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Given that $(X_n)_{n \in \mathbb(N)}$ converges in distribution to $X$, I am trying to show that $$E(|X|) \le \lim_{n \to \infty} \inf (E|X_n|).$$

I though of using Portmanteau theorem since part e) in my notes claims the following

  • for each $U \subset S $ open $\lim_{n \to \infty} \inf \mu_n(U) \ge \mu(U)$.

But how do I come to expectations from probability measures ?

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    What is your definition of "converges in distribution"? – 2017-01-14
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    I guess I mixed up the things here. In this case $X_n$ and $X$ are real-valued but Portmanteau theorem works for an arbitrary metric space $(S,d)$ (that's where that $S$ in the statement comes from). For real-valued random variables convergence in law follow from the weak convergence of the corresponding probability measures. Is that what you asked ? – 2017-01-14
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    **Hint:** Apply the appropriate part of the Portmanteau theorem using the bounded continuous function $f_k(x):=\min(|x|,k)$, and then let $k\to\infty$. – 2017-01-14
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    @JohnDawkins Thanks for the hint. However, I'm still struggling with this problem. So for a fixed $x$, the limit for $k \to \infty$ of the given sequence of functions will be $|x|$, right ? And also which part of the Theorem should I apply. I'm at the statement (i) now (which is basically the definition of weak convergence). The inequality I have to prove is similar to part (v) (or e) while introducing a sequence of functions leads us to part (iii) (or c) but those have to converge to the indicator functions from below which doesn't seem to be our case. – 2017-01-16
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    Different authors state the theorem in slightly different ways, so "part (iii) (or c)" is not of much help without context. I had in mind that you use the part that says $\mu_n$ converges weakly to $\mu$ if and only if $\int f\,d\mu_n\to\int f\,d\mu$ for each bounded continuous function $f$. Thus, for each $k$, $E(|X_n|)\ge E(f_k(X_n))=\int f_k\,d\mu_n$. Let $n\to\infty$ in this inequality, and then let $k\to\infty$. – 2017-01-16

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