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On a topological space $X$ with Borel sigma-algebra defined some measure $\mu$. I am wondering about measure with the following property

$A$ is non-empty and open $\Rightarrow \mu(A)>0$.

As far as I can see in the literature, the term "Borel regular" is used.

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    Gortaur: You can see what was edited by clicking on edited xx hours ago above Henno's name. The downvote may have to do with omitting the non-emptiness again...2011-06-08

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In measure theory, volume 4I (topological measure theory, part 1), by Fremlin, which can be found here, 411N(f), such a measure is called strictly positive, and this is indeed standard (I've seen it in several papers, at least in topological measure theory). One could also say that $\mu$ has full support. Borel regular is another property, and I refer to the Fremlin books for a very in depth treatment of regularity properties for measures on a topological space. The existence of a ($\sigma$-)finite strictly positive Borel measure on a space $X$ has some topological implications (like ccc and property (K)), and has been studied in several papers.

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    ccc = every family of pairwise disjoint non-empty open sets is at most countable; property (K) = every uncountable family of non-empty open sets has an uncountable subfamily that pairwise all intersect.2011-06-08
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The appropriate question and answer can be found here.

Short version: For $ X = [0,1]^n $ you may be interested on Oxtoby-Ulam measures. Turns out that, in this case, they are just pull-backs of the Lebesgue measure by a homeomorphism of $ [0,1]^n $.

This problem is much more complicated for other topological spaces $ X $, as the link above suggests.

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    I support Bazi$n$ga's suggestion to rephrase the question.2011-06-07
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Such measure cannot exists on any reasonable topological space (e.g. Euclidean space or any space with no isolated points) for the following reason:

By definition a measure has to be (finitely) additive: Measure of the disjoint union of two sets in the sigma algebra must be the sum of their measures.

Now given an open set, can we decompose it into two disjoint sets both with empty interior? The answer is yes for any reasonable spaces, right? (For example, we write R as rationals and irrationals.) If such decomposition exists, we have both sets with measure $0$ but their union has positive measure: $\Rightarrow\Leftarrow $ can't happen~