\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 vector 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 vector 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:bornological-continuous-complete} and \autoref{proposition:metrisable-bornological}, $E^*$ 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}

\begin{proposition}
\label{proposition:vector-measure-properties}
    Let $(X, \cm)$ be a measurable, $E$ be a normed vector space over $K \in \RC$, and $\mu: \cm \to E$ be a vector measurethen:
    \begin{enumerate}
        \item For any $\seq{A_n} \subset \cm$ with $A_n \subset A_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}A_n} = \limv{n}\mu(A_n)$.
        \item For any $\seq{A_n} \subset \cm$ with $A_n \supset A_{n+1}$ for all $n \in \nat$, $\mu(\bigcap_{n \in \natp}A_n) = \limv{n}\mu(A_n)$.
    \end{enumerate}
\end{proposition}
\begin{proof}
    (1): Let $A_0 = \emptyset$. For each $n \in \natp$, let $B_n = A_n \setminus B_{n-1}$, then $\bigcup_{n \in \natp}A_n = \bigsqcup_{n \in \natp}B_n$ and
    \[
    \limv{n}\mu(A_n) = \sum_{n = 1}^\infty \mu(B_n) = \mu\paren{\bigcup_{n \in \natp}A_n}
    \]

    (2): By (1), 
    \begin{align*}
        \limv{n}\mu(A_n) &= \mu(A_1) - \limv{n}\mu(A_1 \setminus A_n) \\
        &= \mu(A_1) - \mu\paren{A_1 \setminus \bigcup_{n \in \natp}A_n} = \mu\paren{\bigcap_{n \in \natp}A_n}
    \end{align*}
     
\end{proof}