Added the algebraic tensor product.

src/measure/bochner-integral/bochner.tex

@@ -44,6 +44,8 @@
     
 \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
@@ -58,3 +60,48 @@
     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 $f \in L^1(X; E)$, $I_\lambda f = \int f d\lambda\mu$ 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 $\int f d\lambda\mu = \limv{n}\int f_n d\lambda\mu$. 
+\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 $h \mapsto \int h d\lambda\mu$ is a bounded linear operator, $\int f d\lambda\mu = \limv{n}\int f_n d\lambda\mu$. 
+\end{proof}
+
+