Moments and weak convergence of probability measures II

Suppose a sequence of probability measures $\mu_n \Rightarrow \mu$ converges weakly to a limit, and suppose moreover that $$\lim_{n \rightarrow \infty} \int x^k \mu_n (dx) = m_k \in \mathbb{R}$$ for some sequence of numbers $\{m_k\}_{k=1}^{\infty}$. Is it true that $$\int x^k \mu (dx) = m_k?$$ I believe so, but I can't seem to prove it, since the functions $x^k$ are unbounded. If anyone could offer any insight, I'd greatly appreciate it!

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Where does the "II" in the title come from? –  Michael Greinecker Jan 30 '13 at 8:37
Homework?   –  Did Jan 30 '13 at 9:23

By the Skorohod representation theorem, we can assume that the $\mu_n$ are the laws of a sequence of random variables $\{X_n\}$ on some probability space, and the $X_n$ converge almost surely to some $X$ whose law is $\mu$. Now we have to show that if $E[X_n^k] \to m_k$ for all $k$, then $E[X^k] = m_k$.
Let's do $k=1$ first. Since $E[X_n^2] \to m_2$, in particular we have that $\{X_n\}$ is bounded in $L^2$. There is a fact, sometimes called the "crystal ball condition", that if a sequence of random variables is bounded in $L^p$ for some $p > 1$, then it is uniformly integrable. So we have that $\{X_n\}$ is uniformly integrable and converges to $X$ almost surely. By the Vitali convergence theorem, we have $X_n \to X$ in $L^1$, i.e. $E X_n \to EX$. This shows $EX = m_1$.
For general $k$, choose any even integer $r > k$, and set $p=r/k > 1$. Then we have that $E [|X_n^k|^p] = E[X_n^r]$ is bounded, so $\{X_n^k\}$ is bounded in $L^p$. Since $X_n^k \to X^k$ almost surely, as before we have $E[X_n^k] \to E[X^k]$, which is to say $E[X^k] = m_k$.