Not every compact set is conformally removable

By [F1], $\bar{\mathbb{D}}$ is compact.

By [F3], its boundary ${\mathbb{S}}^1$ is globally conformally removable.

By [F2], every compact sphere set with positive area in its finite chart is globally conformally nonremovable.

For $0\le r\le 1$, put $\rho (r)=r(1+r)\mathrm{/}2$. This function is continuous and strictly increasing from $[0,1]$ onto $[0,1]$, with inverse $q(s)=(\sqrt{1+8s}-1)\mathrm{/}2$. The map $F$ in the Statement sends each radius $r\le 1$ to $\rho (r)$ with the same argument and is the identity for $r\ge 1$; the inside and outside formulas agree at $r=1$. Its inverse uses $q$ on radii in $[0,1]$ and is the identity outside. Both maps are continuous at $0$, at radius $1$, and at $\mathrm{\infty }$, so [F6] makes $F$ a sphere homeomorphism.

The map $F$ fixes every point of ${\mathbb{S}}^1$, is the identity and hence conformal on $\widehat{\mathbb{C}}\setminus \bar{\mathbb{D}}$ by [F6], but $F(1\mathrm{/}2)=3\mathrm{/}8$. By [F4], a Möbius map fixing the three distinct points $1,-1,i$ would be the identity, so $F$ is not Möbius. The compactness in step 1.1 and the global definition therefore show that $\bar{\mathbb{D}}$ is not conformally removable.

Construct $C\subseteq [0,1]$ by starting with $[0,1]$ and, at stage $n\ge 1$, removing the middle open interval of length $4^{-n}$ from each of the $2^{n-1}$ remaining intervals. The preceding intervals have length $2^{-n}+2^{-2n+1}>4^{-n}$, so each removal fits. The total length removed at stage $n$ is $2^{n-1}{4}^{-n}=2^{-n-1}$, and the sum over all stages is $1\mathrm{/}2$. The remaining intervals at stage $n$ have length $2^{-n-1}+2^{-2n-1}$, which tends to zero; hence the intersection $C$ is compact, has empty interior, and has Lebesgue measure $1\mathrm{/}2$. Put $K=C\times C\subseteq {\mathbb{R}}^2\cong \mathbb{C}$. It is closed and bounded, hence compact by [F1], and has empty interior because its first-coordinate projection is contained in the nowhere-dense set $C$. Since $K$ is Borel, the product rectangle formula and the agreement of product and Euclidean Lebesgue measure on Borel sets give ${\lambda }_2(K)={\lambda }_1(C{)}^2=1\mathrm{/}4$. By step 1.3, $K$ is not conformally removable.

Apply [F7] with $h(t)=t^2$, which satisfies its gauge hypotheses. For every $\delta ,\epsilon >0$, the zero value of ${\mathrm{\Lambda }}_{t^2}(C_h)$ gives a finite or countable disk cover with radii $r_j\le \delta \mathrm{/}2$ and ${\sum }_j{r}_j^2<\epsilon \mathrm{/}4$. Each disk has diameter at most $2r_j$, so [F8] gives ${\mathcal{H}}_{\delta }^2(C_h)\le 4{\sum }_j{r}_j^2<\epsilon$. As $\epsilon$ is arbitrary, every scale content is zero, hence ${\mathcal{H}}^2(C_h)=0$. Each disk is also contained in a closed square of side $2r_j$, so [F5]'s box formula and countable subadditivity give planar area at most $4{\sum }_j{r}_j^2<\epsilon$. By [F1], the closed Jordan curve is compact and hence closed and Borel, so its area is zero. The same [F7] supplies a non-Möbius sphere homeomorphism conformal off this curve; the global definition therefore makes it nonremovable. Only that existence input is cited, and none of the local witnesses above uses it.