# When does the boundary have measure zero?

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?

• – David Mitra Dec 19 '13 at 22:17
• $U$ open and convex implies $\partial U$ has measure zero. – GEdgar Dec 19 '13 at 22:41
• @GEdgar Convexity alone suffices. – Michael Greinecker Dec 20 '13 at 1:20
• @MichaelGreinecker Thanks for the reference! – Daniel Robert-Nicoud Dec 20 '13 at 12:17

Is the stronger statement: "$U\subset \mathbb{R}^n$ open $\Rightarrow \partial U$ has measure zero" true as well?
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.
• ... and isn't an open set in $\mathbb R^n$ trivially a "manifold" ? – GEdgar Dec 19 '13 at 22:42