Convergence of expectations under almost everywhere convergence and uniform integrability

Question

Prove that if a sequence of random variables $X_n$ converges in distribution to $X$, and if the $X_n$ are uniformly integrable (UI), then $$ \lim_{n\rightarrow\infty} E[X_n] = E[X]. $$ Can you please help with this question?

Converges in Distribution means $$ \lim_{n\rightarrow\infty} P(X_n \leq a) = P(X \leq a) $$

$X_n$ is UI if $\sup_nE(|X_n|^{1+\epsilon} < \infty$

I have to use Skorohod Theorem to construct $Y_n$ and Y then $Y_n \rightarrow Y$ a.s

so I have to show $$ \lim_{n\rightarrow\infty} E[Y_n] = E[Y]. $$

How can I show that in details?

Thank you

Answer 1

If you want to use Skorohod theorem, then you are actually reduced to show the following lemma:

If $(Z_n)_{n\geqslant 1}$ is a uniformly integrable sequence which converges almost everywhere to $Z$, then $\mathbb E(Z_n)\to\mathbb E(Z)$.

This is an extension of dominated convergence theorem. To the lemma is true, use dominated convergence theorem with $(Z_n\chi_{|Z_n|\leqslant R})_{n\geqslant 1}$ for a well chosen $R$.