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!
 A: Here's one way to show it, though it may be possible to do it with less machinery.
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 $E[X^k] = m_k$.
