Showing $\{X_n\}$ is uniformly integrable when $\sup _{n} \mathbb{E}\left[X_{n}^{2}\right]<\infty$

Question

I got a question that show that a family of rvs $\left\{X_{n}\right\}$ is uniformly integrable when $\sup _{n} \mathbb{E}\left[X_{n}^{2}\right]<\infty$

What I have tried:

$$\sup_n\mathbb{E}[|X|] = \sup_n\{\mathbb{E}[|X|\cdot 1_E] + \mathbb{E}[|X| \cdot 1_{E^c}] \}\leq \sup_n\{\mathbb{E}[1 \cdot1_E] + \mathbb{E}[X^2 \cdot 1_{E^c}] \}\leq\sup_n\{1+\mathbb{E}[X^2]\} < \infty.$$

However, it seems like that it $\sup_n\mathbb{E}[|X|]< \infty$ cannot imply uniformly integrable.

How can I prove $\lim _{\alpha \rightarrow \infty} \sup _{n} \mathbf{E}\left(\left|X_{n}\right| \mathbf{1}_{\left|X_{n}\right| \geq \alpha}\right)=0$?

Thanks

Answer 1

$E|X_n|1_{|X_N| \geq \alpha} \leq \sqrt {EX_n^{2}} \sqrt {P(|X_n|\geq \alpha)}$. Hence it is enough to show that $\sup_n {P(|X_n|\geq \alpha)} \to 0$ as $\alpha \to \infty$. But ${P(|X_n|\geq \alpha)=P(X_n^{2}\geq \alpha^{2})}\leq \alpha^{-2} EX_n^{2}$. I will let you finish the proof.

Answer 2

Let $M^2=\sup_n E[|X_n|^2]$. Then $E[|X_n|\mathbb{1}(|X_n|>a)\leq \sqrt{P(|X_n|\geq a)}M\leq\frac{1}{a}M^2 $