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A set like the rationals has Lebesgue measure 0 and its closure are the real numbers which has positive Lebesgue measure. However Q is dense. If a set where to be nowhere dense and also null what would the measure of its closure be in R?

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  • $\begingroup$ Anything whatever. $\endgroup$ – David C. Ullrich Sep 17 at 1:09
  • $\begingroup$ How can it be anything? $\endgroup$ – John Castro Sep 17 at 1:11
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    $\begingroup$ Hint: In fact any closed subset of the line is the closure of a countable subset. $\endgroup$ – David C. Ullrich Sep 17 at 1:20
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    $\begingroup$ ??? $E=\overline E$ and $m(E)>0$, so what? The question is whether there exists a null set $A$ with $E=\overline A$. $\endgroup$ – David C. Ullrich Sep 17 at 1:22
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    $\begingroup$ The "fat Cantor set" is the closure of the set of the endpoints of the intervals formed at each stage of its construction. $\endgroup$ – Angina Seng Sep 17 at 1:25
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A fat Cantor set, a small modification of the construction of the middle-third Cantor set in $[0,1]$ is nowhere dense and compact and can be given any measure $m>0$. A countable dense set $D$ of this fact Cantor set is then a nowhere dense null-set of measure $m$.

Nowhere dense and null-set are unrelated. (But in the reals a closed null set is nowhere dense).

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  • $\begingroup$ Nowhere dense and null-set are unrelated --- There is one sometimes useful relation when closures are done first, namely a set whose closure is null is automatically nowhere dense. $\endgroup$ – Dave L. Renfro Sep 17 at 12:16
  • $\begingroup$ @DaveL.Renfro True, because non-empty open sets have measure $>0$. This is not the case in all spaces but an artefact of Euclidean spaces. $\endgroup$ – Henno Brandsma Sep 17 at 12:27

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