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We know one of the definitions of uniform integrability is:

$$\lim_{M\to\infty} \sup_{n\in N} E[|X_n|\cdot 1_{|X_n>M|}]\to 0$$

I wanna know how to prove the equivalence to

$$\lim_{M\to\infty} \limsup_{n\to\infty} E[|X_n|\cdot 1_{|X_n>M|}]\to 0$$

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    $\begingroup$ Enclose in $...$ for inline formula and $$...$$ for displayed formula instead of that. $\endgroup$ – UmbQbify -Key20- Aug 1 at 10:31
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One implication is obvious. For the other part note that if $\lim \sup E[|X_n|1_{|X_n|>M_o}] <\epsilon$ for some $M_0$ then there exists $n_0$ such that $E[|X_n|1_{|X_n|>M_o}|<\epsilon$ for all $n \geq n_0$. For $n=1,2..,n_0-1$ $E[|X_n|1_{|X_n|>M}] \to 0$ as $ M \to \infty$. Hence there exist $M_1,M_2,...,M_{n_0-1}$ such that $E[|X_n|1_{|X_n|>M_o}] <\epsilon$ for each $n <n_0$. Now take the maximum of $M_1,M_1,M_2,...,M_{n_0-1}$.

[I have used the fact that if $X$ is any integrable random variable then $E[X1_{|X|>M}] \to 0$ as $ M \to \infty$. This follows by DCT].

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