# Does every bounded Jordan measurable set have porous boundary?

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?

• What exactly do you mean by "porous"? I have never heard that word in a mathematical context. – PhoemueX Sep 17 '15 at 12:05
• I mean it in this sense: en.wikipedia.org/wiki/Porous_set – Neznajka Sep 17 '15 at 12:18
• 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. – 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.