Riemann maps of Jordan domains extend to homeomorphisms of the closures

If $\mathrm{\infty }\in \mathrm{\Gamma }$, choose $q\mathrm{\notin }\mathrm{\Gamma }$ and a Möbius map $M$ with pole at $q$; then $M(\mathrm{\Gamma })\subset \mathbb{C}$. Prove the assertion in this normalized coordinate and transport maps and closures back by $M^{-1}$. Hence assume $\mathrm{\Gamma }\subset \mathbb{C}$.

By [F1], the finite plane complement has bounded component Omega and unbounded component Ustar. Set ${\mathrm{\Omega }}^{\ast }=Ustar\cup \{\mathrm{\infty }\}$, the corresponding sphere domain; its closure is a topological closed disk. Choose $a\in \mathrm{\Omega }$ and ${\tau }_a(z)=1\mathrm{/}(z-a)$; then $V={\tau }_a({\mathrm{\Omega }}^{\ast })$ is the bounded Jordan component containing0. Omega and V are open connected plane domains; Omega-star is an open connected sphere domain.

The complements $\mathbb{C}\setminus \mathrm{\Omega }$ and $\mathbb{C}\setminus V$ are connected and unbounded because each is the closure of the unbounded Jordan component, while $\mathrm{\Omega }$ and $V$ are bounded. For every cycle in either domain its index is locally constant off its trace and vanishes far away by [F2]; unboundedness supplies an omitted point with index zero, and connectedness then makes the index zero at every omitted point. Thus $\mathrm{\Omega }$ and $V$ are proper homologically simply connected domains.

Apply [F3] to $\mathrm{\Omega }$ with base point $a$ and invert the resulting map to obtain $\psi :\mathbb{D}\to \mathrm{\Omega }$. Apply [F3] to $V$ with base point $0$ to obtain $g:V\to \mathbb{D}$ with $g(0)=0$. With $I(z)=1\mathrm{/}z$, the map ${\psi }^{\ast }(z):={\tau }_a^{-1}(g^{-1}(I(z)))$ is a conformal equivalence from $\widehat{\mathbb{C}}\setminus \bar{\mathbb{D}}$ onto ${\mathrm{\Omega }}^{\ast }$ and satisfies ${\psi }^{\ast }(\mathrm{\infty })=\mathrm{\infty }$.

Fix any conformal equivalence $h:\mathbb{D}\to \mathrm{\Omega }$ and any $\xi \in \mathrm{\partial }\mathbb{D}$. The Möbius map ${\alpha }_{\xi }(z)=\xi (i-z)\mathrm{/}(i+z)$ sends $\mathbb{H}$ to $\mathbb{D}$ and $0\in \mathrm{\partial }\mathbb{H}$ to $\xi$; put $\mathrm{\Phi }=h\circ {\alpha }_{\xi }$. For $r>0$, let ${\mathrm{\Pi }}_r=\{r\mathrm{/}2<\mathrm{\mid }z\mathrm{\mid }<r,\mathrm{Im}z>0\}$ and let ${\mathcal{G}}_r$ be its half-circle family.

For each $r>0$ and $r\mathrm{/}2<\rho <r$, define the open semicircle image length $L(\rho )={\int }_0^{\pi }\mathrm{\mid }{\mathrm{\Phi }}^{\mathrm{\prime }}(\rho e^{i\theta })\mathrm{\mid }\rho d\theta$, possibly infinite. On compact angle subintervals this is exactly its C1 length by [F6], and exhaustion gives the improper length. Cauchy–Schwarz gives $L(\rho {)}^2\le \pi \rho {\int }_{S_{\rho }}\mathrm{\mid }{\mathrm{\Phi }}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}ds$. Integrating $d\rho \mathrm{/}\rho$ and applying polar Fubini yields ${\int }_{r\mathrm{/}2}^{r}L(\rho {)}^{2}d\rho \mathrm{/}\rho \le \pi {\int }_{{\mathrm{\Pi }}_r}\mathrm{\mid }{\mathrm{\Phi }}^{\mathrm{\prime }}{\mathrm{\mid }}^{2}dA=\pi \mathrm{\mid }\mathrm{\Phi }({\mathrm{\Pi }}_r)\mathrm{\mid }$ by [F5]. No conformal invariance theorem for a not-yet-extended boundary family is used.

For dyadic $r_n=2^{-n}$ the half-annuli ${\mathrm{\Pi }}_{r_n}$ are disjoint. Their images are disjoint measurable subsets of bounded Omega, so [F7] gives $A_n=\mathrm{\mid }\mathrm{\Phi }({\mathrm{\Pi }}_{r_n})\mathrm{\mid }\to 0$. Step 6.1 and ${\int }_{r_n\mathrm{/}2}^{r_n}d\rho \mathrm{/}\rho =\log 2$ allow a radius ${\rho }_n$ with $L({\rho }_n{)}^2\le \pi A_n\mathrm{/}\log 2+4^{-n}$. Countable Choice selects one at each scale. Hence these image arcs have finite length tending to zero and their source caps shrink to xi. Their Euclidean diameters are at most their lengths, and their spherical diameters also tend to zero by [F9]. Finite image length supplies endpoint limits in the next step, before any boundary family is invoked.

Each finite-length image arc in step 7.1 has endpoint limits, since the remaining variation of a rectifiable arc tends to zero. Those limits lie on $\mathrm{\Gamma }$: an interior limit, transported by the continuous inverse map, would make a sequence approaching the source boundary converge to an interior point. The arc interior is embedded; if the endpoints coincide it is a Jordan loop based at that boundary point. If they differ, uniform continuity of the inverse of a Jordan parameterization shows that they are joined by one $\mathrm{\Gamma }$ subarc of diameter tending to zero. Join this small subarc to the image arc to form a small Jordan loop $L_n$; in the coincident case take the image arc itself. Its bounded interior has diameter tending to zero, since it lies in the convex hull of its boundary (every affine coordinate attains its extreme on that boundary). The connected exterior ${\mathrm{\Omega }}^{\ast }$ contains infinity and avoids $L_n$, hence lies on its unbounded side. Every point of the other $\mathrm{\Gamma }$ subarc, or of $\mathrm{\Gamma }$ minus the coincident endpoint, is approached from ${\mathrm{\Omega }}^{\ast }$ and misses $L_n$, so also lies on that unbounded side. Thus the bounded interior of $L_n$ is contained in $\mathrm{\Omega }$. For large $n$ it excludes the fixed point $h(0)$. Crosscut separation in the topological Jordan disk therefore identifies it with the image of the source cap adjacent to $\xi$. These image caps have diameter tending to zero. Nested source caps then show that $h$ has a unique limit at $\xi$, and that its continuous extension is continuous there. Apply this at every boundary point.

Suppose two distinct boundary points have the same image $a$. The image of their straight source chord is a Jordan loop $L$ meeting $\mathrm{\Gamma }$ only at $a$. Its bounded interior is contained in $\mathrm{\Omega }$: the connected exterior of $\mathrm{\Omega }$ contains infinity and avoids $L$, and exterior approximation puts every point of $\mathrm{\Gamma }\setminus \{a\}$ on the unbounded side of $L$. One source chord cap maps to this bounded interior. Continuity forces its source boundary-circle arc to map into $\mathrm{\Gamma }\cap \overline{\mathrm{int}L}=\{a\}$. Flatten a subarc by a Möbius coordinate; Schwarz reflection [F10] extends $h-a$ across the interval where it is zero, and the identity theorem forces $h$ constant, a contradiction. Hence the boundary extension is injective. It is onto $\mathrm{\Gamma }$, because inverse images of a sequence in $\mathrm{\Omega }$ approaching any given boundary point have a subsequence in the compact closed disk, and its limit cannot be interior. The extension is a continuous bijection of compact Hausdorff closures, hence a homeomorphism.

For any conformal equivalence $h^{\ast }:\widehat{\mathbb{C}}\setminus \bar{\mathbb{D}}\to {\mathrm{\Omega }}^{\ast }$, the map ${\tau }_a\circ h^{\ast }\circ I$ is a conformal equivalence $\mathbb{D}\to V$; repeating the crosscut and reflection argument of steps 5.1–9.1 for $V={\tau }_a({\mathrm{\Omega }}^{\ast })$ gives a homeomorphic extension, and conjugating back by ${\tau }_a^{-1}$ and $I$ gives the required extension of $h^{\ast }$.

Any two continuous extensions agree on the dense open domain of their conformal equivalence, so agree on its closure because the sphere is Hausdorff; the extension is unique. Transporting the normalized conclusions from step 1.1 proves the theorem for the original sphere curve; the exterior normalization at $\mathrm{\infty }$ is asserted only in a coordinate where $\mathrm{\infty }\mathrm{\notin }\mathrm{\Gamma }$.