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Let $X=\mathbb{R}^n$ ($n\ge 1$) be equipped with the natural topology (the one coming from the norms) and let $\lambda$ denote the Lebesgue-measure on $X$.

Let $M\subset X$ be a null set: $\lambda(M)=0$. Is it possible that the closure of $M$ has positive measure, i.e. $\lambda(\overline{M}) > 0$?

Thank you!

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up vote 6 down vote accepted

Actually, more is true: the closure of a null can have the measure you want. We just deal with the case $n=1$. For example, taking the rational numbers, the measure of the closure is infinite as this set is dense. Now let $a>0$. Then $\Bbb Q\cap (0,a)$ has measure $0$ but it's closure is $[0,a]$ and has measure $a$.

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Yep. The points with rational coordinates are countable and so have zero measure, and their closure is all of $\mathbb{R}^n$!

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$\mathbb{Q}^n\subset \mathbb{R}^n$ is a closed set (a countable union of points, which are closed sets), so its closure is $\mathbb{Q}^n$ itself, which is a zero set. –  Sh4pe Oct 25 '12 at 14:53
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$\mathbb{Q}$ is not closed in $\mathbb{R}$... For instance, choose rational points converging to $\sqrt{2}$, which is not rational... –  Zach L. Oct 25 '12 at 14:56
    
@Sh4pe Closed sets are not closed under infinite union, just intersection. –  Joe Johnson 126 Oct 25 '12 at 14:58
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Damn, you are right. What was I thinking... –  Sh4pe Oct 25 '12 at 15:07
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