Is a Radon measure determined by its action on $C_b(X;\mathbb{R})$? Let $\mu$ and $\nu$ be  finite Radon measures on $\left(X,\mathscr{B}(X)\right)$, where $X$ is a topological space (not necessarily locally compact) and $\mathscr{B}(X)$ the Borel sigma field. Then if
$$\int_X f(x)\,\mu(\mathrm{d}x)=\int_X f(x)\,\nu(\mathrm{d}x)$$
for all bounded continuous functions $f\in C_\mathrm{b}\left(X;\mathbb{R}\right)$, must we have $\mu=\nu$? I strongly suspect that this is the case, but cannot seem to prove it.
Any hints/partial answers would be much appreciated!
EDIT: Suppose further that $C_\mathrm{b}\left(X;\mathbb{R}\right)$ is an algebra which separates points.
 A: Yes.
Let $T : X \to \beta X$ be the Stone-Cech compactification of $X$, so that $T$ is continuous (in particular Borel) and $\beta X$ is compact Hausdorff.  I claim first that $T$ is injective.  For if $x,y$ are distinct points of $X$, there is by assumption a continuous $f : X \to [0,1]$ with $f(x) \ne f(y)$.  Since $[0,1]$ is compact Hausdorff, by the universal property, $f$ factors through $T$, meaning $T(x) \ne T(y)$.
Let $\bar{\mu} = \mu \circ T^{-1}$, $\bar{\nu} = \nu \circ T^{-1}$ be the pushforwards of $\mu,\nu$ onto $\beta X$.  These are finite Radon measures on $\beta X$.  To see the inner regularity, let $A \subset \beta X$ be Borel and fix $\epsilon > 0$. Then $T^{-1}(A)$ is Borel in $X$.  Since $\mu$ is Radon, there exists a compact $K \subset X$ such that $\mu(T^{-1}(A) \setminus K) < \epsilon$.  Since $T$ is continuous, $T(K)$ is compact in $\beta X$.  Since $T$ is injective, $T^{-1}(A) \setminus K = T^{-1}(A \setminus T(K))$, so 
$$\bar{\mu}(A \setminus T(K)) = \mu(T^{-1}(A \setminus T(K))) = \mu(T^{-1}(A) \setminus K) < \epsilon.$$
Now if $g : \beta X \to \mathbb{R}$ is continuous, then $g \circ T \in C_b(X; \mathbb{R})$, so by assumption $\int_X (g \circ T) \,d\mu = \int_X (g \circ T) \,d\nu$.  By change of variables, this means $\int_{\beta X} g\,d\bar{\mu} = \int_{\beta X} g\,d\bar{\nu}$.  But $g$ was an arbitrary continuous function and $\bar{\mu}, \bar{\nu}$ are Radon measures on a compact Hausdorff space, so $\bar{\mu} = \bar{\nu}$.  
So if $K \subset X$ is compact, then $T(K)$ is compact in $\beta X$ and in particular Borel.  Then
$$\mu(K) = \bar{\mu}(T(K)) = \bar{\nu}(T(K)) = \nu(K).$$
Thus $\mu, \nu$ agree on all compact sets.  Since they are both Radon, they agree on all Borel sets.
(This solution loosely follows a hint given in Exercise 7.14.79 of Bogachev's Measure Theory.)
