I'm preparing myself for the final exam of my graduate Probability Theory course and was stuck with another one of the exercises our professor gave us.
Let $X_n, n=1,2,\ldots,$ and $X$ be nonnegative random variables such that $X_n \rightarrow X$ in distribution as $n \rightarrow \infty.$
Prove or disprove: If $X_n, n=1,2,\ldots,$ and X have finite means $\mu_n,n=1,2,\ldots,$ and $\mu$, respectively, then $\mu_n \rightarrow \mu$ as $n \rightarrow \infty$.
Here is my current work:
$\mathbb{E}[X_n]=\mu_n < \infty, \mathbb{E}[X] = \mu < \infty.$ We want to show that $\lim_{n\rightarrow \infty}\mathbb{E}[X_n] = \mathbb{E}[X],$ i.e. $\mathbb{E}[X_n - X]=0.$
I think the statement is wrong, and made the following counterexample to disprove it:
Take $X \neq Y$ almost surely, and $X = Y$ in distribution. Then $X_n \rightarrow X = Y$ in distribution, but $\mathbb{P}(|X_n - Y|>\varepsilon) = \mathbb{P}(|X-Y| > \varepsilon) > 0.$
Is this correct?
As @RodrigoRibeiro pointed out, your counterexample doesn't work because $X=Y$ in distribution implies $\mathbb{E}X= \mathbb{E}Y$ and therefore the sequence $\mu_n = \mathbb{E}X$ converges to $\mu = \mathbb{E}Y$.
Nevertheless, the statement is wrong. Consider $((0,1),\mathcal{B}((0,1)))$ endowed with the Lebesgue measure and
$$X_n(\omega) := n^2 1_{(0,1/n)}(\omega), \qquad \omega \in (0,1).$$
Then $X_n \to X:=0$ almost surely (hence, in particular, $X_n \to X$ in distribution), but
$$\mathbb{E}X_n = n$$
does not converge to $\mathbb{E}X=0$.