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.
Measure Theory – Does Boundary of Open Set Have Measure Zero?
analysismeasure-theory
Best Answer
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.