Is a simply connected subset of $\mathbb{R}^n$ necessarily measurable?

Question

Let $S\subset\mathbb{R}^n$ be a subset that is simply connected as a topological space.

Is $S$ necessarily Lebesgue measurable?

Is it necessarily Borel measurable?

What's the proof?

(Intuitively I think the answer should be "yes" to both, but I would have no idea how to prove. I mean, $S$ could be pretty wild.)

Answer 1

Let $V\subseteq [0,1]$ be a nonmeasurable subset of $\Bbb R$, such as the Vitali set, and consider the "comb" $(V\times[0,1])\cup ([0,1]\times\{0\})$, which is a simply connected nonmeasurable subset of $\Bbb R^2$.

To see that this set is nonmeasurable remember that if $(X,\mathcal A,\mu)$ and $(Y,\mathcal B,\nu)$ are two measure spaces then for a measurable $E\subseteq X\times Y$ we have that the slices $E_y=\{x\in X\mid (x,y)\in E\}$ are $\mu$-measurable for $\nu$-almost every $y$ and that the slices $E_x=\{y\in Y\mid (x,y)\in E\}$ are $\nu$-measurable for $\mu$-almost every $x$. If you slice this set in the right direction you'll get Vitali sets for all slices except a single one, so it cannot be measurable.

To see that this set is simply connected is best to just draw it.