# Convergence of conditional expectations equivalence

Suppose that as sequence of random variables $\{X_n\}$ and $X$ are defined on $(\Omega,\mathcal{F},P)$ and $\mathcal{G}\subset \mathcal{F}$. I know that for any $G\in \mathcal{G}$,

$$E[f(X_n)1_G]\to E[f(X)1_G]$$

for some bounded function $f$. Does it imply that

$$E[f(X_n)\mid \mathcal{G}]\to E[f(X)\mid \mathcal{G}] \quad\text{a.s.}?$$

The other way is trivial by an application of the dominated convergence theorem, i.e.

$$\lim_n E[f(X_n)1_G]=E[\lim_n E[f(X_n)\mid\mathcal{G}]1_G]= E[f(X)1_G].$$

No. Take $\mathcal G =\mathcal F$. Let $X_n \to X$ in probability. Then the hypothesis is satisfied for any bounded measurable $f$, But well known examples show that $f(X_n)$ need not converge almost surely. If you want $\mathcal G$ to be properly contained in $\mathcal F$ take the latter to be Lebesgue measurable sets in $(0,1)$, the former to be Borel sigma algebra, $F$ to be the identity function and consider the standard example of $X_n$ converging in probability but not almost surely.
• I'm aware of this example, so I particularly specified that $\mathcal{G}$ is a strict subset of $\mathcal{F}$. Does it change anything? Jul 28 '18 at 0:30
• @akm47 I have edited the answer. I should point out that not every Mathematician uses $\subset$ for proper subset. You should have added $\mathcal G \neq \mathcal F$ in the statement. Jul 28 '18 at 1:54