Added the Bochner integral.

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}
+