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When studying weak border conditions (in Sobolev Spaces), the usual motivation for the weak meaning of inequalities is that the frontier 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 frontier has positive Lebesgue measure? I really can't think of anything like that.

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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
up vote 6 down vote accepted

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.

However, the open sets you can visualize are far more regular than the average open set, so it's not easy (if at all possible) to get a good mental picture of an open set whose boundary has positive Lebesgue measure. And the open sets one does analysis on, typically also are quite regular and have nice boundaries.

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.

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