Suppose that $W_n \to W_{\infty}$ a.s. where $W_{\infty}$ is independent of random variable $V$. Moreover, suppose that $E[|V|]<\infty$.
Is it true that
\begin{align}
\lim_{n \to\infty} E[V|W_n]= E[V|W_\infty]=E[V] \text{ a.s.}
\end{align}
The last equality is of course trivial. Therefore, I am looking the proof of the first. This looks like some kind of continuity result, and I am actually not sure if it holds. If the result doesn't hold, I would like to know what extra conditions can be added for it to hold.
Best Answer
Let $V$ be any integrable random variable, $Y=V$ and $W_n=Y/n$. Then $W_n\to 0=:W_\infty$ which is independent of $V$ but the $\sigma$-algebra generated by $W_n$ is the same as that generated by $V$ hence $\mathbb E\left[V\mid W_n\right]=V$ and we get a counter-example for any non-degenerated $V$.