Convergence in distribution and limit inferior of expectation

Question

I am trying to solve this exercise in Probability Theory by A. Klenke (3rd version) by applying the continuous mapping theorem or the portemanteau theorem but with no results:

Let $X,X_1,X_2,...$ be real random variables with $X_n$ converging in distribution to $X$. Show that $E(|X|)\leq \liminf_{n\to \infty} E(|X_n|)$.

In order to apply the continuous mapping theorem I think I need to know if $P_X({0})=0$, as it requires, as a premise, that the set of points of discontinuity has measure zero. But I don't know anything about it.

Any suggestions?

Thank you.

Answer 1

For a fixed $M>0$, let $\varphi_M$ be the function defined by $$\varphi_M(x)=\begin{cases}-M&\mbox{ if }x\leqslant -M\\ x&\mbox{ if }-M<x\leqslant M\\ M&\mbox{ if } x>M\end{cases}.$$ Since $x\mapsto\varphi_M(\lvert x\rvert)$ is continuous and bounded, it follows that for each fixed $M$, $$ \mathbb E\left[\varphi_M\left(\lvert X\rvert\right)\right]=\lim_{n\to\infty} \mathbb E\left[\varphi_M\left(\lvert X_n\rvert\right)\right]=\liminf_{n\to\infty} \mathbb E\left[\varphi_M\left(\lvert X_n\rvert\right)\right]. $$ Moreover, since $\varphi_M(\lvert x\rvert)\leqslant \lvert x\rvert$, we get that $$ \mathbb E\left[\varphi_M\left(\lvert X\rvert\right)\right]\leqslant \liminf_{n\to\infty} \mathbb E\left[ \lvert X_n\rvert \right]. $$ We conclude by monotone convergence, letting $M\to\infty$.