Potential Theory of Signed Riesz Kernels : Capacity and Hausdor ff Measure

where the supremum is taken over those analytic functions on C \ E such that |f(z)| ≤ 1, for z / ∈ E. It is easily shown that sets of zero analytic capacity are the removable sets for bounded analytic functions. In [4], one proves Vitushkin’s conjecture, namely, the statement that among compact sets of finite length (one-dimensional Hausdorff measure), the sets of zero analytic capacity are precisely those that project into sets of zero length in almost all directions. Equivalently, by Besicovitch theory, these are the purely unrectifiable sets, that is, the sets that intersect each rectifiable curve in zero length. In [11], the Cantor sets of vanishing analytic capacity are characterized, and in [21], the semiadditivity of analytic capacity is proven. When dealing with analytic capacity, one very often finds oneself working with the Cauchy kernel 1/z and not using analyticity at all. Indeed, analytic capacity itself can


Potential Theory of Signed Riesz Kernels: Capacity and Hausdorff Measure Laura Prat 1 Introduction
There has recently been substantial progress in the problem of understanding the nature of analytic capacity (see [4,11,21]).Recall that the analytic capacity of a compact subset E of the plane is defined by where the supremum is taken over those analytic functions on C \ E such that |f(z)| ≤ 1, for z / ∈ E. It is easily shown that sets of zero analytic capacity are the removable sets for bounded analytic functions.
In [4], one proves Vitushkin's conjecture, namely, the statement that among compact sets of finite length (one-dimensional Hausdorff measure), the sets of zero analytic capacity are precisely those that project into sets of zero length in almost all directions.
Equivalently, by Besicovitch theory, these are the purely unrectifiable sets, that is, the sets that intersect each rectifiable curve in zero length.In [11], the Cantor sets of vanishing analytic capacity are characterized, and in [21], the semiadditivity of analytic capacity is proven.
When dealing with analytic capacity, one very often finds oneself working with the Cauchy kernel 1/z and not using analyticity at all.Indeed, analytic capacity itself can easily be expressed without making any reference to analyticity in the form where the supremum is taken over all complex distributions T supported on E such that the Cauchy potential of T , f = 1/z * T , is a function in L ∞ (C) satisfying f ∞ ≤ 1.Then, it seems interesting to try to isolate properties of analytic capacity that depend only on the basic characteristics of the Cauchy kernel such as oddness or homogeneity.With this purpose in mind, we start in this paper the study of certain real variable versions of analytic capacity related to the Riesz kernels in R n .Their definition is as follows.Given 0 < α < n and a compact subset E of R n , set where the supremum is taken over all real distributions T supported on E such that, for 1 ≤ i ≤ n, the ith α-Riesz potential T * x i /|x| 1+α of T is a function in L ∞ (R n ) and sup 1≤i≤n T * x i /|x| 1+α ∞ ≤ 1.When n = 2 and α = 1, writing 1/z = x/|z| 2 − i(y/|z| 2 ) with z = x + iy, we obtain γ 1 (E) ≤ γ(E), for all compact sets E. According to Tolsa's Theorem [21], one has for all compact sets E, where γ + (E) is defined by the supremum in (1.2) where one now requires T to be a positive measure supported on E (with Cauchy potential bounded almost everywhere by 1 on C).Thus, on compact subsets of the plane, γ and γ 1 are comparable in the sense that, for some positive constant C, one has Therefore, our set function γ α can be viewed as a real variable version of analytic capacity associated to the vector-valued kernel x/|x| 1+α .Of course, one can think of other possibilities; for example, one can associate in a similar fashion a capacity γ Ω to a scalar kernel of the form K(x) = Ω(x)/|x| α , where Ω is a real-valued smooth function on R n , homogeneous of degree zero.We will not pursue this issue here.
In Section 3, we compare the capacity γ α to Hausdorff content.We obtain quantitative statements that, in particular, imply that if E has zero α-dimensional Hausdorff measure, then it has also zero γ α capacity.In the other direction, one gets that if E has Hausdorff dimension larger than α, then γ α is positive.Then, the critical situation occurs in dimension α, in accordance with the classical case.
The main contribution of this paper is the discovery of an interesting special behaviour of γ α for noninteger indexes α.When α is an integer and E is a compact subset of an α-dimensional smooth surface, then one can see that γ α (E) > 0 provided that H α (E) > 0, with H α being α-dimensional Hausdorff measure (see [14], where it is shown that if E lies on a Lipschitz graph, then γ n−1 (E) is comparable to the (n − 1)-Hausdorff measure H n−1 (E)).In particular, there are sets of finite α-dimensional Hausdorff measure H α (E) and positive γ α (E).It turns out that this cannot happen when 0 < α < 1.
Theorem 1.1.Let 0 < α < 1 and let E ⊂ R n be a compact set with H α (E) < ∞.Then, Notice that the analogue of the above result in the limiting case α = 1 is the difficult part of Vitushkin's conjecture: if E is a purely unrectifiable planar compact set of finite length, then γ(E) = 0. We do not know how to prove Theorem 1.1 for a noninteger α > 1.Even for an integer α > 1, we do not know if the natural analogue of Vitushkin's conjecture is true.However, we do have a result in the Ahlfors-David regular case.Recall that a closed subset E of R n is said to be Ahlfors-David regular of dimension d if it has locally finite and positive d-dimensional Hausdorff measure in a uniform way: where B(x, r) is the open ball centered at x of radius r and d(E) is the diameter of E. Notice that if E is a compact Ahlfors-David regular set of dimension d, then H d (E) < ∞.
In proving Theorem 1.1, we use a deep recent result of Nazarov, Treil, and Volberg [18] on the L 2 -boundedness of singular integrals with respect to very general measures (see Section 2 for a statement).As a technical tool, we also need a variant of the wellknown symmetrization method relating Menger curvature (see Section 2 for a definition) and the Cauchy kernel (see [13,15,16]).Symmetrization of the kernel x/|x| 1+α leads to a nonnegative quantity, only for 0 < α ≤ 1.For α = 1, this is Menger curvature and, for 0 < α < 1, a description can be found in Lemma 4.2.However, nonnegativity and homogeneity seem to be more relevant facts than having exact expressions for the symmetrized quantity.The lack of nonnegativity, for α > 1, is the reason that explains the restriction on α in Theorem 1.1.
The proof of Theorem 1.2 follows the line of reasoning of a well-known result of Christ [3] stating that if an Ahlfors-David regular set E of dimension one in the plane has positive analytic capacity, then the Cauchy integral operator is bounded in L 2 (F, H 1 ), where F is another Ahlfors-David regular set such that H 1 (E ∩ F) > 0. The main difficulty for us lies in the fact that if α is noninteger, then, according to a result of Vihtil ä [24], there are no Ahlfors-David regular sets E on which the α-dimensional Riesz operator is bounded in the space L 2 (E, H α ).This prevents us from directly adapting Christ's arguments.
Throughout the paper, the letter C will stand for an absolute constant that may change at different occurrences.
If A(X) and B(X) are two quantities depending on the same variable (or variables) X, we will say that A(X) ≈ B(X) if there exists C ≥ 1 independent of X such that In Section 2, one can find statements of some auxiliary results and the basic notation and terminology that will be used throughout the paper.As we have already mentioned above, in Section 3, we compare γ α to Hausdorff content.Theorem 1.1 is proven in Section 4 and Theorem 1.2 in Section 5.

L -boundedness of singular integral operators
if the following holds: (1) |K(x, y)| ≤ C|x − y| −α , for some 0 < α < n (with α not necessarily integer) and some positive constant C < ∞, (2) there exists 0 < ≤ 1 such that, for some constant Let µ be a Radon measure on R n .Then, the Calder ón-Zygmund operator T associated to the kernel K and the measure µ is formally defined as This integral may not converge for many functions f because for x = y the kernel K may have a singularity.For this reason, we introduce the truncated operators T , > 0: We say that the singular integral operator T is bounded in L 2 (µ) if the operators T are bounded in L 2 (µ) uniformly in .
The maximal operator T * is defined as (2.4) Let 0 < α < n and consider the Calder ón-Zygmund operator R α associated to the antisymmetric vector-valued Riesz kernel x/|x| 1+α .
For the proof of Theorem 1.1, a deep result of Nazarov, Treil, and Volberg will be needed.First, we introduce some more notation.We say that B(x, r) is a non-Ahlfors disk with respect to some constant M > 0 if µ(B(x, r)) > Mr. Let b be a bounded function.We say that a disk B(x, r) is nonaccretive with respect to b if, for some fixed positive constant , we have Let φ be some nonnegative Lipschitz function with Lipschitz constant 1 and consider the antisymmetric Calder ón-Zygmund operator K φ associated to the suppressed kernel k φ : The kernel k φ has the very important property of being well suppressed (we are borrowing the terminology from [18]) at the points where φ > 0, that is, We will state now a T (b) theorem of [18] for the Cauchy kernel.
K * θ b(x) ≤ B, for µ-almost all x and for every Lipschitz function θ with constant 1 such that θ ≥ φ.
One can use this result to give an alternative proof of Vitushkin's conjecture (see [18]).
To use their result for the α-Riesz transform R α , 0 < α < n, we need an appropriate version of the suppressed kernels associated to the Riesz α-operator R α .We have found that the following kernel does the job: where denotes the integer part of α.Notice that k φ,1 = k φ .
For the sake of completeness, we state the properties of the kernel k φ,α in a separate lemma.
Using this operators and adapting Theorem 2.1, one obtains the following result for the α-Riesz transform R α .
Theorem 2.3.Let µ be a positive measure on R n such that lim sup r→ 0 µ(B(x, r))/r α < +∞, for µ-almost all x, and b an Remark 2.4.The set F in Theorem 2.3 corresponds to C \ H. Namely, F is the set where there are no problems (every disk is Ahlfors and accretive and the maximal operator is uniformly bounded).Remark 2.5 (Volberg, personal communication).Instead of using the Calder ón-Zygmund operator related to the suppressed kernel defined in (2.7), one can also use the operator related to the following suppressed kernel: ) For the proof of Theorem 1.2, we need to define some sets Q k β that will be the analogues of the Euclidean dyadic cubes.These "dyadic cubes" were introduced by Christ in [3].
Theorem 2.6 [3].For a space of homogeneous type (E, ρ, µ) with µ as above, there exists a collection of Borel sets ) for each (k, β) and each l < k, there is a unique γ such that , for every k, β and for every t > 0.
We denote by For the variant of the T (b) theorem that we need (see [3,Theorem 20]), we require the definitions of a dyadic para-accretive function and a dyadic BMO function.
for some fixed constants c > 0 and N ∈ N.
Definition 2.8.A locally µ integrable function f belongs to dyadic BMO(µ) if where the supremum is taken over all dyadic cubes Q ∈ Q(E).
At the beginning of this section, we have defined Calder ón-Zygmund operators and standard kernels in the Euclidean case.In the context of spaces of homogeneous type, one has a slightly different definition for them (see [2, pages 93-94]).
Theorem 2.9 [3].Let E be a space of homogeneous type with underlying doubling measure µ, b a dyadic para-accretive function, and T a Calder ón-Zygmund operator associated to an antisymmetric standard kernel.Suppose that T (b) belongs to dyadic BMO(µ).
Then, T is a bounded operator in L 2 (µ).
A recent new approach to a variety of T (b) theorems can be found in [1].
For the proof of Theorem 1.2, the following result of Vihtil ä will also be needed.
Theorem 2.10 [24].Let µ be a nonzero Radon measure in R n for which there exist con- for all x ∈ spt(µ) and 0 < r < d(spt µ).If R α is a bounded operator in L 2 (µ), then α is an integer.
This theorem was proved by using an approach based on tangent measures.

Relation between γ α and Hausdorff content
We need the following lemma.
Lemma 3.1.If a function f(x) has compact support and has continuous derivatives up to order n, then it is representable, for 0 < α < n, in the form where ϕ i , i = 1, . . ., n, are defined by the formulas in which c n,α is a constant depending on n and α.
Proof.Assume first that n = 2k + 1. Taking Fourier transform of the right-hand side of (3.1), we get, for appropriate numbers a n,α and b n,α , A similar argument proves (3.1) in the case n = 2k.
We are now ready to describe the basic relationship between γ α and Hausdorff content (the d-dimensional Hausdorff content will be denoted by M d (see [9] for the definition and basic properties)).Lemma 3.2.If 0 < α < n, then there exist constants C and C such that for any compact set E ⊂ R n and > 0.
Proof.We proof first the second inequality.Let {Q j } j be a covering of E by dyadic cubes Let T be a distribution with compact support contained in E such that the ith α- Applying Lemma 3.1 to each g j , we obtain functions ϕ i j satisfying (3.1) with f and ϕ i replaced by g j and ϕ j i , respectively.Thus, Take n = 2k + 1 (for n = 2k, the argument is similar) and write Let Q 0 be the unit cube centered at 0. Integrating by parts to bring the ∆ k ∂ i derivatives from g j to the kernel k α , changing variables, and using (3.6) For the reverse inequality, we use a standard argument that we reproduce for the reader's convenience.Suppose that M α+ (E) > 0, for some > 0. By Frostman's Lemma (see [12,Theorem 8.8]), there exists a measure µ supported on E such that µ(E) ≥ CM α+ (E) > 0 and µ(B(x, r)) ≤ r α+ , x ∈ R n and r > 0.Then, by a change of variables, we obtain Using this estimate, we get the desired inequality, namely, Let dim(E) be the Hausdorff dimension of the set E. A qualitative version of Lemma 3.2 is the following corollary. ( 4 Proof of Theorem 1.1

Distributions that are measures
We start by a lemma that shows that certain distributions are actually measures.
Then, T is a measure which is absolutely continuous with respect to the restriction of H α to E and has a bounded density, that is, Proof.We first show that T is a measure.For this, it is enough to prove that Given > 0, we can cover the compact set E with open balls B j of radius r j , j = 1, . . ., k, such that B j ∩ E = ∅, r j < , and Let ψ be a function in C ∞ 0 with spt ψ ⊂ B(0, 1) and ψ(x)dx = 1.Define To prove (4.2), we can assume without loss of generality that spt(f) ⊂ ∪ j B j .This Assume that n = 2k + 1 (the argument for even dimensions is similar).Applying Lemma 3.1 to ψ , using the boundedness of T * x i /|x| 1+α , for 1 ≤ i ≤ n, and setting We will show that where C is a constant depending on the L 1 -norm of ψ and k ∂ i ψ but not on .
To prove (4.6), we use Fubini's Theorem and a change of variables: Let B 0 be an open ball and let B 0 denote its closure.Let H α E stand for the restriction of H α to E. If we show that then, taking a sequence of open balls B i 0 ↓ B 0 and applying (4.9) to these balls, we will have It is shown in [12, page 271] that, for α = 1, (4.10) implies The argument extends verbatim to any α and thus we can take (4.11) for granted, which gives (4.1) by Radon-Nikodym's Theorem.
It remains to prove (4.9).We know that, for every δ > 0, there exists a compact Let Recall that the radii of the balls B j satisfy r j < .For an appropriate > 0, the following holds: This last condition implies that, for j 1 ∈ J 1 and j 2 ∈ J 2 , we have B j 1 ∩ B j 2 = ∅.So, using inequalities (4.3), (4.14), and (4.12), If χ B 0 denotes the characteristic function of the ball B 0 , then Arguing as in (4.5), (4.6), and (4.7), we get and letting and δ tend to zero, we get (4.9).

Symmetrization of the Riesz kernel
The symmetrization process of the Cauchy kernel introduced in [15] has been successfully applied in the last years to many problems of analytic capacity and L 2 -boundedness of the Cauchy integral operator (see, e.g., [13,16,23]; the survey papers [5,22] contain many other references).Given three distinct points z 1 , z 2 , and z 3 in the plane, one finds out, by an elementary computation, that where the sum is taken over the six permutations of the set {1, 2, 3} and c(z Menger curvature, that is, the inverse of the radius of the circle through z 1 , z 2 , and z 3 . In particular, (4.19) shows that the sum on the right-hand side is a nonnegative quantity.
On the other hand, it has been proved in [7] that nothing similar occurs for the Riesz kernel k α = x/|x| 1+α with α integer and 1 < α ≤ n.In this section, we show that, for 0 < α < 1, we recover an explicit expression for the symmetrization of the Riesz kernel k α and that the quantity one gets is also nonnegative.For α > 1, the phenomenon of change of signs appears again.
Lemma 4.2.Let 0 < α < 1, and let x 1 , x 2 , and x 3 be three distinct points in R n .Then, where L(x 1 , x 2 , x 3 ) is the largest side of the triangle determined by x 1 , x 2 , and x 3 .In particular, p α (x 1 , x 2 , x 3 ) is a nonnegative quantity.
Proof.If n = 1 and x 1 < x 2 < x 3 , then where a = x 2 − x 1 and b = x 3 − x 2 .An elementary estimate shows that (4.22) holds in this case, even with 2 1+α replaced by 2 α in the numerator of the last term.
Note that if x 1 , x 2 , x 3 ∈ R n , one can write where θ ij is the angle opposite to the side x i x j in the triangle determined by x 1 , x 2 , and x 3 .Without loss of generality, we can assume that We consider two different cases.
Finally, for each t, the function has a minimum at y = 0, and this proves the claim and thus the first inequality in (4.22).

The main step
Let 0 < α < n and suppose that µ is a measure such that µ(B(x, r)) ≤ C 0 r α , for some constant C 0 and for all balls B(x, r) ⊂ R n .We will now analyze what happens in a ball B(x, r) satisfying the lower-density condition µ(B(x, r)) ≥ r α for a given number > 0. and Proof.Without loss of generality, we may assume that B 0 = B(0, 1).Let a ≥ 1 and b ≥ 1 be two constants to be chosen at the end of the construction and suppose that the lemma is not true.This means that given any pair of closed balls B 1 and B 2 of radius a −1 centered at spt µ ∩ B 0 , then either or one of the two balls, say B i , satisfies Notice that a simple estimate of the volume of the union of the balls in a given family reveals that each family contains no more than (2a) n balls.We have which means that there exists at least one family B i such that

.39)
Consider the set The fact that each family B i contains no more than (2a) n balls implies that and so we get If a and b are appropriately chosen, this inequality gives a contradiction.
Let 0 ≤ α < ∞ and let µ be a positive Borel measure on R n .The upper and lower α-densities of µ at x ∈ R n are defined by respectively.
By the 5-covering Theorem (see [12, Theorem 2.1]), for each i ∈ N, there are dis- that is, Fix i = 1 and consider the disjoint balls B 1 j , for 1 ≤ j ≤ m 1 .For every B 1 j , we can use Lemma 4.3 twice to find three balls B 1 , B 2 , and B 3 centered at spt(µ) ∩ B 1 j enjoying the following properties: their mutual distances and their radii are comparable to r(a j ) and the mass µ(B 1 j ∩ B l ) is also comparable to r(a j ) α .The comparability constants in the above statements depend only on c 1 , c 2 , and n.Define a set of triples by Applying Lemma 4.2, we obtain ≥ Cr 1 a j α . (4.49)

.51)
Let q be such that sup and consider the balls of the qth generation, namely, B q j , for 1 ≤ j ≤ m q .Repeat the process described above, replacing B 1 j by B q j .We then find balls B 1 , B 2 , and B 3 centered at points in spt µ ∩ B q j , whose mutual distances and radii are comparable to r q (a j ) and such that µ(B q j ∩ B l ) is also comparable to r q (a j ) α , l = 1, 2, 3. Set Hence, again by (4.52), Notice that the sets of triples A 1 and A 2 are disjoint because of the definition of q.Define t 2 as we did before for t 1 so that, for (x 1 , x 2 , x 3 ) ∈ A 2 , one has |x i − x j | > t 2 , for i, j ∈ {1, 2, 3}, i = j.It becomes now clear that we can inductively construct disjoint sets of triples A k , k = 1, 2, . . ., such that and therefore, Suppose that γ α (E) > 0, for 0 < α < 1. Applying Lemma 4.1, we find a measure of the We can apply now Theorem 2.3 to get a set F ⊂ E of positive H α -measure such that the operator R α is bounded in L 2 (H α , F).On the other hand, since ).This means that we can apply Theorem 4.4 to obtain This last fact contradicts the L 2 -boundedness of R α on L 2 (H α , F) by a well-known argument that we now outline briefly (see [15,16]). where Interchanging the roles of x and y, and then of x and z, and estimating the error terms in a standard way, we obtain Letting → 0, we get the promised contradiction.
Potential Theory of Signed Riesz Kernels 961 Remark 4.5.Notice that if we knew that, for some 0 < α < n, the α-Riesz kernel never defines a bounded operator on a set of finite α-Hausdorff measure, then Theorem 1.1 would extend to this value of α.For any 0 < α < 1, this follows from the symmetrization method, as it is shown above.For 1 < α < n, to get such a result, we have to restrict ourselves to α-dimensional Ahlfors-David regular sets and noninteger α (see Theorem 2.10).

Proof of Theorem 1.2
As a tool to prove Theorem 1.2, consider the tangent measures that were introduced by Preiss in [20].
Let T a,r be the map that blows up B(a, r) to B(0, 1), that is, The image of µ under T a,r is given by (5.2) Definition 5.1.Let µ be a Radon measure on R n .The measure σ is said to be a tangent measure of µ at a point a ∈ R n if σ is a nonzero Radon measure on R n and if there exist sequences {r i } and {c i } of positive numbers such that r i → 0 and c i T a,r i µ → σ weakly, as The set of all tangent measures to µ at a is denoted by Tan(µ, a).
Proof of Theorem 1.2.Let 0 < α < n and let E ⊂ R n be a compact Ahlfors-David regular set of dimension α.Suppose that γ α (E) > 0.Then, there exists a distribution S with compact support contained in E, whose α-Riesz potential S * x/|x| 1+α is in L ∞ (R n ) and such that S, 1 = 0.
We will now construct an α-dimensional Ahlfors-David regular measure σ, that is, a measure such that, for some constant C, whose α-Riesz operator R α is bounded in L 2 (σ).Then, applying Theorem 2.10, we will conclude that α must be an integer.
We first sketch briefly the main ideas involved in the construction of the measure σ.The first step will be to construct a set E with H α (E ∩ E ) > 0 and a doubling measure µ on E .The pair (E , µ) is then endowed with a system of dyadic cubes Q(E ) satisfying the properties of Theorem 2.6.We also define a bounded function b on E , which will be dyadic para-accretive with respect to the system of dyadic cubes Q(E ) and such that the function R α (bµ) belongs to dyadic BMO(µ).Therefore, the α-Riesz transform R α associated to µ will be bounded on L 2 (E , µ) by the T (b) theorem on a space of homogeneous type (Theorem 2.9).The required Ahlfors-David regular measure σ will be a tangent measure of µ at some point of density of E inside E .The fact that the α-Riesz transform R α , associated to σ defines a bounded operator on L 2 (σ) will follow from the L 2 (µ)-boundedness of R α associated to µ by taking weak limits.Now, we turn to the construction of the set E and the measures µ and σ.Let Q(E) be a system of dyadic cubes on E satisfying the properties (1) through (6) in Theorem 2.6.
The first dyadic cube of E to examine is E itself.By hypothesis, there exists a function h ∈ L ∞ (E) such that E h dH α = 0. Let 0 > 0 be a sufficiently small constant to be fixed Then, for every positive integer k, there exists at which is a contradiction.
We now run a stopping-time procedure.Let > 0 be another constant, much smaller than 0 , to be chosen later.Take a dyadic cube Q ∈ Q 1 (E) and check whether or not the condition is satisfied.If (5.6) holds for that cube Q, and Q has more than one child, we call it a stopping-time cube.If (5.6) holds but Q has only one child, then we look for the first descendent of Q with more than one child and we call it a stopping-time cube.Notice that (5.6) remains true for this descendent.
If (5.6) does not hold for Q, then we examine each child of Q and repeat the above procedure.After possibly infinitely many steps and possibly passing through all generations, we obtain a collection of pairwise disjoint stopping-time cubes {P γ } in E. Each P γ has at least two children and satisfies the nonaccretivity condition (5.6) with Q replaced by P γ .
We want to construct the set E by excising from E the union of the stopping-time cubes P γ and replacing each child R β of P γ by a certain ball B β .
Property (5) of Theorem 2.6 gives us a constant 0 < a 1 < 1, such that, for each For each stopping-time cube In what follows, set δ k = r β , where k is the generation of R β .That is, for each γ, the sets F γ replace the stopping-time cubes P γ in the new set E .In other words, (5.9) We will define now a measure µ on this set E as follows: where L n is the n-dimensional Lebesgue measure.
We will now check that there exist some positive constants M 0 and M 1 such that, for every x ∈ E and r > 0, (1) the measure µ has α-growth, that is, (2) the measure µ is doubling, that is, To prove that µ has α-growth, first let x ∈ E \ ∪ β B β , r > 0, and let β be such that On the other hand, since x ∈ E \ B β , then x / ∈ R β and property (5) in Theorem 2.6 gives us |x − z β | > a 1 r β .Thus, by the definition of r β and property (4) in Theorem 2.6, we which implies that R β ⊂ B(x, 5r/a 1 ).Since our initial set E is Ahlfors-David regular and (5.15) If, for some β, x ∈ B β , then the above inequality follows in the same way because the diameter of B β is less than the distance to its complement in E .Thus, the measure µ satisfies (5.11).
To prove that µ is a doubling measure, take any x ∈ E \ ∪ β B β and r > 0.Then, arguing as above, but with r replaced by 2r, we obtain because our initial measure H α is doubling on E.
We claim that, for some positive constant M 1 , the following holds: which proves (5.12).
To prove (5.17 r), and due to the definition of µ, (5.34)).Hence, the doubling property for H α on E gives that and proves claim (5.17).
If x ∈ B β , for some β and r ≤ r β /2, then the doubling property for µ holds clearly.
If r > r β /2.Then, arguing as above, one gets the doubling property for µ on E .Therefore, (5.12) holds.
For a system of dyadic cubes Q(E ) on E satisfying the properties of Theorem 2.6 with respect to the doubling measure µ, take all dyadic cubes Q ∈ Q(E) which are not contained in any stopping-time cube P γ , together with each F γ = ∪ β B β and with the dyadic cubes of Q(B β ) in each F γ .Namely, where the dyadic systems Q(B β ) associated to the balls B β coming from all the F γ .Hence, each After defining the set E , the doubling measure µ, and the system of dyadic cubes Q(E ), our next step will consist in modifying the function h on the union ∪ γ F γ in order to obtain a new function b, defined on E , bounded and dyadic para-accretive with respect to the system of dyadic cubes Q(E ).In fact, we want b to satisfy (5.20) Condition (5.20) does not seem to contribute to the accretivity of the new function b with respect to the measure µ because the cubes P γ were chosen precisely because the mean of h on them became too small.But although our b has a small mean on F γ , as h does on P γ , we will have a satisfactory lower bound on the integral of b over each child In this way, b becomes "more" accretive than h.
where the coefficients c β are defined below to get the boundedness of the function b and ( Notice first that due to properties (5) and (6) of Theorem 2.6, B β ∩ B η = ∅, for β = η, and B β ∩ (E \ ∪ γ P γ ) = ∅ so that the function b is well defined on E .
To define the coefficients c β , fix P γ and let N γ = {β : R β is a child of P γ }.The number of children of the dyadic cubes is in between 2 and a fixed upper bound, that is, where c 1 is some constant independent of γ.
Order the children {R β } of P γ starting with the cube R β with the smallest H αmeasure and ending with the cube R β with the biggest one.Write , where R j β stands for the jth child R β in this ordering.We want to divide the children of P γ into two nonempty collections I and II, each with the same number of elements (plus or minus one) in the following way: Clearly, β∈II Define the coefficients c β as where the c β satisfy 0 ≤ c β ≤ 1 and, moreover, a certain constraint specified below.
Notice that the above-defined function b is bounded: Moreover, integrating b on F γ with respect to the measure µ, we get We claim that we can choose 0 > 0 sufficiently small so that there exist numbers Once (5.27) is proved, we get the desired expression for the integral of b over F γ , namely, To show (5.27), let N 2 = {β : β ∈ II} and define (5.29) With this choice of the coefficients c η , equality (5.27) clearly holds.Thus, we only have to show that there exists 0 > 0 such that 0 ≤ c η ≤ 1, for all η.
The inequality c η ≤ 1 is equivalent to (5.30) Notice that, by the way the indexes were ordered, for all η ∈ II, which implies c η ≤ 1.
For the lower inequality (5.32), we have to choose 0 such that c η ≥ 0 .Recall that, for P γ , the stopping-time condition (5.6) holds with Q replaced by P γ , and that the children of P γ have comparable measures.Moreover, we know that there exists some (small) positive constant 0 < c < 1/2 such that β∈I H α (R β ) ≥ cH α (P γ ).Then, we have where c 1 > 0 is the upper bound for the number of children of a dyadic cube.
In order to construct the function b, we have to carry out this procedure for each ( Thus, the function b satisfies the para-accretivity condition on the cubes F γ . For future reference, note that, for every cube Q ∈ Q(E ), such that Q F γ for all γ, there is a nonstopping time cube Q * ∈ Q(E) uniquely associated to Q by the identity (5.33) Moreover, one has (5.35) We will check now that, by construction, the function b is dyadic para-accretive with respect to Q(E ). (1) In both cases, the paraaccretivity of b follows as above due to the lower bound of |c β |. (2) If Q ∈ Q 1 (E ), the case Q = F γ has already been discussed, so we are only left Let Q * ∈ Q(E) be the cube defined in (5.33).Recall that Q * is a nonstopping time cube.Then, due to (5.20) and (5.34), we can write (5.36) Hence, b is a dyadic para-accretive function with respect to Q(E ).
We are still left with the fact that R α (bµ) belongs to dyadic BMO(µ).We postpone the proof of the BMO-boundedness and we continue with the argument.
At this point, we have constructed a set E with a system of dyadic cubes Q(E ), a function b dyadic para-accretive with respect to this system of dyadic cubes, and a measure µ which is doubling and has α-growth.Moreover, we are assuming that the function R α (bµ) belongs to dyadic BMO(µ).Therefore, by the T (b) theorem (see Theorem 2.9), the Riesz α-operator R α associated to the measure µ is bounded in L 2 (µ).

Notice that since
from the choice of 0 and .This shows that H α (E \ ∪ γ P γ ) > 0, and therefore, that In fact, from (5.8), we get the better lower bound (5.39) By density (see, e.g., [12,Corollary 2.14]), for H α E -almost all x ∈ E good , the limit lim sup r→ 0 H α E bad ∩ B(x, r) r α = 0. (5.40) Therefore, for such x, using the lower bound from the Ahlfors-David regularity of the set E, we obtain lim inf r→ 0 (5.41) Moreover, the upper bound coming from the Ahlfors-David regularity of the set E implies that, for every x ∈ E, we have lim sup r→ 0 (5.42) Let x 0 ∈ E good be a point satisfying (5.41) and (5.42) and let σ ∈ Tan(H α E good , x 0 ).Then, [12,Lemma 14.7] shows that there is a positive number C such that which is the same as to say that σ is an Ahlfors-David regular measure.Now, we only have to show that the α-Riesz operator associated to σ is bounded in L 2 (σ).
Notice first that due to Remark 5.2, there exists a sequence r i → 0 such that, for some positive number d, where the last identity is the definition of the measures H α r i .
In what follows, we let T stand for the α-Riesz operator R α .
Fix a radial function ϕ ∈ C ∞ such that 0 ≤ ϕ ≤ 1, ϕ = 0 on B(0, 1/2), and ϕ = 1 on R n \ B(0, 1).For > 0, define the regularized operators T as follows: for complex Radon measures ν in R n .One can easily check that, for > 0, where M(fν) is the Hardy-Littlewood maximal function: f(y)dν(y). (5.47) It is well known that M is bounded in L 2 .Thus, the L 2 -boundedness of the truncated operators T is equivalent to that of T .If the measure we are considering is nondoubling, then the maximal function in (5.47) does not work, but instead of M, one can consider a modified maximal operator introduced in [17] to get the same equivalence.
Notice that the fact that the operator T with respect to H α E good is bounded in L 2 (H α E good ) implies that, for each r > 0, the operator T with respect to H α r is bounded in L 2 (H α r ); namely, for f and g test functions, we have (5.48) Therefore, T (fσ), gσ = lim which means that T is bounded in L 2 (σ).
We still have to show that T (bµ) is a BMO function.We claim that since the function b ∈ L ∞ (E ), it is enough to show the following L 1 -inequality: for every Q ∈ Q(E ), where µ Q denotes the restriction of the measure µ to Q.
Suppose (5.50) holds for every Q ∈ Q(E ), and let, for some positive constant A, As a consequence of the "small boundary condition" for the dyadic cubes (see Theorem 2.6, property (6)), we have (see the bound for the second integral in (5.64)).The standard estimates for the Calder ón-Zygmund operators show that ) which proves the claim.
To prove (5.50), let Q ∈ Q(E ) be some dyadic cube of E .We distinguish now between two cases.
(1) For some β, let Then, Fubini and a change of variables give us, for some constant c, where at the last step, we have used the fact that E is Ahlfors-David regular, and so for some γ, then one argues as in the previous case because, for each γ, the number of B β involved in β B β = F γ is bounded above by some constant independent of γ.Thus, let Q ∈ Q 1 (E ) \ {F γ } γ and let Q * be the uniquely associated nonstopping dyadic cube in Q(E) defined before.Using (5.20), we can write (5.56) We claim that the following estimates hold for each β: (5.61) which is (5.50) provided that inequalities (5.60), (5.57), (5.58), and (5.59) hold.
Hence, the standard estimates for the Calder ón-Zygmund operators and the α-growth of the measure µ give (5.64) The first integral in (5.64) may be estimated in the same way as (5.63).Thus, we get x − y |x − y| 1+α dH α (x)dµ(y) ≤ CH α R β . (5.65) To deal with the second integral in (5.64), let j ∈ Z and define the set  which is (5.58).
To show (5.59), let R c β be the complement of R β .Then, using that hH α * K is a bounded function, we can write   by arguing similarly as in the proof of (5.63).
We are now left with the proof of (5.60).Notice that we can write (5.77) To deal with the first integral in the last line of (5.77), set g = hχ E\2Q * .Then, one has a BMO estimate for T (g) restricted to Q * ; namely, there exists some constant c, depending on g and Q * , such that (something similar was done before (5.52) to show that (5.50) suffices for the BMO estimate).Using the small boundary condition (see Theorem 2.6, property (6)), we also have (see the estimates for the second integral in (5.64)).
Thus, if we write (5.81) The upper bound on |c| is obtained by using the fact that Q * ∈ Q(E) is not a stopping-time cube, namely, that Q * h dH α > H α (Q * ).Therefore, using that H α (Q * ) = µ(Q), we get (5.82)

Potential Theory of Signed Riesz Kernels 979
To estimate the second integral in (5.77), notice that, for each β, we can write T hχ Q * \R β dµ. (5.83) The first integral has been already estimated in (5.59).To deal with the second one, write Q * as a finite union of cubes R γ of the same generation as R β , that is, such that d(R γ ) ≈ r β .Then, for each γ = β, if x ∈ B β and y ∈ R γ , one has |x − y| ≥ Cr β .Hence, because of the facts that E is Ahlfors-David regular, and so H α (R γ ) ≈ d(R γ ) α ≈ r α β and h is a bounded function.Plugging all these estimates in (5.77) and using that, for each β, H α (R β ) = µ(B β ), we obtain (5.85) which finishes the proof of (5.60).Therefore, T (bµ) is a dyadic BMO function.Remark 5.3.For a different proof of Theorem 1.2 without using tangent measures, see [19].

Lemma 4 . 3 .
There exist constants a ≥ 1 and b ≥ 1 depending only on C 0 and such that given any ball B 0 = B(x, r) satisfying µ(B 0 ) ≥ r α , there exist two balls B 1 = B(x 1 , r/a)

. 36 )
Consider the covering of spt µ ∩ B 0 by balls of radius a −1 centered at spt µ ∩ B 0 .Apply Besicovitch's covering lemma to this covering to obtain N = N(n) families B i of disjoint balls such that spt µ ∩ B 0 ⊂ .40) Condition (4.35) implies that all balls in M are contained in a ball of radius 8/a, using that µ(B(x, r)) ≤ C 0 r α holds for any ball B(x, r) in R n .

1 )
stopping-time cube P γ .The P γ are the cubes where the accretivity condition for h fails.The function h 1 has the advantage that although Pγ h dH α = Fγ h 1 dH α , we have a satisfactory lower bound on the integral over each child B β of F γ .This is due to the definition of the coefficients c β .(If β ∈ I, then | B β b dµ| = |c β |µ(B β ) ≥ 0 µ(B β ).