Given a Lebesgue measurable set A with strictly positive measure, can we find an open interval (a,b) such that x belongs to A for almost every x in (a,b)?
Thanks in advance for any comments!
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6$\begingroup$ the cantor set of positive measure is a counterexample. $\endgroup$– rpotrieCommented Sep 9, 2010 at 15:12
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2$\begingroup$ There's also a measurable set $A\subset\mathbb{R}$ such that for all nonempty open interval $I$ one has $0 < m(A\cap I) < m(I).$ It's a well-known, old exercise. See Rudin's book. $\endgroup$– Pietro MajerCommented Sep 9, 2010 at 15:38
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3$\begingroup$ See, for example en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set, You may be interested in a weaker statement that what you ask for, see en.wikipedia.org/wiki/Lebesgue's_density_theorem $\endgroup$– rpotrieCommented Sep 9, 2010 at 15:38
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1$\begingroup$ The question seems to have been answered. Voting to close as no longer relevant. $\endgroup$– Victor ProtsakCommented Sep 9, 2010 at 16:43
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1$\begingroup$ We don't generally close questions on MO just because they have been answered---after all, there are hundreds of non-closed questions here with excellent answers. Rather, I think what should happen is that rpotrie should answer the question with an answer about the fat Cantor set, rather than merely a comment (or someone else should if he doesn't). $\endgroup$– Joel David HamkinsCommented Sep 9, 2010 at 20:19
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3 Answers
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The usual Cantor set constructed by removing 1/3 at each step is nowhere dense but has measure 0. However, there exist nowhere dense sets which have positive measure. The trick is to try to remove less, for instance you remove 1/4 from each side of [0,1] during the first step then 1/16 from each pieces etc...
The resulting set is the fat Cantor set: it is nowhere dense and it has positive measure.
answered Sep 11, 2010 at 22:20
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I actually worked this problem out during my Measure Theory course a couple years ago.
http://www.austinmohr.com/Work_files/hw2_3.pdf
Therein, you'll find a construction of a Cantor-like set having any measure strictly between 0 and 1. As with the Cantor set, you cannot find an interval contained in this Cantor-like set.
I don't suggest you accept my solution without checking it yourself, as it was written by one still coming to grips with the material. It should give you a good idea of how to construct your proof, however.
answered Sep 11, 2010 at 22:40
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$\begingroup$ ThankS Austin Mohr, both your comments and pdf file are very useful. It is really hard for me to choose an answer between you and Alephomega's and also Rpotrie's one. I just chose the one with highest mark. Thank you again. :-) $\endgroup$– AnandCommented Sep 15, 2010 at 12:06
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From the positive side, as I've mentioned on the comments, you should look at Lebesgue's density theorem. It says that you will get intervals where the measure of the set intersected with the interval is as close to full as you like, in fact, this can be done for intervals around almost every point of the set, and it holds in other contexts also.
