The Singular Strata of a Free-Boundary problem for harmonic measure

In this paper, we obtain \textit{quantitative} estimates on the fine structure of the singular set of the mutual boundary $\partial \Omega^{\pm}$ for pairs of complementary domains, $\Omega^+, \Omega^- \subset \mathbb{R}^n$ which arise in a class of two-sided free boundary problems for harmonic measure. These estimates give new insight into the structure of the mutual boundary, $\partial \Omega^{\pm}.$


Introduction
The focus of this paper is the study of a class of two-phase free boundary problems for harmonic measure.For n ≥ 3, let + ⊂ ‫ޒ‬ n and − = + c be unbounded nontangentially accessible (NTA) domains (see Definition 2.1), let ω ± be their associated harmonic measures, and let u ± be the associated Green's functions with poles at infinity.Let ω − ≪ ω + ≪ ω − , and let h = dω − /dω + satisfy ln(h) ∈ C 0,α for some 0 < α < 1.We obtain new results on the structure of the geometric singular set of the boundary ∂ ± .
This problem was introduced without the regularity assumption on ω ± by Kenig, Preiss, and Toro [ Kenig et al. 2009], with other work under the assumption that ln(h) ∈ VMO(∂ ± ) by Kenig and Toro [2006], Badger [2011;2013], and Badger, Engelstein, and Toro [Badger et al. 2017].Questions about the structure of the free boundary and the singular set when ln(h) ∈ C 0,α for 0 < α < 1 have been addressed by Engelstein [2016] and Badger, Engelstein, Toro [Badger et al. 2020], respectively.Engelstein [2016] shows that under the additional assumption that the boundary is sufficiently flat in the sense of Reifenberg, the boundary is locally C 1,α .In [Badger et al. 2020], the authors remove the assumption of flatness and prove that the geometric singular set is contained in countably many C 1,β submanifolds of the appropriate dimension.See [Kenig et al. 2009] for an overview of this problem in lower dimensions and [Badger et al. 2017; 2020] for further background.
Until recently, almost all work on the two-sided free boundary problem for harmonic measure in higher dimensions has operated under the assumption that ± are NTA domains because the NTA conditions allow for scale-invariant estimates of harmonic measure.However, Azzam, Mourgoglou, Tolsa, and Volberg [Azzam et al. 2019] proved, among other things, that if we relax the assumption that the domains are NTA, then ω − ≪ ω + ≪ ω − on G ⊂ ∂ ± implies that G can be decomposed into G = R ∪ B, where R is (n−1)-rectifiable and ω ± (B) = 0.However, we shall work under the assumption that ± are NTA domains.
Based upon [Badger et al. 2020], we know that when ln(h) ∈ C 0,α , the singular set of ∂ ± is countably C 1,β -rectifiable where β depends on but is not equal to α.This leaves open the question of whether or not the singular set is dense, or more generally how it sits in space.In this paper, we answer the question of how the singular set "sits in space".In particular, we provide upper Minkowski content bounds upon the quantitative strata of the singular set (see Theorem 2.15).The main approach will be to follow [ .While these tools are common for problems in calculus of variations, it is important to note that the jump functions v are not minimizers of any energy, nor do they satisfy any global PDE.

Definitions and statement of main results
2A. Domains and their Green's functions.Nontangentially accessible (NTA) domains were formally introduced by Jerison and Kenig [1982] to study the boundary behavior of PDEs on nonsmooth domains.Definition 2.1.A domain ⊂ ‫ޒ‬ n is a nontangentially accessible (NTA) domain if there exist constants M > 1 and R 0 > 0 such that the following holds: (1) satisfies the corkscrew condition.That is, for any Q ∈ ∂ and 0 < r < R 0 , there exists a point A r (Q) ∈ with the following two properties: |A r (Q) − Q| < r and B r/M (A r (Q)) ⊂ .
(2) c also satisfies the corkscrew condition.
(3) satisfies the Harnack chain condition.That is, for any ϵ > 0 and Q ∈ ∂ , if then there exists a "Harnack chain" of balls {B r i (y i )} N i=1 satisfying: (a) x 1 ∈ B r i (y 1 ) and x 1 ∈ B r N (y N ).Note that by increasing the radii if necessary, we may assume that r i ∼ M dist(y i , ∂ ).
We say that + is a two-sided NTA domain if both + and − := c are NTA domains.We shall refer to the complementary pair ± of domains as complementary two-sided NTA domains and denote their mutual boundary by ∂ ± .
In this paper, we shall only deal with unbounded two-sided NTA domains.That is, we shall assume that R 0 = ∞.However, the results are essentially local.
Definition 2.2 (Green's functions).For ± ⊂ ‫ޒ‬ n a pair of complementary two-sided NTA domains, we shall use u ± to denote the Green's function with pole at infinity corresponding to ± , respectively.
Recall that u ± are unique up to scalar multiplication and that to each u ± is associated the harmonic measure ω ± , defined by the property that, for all φ ∈ C ∞ c ‫ޒ(‬ n ), See [Garnett and Marshall 2005] for more details about harmonic measures.
Observe that if ω + is the harmonic measure associated to u + , then cω + is the harmonic measure associated to cu + for any c > 0.

2B.
A class of functions and their rescalings.
Definition 2.4.Let ± ⊂ ‫ޒ‬ n be a pair of complementary two-sided NTA domains with mutual boundary ∂ ± .For any Q ∈ ∂ ± and any Green's functions u ± we define the jump function (2-1) The scaling h(Q)u + normalizes the Radon-Nikodym derivative of the harmonic measure associated to h(Q)u + and u − at Q ∈ ∂ ± .
Definition 2.5.Let ± ∈ D(n, α, M 0 ) and Q ∈ ∂ ± .For scales 0 < r , we define the rescaling of the function v Q at scale r at the point and the corresponding rescaled measure as 2C. Quantitative symmetry.The geometry we wish to capture with the blow-ups T x f is encoded in their translational symmetries.
Definition 2.10.Let f : ‫ޒ‬ n → ‫ޒ‬ be a continuous function.We say f is 0-symmetric if f (x) := c P + (x) − P − (x) (2-4) for some c > 0, where P ± are the positive and negative parts of a homogeneous harmonic polynomial P.
We will say that f is k-symmetric if f is 0-symmetric and there exists a k-dimensional subspace V such that f (x + y) = f (x) for all x ∈ ‫ޒ‬ n and all y ∈ V.
The constant c > 0 is there to allow for the function to "hinge" along its zero set.We must allow this kind of "hinging" to accommodate for the "nonalignment" issue in the blow-ups at We now define a quantitative version of symmetry.
Definition 2.11.For any f ∈ C(‫ޒ‬ n ), f will be called (k, ϵ, r, p)-symmetric if there exists a k-symmetric function P such that Definition 2.12 (quantitative singular strata).Let v ∈ A(n, α, M 0 ) and 0 < r ≤ 1.We denote the (k, ϵ, r )-singular stratum of v by S k ϵ,r (v), and we define it by We shall also use the notation , then there exists an ϵ > 0 such that x ∈ S k ϵ (v).Remark 2.13.Note that the singular set and its strata are all stable under the operations for all c ̸ = 0.The former is a trivial consequence of the fact that T p,r f = T p,r (c f ).The latter follows from Definition 2.10 and Theorem 2.6.In particular, for all v ∈ A(n, α, M 0 ), we have Previous results on the singular set are summed up in the following theorem.
and there exists a constant such that (2-8) Thanks to an ϵ-regularity result due to [Engelstein 2016] we are able to strengthen the conclusion of Theorem 2.15 when we consider the full singular set.
There exists 0 < ϵ = ϵ(M 0 , , α) such that sing(∂ ± ) ⊂ S n−3 ϵ ; see Lemma 12.1.Thus, there is a constant (2-9) Proof.This follows immediately from Lemma 12.1 and Theorem 2.15.□ 2E.Outline of the proof of Theorem 2.15.In order to prove a theorem of this kind, we must build a cover of S k ϵ,r (v), and we must count how many balls we use.Therefore two things are critical: getting geometric information about S k ϵ,r (v) and keeping track of how the balls pack.The overall strategy of proof is similar to that of [De Lellis et al. 2018; Edelen and Engelstein 2019].However, there are several major differences.First, the functions v ∈ A(n, α, M 0 ) considered here are not harmonic functions or minimizers of an energy.Sections 3-5 are devoted to showing that the relevant analogs of harmonic results (e.g., compactness, almost monotonicity of the Almgren frequency, local uniform boundedness of the Almgren frequency, quantitative rigidity for the Almgren frequency, cone-splitting, etc.) hold for v ∈ A(n, α, M 0 ).In particular, we prove an estimate on the nondegeneracy of the almost monotonicity for Almgren frequency in Lemma 4.9.Local geometric control on S k ϵ (v) is obtained in Section 6.
However, geometric control is not enough to obtain Theorem 2.15.To obtain finite upper Minkowski content bounds we need the discrete Reifenberg theorem from [Naber and Valtorta 2017]; see Theorem 9.1.This requires that we prove a "frequency pinching" result (Lemma 8.2) in which we connect the drop in the Almgren frequency over small scales with the β-numbers.The main challenge is to connect the lower bound on the derivative of the Almgren frequency (Lemma 4.9) and employ the techniques of [De Lellis et al. 2018] to obtain the necessary estimates on In Section 9, we obtain the necessary packing estimates, following the framework of [Naber and Valtorta 2017] to accommodate the estimates of Section 8.Sections 10 and 11 construct the covering which proves the theorem according to the program laid out by [Naber and Valtorta 2017].These are included for completeness.

Compactness
The main goal of this section is to show that A(n, α, M 0 ) enjoys sufficient compactness to allow for limit-compactness arguments.Namely, we wish to establish that, for any sequence v i ∈ A(n, α, M 0 ), we can extract a subsequence which converges to a function v ∞ and that N ( p, r, v i ) → N ( p, r, v ∞ ); see Corollary 4.3.This requires strong convergence in W 1,2  loc ‫ޒ(‬ n ); see Lemmas 3.10 and 3.6.On a technical level, we must extend the compactness implied by Theorem 2.6 for and T Q ′ ,r v. Throughout, we shall make essential use of "standard NTA results" such as the doubling of harmonic measure and various comparability results, all of which may be found in [Jerison and Kenig 1982].
Using (3-1) in the above integral equations implies that in any compact set Remark 3.2.By Theorem 2.6, we know that subsequential limits as r → 0 of the functions v Q Q,r converge to homogeneous harmonic polynomials.However, for r converges to a homogeneous harmonic polynomial.As r → 0, the function v Q Q ′ ,r will converge to a 0-symmetric function (see Definition 2.10) where c = h(Q).Definition 3.3.We shall abuse the notation T Q,r from Definition 2.7 to denote translated and scaled versions of various objects.For example, for sets this is the usual push-forward However, for the measures ω ± , we will denote by T Q,r ω ± the harmonic measures associated to the positive and negative parts of T Q,r v.The corkscrew points A ± R (Q) will always denote the corkscrew point associated to Q at scale R in the appropriate domain ± .We shall use 0) and all radii 0 < r ≤ 2, the function T Q,r v is locally Lipschitz with uniform constants depending only upon M 0 , , α.
Proof.Recall that by Definition 2.5, By NTA estimates, for all 0 < r , we have is bounded above and below by constants which only depend upon M 0 .By constructing Harnack chains from T Applying Harnack's inequality to the function −v in a chain of balls which connect T Q,r A − r (Q) and y in − , we have y)| is bounded above and below by constants that only depend upon M 0 .Thus, by the uniform Lipschitz property of v Q,r guaranteed by Theorem 2.6, we can find a ball of radius 0 < c such that |v Q,r | ≥ c(M 0 ) on ∂ B 1 (0)∩ B c (y).Thus, H (0, 1, v Q,r ) ≥ c(M 0 ).Now, recalling Definition 2.7 and the fact that T 0,1 v = T 0,1 (cv) for any constant c > 0, we have T Q,r v = T 0,1 v Q,r .Since we assumed ∥ln(h)∥ α ≤ , Q ∈ B 2 (0), and 0 < r ≤ 2, the v Q,r are locally uniformly Lipschitz by Theorem 2.6.Thus H (0, is bounded above and below by constants that only depend upon the NTA constant M 0 , , and R. Thus, by Harnack chains between T Q,r A − r (Q) and Note that C only depends upon R, M 0 , and ϵ.
To get the same inequality for p ∈ T Q,r ) by constants which only depend upon R, , α, and the NTA constants in the definition of the class A(n, α, M 0 ).Applying the same Harnack chain and Harnack inequality argument as above gives the lemma.□ Lemma 3.6 (compactness).Let {v i } be a sequence of functions in A(n, α, M 0 ) such that ∥ln(h)∥ α ≤ .
There is a subsequence {v j } and a Lipschitz function v ∞ ∈ W 1,2 loc such that T Q j ,r j v j → v ∞ in the following senses: (1) ( To see (1), we recall Lemma 3.4 and the fact that T Q i ,r i v i (0) = 0.By the Arzelà-Ascoli theorem there exists a subsequence such that Being uniformly locally Lipschitz and uniformly bounded also implies that the functions {T Q j ,r j v i } are bounded in W 1,2  loc ‫ޒ(‬ n ).By Rellich compactness, there exists a further subsequence such that locally in the Hausdorff metric on compact subsets.Proof.We argue by contradiction.Suppose that there exists an ϵ > 0, a radius 0 < R, and a sequence of functions T Q i ,r i v i for which we can find a sequence of points Taking a subsequence which converges in C loc ‫ޒ(‬ n ), we may assume that Furthermore, since the T Q i ,r i v i are uniformly locally Lipschitz, x i → x ∞ , and x i ∈ {T Q i ,r i v i = 0}, we have This implies x ∞ ∈ {v ∞ = 0}, which contradicts our previous assertion that The other direction goes the same way.Suppose that we could find a subsequence of We must now bound the upper Minkowski dimension of A = {v ∞ = 0}.We do so crudely, using only that A is the mutual boundary of a pair of two-sided NTA domains.That is, using the machinery of porous sets we are able to prove the following lemma.Lemma 3.9.Let ⊂ ‫ޒ‬ n be the mutual boundary of a pair of unbounded two-sided NTA domains with NTA constant 1 < M 0 .Then, there exists This is an elementary fact which seems to be omitted in the literature.We defer the proof to the Appendix.We now prove strong convergence.Lemma 3.10 (strong compactness).Let {v i } be a sequence of functions in A(n, α, M 0 ) such that There is a subsequence {v j } and a Lipschitz function v ∞ ∈ W 1,2  loc such that T Q j ,r j v j → v ∞ in the following senses: By Lemma 3.7, Theorem 3.8, and Lemma 3.9, we have that dim M ({v ∞ = 0}) ≤ n − ϵ.In particular, then, H n (B r ({v ∞ = 0} ∩ B R (0))) → 0 as r → 0 (see [Mattila 1995] for fundamental facts about Minkowski content, dimension and Hausdorff measure).Thus, for any θ > 0 we can find an r (θ ) > 0 such that where the penultimate inequality uses the fact that v j are uniformly Lipschitz, and the last equality follows from convergence in C(B R+r (0) )) convergence because the T Q j ,r j v j are harmonic functions in this region.Since θ > 0 was arbitrary, we have that The other inequality follows from the same trick or from lower semicontinuity.Therefore, we have the equality Thus, by weak convergence and norm convergence we have Because the functions v Q p,r are merely Lipschitz, we will often need to work with a mollified version of them.We will use the convention that

Almost monotonicity of the Almgren frequency function
One of the key tools of this paper will be the Almgren frequency function (introduced in [Almgren 1979]).
Definition 4.1 (Almgren frequency function).For any Lipschitz function v : ‫ޒ‬ n → ‫,ޒ‬ radius r > 0, and point Q ∈ ∂ ± , the Almgren frequency function is defined as where If u is harmonic then N ( p, r, u) is monotonically nondecreasing.If additionally one assumes that u( p) = 0 then lim r →0 N ( p, r, u) = N ( p, 0, u) ≥ 1 is the degree of the leading homogeneous harmonic polynomial in the Taylor expansion of u at the point p.
4A. Consequences of Section 3 for the Almgren frequency function.Before turning to the main results of this section, we note that the results of Section 3 immediately imply the following corollaries.
Corollary 4.3.Under the hypotheses of Lemma 3.6, there exists a subsequence such that, for all r ∈ (0, 2], Moreover, if v ϵ = v ⋆φ for a mollifier φ as in Corollary 3.11 then, for all Q ∈ B 1 (0)∩∂ ± and 0 Proof.This follows from the convergence of the numerator and the denominator; the former follows from Lemma 3.6 (2) and the latter from Lemma 3.6 (1).For the convolution, both follow from Corollary 3.11.
Proof.We recall that the Almgren frequency function is invariant under rescalings of the function v. Therefore, ) is bounded by Lemma 3.4 and the constant only depends upon M 0 , , and α. □ 4B.Quantitative almost monotonicity.This section is dedicated to providing a quantitative version of the following result of Engelstein [2016].
There exists a constant C < ∞ (which can be taken uniformly over K and r ∈ (0, 1]) such that The quantitative version of this result which we prove below in Lemma 4.9 is essential for connecting the Almgren frequency to Jones' beta numbers in the "frequency pinching" result later in Lemma 8.2.It comes from examining the derivative of the Almgren frequency function in the r variable.
Throughout this section, we shall use the notation (v ϵ ) ν (y) = ∇v ϵ (y) • ν(y), where ν(y) is the unit normal to ∂ B r (Q) at y.By differentiation (see [Engelstein 2016, Section 5.1] for details of the derivation), We write the decomposition We call what remains N ′ 2 (Q, r, v ϵ ): Note that by the Cauchy-Schwarz inequality, Proof.Recall that for the Cauchy-Schwarz inequality, we have, for λ = ⟨u, v⟩/∥v∥ 2 , using the divergence theorem on λ, and letting ϵ → 0, we have This proves the equality.To prove the lower bound, we let C = Lip(T 0,1 v| B 2 (0) ) and observe that H (Q, r, T 0,1 v ϵ ) ≤ Cr n+1 .Plugging this into the above equation, we get the desired inequality dσ (y).□ In order to bound the parts of where C = C(α, M 0 , ).
Engelstein [2016, Lemmata 5.4, 5.5, and 5.6] proves the claim for the functions v Q Q,1 .Hence, for any such v and any such Q, the integral estimates hold for u where We now state the main result of this section.
Lemma 4.9. where Proof.We begin by normalizing v. Since N (r, p, v) = N (r, p, cv) for any c ̸ = 0, we may work with T 0,1 v.Note that by Remark 4.8 and (4-5), N (Q, r, v) is continuous in r and hence we may find an 0 ≤ s < s 1 such that By Corollaries 3.11 and 4.3 we can find an ϵ ≪ s small enough that Thus, we reduce to estimating Recalling Remark 4.8, Lemma 4.6, and letting ϵ → 0 gives the lemma.□ Using these estimates it is possible to control the drop across scales from the total drop.
Proof.This is essentially a "rays of the sun" argument.To wit, The bounds in Remark 4.8 give the desired statement.□ We now turn our attention to proving a "doubling" property for H ( p, r, v).This is an analog of classical harmonic results for the Almgren frequency function, modified for our almost harmonic functions v ∈ A(n, α, M 0 ).
Proof.First, observe that Next, we consider the identity ln We bound N (r, Q, v ϵ ) by Lemma 4.9.We bound the last term using Lemma 4.7.Plugging in these bounds, we have, for ϵ ≪ s, Evaluating and exponentiating gives the desired result.
(a consequence of Corollary 3.11), we have the following inequality.For all v ∈ A(n, α, M 0 ) with

Quantitative rigidity
Throughout the rest of the paper, we shall need to use limit-compactness arguments.The key will be that v → u for some harmonic function u as ∥ln(h)∥ α → 0. We make this rigorous in the following lemma.
Then there exists a function v ∞ and a subsequence v j such that T Q j ,r j v j → v ∞ in the sense of Lemma 3.10 and v ∞ is harmonic.
Proof.Lemma 3.10 gives a subsequence T Q j ,r j v j which converges strongly in W 1,2 loc ‫ޒ(‬ n ) to a function v ∞ .We claim that v ∞ is harmonic.To see this, we investigate the behavior of its mollifications v ∞,ϵ = v ∞ ⋆φ ϵ .Observe that by Young's inequality, Thus, for any ϵ > 0 we have T Q j ,r j v j,ϵ → v ∞,ϵ as j → ∞ strongly in L 2 (B 2 (0)).By a similar argument applied to ∇T Q j ,r j v j,ϵ , we also have that We will show that for ϵ ≪ 1 the function v ∞,ϵ is harmonic.First, for any test function ξ ∈ C ∞ c (B 2 (0)), we have However, by assumption, we also have where T Q j ,r j ω ± are the interior and exterior harmonic measures associated to T Q j ,r j v j .Note that T Q j ,r j ω − ̸ = ω − Q j ,r j , but, by Definitions 2.4 and 2.7 and Lemma 3.4, there is a constant ) are uniformly bounded by Theorem 2.6, the T Q j ,r j ω − (B 3 (0)) are, too.Thus, as j → ∞, we have that T Q j ,r j v j,ϵ ⇀ 0 in L 2 (B 2 (0)) as well.Thus, as ϵ → 0. Thus, v ∞ must satisfy the mean value property and is therefore harmonic. □ Now that we have Lemma 5.1, we can prove a quantitative rigidity result.Loosely speaking, it says that if a function v ∈ A(n, α, M 0 ) behaves like a homogeneous harmonic polynomial with respect to the Almgren frequency (in the sense that it has small drop across scales), then it must be close to being a homogeneous harmonic polynomial.This will connect the behavior of the Almgren frequency to our quantitative stratification.

Lemma 5.2 (quantitative rigidity). Let
Proof.We argue by contradiction.Assume that there exists a δ > 0 such that there is a sequence of functions v i ∈ A(n, α, M 0 ) with ∥ln(h i )∥ α ≤ 2 −i for which there exists a point By Lemma 5.1 there exists a subsequence T Q j ,1 v j which converges strongly in W 1,2 loc to a harmonic function v ∞ .Therefore N (Q, r, v ∞ ) is monotone increasing.Further, by Corollary 4.3 we know that lim j→∞ N (0, r, T Q j ,1 v j ) = N (0, r, v ∞ ) for all 0 < r ≤ 1.By Lemma 4.10 and the aforementioned convergence, we have that This implies that v ∞ is a homogeneous harmonic polynomial (see, for example, the proof of [Han and Lin 1994, Theorem 2.2.3]).Thus, we arrive at our contradiction, since the T Q j ,1 v j were assumed to stay away from all such functions in L 2 (B 1 (0)).

A dichotomy
The proof technique in the rest of the paper is an adaptation of techniques developed by Naber and Valtorta [2017].This section is dedicated to proving a lemma that gives us geometric information on the quantitative strata.Roughly, it says that if we can find (k+1) points that are well-separated and the Almgren frequency has very small drop at these points, then the quantitative strata is contained in a neighborhood of the affine k-plane which contains them and we have control on the Almgren frequency for all points in that neighborhood.This is a quantitative analog of the following classical result.Proposition 6.1.Let P : ‫ޒ‬ n → ‫ޒ‬ be a homogeneous harmonic polynomial.Let 0 ≤ k ≤ n − 2. If P is translation-invariant with respect to some k-dimensional subspace V and P is homogeneous with respect to some point x ̸ ∈ V, then P is (k+1)-symmetric with respect to span{x, V }.We shall use the notation ⟨y 0 , . . ., y k ⟩ to denote the k-dimensional affine linear subspace which passes through y 0 , . . ., y k .Lemma 6.2.Let v ∈ A(n, α, M 0 ) and 0 < ϵ be fixed.Let γ , η ′ , ρ > 0 be fixed, then there exist constants 0 < η 0 (n, α, E 0 , ϵ, η ′ , γ , ρ) ≪ ρ and 0 < β(n, α, E 0 , ϵ, η ′ , ρ) < 1 such that, if (2) there exist points {y 0 , y 1 , . . ., y k } ⊂ B 1 (0) ∩ ∂ ± satisfying y i ̸ ∈ B ρ (⟨y 0 , . . ., y i−1 , y i+1 , . . ., y k ⟩) and N (y i , γρ, v) ≥ E − η 0 for all i = 0, 1, . . ., k, and (3) ∥ln(h)∥ α ≤ η 0 , then, writing ⟨y 0 , . . ., Proof.There are two conclusions.We argue by contradiction for both.Suppose that the first claim fails.
That is, assume that there exist constants γ , ρ, η ′ > 0 for which there exists a sequence and points {y i, j } j satisfying (2) above, with ∥ln(h i )∥ α ≤ 2 −i , η 0 < 2 −i , and a sequence β i ≤ 2 −i such that, for each i, there exists a point By Lemma 5.1, there exists a subsequence v i such that T 0,1 v i converges to a harmonic function v ∞ in the senses outlined in the lemma.Further, by the compactness of [0, E 0 ], B 1 (0), and the Grassmannian, we may assume that 0} is a two-sided NTA domain with constant 2M 0 by Theorem 3.8.Note that the convergence given by Lemma 5.1 implies sup and for all y i and all r ∈ [γρ, 2].Thus, v ∞ is a 0-symmetric function in B 2 (y j ) \ B γρ (y j ) for each y j .By unique continuation, v ∞ is 0-symmetric with respect to y j for each j.
Because the y j ∈ B 1 (0) are in general position, by Proposition 6 ) must be nondecreasing in r .This proves the first claim.Now assume that the second claim fails.That is, fix β > 0 and assume that there is a sequence and points {y i, j } j satisfying (2) above, with ∥ln(h i )∥ α ≤ 2 −i and a sequence η i → 0 such that for each i there exists a point Again, we extract a subsequence as above.The function v ∞ will be harmonic and k-symmetric in B 1+δ (0), as above, and x i → x ∈ B 1 (0) \ B β (L).Note that by our definition of S k ϵ,η i (v i ) and the convergence in Lemma 5.1, x ∈ S k ϵ/2 (v ∞ ).Since v ∞ is k-symmetric and L is its k-dimensional spine, every blow-up at a point in B 1 (0) \ B β (L) will be (k+1)-symmetric.Thus, there must exist a radius r for which v ∞ is k + 1, 1  4 ϵ, r, x -symmetric.This contradicts the conclusion that x ∈ S k ϵ/2 (v ∞ ).□ Consider the following dichotomy: either we can find (k+1) well-separated points y i j with very small drop in frequency or we cannot.In the former case, Lemma 6.2 implies that the Almgren frequency has small drop on all of S k ϵ,η (v) (and we also get good geometric control).In the latter case, the set on which the Almgren frequency has small drop is close to a (k−1)-plane.In this case, even though we have no geometric control on S k ϵ,η (v), we have very good packing control on the part with small drop in frequency.We make this formal in the following corollary.Corollary 6.3 (key dichotomy).Let γ , ρ, η ′ ∈ (0, 1) and 0 < ϵ be fixed.There exist and ∥ln(h)∥ α ≤ η, then one of the following possibilities must occur: (2) There exists a (k−1)-dimensional affine plane L k−1 such that Remark 6.4.The former case is simply the conclusion of Lemma 6.2.In the latter case of the dichotomy, we know that all points in ) is almost monotonic and uniformly bounded, this can happen for each Q only finitely many times.

Spatial derivatives of the Almgren frequency
The main result of this section is Corollary 7.7, in which we estimate the difference between the Almgren frequency at nearby points.First, we need a preliminary estimate which extends one of the results of Lemma 4.7 to points , and p ∈ B s/3 (Q).Then we have the estimate Furthermore, for all 0 < s ≤ 1 2 and all 1 2 s ≤ r ≤ 2s, If we can show that, for all p and all 0 < s ≤ 1 2 , and that dist(x max ( p, s), ∂ ± ) ≥ δ(M 0 ) > 0, then If this can be shown, then recalling the doubling of harmonic measure on NTA domains, the above string of inequalities proves that H ( p, s, v ϵ ), H (Q, s, v ϵ ), and H Q, 1 2 s, v ϵ share a common lower bound.The common upper bound follows from a similar argument using Remark 3.1.That is, if we can show, for all p and all 0 < s ≤ 1 2 , that dist(x max ( p, s), ∂ ± ) ≥ δ(α, M 0 , ) > 0, then by Harnack chains we know that s n−2 and that Recalling the doubling of harmonic measure on NTA domains, the above string of inequalities proves that H ( p, s, v ϵ ), H (Q, s, v ϵ ), and H Q, 1 2 s, v ϵ share a common lower bound.This would prove the lemma.Let Q, p, and s be given.By the maximum principle for harmonic functions applied to v − in − , we have |v(x max ( p, s) By NTA estimates [Engelstein 2016, Lemma 5.4], we have Therefore, by the uniform Lipschitz estimates of Theorem 2.6 and Remark 3.1 we infer that dist(x max ( p, s) ± , ∂ ± ) ≳ M 0 , ,α s.
Therefore, we may use Harnack chains and estimate Thus, by the doubling of harmonic measure on NTA domains (see [Jerison and Kenig 1982]), we infer that |v(x max ( p, s) This proves the lemma.□ Remark 7.2.As a consequence of Lemma 7.1 and Corollary 4.4, we observe that if v ∈ A(n, α, M 0 ) then, for every 0 < r ≤ 1 2 and every point p ∈ B 1 (0) such that dist( p, ∂ ± ) ≤ 1 3 r and for any 0 where C = C(M 0 ).
Proof.Let p, Q, and s be as above.For sufficiently small 0 < ϵ, Chasing through the change of variables x = r y + Q, we see that Thus, we calculate that for the change of variables where the last two inequalities are because the v Q,r are uniformly locally Lipschitz, 1 + ϵ/r < 2, the ω − Q,r (B 2 (0)) are uniformly bounded for Q ∈ B 1 (0) and r < 2, and the doubling of harmonic measure on NTA domains.□ Lemma 7.4.Let v ∈ A(n, α, M 0 ), Q ∈ ∂ ± ∩ B 1 (0), and 0 < r .Then for p ∈ B r/3 (Q) and ⃗ v ∈ ‫ޒ‬ n such that |⃗ v| ≤ r , we calculate the spatial directional derivatives as follows: . (7-4) Proof.Equation (7-4) follows immediately from the preceding equations.The spatial derivative for H (Q, r, u) follows from differentiating inside the integral.To obtain the spatial derivative for D(Q, r, v), we recall the divergence theorem: Now, we focus upon the last term.Recalling Green's theorem and the fact that partial derivatives of harmonic functions are themselves harmonic, For the sake of concision, we define the following notation for v ∈ A(n, α, M 0 ), y ∈ , and radii 0 < r, R ≤ 2.
W r,R (y Proof.We begin by noting that Lemmas 7.3 and 7.1 give Now, we write the decomposition Therefore, plugging this into (7-4), we obtain for where for the purposes of this lemma We begin by estimating A. We rewrite } and all z ∈ ∂ B r ( p).Therefore, by Lemma 4.10, for all z ∈ ∂ B r ( p), Furthermore, we estimate by the divergence theorem Thus, we may give the following preliminary estimate on A: Focusing upon the term 2 Now we estimate B using Cauchy-Schwartz: For |C|, the same Cauchy-Schwartz argument plus Corollary 4.4 shows that .
This proves the pointwise estimate We prove the lemma by reversing the roles of Q and Proof.First, since for any 0 ̸ = c we have N (Q, r, v) = N (Q, r, cv), we shall assume for the purposes of this lemma that v = T 0,1 v.We shall show that where We estimate each term separately.
Bounding A. We begin by rewriting A: Observe that by Lemma 7.1 and Remark 4.12 we may using Hölder's inequality to estimate . Now, divide the spheres as follows: , where Notice that max Then, use the coarea formula for the function φ , where Thus, we obtain . Note that a simple calculation gives, for any 1 ≤ p < 2, Since T Q,r v ϵ is uniformly locally Lipschitz by Lemma 3.4, recalling Definition 2.7 and choosing p = 1 above, we see Bounding C. By Hölder's inequality (or Jensen's inequality for concave functions) and Lemma 7.1, we may reduce to considering Now, we change variables using Definition 2.7 and Lemma 7.1, and use Young's inequality to get Now, by Corollary 4.4 and Lemma 3.4, the numerator is bounded by a constant.Whence, by a calculation similar to (7-7), we obtain An identical argument holds for Q ′ in the place of Q.Thus, we have that |C| ≲ n,α,M 0 , 1.
Bounding B. Using Cauchy-Schwartz, Lemma 7.1, and the estimates of the term |C| above, we obtain .
Bounding D. Note that Chasing through the change of variables x = r y + Q, we see that Thus, we calculate that, for the change of variables where the last two inequalities are because the v Q,r are uniformly locally Lipschitz, 1 + ϵ/r < 2, the ω − Q,r (B 2 (0)) are uniformly bounded for Q ∈ B 1 (0) and r < 2, and the doubling of harmonic measure on NTA domains.Thus, by Lemma 7.1 we have Letting ϵ → 0, we see that D vanishes.Thus, putting together the estimates for A, B, C, D we have . This proves the lemma.□

Frequency pinching
In this section, we prove a "frequency pinching" result (Lemma 8.2) in the style of [De Lellis et al. 2018].This kind of result relates Jones' beta numbers to the drop in Almgren frequency.
Definition 8.1 (Jones' beta numbers).For µ a Borel measure, we define β k µ,2 (Q, r ) 2 as follows: where the infimum is taken over all affine k-planes.
Taking the infimum here -as opposed to the minimum -is a convention.The space of admissible planes is compact, so a minimizing plane exists.Let V k µ (Q, r ) denote a k-plane which minimizes the infimum in the definition of β k µ (Q, r ) 2 .Note that this k-plane is not a priori unique.Lemma 8.2 (frequency pinching).There exists a constant δ 0 = δ 0 (n, α, M 0 , ) > 0 such that, for any Before proving Lemma 8.2, we prove a few preliminary lemmas.We begin by noting that for any finite Borel measure µ and any B r (Q) we can define the µ center of mass by X = / B r (Q) x dµ(x) and define the covariance matrix of the mass distribution in B r (Q) by With this matrix, we may naturally define a symmetric, nonnegative bilinear form v n be an orthonormal eigenbasis and λ 1 ≥ • • • ≥ λ n ≥ 0 their associated eigenvalues.These objects enjoy the relationships Lemma 8.3.Let v ∈ A(n, α, M 0 ), and let Q ∈ ∂ ∩ B 1 (0) and 0 < r ≤ 1 4 .Let µ, Q, λ i , ⃗ v i be defined as above.For any i and any scalar c ∈ ‫,ޒ‬ Proof.Observe that by the definition of center of mass, for any ⃗ w ∈ ‫ޒ‬ n .Therefore, for any z for which ∇v(z) is defined, .

Squaring both sides and integrating over
Proof.First, we observe that Therefore, by the triangle inequality, Now, using Corollary 7.7 at scale r = |z − Q ′ | and the almost monotonicity of the Almgren frequency, we estimate where the term C r (Q, Q ′ ) is defined by We finish the proof by observing that Proof.We argue by contradiction.Assume that there is a sequence of functions v i ∈ A(n, α, M 0 ), Q i ∈ B 1/16 (0)∩∂ ± i , and 0 < r i ≤ 1 16 such that v i is (0, 2 − j , 8r i , Q i )-symmetric but not (k +1, ϵ, 8r i , Q i )symmetric.And, for each i, there exists an orthonormal collection of vectors {⃗ v i j } such that By Lemma 3.6, we may extract a subsequence T Q j ,r j v j for which T Q j ,r j v j converges to a nondegenerate function v ∞ .Similarly, {⃗ v i j } converges to an orthonormal collection {⃗ v i }.Given the assumptions above, v ∞ is also 0-symmetric in B 8 (0) and ∇v ∞ • ⃗ v i = 0 for all i = 1, . . ., k +1.Thus, v ∞ is (k+1, 0)-symmetric in B 8 (0).But, this is a contradiction, since the T Q j ,r j v j were supposed to stay away from (k+1)-symmetric functions in L 2 (B 1 (0)).□ 8A.The proof of Lemma 8.2.By Lemma 8.5 and properties of the Jones' beta numbers, we have, for {⃗ v i } the orthonormal basis and λ i the associated eigenvalues of the quadratic form in Lemma 8.3, By choosing c = N (Q, 7r, v) in Lemma 8.3 and recalling Lemma 8.4, we have Therefore, collecting constants and using Lemmas 4.9 and 7.1, we have y) 2 dµ(y).□

Packing
The following theorem of Naber and Valtorta [2017] is a powerful tool which links the sum of the β k µ (Q, r ) 2 over all points and scales to packing estimates.Theorem 9.1 [Naber and Valtorta 2017, discrete Reifenberg].Let {B τ i (x i )} i be a collection of disjoint balls such that, for all i = 1, 2, . . ., we have τ i ≤ 1.Let ϵ k > 0 be fixed.Define a measure and suppose that, for any x ∈ B 2 (0) and any scale l ∈ {0, 1, 2, . . .}, if B r l (x) ⊂ B 2 (0) and µ(B Then there exists a Now we are ready to prove the crucial packing lemma.
Lemma 9.2.Fix 0 < ϵ, and let v ∈ A(n, α, M 0 ) satisfy ∥ln(h)∥ α ≤ η and sup Q∈B 1 (0 There is an η 1 (n, α, M 0 , ϵ) > 0 such that if η ≤ η 1 , then for any r we have the packing estimate Proof.Choose δ 0 (n, α, M 0 , ϵ) as in Lemma 8.2, and γ (n, α, M 0 , δ 0 ) as in Lemma 5.2.Note that we may assume without loss of generality that η 1 ≤ 1, and so for C 1 (α, M 0 , 1) the constant in Lemma 4.9, let We will employ the convention that r i = 2 −i .For each i ∈ ‫,ގ‬ define the truncated measure We will write β i (x, r ) = β k µ i ,2 (x, r ).Observe that the β i enjoy the following properties.First, because the balls are disjoint, for all j ≥ i, Furthermore, for r i ≤ 2 −4 , recalling Lemma 4.10 our assumption of the Almgren frequency gives that Thus, for 0 < η small enough depending only upon α and M 0 , we have |W r j /2,16r j (Q ′ )| ≤ 1. Therefore In particular, by Lemmas 5.2 and 8.2 and our choice of η ≤ η 1 , The claim of the lemma is that µ 0 (B 1 (0)) ≤ C(n, α, M 0 , ϵ).We prove the claim inductively.That is, we shall argue that there is a fixed scale 0 < R = 2 −ℓ (depending only upon n, α, M 0 , ϵ) such that, for r i ≤ R and all x ∈ B 1 (0), Observe that since r Q ′ ≥ r > 0, for r i < r the claim is trivially satisfied because µ i = 0. Assume, then, that the inductive hypothesis holds for all j ≥ i + 1.Let x ∈ B 1 (0).We consider µ i (B 4r i (x)).Observe that we can get a course bound by writing µ j (B 4r j (x)) = µ j+2 (B 4r j (x)) + r k Q ′ , where the sum is taken over all Q ′ ∈ B 4r j (x) with r j+2 < r Q ′ ≤ r j .Since the balls B r Q ′ (Q ′ ) are disjoint, there is a dimensional constant c(n) which bounds the number of such points.Thus, we may take β j (z, r j ) 2 dµ j (z) |W r j /2,16r j (y)|dµ j (y) dµ j (z) µ j (B r j (y)) Therefore, recalling r i = 2 −i we see that Therefore Thus, for η ≤ η 1 (n, α, M 0 , ϵ) sufficiently small and r i ≤ R(n, α, M 0 , ϵ) = 2 −ℓ sufficiently small, For such i and µ i satisfying the hypotheses of Theorem 9.1, The discreet Reifenberg theorem therefore implies that µ i (B r i (x)) ≤ C D R r k i .Thus, by induction, the claim holds for r i ≤ R = 2 −ℓ .We may use a packing argument using balls of radius 2 −ℓ to obtain estimates at larger scales.That is, Covering control (C) follows from our choice of a maximal 2 5 r i -net at each scale i.If i is the first scale at which a point x ∈ S k ϵ,η R (v) was not contained in our inductively refined cover, it would violate the maximality assumption.
The last condition (D) follows because we stop only if j is the first scale for which r j ≤ R. Since we decrease by a factor of ρ at each scale, (D) follows.□ 10C.Bad trees.Let B r A (x) be a bad ball.Note that for every bad ball, there is a (k−1)-dimensional affine plane L k−1 associated to it which satisfies the properties elaborated in Corollary 6.3.Our construction of bad trees will differ in several respects from our construction of good trees.The idea is still to define an inductively refined cover at decreasing scales of B r A (x) ∩ S k ϵ,η R (v).We shall again sort balls at each step into "good", "bad", and "stop" balls.But these balls will play slightly different roles and the "stop" balls will have different radii.
We reuse the notation G i to denote the collection of centers of good balls of scale r i , B i to denote the collection of centers of bad balls of scale r i , and S i to denote the collection of centers of stop balls of scale r i .
At scale i = A, we set B A = x, since B r A (x) is a bad ball, and set S A = G A = ∅.Suppose, now that we have constructed good, bad, and stop balls for scale i − 1 ≥ A. If r i > R, then define S i to be a maximal 2 Note that η ≪ ρ, so ηr i−1 < r i .We then let {z} be a maximal 2 5 r i -net in We then sort {z} into the disjoint union G i ∪ B i depending on whether B r i (z) is a good ball or a bad ball.If r i ≤ R, we terminate the process by defining G i = B i = ∅ and letting S i be a maximal 2 .
The covering at which we arrive at the end of this process shall be called the "bad tree at B r A (x)".We shall follow [Edelen and Engelstein 2019] and denote this by T B = T B (B r A (x)).We shall call the collection of "good" ball centers, i G i , the "leaves of the tree" and denote this collection by F(T B ).We shall denote the collection of "stop" ball centers by S(T B ) = i S i .
As before, we shall use the convention that for g ∈ F(T B ) we let r g = r i for i such that g ∈ G i .However, note that now, if s ∈ S i ⊂ S(T B ), we let r s = ηr i−1 .
Theorem 10.4.A bad tree T B (B r A (x)) enjoys the following properties: B r g (g).
(D) Size control: for any s ∈ S(T B ), at least one of the following holds: Proof.Conclusion (C) follows identically as in Theorem 10.3.Next we consider the packing estimates.Let r i > R.Then, by construction, for any b ∈ B i−1 , we have Thus, since the points G i ∪ B i are 2 5 r i disjoint, we calculate We can push this estimate up the scales as follows: Summing over all i ≥ A, then, we have that Since we chose c 2 ρ ≤ 1 2 , we have that the sum converges and This proves (A).
To see (B), we observe that for any given scale i ≥ A + 1, the collection of stop balls {B ηr i−1 (s)} s∈S i form a Vitali collection centered in B r i−1 (B i−1 ).Thus, we have Since by construction there are no stop balls at the initial scale A, we compute that This is (B).
We now argue (D).For s ∈ S i where r i > R, by construction s ∈ B r i−1 (b) \ B 2ρr i−1 (L k−1 ) for some b ∈ B i−1 .By Corollary 6.3, the construction, and our choice of η ≤ 1 2 ρ, we have sup On the other hand, if This proves (D).□

The covering
Assuming that ∥ln(h)∥ α ≤ η/(2C 1 + 1), for 0 < η ≤ η 0 (n, α, E + 1, ϵ, η ′ , γ 0 , ρ) as in Section 10, we now wish to build the covering of S k ϵ,η R ∩ B 1 (0) using the tree constructions above.Note that B 1 (0) is either a good ball or a bad ball.Therefore, we can construct a tree with B 1 (0) as the root.Then in each of the leaves, we construct either good trees or bad trees, depending upon the type of the leaf.Since in each construction we decrease the size of the leaves by a factor of ρ < 1 10 , we can continue alternating tree types until the process terminates in finite time.
Explicitly, we let F 0 = {0} and let B 1 (0) be the only leaf.We set S 0 = ∅.Now, assume that we have defined the leaves and stop balls up to stage i − 1.Since by hypothesis, the leaves in F i are all good balls or bad balls, if they are good, we define for each f ∈ F i−1 the good tree T G (B r f ( f )).We then set Since all the leaves of good trees are bad balls, all the leaves of F i are bad.
If, on the other hand, leaves of F i−1 are bad, then for each f ∈ F i−1 we construct a bad tree T B (B r f ( f )).In this case, we set Since all the leaves of bad trees are good balls, all the leaves of F i are good.
This construction gives the following estimates.
Lemma 11.1.For the construction described above, there is an N ∈ ‫ގ‬ such that F N = ∅ with the following properties: (A) Leaf packing: (D) Size control: for any s ∈ S N , at least one of the following holds: Proof.By construction, each of the leaves of a good or bad tree satisfy r f ≤ r i .Thus, there is an i sufficiently large such that r i < R. Thus, N is finite.
To see (A), we use the previous theorems.That is, if the leaves F i are good, then they are the leaves of bad trees rooted in F i−1 .Thus, we calculate by Theorem 10.4 On the other hand, if the leaves F i are bad, then they are the leaves of good trees rooted in F i−1 .Thus, we calculate by Theorem 10.3 Concatenating the estimates, since we alternate between good and bad leaves, we have By our choice of ρ, The estimate (A) follows immediately.We now turn our attention to (B).Each stop ball s ∈ S N is a stop ball coming from a good or a bad tree rooted in one of the leaves of a bad tree or good tree.We have the estimates from Theorems 10.3 and 10.4, which give bounds packing both leaves and stop balls.Combining these, we get Recalling the dependencies of η gives the desired result.Property (C) follows inductively from the analogous covering control in Theorems 10.3 and 10.4 applied to each tree constructed.Property (D) is immediate from these theorems as well.
We now wish to apply Corollary 11.2 to the rescaled functions ṽi (x) = v(κ x + Q i ) in B 1 (0).However, a careful reader may object that ṽi is not in A(n, α, M 0 ), since it is possible that h(0) ̸ = 1.However, ṽi (x) = c h(0)u + (xκ + Q) − u − (xκ + Q), where by Remark 3.1 we can control 0 < c < ∞ by constants that only depend upon κ and α.Thus, by multiplying the positive part by a constant controlled by , α, and M 0 , we obtain a new function (which we also label ṽi ) which is in A(n, α, M 0 ).
We now construct the desired covering in B 1 (0) for each ṽi .Ensuring that c(n, α, M 0 , , ϵ) is sufficiently large, we may reduce to arguing for r < η.We now use Corollary 11.2 to build a covering U 1 .If every r x equals R, then the packing and covering estimates give the claim directly, since x ≤ c(n, α, M 0 , , ϵ).
If there exists an r x ̸ = R, we use Corollary 11.2 to build a finite sequence of refined covers U 1 , U 2 , U 3 , . . .such that, for each for each i, the covering satisfies the following properties: (A i ) Packing: If we can construct such a sequence of covers, then we claim that this process will terminate in finite time, independent of R. Recall that blow-ups of ṽi are homogeneous harmonic polynomials.Therefore N (Q, 0, ṽi ) = lim r →∞ N (Q, r, ṽi ) ≥ 1 for all Q ∈ ∂ ± .By Remark 4.8 we have that, for all 0 < r ≤ 1, N (Q, r, ṽi ) ≥ 1 − C(n, α, M 0 , , ϵ) for all p ∈ B 1 (0).Therefore, we know that, for i large enough that i > (C(n, α, M 0 , , ϵ) + C(n, α, M 0 , , ϵ) − 1) 2 η 0 , it must be the case that r x = R for all x ∈ U i .In this case, we will have the claim with a bound of the form R k−n Vol(B R (S k ϵ,η 0 R ( ṽi ) ∩ B 1 (0))) ≤ c(n, α, M 0 , , ϵ) C(n,α,M 0 , ,ϵ) .
Thus, we reduce to inductively constructing the required covers.Suppose we have already constructed U i−1 as desired.For each x ∈ U i−1 with r x > R, we apply Corollary 11.2 at scale B r x (x) to obtain a new collection of balls U i,x .From the assumption that r x ≤ 1 10 and the way that Hölder norms scale, it is clear that ṽi satisfies the hypotheses of Corollary 11.2 in B r x (x) with the same constants.To check packing control, we have that y∈U i,x r k y ≤ c(n, α, M 0 , , ϵ)r k x .
To obtain the cover which proves the theorem, we simply scale each covering of S k ϵ,η 0 R/κ ( ṽi ) ∩ B 1 (0) to a covering of S k ϵ,η 0 R (v) ∩ B κ (y i ) and sum over the c(n, α, M 0 , , ϵ) such balls which cover S k ϵ,η 0 R (v) ∩ B 1/4 (0).This completes the proof.□ by increasing our constant C(n, α, M 0 , , ϵ).This concludes the proof of Theorem 2.15.□ By Lemma 3.6 there exists a subsequence for which T Q j ,r j ∂ ± j converge locally in the Hausdorff metric to a limit set A ⊂ ‫ޒ‬ n .By Theorem 3.8, A must be the mutual boundary of a pair of two-sided NTA domains ± ∞ with constant 2M 0 .Furthermore, up to scaling and rotation, the space of homogeneous harmonic functions of two variables in ‫ޒ‬ n with 2 ≤ degree(P) ≤ k(M 0 ) is finite-dimensional.Since the space of rotations is compact, we may find a subsequence P j which converges to P ∞ locally in the Hausdorff metric for some (n−2)-symmetric P ∞ .This implies that (12-1) Indeed, if there were p ∈ i {W i,∞ : P ∞ > 0 on W i,∞ } ∩ B 1 (0) such that p ∈ − ∞ , since W i,∞ and − are open, there would exist a ball B δ ( p) ⊂ − ∩ W i,∞ .Therefore, since P j → P ∞ and T Q j ,r j ∂ ± j → A locally in the Hausdorff metric, for all j sufficiently large, B δ/2 ( p) ⊂ W i, j ∩ T Q j ,r j ∂ − j .This is a contradiction.The other equation follows identically.Now we claim that A ∩ B 1 (0) = P ∞ ∩ B 1 (0).Suppose not, then there exists a point p ∈ P ∞ with p ̸ ∈ A or there exists a point Q ∈ A such that Q ̸ ∈ P ∞ .In the former case, suppose dist(Q, A) > δ.Then B δ ( p) must intersect at least two contiguous connected components, W i,∞ and W j,∞ .Since they are contiguous, the sign of P ∞ must be positive on one and negative on the other.This contradicts (12-1).Similarly, if there exists a point Q ∈ A such that Q ̸ ∈ P ∞ then there exists a ball B δ (Q) which intersects both ± ∞ but which is contained in a single W i,∞ .This also contradicts (12-1).However, if P ∞ is (n−2)-symmetric with degree ≥ 2, then P ∞ does not divide ‫ޒ‬ n into two connected components.This contradicts our assumption that A = P ∞ was the mutual boundary of a pair of two-sided NTA domains with constant 2M 0 .Therefore, such a constant 0 < c = c(M 0 , , α) must exist.
This quantitative control allows us to define a quantitative stratification following [Cheeger and Naber 2013].