Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there some simple conditions which would ensure that the boundary of an open set in $\mathbb{R}^n$ has measure zero? Also, is it true that the boundary of a closed set in $\mathbb{R}^n$ has measure zero?

share|cite|improve this question
up vote 2 down vote accepted

The boundary of a closed or open set can have positive measure. In $\Bbb R$, for instance, let $C$ be a so-called fat Cantor set; then $C$ is closed and has empty interior, so it is its own boundary and the boundary of its complement.

I don’t immediately see a proof, but I suspect that the boundary of every regular open set in $\Bbb R^n$ has measure $0$; a set $U$ in a topological space $X$ is regular open if $U=\operatorname{int}_X\operatorname{cl}_XU$. Added: Incorrectly, as it appears; see LVK’s comment below.

share|cite|improve this answer
Thanks; but are there any conditions under which the boundary does have measure zero? – user38404 Sep 22 '12 at 4:37
A Jordan curve of positive measure divides the plane in two regular open sets. – user31373 Sep 22 '12 at 5:03
There are in fact regular open sets in one dimension whose boundary is a fat Cantor set. For instance, one can construct a dense open set $U$ which is a disjoint union $\bigcup_{n\in\mathbb{N}}U_{n}$ of open intervals such that between any two open intervals in $\{U_{n}|n\in\mathbb{N}\}$ there is another open interval. One can therefore partition the set $\{U_{n}|n\in\mathbb{N}\}$ of open intervals into two sets $A,B$ so that between any two open intervals there are intervals in $A$ and $B$ between those intervals. – Joseph Van Name Jan 22 '15 at 3:01
The sets $\bigcup A$ and $\bigcup B$ will be regular open sets, but their boundaries could have positive measure. – Joseph Van Name Jan 22 '15 at 3:01

Here is a sufficient geometric condition for a general set $E$ to have measure zero: for every $x\in E$ there is $c>0$ such that for all sufficiently small $r>0$ the $r$-neighborhood of $x$ contains a ball of radius $cr$ that is disjoint from $E$. This is a weak form of condition known as porosity. To see that it implies having measure zero, use the Lebesgue density theorem.

Porosity is easy to verify for the boundary of a given open set: it suffices to find, for every boundary point, a subset in the shape of a cone (possible twisted) with a vertex at that point. The shape and size are allowed to depend on the point. Smooth, Lipschitz and uniform domains are covered by this condition.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.