Let $\mu$ be a finite measure over $X\times Y$ and suppose that we have $\mu(x,y)=\mu_x(x|y)\times \mu^y(y)$ (regular conditional distribution). Let $\mu^x$ denote the marginal of $\mu$ over $x$.
Suppose $f_n,f^*\in L^2(X,\mu^x)$ and $\lVert f_n\rVert_{\infty}\leq 1$ for each $n$.
Now, suppose that we know that, $$\int (f_n(x)-f^*(x)) g(x) \,d\mu^x(x)\to 0 $$ for every $g\in L^2(X,\mu^x)$ (i.e. $f_n$ converges to $f^*$ in the weak topology).
What assumption can we make on $\mu_x(x|y)$ and/or $\mu_y(y)$ so that we know also that (possibly passing to a subsequence) $$\int (f_n(x)-f^*(x)) g(x) \,d\mu_x(x|y)\to 0 $$ for every $g\in L^2(X,\mu)$ $\mu^y$-almost surely in $y$? I understand that this works if $\mu$ is absolutely continuous with respect to with respect to $\mu^x\times \mu^y$.
Is there a weaker condition that would also guarantee it?
