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The boundary of a subset of $\mathbb{R}^n$ doesn't necessary have (Lebesgue) measure zero, think for example to $\mathbb{Q}^n\subset\mathbb{R}^n$, which satisfies $\partial\mathbb{Q}^n=\mathbb{R}^n$.

My question is: are there other "nice" conditions for a subset $U\subseteq\mathbb{R}^n$ to have boundary $\partial U$ of measure zero?

For example we could have $U\subset\mathbb{R}^n$ open the interior of the embedding of a manifold with boundary, then $\partial U$ is the manifold boundary, which is itself a manifold of dimension $n-1$. Then Sard's theorem implies that $\partial U$ has measure zero.

Is the stronger statement: "$U\subset\mathbb{R}^n$ open $\Rightarrow$ $\partial U$ has measure zero" true as well?

Daniel Robert-Nicoud
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1 Answers1

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Is the stronger statement: "$U\subset \mathbb{R}^n$ open $\Rightarrow \partial U$ has measure zero" true as well?

No. In 1902 Wiliam F. Osgood presented his construction of "A Jordan curve of positive Area".

That provides an open set $U \subset \mathbb{R}^2$ such that $\partial U$ - the Jordan curve of positive area - has positive Lebesgue measure. Barring better ideas, you can use a product of such an open set with a (hyper)cuboid to have higher-dimensional examples.

The complement of a fat Cantor set provides a one-dimensional example.

Daniel Fischer
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    ... and isn't an open set in $\mathbb R^n$ trivially a "manifold" ? – GEdgar Dec 19 '13 at 22:42
  • @GEdgar Yes, sorry, i edited the question to better explain what I meant. Now it should work. – Daniel Robert-Nicoud Dec 19 '13 at 22:46
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    @Daniel Fischer Really I appreciate your knowledgeability, patience and contributions to the community. I know you don't need more points, but it's still nice of you to make this kind of thing wiki. – Betty Mock Dec 19 '13 at 23:01