Equality of expectations in martingale

Question

Let $(X_n)_{n \geq 0}$ be an $\mathcal{F}_n$-martingale that converges to $X_{\infty}$ in $L^p$. Why is it true that for $A \in \mathcal{F}_n$ and $n\leq m$

$$ E[X_{\infty}\mathbb{1}_A]=\lim_{m\to\infty}E[X_m \mathbb{1}_A] = E[X_n\mathbb{1}_A]$$

I know that martingales have constant expectation, but I didn't think

1. it stayed true for $n \to \infty$
2. $E[X_n]=E[X_m]$ implied $E[X_m \mathbb{1}_A] = E[X_n\mathbb{1}_A]$ for all $A\in \mathcal{F}_n$

Answer 1

I suppose you are taking $p \geq 1$. $|EX_{\infty}1_A-EX_m1_A| \leq E|X_{\infty}-X_m| \to 0$ as $ m\to \infty$ since convergence in $L^{p}$ implies convergence in $L^{1}$.

Since $EX_m1_A=EX_n1_A$ for all $m \geq n$ when $A \in \mathcal F_n$ the equality of the limit with $EX_n1_A$ follows.