Riemann mapping theorem with pathological boundary

From wikipedia: In complex analysis, the Riemann mapping theorem states that if $$U$$ is a non-empty simply connected open subset of the complex number plane $$\mathbb{C}$$ which is not all of $$\mathbb{C}$$, then there exists a biholomorphic mapping $$f$$ (i.e. a bijective holomorphic mapping whose inverse is also holomorphic) from $$U$$ onto the open unit disk.

Is this also true for the boundary is not smooth? Even the boundary is a Jordan curve? Because we have to transform holomorphically from that to a smooth boundary which for me is not very possible. Or there are some condition omitted in the statement?

• The wikipedia statement is true in all generality. Boundary smoothness has nothing to do with it. – zhw. Nov 30 '18 at 23:54

It is true without any assumption on the boundary. It is another question whether the biholomorphic $$h : U \to U_1(0)$$ extends to homeomorphism $$\bar{h} : \overline{U} \to \overline{U_1(0)}$$. See for example https://arxiv.org/pdf/1307.0439.pdf.