Imagine we have a sequence of finite measures $\nu\_n << \mu\_n$ (on the torus $\mathbb{T}^2\subseteq \mathbb{R}^2$) converging weakly to some measures $\nu << \mu$. Do we automatically have that the Radon-Nikodým derivatives $h\_n$ of $\nu\_n$ wrt. $\mu\_n$ converge to the Radon-Nikodým derivative $h$ of $\nu$ wrt. $\mu$?


Does the situation becomes easier if we assume that $0\leq h\_n \leq 1$ and $0\leq h\leq 1$?


Does the results carry over if we consider finite random measures instead of deterministic measures? And is the result still available if the convergence is only in the distributional sense, i.e. for compactly supported smooth functions?


Thanks a lot for your help!



Convergence of Radon-Nikodým derivative
The answer is negative for a.e. pointwise convergence as well as $L^1$ convergence with respect to $\mu$. Let $\mu\_n=\mu$ be half the Lebesgue measure, color the torus as a checkerboard with $2^n$ cells, and assume $h\_n$ takes value $1$ on black cells and $0$ on white cells. Then $\nu\_n=h\_n\mu \to \mu$ $\*$-weakly, but for all $n$ the function $h\_n-h$ is everywhere $\pm\frac12$.


To control densities, you want convergence in total variation (or variations on this).


