# Are there many “cusps” in a rectfiable star-shaped set?

Let me first recall the definition of density with respect to a measurable set $E$ as follows:

A point $x \in \mathbb{R}^n$ is a point of density $\alpha$ for $E$ if

$$\lim_{r \rightarrow 0} \frac{|E \cap B_r(x)|}{|B_r(x)|}=\alpha$$

Motivation: Clearly, by Lebesgue differentiation theorem, a.e. $x \in E$ has density $1$, and a.e. $x \in E^c$ has density $0$. Let $E^{(\alpha)}$ denote the set of points of density $\alpha$ w.r.t. $E$. Intuitively, $E^{(1)}$ is the measure-theoretical interior of $E$, and $E^{(0)}$ is the measure theoretical exterior of $E$.

If $E$ is a smooth set, then only boundary point of $E$ has density $\frac{1}{2}$, that is, $E^{(\frac{1}{2})}=\partial E$. More generally if $E$ is a rectifiable set, or a set of finite perimeter, see https://en.wikipedia.org/wiki/Caccioppoli_set, then a.e. $x \in \partial^*E$ is a point of density $\frac{1}{2}$.

Based on the background in the above, it's very natural to define the cusp of a set $E$ by $\partial E \cap E^{(0)}$. This should match the intuition of a cusp, the points of density $0$ on the boundary. In the definition of cusp, we don't need to assume regularity of boundary of $E$.

Now I have a very naive question: If $E$ is a star-shaped set in $\mathbb{R}^n$, and to ensure there is no pathological issue let's further assume $\partial E$ is rectifiable, then is it true that if $E$ is star-shaped, then the cusp of $E$ is countable? If not, can one prove $\mathcal{H}^{n-1}(\mbox{cusp of } E)=0$? Here $\mathcal{H}^{n-1}$ is the Hausdorff measure, or surface measure.

I had this question couple of days ago. When I was trying to explain to my friends the ideas of geometric measure theory during a cookie hour, naturally I drew a lot of pictures and suddenly came up with this question. So far I've no idea how to solve it. I think it doesn't require strong GMT background, but a smart observation should solve my question. Can anyone here want to have a try and share any ideas? Thanks very much!

## 1 Answer

The set of cusps need not be countable. Take a bell-shaped even function $f\in C_0^\infty(\mathbb{R})$ with ${\rm supp}\, f =[-1,1]$ that is positive inside the interval. Take the union of graphs of $f$ and $-f$ and rotate it along the $z$-axis. The resulting surface has a cusp on the equator and the domain bounded by the surface is star-shaped. However, I believe that the $\mathcal{H}^{n-1}$ measure of the set of cusps equals zero, at least the radial projection of the set on the unit sphere has measure zero as explained below.

Using spherical coordinates we can represent the boundary as the graph of a function. Then, if the set above the graph has locally finite perimeter, I believe, the set of cusps on the graph has $(n-1)$ measure zero. At least it is true that the projection on the $n-1$-dimensional coordinate plane above which the graph is defined has measure zero. It follows from the integration by parts formula.