# nowhere dense set in [0,1] has Lebesgue measure 1

Is it possible that a nowhere dense set in [0,1] has Lebesgue measure 1? Fat Cantor has positive measure strictly between 0,1.

Could anyone please provide me with a link to the proof? Many thanks,

• Assume there is such a set and call it $F$. Then $F^c$ is a (dense) set with nonempty interior in $[0,1]$ with zero measure, which is absurd. – Ian Coley Sep 15 '13 at 4:42

## 1 Answer

The intersection of two subsets of $[0,1]$ with full measure has also a full measure.

You can also so this property is hereditary. Namely if $A$ is a set of full measure, then $A\cap(a, b)$ has full measure as well.

Groom this you can show that a set of measure $1$ must meet all open sets, and therefore cannot be nowhere dense.

• thanks so much for answering. how come A \cap (a,b) has full measure? (a,b) is any open set in [0,1]? – 90b56587 Sep 15 '13 at 5:14
• Suppose it didn't. Write $A=A\cap[0,a]\cup A\cap(a,b)\cup A\cap[b,1]$. The sum of these three sets is a full measure. But the first set cannot have measure larger than $a$ and the third cannot have a measure larger than $1-b$. So the remainder must have measure $b-a$, which is the full measure of the interval. – Asaf Karagila Sep 15 '13 at 5:19
• Thanks for clarifying. I though by full measure you mean measure 1. Apparently you mean the full measure of that interval. the proof makes complete sense. Thank you! – 90b56587 Sep 15 '13 at 5:22