From moments convergence to weak convergence

Let ${\mathcal{P}}$ be the set of probability measures ${\mu}$ on ${\mathbb{R}}$ such that ${\mathbb{R}[X]\subset\mathrm{L}^1(\mu)}$. Let us consider the equivalent relation ${\sim}$ on ${\mathcal{P}}$ given by ${\mu_1\sim\mu_2}$ if and only if for every ${P\in\mathbb{R}[X]}$,

$$\int\!P\,d\mu_1=\int\!P\,d\mu_2$$

(i.e. ${\mu_1}$ and ${\mu_2}$ share the same sequence of moments). We say that ${\mu\in\mathcal{P}}$ is characterized by its moments when its equivalent class is a singleton. Every compactly supported probability measure on ${\mathbb{R}}$ belongs to ${\mathcal{P}}$ and is indeed characterized by its moments thanks to the Weierstrass density theorem. Beyond compactly supported probability measures, one may use the Carleman criterion. The following classical lemma is useful when using the moments method, for instance when proving the Wigner semicircle theorem.

Lemma 1 (From moments convergence to weak convergence) If ${\mu,\mu_1,\mu_2,\ldots}$ belong to ${\mathcal{P}}$ with

$$\lim_{n\rightarrow\infty}\int\!P\,d\mu_n=\int\!P\,d\mu$$

for every ${P\in\mathbb{R}[X]}$ and if ${\mu}$ is characterized by its moments then for every $f\in\mathcal{C}_b(\mathbb{R},\mathbb{R})$

$$\lim_{n\rightarrow\infty}\int\!f\,d\mu_n=\int\!f\,d\mu.$$

Proof: The convergence assumption implies that for every polynomial ${P}$

$$C_P:=\sup_{n\geq1}\int\!P\,d\mu_n<\infty.$$

Thus, by Markov's inequality, for every real ${R>0}$,

$$\mu_n([-R,R]^c)\leq \frac{C_{X^2}}{R^2}$$

and therefore ${(\mu_n)_{n\geq1}}$ is tight. As a consequence, by Prokhorov's theorem, it suffices now to show that if ${(\mu_{n_k})_{k\geq1}}$ converges weakly to ${\nu}$ then ${\nu=\mu}$. Recall that the weak convergence here is also known as the narrow convergence and corresponds to the convergence for continuous bounded functions.

Let us show that ${\mu=\nu}$. Let us fix some ${P\in\mathbb{R}[X]}$ and a real number ${R>0}$. Let ${\varphi_R:\mathbb{R}\rightarrow[0,1]}$ be continuous with ${\mathbf{1}_{[-R,R]}\leq\varphi_R\leq\mathbf{1}_{[-R-1,R+1]}}$. We start from the decomposition

$$\int\!P\,d\mu_{n_k}=\int\!\varphi_RP\,d\mu_{n_k}+\int\!(1-\varphi_R)P\,d\mu_{n_k}.$$

Since ${(\mu_{n_k})_{k\geq1}}$ tends weakly to ${\nu}$ we have

$$\lim_{k\rightarrow\infty}\int\!\varphi_RP\,d\mu_{n_k}=\int\!\varphi_RP\,d\nu.$$

Additionally, by Cauchy-Schwarz's and Markov's inequalities,

$$\left|\int\!(1-\varphi_R)P\,d\mu_{n_k}\right|^2 \leq \mu_{n_k}([-R,R]^c)\int\!P^2\,d\mu_{n_k} \leq \frac{C_{X^2}C_{P^2}}{R^2}.$$

On the other hand, we know that

$$\lim_{k\rightarrow\infty}\int\!P\,d\mu_{n_k}=\int\!P\,d\mu.$$

Therefore, we obtain

$$\lim_{R\rightarrow\infty}\int\!\varphi_RP\,d\nu=\int\!P\,d\mu.$$

Using this for ${P^2}$ we obtain by monotone convergence that ${P\in\mathrm{L}^2(\nu)\subset\mathrm{L}^1(\nu)}$ and then by dominated convergence that

$$\int\!P\,d\nu=\int\!P\,d\mu.$$

Since ${P}$ is arbitrary and ${\mu}$ is characterized by its moments, it follows that ${\mu=\nu}$. $\Box$