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

Thanks for any help

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It could even be a closed set in $[0,1]$. –  GEdgar Oct 26 '12 at 21:46
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3 Answers

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.

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Thank you Brian –  Tomás Oct 26 '12 at 21:49
    
@Tomás: You’re welcome. –  Brian M. Scott Oct 26 '12 at 21:49
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Yes. $\mathbb R\setminus \mathbb Q$.

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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
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Yes.

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.

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