Compact subsets of lines and round circles are removable for quasiconformal maps

If $K=\mathrm{\varnothing }$, the assertion is the hypothesis. Otherwise, [F10] makes the image under $h$ of each component of $U\setminus K$ open, so both it and the source component are complex domains. By [F1], the map on each component is analytically $K_0$-quasiconformal. Its weak coordinate derivatives are locally square integrable there and satisfy the same Beltrami bound.

The set $K$ has planar measure zero by [F3]. Fix an open square $R$ with $\bar{R}\subset U$ and put $O=R\setminus K$. Choose the maximal dyadic squares $Q_j$ whose closures lie in $O$. They have disjoint interiors and cover $O$ except for the countable union of dyadic grid lines; each grid-line segment inside $R$ is null because it lies in a degenerate rectangle of measure zero. Every point of $O$ off those grid lines belongs to a sufficiently small dyadic square with closure in $O$, and then to a maximal such square. Each $\overline{Q_j}$ lies in one component of $U\setminus K$, so [F2] gives ${\int }_{Q_j}\mathrm{\mid }Dh{\mathrm{\mid }}_{\mathrm{H}\mathrm{S}}^{2}dA\le C(K_0){\lambda }_2(h(Q_j))$. The images $h(Q_j)$ are pairwise disjoint open sets because $h$ is a homeomorphic embedding and [F10] makes it open; they lie in $h(\bar{R})$, which is bounded by [F11]. Hence [F9] gives ${\sum }_j{\lambda }_2(h(Q_j))={\lambda }_2\text{\hspace{0.17em}⁣}({\bigcup }_{j}h(Q_j))\le {\lambda }_2(h(R))<\mathrm{\infty }\mathrm{.}$ Countable additivity and the null grid lines therefore give ${\int }_{R\setminus K}\mathrm{\mid }Dh{\mathrm{\mid }}_{\mathrm{H}\mathrm{S}}^{2}dA\le C(K_0){\lambda }_2(h(R))<\mathrm{\infty }\mathrm{.}$

Extend each weak coordinate derivative $g_i$ from $O$ by $0$ on $K\cap R$. It is measurable, and step 1.2 gives $g_i\in L^2(R)$. By [F4], for almost every horizontal line and almost every vertical line through $R$, the restriction of $g_i$ is in $L^2$; [F8] then puts it in $L^1$ on that bounded line interval.

Also discard the null line families on a countable rational-box cover of $O$ where the ACL representative or its agreement almost everywhere with $h$ fails; [F4]–[F5] justify this common exceptional family. Fix one of the remaining good coordinate lines. Its intersection with $K$ is finite except for at most one exceptional line when $K$ lies in a straight line parallel to the chosen direction; that one line is a null member of the parallel family. For a circle there are at most two intersection points on every coordinate line. On each open interval left after removing these finitely many points, [F1] and [F5] give an absolutely continuous representative with derivative $g_i$ a.e. The representative agrees a.e. with the continuous restriction of $h$, so the two agree everywhere on each such interval. Since $g_i\in L^1$ on the whole line interval, [F6] and continuity of $h$ at the finitely many missing points show that the restriction of $h$ on the full interval is the indefinite integral of $g_i$ plus a constant. It is therefore absolutely continuous across every point of $K$.

Applying step 2.1 in both coordinate directions on a countable rational-box cover of $U$ shows that $h$ itself is ACL on $U$. On every relatively compact rational box, the derivatives $g_i$ belong to $L^2$ by step 1.2; the ACL characterization [F5] therefore identifies them as the weak derivatives of $h$, so $h\in W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}(U)$. Since $K$ has planar measure zero, the inequality $\mathrm{\mid }{h}_{\bar{z}}\mathrm{\mid }\le k_0\mathrm{\mid }{h}_z\mathrm{\mid }$ continues to hold almost everywhere on $U$. Hence $h$ is analytically $K_0$-quasiconformal on $U$, and [F1] gives geometric $K_0$-quasiconformality with the same bound.

Let $H$ be a sphere homeomorphism that is $K_0$-quasiconformal off a generalized circle $\mathrm{\Sigma }$, in the chartwise sense of the Statement. Choose a finite point $p\mathrm{\notin }\mathrm{\Sigma }$ and Möbius charts $\varphi ,\psi$ sending $p,H(p)$ to $\mathrm{\infty }$, respectively (if $H(p)=\mathrm{\infty }$, take $\psi$ to be the finite chart). Then $g=\psi \circ H\circ {\varphi }^{-1}:\mathbb{C}\to \mathbb{C}$ is a homeomorphism. The set $K=\varphi (\mathrm{\Sigma })$ is a finite round circle: write $\mathrm{\Sigma }$ as $\alpha \mathrm{\mid }z{\mathrm{\mid }}^2+2\mathrm{Re}(\beta z)+\gamma =0$ with $\alpha ,\gamma \in \mathbb{R}$; under $w=1\mathrm{/}(z-p)$, multiplication by $\mathrm{\mid }w{\mathrm{\mid }}^2$ gives $P(p)\mathrm{\mid }w{\mathrm{\mid }}^2+2\mathrm{Re}((\alpha p+\bar{\beta })w)+\alpha =0$, where $P(p)=\alpha \mathrm{\mid }p{\mathrm{\mid }}^2+2\mathrm{Re}(\beta p)+\gamma \mathrm{\ne }0$. This is a Euclidean circle, hence compact by [F11]. By [F7] and the quasiconformal composition interface, $g$ is $K_0$-quasiconformal off $K$. The planar assertion applies to compact $K\subset \mathbb{C}$ with $U=\mathbb{C}$, so $g$ is $K_0$-quasiconformal on the whole finite chart. The omitted source point $p$ lies off $\mathrm{\Sigma }$, where $H$ was already quasiconformal. This proves the sphere assertion with the same bound.