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Let $X$ be a topological space and $\mu$ an "outer" Borel regular measure on $X$ (for all $A\subset X$, there is $B$ Borel with $\mu(A)=\mu(B)$).

Assume that $X=\cup _{i=1}^\infty U_i$, where each $U_i$ is open and $\mu (U_i)<\infty$. What is the minimum regularity that we must ask of $X$, so that $\mu$ is a regular measure?

There is a claim in a book I am reading, which say that only with these hypothesis, we can conclude that $\mu$ is regular, however I don't think it is true (or it is?).

If each closed set can be written as a countable union of open sets, then I think it is true, however, I don't think this is true in every topological space.

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  • $\begingroup$ Related: math.stackexchange.com/q/258407/8157 $\endgroup$ – Giuseppe Negro Jul 31 '14 at 14:48
  • $\begingroup$ @GiuseppeNegro, if every open set can be written as a union of closed sets then, I think that Evans's proof does work here. If this is not the case and if, for example, the only closed set contained in a given open set, is the empty set, then the measure of this set must be zero, however, I think we can constructo a example where this does not occur. $\endgroup$ – Tomás Jul 31 '14 at 15:52
  • $\begingroup$ No idea. It has been a while since I last thought on those things. I am sure that, in the ordinary "finite-dimensional" topological spaces ($\mathbb{R}^n$, smooth manifolds...), the proposition is true. I cannot guarantee for pathological monsters, though, and I am not going to think about it anytime soon, frankly. No time for that, sorry $\endgroup$ – Giuseppe Negro Jul 31 '14 at 17:43

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