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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?

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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
    
math.stackexchange.com/questions/182870/… 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

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