16.2 Vector Measures
Definition 16.2.1 (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 vector measure if:
$\mu(\emptyset) = 0$.
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 16.2.2. 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 9.3.4 and Proposition 9.3.2, $E^{*}$ is a Banach space. The Uniform Boundedness Principle implies that