Countable separating family for measures on a Polish space.

Question

Let $(E, \mathcal{A})$ be a Polish space. Let $\mathcal{M}_1$ be the space of measures $\mu$ on $( E, \mathcal{A})$ with $\mu(E) \leq 1$. A family of bounded continuous function $\mathcal{E}$ is called separating family for $\mathcal{M}_1$ iff for all $\mu, \gamma \in \mathcal{M}_1$ we have \begin{equation} \forall f \in \mathcal{E}: \int f d\mu = \int g d \gamma \quad \implies \mu = \gamma. \end{equation} Does there exist a countable separating family for $\mathcal{M}_1$?

Answer 1

Let $\{x_n\}$ be a countable dense set in E. For any $n,m \geq 1, r \in \mathbb Q$ let $f_{n,m,r}$ be a continuous function with values in $[0,1]$ such that $f_{n,m,r}=1$ on $B(x_n,r)$ and $0$ off $B(x_n,r+\frac 1 m)$. If $\int f d\mu=\int f d\nu$ for all these functions we can let $m \to \infty$ to get $\mu (B(x_n,r))=\nu (B(x_n,r))$ for all n and r. If we consider all finite products of the functions just described we get another countable family and for this family $\int f d\mu=\int f d\nu$ for all these functions implies that $\mu$ and $\nu$ assign the same measure to finite intersections of the balls $B(x_n,r)$. An application of the $\pi - \lambda$ theorem should now complete the proof. [I am almost sure but not fully sure that this works :-). Let me know if you find a hole in the argument].