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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?

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up vote 3 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.

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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.

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