Proof, [Theorem II.2.1, Yos12]. Let $\cm_{0}$ be the euivalence classes of essentially equal sets in $\cm$, equipped with the Fréchet-Nikodym metric with respect to $\mu$. For each $n \in \natp$, $\nu_{n} \ll \mu$ by assumption (a), and $\nu_{n} \in UC(\cm_{0}; E)$.

(1): Let $\eps > 0$. For each $N \in \natp$, let

then since $\seq{\nu_n}\subset UC(\cm_{0}; E)$, $A_{N} \subset \cm_{0}$ is closed. By assumption (b), $\cm_{0} = \bigcup_{N \in \natp}A_{N}$. By the Baire Category Theorem, there exists $N \in \natp$, $A \in \cm_{0}$, and $\delta > 0$ such that for every $m, n \ge N$ and $B \in \cm_{0}$ with $\mu(A \Delta B) \le \delta$, $\norm{\nu_m(B) - \nu_n(B)}_{E} \le \eps$.

Let $B \in \cm$ with $\mu(B) = \mu(\emptyset \Delta B) \le \delta$ and write $B = (A \cup B) \setminus (A \setminus B)$, then $\mu((A \cup B) \Delta B) \le \delta$ and $\mu((A \setminus B) \Delta B) \le \delta$. Thus

and $\seq{\mu_n}$ is uniformly absolutely continuous with respect to $\mu$.

(2): Let $\seq{A_n}\subset \cm$ and $A \in \cm$ with $A = \bigsqcup_{n \in \natp}A_{n}$, then for each $N \in \natp$,

Since $\seq{\nu_k}$ is uniformly absolutely continuous,

so $\nu$ is a vector measure.

Absolute continuity is equivalent to uniform continuity as a mapping $\cm_{n} \to E$, and uniform absolute continuity is equivalent to uniform equicontinuity as mappings $\cm \to E$. By the Arzelà-Ascoli Theorem, $\nu \in UC(\cm_{0}; E)$, so $\nu \ll \mu$.$\square$