Is the limit of a $L^2$-convergent sequence of random variables unique up to a.e.? Is the limit of a $L^2$-convergent sequence of random variables unique up to a.e.? In other words, if $X$ and $Y$ are both limits, will $X=Y$ a.e.? If yes, is Ito integral, which is defined as $L^2$ limit of a sequence of Ito integrals of simple processes, defined only up to a.e.?
Conversely, if the sequence converges to a random variable $X$, and $Y$ is another random variable same as $X$ a.e., will $Y$ also be the limit of the $L^2$-convergent sequence? 
Similar questions for a sequence of random variables that converges in probability. 
Thanks in advance!
 A: Hint: $\|X-Y\|_2\leqslant\|X_n-X\|_2+\|X_n-Y\|_2$. And some random variables $X$ and $Y$ such that $\|X-Y\|_2=0$ are such that...
Application: If $X_n\to X$ in $L^2$ and $X_n\to Y$ in $L^2$, then $\|X_n-X\|_2+\|X_n-Y\|_2\to0$ hence $\|X-Y\|_2=0$. For every positive $u$, $$
(X-Y)^2\geqslant u^2\cdot[|X-Y|\geqslant u],
$$ 
hence 
$$
\|X-Y\|_2^2\geqslant u^2\cdot\mathrm P(|X-Y|\geqslant u).
$$
If $\|X-Y\|_2=0$, this shows that $\mathrm P(|X-Y|\geqslant u)=0$ for every positive $u$. That is, $\mathrm P(|X-Y|\gt0)=0$, which is equivalent to $X=Y$ almost surely.
One can adapt this proof to the convergence in probability. For every positive $u$, 
$$
[|X-Y|\geqslant2u]\subseteq[|X_n-X|\geqslant u]\cup[|X_n-Y|\geqslant u],
$$
hence
$$
\mathrm P(|X-Y|\geqslant2u)\leqslant\mathrm P(|X_n-X|\geqslant u)+\mathrm P(|X_n-Y|\geqslant u).
$$
Maybe you can continue...
Edit Recall that the limits almost sure, in $L^p$ and in probability are only defined almost surely, that is, if $X_n\to X$ in either one of these three acceptions and if $X=Y$ almost surely, then $X_n\to Y$ as well. If $X_n\to X$ in $L^2$ for example, use $\|X_n-Y\|_2\leqslant\|X_n-X\|_2+\|X-Y\|_2$ and the fact that $\|X-Y\|_2=0$ if (and only if) $X=Y$ almost surely. Thus  $\|X_n-Y\|_2\leqslant\|X_n-X\|_2$ and  $\|X_n-X\|_2\to0$, which implies that  $\|X_n-Y\|_2\to0$.
