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The question is presented in

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}


\begin{eqnarray} \lim_{n\to\infty}E[|X^n_2|]=E[|X_2|] \end{eqnarray}

Can we prove


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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 $c>0$,

$$0\leq f_c(x):=|x|1_{|x|\geq c}\leq 2(|x|-\frac{c}{2})_+:=g_c(x)$$


$$E[f_c(X_1^n)]\leq E[g_c(X_1^n)]$$

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,

$$\lim_{c\to\infty}\sup_n E[g_c(X_2^n)]=0$$


$$\lim_{c\to\infty}\sup_n E[|X_1^n|1_{|X_1^n|\geq c}]=\lim_{c\to\infty}\sup_n E[f_c(X_1^n)]=0$$

Therefore, we get


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@GodeGolf: You could even "accept" your answer on your own question, I think. – UwF Apr 13 '14 at 13:55

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