Is a sequence of identical distributed integrable random variables even uniformly integrable?

Question

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:

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

Answer 1

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.