Let $X_n, X$ be random variables with $X_n\xrightarrow{d} X$. Show that then $$ \mathbb{E}\lvert X\rvert\leqslant\liminf_n \mathbb{E}\lvert X_n\rvert. $$
So let $X_n\xrightarrow{d} X$; in our lecture this means that for all bounded continious functions $f$ it is $$ \lim_n\int f\circ X_n\, d\mathbb{P}=\int f\circ X\, d\mathbb{P}.~~~~~~~~~(*) $$
I do not know, how to make the proof. Nonetheless, my first idea was, to work with Fatou's lemma, i.e. $$ \liminf_n\mathbb{E}\lvert X_n\rvert\geqslant\mathbb{E}\liminf_n\lvert X_n\rvert=\int\liminf_n\lvert X_n\rvert\, d\mathbb{P}. $$
Now I do not know how to continue (in case that this is the right start).
Maybe you can help me.
With greetings
math12
Fatou's lemma is true when we have almost sure convergence but here we only have convergence in distribution. The problem can be settled in this way plus Skorohod representation but it's an overkill way.
However, we can use the definition of convergence in distribution for the map $$f_R(x):=\begin{cases} x&\mbox{ if }-R\leqslant x\leqslant R;\\ R&\mbox{ if }x\gt R\\ -R&\mbox{ if }x\lt R. \end{cases}$$ (that is, $f_R(x)=\min\{|x|,R\}\leqslant x$) This function is continuous and bounded, hence for each $R$, $\mathbb E[f_R(X_n)]\to\mathbb E[f_R(X)]$. Since $f_R(x)\leqslant |x|$ for any $x$, $$\mathbb E[f_R(X)]=\liminf_{n\to+\infty}\mathbb E[f_R(X_n)] \leqslant \liminf_{n\to+\infty}\mathbb E|X_n|.$$ We conclude as $R$ was arbitrary.