TITLE: Is a sequence of identical distributed integrable random variables even uniformly integrable?
QUESTION [5 upvotes]: Let $\{X_n\}_{n \geq 1}$ be a sequence of identically distributed random variables in $L^1$. Is this sequence uniformly integrable? 
I have to show that
$$
\lim_{c \to \infty} \sup_{n \geq 1} E\big[\mathbf{1}_{\{ |X_n| > c\}} \, |X_n|\big] = 0.
$$
By Chebyshev's inequality we have 
$$
E\big[\mathbf{1}_{\{ |X_n| > c\}}\big] \leq \frac{E[|X_n|]}{c}
$$
and since $E[X_n]$ is finite, this converges to zero as $c$ converges to infinity. 
Now there are two questions remaining: 

Does $E[\mathbf{1}_{\{ |X_n| > c\}}] \to 0$ imply also $E[\mathbf{1}_{\{ |X_n| > c\}} \, |X_n|] \to 0$?
Does the assumption of same distributions imply that $E[\mathbf{1}_{\{ |X_n| > c\}} \, |X_n|] = E[\mathbf{1}_{\{ |X_1| > c\}} \, |X_1|]$?

REPLY [4 votes]: Since the random variables are identically distributed, we have
$$\mathbb{E}(1_{\{|X_n| > c\}} |X_n|) = \mathbb{E}(1_{\{|X_1|>c\}} |X_1|)$$
for all $n \geq 1$ and therefore
$$\sup_{n \geq 1} \mathbb{E}(1_{\{|X_n| > c\}} |X_n|) = \mathbb{E}(1_{\{|X_1|>c\}} |X_1|).$$
As $X_1$ is integrable, the right-hand side converges to $0$ as $c \to \infty$; this follows for instance from the dominated convergence theorem or the monotone convergence theorem.