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Compact sets of positive area are not conformally removable

Statement

Assume the Axiom of Choice. Let K⊆C^ be compact and suppose its finite-chart part E:=K∩C has positive planar Lebesgue area, λ2(E)>0, using C≅R2 (Lebesgue measurable sets, the family L(Rn), and the restricted set function λn, C=R[x]/(x2+1) as the Euclidean plane and as a normed real algebra: what the identification preserves). Then there is a homeomorphism F:C^→C^ conformal on C^∖K that is not a Möbius transformation. Thus K is not globally conformally removable (Conformal removability of compact sets), and every globally conformally removable compact set has zero area in the finite chart.

Facts & Assumptions

Given: AC, a compact set K⊆C^, and E=K∩C with λ2(E)>0.

[F2]

Under Countable Choice, Borel subsets of R2 are Lebesgue measurable (Assuming countable choice, every Borel subset of Rn is Lebesgue measurable, Lebesgue measurable sets, the family L(Rn), and the restricted set function λn). Thus the indicator of E is a measurable function in the finite chart.

[F3]

A Beltrami coefficient on the sphere is an almost-everywhere class determined by its finite-chart representative, with its infinity-chart expression fixed by the holomorphic transition rule; its norm is the essential supremum of the modulus (Measurable Beltrami coefficients and measurable conformal structures).

[F4]

A weak solution on the sphere is locally W1,2 in holomorphic charts and satisfies Fzˉ=μFz almost everywhere; a sphere Beltrami coefficient of norm below 1 has a quasiconformal homeomorphic solution with that coefficient (Weak solutions of the Beltrami equation, The measurable Riemann mapping theorem on the sphere).

[F5]

If a homeomorphism of complex domains lies in Wloc1,2 and has weak Wirtinger derivative fzˉ=0 almost everywhere, then it is conformal (Every 1-quasiconformal homeomorphism is conformal).

[F6]

Möbius transformations are biholomorphic in the sphere charts, so their Beltrami coefficient is zero almost everywhere (Every Möbius transformation is a biholomorphism of the Riemann sphere, The Beltrami coefficient and the maximal dilatation).

[F7]

A compact sphere set is globally conformally removable exactly when every sphere homeomorphism conformal off it is Möbius (Conformal removability of compact sets).

[F8]

AC implies Countable Choice (AC implies DC implies countable choice); the Borel/Lebesgue, coefficient, weak-solution and normalized measurable-Riemann-mapping interfaces use the stated choice assumptions (The Axiom of Choice, The Axiom of Countable Choice (ACω)).

Proof

technique · prescribe a nonzero measurable Beltrami coefficient on the positive-area set and solve it on the sphere
1.1F1F2F8given

Let E:=K∩C. By [F1], E is Borel in the finite chart, and [F2] makes it Lebesgue measurable. The Countable Choice assumption of the Borel/Lebesgue interface follows from AC by [F8].

2.1F2F3step 1.1givenalgebra

Define the finite-chart function μ0(z)=12 for z∈E and μ0(z)=0 for z∉E. By [F2] it is measurable. Since λ2(E)>0, its essential supremum is exactly ∥μ0∥∞=12: the pointwise bound gives at most 12, while for every t<12 the set {∣μ0∣>t} contains E and has positive measure. The sphere-chart rule [F3] therefore defines a Beltrami coefficient μ with ∥μ∥∞=12<1.

3.1F3F4step 2.1given

Apply the existence clause of [F4] with k=12. It gives an orientation-preserving sphere homeomorphism F that is a weak solution for μ and has Beltrami coefficient μF=μ almost everywhere.

4.1F3F4F5F8step 1.1step 3.1

On every local chart in C^∖K, the coefficient μ is zero almost everywhere because its finite-chart support is E⊆K and the transition rule preserves zero. Hence the weak Beltrami equation from [F4] gives Fzˉ=0 almost everywhere there. The local coordinate maps belong to Wloc1,2 by [F4], so [F5] makes them conformal. The Countable Choice assumptions of these measurable and weak-solution interfaces follow from AC by [F8]. Thus F is conformal on C^∖K.

5.1F3F6F7F8step 2.1step 3.1step 4.1given∎

If F were Möbius, [F6] would give μF=0 almost everywhere. This contradicts μF=μ=12 on the positive-area set E by step 2.1. Therefore F is not Möbius, and [F7] says K is not globally conformally removable. The same argument for any compact K of positive area proves that every globally conformally removable compact set has zero area. The normalized measurable-Riemann-mapping and measure interfaces use the choice assumptions recorded in [F8].

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