This is a question from here:
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 the following true or false?
If $|\nu_n - \nu|(X) \rightarrow 0$, then $\nu \ll \mu$.
I have a few scattered thoughts about it, if decompose $\nu$ into $\nu = \nu_{ac} + \nu_s$, such that $\nu_{ac} \ll \mu$, and $\nu_s \perp \mu$. Then it suffices to show that $\nu_s = 0$ to prove the statement to be true. Assume that there is some $E \in \mathcal{F}$ such that $\nu_s(E) > 0$, that is, there is some $X_s \in \mathcal{F}$ such that $\nu_s(E) = \nu_s(E \cap X_s)$ and $\mu(X_s) = 0$. So by absolute continuity of the $\nu_n$’s, $\nu_n(X_s) = 0$. So clearly $\nu$ cannot be the limit of the $\nu_n$’s.
Here’s where I’m stuck. I feel like in general this statement is false, because I should be able to construct a $\nu$ that agrees with the limit of $\nu_n(X)$, but I want it to disagree with the limit of the $\nu_n(E)$ for some $E$ such that $\mu(E) = 0$. Does the fact that everything here is a finite measures here give me anything more to go off? What properties is relevant here?
The basic inequality $$\nu(E) \leq |\nu-\nu_n|(E)+\nu_n(E) $$ implies that if $\mu(E) = 0$, then $\nu(E) = 0$.
For completeness, if $\mu$ and $\nu$ are finite signed measures, then $$|\mu+\nu|(A)\leq |\mu|(A)+|\nu|(A)$$