A single point is conformally removable

If $P=\mathrm{\varnothing }$, $F$ is already holomorphic on the whole sphere. Otherwise fix an arbitrary $p\in P$, put $q=F(p)$, and choose Möbius maps $A,B$ by $A(z)=z-p$ when $p\in \mathbb{C}$, $A(z)=1\mathrm{/}z$ when $p=\mathrm{\infty }$, and $B(w)=w-q$ when $q\in \mathbb{C}$, $B(w)=1\mathrm{/}w$ when $q=\mathrm{\infty }$; then $G:=B\circ F\circ A^{-1}$ is a sphere homeomorphism, holomorphic off the finite set $A(P)$, with $G(0)=0$.

By continuity of $G$ at $0$ and $G(0)=0$, choose $r>0$ so that $\{\mathrm{\mid }z\mathrm{\mid }<r\}\cap A(P)=\{0\}$ and on $\mathrm{\mid }z\mathrm{\mid }<r$ the map $G$ takes values in the finite target chart and $\mathrm{\mid }G(z)\mathrm{\mid }<1$. Thus the scalar chart expression $g(z)=G(z)$ is holomorphic on $0<\mathrm{\mid }z\mathrm{\mid }<r$, bounded there, and has limit $0$ at the puncture.

Apply [F3] to extend $g$ holomorphically across $0$ with value $0=G(0)$. The extension agrees with the original map by continuity, so $G$ is holomorphic at $0$ as a sphere map. Since $A$ and $B$ are biholomorphic, $F=B^{-1}\circ G\circ A$ is holomorphic at the arbitrary point $p$.

Repeating the pointwise argument for every $p\in P$ shows that $F$ is holomorphic on the whole sphere. In any source and target charts, a sufficiently small connected chart neighborhood gives an injective holomorphic map; [F5] makes its local inverse holomorphic. These local inverses are the chart expressions of the global inverse homeomorphism, so $F$ is biholomorphic.

By [F6], $F$ is Möbius; [F1] therefore says that the finite compact set $P$ is globally conformally removable. This includes the singleton case, and the empty-set case from step 1.1.

If $P=\mathrm{\varnothing }$, its Hausdorff measure is zero by the empty cover. If $P$ has $n\ge 1$ points, then for every $\delta ,\epsilon >0$ cover each point by a chordal ball of radius $r<\min (\delta \mathrm{/}3,\epsilon \mathrm{/}(3n))$; each ball has diameter at most $2r<\delta$ and the sum of the $n$ diameters is less than $\epsilon$. By [F7] and the definition of ${\mathcal{H}}^1$, ${\mathcal{H}}_{\chi }^1(P)=0$.