Weak convergence of random variables implies $\mathbb E | X| \le \liminf_n \mathbb E|X_n|$

Question

Proof that, if $X_n \rightarrow X$ weakly, then $\mathbb E | X| \le \liminf_n \mathbb E|X_n|$.

I know, that I should use Fatou's lemma but I don't know what can I do first.

Answer 1

Let $$ Y_n = \sqrt{|X_n|}\\ Y = \sqrt{|X|} $$ Then $Y_n\to Y$ weakly. Indeed: $$ Ef(Y_n) = Ef(\sqrt{|X_n|}) \to Ef(\sqrt{|X|}) = Ef(Y) $$because if $f$ is bounded continuous so is $ x\to f(\sqrt{|x|}) $.

Then, consider a subsequence $X'_n$ such as $E|X'_n| \to \liminf E|X_n|$ $$ E(Y_n'^2) \to \liminf E|X_n| \\ \begin{align} 0&\le E((Y_n' - Y)^2) = E(Y_n'^2) + E(Y^2) - 2 E(Y_n' Y) \\&\to \liminf E|X_n| + E(Y^2) - 2E(YY) = \liminf E|X_n| - E(Y^2) \end{align} \\ \implies \liminf E|X_n| \ge E(Y^2) $$