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This question tortures me for a while: if a set $E$ has positive Lebesgue measure, does it necessarily contain an interval? I would be truly grateful for help.

Marina
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The irrational numbers are such an example.

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    +1. Small comment: it's possible to come away from this problem, and facts like regularity, with the impression that a positive measure set must almost contain an interval: that is, for all $E$ measurable with positive measure, there is a nontrivial interval $I$ such that $I\subseteq E\cup Z$ for some measure-zero set $Z$. However, this is false: see http://math.stackexchange.com/questions/57317/construction-of-a-borel-set-with-positive-but-not-full-measure-in-each-interval. – Noah Schweber Feb 14 '16 at 04:24
  • @NoahSchweber Good point; that's a good example of how Littlewood's three principles can sometimes be a bit misleading. –  Feb 14 '16 at 04:25
  • I feel totally mislead. With the "fat" Cantor set, for example, they say that it is the boundary of the set which brings that positive measure, not the set itself. But since the set is closed, is not the boundary included into the set? – Marina Feb 14 '16 at 04:30
  • @Marina Yes, that set is closed and is equal to its own boundary. –  Feb 14 '16 at 04:32
  • Things like this make me believe that the real numbers are not continuous. – Marina Feb 14 '16 at 04:34
  • @Marina : ​ Bear in mind that the fat Cantor set's interior is empty, so everything in it is in its boundary. ​ ​ ​ ​ –  Feb 14 '16 at 07:42
  • @user296602 , can you explain how the irrationals (as a subset in $R$) have a finite positive Lebesgue measure but don't contain any inerval (not singletons)? – user652838 Nov 10 '20 at 17:10