If two measures coincide on bounded continuous functions, do they coincide on Borel subsets? Let $X$ be a Hausdorff space, endowed with the Borel $\sigma$-algebra, and let $\mu, \nu$ be two regular Borel probabilities on $X$. Assume that $\int _X f \ \mathrm d \mu = \int _X f \ \mathrm d \nu$ for all $f \in C_b (X)$ (the bounded continuous complex functions).

Does it follow that $\mu (B) = \nu (B)$ for all $B \in \mathcal B (X)$ (the Borel subsets)?

It clearly happens for locally-compact spaces: taking $f \in C_0 (X)$ (the functions that vanish at infinity), we have that $\mu = \nu$ in $C_0 (X) ^*$ (with the Riesz-Markov theorem). I believe that this is true on completely regular spaces too, with a similar argument and by endowing $C_b (X)$ with the strict topology.

It is true with perfectly normal spaces at least.
Let $\mathbb{B}(X)$ be all bounded measurable functions.
Take $\mathcal{H}=\{f\in \mathbb{B}(X)\ :\ \int fd\mu=\int fd\nu\}$.
$\mathcal{H}$ is a vectorial space which contains $1$ and is closed by monotone limits : if $(f_n)_{n\in\mathbb{N}}\subset\mathcal{H}$ and $f_n\nearrow f\in \mathbb{B}(X)$ pointwise then $f\in\mathcal{H}$ (Beppo-Levy).
If $\mathcal{H}_0\subset\mathcal{H}$ is closed by multiplication then $\mathcal{H}$ contains all $\sigma(\mathcal{H}_0)$-measurable functions by the functional monotone class theorem.
If $X$ is perfectly normal (edit) you can take $\mathcal{H}_0=\mathcal{C}_b(X)$ since $\sigma(\mathcal{C}_b(X))=\mathcal{B}(X)$.
