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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.

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Yes. This can be proved using Dynkin's $\pi$-$\lambda$ theorem. 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.

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  • $\begingroup$ @Nate: Thank you very much. (The key step that I'd been missing was how to obtain a countable $\pi$-system that generates $\Sigma$.) When writing a paper, do you think I can assume (without proof or reference) the answer to my question? In other words, is the fact which I asked about sufficiently "standard" that I don't need to justify it? Would you want the paper to give acknowledgement of your help? $\endgroup$
    – Julian Newman
    Commented Apr 11, 2013 at 20:23
  • $\begingroup$ @Julian: Well, I for one had to think about it for a couple of minutes. If it were my paper, I'd probably at least put in a few words to indicate to the reader how it goes. If you'd like to acknowledge me, my "publication name" is Nathaniel Eldredge. Glad to help! $\endgroup$
    – Nate Eldredge
    Commented Apr 12, 2013 at 1:00
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    $\begingroup$ "Dynkin's theorem $\pi$-$\lambda$-theorem" is, possibly up to a minor variation, due Sierpi\'nski from about 1927. I don't have the precise reference(s) at hand, but see Bogachev's (2007) Measure Theory 1, p. 419, or Elstrodt's (2009) Ma\ss- und Integrationstheorie, 6. Auflage, p. 25. So, calling the result "Dynkin $\pi$-$\lambda$ theorem", honouring a popularizer of it, might be defensible, but writing here "Dynkin's" seems wrong to me. $\endgroup$
    – Lutz Mattner
    Commented Apr 28, 2013 at 8:55

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