TITLE: Equivalent measures if integral of $C_b$ functions is equal
QUESTION [6 upvotes]: Is it true that if $X$ is a measure space and $\mu, \nu$ are Borel probability measures on $X$ if 
$$ \int_X \phi \ d \mu = \int_X \phi \ d \nu \qquad \forall \phi \in C_b(X) \text{ (continuous and bounded functions)} $$
then $$ \mu = \nu  \text{ ?}$$
If $E$ is a measureable set s.t. $1_E \in C_b(X)$ then $\mu(E) = \nu(E)$, but if it's not?

REPLY [3 votes]: Assume that $\left(X,d\right)$ is a metric space and let $F$ be a closed subset of $X$. For $S\subset X$ and $x\in S$, define $d(x,S):=\inf\{d(x,y),y\in S\}$.
Let $O_n:=\left\{x\in X,d(x,F)<n^{-1}\right\}$. Then $O_n$ is open and the map $$f_n\colon x\mapsto \frac{d\left(x,X\setminus O_n\right)}{d\left(x,X\setminus O_n\right)+d\left(x,F\right)}$$ is continuous and bounded. It converges pointwise and monotonically to the characteristic function of $F$. So we get by monotone convergence that $\mu(F)=\nu(F)$ for all closed set $F$. 
Now given a Borel set $B$ and $\varepsilon>0$, we can find a closed set $F$ and an open set $O$ such that $F\subset S\subset O$ and $\mu(O\setminus S)\leqslant\varepsilon$. Conclude.