $C_c(X; E)$ is dense in $L^p(X; E)$.
 \end{proposition}
 \begin{proof}
     By \autoref{proposition:lp-simple-dense}, $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$. Using linearity, it is sufficient to approximate indicator functions of Borel sets with finite measure.

src/measure/sets/algebra.tex

@@ -72,90 +72,14 @@
     Let $X$ be a set and $\ce \subset 2^X$, then the $\sigma$-algebra $\sigma(\ce)$ \textbf{generated by} $\ce$ is the smallest $\sigma$-algebra on $X$ containing $\ce$.
 \end{definition}
 
+\begin{definition}[Induced $\sigma$-Algebra]
+\label{definition:}
+    Let $X$ be a set, $\cm \subset 2^X$ be a $\sigma$-algebra over $X$, and $E \subset X$, then the collection
+    \[
+    \cm_E = \bracs{A \cap E|A \in \cm}
+    \]
 
-\begin{definition}[Borel $\sigma$-Algebra]
+    is the \textbf{$\sigma$-algebra on $E$ induced by $\cm$}.
-\label{definition:borel-sigma-algebra}
-    Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$.
 \end{definition}
 
 
-\begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$]
-\label{definition:borel-sigma-algebra-extended}
-    The family
-    \[
-    \cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real}
-    \]
-
-    is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$.
-\end{definition}
-
-
-\begin{proposition}
-\label{proposition:borel-sigma-real-generators}
-    The following families of sets generate the Borel $\sigma$-algebra on $\real$:
-    \begin{enumerate}
-        \item $\bracs{(-\infty, a]| a \in \real}$.
-        \item $\bracs{(a, \infty)|a \in \real}$.
-        \item $\bracs{[a, \infty)| a \in \real}$.
-        \item $\bracs{(-\infty, a)| a \in \real}$.
-        \item $\bracs{[a, b)| -\infty < a < b < \infty}$.
-        \item $\bracs{[a, b]| -\infty < a < b < \infty}$.
-        \item $\bracs{(a, b]| -\infty < a < b < \infty}$.
-        \item $\bracs{(a, b)| -\infty < a < b < \infty}$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    It is sufficient to show that the $\sigma$-algebra generated by any of the above two families coincide, and that the resulting $\sigma$-algebra is the Borel $\sigma$-algebra on $\real$.
-
-    (1) $\to$ (2): For any $a \in \real$, $(a, \infty) = (-\infty, a)^c$.
-
-    (2) $\to$ (3): For any $a \in \real$, $[a, \infty) = \bigcap_{n \in \natp}(a - 1/n, \infty)$.
-
-    (3) $\to$ (4): For any $a \in \real$, $(-\infty, a) = [a, \infty)^c$.
-
-    (4) $\to$ (5): For any $a, b \in \real$, $[a, b) = (-\infty, b) \cap (-\infty, a)^c$.
-
-    (5) $\to$ (6): For any $a, b \in \real$, $[a, b]= \bigcap_{n \in \natp}(a - 1/n, b]$.
-
-    (6) $\to$ (7): For any $a, b \in \real$, $(a, b] = \bigcup_{n \in \natp}[a + 1/n, b]$.
-
-    (7) $\to$ (8): For any $a, b \in \real$, $(a, b) = \bigcup_{n \in \natp}(a, b - 1/n]$.
-
-    (8) $\to$ (1): For any $a \in \real$, $(-\infty, a] = \bigcup_{n \in \natp}\bigcap_{k \in \natp}(-n, a + 1/k]$.
-
-    For any $U \subset X$ open and $q \in U \cap \rational$, there exists $r_q > 0$ such that $(q - r_q, q + r_q) \subset U$. In which case,
-    \[
-    U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q)
-    \]
-
-    so (8) generates all open sets in $\real$. Conversely, every element of (8) is open, so the $\sigma$-algebra generated by (8) is the Borel $\sigma$-algebra on $\real$.
-\end{proof}
-
-\begin{proposition}
-\label{proposition:borel-sigma-extended-generators}
-    The following families of sets generate the Borel $\sigma$-algebra on $\ol \real$:
-    \begin{enumerate}
-        \item $\bracs{[-\infty, a]| a \in \real}$.
-        \item $\bracs{(a, \infty]|a \in \real}$.
-        \item $\bracs{[a, \infty]| a \in \real}$.
-        \item $\bracs{[-\infty, a)| a \in \real}$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1) $\to$ (2): For any $a \in \real$, $(a, \infty] = [-\infty, a]^c$.
-
-    (2) $\to$ (3): For any $a \in \real$, $[a, \infty] = \bigcap_{n \in \natp}(a - 1/n, \infty]$.
-
-    (3) $\to$ (4): For any $a \in \real$, $[-\infty, a) = [a, \infty]^c$.
-
-    (4) $\to$ (1): For any $a \in \real$, $[-\infty, a] = \bigcap_{n \in \natp}[-\infty, a + 1/n)$.
-
-    By definition, all elements of (1), (2), (3), and (4) belong to $\cb_{\ol{\real}}$. Let $\cm$ be the $\sigma$-algebra generated by (1), (2), (3), and (4), then
-    \[
-    \bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n]
-    \]
-
-    are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \autoref{proposition:borel-sigma-real-generators}.
-
-    In addition, for any $E \in \cb_{\ol{\real}}$, $E = (E \cap \real) \cup (E \setminus \real)$, where $E \cap \real \in \cb_\real$. Since $\bracs{\infty}, \bracs{-\infty} \in \cm$ and $\cb_\real \subset \cm$, $E \in \cm$ and $\cm = \cb_{\ol \real}$.
-\end{proof}

src/measure/sets/borel.tex

@@ -0,0 +1,125 @@
+\section{The Borel $\sigma$-Algebra}
+\label{section:borel-sigma-algebra}
+
+
+\begin{definition}[Borel $\sigma$-Algebra]
+\label{definition:borel-sigma-algebra}
+    Let $(X, \topo)$ be a topological space, then the \textbf{Borel $\sigma$-algebra} $\cb_X$ on $X$ is the $\sigma$-algebra generated by $\topo$.
+\end{definition}
+
+
+\begin{definition}[Borel $\sigma$-Algebra on $\ol{\real}$]
+\label{definition:borel-sigma-algebra-extended}
+    The family
+    \[
+    \cb_{\ol{\real}} = \bracsn{E \subset \ol \real| E \cap \real \in \cb_\real}
+    \]
+
+    is the \textbf{Borel $\sigma$-algebra} on $\ol{\real}$.
+\end{definition}
+
+
+\begin{proposition}
+\label{proposition:borel-sigma-real-generators}
+    The following families of sets generate the Borel $\sigma$-algebra on $\real$:
+    \begin{enumerate}
+        \item $\bracs{(-\infty, a]| a \in \real}$.
+        \item $\bracs{(a, \infty)|a \in \real}$.
+        \item $\bracs{[a, \infty)| a \in \real}$.
+        \item $\bracs{(-\infty, a)| a \in \real}$.
+        \item $\bracs{[a, b)| -\infty < a < b < \infty}$.
+        \item $\bracs{[a, b]| -\infty < a < b < \infty}$.
+        \item $\bracs{(a, b]| -\infty < a < b < \infty}$.
+        \item $\bracs{(a, b)| -\infty < a < b < \infty}$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    It is sufficient to show that the $\sigma$-algebra generated by any of the above two families coincide, and that the resulting $\sigma$-algebra is the Borel $\sigma$-algebra on $\real$.
+
+    (1) $\to$ (2): For any $a \in \real$, $(a, \infty) = (-\infty, a)^c$.
+
+    (2) $\to$ (3): For any $a \in \real$, $[a, \infty) = \bigcap_{n \in \natp}(a - 1/n, \infty)$.
+
+    (3) $\to$ (4): For any $a \in \real$, $(-\infty, a) = [a, \infty)^c$.
+
+    (4) $\to$ (5): For any $a, b \in \real$, $[a, b) = (-\infty, b) \cap (-\infty, a)^c$.
+
+    (5) $\to$ (6): For any $a, b \in \real$, $[a, b]= \bigcap_{n \in \natp}(a - 1/n, b]$.
+
+    (6) $\to$ (7): For any $a, b \in \real$, $(a, b] = \bigcup_{n \in \natp}[a + 1/n, b]$.
+
+    (7) $\to$ (8): For any $a, b \in \real$, $(a, b) = \bigcup_{n \in \natp}(a, b - 1/n]$.
+
+    (8) $\to$ (1): For any $a \in \real$, $(-\infty, a] = \bigcup_{n \in \natp}\bigcap_{k \in \natp}(-n, a + 1/k]$.
+
+    For any $U \subset X$ open and $q \in U \cap \rational$, there exists $r_q > 0$ such that $(q - r_q, q + r_q) \subset U$. In which case,
+    \[
+    U = \bigcup_{q \in U \cap \rational}(q - r_q, q + r_q)
+    \]
+
+    so (8) generates all open sets in $\real$. Conversely, every element of (8) is open, so the $\sigma$-algebra generated by (8) is the Borel $\sigma$-algebra on $\real$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:borel-sigma-extended-generators}
+    The following families of sets generate the Borel $\sigma$-algebra on $\ol \real$:
+    \begin{enumerate}
+        \item $\bracs{[-\infty, a]| a \in \real}$.
+        \item $\bracs{(a, \infty]|a \in \real}$.
+        \item $\bracs{[a, \infty]| a \in \real}$.
+        \item $\bracs{[-\infty, a)| a \in \real}$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1) $\to$ (2): For any $a \in \real$, $(a, \infty] = [-\infty, a]^c$.
+
+    (2) $\to$ (3): For any $a \in \real$, $[a, \infty] = \bigcap_{n \in \natp}(a - 1/n, \infty]$.
+
+    (3) $\to$ (4): For any $a \in \real$, $[-\infty, a) = [a, \infty]^c$.
+
+    (4) $\to$ (1): For any $a \in \real$, $[-\infty, a] = \bigcap_{n \in \natp}[-\infty, a + 1/n)$.
+
+    By definition, all elements of (1), (2), (3), and (4) belong to $\cb_{\ol{\real}}$. Let $\cm$ be the $\sigma$-algebra generated by (1), (2), (3), and (4), then
+    \[
+    \bracs{\infty} = \bigcap_{n \in \nat}[n, \infty] \quad \bracs{-\infty} = \bigcap_{n \in \nat}[-\infty, n]
+    \]
+
+    are elements of $\cm$. For any $a \in \real$, $(a, \infty) = (a, \infty] \setminus \bracs{\infty}$, so $\cm \supset \cb_\real$ by \autoref{proposition:borel-sigma-real-generators}.
+
+    In addition, for any $E \in \cb_{\ol{\real}}$, $E = (E \cap \real) \cup (E \setminus \real)$, where $E \cap \real \in \cb_\real$. Since $\bracs{\infty}, \bracs{-\infty} \in \cm$ and $\cb_\real \subset \cm$, $E \in \cm$ and $\cm = \cb_{\ol \real}$.
+\end{proof}
+
+
+
+\begin{proposition}
+\label{proposition:separable-metric-borel-sigma-algebra}
+    Let $X$ be a separable metric space, then the Borel $\sigma$-algebra on $X$ is generated by the following families of sets:
+    \begin{enumerate}
+        \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{proposition}
+\begin{proof}
+    (1) $\subset$ (2): Let $U \subset X$ be open. By \autoref{definition:dense}, there exists a countable dense subset $S \subset U$. For each $x \in S$, let $r_x > 0$ such that $B(x, r) \subset U$, then $U = \bigcup_{x \in S}B(x, r_x)$ is a countable union of open balls.
+
+    (2) $\subset$ (3): For any $x \in X$ and $r > 0$, $B(x, r) = \bigcup_{n \in \natp}\overline{B(x, r - 1/n)}$ is a countable union of closed balls. 
+
+    (3) $\subset$ (1): For each $x \in X$ and $r > 0$, $\overline{B(x, r)}$ is closed. 
+\end{proof}
+
+
+\begin{lemma}
+\label{lemma:borel-induced}
+    Let $X$ be a topological space and $Y \subset X$ be a subspace, then the Borel $\sigma$-algebra on $Y$ coincides with the $\sigma$-algebra on $Y$ induced by $\cb_X$.
+\end{lemma}
+\begin{proof}
+    Since $\bracsn{A \in \cb_X|A \cap Y \in \cb_Y}$ is a $\sigma$-algebra that contains all open sets in $X$, $\cb_Y$ contains the induced $\sigma$-algebra. 
+
+    On the other hand, the induced $\sigma$-algebra contains all open sets in $Y$, so it contains $\cb_Y$. 
+
+    Therefore the two $\sigma$-algebras coincide. 
+\end{proof}
+
+

src/measure/sets/index.tex

@@ -2,6 +2,7 @@
 \label{chap:set-system}
 
 \input{./algebra.tex}
+\input{./borel.tex}
 \input{./lambda.tex}
 \input{./elementary.tex}
 \input{./limits.tex}

src/measure/vector/variation.tex

@@ -3,7 +3,7 @@
 
 \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
+    Let $(X, \cm)$ be a measure space, $\mu$ be a vector measure taking values in a normed vector space $E$, and
     \[
     |\mu|(A) = \sup\bracs{\sum_{j = 1}^n \norm{\mu(A_j)}_E \bigg | \seqf{A_j} \subset \cm, A = \bigsqcup_{j = 1}^n A_j}
     \]

src/measure/vector/vector.tex

@@ -3,7 +3,7 @@
 
 \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:
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector 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. 
@@ -13,7 +13,7 @@
 
 \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
+    Let $(X, \cm)$ be a measurable space, $E$ be a normed vector space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measure, then
     \[
     \sup_{A \in \cm}\norm{\mu(A)}_E < \infty
     \]