Proposition 21.4.3.label Let $(X, \cm)$ be a measurable space, $\mu$ be a positive measure on $(X, \cm)$, $E$ be a normed space over $K \in \RC$, and $\nu: \cm \to E$ be a vector measure, then the following are equivalent:
- (1)
$\nu$ is absolutely continuous with respect to $\mu$.
- (2)
For each $\eps > 0$, there exists $\delta > 0$ such that for any $A \in \cm$ with $\mu(A) < \delta$, $\norm{\nu(A)}_{E} < \eps$.
Proof, [Theorem 3.15, Fol99]. $(\neg 2) \Rightarrow (\neg 1)$: Using Lemma 21.4.2, assume without loss of generality that $\nu$ is a positive finite measure.
By assumption, there exists $\eps > 0$ and $\seq{A_n}\subset \cm$ such that for each $n \in \natp$, $\mu(A_{n}) < 2^{-n}$ and $\norm{\nu(A_n)}_{E} \ge \eps$. For each $N \in \natp$, let $B_{N} = \bigcup_{n \ge N}A_{n}$, then $\mu(B_{N}) < 2^{-N+1}$. By Proposition 20.1.5,
but by Proposition 21.2.3,
so (1) does not hold.$\square$
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