Let $(\mathcal{X} , \mathcal{B})$ be a measurable space. Let $(\nu_n)_{n=1}^\infty$ be a sequence of signed measures on $\mathcal{B}$ which converge setwise to a set function $\nu$. Suppose there exists a finite measure $\mu$ on $\mathcal{B}$ such that $|\nu_n(B)| \leq \mu(B)$ for every $B \in \mathcal{B}$ and $n \in \mathbb{N}$. Prove that $\nu$ is a signed measure.
I've been banging my head against the wall on this for a while, and so far this is the only progress I've made:
Let $\bigcup_{i=1}^\infty A_i$ be a union of pairwise disjoint members of $\mathcal{B}$. To show $\nu$ is a signed measure, it is enough to show that $\nu(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty \nu(A_i)$. Notice that each $\nu_n$ is absolutely continuous with respect to $\mu$, so we can write $d \nu_n = f_n d\mu$ for some integrable function $f_n : \mathcal{X} \to \mathbb{R}$ (this is a result from a previous exercise/Radon-Nikodym theorem). So:
$\nu(\bigcup_{i=1}^\infty A_i) = \lim_{n \to \infty} \int_{\bigcup A_i} f_n d\mu = \lim_{n \to \infty} (\sum_{i=1}^\infty \int_{A_i} f_n d\mu)$
$\sum_{i=1}^\infty \nu(A_i) = \sum_{i=1}^\infty (\lim_{n \to \infty} \int_{A_i} f_n d\mu)$
So I've managed to reduce this question to the interchange of this limit and infinite sum. The most natural thing to try is Dominated Convergence Theorem, but I can't seem to find an integral dominator for all the $f_n$'s. Could someone please point me in the right direction here?