# Proving of uniform integrability [closed]

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

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

• Enclose in $...$ for inline formula and $$...$$ for displayed formula instead of that. – UmbQbify -Key20- Aug 1 at 10:31

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 $$n . Now take the maximum of $$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].