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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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    1. and 2. don't hold for Lebesgue measure, [fat Cantor sets](http://en.wikipedia.org/wiki/Smith%2DVolterra%2DCantor_set) have empty interior but positive measure.2011-06-06
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    Lebesgue measure doesn't have properties 1 or 2. Consider a fat Cantor set (http://en.wikipedia.org/wiki/Fat_Cantor_set). Property 3 is often referred to as "the measure has full support". (http://en.wikipedia.org/wiki/Support_of_a_measure)2011-06-06
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    Re 3., remember that the empty set is always open. Hence 3. does not hold for the Lebesgue measure (nor for any other measure).2011-06-06
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    Good point. A measure has full support if and only if every *nonempty* open set has positive measure.2011-06-06
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    Another counterexample for 1 and 2 is the set of irrationals. Maybe more elementary than a fat Cantor set?2011-06-06
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    @GEdgar, thanks, but fat Cantor set is better since it's also closed and that fact killed me. I should remember it.2011-06-06
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    What was edited and why vote down?2011-06-08
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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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    What is ccc and (K)?2011-06-08
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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 don't think you understand my question correctly.2011-06-06
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    As Didier Piau pointed out, your question does not make sense as its written. To try to help you, I offered a version of your question that makes sense and is coherent with the title (yours isn't as pointed out by Theo, Nate and GEdgar). If you rephrase your question perhaps me and others can help you more.2011-06-06
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    I support Bazinga'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~