A Jordan curve is a continuous closed curve in the plane with no self-intersections. My question is, is the interior of a Jordan curve always a Borel set?
If not, is the interior of a convex Jordan curve at least a Borel set. I know that convex sets need not be Borel sets, but maybe the combination implies Borel.
If not, does anyone know of a counterexample?
The image of a Jordan curve is a compact set, so that its complement is open.
The interior of the Jordan curve is one of the two connected components of that complement, and therefore open as well.
Every open set is a Borel set.