68 lines
2.6 KiB
TeX
68 lines
2.6 KiB
TeX
\section{Vector Measures}
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\label{section:vector-measures}
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\begin{definition}[Vector Measure]
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\label{definition:vector-measure}
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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:
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\begin{enumerate}
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\item[(M1)] $\mu(\emptyset) = 0$.
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\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.
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\end{enumerate}
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\end{definition}
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\begin{proposition}
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\label{proposition:vector-measure-bounded}
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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
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\[
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\sup_{A \in \cm}\norm{\mu(A)}_E < \infty
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\]
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\end{proposition}
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\begin{proof}
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For each $\phi \in E^*$, the mapping
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\[
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\mu_\phi: \cm \to \complex \quad A \mapsto \dpn{\mu(A), \phi}{E}
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\]
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is a complex measure. For each $A \in \cm$, let
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\[
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x_A: E^* \to \complex \quad \phi \mapsto \dpn{\mu(A), \phi}{E}
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\]
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then for each $\phi \in E^*$,
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\[
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\sup_{A \in \cm}|\dpn{\phi, x_A}{E^{**}}| = \sup_{A \in \cm}|\dpn{\mu(A),\phi}{E}| < \infty
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\]
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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
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\[
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\sup_{A \in \cm}\norm{\mu(A)}_{E} = \sup_{A \in \cm}\norm{x_A}_{E^{**}} < \infty
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\]
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\end{proof}
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\begin{proposition}
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\label{proposition:vector-measure-properties}
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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:
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\begin{enumerate}
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\item For any $\seq{A_n} \subset \cm$ with $A_n \subset A_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}A_n} = \limv{n}\mu(A_n)$.
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\item For any $\seq{A_n} \subset \cm$ with $A_n \supset A_{n+1}$ for all $n \in \nat$, $\mu(\bigcap_{n \in \natp}A_n) = \limv{n}\mu(A_n)$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(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
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\[
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\limv{n}\mu(A_n) = \sum_{n = 1}^\infty \mu(B_n) = \mu\paren{\bigcup_{n \in \natp}A_n}
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\]
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(2): By (1),
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\begin{align*}
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\limv{n}\mu(A_n) &= \mu(A_1) - \limv{n}\mu(A_1 \setminus A_n) \\
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&= \mu(A_1) - \mu\paren{A_1 \setminus \bigcup_{n \in \natp}A_n} = \mu\paren{\bigcap_{n \in \natp}A_n}
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\end{align*}
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\end{proof}
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