\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}

\subsection{Vector Measure Version}
\label{subsection:bochner-vector}



\begin{definition}[Bochner Integral]
\label{definition:bochner-integral-vector}
    Let $(X, \cm)$ be a measurable space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$,
    \[
    \lambda: E \times F \to G \quad (x, y) \mapsto xy
    \]

    be a bounded bilinear map, and $\mu: \cm \to F$ be a vector measure, then there exists a unique $I_\lambda \in L(L^1(X, |\mu|; E); G)$ such that:
    \begin{enumerate}
        \item For any $x \in E$ and $A \in \cm$, $I_\lambda(x \cdot \one_A) = x \mu(A)$.
        \item For any $f \in L^1(X, |\mu|; E)$, $\normn{I_\lambda f}_{G} \le \norm{\lambda}_{L^2(E, F; G)} \cdot \norm{f}_{L^1(X, |\mu|; E)}$.
    \end{enumerate}

    
    For any $f \in L^1(X; E)$, $I_\lambda f = \int \lambda(f, d\mu)$ is the \textbf{Bochner integral} of $f$ with respect to $\mu$ and $\lambda$. 
\end{definition}
\begin{proof}
    Same as \autoref{definition:bochner-integral}.
\end{proof}


\begin{theorem}[Dominated Convergence Theorem]
\label{theorem:dct-bochner-vector}
    Let $(X, \cm, \mu)$ be a measure space, $E, F$ be normed spaces over $K \in \RC$, $G$ be a Banach space over $K$, 
    \[
    \lambda: E \times F \to G \quad (x, y) \mapsto xy
    \]

    be a bounded bilinear map, $\mu: \cm \to F$ be a vector measure, $\seq{f_n} \subset L^1(X, |\mu|; E)$, and $f \in L^1(X, |\mu|; E)$. 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 $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. 
\end{theorem}
\begin{proof}
    By the classical \hyperref[Dominated Convergence Theorem]{proposition:dct-lp}, $f_n \to f$ in $L^1(X, |\mu|; E)$. Since $h \mapsto \int \lambda(f , d\mu)$ is a bounded linear operator, $\int \lambda(f , d\mu) = \limv{n}\int \lambda(f_n, d\mu)$. 
\end{proof}


