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]} \\),<\/p>\n
\\[ \\int\\!P\\,d\\mu_1=\\int\\!P\\,d\\mu_2 \\]<\/p>\n
(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<\/strong> 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<\/a>. The following classical lemma is useful when using the moments method, for instance when proving the Wigner semicircle theorem.<\/p>\n Lemma 1 (From moments convergence to weak convergence)<\/strong> If \\( {\\mu,\\mu_1,\\mu_2,\\ldots} \\) belong to \\( {\\mathcal{P}} \\) with <\/em><\/p>\n \\[ \\lim_{n\\rightarrow\\infty}\\int\\!P\\,d\\mu_n=\\int\\!P\\,d\\mu \\]<\/em><\/p>\n for every \\( {P\\in\\mathbb{R}[X]} \\) and if <\/em>\\( {\\mu} \\) is characterized by its moments <\/em>then for every \\(f\\in\\mathcal{C}_b(\\mathbb{R},\\mathbb{R})\\)<\/p>\n \\[ \\lim_{n\\rightarrow\\infty}\\int\\!f\\,d\\mu_n=\\int\\!f\\,d\\mu. \\]<\/em><\/p>\n<\/blockquote>\n Proof:<\/em> The convergence assumption implies that for every polynomial \\( {P} \\)<\/p>\n \\[ C_P:=\\sup_{n\\geq1}\\int\\!P\\,d\\mu_n<\\infty. \\]<\/p>\n Thus, by Markov's inequality, for every real \\( {R>0} \\),<\/p>\n \\[ \\mu_n([-R,R]^c)\\leq \\frac{C_{X^2}}{R^2} \\]<\/p>\n 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.<\/p>\n 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<\/p>\n \\[ \\int\\!P\\,d\\mu_{n_k}=\\int\\!\\varphi_RP\\,d\\mu_{n_k}+\\int\\!(1-\\varphi_R)P\\,d\\mu_{n_k}. \\]<\/p>\n Since \\( {(\\mu_{n_k})_{k\\geq1}} \\) tends weakly to \\( {\\nu} \\) we have<\/p>\n \\[ \\lim_{k\\rightarrow\\infty}\\int\\!\\varphi_RP\\,d\\mu_{n_k}=\\int\\!\\varphi_RP\\,d\\nu. \\]<\/p>\n Additionally, by Cauchy-Schwarz's and Markov's inequalities,<\/p>\n \\[ \\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}. \\]<\/p>\n On the other hand, we know that<\/p>\n \\[ \\lim_{k\\rightarrow\\infty}\\int\\!P\\,d\\mu_{n_k}=\\int\\!P\\,d\\mu. \\]<\/p>\n Therefore, we obtain<\/p>\n \\[ \\lim_{R\\rightarrow\\infty}\\int\\!\\varphi_RP\\,d\\nu=\\int\\!P\\,d\\mu. \\]<\/p>\n 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<\/p>\n \\[ \\int\\!P\\,d\\nu=\\int\\!P\\,d\\mu. \\]<\/p>\n Since \\( {P} \\) is arbitrary and \\( {\\mu} \\) is characterized by its moments, it follows that \\( {\\mu=\\nu} \\). $\\Box$<\/p>\n","protected":false},"excerpt":{"rendered":" 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…<\/p>\n