The measurable Riemann mapping theorem on the sphere

Let $f,g$ be two solutions and set $h=g\circ f^{-1}$. By [F2], the composition is a 1-quasiconformal homeomorphism and its Beltrami coefficient vanishes almost everywhere. Localize in source and target charts of the sphere and apply [F3]'s one-quasiconformal theorem to each chart expression; both directions are holomorphic, so $h$ is a biholomorphic self-map of the sphere and therefore Möbius. Conversely, postcomposition by a Möbius map leaves the coefficient unchanged by [F2]. If $f,g$ both fix $0,1,\mathrm{\infty }$, the Möbius map $g\circ f^{-1}$ fixes three distinct points and is the identity by [F4].

Write $U_0=\widehat{\mathbb{C}}\setminus \{\mathrm{\infty }\}$ and $U_{\mathrm{\infty }}=\widehat{\mathbb{C}}\setminus \{0\}$. By [F6] the standard sphere is a smooth manifold, so [F7] gives a smooth partition $({\rho }_0,{\rho }_{\mathrm{\infty }})$ subordinate to these two chart domains. Choose in [F8] a nonnegative bump supported in the unit disk and equal to one on the half-unit disk, and normalize it to a unit-mass kernel $\phi$ and set ${\phi }_{{ϵ}_n}(z)={ϵ}_n^{-2}\phi (z\mathrm{/}{ϵ}_n)$ with ${ϵ}_n=1\mathrm{/}(n+1)$ for $n\in \mathbb{N}$. These are fixed chart and kernel choices; no choice of a family is needed here.

Let ${\mu }_0$ and ${\mu }_{\mathrm{\infty }}$ be chart representatives of $\mu$, and define ${\nu }_{0,n}={\mu }_0\ast {\phi }_{{ϵ}_n}$ and ${\nu }_{\mathrm{\infty },n}={\mu }_{\mathrm{\infty }}\ast {\phi }_{{ϵ}_n}$. Each is smooth by [F8] and satisfies $\mathrm{\mid }{\nu }_{j,n}\mathrm{\mid }\le k$, since it averages a representative bounded by $k$ against a nonnegative unit-mass kernel. Glue the chartwise sections ${\rho }_0{\nu }_{0,n}$ and ${\rho }_{\mathrm{\infty }}{\nu }_{\mathrm{\infty },n}$, extending each by zero outside its subordinate chart support, and call the sum ${\mu }_n$. The pullback law [F1] makes this a smooth sphere coefficient; the weights are nonnegative and sum to one, so $\mathrm{\parallel }{\mu }_n{\mathrm{\parallel }}_{\mathrm{\infty }}\le k$. At a Lebesgue point $x$ of a chart representative, $\mathrm{\mid }{\nu }_{j,n}(x)-{\mu }_j(x)\mathrm{\mid }\le \mathrm{\parallel }\phi {\mathrm{\parallel }}_{\mathrm{\infty }}{ϵ}_n^{-2}{\int }_{B(x,{ϵ}_n)}\mathrm{\mid }{\mu }_j(y)-{\mu }_j(x)\mathrm{\mid }dy\to 0$ by [F9]. The inversion transition carries its exceptional null set to a null set, so ${\mu }_n\to \mu$ almost everywhere in both charts.

For each $n$, choose an orientation-preserving quasiconformal solution $F_n$ for ${\mu }_n$ by [F5]; Countable Choice, supplied by [F14], permits these countably many selections. The pointwise equation and $\mathrm{\mid }{\mu }_n\mathrm{\mid }\le k$ give the common analytic bound $K_0=(1+k)\mathrm{/}(1-k)$, and [F4] normalizes $F_n$ by postcomposing with a Möbius map to obtain $f_n(0)=0$, $f_n(1)=1$, $f_n(\mathrm{\infty })=\mathrm{\infty }$. The composition formula [F2] preserves the coefficient and orientation. By the geometric/analytic equivalence in [F10], each $f_n$ belongs to the normalized geometric $K_0$-quasiconformal family.

By [F10], pass to a subsequence (still written $f_n$) converging uniformly in chordal distance to a normalized orientation-preserving $K_0$-quasiconformal homeomorphism $f$. Fix a bounded open disk $B⋐\mathbb{C}$; its closure is compact by [F16]. By [F16], $f(\bar{B})$ is compact in the chordal metric. It avoids $\mathrm{\infty }$, since $f$ is injective and fixes $\mathrm{\infty }$. The continuous function $w\mapsto \chi (w,\mathrm{\infty })$ therefore has a positive minimum $\delta$ on $f(\bar{B})$. The set $C_{\delta }=\{w\in \widehat{\mathbb{C}}:\chi (w,\mathrm{\infty })\ge \delta \mathrm{/}2\}$ is closed in the compact chordal sphere, hence compact by [F16]; it lies in the finite chart and is Euclidean bounded by [F16]. Uniform chordal convergence and the triangle inequality put every sufficiently late $f_n(\bar{B})$ inside $C_{\delta }$. Each of the finitely many earlier images is compact, avoids $\mathrm{\infty }$ because $f_n$ fixes $\mathrm{\infty }$, and is therefore Euclidean bounded by [F16]. Thus ${\sup }_n\mathrm{\mid }{f}_n(B)\mathrm{\mid }<\mathrm{\infty }$.

The area estimate [F11] now gives ${\sup }_n{\int }_B\mathrm{\mid }D{f}_n{\mathrm{\mid }}_{HS}^2<\mathrm{\infty }$, hence both sequences $(f_n{)}_z$ and $(f_n{)}_{\bar{z}}$ are bounded in $L^2(B)$. By [F12] and [F14], take a subsequence weakly convergent for the first sequence and then a subsubsequence weakly convergent for the second. Uniform convergence in the finite chart lets every compactly supported smooth test function pass through the weak-derivative identity, so the weak limits are the distributional derivatives $f_z$ and $f_{\bar{z}}$ of $f$. Thus $f\in W_{loc}^{1,2}(B)$.

For any $\eta \in L^2(B)$, split the weak pairing as ${\int }_B\eta ({\mu }_n(f_n{)}_z-\mu f_z)={\int }_B\eta \mu ((f_n{)}_z-f_z)+{\int }_B\eta ({\mu }_n-\mu )(f_n{)}_z$. The first term tends to zero by weak convergence; by [F13], the second is at most $\mathrm{\parallel }\eta ({\mu }_n-\mu ){\mathrm{\parallel }}_2\mathrm{\parallel }(f_n{)}_z{\mathrm{\parallel }}_2$, which tends to zero by [F9], dominated convergence and the uniform $L^2$ bound. Since $(f_n{)}_{\bar{z}}={\mu }_n(f_n{)}_z$, passage to weak limits gives $f_{\bar{z}}=\mu f_z$ almost everywhere on $B$. Taking the countable exhaustion $B=D(0,m)$, [F9] assembles a global full-measure set in the finite chart; [F1] transports the equation to the infinity chart. Thus $f$ is a sphere weak solution.

For each $m$, let $Z_m$ be the Borel set in $D(0,m)$ where a finite Borel representative of $f_z$ vanishes, and let $N$ be a Borel null set outside which the equation from step 3.1 holds. On the relatively compact Borel set $E_m=Z_m\setminus N$ one has $f_{\bar{z}}=f_z=0$, hence $J_f=0$. The established area formula in [F15] gives $\mathrm{\mid }f(E_m)\mathrm{\mid }={\int }_{E_m}{J}_f=0$; its inverse N-property then makes $E_m$ null. Countable additivity over $m$, together with $N$ being null, shows $f_z\mathrm{\ne }0$ almost everywhere. The definition of ${\mu }_f$ now gives ${\mu }_f=\mu$ almost everywhere and $K_f=K(\mu )\le K_0$. This uses both the area formula and independently proved inverse-N clause; the lower area inequality alone would not suffice.

Step 4.1 gives the normalized solution for (i), and step 1.1 proves its uniqueness. For any distinct target triple $(a,b,c)$, [F4] gives the unique Möbius map carrying $(0,1,\mathrm{\infty })$ to $(a,b,c)$; postcomposing the normalized solution preserves its Beltrami coefficient by [F2]. Any other solution with those three values differs by a Möbius map fixing the triple, hence is equal to it.