TITLE: A moment problem on $[0,1]$ in which infinitely many moments are equal
QUESTION [17 upvotes]: Suppose $\mu$ and $\nu$ are two probability measures on $[0,1]$. Let their $n$-th moments be denoted by $\mu_n$ and $\nu_n$, respectively, for $n \in \mathbb{N}$.
If we know that $\mu_n=\nu_n$ for infinitely many $n$, can we conclude that $\mu=\nu$?
One way to resolve this would be to see if the span of $\{ x^n \mid n \in S \}$ with $|S|=\infty$ is dense in the set of continuous functions $C[0,1]$. Is such a set always dense?

REPLY [15 votes]: We actually have that $\mu=\nu$ is guaranteed if and only if $\sum_{n\in S}\frac 1n$ is divergent. It's a condition which translates the fact that the set of indexed $k$ such that $\mu_k=\nu_k$ has to be large enough.
Recall Müntz-Szász theorem, which states (in particular) the following:

Theorem: If $(n_k,k\geqslant 0)$ is an increasing sequence of integers with $n_0=0$, then the following conditions are equivalent:

the vector space generated by $\{x^{n_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.

the series $\sum_{k=1}^\infty\frac 1{n_k}$ is divergent.



If $\sum_{n\in S}\frac 1n$ is divergent, we can conclude by density that $\mu$ and $\nu$ coincide.
If the series is convergent, we can find $F\in (C[0,1])'$ such that $F(x^n)=0$ for all $n\in \{0\}\cup S$, but $F$ is not identically vanishing (this comes from Hahn-Banach theorem). We represent $F$ as a (non-zero) signed measure $m:=m^+-m^-$ (Hahn decomposition). Since $m^+[0,1]=m^-[0,1]$, we can rescale these measures in order to get probability measures. Then with the same notations as in the OP, $m^+_n=m^-_n$ for all $n\in \{0\}\cup S$, but $m^+\neq m^-$.