Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is like in the title. $G_\delta\subset\mathbb{R}^n$ with Lebesgue measure.

Thanks for any help

share|cite|improve this question
It could even be a closed set in $[0,1]$. – GEdgar Oct 26 '12 at 21:46
up vote 6 down vote accepted

Yes: any fat Cantor set in $\Bbb R$ is an example, since all closed sets in $\Bbb R$ are $G_\delta$’s.

share|cite|improve this answer
Thank you Brian – Tomás Oct 26 '12 at 21:49
@Tomás: You’re welcome. – Brian M. Scott Oct 26 '12 at 21:49

Yes. $\mathbb R\setminus \mathbb Q$.

share|cite|improve this answer
Aw, that one’s boring! :-) +1 – Brian M. Scott Oct 26 '12 at 21:48
Thanks you kahen – Tomás Oct 26 '12 at 21:53
His example make me feel like a dumb @BrianM.Scott haha, but i like you better, because it is a nowhere dense set. – Tomás Oct 26 '12 at 22:17


For $n\in\mathbb N$ let $F_n = \{x\in(0,1)|\;\exists m\in\mathbb Z:x=\frac{m}{n}\}$ and $U_n=(0,1)\setminus F_n$. The sets $U_n$ are open and have measure $1$, but the intersection $A=\bigcap_{n=1}^{\infty}U_n$ is simply the set of all irrational numbers in $(0,1)$. It is thus a $G_\delta$ set with measure $1$ and empty interior.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.