Suppose that $X_N,X$ are defined on $(\Omega,\mathcal F, P)$ and that $X_N\to X$ in distribution. Assume $X_N,X$ are integrable. Let $\mathcal G$ be a sub sigma algebra. Suppose that we know $E[X_N\mid\mathcal G]$ converges in distribution. Does this imply that $E[X_N\mid\mathcal G]$ converge in distribution to $E[X\mid\mathcal G]$?
It seems like an intuitive claim but I am struggling to prove it or find a counterexample. I know that if $A,B$ are continuity sets of $X$ then
\begin{align} P(X_N\in A\mid B)&=P(X_N\in A\cap B)/P(X_N \in B) \\&\to P(X\in A\cap B)/P(X\in B) \\&=P(X\in A\mid B). \end{align}
Can this conclude?
If you take $\mathcal{G}$ to be the trivial $\sigma$-algebra, then you are just asking whether, if $E[X_N]$ converges, it must converge to $E[X]$. This is certainly false.
Let $X_N$ be a random variable for which $X_N = N$ with probability $1/N$, and $0$ otherwise. Let $X=0$, then clearly $X_N \to X$ in distribution. But $E[X_N] = 1$ for all $N$ whereas $E[X]=0$.