Let $\Omega \subset \mathbb{R}^d$ be a region ( bounded, simply connected, open set ). What are some regularity conditions to assure the boundary $\partial\Omega$ is a set of (lebesgue-)measure zero? Is there any geometric / topological condition, which is equivalent to the statement that $\mu(\partial\Omega) = 0$?

I am particularly interested in some weak conditions, in a sense of not being too restrictive. I'm not interested in statements as strong as "if $\partial\Omega$ is a submanifold, ...".

  • 2
    $\begingroup$ The first condition that comes into my mind which implies Lebesgue measure zero is upper porosity. Suppose that there exists a constant $0 < c < 1$ so that for all $x \in \partial\Omega$ there are arbitrarily small radii $r$ so that $B(y,cr) \subset B(x,r)\setminus \partial\Omega$ for some $y\in \mathbb{R}^n$. Then $\mathcal{L}(\partial\Omega)=0$, where $\mathcal{L}$ is the Lebesgue measure. (This follows immediately by considering density points.) If one assumes that this holds for all small enough radii, one also gets an estimate on the dimension. $\endgroup$ – Tapio Rajala Jan 23 '11 at 14:06
  • $\begingroup$ As this condition seems very natural to protect onesself from spacefillung boundaries, I do not see that it makes any use of $\Omega$ being a region. On the other hand I do not now how to use this information without stating very strong requirements. $\endgroup$ – Alexander Thumm Jan 24 '11 at 9:02
  • $\begingroup$ Well, $\Omega$ being a region isn't that restrictive, so it is hard to guess what kind of conditions you are looking for. The next set of conditions that come into my mind have to do with curves inside the region, but these are already quite strong. For example if you assume that the region is a John-domain you already get an upper bound on the dimension. However, under any such condition I would prove the boundary to have zero measure via porosity (the holes being inside the region in this case)... $\endgroup$ – Tapio Rajala Jan 24 '11 at 10:44
  • $\begingroup$ I vaguely remember a theorem that says that a bounded set is Jordan-measurable if and only if its boundary is a set of Jordan-measure zero. Since Jordan null sets are Lebesgue null sets, this is a criterion if I am right. $\endgroup$ – Florian Jan 27 '11 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.