Added facts about vector measures.

src/measure/vector/vector.tex

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+\section{Vector Measures}
+\label{section:vector-measures}
+
+\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:
+        \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. 
+    \end{enumerate}
+    
+\end{definition}
+
+\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
+    \[
+    \sup_{A \in \cm}\norm{\mu(A)}_E < \infty
+    \]
+\end{proposition}
+\begin{proof}
+    For each $\phi \in E^*$, the mapping
+    \[
+    \mu_\phi: \cm \to \complex \quad A \mapsto \dpn{\mu(A), \phi}{E}
+    \]
+
+    is a complex measure. For each $A \in \cm$, let
+    \[
+    x_A: E^* \to \complex \quad \phi \mapsto \dpn{\mu(A), \phi}{E}
+    \]
+
+    then for each $\phi \in E^*$,
+    \[
+    \sup_{A \in \cm}|\dpn{\phi, x_A}{E^{**}}| = \sup_{A \in \cm}|\dpn{\mu(A),\phi}{E}| < \infty
+    \]
+
+    By \autoref{proposition:bornologic-continuous-complete} and \autoref{proposition:metrisable-bornologic}, $E^*$ is a Banach space. The \hyperref[Uniform Boundedness Principle]{theorem:uniform-boundedness} implies that
+    \[
+    \sup_{A \in \cm}\norm{\mu(A)}_{E} = \sup_{A \in \cm}\norm{x_A}_{E^{**}} < \infty
+    \]
+
+\end{proof}
+
+
+