Could an open star-shaped set have boundary of positive Lebesgue measure?

Question

Previously I asked a question Can a star-shaped set have boundary of positive Lebesgue measure?, later someone has kindly answered it in a comment. It turned out I asked a trivial question.

So I'm asking a further question because I really want to understand the phenomenon. What if the star-shaped set is open? Is it true then the boundary has zero Lebesgue measure? The reason I'm curious is because I want to see how "thick" the boundary of an open star-shaped set can be. This is something I'm curious about.

Maybe here I'm asking another trivial question. If so, can anyone tell me what is the right formulation of the question of understanding how "think" can the boundary of a star-shaped open set be? I'm not even sure what should be the appropriate tags to classify this question. Are there further useful references of similar flavor?

Any comment is really appreciated! Thank you very much!

Answer 1

Let $D$ be the open unit disc in the plane centered at $0.$ Let $U\subset \partial D$ be an open dense subset of $\partial D$ having small arc length measure. Define

$$\Omega = D \cup \left(\cup_{\zeta \in U} R(\zeta)\right ),$$

where $R(\zeta ) = \{r\zeta : r>0\}.$ Then $\Omega $ is open and star-shaped with respect to the origin. Since every point of $\partial D \setminus U$ is a limit point of $U,$

$$\partial \Omega =\left (\cup_{\zeta \in \partial D \setminus U}\, R(\zeta)\right ) \cap \{z\in \mathbb R^2 :|z|\ge 1\}.$$

Because the arc length measure of $\partial D \setminus U$ is positive, we can see by polar coordinates that $\partial \Omega$ has infinite area measure.

Answer 2

Here is a version of your question one can ask: Suppose that $D$ is an open subset of $R^n$ containing the origin $0$ such that every ray $\rho$ emanating from $0$ intersects the boundary of $D$ in at most one point. (In particular, $D$ is starlike.) Is it true that $\partial D$ has the Lebesgue measure 0? Indeed, this is (more or less trivially) true, and it follows, by applying the spherical coordinates, from the answer to this question. More interestingly, $\partial D$ can have positive outer measure, see the same link.