Closeness of probability measures Consider a set of probability measures $\{P_n\}$. Suppose $P_n$ converges to $P^*$ weakly and
$$
\int \xi^2 P_n(d\xi)< \infty.
$$
Can we claim
$$
\int \xi^2 P^*(d\xi)<\infty
$$
and
$$
\lim_{n\to \infty} \int \xi^2 P_n(d\xi) =\int \xi^2 P^*(d\xi)?
$$
 A: The question is: is the set of probability measures having a moment of order $2$ sequentially closed for convergence in law?
The answer is no. Using Glivenko-Cantelli theorem, we can show that any probability measure on $\Bbb R$ endowed with its Borel $\sigma$-algebra for usual topology, is the limit in law of probability measures which have a finite support.
The result remains true when we consider a separable metric space instead of $\Bbb R$. For example, we can avoid Glivenko-Cantelli showing that 
$$\left\{\sum_{j=1}^na_j\delta_{x_{k_j}},a_j\geqslant 0,\sum_{j=1}^na_j=1,n\in\Bbb N\right\}$$
is dense for the topology of convergence in law, where $\{x_j,j\geqslant 1\}$ is a countable subset. 
A: Assume that $P_n=p_n\delta_{x_n}+(1-p_n)\delta_0$ for some $(x_n)$ and $(p_n)$ with $p_n\to0$, then $P_n$ is square integrable and $P_n\to P^*=\delta_0$ but $\int\xi^2P_n(\mathrm d\xi)=p_nx_n^2$ while $\int\xi^2P^*(\mathrm d\xi)=0$ hence the square moments can diverge (consider $p_n=1/n$ and $x_n=2^n$).
