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Let $X_n$ be a family of random variables. Then, $X_n$ is uniformly integrable if $$\lim_{M\to \infty} \sup_n E[|X_n|:|X_n| > M] = 0 \tag{1}$$ or equivalently

$$ \text{For any }\epsilon > 0,\text{ there exists }M>0\text{ such that }E[|X_n|:|X_n| > M] < \epsilon\text{ for all }n \geq 1 \tag{2}$$

It is clear to me that (1) implies (2), but I do not see why (2) implies (1)?

(2) would imply (1) if $E[|X_n|:|X_n| > M]$ decreases as $M$ increases, but I don't see why.

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  • $\begingroup$ Recall that $X\in L^1$ if and only if $\lim_{M\to\infty}\mathbb E[|X|\mathsf 1_{|X|>M}]=0$. $\endgroup$ – Math1000 Oct 18 '15 at 1:54
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(2) would imply (1) if $\mathbb{E}[\vert X_n\vert:\vert X_n\vert >M]$ decreases as $M$ increases, but I don't see why.

Note that the variables $1_{\vert X_n\vert>M}$ do indeed decrease when $M$ increases for all $n$ (the sets $\{\vert X_n\vert>M\}$ become smaller as $M$ becomes larger). This implies $\mathbb{E}[\vert X_n\vert:\vert X_n\vert >M]$ decreases as $M$ increases.

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You just need to replace the $\lim$ and $\sup$ with appropriate substitutes. Let me break it down for you. First, the limit expression:

$$\lim_{M\to \infty} \sup_n E[|X_n|:|X_n| > M] = 0 \Leftrightarrow \forall{\varepsilon} > 0,\exists{M>0}\text{ s.t. }\sup_n E[|X_n|:|X_n| > M] < \varepsilon$$

Now the supremum:

$$\sup_n E[|X_n|:|X_n| > M] < \varepsilon \Leftrightarrow E[|X_n|:|X_n| > M]<\varepsilon \quad \forall{n}$$

Combine the two equivalencies above to finish the proof.

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