> [!theorem] > > Let $(X, \cm, \mu)$ be a finite [[Measure Space|measure space]], and $\seq{\nu_n}$ be a family of [[Complex Measure|complex measures]] on $(X, \cm)$ such that > 1. $\nu_n$ is [[Absolute Continuity|absolutely continuous]] with respect to $\mu$ for all $n \in \nat$. > 2. For every $E \in \cm$, $\limv{n}\nu_n(E)$ exists. > > then $\seq{\nu_n}$ is [[Uniform Absolute Continuity|uniformly absolutely continuous]] with respect to $\mu$, and the pointwise limit $\nu = \limv{n}\nu_n$ is a complex measure on $(X, \cm)$ with $\nu \ll \mu$. > > *Proof*. Let $\cm_0$ be the equivalence classes of essentially equal measurable sets, equipped with the [[Fréchet-Nikodym Metric|Fréchet-Nikodym metric]]. For each $n \in \nat$, since $\nu_n \ll \mu$, $\nu_n$ assigns the same value to each equivalence class, and defines a *continuous* function on $\cm_0$. > > Fix $\eps > 0$. For each $m \in \nat$, let > $ > E_m = \bracs{E \in \cm_0: \sup_{n \ge m}\abs{\nu_m(E) - \nu_n(E)} \le \eps} > $ > then by continuity of each $\nu_n$, $E_m \subset \cm_0$ is closed. Since $\nu_n$ converges pointwise, $\cm_0 = \bigcup_{m \in \nat}E_m$. By the [[Baire Category Theorem]], there exists $M \in \nat$, $E \in \cm_0$, and $\delta > 0$ such that $\abs{\nu_m(F) - \nu_n(F)} \le \eps$ for all $F \in B(E, \delta)$ and $m, n \ge M$. > > Let $F \in \cm$ such that $d(\emptyset, F) \le \eps$ and write > $ > F = (E \cup F) \setminus (E \setminus F) > $ > then $d(E \cup F, E), d(E \setminus F, E) \le \delta$, so > $ > \begin{align*} > \abs{\nu_M(F) - \nu_n(F)} &= \abs{\nu_M(E \cup F) - \nu_M(E \setminus F) - \nu_n(E \cup F) + \nu_n(E \setminus F)} \\ > &\le \abs{\nu_M(E \cup F) - \nu_n(E \cup F)} + \abs{\nu_M(E \setminus F) + \nu_n(E \setminus F)} \\ > \nu_n(F) &\le \nu_M(F) + 2\delta > \end{align*} > $ > and $\seq{\nu_n}$ is uniformly absolutely continuous. > > Let $\seq{E_n} \subset \cm$ and $E \in \cm$ such that $E = \bigsqcup_{n \in \nat}E_n$, then > $ > \begin{align*} > \abs{\nu(E) - \sum_{n \le N}\nu(E_n)} &= \nu\paren{E \setminus \bigsqcup_{n \le N}E_n} \\ > &= \limv{m}\nu_m\paren{E \setminus \bigsqcup_{n \le N}E_n} > \end{align*} > $ > where by uniform absolute continuity, the above difference goes to $0$ as $N \to \infty$.