The question is presented in
https://mathoverflow.net/questions/155392/a-dominated-convergence-theorem-for-martingale
Let $\{(X_1^n, X_2^n)\}_n$ be a sequence of martingales defined some probability space. (which means $E[X_2^n|X_1^n]=X_1^n$)
Suppose there exists $(X_1, X_2)$ such that
\begin{eqnarray} X_1^n&\to& X_1,~ a.s \\ X_2^n&\to& X_2,~ a.s \end{eqnarray}
and
\begin{eqnarray} \lim_{n\to\infty}E[|X^n_2|]=E[|X_2|] \end{eqnarray}
Can we prove
$$\lim_{n\to\infty}E[|X_1^n-X_1|]=0$$
I have found the proof to this problem, I do not know whether this question is put on hold.
It is enough to show that the family $\{X_1^n\}$ is uniformly integrable. By the following inequality: for any $c>0$,
$$0\leq f_c(x):=|x|1_{|x|\geq c}\leq 2(|x|-\frac{c}{2})_+:=g_c(x)$$
Thus
$$E[f_c(X_1^n)]\leq E[g_c(X_1^n)]$$
And by Jensen's inequality we get
$$E[f_c(X_1^n)]\leq E[g_c(X_1^n)]\leq E[g_c(X_2^n)]$$
Moreover, $X_2^n\to X_2$ a.s. and $E[|X_2^n|]\to E[|X_2|]$ impliy in particular $\{X_2^n\}$ is uniformly integrable (classical result found in Foundation of modern probability). Hence,
$$\lim_{c\to\infty}\sup_n E[g_c(X_2^n)]=0$$
and
$$\lim_{c\to\infty}\sup_n E[|X_1^n|1_{|X_1^n|\geq c}]=\lim_{c\to\infty}\sup_n E[f_c(X_1^n)]=0$$
Therefore, we get
$$\lim_{n\to\infty}E[|X_1^n-X_1|]=0$$
