Show that there is a probability such that $P_n$ converges weakly/in distribution as $n \to \infty$. Suppose that $P_n$ $n \ge 1$ is a sequence of probabilities concentrated on $[a,b]$. Suppose that one may show for each positive integer $r$ that $\int_{[a,b]}x^rP_n(dx) \to m_r \in R$ as $n \to \infty$. Show that there is a probability $P$ such that $P_n \Rightarrow P$ as $n \to \infty$ and $\int_{[a,b]}x^rP(dx) = m_r$ for each $r \ge 1$.
My attempt at a solution:
I believe that because we are looking at probabilities on $[a,b]$, we automatically have that the sequence is tight. At which point I would like to apply Prohorov's Theorem, which would implies that the weak closure of the sequence is compact in the weak topology. But I have no idea what to do from here.
Edit 1: I have made some progress. So, for a subsequence that converges weakly to $P$, say, $P_{n_k}$, whose existence we are guaranteed by Prohorov's Theorem, that subsequence converges weakly to some $P_k$. Therefore, for all continuous, bounded functions $f$ on $[a,b]$, we have that 
$$\int_{[a,b]} f(x)P_{n_k}(dx) \to \int_{[a,b]}f(x)P_k(dx)$$
I'm thinking that from here we'll want to apply Weierstrass (all continuous, bounded functions can be approximated by polynomials) but I don't quite know how to do it.
 A: Indeed, a subsequence convergences in distribution. In order to exhibit the existence of $\mathbb P$, we have to show that there is at most one limit point. 
Since the map $x\mapsto x^j$ is continuous and bounded on $[a,b]$, we have 
$$\tag{conv} \lim_{ l\to\infty } \int_{[a,b]}x^j\mathrm d\mathbb P_{n_l}(x)= \int_{[a,b]}x^j\mathrm d\mathbb Q(x)  $$
if $\mathbb P_{n_l}\to \mathbb Q$. Assume that $\mathbb P_{n_l}\to \mathbb Q$ and $\mathbb P_{m_l}\to \mathbb Q'$, then by the assumption we have 
$$\int_{[a,b]}x^j\mathrm d\mathbb Q(x) =m_j=\int_{[a,b]}x^j\mathrm d\mathbb Q'(x) $$
and it follows by the Stone-Weierstrass theorem that $\mathbb Q=\mathbb Q'$ hence that $\mathbb P_n\to \mathbb P$ weakly.
The last part follows by applying (conv) to the whole sequence.
A: Hints:


*

*As you already noted weak convergence of the sequence $(P_n)_n$ is equivalent to the following statement: There exists a measure $P$ such that for any subsequence $(P_{n_k})_{k \in \mathbb{N}}$ there exists a subsequence $(P_{n_{k_\ell}})_{\ell}$ such that $P_{n_{k_\ell}}$ converges weakly to $P$.

*Let $(P_{n_k})_k$ be an arbitrary subsequence. By Prokohorov's theorem, there exists a subsequence $P_{n_{k_\ell}}$ which converges weakly to some measure $P$. It remains to show that $P$ does not depend on the chosen subsequence.

*Since $[a,b] \ni x \mapsto x^r$ is bounded for each $r \in \mathbb{N}_0$, we have $$\int_{[a,b]} x^r  \, dP_{n_{k_\ell}}(x) \to \int_{[a,b]} x^r \, dP(x)$$ by weak convergence. On the other hand, by assumption, the left-hand side converges to $m_r$. Consequently, $$m_r = \int_{[a,b]} x^r \, dP(x).$$

*Any measure $Q$ on $[a,b]$ is uniquely determined by its moments $$\int_{[a,b]} x^r \, dQ(x), \qquad r \in \mathbb{N}_0.$$ This follows from the fact that any measure is uniquely determined by its characteristic function and $$\int e^{\imath \, t x} \, dQ(x) = \sum_{r \in \mathbb{N}_0} (\imath \, t)^r \int_{[a,b]} x^r \, dQ(x).$$ (Note that the dominated convergence theorem is applicable since $$\int_{[a,b]} e^{|t| x} \, dQ(x) \leq e^{|t| \max\{|a|,|b|\}} Q([a,b]) < \infty.$$

*Conclude.

