Hello all.


I have the following (perhaps basic) question: Let $X$ be a separable metric space. Does there necessarily exist a countable set $\mathcal{C}$ of Borel sets in $X$ such that any two probability measures which agree on $\mathcal{C}$ must agree on the whole of $\mathcal{B}(X)$?


(And slightly more generally: Let $(X,\Sigma)$ be a countably generated measurable space. Then does there necessarily exist a countable set $\mathcal{C}$ of $\Sigma$-measurable sets such that any two probability measures which agree on $\mathcal{C}$ must agree on the whole of $\Sigma$?)


Thanks,
Julian.



"Uniqueness of extension" results for measures on separable spaces
Yes. This can be proved using [Dynkin's $\pi$-$\lambda$ theorem](http://en.wikipedia.org/wiki/Dynkin_system). The collection $\mathcal{L} := \{ B \in \Sigma : \mu(B) = \nu(B)\}$ is a $\lambda$-system. By Dynkin's theorem, if $\mathcal{L}$ contains a $\pi$-system which generates $\Sigma$ then $\mathcal{L} = \Sigma$, i.e. $\mu=\nu$. If $\mathcal{C}\_0$ is countable and generates $\Sigma$, then the collection $\mathcal{C}$ of all finite intersections of sets from $\mathcal{C}\_0$ is a countable $\pi$-system which generates $\Sigma$. So any two probability measures that agree on $\mathcal{C}$ must be equal.


