Conformal removability is invariant under quasiconformal maps

First suppose $K$ is removable. If $h(\mathrm{\infty })=p\mathrm{\ne }\mathrm{\infty }$, put $A(z)=1\mathrm{/}(z-p)$; if $h(\mathrm{\infty })=\mathrm{\infty }$, put $A=\mathrm{id}$. Then $h_0=A\circ h$ fixes $\mathrm{\infty }$. By [F1] and [F5], $h_0(K)$ is removable exactly when $h(K)$ is, and $h_0$ is quasiconformal, so it suffices to treat the case $h(\mathrm{\infty })=\mathrm{\infty }$. By [F2], $K$ has area zero; [F3] then gives area zero for $S:=h(K)$.

Let $G:\widehat{\mathbb{C}}\to \widehat{\mathbb{C}}$ be any homeomorphism conformal off $S$, and define $u:=h^{-1}\circ G^{-1}$. On $\widehat{\mathbb{C}}\setminus G(S)$, $G^{-1}$ is conformal and $h^{-1}$ is quasiconformal, so $u$ is quasiconformal there with dilatation bounded by that of $h^{-1}$. Extend its Beltrami coefficient by zero on the compact set $G(S)$; this gives a measurable sphere coefficient $\mu$ with $\mathrm{\parallel }\mu {\mathrm{\parallel }}_{\mathrm{\infty }}<1$. Countable Choice for the measurable-coefficient interface follows from [F7] and the assumed AC. By [F4], choose a quasiconformal sphere homeomorphism $F$ with ${\mu }_F=\mu$ almost everywhere.

The equality ${\mu }_F={\mu }_u$ holds on $\widehat{\mathbb{C}}\setminus G(S)$. The composition formula [F5] therefore gives zero Beltrami coefficient for $F\circ u^{-1}$ on $u(\widehat{\mathbb{C}}\setminus G(S))=\widehat{\mathbb{C}}\setminus K$, since $u^{-1}=G\circ h$. Thus $F\circ u^{-1}=F\circ G\circ h$ is locally $1$-quasiconformal there; [F6] makes it conformal on every component of $\widehat{\mathbb{C}}\setminus K$. Removability of $K$ and [F1] imply that $\mathrm{\Phi }:=F\circ G\circ h$ is Möbius.

Rearranging gives $G=F^{-1}\circ \mathrm{\Phi }\circ h^{-1}$, which is quasiconformal on the whole sphere by [F5]. It is conformal off $S$, and $S$ has area zero by step 1.1; hence its Beltrami coefficient vanishes almost everywhere. Thus $G$ is locally $1$-quasiconformal in sphere charts, and [F6] makes it conformal everywhere and Möbius. This proves that $S=h(K)$ is removable.

Conversely, if $h(K)$ is removable, apply the implication just proved to the quasiconformal map $h^{-1}$ and the compact set $h(K)$; this shows that $K$ is removable. Therefore removability is equivalent for $K$ and $h(K)$.