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Is it true that the Lebesgue measure of the boundary of any set is zero? If no, then what are the counterexample and what are the conditions under which the above statement is true?

Like for example, this book says that dimension of the boundary of a manifold has dimension one less than the dimension of the manifold. Then in this case, the Lebesgue measure of the boundary is zero.

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    No, take $\Bbb Q$ for example. – Wojowu Jan 17 '16 at 16:55
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    The Jordan measure of the boundary of a Jordan measurable set is zero. But there are Lebesgue measurable sets with quite large boundaries; for instance the boundary of $\mathbb{Q}$ (in $\mathbb{R}$) is $\mathbb{R}$ itself. – Ian Jan 17 '16 at 16:58
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    Note that one corollary is that not all subsets of $\mathbb{R}^n$ are manifolds. – Ian Jan 17 '16 at 17:04
  • Another interesting counter-example is the fat Cantor set which is nowhere dense and yet its boundary has positive Lebesgue measure. – achille hui Jan 17 '16 at 17:18

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As stated by Wojowu above: $\mathbb{Q}$ is an example of a set whose boundary has nonzero outer measure. The boundary of $\mathbb{Q}$ is $\mathbb{R}$. $\mathbb{R}$ is, as we know, not a zero set.

Alekos Robotis
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