Here I state the problem:
Let $\mu, \nu, \nu_n$ be finite measures on $(X, \mathcal{F})$, and suppose that $\nu_n \ll \mu$ for all $n \in \mathbb{N}$. Is it true that if $|\nu_n - \nu|(X) \to 0$ as $n \to \infty$, then $\nu \ll \mu$?
My Attempt: We show the statement is true. By Radon Nikodym, since $\nu_n \ll \mu$ for all $n \in \mathbb{N}$, we can find functions $f_n: X \to [0, \infty)$ such that $\nu_n(E) = \int_E f_n d\mu$ with $E \in \mathcal{F}$. Recall that we have by the Jordan Decomposition $$ |\nu_n - \nu| = (\nu_n - \nu)^+ + (\nu_n - \nu)^-. $$ Therefore, we have $(\nu_n - \nu)^+(X) \to 0$ and $(\nu_n - \nu)^-(X) \to 0$.
I am then a little lost about the information we have here. Maybe the use of Radon Nikodym Theorem is not helpful here? I don't see how we can put the $f_n$ to use. On the other hand, we may also show for all $\epsilon > 0$, there exists $\delta > 0$ such that if $\mu(E) < \delta$, then $\nu(E) < \epsilon$ as $\mu, \nu$ are finite measures. However, it is also not clear why this would help for me. Any hints would be appreciated.
For any real or complex measure $\tau$ we have $| \tau(E)| \leq |\tau|(E)$. You can find this in Rudin's RCA.
If $\mu (A)=0$ then $\nu_n (A)=0$ for each $n$ and $|\nu_n(A)-\nu(A)| \leq |\nu_n-\nu| (A) \to 0$ so $\nu (A)=0$. Hence, $\nu << \mu$.
[If $\tau$ is a real valued signed measure then $|\tau(A)|=|\tau^{+}(A)-\tau^{-}(A)|\leq \tau^{+}(A)+\tau^{-}(A)=|\tau|(A)$].