$\Omega$-symmetric measures and related singular integrals

Let $\mathbb{S} \subset \mathbb{C}$ be the circle in the plane, and let $\Omega: \mathbb{S} \to \mathbb{S}$ be an odd bi-Lipschitz map with constant $1+\delta_\Omega$, where $\delta_\Omega>0$ is small. Assume also that $\Omega$ is twice continuously differentiable. Motivated by a question raised by Mattila and Preiss in [MP95], we prove the following: if a Radon measure $\mu$ has positive lower density and finte upper density almost everywhere, and the limit $$ \lim_{\epsilon \downarrow 0} \int_{\mathbb{C} \setminus B(x,\epsilon)} \frac{\Omega\left((x-y)/|x-y|\right)}{|x-y|} \, d\mu(y) $$ exists $\mu$-almost everywhere, then $\mu$ is $1$-rectifiable. To achieve this, we prove first that if an Ahlfors-David 1-regular measure $\mu$ is symmetric with respect to $\Omega$, that is, if $$ \int_{B(x,r)} |x-y|\Omega\left(\frac{x-y}{|x-y|}\right) \, d\mu(y) = 0 \mbox{ for all } x \in \mbox{spt}(\mu) \mbox{ and } r>0, $$ then $\mu$ is flat, or, in other words, there exists a constant $c>0$ and a line $L$ so that $\mu= c \mathcal{H}^{1}|_{L}$.

We say that a measure is symmetric if it satisfies C(x, r) = 0 for all x ∈ spt(µ) and r > 0.
Here spt(µ) denotes the support of µ.Mattila showed that any symmetric locally finite Borel measure on C is either discrete or continuous.In the latter case, it is either the 2-dimensional Lebesgue measure (up to a multiplicative constant) or a countable sum of 1-dimensional Hausdorff measures restricted to equidistant affine lines.Mattila needed such characterisation to understand the geometry of tangent measures of a measure µ for which the Cauchy transform exists µ-almost everywhere (in the sense of principal values) -thus to understand the geometry of µ itself.Briefly after, Mattila and Preiss (see [MP95]) generalised this to the higher dimensional equivalent.Shortly after, Huovinen in [H97] proved a similar result for measures µ symmetric with respect to kernels of the type K(z) = z k |z| k+1 , where k is an odd integer.Measure symmetric with respect to more general kernels appeared recently in the works of Jaye and Merchán (see [JM18a] and [JM18b]), where the authors give some necessary and sufficient conditions for the principal value integral of the corresponding kernel to exist.
Let us now introduce the main results of this note; after we will give a few remarks on its possible applications.Consider a measure µ in C, and let K be a measurable map C \ {0} → C \ {0} given by K Ω (x) = K(x) = |x|Ω x |x| , where Ω : S → S.
Remark 1.2.One could similarly consider kernels K given by K(x) = ρ(|x|)Ω(x/|x|).However Lemma 2.5 below says that this would give an effectively more general result.
Definition 1.3 (Ahlfors regularity).A measure µ on C is called Ahlfors 1-regular (or Ahlfors-David regular) with constant C 0 > 0 if for all x ∈ spt(µ) and all r > 0, we have that Our aim here is to prove the following result.
Theorem 1.4.Let K : C \ {0} → C \ {0} be given by K(x) = |x|Ω x |x| , where Ω : S → S is an odd, twice continuously differentiable map which is also bi-Lipschitz with constant 1 + δ Ω (with respect to the geodesic distance on the 1-sphere).Let µ be an Ahlfors 1-regular, Ω-symmetric measure in C.Then, if δ Ω is sufficiently small (smaller than some absolute constant), µ = cH 1 | L for some line L.This result can be considered a first step towards answering a question posed my Mattila and Preiss in [MP95]: given a measure µ in R d with positive n-dimensional density, the existence of the principal values for singular integrals with kernels like K(x) = Ω(x/|x|) |x| n implies that the tangent measures ν of µ satisfy an Ω-symmetricity condition: ˆB(x,r) |x − y|Ω((x − y)/|x − y|) dν(y) for x ∈ spt(ν) and r > 0. (1.4)Note however, that our assumption of Ahlfors-David regularity is really strong.Hence a more significant step in answering the question of Mattila and Preiss would be to get rid of such an assumption.In any case, in the last section of this note we prove the following application.
Corollary 1.5.Let K be as in the statement of Theorem 1.4 and let µ be a Radon measure in the plane such that for µ-almost all x ∈ C the following conditions hold.
• The lower density is positive and the upper density is finite, that is exists and is finite.Then µ is 1-rectifiable.
Remark 1.6.The assumption on the lower density is somewhat unsatisfactory.A natural question is therefore whether Corollary 1.5 still holds without such an assumption.
Remark 1.7.The converse of Corollary 1.5 holds, too, and it is a result of A. Mas ([Mas13], Corollary 1.6).Mas proved that if a measure on R d is n-rectifiable, then the principal values exists for a large class of kernels, namely, for odd kernels with derivative bounds |∇ j K(x)| 1 |x| n+j , j = 0, 1, 2 and for all x ∈ R d \ {0}.
A second application will be presented in an upcoming paper [V19a], where we give a new characterisation of uniform rectifiability.In this context, the assumption of Ahlfors regularity is natural.We refer the reader to [V19a] for more details.
1.1.Outline.The core of this note is showing that if a measure µ is Ω-symmetric then its support lies in some line L. To do so, we will use techniques from [T08a] and [T08b] to obtain a bound for the sum of the Jones β over all cubes contained in some top cube Q 0 .Hence, we will show that such sum goes to zero as we increase the size of Q 0 to infinity.Such a strategy is a modification of some ideas in [MP95], Section 6. 1.2.Acknowledgements.I would like to warmly thank Jonas Azzam for his patience and kind support.I also thank Xavier Tolsa and Joan Mateu for several useful conversations.

Preliminaries
2.1.Notation.We gather here some notation and some results which will be used later on.We write a b if there exists a constant C such that a ≤ Cb.By a ∼ b we mean a b a.
For sets For a point x ∈ R n and a subset We write We will call S the 1-sphere in C; for two points x, y ∈ S, we denote by d S the geodesic distance on S. (2.1) Remark 2.1.Throughout this note, the notation M • x will be used to indicate a matrix M acting on a vector x -this will never be used to denote the dot product (which we will denote with the standard •, • ).
The 1-dimensional Hausdorff measure of A is then defined by 2.2.Jones β numbers.Let µ be a 1-Ahlfors regular measure (see (1.3)).We define were the infimum is taken over all affine lines P .This quantity measures how far the support of µ is from being a line.Its relation with problems involving singular integrals and rectifiability has a fairly long history.Variants of such coefficients were firstly introduced by Jones in [J90] (hence Jones-β-numbers) while working on the Analyst's traveling salesman problem in the plane (see also [Ok92] and [S07] application of these coefficients in the euclidean space and in Hilbert space); shortly after David and Semmes introduced a variant of these coefficients (and this is the variant which we will use) to develop their theory of uniformly rectifiable sets (see for example [DS91] and [DS93]); the Jones coefficients have been extensively used in recent years within Geometric Measure Theory; see for example the Reifenbergtype parameterisation results by David and Toro in [DT12] (see also Ghinassi in [G17]) and by Edelen, Naber and Valtorta [ENV16]; the series of Tolsa and Azzam and Tolsa [T15], [T17] and [AT17]; the series of Badger and Schul [BS15], [BS16], [BS17].
2.3.Intrinsic cubes with small boundaries.The following construction, due to David in [Dav06], provides us with a dyadic decomposition of the support of an AD-regular measure.Such construction has been extended by Christ in [C90] to spaces of homogeneous type and further refined by Hytönen and Martikainen in [HM12].Here is the construction.
Theorem 2.2.Let µ be an n-AD regular measure in R d .There exists a collection D µ of subsets Q ⊂ spt(µ) with the following properties.
(1) We have where D j µ can be thought as the collection of cubes of sidelength 2 −j .(2) For each j ∈ Z, For a proof of this, see Appendix 1 in [Dav06].
We will denote the center of Q by z Q .Furthermore, we set For a cube Q ∈ D µ , we set (2.4) 2.4.Balanced points.The following Lemma will be very useful; it holds for a general n-ADR measure in R d but we state it taylored to our context.Lemma 2.4 ([DS91], Lemma 5.8).Let µ be an Ahlfors 1-regular measure with constant C 0 in R d .There exists a constant η = η(C 0 ) < 1 so that for each cube Q ∈ D µ we can find two points x 0 , x 1 ∈ Q so that We may refer to x 0 and x 1 as the 'balanced points', and the unique line they span as the 'balanced line'; we will denote such line by L Q .
(2.6) Note that the same lemma holds for any ball B(x, r) centered on spt(µ); in this instance we will denote the balanced line as L x,r .

Splitting the kernel
Let us fix a cube Q ∈ D µ and a constant A > 1 which will be bounded above later.In the following, we will assume that Remark 3.1.An obvious fact, which we will use over and over, is that the map K(x) = |x|Ω(x), because it is anti-symmetric, maps a line trough the origin to another line through the origin; this is false d-dimensional affine planes in R n , say, and it is one of the difficulties to go through to generalise the arguments which we use here beyond the plane.
Remark 3.2.From now one, we take Ω so that it satisfies all the hypotheses of Theorem 1.4, that is, Ω is an odd, C 2 maps from the circle to the circle, which is moreover bi-Lipschitz with constant 1 + δ Ω , and δ Ω is sufficiently small.We start with an elementary lemma.
Lemma 3.3.Let Ω : S → S be a bi-Lipschitz function of the sphere.Then there exists a unique bi-Lipschitz function ω : Proof.It is well known that there exists one such continuous function ω whenever Ω is continuous.We show that ω is also bi-Lipschitz.For t = s points in [0, 2π], we have that (recall the definition of d S in (2.1)), The reverse inequality is shown in the same way.
The next two lemmata simply say that the dot products change little under Ω.
Lemma 3.4.Let y ∈ C and let L be a line through the origin; denote by ν a normal unit vector to L and by ν a normal unit vector to K(L).Then Proof.Note that if one term in (3.3) is zero, the other one must also be zero; if | y, ν | = 0, then y ∈ L, and since K(L) is perpendicular to ν, | K(y), ν | = 0. Similarly, if | K(y), ν | = 0, then K(y) ∈ K(L), and thus y ∈ L. Hence we may assume that neither of the terms in (3.3) vanishes.Let us denote by e L the unit vector which spans L and set θ L to be the angle that e L makes with the positive real axis; we let θ y denote the same thing for some other vector y ∈ C. We also set α := |θ y − θ L |; using symmetry, it is enough to consider α such that 0 ≤ α ≤ π/2.Let us consider two cases separately.
Case 1. Suppose first that Using basic properties of trigonometric functions we see that Case 2. We consider the case where We have , and so in particular, if δ Ω < 1, sin(|ω(θ y ) − ω(θ L )|) is nonzero.This is one of the reasone we need δ Ω to be small.If it is too large, the comparability of dot products as in (3.3) fails.Moreover, as expected, the constant in (3.3) worsen as δ Ω → 1, see (3.8).By Taylor's expansion, we have for some constant C; to satisfy (3.7), we require Note that for α ∈ [0, π/2], α −1 tan(α) ≥ 1; hence to make δ Ω close to one, we need to make C very large.If, on the other hand π > ξ > π/2, then it is immediate that (3.7) holds (with C = 1).
One can prove in the same manner.
Lemma 3.6.Let y ∈ C and let L be a line through the origin; denote by ν a normal unit vector to L and by ν a normal unit vector to K(L).Suppose that, for some Then there exists a constant C, depending on δ Ω , so that Moreover, if e L is a unit vector so that span(e L ) = L, and if |y| ≥ y, e L ≥ |y|δ 0 , (3.11) then there exists a constant C, depending on δ Ω so that Proof.The arguments will be similar to those used in Lemma 3.4, the only difference being that we do not have absolute values, thus one must be a little careful about changes of sign.Let us sketch a proof.Suppose first that the angle made by y and L is between 0 and π 2 .This can be done without loss of generality by symmetry.Using the same notation as in Lemma 3.4, suppose first that The because y, ν ≥ |y|δ 0 , we can use the computation (3.4), and obtain (3.10).
On the other hand, if then we can compute as in (3.5)-(3.8) to obtain (3.10) also in this case.The inequalities in (3.12) can be proven in exactly the same way if we assume (3.11) instead of (3.9).
We now use Taylor expansion to split the difference of C Ω,φ (x, r) and C Ω,φ (0, r) into two terms, one linear and the other which can be controlled in an appropriate manner.
Lemma 3.7.Let 0, x ∈ Q; let φ : R → R + be a C 2 , radial function; we have where T (x) is a term linear in x and E(x) is controlled approriately.Remark 3.8.We have exact expressions for T and E, see (3.15) and (3.16) below; we will refer to T as 'the linear term' and to E as 'the error term'.
Proof.We Taylor expand both the smooth cut off and the kernel around 0 as follows.First, we see that where ξ x,0 is contained in the line segment joining x and 0 and Now, we let Again by Taylor's expansion, we have that Applying this to the above difference, we get the following terms: 3.0.1.Splitting of A. We now expand the kernel to obtain To ease the notation, set Similarly, we expand K(x − y) in B to obtain the terms We further split B 1 as Moreover, we split B 1,2 as Now we set (for fixed r > 0),

Case when β's are small
In this section and in the next one we will obtains certain bounds on the β numbers (as defined in (2.2), with p = 2).The method that we use come from a paper of Tolsa, [T08b].We have a few more terms to deal with because our kernel is not linear.
We first consider the case where our measure is already quite flat, that is, the β numbers are small.In the next section we will look at the case where the β numbers are instead large.
Let A be as in (3.1), let N ∼ log(A), and let Q ∈ D µ so that r ∼ 2 N ℓ(Q) ∼ Aℓ(Q); let x 0 and x 1 be the balanced points in Q guaranteed by Lemma 2.4.We will assume throughout this section that where τ > 0 is a small constant which will be fixed later.
It will be convenient for us to define the C Ω,φ quantity (recall the definition in (2.7)) in terms of a smooth cut off of the annulus A 0, 1 2 , 2 : let χ 1 2 be a smooth radial function supported on B(0, 1) and so that Then we see that, for r > 0, φ x−• r 2 is supported on A x, r 2 , 2r .Recall Lemma 2.5 and Remark 2.6: C Ω,φ (x, r) = 0 if and only if C Ω (x, r) = 0; throughout this section we will work with the quantity C Ω,φ with φ as in (4.2).

Lower bounds on the linear component T.
We denote by Π L the standard orthogonal projection onto the line L.
Lemma 4.1.Fix a cube Q and let x 0 and x 1 be two balanced points of Q and let L Q be the corresponding balanced line (see (2.6) for definitions).Let T be defined as in (3.15), with φ as in (4.2).For any z ∈ 3Q \ L Q , set We will prove this lemma through some sublemmata.But first, note that without loss of generality, we can work with x 0 = 0; (4.5) note also that e z is perpendicular to L Q .
Let z ∈ 3Q and set z := Π LQ (z).Let us look at the quantity (Recall the definition of T in (3.15)).We will first take care of the term containing A 2 and later the one containing B 1,2,1 .
4.1.1.Lower bounds on A 2 .Recall that Remark 4.2.Consider any v ∈ C. If by the product rule we write we see the following.
• First, Now if we split v into the component parallel to the span of y and its orthogonal complement (let us denote them by v (y) and v ⊥(y) , respectively -we may just write v and v ⊥ when the y-dependence is obvious), then clearly • Second, let us consider the second term on the right hand side of (4.6); the differential DΩ is a map therefore we have that We choose this notation to emphasise that e z (as defined in (4.3)) and e x1 together span C; they are in fact an orthonormal basis.On the other hand, we defined y/|y| as ŷ to emphasise the fact that it is an element of S.
For future reference, we summarise these remarks in the following Sublemma; it follows immediately from (4.6), (4.7) and (4.10).
Sublemma 4.4.Keep the notation as above.Then Our next short term goal is to find a lower bound for the first term of T (as in (3.15)).Sublemma 4.5.Let δ 0 be a constant sufficiently small; recall that x 0 = 0 and x 1 are two balanced points in Q and L Q is the corresponding balanced line (see (2.6)).Let ν denote the vector Ω(x 1 ) rotated by 90 degrees counter-clockwise (we are just choosing the orientation of the normal to K(L Q ), so that it agrees with (4.8)).Then For the sake of clarity, we consider a few cases separately.Denote C + the half space 'above' L, and C − the half space below.Denote Q 1 the first quadrant in the plane with basis e x1 , e z , and Q 2 the second quadrant.
• Suppose that y, z ∈ C + .Under these assumptions, we look first at the integrand of I: suppose that |y| ≥ y, e z ≥ |y|δ 0 , (4.15) From Lemma 3.6, in particular (3.10), one can see that for an appropriate choice of δ Ω (for example δ Ω < 1 10 will be more than enough), we have where C is a constant that increases as we make δ Ω close to one.
With the same arguments as above, we obtain that Thus, whenever y, z ∈ C + , and we have that y, z are so that (4.15) and (4.17 where we are still integrating with respect to φ(|y| 2 /r 2 ) dµ(y) r 2 .The first integral can be bounded below using the previous discussion (namely (4.16), (4.20), (4.21), (4.22)), and using symmetry: we obtain As for the other integrals, we see that This implies that, for y ∈ F and δ 0 sufficiently small, we have that ŷ, e z Ω(y), ν + ŷ⊥ , e z d ŷ Ω • ŷ⊥ , ν (4.25) where we also used (4.18).Thus, and similarly for ´H .Now using the lower regularity of µ, we obtain the lemma.
Sublemma 4.6.Keep the notation as above.In particular, recall that x 0 = 0 and x 1 are balanced points of Q (and so 0, x 1 ∈ L Q -which is the balanced line), that z ∈ 3Q, e z , as given in (4.3) is perpendicular to L Q and that ν is the normal vector to (4.27) Proof.As before, we split the integral into the radial derivative and the spherical one and we apply Lemma 4.4: Note that since d ŷ Ω • ŷ⊥ ∈ T Ω(y) S and Ω(y) ∈ T Ω(y) S ⊥ , we have Let us now compute I(e z ), ν .We have The second inner product in the integral can be re-written as ŷ, e z r −1 K(y), r −1 Ω(y) = ŷ, e z |y| r 2 = y, e z r 2 , and so we obtain It immediately follows from Lemma 3.4 that This proves the Sublemma.
Proof.This follows at once from Sublemma 4.5 and Sublemma 4.6.
We divert for a moment from the main argument, to delve a little more in the choice of the balanced cubes x 0 and x 1 .The following definition will be used later on.For a point y ∈ spt(µ), set Lemma 4.8.There exists a constant c * > 0 such that the following holds.Let Q ∈ D µ and let P (Q) be the line minimising β µ,2 (Q); we can choose x 0 , x 1 as in Lemma 2.4 so that first, and second, Proof.Let y 0 , y 1 ∈ Q be the balanced points guaranteed by Lemma 2.4 and, for a constant 0 < η < 1 to be fixed later, let B j denote B(y j , ηℓ(Q)) ∩ Q.For two constants c, c ′ > 0, we set By Chebyshev inequality and Cauchy -Schwarz, we have, for j = 0, 1, Note that by Ahlfors regularity, we have that µ(B Q ) 1 η µ(B j ).Similarly, we have We want to show that, for c, c ′ large enough, we have that F j ∩ G j = ∅; we argue by contradiction and we assume that for all c, c ′ the intersection is empty.By (4.36) and (4.37), we have This is clearly a contradiction.Thus for c, c ′ large enough (depending only on η and c 0 ), we see that F j ∩ G j = ∅.Taking c * = c * (η, c 0 ) := max{c, c ′ } we end the proof of the lemma.
Proof of Lemma 4.1.Recall the notation for balanced lines as in (2.6) and the lines below it.We see that and, using Lemma 4.8, that dist with N ∼ log 2 (A).Recalling that µ is Ahlfors 1-regular, we have that Using Corollary 4.7, we finally obtain An appropriate choice of τ and δ 0 gives the lemma.4.2.Upper bound on the (nonlinear) term E. Keep the notation as above.In this subsection we will prove an upper bound for the error term E in terms of distance to the balanced plane L Q .
Lemma 4.9.Let x 0 , x 1 be the balanced points given by Lemma 2.4 and chosen as in Lemma 4.8.We have that We will prove this lemma in the following several paragraphs; each paragraph corresponds to a piece of E. Let us recall that where the various terms were defined in Section 3.
Remark 4.10.Without loss of generality, we may let x 0 = 0; also, to ease the notation we let x 1 = x.

4.2.1.
Estimates on A 3 .Recall that To estimate this term, we want to understand the integrand For each i = 1, 2, and writing ξ := ξ x,0 − y, we have We write we write II and III in the same manner.
Now, on one hand, using Lemma 3.4 we see that , and therefore, 1 r Note that here we have used that by the definition of the smooth cut off φ (see (4.2). 4.2.3.Bounds for II.We now look at II(ξ, x), ν .To do so, we consider first the term (D|ξ|) T DΩ i (ξ).The other one can be dealt with in the same fashion.First, notice that, for i = 1, 2, Thus we have that (4.47) Let us look at the second inner product above: we may write Then recall Remark 4.2 and in particular the expression for DΩ(ξ) in (4.10): we see that since | ξ| = 1 by definition, and Ω has Lipschitz constant close to one.Then, denoting by x ⊥ the vector x rotated by 90 degrees clockwise (so that it is parallel to e z , as defined in (4.3)), we obtain On the other hand, as far as the first inner product in the right hand side of (4.47) is concerned, we immediately see that it's bounded above by |x|.Thus, using again (4.46), Since ξ = ξ 0,x − y, and ξ 0,x ∈ [0, x] ⊂ L Q , we also have that dist(ξ, L Q ) = dist(y, L Q ).We therefore conclude (together with the remark that one can deal with the second term in II(ξ, x) in exactly the same way) that 1 r For this part, we will distinguish between real and imaginary part, rather than using the indices 1 and 2; also, we will do the computations using y as variable (instead of ξ), to make the notation less cumbersome.
Using Lemma 3.3, we can write The same holds for D 2 ℑΩ(y) when we replace cos with sin.We compute 2 + (D cos(ω(arg(y))))ω ′ (arg(y))∂ 2 1 arg(y).The same holds for ∂ 2 2 cos(ω(arg(y))) if we replace ∂ 1 with ∂ 2 .We also have the mixed derivatives Clearly the same calculations hold when computing D 2 ℑΩ(y): we replace cos with sin.Now we split the matrix D 2 ℜΩ(y) into three matrices, corresponding to the three terms in the sum above.We denote the first one as M ℜ 1 , the second one as M ℜ 2 and the third one as And also And also 3 ) 2,2 = (D cos(ω(arg(y))))ω ′ (arg(y))∂ 2 2 arg(y).We will denote the corresponding matrices for ℑΩ(y) by M ℑ i , i = 1, 2, 3. Thus we may split the integral in (4.50) as Estimates for A 3,1 .Let us introduce the following notation. [11] Then we can write where the dots stand for ω(arg(y)).We compute: where we used again that dist(ξ, L Q ) ≤ dist(y, L Q ).
By Taylor's theorem, we have that where ζ 1 , ζ 2 ∈ [θ y , θ x ] and θ y (resp.θ x ) is the angle that y (resp.x) makes with the real axis.Hence we may write And also we have

y). (4.57)
Estimates for A 3,3 .We see that we may re write If we expand as above y 1 and y 2 in terms of x 1 and x 2 respectively, we obtain that It is immediate to see that B has the very same structure.Notice that for i = 1, 2, 4 we have that |A i |, |B i | |y||x| 2 dist(y, L Q ); on the other hand, for i = 3, 5, we have that This shows the desired bound for A 3,3 and so for A 3 .
4.2.5.Estimates on B 1,1 .Recall that also recall that since x ∈ Q and y ∈ A(0, r/2, 2r), where r ∼ Aℓ(Q), and A > 0 is a large constant, then |x| |y| ≤ 1, and so |x| 4 |y| 2 ≤ |x| 2 .Finally, we see that 4.2.6.Estimates on B 1,2,2 .We want to compute On the other hand, using the same computation as for the estimate (4.58), and in particular the bound |D 2 K(y)| ≤ |y| −1 , and also (4.46), we see that Hence we obtain that Recall the definition of [SO] in (3.13) and that of [DP ] in (3.14); we first estimate the common term DK(−y) • x, ν .As before, we write it explicitly: On one hand, we see that Recall that, because ν is the normal unit vector to K(L Q ), we can apply Lemma 3.4, so to see that On the other hand, using once more Remark 4.2, an in particular (4.10), we see that DΩ(y) ; and so, arguing as in (4.48), Putting together (4.60), (4.61) and (4.62) for one term, and (4.63) for the other, we see that Thus the estimate (4.58) of the term SO let us conclude that The same bound can be obtained for | B 2,2 (x), ν by using the estimate (4.59) on the term DP .All in all we obtain ˆB(0,r) dist(y, L Q ) dµ(y).(4.65) 4.2.8.Estimates for B 3 .This term has the same form as A 3 , the only difference being the presence of DP and SO.However, from (4.58) and (4.59), we see that both are ≤ 1. Indeed recall that, for example, |DP | ≤ |x| 2 r 2 ; but x ∈ Q, and r ∼ Aℓ(Q), where A is large.Hence |DP | ≤ 1 and the same holds for |SO|.Thus we obtain the bound as in Subsection 4.2.1.4.2.9.Estimates for C. We Taylor expand K(x − y) as before and split up C(x) consequently so that The proofs that we gave for the previous terms hold for these ones, too.Let us briefly sketch them in this situation.
• The term C 1 can be estimated as it was done for B 1,1 , see Subsection 4.2.5;indeed, it is immediate from Lemma 3.4 that Recall also that the definition of φ in (4.2) implies therefore that φ ′′ (ξ s0,s ) = 0 only if y ∈ A 0, r 4 , 4r .Thus one can estimate DP and SO as it was done in (4.58) and (4.58).This gives the desired bound • Let us now look at C 2 (x): this can be estimated like B 2 , see Subsection 4.2.7.Note the similarities: both C 2 and B 2 have as main integrand DK(y)• x, which can be estimated as in (4.60), (4.61), (4.62) and (4.63).Moreover, we have seen in the remark above that the presence of ξ s0,s in the argument of the smooth cut off does not cause trouble, and therefore SO and DP have the same estimates as in (4.58) and (4.59).• Finally C 3 can be estimated as B 3 (i.e. as A 3 ), see Subsection 4.2.1.
Hence we have the bound These estimates together prove Lemma 4.9.4.2.10.A further estimate on the error term.We now want to prove a bound on E when evaluated at any point in 3Q (not just at a balanced point).
Lemma 4.12.Keep the notation as above; in particular z ∈ 3Q, x 0 , x 1 are balanced points and L Q is the balanced line.Then Proof.Again, we assume x 0 = 0.The proof of this lemma is very similar to the one for Lemma 4.9.
• For the terms B 1,1 , B 1,2,2 , C 1 the proof goes through verbatim, since we only used that |x 1 | ≤ r, and the same holds for |z| (recall that r ∈ [Aℓ(Q), 2Aℓ(A)], and A is a large constant).• On the other hand, let us consider the term B 2 (z), ν : Now the first term is easily taken care of, as D|y|Ω(y) • z, ν = D|y|, z Ω(y), ν .On the other hand, recalling (4.10), we have Denote by z the projection of z onto L Q .We see that since z ∈ 3Q and z ∈ L Q , which is the span of x 1 .Moreover, as in (4.63), we have that | x 1 , ŷ⊥ | dist(y, L Q ).With these considerations, and recalling that |SO| |z| 2 r −2 (see (4.58)), we obtain the bound where in the last term we used the Ahlfors regularity of µ. • Let us sketch a proof of the bound for | A 3 (z), ν |; it is almost verbatim the same as that for | A 3 (z), ν |.First, recall that Put ξ := ξ z,0 − y.As we did in Subsection 4.2.1, we split the integrand z T D 2 (ξ)z into three pieces, so to have See (4.43) and the expression above it.First, we can see that Moreover, using Lemma 3. Recalling that z ∈ 3Q (and so |z| ∼ r/A)), that |ξ| ∼ r and the Ahlfors regularity of µ, we then see that This is as far as the term I(ξ, z) is concerned.
Let us now look at | II(ξ, z), ν |.Once again, the proof given in Subsection 4.2.3 works here as well: the only difference is that one ends up with an extra term involving dist(z, L Q ) as for I(ξ, z), ν .We spare the reader the details.
Finally, we look at the term III(ξ, z), ν .Recall that this term was subsequently split into three further terms, which we called A 3,1 , A 3,2 and A 3,3 .
Estimate for A 3,1 .If we follow the same computation as for the corresponding term in Subsection 4.2.4,we end up considering again the quantity | Ω(ξ), ν |, as in (4.53).We can deal with this term as we did in (4.70) and then obtain an estimate like (4.71).
Estimates for A 3,2 .We want to carry out the same computations that were carried out in the corresponding part of Subsection 4.2.4.Note however that to do so, we need to assume that |z| = 0 (in particular, see (4.55)).This can be done without loss of generality: µ is Ahlfors regular, and thus µ({z}) = 0. We can then follow through the computations up to equation (4.56), which in our present situation looks like By triangle inequality, we have that With the second term, one can carry on as above, see (4.57) and the short paragraph above it.We can deal with the first term as follows.First, note that by the assumption (4.1), we see that But recall that ξ = ξ z,0 − y, and thus (as in (4.70)) | ξ, e z | | z, e z | + dist(y, L Q ).
One can then proceed as below (4.70).• The proof for the term B 3 (z), ν is just the same as for the term B 3 (x 1 ), ν , taking also into account the slight modifications used above.• Finally, the term C(z), ν may be split into C 1 (z), ν , C 2 (z), ν and C 3 (z), ν ; these ones may be dealt with as in the previous sections (bearing in mind the remarks made above).
Lemma 4.13.If x 0 and x 1 are balanced points of Q, with balanced plane L Q , then, for z ∈ 3Q, we have Proof.Again, let us assume that x 0 = 0. Since µ is an Ω-symmetric measure and 0, x 1 ∈ spt(µ), we have that where N ∼ log(A).Hence we deduce that 1 r ˆB(x0,r)

Case when β's are large
In this section we will assume that (4.1) doesn't hold, and so that we have where N ∼ log(A), and A ≥ 1 is a (large) constant to be fixed later on.Let ϕ be a function so that ϕ is twice continuously differentiable (5.2) Let us fix some notation which will be used throught the sections below.Let ϕ, R Ω µ as above, and let µ be an Ahlfors 1-regular Ω-symmetric measure in C; fix a µ-cube Q with side length ℓ(Q) and let r ∈ [Aℓ(Q), 2Aℓ(Q)], where A is as above; denote by L Q the balanced line given in Lemma 2.4 and x 0 , x 1 the two corresponding balanced points.
Lemma 5.2.With the notation above (in particular r ∼ ℓ(Q)A, A > 1 a large constant ), we have that for any z ∈ 3Q, (5.6) We will prove Lemma 5.2 in a few subsections.Without loss of generality, we will assume that x 0 = 0, and we will also set x 1 = x and e x := x |x| .
Recall also the notation e z defined in (4.3): if z ∈ 3Q and z is the orthogonal projection of z onto L Q , then we put The reader should also keep in mind the notation used in Remark 4.2 (in particular that of ŷ and ŷ⊥ ).
As in Lemma 3.7, we split the difference R(0, r) − R(x, r) by Taylor expansion, where x ∈ Q \ ηQ belongs to the line L Q (here η is as in (2.4)).Note however that we do not expand the denominator 1 |x−y| .Thus we obtain several terms, with which we then construct T (x) and E(x) so that 0 = R(0, r) − R(x, r) = T (x) + E(x).
(5.7) See (3.15), (3.16) and the terms before for definitions; the reader should bear in mind however, that rather than having a weight of the form 1 r 2 we now have one of the form 5.1.Lower bounds on T. Recall from (3.15) that this is given by , where (5.8) y) • x r dµ(y).(5.10) 5.1.1.Bounds for B 1,2,1 .Note that for any vector z ∈ C, we can split (as usual using (4.12)) the dot product in the square brackets in (5.10) as follows: where the constant behind depends on the Lipschitz constant δ Ω of Ω.
Proof.The proof of this is the same as in Sublemma 4.5, i.e. it mainly follows from Lemma 3.6.Let us give a couple of remarks.First, even if Ω(e x ) = ν (where ν is the unit normal to K(L Q )), the proof given for Sublemma (4.5) works for Ω(e x ) as well.Indeed the reason why we used ν in Section 4 is to obtain a bound in terms of dist(y, L Q ) for the error term -in that case we needed ν, and Ω(e z ) would not work.Second, note that with for ϕ as in (5.2) and (5.3), the derivative ϕ ′ will be non negative and supported on an annulus A 0, 1 2 , 2 .Thus ϕ ′ effectively has all the property of φ (as in (4.2)) that were used in the proof of Sublemma 4.5.5.1.2.Bounds for A 2 .Keeping the notation as above, let us prove the following.Sublemma 5.4.
Once again, we split the domain of integration into subsets G, F and H, as defined in (4.24) and we argue as for I. Thus This together with the estimate for I, (5.16), proves the sublemma.
We now have that | T (e x ), Ω(e x ) + T (e z ), Ω(e z ) | δ 2 0 r −1 .(5.18) 5.2.Upper bounds on the error term E. Lemma 5.5.Keep the notation as in the preceding subsection.Then Proof.The proof of this is similar to the one given for Lemma 4.9 -actually it's easier: we do not need to bound in terms of the distance to L Q , but only in terms of absolute values.Let us give a sketch of the proofs.Recall from (3.16) that • The term A 3 can be bounded as follows.
In this instance, we are integrating against ϕ ′ , which is supported on the annulus A(0, 1/2, 2).Thus Recall from (4.58) that |SO| |x| 2 r 2 ; this estimate is still valid here since spt(ϕ ′ ) is basically the same as spt(φ) (as in (4.2)).It is immediate to see then that In this section we will show that µ lies in a line, from which we derive that µ is, in fact, flat.Recall the notation (2.4).Lemma 6.1.Let Q ∈ D µ be a cube so that ℓ(Q) ∼ 1 A r, where A > 1 is a (possibly large) constant.Then we have where Q ∈ D µ is the unique cube such that Q ⊂ Q and ℓ( Q) ∼ Aℓ(Q).
Proof.Suppose first that x 0 is so that N k=0 β(x 0 , ℓ(Q)2 k ) < τ (where N ∼ log(A)).Then (6.1) follows from Lemma 4.14 and the fact that On the other hand, suppose that x 0 is so that N k=0 β 2 (x 0 , ℓ(Q)2 k ) > τ .Then using Lemma 5.2, we see that This proves the lemma also in this case.Lemma 6.2.For A large enough, and δ 0 , τ chosen appropriately, the support of µ lies in a line L.
Proof.We see that (6.1) implies that for some cube S ∈ D µ , and 1 > γ > 0, Let N A ∈ N be so that A ∼ 2 NA and pick γ < 1 sufficiently small.Using the Ahlfors regularity of µ, we see that Hence, writing out the sum on the right hand side of (6.4) as sum of the two sums in (6.5) and (6.6), we obtain A log(A) 2 ℓ(S) γ .(6.7) • The lower density is positive and the upper density is finite, that is θ 1, * (µ, x) := lim sup To prove this we will follow Mattila (see [Mat95]) in using Preiss' Theorem, which states that if a measure µ satisfies certain density conditions, and moreover its tangent measures are all flat, then µ is rectifiable.Before stating this precisely, let us remind the reader what tangent measures are.
For a measure µ on C, and a map F : C → C, we denote the push forward measure under F as F [µ].
Definition 7.2.A locally finite Borel measure ν is called a tangent measure of µ at z 0 ∈ C, if it is a weak limit of some sequence c i T z0,ri [µ], where 0 < c i < ∞, r i ↓ 0 and T z0,ri (A) = µ(r i A + z 0 ).We will denote the set of tangent measures of µ at z 0 by Tan(µ, z 0 ).We also put Tan(µ) := ∪ z0∈spt(µ) Tan(µ, z 0 ).Remark 7.3.Tangent measures were originally introduced by Preiss in [Pr87]; in this work, Preiss completed a theory started from Besicovitch in the 1920's which had as fundamental question the following: what is the relation between limits of density ratios (such as (7.2) and (7.1)) and rectifiability?We refer the interested reader to the very readable monograph [DeL] of De Lellis on Preiss' result.
Remark 7.5.Preiss' theorem holds much more generally, that is, it holds for measures in R n satisfying the d-dimensional version of the density condition above.Moreover, Mattila proves in [Mat95], Theorem 4.19, that for the case of Radon measures in the plane, assuming lower density positive is enough in the above theorem.
With Preiss' Theorem at hand, our tasks are clear: if we show that any tangent measure ν (of a Borel measure µ which satisfies (7.1), (7.2) and (7.3)) is Ω-symmetric and Ahlfors 1-regular, then we are done.Lemma 7.8, each of them is Ahlfors regular.Thus, any ν ∈ Tan(µ) satisfies the hypotheses of Theorem 1.4, and therefore we can conclude that ν = cH 1 | L whenever ν ∈ Tan(µ), where c > 0 and L is a line.Hence Preiss' Theorem (Theorem 7.4) let us conclude that µ is 1-rectifiable.

18 5 .
Case when β's are large 32 1.Introduction While investigating for what kind of measures µ in C does the Cauchy transform exists µ-almost everywhere (in the sense of principal values), Mattila (see [Mat95]) gave a complete characterisation of what he termed symmetric measures; for a measure µ, let us set C(x, r) := ˆB(x,r) (x − y) dµ(y).(1.1)