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what is the exterior measure of $\mathcal{N}$, which is defined by picking a representation of the coset $\mathbb{R}$/$\mathbb{Q}$ in the interval [0,1]. Is it zero? How to prove?

Alex
  • 2,583
  • I do not believe it's $0$. Were it, for each $n$ we could find an open set $U_n$ containing $\mathcal{N}$ satisfying $m^*(U_n) < \frac{1}{n}$. Then $\mathcal{N}$ is contained in the intersection of all the $U_n$, and this intersection is a $G_{\delta}$ set of measure $0$. By completeness of Lebesgue measure, $\mathcal{N}$ is measurable. – anonymous Dec 28 '12 at 21:37
  • http://math.stackexchange.com/questions/182870/what-is-the-outer-measure-of-vitali-set

    Answer here

     – anonymous Dec 28 '12 at 21:42
  • then what is the exterior measure of $\mathcal{N}$,is it undetermined? – Alex Dec 28 '12 at 21:59
  • but if for every given n, we can find a representation that is in (0,$\frac{1}{n}$),then this $\mathcal{N}$is contained in (0,$\frac{1}{n}$). Is there a logical mistake here? – Alex Dec 28 '12 at 22:01
  • No, why should there be? For every $n$ you have a different $\mathcal{N}_n$ contained in $(0,\frac1n)$. – martin.koeberl Dec 28 '12 at 23:03
  • If so, then the outer measure of $\mathcal{N}$ can be any value. But how to construct such a set with a given value $a$ as its outer measure? – Alex Dec 29 '12 at 00:01
  • To answer your last question, see this – leo Dec 29 '12 at 17:58

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