5
$\begingroup$

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

Thanks for any help

  • 0
    It could even be a closed set in $[0,1]$.2012-10-26

3 Answers 3

8

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

  • 0
    @Tomás: You’re welcome.2012-10-26
6

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

  • 1
    Aw, that one’s boring! :-) +12012-10-26
2

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.