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Inspired by this question here and in particular the answer I was wondering:

Do most nowhere dense sets have measure zero?

By "most" I mean in the sense of the "measure" of the set of all nowhere dense sets of measure zero. Unfortunately, the set of my measure theory knowledge is of zero measure so I hope this makes sense. Is it possible to define a natural measure on the set of all nowhere dense sets and then to prove that the set of nowhere dense sets of measure zero has measure zero in that measure?

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Rudy the Reindeer
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    It's certainly possible to define a measure on any set of sets; for instance, the zero measure. The more difficult question is whether there is a natural measure to use. – Nate Eldredge Apr 24 '14 at 16:38
  • @NateEldredge Then perhaps "natural measure" might be what I meant. Thank you for the comment! – Rudy the Reindeer Apr 24 '14 at 16:39
  • @Bryan Which section am I looking at? – Rudy the Reindeer Apr 29 '14 at 17:29
  • http://en.wikipedia.org/wiki/Fat_Cantor_set –  Apr 29 '14 at 17:32
  • @Bryan Yes. Which section in this article am I looking at? – Rudy the Reindeer Apr 29 '14 at 17:36
  • Well, the first. First, what kind of measure are you proposing to put on the set of all nowhere dense subsets? For an arbitrary topological space, you won't have that the nowhere dense sets form a $\sigma$-algebra, and you will have to restrict your measure to special subsets. And it isn't immediately clear that the set of all positive measure nowhere dense sets will be measurable, much less have measure zero. –  Apr 29 '14 at 17:44
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    @Bryan: I think how to choose the measure is part of the question. It's true that the set $\mathcal{N}$ of nowhere dense sets is not a $\sigma$-algebra; I think the idea would be to somehow choose an appropriate $\sigma$-algebra $\mathcal{F} \subset 2^{\mathcal{N}}$ and define an appropriate measure on that. It's not immediately clear how to do any of this, but that's why it's a question! – Nate Eldredge Apr 29 '14 at 17:46
  • The collection of nonempty compact sets in ${\mathbb R}^n$ is a complete metric space under the Hausdorff metric, and thus the size of collections of compact subsets can be measured in the sense of Baire category. Almost all (Baire category sense) compact sets have Hausdorff dimension zero (way smaller than measure zero), are perfect, and are nowhere dense. For a concise list of references see my two comments here. – Dave L. Renfro Apr 29 '14 at 18:10

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