I believe that the following is true, and I'd like to make sure that it is and to have a reference. Suppose that $\mu\_N$ are a sequence of measures on $\mathbb{R}$. Let $m\_{N,k}$ be the $k$-th moment of $\mu\_N$. Suppose that, for each $k$, $m\_{N,k}$ converges to a limit $m\_k$ as $N\rightarrow \infty$; no uniformity assumption is made, in that for each $k$ the convergence may be different. Suppose further that $m\_k$ obeys Carleman's condition:


$\sum\_{n=1}^\infty m\_{2n}^{-\frac{1}{2n}}=\infty$


Then, I would like to know that the following 3 results are true:


1) There exists a measure $\mu$ such that the moments of $\mu$ are equal to $m\_k$.


2) Such a measure $\mu$ is unique.


3) The measures $\mu\_N$ weakly converge to $\mu$ as $N \rightarrow \infty$. 


Result (2) I think follows straightforwardly from Carleman's, but I want to verify that (1),(3) are correct.



Converging to moments obeying Carleman's condition
Yes, this works. It's convenient to view the $\mu\_n$ as measures on the compact space $\mathbb R\_{\infty}$, with $\mu\_n(\{\infty\})=0$. By a diagonal process, we then find a subsequence (which I'll write as the original sequence) such that $x^{2k}\, d\mu\_n(x) \to d\rho\_k(x)$ weak $\*$ for certain measures $\rho\_k$ for all $k$; this follows because $\int\_{\mathbb R\_{\infty}} x^{2k}\, d\mu\_n(x)\le C\_k$ by assumption.


From this condition, we also deduce that $\rho\_k(\{\infty\})=0$.
By comparing integrals of compactly supported continuous functions, we then find that $d\rho\_k(x) = x^{2k}\, d\rho\_0(x)$.


This measure $\rho=\rho\_0$ has the correct even moments, by construction, and if $n$ is odd, then we write
$$
\int x^n\, d\rho(x) = \int x^{n-1} g(x)\, d\rho(x) + \int x^{n+1} h(x)\, d\rho(x),
$$
with both $g,h\in C(\mathbb R\_{\infty})$, and $g$ supported near $x=0$ and $h$ supported away from $x=0$. Again, it follows that $\int x^n\, d\rho(x)= m\_n$.


This gives (1). Then (2) is clear because the moment problem with the $m\_n$'s is determinate, and then (3) also follows because, as we saw, the only possible limit point of the $\mu\_n$ is the unique measure $\rho$ with these moments.


