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$R$ is a set of real numbers, m is the Lebesgue measure on $R$.

Does exist a nowhere dense subset $A\subseteq R $, such that $m(A)=+\infty$?

I know that if $A$ is of the first category, there exist such a set.

Thanks a lot.

David Chan
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1 Answers1

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Note that the fat Cantor set is nowhere dense, has measure $1/2$ and is a subset of the unit interval $[0,1]$. If you take the union of the translates of this set by the integers, you will get a nowhere dense set that is of infinite measure.

Rolf Hoyer
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  • you will get a nowhere dense set"——Why this set is still a nowhere dense set, and this set is closed in R? – David Chan Apr 27 '15 at 11:48
  • @David: Yes, it’s closed: it’s the union of a locally finite family of closed sets. Its complement is therefore open, and its complement is obviously dense, so this set must be nowhere dense. – Brian M. Scott Apr 27 '15 at 20:38
  • @Brian M. Scott: Well, I have known why. A perfect answer! A wonderful explanation! – David Chan Apr 28 '15 at 03:36