This question came to me as I was watching this Numberphile video on floating bodies. In the video, Professor Elisabeth Werner gives an example of a subset of $\Bbb R^2$ that is not convex, but in her drawing it is also starlike. Then the idea came to me that I can think of plenty of sets $E$ that are starlike, but they have some "local convexity" here and there, where there is a little disk contained in $E$. Hence such a set satisfies $m(E) > 0$, where $m$ is Lebesgue measure. (Let's assume all our sets are Lebesgue measurable.)
Question: If $E\subset\Bbb R^2$ is starlike (and Lebesgue measurable), and contains no open disk, must $m(E) = 0$?
For example, a union of finitely many line segments centered at a point has measure $0$. I thought maybe if we took all lines through the origin in $\Bbb R^2$ with an irrational slope, this could be a candidate for a set $E$ satisfying the hypothesis, but with $m(E) > 0$. I have not checked whether it is measurable.
Definition. A set $E\subset \Bbb R^2$ is starlike if there is a point $p\in E$ such that for each $q\in E$, the line segment joining $p$ and $q$ is wholly contained in $E$. Such a point $p$ is called a center of $\pmb E$.
It doesn't seems so: take a nowhere dense set of positive (1-dimensional) measure in the unit circle. Then take the union of all lines through origin and points in that set.