Compact sets of finite length are removable for continuous analytic functions

A planar square of area $A>0$ has infinite one-dimensional Hausdorff measure: if sets of diameters $d_j\le \delta$ cover it, each nonempty covering set lies in a half-open square of side $2d_j+ϵ2^{-j}$, where $ϵ>0$. By [F2] and countable subadditivity, $A\le {\sum }_j(2d_j+ϵ2^{-j}{)}^2$; letting $ϵ\downarrow 0$ gives $A\le 4{\sum }_j{d}_j^2\le 4\delta {\sum }_j{d}_j$, so the covering sums tend to infinity as $\delta \downarrow 0$. A chordal chart contains such a square with comparable distances by [F6]; hence ${\mathcal{H}}_{\chi }^1(\widehat{\mathbb{C}})=\mathrm{\infty }$ and $K\mathrm{\ne }\widehat{\mathbb{C}}$.

If $K=\mathrm{\varnothing }$, compactness and Liouville [F8, F10] already make $f$ constant; hence assume $K\mathrm{\ne }\mathrm{\varnothing }$. Choose $p\in \widehat{\mathbb{C}}\setminus K$ and let $M$ be the identity if $p=\mathrm{\infty }$, and $M(z)=1\mathrm{/}(z-p)$ otherwise. Then $K_0:=M(K)$ is compact and avoids $\mathrm{\infty }$. By [F5], $M$ is chordally bi-Lipschitz, so [F1] gives ${\mathcal{H}}_{\chi }^1(K_0)<\mathrm{\infty }$; since $K_0$ lies in a bounded disk, [F6] gives ${\mathcal{H}}_{\mathrm{E}\mathrm{u}\mathrm{c}\mathrm{l}}^1(K_0)<\mathrm{\infty }$. Put $g=f\circ M^{-1}$. Möbius maps and their inverses are conformal off their poles, so $g$ is continuous on the sphere and holomorphic on $\mathbb{C}\setminus K_0$.

Choose a closed square $S$ whose interior contains $K_0$. Fix $z\in S^{\circ }\setminus K_0$ and put $d=\mathrm{dist}(z,K_0)>0$. At scales ${\eta }_n\downarrow 0$, apply [F4] to obtain finite open-square covers of $K_0$, discarding squares missing it. Their side sums are bounded by a constant $B$, their diameters tend to zero, and every square lies within $\sqrt{2}{\eta }_n$ of $K_0$. For large $n$ their closures lie in $S^{\circ }$ and miss $z$ by distance at least $d\mathrm{/}2$. Write $O_n$ for their union. Its boundary is polygonal and misses $K_0$, since the open squares cover that compact set.

Partition $O_n$ by assigning its points to the first covering square containing them and removing all earlier squares. Subdividing the finitely many remaining pieces gives polygonal cells with disjoint interiors, each inside one original square. Their boundary edges are subsegments of the original square edges; each such edge segment occurs at most twice, once on each side. Thus the total cell perimeter is at most twice the sum of square perimeters, at most $8B$. Coincident edges are consolidated, zero-area pieces omitted, and holes carry the negative orientation. Internal edges cancel in the sum of the oriented cell boundaries. The construction does not require a cell boundary to miss $K_0$, because only continuity is used on those boundaries.

The function $H(\zeta )=g(\zeta )\mathrm{/}(\zeta -z)$ is holomorphic on the region between $\mathrm{\partial }S$, $\mathrm{\partial }{O}_n$ and a small circle about $z$. Its compact boundary misses $K_0$. Triangulate after deleting that circle and subdividing away from $K_0$; Cauchy's theorem [F7] cancels internal edges. Let the small circle shrink; continuity of $g$ gives its integral tending to $2\pi ig(z)$. Hence $2\pi ig(z)={\int }_{\mathrm{\partial }S}Hd\zeta -{\int }_{\mathrm{\partial }{O}_n}Hd\zeta$. The last integral equals the sum of the oriented cell integrals by step 3.1. On each cell choose a point $a$; the integral of the constant $H(a)$ around its complete polygonal boundary is zero. Uniform continuity of $H$ on a fixed compact neighborhood of $K_0$ therefore bounds the sum by $8B{\omega }_H(\sqrt{2}{\eta }_n)$, which tends to zero. No holomorphicity inside a covering square or cell is assumed.

It follows that $g(z)=(2\pi i{)}^{-1}{\int }_{\mathrm{\partial }S}g(\zeta )\mathrm{/}(\zeta -z)d\zeta$ for every $z\in S^{\circ }\setminus K_0$. The right-hand side is holomorphic on $S^{\circ }$: on each compact subset the kernel and its difference quotients converge uniformly on the finite contour, so differentiation under its integral is justified. Finite ${\mathcal{H}}^1(K_0)$ implies planar area zero, since a cover of diameter at most $\delta$ and bounded diameter sum has area cost at most $C\delta ({\mathcal{H}}^1(K_0)+1)$ by [F2]. Thus its complement is dense, and continuity extends this equality across $K_0$. Hence $g$ is entire.

The function $g$ is entire and continuous on the compact sphere, so [F8] makes its image compact in $\mathbb{C}$ and bounded. Liouville's theorem [F10] makes $g$ constant. Since $M$ is bijective, $f=g\circ M$ is constant.