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?



A moment problem on $[0,1]$ in which infinitely many moments are equal
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](http://en.wikipedia.org/wiki/M%C3%BCntz%E2%80%93Sz%C3%A1sz_theorem) 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:
> 
> 
> 1. the vector space generated by $\{x^{n\_k},k\in\mathbb N\}$ is dense in $C[0,1]$ endowed with the uniform norm.
> 2. 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 $m^+\neq m^-$.


