Added the Bochner integral.

src/fa/norm/separable.tex

@@ -0,0 +1,48 @@
+\section{Separable Normed Spaces}
+\label{section:separable-banach-space}
+
+\begin{proposition}
+\label{proposition:separable-dual}
+    Let $E$ be a separable normed 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 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}
+
+