Let $A\subset \mathbb{R}^n$ be a bounded Jordan measurable set. I wonder if its boundary $\partial A$ is necessarily porous.

I know that $\partial A$ has Lebesgue measure zero. I also think that one can construct an example of a null Lebesgue set that is not porous. But is it possible to construct such a set in a way that it is also the boundary of a Jordan measurable set?

  • 1
    $\begingroup$ What exactly do you mean by "porous"? I have never heard that word in a mathematical context. $\endgroup$ – PhoemueX Sep 17 '15 at 12:05
  • 3
    $\begingroup$ I mean it in this sense: en.wikipedia.org/wiki/Porous_set $\endgroup$ – Neznajka Sep 17 '15 at 12:18
  • 2
    $\begingroup$ If you can show that there is a bounded, closed, non-porous null-set, then there are Jordan measurable sets with non-porous boundary. Indeed, let $B$ be such a set. Let $A := B$. Since $A = B$ is a null-set, it has empty interior. Since $A=B$ is closed, we thus have $\partial A = A =B$ which is a null-set. Hence, $A$ is Jordan measurable with $\partial A = B$, a non-porous set. Conversely, if there is no bounded, closed non-porous null-set, then a bounded Jordan measurable set must have porous boundary, since the boundary is a closed, bounded null-set. $\endgroup$ – PhoemueX Sep 17 '15 at 14:36

The answer is negative. Let's consider a Cantor-type set $K\subset [0,1]$ built by removing the $c_j$ proportion of each interval at $j$th step ($c_j<1$, $j=1,2,\dots$). It's a compact set with empty interior, so $\partial K=K$. The measure of $K$ is zero provided that $$ \prod_{j=1}^\infty (1-c_j)=0 $$ Hence, $K$ is Jordan measurable in this case.

On the other hand, if $c_j\to 0$, then $K$ is not porous in $\mathbb{R}$: the gaps become relatively small on smaller scales.

For example, $c_j=1/(j+1)$ satisfies both properties: zero measure and non-porosity.

To obtain an example in $\mathbb{R}^n$, take the Cartesian product.

  • $\begingroup$ Thank you! That answers it but I wonder if an example could still be constructed if the question was "Does every open bounded Jordan measurable set have porous boundary?" $\endgroup$ – Neznajka Sep 19 '15 at 4:57
  • 2
    $\begingroup$ Subtract the above set from an open disk containing it. $\endgroup$ – user147263 Sep 19 '15 at 4:59

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.