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?
Best Answer
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.