I am working at the moment with a sequence of finite measures $\nu_n << \mu_n$ converging weakly to $\nu << \mu$. I have seen that if the Radon-Nikodým derivatives $h_n := \dfrac{d\nu_n}{d\mu_n}$ converge pointwise to some $h$, then $h$ is the Radon-Nikodým derivative of $\nu$ wrt. $\mu$. I am interested of a somehow similar question.
Imagine $h_n$ is uniformly bounded in $n$, e.g. $0\leq h_n\leq 1$. Suppose also that $\mu_n$ is aboslutely continuous wrt. Lebesgue measure with strictly positive density. That means in particular, that $h_n$ is defined Lebsegue-a.e. We thus may define the measure $d\Lambda_n := h_n(x) dx$. Because of the boundedness, this sequence is tight and we get at least a converging subsequence. Now, under which condition on the sequence of measures do we get that $\Lambda_n$ converges to $\Lambda$ given by $d\Lambda = \dfrac{d\nu}{d\mu}(x)dx$?
I cross-reference a question I asked on MO and which is the origin of my question:
The question therein is somewhat different, because we have a specific scenario in which we may reformulate the problem in other ways, but I am hoping that solutions to the problem above may easily be modified to give solutions to the latter.