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Let's suppose that (A, F) is a measurable space (A underlying set, F the sigma algebra), and that F arises from some topology T.

I would like a theorem of the form:

All singletons belong to F if and only if (A, F, T) has properties P1,P2,...

where P1,P2,... may be measure theoretic or topological conditions on the triple (A, T, F).

Thanks a lot for your help.

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up vote 0 down vote accepted

A necessary condition is that the underlying topological space is $T_0$. This is sufficient if the space has a countable basis.

A sufficient condition is that the underlying space is $T_1$.

If the $\sigma$-algebra is generated by an arbitrary family $\mathcal{F}$, a necessary and sufficient condition for $\{x\}$ to be measurable is that there is a countable subset $\mathcal{C}\subseteq\mathcal{F}$ such that for all $y\neq x$ there is $F\in\mathcal{C}$ such that $x\in F$ and $y\notin F$ or $y\in F$ and $x\notin F$.

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Well yes of course... I am thinking more along the lines of 'locally compact second countable Hausdorff'... which I think may be necessary and sufficient. But I'm not sure... – Frank Feb 14 '14 at 13:21
@Frank Note that being $T_1$ weakens your sufficiency condition, but it is still not necessary. – Michael Greinecker Feb 14 '14 at 13:25
Being $T_0$ and second countable is a different sufficient weakening of your condition. – Michael Greinecker Feb 14 '14 at 13:36

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