Determining measures by integrals

Let $(S,d)$ be a metric space and let $\mu$, $\nu$ be two Borel probability measures on $S$. Then $$ \int f\,\mathrm d\mu=\int f\,\mathrm d\nu,\;\forall\,f\in \mathrm{bC}(S)_+\iff\mu=\nu. $$

Proof: Assume that the integrals of every function in $\mathrm{bC}(S)_+$ are identical. Then to show $\mu=\nu$, it is enough to show that $\mu(U)=\nu(U)$ for all $U\subseteq S$ open. For each $n\geq 1$ we define $$ f_n(x):=n\cdot d(x,U^c)\wedge 1,\quad n\geq 1. $$ Then the $f_n$'s are bounded, non-negative and continuous (they are in fact Lipschitz continuous) and $f_n\uparrow 1_U$ pointwise. Thus, $$ \mu(U)=\int 1_U\,\mathrm d\mu=\lim_{n\to\infty}\int f_n\,\mathrm d\mu=\lim_{n\to\infty}\int f_n\,\mathrm d\nu=\nu(U). $$

We can relax the assumption that $\mu$ and $\nu$ need to be probability measures. The above also holds in the more general setting where $\mu$ and $\nu$ are just measures such that there exists a sequence $(A_n)_{n\geq 1}$ of open Borel sets such that $$ S=\bigcup_{n\geq 1}A_n,\quad \text{and}\quad \mu(A_n)=\nu(A_n)<\infty,\quad n\geq 1. $$

If is enough to check the equality of integrals with $f(x) = e^{i\langle x,\theta\rangle}$ for every $\theta \in \Bbb R^d$. This is the Fourier transform.

The proofs of both this and the general result given by Stefan can be found in Billingsley's Convergence of Probability Measures.