# Does the boundary of an open set have measure zero (in $\mathbb{R}^n$)?

When studying weak border conditions (in Sobolev Spaces), the usual motivation for the weak meaning of inequalities is that the boundary of most open sets in $\mathbb{R}^n$ has zero Lebesgue measure. But is there any open set $U\subseteq\mathbb{R}^n$ such that it's boundary has positive Lebesgue measure? I really can't think of anything like that.

• Take the complement of a thick Cantor set. – Daniel Fischer Aug 3 '13 at 19:01
• @DanielFischer You're being a gentleman. We just call them "fat". – Pedro Tamaroff Aug 3 '13 at 19:07
• @PeterTamaroff A gentleman would've said "adipose". – Daniel Fischer Aug 3 '13 at 19:08
• @DanielFischer That's what a physiologist would have said! – Pedro Tamaroff Aug 3 '13 at 19:10
• Also note "thick" can mean dumb. Poor Cantor set. – zhw. Feb 14 '18 at 18:37

There are a lot of open sets in $\mathbb{R}^n$ whose boundary has positive Lebesgue measure. I wouldn't be surprised if "most" open sets have boundaries with positive Lebesgue measure, where "most" might be referring to cardinality, or some topological or measure-theoretic size.
Examples of open sets whose boundary is not a null set are for example complements of a thick Cantor set in dimension $1$ (products where at least one factor is such in higher dimensions).
Somewhat similar, let $(r_k)_{k \in \mathbb{N}}$ be an enumeration of the points with rational coordinates, and let
$$U = \bigcup_{k\in \mathbb{N}} B_{\varepsilon_k}(r_k)$$
for a sequence $\varepsilon_k \searrow 0$ such that $\sum {\varepsilon_k}^n$ converges. Then you have a dense open set $U$ with finite Lebesgue measure, its boundary is its complement and has infinite Lebesgue measure.