Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, $\mathcal{G} \subseteq \mathcal{F}$ be a subalgebra, $X$ be a real-valued random variable and $X_1,X_2,..$ be $\mathcal{G}$-measurable random variables s.t. $$ \mathbb{E}[Uf(X_n)] \rightarrow \mathbb{E}[Uf(X)] $$ holds for all continuous bounded $f:\mathbb{R} \rightarrow \mathbb{R}$ and $\mathcal{G}$-measurable bounded $U:\Omega \rightarrow \mathbb{R}$.
Does this imply $$ \mathbb{E}[Uf(X_n)] \rightarrow \mathbb{E}[Uf(X)] $$ for all continuous bounded $f:\mathbb{R} \rightarrow \mathbb{R}$ and $\mathcal{F}$-measurable bounded $U:\Omega \rightarrow \mathbb{R}$?
It would be very useful for me if this holds, but I can find neither a proof nor a counter example. I would be greatful for a help.
The statement is true iff $X$ is also $\mathcal G- $ measurable (w.r.t to its completion to be more precise).
Suppose $X$ is $\mathcal G- $ measurable. $E[Uf(X_n)]=E[E[Uf(X_n)|\mathcal G]]=E[f(X_n)E[U|\mathcal G]]=E[Vf(X_n)]$ where $V=E[U|\mathcal G]$ is $\mathcal G-$ measurable. Similarly, $E[Uf(X)]=E[Vf(X)]$. Just apply the hypothesis now.
Without loss of generality we may suppose $f(x)=x$ for all $x$. If the conclusion holds for a general $X$ then we have $E[UX]=E[VX]$. But $EVX=E[E[(VX)|\mathcal G]]=E[VE[X|\mathcal G]]=E[UE[X|\mathcal G]]$. Now $E[UX]=E[UE[X|\mathcal G]]$. The validity of this equation for all $\mathcal F-$ measurable $U$ implies that $X=E[X|\mathcal G]$ a.e., so $X$ is necessarily measurable w.r.t. (the completion of) $\mathcal G$.