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
\[\sup_{A \in \cm}\norm{\mu(A)}_{E} < \infty\]
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 Proposition 9.3.4 and Proposition 9.3.2, $E^{*}$ is a Banach space. The Uniform Boundedness Principle implies that
\[\sup_{A \in \cm}\norm{\mu(A)}_{E}= \sup_{A \in \cm}\norm{x_A}_{E^{**}}< \infty\]
$\square$