23.2 Vector Measures
Definition 23.2.1 (Vector Measure).label 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 vector measure if:
- (M1)
$\mu(\emptyset) = 0$.
- (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.
Proposition 23.2.2.label 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
Proof. For each $\phi \in E^{*}$, the mapping
is a complex measure. For each $A \in \cm$, let
then for each $\phi \in E^{*}$,
By Proposition 12.5.5 and Proposition 12.5.3, $E^{*}$ is a Banach space. The Uniform Boundedness Principle implies that
$\square$
Proposition 23.2.3.label 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:
- (1)
For any $\seq{A_n}\subset \cm$ with $A_{n} \subset A_{n+1}$ for all $n \in \nat$, $\mu\paren{\bigcup_{n \in \nat}A_n}= \limv{n}\mu(A_{n})$.
- (2)
For any $\seq{A_n}\subset \cm$ with $A_{n} \supset A_{n+1}$ for all $n \in \nat$, $\mu(\bigcap_{n \in \natp}A_{n}) = \limv{n}\mu(A_{n})$.
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
(2): By (1),
$\square$
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