Martingales: Expectation of almost-sure limit Let $X_n, n\geq 0$ be a martingale. We know that $E[X_n]=E[X_m]$ for all $m, n \geq 0$. Moreover suppose that $X_n \rightarrow X$ P-a.s. 
What do we know about $E[X]$? Is it clear that $E[X]=E[X_0]$?
What if $X_n$ is a submartingale? Is it clear that $E[X] \geq E[X_0]$? And the analogous result for a supermartingale?
What if the convergence is not P-a.s. but in $L^p$ for some $p \geq 1$?
 A: $X := \lim X_n$ right?
Well we want to know if $E[\lim X_n] = \lim E[X_n] (= \lim E[X_0] = E[X_0])$.
Depending on the $X_n$, we may or may not have $E[\lim X_n] = \lim E[X_n]$.
We could use Fatou's Lemmas to say that
$$E[\liminf X_n] \le \liminf E[X_n] \le \limsup E[X_n] \le E[\limsup X_n] \to E[\lim X_n] = \lim E[X_n]$$
but we need the $X_n$'s to satisfy the assumptions of Fatou's Lemmas

For a super/sub mart, we have:
$$E[X] := E[\lim X_n] \stackrel{if}{=} \lim E[X_n] \le / \ge \lim E[X_0] = E[X_0]$$
A: Consider the product-martingale 
$$X_n = \prod_{1\le m \le n}Y_m$$
with $Y$, $Y_1$, $Y_2$, $\dots$, be non-negative, non-degenerate i.i.d random variables with mean $1$. Then you can prove that $\lim\limits_{n\to\infty}X_n = 0$ almost surely (see here). But, obviously $$E[X_n]=E\left[\prod_{1\le m \le n}Y_m\right]=\prod_{1\le m \le n}E[Y_m]=1$$ by the independence of $Y_i$, for any $n\in \mathbb N$, so that $$\lim_{n\to \infty}E[X_n]=1=E[X_1]\neq0=E[X]$$ 
Trivially $X_n$ is a submartingale (and a supermartingale), so that $E[X]=0\not\ge 1=E[X_1]$ even for submartingales. Now, if $$X_n\overset{\mathcal L^p}\longrightarrow X$$ (which is the case for example if the Martingale convergence theorem applies) then you know that $$\lim_{n\to \infty}E[|X_n|^r]=E[|X|^r]$$ for all $1\le r\le p$ (this is a direction implication of $\mathcal L^p$ convergence, see here).
