# The closure of a null nowhere dense set is null?

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?

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

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 $$m$$.
• @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. – Henno Brandsma Sep 17 at 12:27