2

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 $\mathcal{M}_1$?

White
  • 183
  • 9

1 Answers1

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].

Kavi Rama Murthy
  • 359,332
  • It seems that there are counter-examples showing that there are measure which agree only on ball but not on the Borel $/sigma$-algebra : https://math.stackexchange.com/questions/2022647/equality-of-two-borel-measures. But I do not know for finite intersections of balls. You can also take finite sums of finite products to see that if two measures agree on this family, then they agree on finite unions of closed balls centered at $x_n$'s. – Davide Giraudo Mar 15 '18 at 16:39
  • @ Davide Giraudo values on finite intersections of balls do determine Borel probability measures, but values on balls don't. – Kavi Rama Murthy Mar 16 '18 at 08:43
  • Thanks a lot for the answer, i was busy this week and could not think about it. But I think its correct and I do not see any holes now. Will have a closer look now and post the full answer. – White Mar 16 '18 at 10:22