Set $E\subset \mathbb{R}^n$ of positive Lebesgue measure such that the Lebesgue measure of its boundary is zero

Solution 1:

It is true for compact convex sets. Look here math.stackexchange.com/questions/207609/… -- MotylaNogaTomkaMazura

There are a few natural conditions that are weaker than convexity condition. I list them below, and comment on whether they yield the desired conclusion

Star-shaped compact sets

For these, the boundary may be of positive measure.

Let $C\subset \mathbb{R}$ be a fat Cantor set. Consider $\mathbb{R}$ as the $x$-axis in $\mathbb{R}^2$. Define a compact set $K\subset \mathbb{R}^2$ as the union of all line segments beginning at $(0,1)$ and ending at a point of $C$. This is a set of compact set of positive measure with empty interior; all these properties are inherited from $C$. It is star-shaped with respect to the point $(0,1)$.

Quasiconvex compact sets

For these, the boundary may be of positive measure.

A set $K$ is quasiconvex if there is a constant $c$ such that any two points $a,b\in K$ can be joined by a curve of length at most $c|a-b|$ lying in $K$. Unfortunately, such a set can still have the boundary of positive measure. Consider the following modification of the Sierpiński carpet: at $k$th step of construction, the square is divided into $n_k\times n_k$ equal squares and the middle one is removed. Here $n_k$ is an odd number $\ge 3$; the original construction has $n_k\equiv 3$.

The generalized Sierpiński carpet has positive area when $\sum n_k^{-2}<\infty$; specifically, the area is $\prod_k(1-n_k^{-2})$. It has empty interior. The quasiconvexity follows from the fact that when going from $a$ to $b$, having to go around a removed square multiplies the traveled path at most by the factor of $2$.

Sets with a uniform exterior or interior cone condition

For these, the boundary has measure zero.

I mean the following condition: there exists a circular cone $C$ (radius $r$, height $h$) such that for every point $p\in\partial K$, there is an isometric copy $C_p$ of $C$ that has vertex at $p$ and such that

• $C_p\cap K=\{p\}$, for the exterior cone condition
• $C_p\cap K^c=\{p\}$, for the interior cone condition

Note that every convex set satisfies the exterior cone condition: one can use any cone $C$ whatsoever, since there is a whole halfspace that lies outside of $K$.

Either exterior or interior condition implies that the boundary is porous and therefore has zero measure.