# When weak convergence implies moment convergence?

Given a sequence $$(\mu_n)_n$$ of probability measures on $$\mathbb R$$, which converges weakly to a probability measure $$\mu$$, when do we have $$\tag{1} \lim_{n}\int x^kd\mu_n(x)=\int x^k d\mu(x) \qquad \forall k\geq 0\;?$$ Is "$$\mu$$ has compact support" a sufficient condition?

Note that $$\mu_n$$ converges to $$\mu$$ weakly if $$\int \varphi d\mu_n \to \int \varphi d\mu$$ for all $$\varphi$$ which is continuous and has compact support. Note that $$x^k$$ are continuous but not of compact support, so (1) is not immediately obvious.

A condition on the limit measure will never be enough. The sequence $\left(1-{1\over n}\right)\delta_0+{1\over n}\delta_{x(n)}$ converges to $\delta_0$ weakly, but we can make its moments behave horribly by choosing $x(n)$ to be very large.
A sufficient condition for your moments to converge is if all the $\mu_n$s have the same compact support.
This is almost 10 years late. Possibly by know you are Student no longer. In any event, if $$\mu_n\stackrel{n}{\Longrightarrow\infty}\mu$$, a sufficient condition for $$E_{\mu_n}[X]\rightarrow E_\mu[X]$$ is that $$\lim_{a\rightarrow\infty}\sup_n\int_{\{|x|>a\}}|x|\,\mu_n(dx)=0$$ This is equivalent to say that a sequence of random variable $$X_n\sim \mu_n$$, defined on a common domain $$(\Omega,\mathscr{F},\mathbb{P})$$ is uniformly integrable. This can be seen for instance in Billingsley, P. Convergence in Probability Measures, John Wiley & Sons, 1968, pp. 32.
From this result, is not difficult to derived sufficient conditions for $$E_{\mu_n}[X^r]\xrightarrow{n\rightarrow\infty}E_{\mu}[X^r]$$ (again, uniform integrability is the key).
In the counterexample given by the responder (user940), $$\mu_n=(1-\tfrac1n)\delta_0+\frac1n\delta_{x_n}$$ with $$x_n\xrightarrow{n\rightarrow\infty}\infty$$ Notice that given $$a>0$$, for all $$n$$ large enough $$x_n>a$$ and so, $$\int\mathbb{1}_{\{|x|>a\}}|x|\mu_n(dx)=\frac{1}{n}x_n$$ If $$x_n\geq n$$ for all $$n$$, then $$\sup_n\int\mathbb{1}_{\{|x|>a\}}|x|\mu_n(dx)\geq1$$ This means that in this case, the measures do not satisfy the uniformintegrability condition.