Does simply-connected imply measurable? The famous examples of non-measurable sets involve a sophisticated selections of points from a ball (or another object). This raises the following question: if a certain object in a Euclidean space is simply-connected, does it imply that it is Lebesgue-measurable?
(BACKGROUND: I am doing a geometry-related research, where most shapes are simple polygons and circles. But when I try to extend the results to more general shapes, I keep running into problems related to measurability. So, I want to know if assuming that my shapes are simply-connected will save me from this trouble).

Simply connected sets can be nonmeasurable.
Let $E\subset[0,1]$ be a nonmeasurable set and let $A=E\times[0,1]\cup[0,1]\times\{0\}$.
If you draw a picture, you see that $A$ resembles a comb and is connected and simply connected.
The set $A$ is not measurable.
(This MSE question might be worth a look for the last fact.)
