# 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 $$L^2$$?

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$$\big|\mathbb{E}[X_n-X]\big|\leq \mathbb{E}\big[|X_n-X|\big]=\left\Vert X_n-X\right\Vert_{L^1}\rightarrow 0$$, 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.