Does weak convergence with uniformly integrable densities imply absolute continuity of the limit. Suppose $\mu_{n}, n \geq 1, \nu$ are probability measures on measurable space $( \Omega , \digamma)$, $\mu_{n} \ll \nu$, Let $f_{n}= \frac{d \mu_{n}}{d \nu}$. Suppose a convex function  $\psi$ on $[0, + \infty]$ satisfy $\lim_{t \rightarrow \infty}$ $\frac{\psi (t)}{t}$= $\infty$, and
$$\mathop{sup}_{n} \int \psi \circ f_{n} d \nu < \infty$$
Suppose $\mu_{n} \xrightarrow{w} \mu$. Prove $\mu \ll \nu$.
 A: As the OP commented and the title of the problem states, the functions $f_n$ are uniformly integrable: the existence of a convex function for which the displayed inequality holds guarantees  uniform integrability by de la Vallée-Poussin's theorem.  By the Dunford-Pettis theorem  the set of functions $\{f_n\}$ is then relatively compact in the weak topology on $L^1(\nu)$. (These are theorems II.22 and II.25 in Dellacherie and Meyer, Probabilities and Potentials, which has a clear exposition of this stuff.) Hence,  $\{f_n\}$ has   weak limits points in $L^1(\nu).$   Suppose  $f_{n}\rightharpoonup f$ weakly [that is, in $\sigma(L^1(\nu),L^\infty(\nu))$] along some subsequence, for some function $f\in L^1(\nu)$.   Then  for each $g\in (L^1(\nu))^* = L^\infty(\nu),$ $$\langle g , f_{n}\rangle=\int g(x) f_{n}(x)\nu(dx) \to \langle g,f\rangle=\int g(x) f(x) \nu(dx)$$ 
along that subsequence. On the one hand we know that $\mu_n\xrightarrow{d} \mu$ implies that for continuous bounded $g$ we have $\int g(x) \mu_{n}(dx)\to \int g(x)\mu(dx)$ along the subsequence.  But on the other hand, since $\int g(x)f_{n}(x)\nu(dx)=\int g(x)\mu_{n}(dx)$, and since the continuous bounded functions are contained in $L^\infty(\nu)$, we know that for all such $g$ we also have $\int g(x) f(x) \nu(dx) = \int g(x) \mu(dx)$. Hence the measure $\mu$ has density $f$ with respect to $\nu$.  This already implies $\mu\ll\nu$, and also implies that there is only one possible limit $f$ for $f_n$, independent of subsequence, namely the R-N derivative of $\mu$ with respect to $\nu$.
