Uniformly Integrable Second Moments and Weak Convergence. Let $\mathcal{P}(\mathbb{R}^d)$ be the space of Borel probability measures, and let $\mathcal{P}_2(\mathbb{R}^d)$ be the space of Borel probability measures with finite second moment. Let $\{\mu_n\} \subset \mathcal{P}_2(\mathbb{R}^d)$ have uniformly bounded second moments, and assume $\mu_n\to \mu \in \mathcal{P}(\mathbb{R}^d)$ weakly.
Is it true that $\mu \in \mathcal{P}_2(\mathbb{R}^d)$ ?
 A: Yes.
By Skorohod representation theorem there exists random vectors $X_n$ and $X$ such that $\mu_n$ is the distribution of $X_n$,  $\mu$ is the distribution of $X$ and $X_n \to X$ a.s.
Denote by $X_n^{(1)}$ the sequence of first coordinates of vectors $X_n$. We know that $X_n^{(1)} \to X^{(1)}$ a.s., where $X^{(1)}$ is the first coorndite of vectors $X$.
As $X_n$ have uniformly bounded second moments $sup_{n \ge 1} E ||X_n||^2 < \infty$ then $sup_{n \ge 1} E |X_n^{(1)}|^2 < \infty$.
By Fatou's lemma $ E (X^{(1)})^2 = E \underline{lim} (X_n^{(1)})^2 \le \underline{lim} E(X_n^{(1)})^2 < \infty$.
Similarly $ E (X^{(i)})^2 < \infty$ for all $1 \le i \le d$. It follows that the limit of $P_n$ is in $\mathcal{P}_2(R^d)$, q.e.d.
Addition
By De la Vallée-Poussin criterion for Uniform Integrability  $\{X_n^{(1)} \}_{n\ge 1}^{\infty}$ is a uniformly integrable class of r.v. Hence $X^{(1)} \in L_1$ and $EX_n^{(1)} \to EX^{(1)}$. But $EX_n^2$ doesn't converges to $EX^2$ in general case.
Example. Consider $X_n \sim \sqrt{n} Bern(\frac{1}n)$, then $X_n \to 0$. Thus $P_n, P \in \mathcal{P}_2(R^d)$, $EX_n^2 \to 1 \ne 0 = EX^2 $.
