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Let $\{X_n \}$ be a sequence of random variables on a probability space $(\Omega,\mathcal{F},P)$.

Then, $\{X_n\}$ is uniformly integrable if $$\lim_{M \to \infty} \sup_n \int_{|X_n| > M } |X_n| = 0 \tag{1}$$

From this, I know that $$\displaystyle \sup_n E(|X_n|) < \infty \tag{2}$$

But, then I am wondering whether (2) implies (1). If not, can you give me a counterexample?

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  • $\begingroup$ There is something missing; probably you mean $$\limsup_{M \to \infty} \sup_n \int_{|X_n|>M} |X_n| = 0, $$ right? $\endgroup$ – saz Aug 14 '15 at 15:08
  • $\begingroup$ I have in my book $$\lim_{M \to \infty} \sup_n \int_{|X_n| > M } |X_n| = 0 $$ and isn't this equivalent to what I have in (1)? $\endgroup$ – user74261 Aug 14 '15 at 15:12
  • $\begingroup$ $\sup_n$ is missing. $\endgroup$ – saz Aug 14 '15 at 15:14
  • $\begingroup$ Oops. I just fixed it $\endgroup$ – user74261 Aug 14 '15 at 15:14
  • $\begingroup$ Being uniformly integrable is equivalent to $\sup E(a(|X_n|))$ being bounded for some function $a$ such that $a(x)/x\to\infty$ when $x\to\infty$. $\endgroup$ – Did Aug 14 '15 at 18:48
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No, the converse is, in general, not true. Just consider $((0,1),\mathcal{B}(0,1))$ endowed with Lebesgue measure and the sequence of random variables $$X_n(x) := 2n \cdot 1_{(0,1/n)}(x).$$ Then $$\mathbb{E}(|X_n|) = 2$$ for all $n \in \mathbb{N}$, but $$\limsup_{M \to \infty} \sup_{n \in \mathbb{N}} \int_{|X_n|>M} |X_n| \, d\mathbb{P} \geq \int_{|X_M|>M} |X_M| \, d\mathbb{P} = 2.$$

Remark: If the sequence $(X_n)_{n \in \mathbb{N}}$ is bounded in $L^p$ for some $p>1$, i.e. $$\sup_{n \in \mathbb{N}} \mathbb{E}(|X_n|^p)<\infty,$$ then $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable.

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