Does product of $L^1$ convergent martingales converge? Suppose $(X_n)$ and $(Y_n)$ are martingales, such that $(X_n)$ converges to $X$ and $(Y_n)$ to Y in $L^1$.
Then does $X_nY_n$ converge to $XY$ in $L^1$? Does $E[X_n]$ converge to $E[X]$? What if $X_n$ and $Y_n$ are $L^2$ bounded and/or $(X_n), (Y_n)$ converge in $L^2$?
 A: The answer to your second question is yes. Indeed:
$\big|\mathbb{E}[X_n-X]\big|\leq \mathbb{E}\big[|X_n-X|\big]=\left\Vert X_n-X\right\Vert_{L^1}\rightarrow 0$, as $n\to\infty$. 
(Note that since $(X_n)$ is a martingale we have: $\mathbb{E}[X_n]=\mathbb{E}[X_1]$, for $n\in\mathbb{N}$)
For your last question the answer is again yes: If $(X_n), (Y_n)$ are $L^2$-bounded, we can invoke the Martingale Convergence Theorem to deduce that they converge a.s. and in $L^2$ to some random variables $X,Y\in L^2$. Furthermore, $ X_nY_n, XY\in L^1$, by an application of the Cauchy-Schwarz inequality and 
$\mathbb{E}\big[|X_nY_n-XY|\big]\leq\mathbb{E}\big[|X_n(Y_n-Y)|\big]+\mathbb{E}\big[|Y(X_n-X)|\big]$ $$\leq\big(\sup_{n}\left\Vert X_n\right\Vert_{L^2}\big)\left\Vert Y_n-Y\right\Vert_{L^2}+\big(\left\Vert Y\right\Vert_{L^2}\big)\left\Vert X_n-X\right\Vert_{L^2}\rightarrow 0$$
As for your first question, I think that we can't deduce that $XY, X_nY_n$ are members of $L^1$ without any further assumption. 
