# A closed subset of $[0,1]$ with no interiors and has measure exactly $1$

I wonder if there is a construction for a closed subset of $[0,1]$ such that it has no interior (contains no open intervals) and has measure exactly $1$ as $[0,1]$ itself.

I know the fat Cantor set, but I think fat Cantor set can be extended to construct any closed subsets of $[0,1]$ with measure $a <1$. I don't think it can be extended to exactly $1$. However, I also cannot find the contradiction that if there is such a set. Any suggestions ?

• Irrationals? ${}$ Oct 22, 2017 at 14:14
• If you mean Lebesgue measure, it's impossible, naturally: the complement is open, and an open interval always has a Lebesgue measure $>0$.
– user436658
Oct 22, 2017 at 14:16
• I think irrationals is not closed. Oct 22, 2017 at 14:17
• Yeah, I mean Lebesgue measure. So suppose there exist such a set $F$. Then $F^c$ is open with measure $0$. Oh since $F^c$ is open, it is a disjoint countable unions of open intervals. So its measure must be greater than $0$. Got it !! Thank you very much. Oct 22, 2017 at 14:19
• @BothHtob Oh yeah of course; I've missed you want it closed :P Oct 22, 2017 at 14:27

There is no such set. If $K$ is that set, then $[0,1]\setminus K$ is open and non-empty. Therefore, its Lebesgue measure is some number $m>0$ and the Lebesgues measure of $K$ will then be $1-m<1$.
Note: You are right about fat Cantor sets. Their measure can be any number $a\in[0,1)$, but not $1$.