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?

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.