Let $\nu$ a probability measure on a space $\Omega$. Suppose that for every function $f\in C(\Omega)$ such that $\int_\Omega f(x)dx=0$ it holds that
$$\int_\Omega f(x)d\nu=0$$.
I would like to prove that the probability measure $\nu(dx)=\frac{1}{|\Omega|}dx$. Is this true? How could I prove this? For sure I need to use that $\nu$ is a probability measure and therefore is always positive. But I don't know how to do it.
Thank you
I will assume that $\Omega$ is a compact subset of $\mathbb R$. By Riesz Representation Theorem the dual of $C(\Omega)$ consists of regular Borel measures on $\Omega$. The measures $\nu$ and $dx$ define two linear functionals $F$ and $G$ on this space. It is given that $G(f)=0$ implies $F(f)=0$. An elementary linear algebra result shows that $G$ is a multiple of $F$. This gives $\int f d\nu =\int f \frac 1 {|\Omega|} dx$ for every continuous function $f$ and this implies $\frac 1 {\Omega} dx=\nu$.
EDIT: This does not require Riesz Representation Theorem. We only have to observe that $F$ and $G$ are linear functionals on $C(\Omega)$ such that $ker (F) \subseteq ker (G)$.