Convergence in $L^p$: $E[X 1_A] = E[X_n 1_A]$

Question

Let $p>1$ and suppose that $X_n \rightarrow X$ in $L^p$ as $n \rightarrow \infty$.

For $A \in \mathcal{F}_n=\sigma(X_0, \dotsc, X_n)$ it is written

$$E[X 1_A] = E[X_n 1_A]$$

Can you explain me why this is true?

What I tried:

So we have $$E[|X_n-X|^p]^{1/p} \rightarrow 0,$$ so probably this implies $$E[|X_n-X|^p] \rightarrow 0$$ and also (as $p>1$) $$E[|X_n-X|] \rightarrow 0.$$

Then we have $$ E[X 1_A]- E[X_n 1_A] = E[(X-X_n) 1_A] \leq E[|X-X_n| 1_A],$$

but I do not get how to conclude the above result. Can you help me with it? Thanks a lot.

another Proof sketch

Another proof sketch starts with $$E[X 1_A] = \lim_n E[X_n 1_A]$$

but I do not get why this is true, as we only have $$\lim_n X_n = X,$$ so I am not sure why we can pull out the Limit.

Answer 1

If we take $A$ with finite measure $\mu(A)$, we can do the following, using Hölder's ineqality:

$$|E[(X-X_n)I_A]| \leq \int_\Omega |(X-X_n) I_A| d\omega =\\ \|(X-X_n)I_A\|_1\leq \|X - X_n\|_p \|I_A \|_q = \|X - X_n\|_p (\mu(A))^{1/q},$$

where $q$ is such that $\frac{1}{p}+\frac{1}{q}=1,$ but ultimatively doesn't matter because $(\mu(A))^{1/q}$ doesn't depend on $n$ so the r.h.s. converges to $0$.