If integration of arbitrary polynomial with respect to Borel measure $\mu$, over $[0,1]$ vanishes, is it true that $\mu$ equals to $0$ on $ [0,1]$? I am having difficulties to deal with following problems;
Assume $ \displaystyle\int_{[0,1]} x^n d \mu =0$ for all $n$, 
then is it true that $\mu=0$ on [0,1]?
I think it is definitely true.. but I don't know how to proceed..
Can anybody help me?
 A: You are not saying if you include $n=0$ in your conditions. If you do, then the equation for $n=0$ is $\mu([0,1])=0$. 
Assuming you don't, it's not true in the generality you are asking. The problem with using the functions $\{x^n\}$ to define your measure is that they are all $0$ at $0$. So if $\mu$ is the Dirac measure, i.e.
$$
\mu(R)=\begin{cases}1,&\ \text{ if } 0\in R\\0,&\ \text{ if} 0\not\in R\end{cases}
$$
Then $\displaystyle\int_{[0,1]}x^n\,d\mu=0$ for all $n$. 
A: The main interest of teh problem is to show that we can't find a finite signed measure satisfying $\int_{[0,1]}t^nd\mu(t)=0$ for all integer $n\geq 0$ (missing $n=0$, as mentioned before, would make the problem wrong with the counter-example $\delta_0$). If we assume $\mu\geq 0$ then the problem is trivial.
Let $\mu=\mu^+-\mu^-$ the Jordan decomposition of $\mu$, where $\mu^+$ and $\mu^-$ are non-negative measures. We have, by Stone-Weierstrass theorem, that $\int_{[0,1]}f(t)d\mu(t)=0$ for all $f$ continuous on $[0,1]$. 
As the characteristic function of a closed set $F$ can be approximated pointwise by continuous functions $f_n$ such that $0\leq f_n\leq 1$, we have by dominated convergence theorem applied to $\mu^+$ and $\mu^-$ that $\mu^+(F)=\mu^-(F)$ for all $F$ closed. This can be extended to all Borel subsets of $[0,1]$, as they generate this $\sigma$-algebra and are stable by (finite) intersections. So $\mu=0$.
