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!