A PROOF OF THE WEAK (1,1) INEQUALITY FOR SINGULAR INTEGRALS WITH NON DOUBLING MEASURES BASED ON A CALDERÓN-ZYGMUND DECOMPOSITION

XAVIER TOLSA

Abstract. Given a doubling measure μ on ℝ^{d}, it is a classical result of harmonic analysis that Calderón-Zygmund operators which are bounded in L^{2}(μ) are also of weak type (1, 1). Recently it has been shown that the same result holds if one substitutes the doubling condition on μ by a mild growth condition on μ. In this paper another proof of this result is given. The proof is very close in spirit to the classical argument for doubling measures and it is based on a new Calderón-Zygmund decomposition adapted to the non doubling situation.

1. Introduction

Let μ be a positive Radon measure on ℝ^{d} satisfying the growth condition

(1.1)  μ(B(x, r)) ≤ C_{0} r^{n} for all x ∈ ℝ^{d}, r > 0,

where n is some fixed number with 0 < n ≤ d. We do not assume that μ is doubling [μ is said to be doubling if there exists some constant C such that μ(B(x, 2r)) ≤ C μ(B(x, r)) for all x ∈ supp(μ), r > 0]. Let us remark that the doubling condition on the underlying measure μ on ℝ^{d} is an essential assumption in most results of classical Calderón-Zygmund theory. However, recently it has been shown that a big part of the classical theory remains valid if the doubling assumption on μ is substituted by the size condition (1.1) (see for example the references cited at the end of the paper).

In this note we will prove that Calderón-Zygmund operators (CZO’s) which are bounded in L^{2}(μ) are also of weak type (1, 1), as in the usual doubling situation. This result has already been proved in [To1] in the particular case of the Cauchy integral operator, and by Nazarov, Treil and Volberg [NTV2] in the general case. The proof that we will show here is different from the one of [NTV2] (and also from the one of [To1], of course) and it is closer in spirit to the classical proof of the corresponding result for doubling measures. The basic tool for the proof is a decomposition of Calderón-Zygmund type for functions in L^{1}(μ) obtained in [To4].

Date: February 4, 2000.

1991 Mathematics Subject Classification. 42B20.

Key words and phrases. Calderón-Zygmund operators, non doubling measures, non homogeneous spaces, weak estimates.

Supported by a postdoctoral grant from the European Comission for the TMR Network “Harmonic Analysis”. Also partially supported by grants DGICYT PB96-1183 and CIRIT 1998-SGR00052 (Spain).

Our purpose in writing this paper is not only to obtain another proof in the non doubling situation of the basic result that CZO’s bounded in L^{2}(μ) are of weak type (1, 1), but to show that the Calderón-Zygmund decompositon of [To4] is a good substitute of its classical doubling version.

Let us introduce some notation and definitions. A kernel k(·, ·) from L_{loc}^{1}((ℝ^{d} × ℝ^{d}) \ {(x, y) : x = y}) is called a Calderón-Zygmund kernel if

(1) |k(x, y)| ≤ \frac{C}{|x − y|^{n}} ,

(2) there exists 0 < δ ≤ 1 such that

|k(x, y) − k(x′, y)| + |k(y, x) − k(y, x′)| ≤ C \frac{|x − x′|^{δ}}{|x − y|^{n+δ}}

if |x − x′| ≤ |x − y|/2. Throughout all the paper we will assume that μ is a Radon measure on ℝ^{d} satisfying (1.1). The CZO associated to the kernel k(·, ·) and the measure μ is defined (at least, formally) as

T f(x) = ∫ k(x, y) f(y) dμ(y).

The above integral may be not convergent for many functions f because k(x, y) may have a singularity for x = y. For this reason, one introduces the truncated operators T_{ε}, ε > 0:

T_{ε}f(x) = ∫_{|x−y|>ε} k(x, y) f(y) dμ(y),

and then one says that T is bounded in L^{p}(μ) if the operators T_{ε} are bounded in L^{p}(μ) uniformly on ε > 0. It is said that T is bounded from L^{1}(μ) into L^{1,∞}(μ) (or of weak type (1, 1)) if

μ\{x : |T_{ε}f(x)| > λ\} ≤ C \frac{∥f∥_{L^{1}(μ)}}{λ}

for all f ∈ L^{1}(μ), uniformly on ε > 0. Also, T is bounded from M(ℂ) (the space of complex Radon measures) into L^{1,∞}(μ) if

μ\{x : |T_{ε}ν(x)| > λ\} ≤ C \frac{∥ν∥}{λ}

for all ν ∈ M(ℂ), uniformly on ε > 0. In the last inequality, T_{ε}ν(x) stands for ∫_{|x−y|>ε} k(x, y) dν(y) and ∥ν∥ ≡ |ν|(ℝ^{d}).

The result that we will prove in this note is the following.

Theorem 1.1. Let μ be a Radon measure on ℝ^{d} satisfying the growth condition (1.1). If T is a Calderón-Zygmund operator which is bounded in L^{2}(μ), then it is also bounded from M(ℂ) into L^{1,∞}(μ). In particular, it is of weak type (1, 1).

2. The proof

First we will introduce some additional notation and terminology. As usual, the letter C will denote a constant which may change its value from one occurrence to another. Constants with subscripts, such as C_{0}, do not change in different occurrences.

By a cube Q ⊂ R^{d} we mean a closed cube with sides parallel to the axes. We denote its side length by ℓ(Q) and its center by x_{Q}. Given α > 1 and β > α^{n}, we say that Q is (α, β)-doubling if μ(αQ) ≤ β μ(Q), where αQ is the cube concentric with Q with side length α ℓ(Q). For definiteness, if α and β are not specified, by a doubling cube we mean a (2, 2^{d+1})-doubling cube.

Before proving Theorem 1.1 we state some remarks about the existence of doubling cubes.

Remark 2.1. Because μ satisfies the growth condition (1.1), there are a lot of “big” doubling cubes. To be precise, given any point x ∈ supp(μ) and c > 0, there exists some (α, β)-doubling cube Q centered at x with l(Q) ≥ c. This follows easily from (1.1) and the fact that β > α^{n}. Indeed, if there are no doubling cubes centered at x with l(Q) ≥ c, then μ(α^{m}Q) > β^{m}μ(Q) for each m, and letting m → ∞ one sees that (1.1) cannot hold.

Remark 2.2. There are a lot of “small” doubling cubes too: if β > α^{d}, then for μ-a.e. x ∈ ℝ^{d} there exists a sequence of (α, β)-doubling cubes {Q_{k}}_{k} centered at x with ℓ(Q_{k}) → 0 as k → ∞. This is a property that any Radon measure on ℝ^{d} satisfies (the growth condition (1.1) is not necessary in this argument). The proof is an easy exercise on geometric measure theory that is left for the reader.

Observe that, by the Lebesgue differentiation theorem, for μ-almost all x ∈ ℝ^{d} one can find a sequence of (2, 2^{d+1})-doubling cubes {Q_{k}}_{k} centered at x with ℓ(Q_{k}) → 0 such that

lim_{k→∞} \frac{1}{μ(Q_{k})} ∫_{Q_{k}} f dμ = f(x).

As a consequence, for any fixed λ > 0, for μ-almost all x ∈ ℝ^{d} such that |f(x)| > λ, there exists a sequence of cubes {Q_{k}}_{k} centered at x with ℓ(Q_{k}) → 0 such that

lim sup_{k→∞} \frac{1}{μ(2Q_{k})} ∫_{Q_{k}} |f| dμ > \frac{λ}{2^{d+1}} .

In the following lemma we will prove an easy but essential estimate which will be used below. This result has already appeared in previous works ([DM], [NTV2]) and it plays a basic role in [To2] and [To4] too.

Lemma 2.3. If Q ⊂ R are concentric cubes such that there are no (α, β)- doubling cubes (with β > α^{n}) of the form α^{k}Q, k ≥ 0, with Q ⊂ α^{k}Q ⊂ R, then,

∫_{R\Q} \frac{1}{|x − x_{Q}|^{n}} dμ(x) ≤ C_{1},

where C_{1} depends only on α, β, n, d and C_{0}.

Proof. Let N be the least integer such that R ⊂ α^{N}Q. For 0 ≤ k ≤ N we have μ(α^{k}Q) ≤ μ(α^{N}Q)/β^{N−k}. Then,

∫_{R\Q} \frac{1}{|x − x_{Q}|^{n}} dμ(x) ≤ ∑_{k=1}^{N} ∫_{α^{k}Q\α^{k−1}Q} \frac{1}{|x − x_{Q}|^{n}} dμ(x)

≤ C ∑_{k=1}^{N} \frac{μ(α^{k}Q)}{ℓ(α^{k}Q)^{n}}

≤ C ∑_{k=1}^{N} \frac{β^{k−N} μ(α^{N}Q)}{α^{(k−N)n} ℓ(α^{N}Q)^{n}}

≤ C \frac{μ(α^{N}Q)}{ℓ(α^{N}Q)^{n}} ∑_{j=0}^{∞} (\frac{α^{n}}{β} )^{j} ≤ C.
□

The Calderón-Zygmund decomposition mentioned above has been obtained in Lemma 7.3 of [To4] and in that paper it has been used to show that if a linear operator is bounded from a suitable space of type H^{1} into L^{1}(μ) and from L^{∞}(μ) into a space of type BMO, then it is bounded in L^{p}(μ). We will use a slight variant of this decompositon to prove Theorem 1.1. Let us state the result that we need in detail.

Lemma 2.4 (Calderón-Zygmund decomposition). Assume that μ satisfies (1.1). For any f ∈ L^{1}(μ) and any λ > 0 (with λ > 2^{d+1} ∥f∥_{L^{1}(μ)}/∥μ∥ if ∥μ∥ < ∞) we have:

(a) There exists a family of almost disjoint cubes {Q_{i}}_{i} (that is, ∑_{i} χ_{Q_{i}} ≤ C) such that

(2.1)  \frac{1}{μ(2Q_{i})} ∫_{Q_{i}} |f| dμ > \frac{λ}{2^{d+1}} ,

(2.2)  \frac{1}{μ(2ηQ_{i})} ∫_{ηQ_{i}} |f| dμ ≤ \frac{λ}{2^{d+1}} for η > 2,

(2.3)  |f| ≤ λ a.e. (μ) on ℝ^{d} \ ⋃_{i} Q_{i}.

(b) For each i, let R_{i} be a (6, 6^{n+1})-doubling cube concentric with Q_{i}, with l(R_{i}) > 4l(Q_{i}) and denote w_{i} = \frac{χ_{Q_{i}}}{∑_{k} χ_{Q_{k}}}. Then, there exists a family of functions φ_{i} with supp(φ_{i}) ⊂ R_{i} and with constant sign satisfying

(2.4)  ∫ φ_{i} dμ = ∫_{Q_{i}} f w_{i} dμ,

(2.5)  ∑_{i} |φ_{i}| ≤ B λ

(where B is some constant), and

(2.6)  ∥φ_{i}∥_{L^{∞}(μ)} μ(R_{i}) ≤ C ∫_{Q_{i}} |f| dμ.

Let us remark that other related decompositons with non doubling measures have been obtained in [NTV2] and [MMNO]. However, these results are not suitable for our purposes.

Although the proof of the lemma can be found in [To4], for the reader’s convenience we have included it in the last section of the present paper.

Proof of Theorem 1.1. We will show that T is of weak type (1, 1). By similar arguments, one gets that T is bounded from M(ℂ) into L^{1,∞}(μ). In this case, one has to use a version of the Calderón-Zygmund decomposition in the lemma above suitable for complex measures (see the end of the proof for more details).

Let f ∈ L^{1}(μ) and λ > 0. It is straightforward to check that we may assume λ > 2^{d+1}∥f∥_{L^{1}(μ)}/∥μ∥. Let {Q_{i}}_{i} be the almost disjoint family of cubes of Lemma 2.4. Let R_{i} be the smallest (6, 6^{n+1})-doubling cube of the form 6^{k}Q_{i}, k ≥ 1. Then we can write f = g + b, with

g = f χ_{ℝ^{d}\⋃_{i} Q_{i}} + ∑_{i} φ_{i}

and

b = ∑_{i} b_{i} := ∑_{i} (w_{i} f − φ_{i}) ,

where the functions φ_{i} satisfy (2.4), (2.5) (2.6) and w_{i} = \frac{χ_{Q_{i}}}{∑_{k} χ_{Q_{k}}}. By (2.1) we have

μ (⋃_{i} 2Q_{i}) ≤ \frac{C}{λ} ∑_{i} ∫_{Q_{i}} |f| dμ ≤ \frac{C}{λ} ∫ |f| dμ.

So we have to show that

(2.7)  μ\{x ∈ ℝ^{d} \ ⋃_{i} 2Q_{i} : |T_{ε}f(x)| > λ\} ≤ \frac{C}{λ} ∫ |f| dμ.

Since ∫ b_{i} dμ = 0, supp(b_{i}) ⊂ R_{i} and ∥b_{i}∥_{L^{1}(μ)} ≤ C ∫_{Q_{i}} |f| dμ, using condition 2 in the definition of a Calderón-Zygmund kernel (which implies Hörmander’s condition), we get

∫_{ℝ^{d}\2R_{i}} |T_{ε}b_{i}| dμ ≤ C ∫ |b_{i}| dμ ≤ C ∫_{Q_{i}} |f| dμ.

Let us see that

(2.8)  ∫_{2R_{i}\2Q_{i}} |T_{ε}b_{i}| dμ ≤ C ∫_{Q_{i}} |f| dμ

too. On the one hand, by (2.6) and using the L^{2}(μ) boundedness of T and that R_{i} is (6, 6^{n+1})-doubling we get

∫_{2R_{i}} |T_{ε}φ_{i}| dμ ≤ (∫_{2R_{i}} |T_{ε}φ_{i}|^{2} dμ)^{1/2} μ(2R_{i})^{1/2}

≤ C (∫ |φ_{i}|^{2} dμ)^{1/2} μ(R_{i})^{1/2}
≤ C ∫_{Q_{i}} |f| dμ.

On the other hand, since supp(w_{i}f) ⊂ Q_{i}, if x ∈ 2R_{i} \ 2Q_{i}, then |T_{ε}f(x)| ≤ C ∫_{Q_{i}} |f| dμ/|x − x_{Q_{i}}|^{n}, and so

∫_{2R_{i}\2Q_{i}} |T_{ε}(w_{i} f)| dμ ≤ C ∫_{2R_{i}\2Q_{i}} \frac{1}{|x − x_{Q_{i}}|^{n}} dμ(x) × ∫_{Q_{i}} |f| dμ,

By Lemma 2.3, the first integral on the right hand side is bounded by some constant independent of Q_{i} and R_{i}, since there are no (6, 6^{n+1})-doubling cubes of the form 6^{k}Q_{i} between 6Q_{i} and R_{i}. Therefore, (2.8) holds.

Then we have

∫_{ℝ^{d}\⋃_{k} 2Q_{k}} |T_{ε}b| dμ ≤ ∑_{i} ∫_{ℝ^{d}\⋃_{k} 2Q_{k}} |T_{ε}b_{i}| dμ

≤ C ∑_{i} ∫_{Q_{i}} |f| dμ ≤ C ∫ |f| dμ.

Therefore,

(2.9)  μ\{x ∈ ℝ^{d} \ ⋃_{i} 2Q_{i} : |T_{ε}b(x)| > λ\} ≤ \frac{C}{λ} ∫ |f| dμ.

The corresponding integral for the function g is easier to estimate. Taking into account that |g| ≤ C λ, we get

(2.10) μ\{x ∈ ℝ^{d} \ ⋃_{i} 2Q_{i} : |T_{ε}g(x)| > λ\} ≤ \frac{C}{λ^{2}} ∫ |g|^{2} dμ ≤ \frac{C}{λ} ∫ |g| dμ.

Also, we have

∫ |g| dμ ≤ ∫_{ℝ^{d}\⋃_{i} Q_{i}} |f| dμ + ∑_{i} ∫ |φ_{i}| dμ

≤ ∫ |f| dμ + ∑_{i} ∫_{Q_{i}} |f| dμ ≤ C ∫ |f| dμ.

Now, by (2.9) and (2.10) we get (2.7).

If we want to show that T is bounded from M(ℂ) into L^{1,∞}(μ), then in Lemma 2.4 and in the arguments above f dμ must be substituted by dν, with ν ∈ M(ℂ), and |f| dμ by d|ν|. Also, condition (2.3) of Lemma 2.4 should be stated as “On ℝ^{d} \ ⋃_{i} Q_{i}, ν is absolutely continuous with respect to μ, that is ν = f dν, and moreover |f(x)| ≤ λ a.e. (μ) x ∈ ℝ^{d} \ ⋃_{i} Q_{i}”. With other minor changes, the arguments and estimates above work in this situation too. □

3. Proof of Lemma 2.4

(a) Taking into account Remark 2.2, for μ-almost all x ∈ ℝ^{d} such that |f(x)| > λ, there exists some cube Q_{x} satisfying

(3.1)  \frac{1}{μ(2Q_{x})} ∫_{Q_{x}} |f| dμ > \frac{λ}{2^{d+1}}

and such that if Q′_{x} is centered at x with l(Q′_{x}) > 2l(Q_{x}), then

\frac{1}{μ(2Q′_{x})} ∫_{Q′_{x}} |f| dμ ≤ \frac{λ}{2^{d+1}} .

Now we can apply Besicovich’s covering theorem (see Remark 3.1 below) to get an almost disjoint subfamily of cubes {Q_{i}}_{i} ⊂ {Q_{x}}_{x} satisfying (2.1), (2.2) and (2.3).

(b) Assume first that the family of cubes {Q_{i}}_{i} is finite. Then we may suppose that this family of cubes is ordered in such a way that the sizes of the cubes R_{i} are non decreasing (i.e. l(R_{i+1}) ≥ l(R_{i})). The functions φ_{i} that we will construct will be of the form φ_{i} = α_{i} χ_{A_{i}}, with α_{i} ∈ ℝ and A_{i} ⊂ R_{i}. We set A_{1} = R_{1} and φ_{1} = α_{1} χ_{R_{1}}, where the constant α_{1} is chosen so that ∫_{Q_{1}} f w_{1} dμ = ∫ φ_{1} dμ.

Suppose that φ_{1}, …, φ_{k−1} have been constructed, satisfy (2.4) and

∑_{i=1}^{k−1} |φ_{i}| ≤ B λ,

where B is some constant which will be fixed below.

Let R_{s_{1}}, …, R_{s_{m}} be the subfamily of R_{1}, …, R_{k−1} such that R_{s_{j}} ∩R_{k} ≠ ∅. As l(R_{s_{j}}) ≤ l(R_{k}) (because of the non decreasing sizes of R_{i}), we have R_{s_{j}} ⊂ 3R_{k}. Taking into account that for i = 1, …, k − 1

∫ |φ_{i}| dμ ≤ ∫_{Q_{i}} |f| dμ

by (2.4), and using that R_{k} is (6, 6^{n+1})-doubling and (2.2), we get

∑_{j} ∫ |φ_{s_{j}}| dμ ≤ ∑_{j} ∫_{Q_{s_{j}}} |f| dμ

≤ C ∫_{3R_{k}} |f| dμ ≤ Cλμ(6R_{k}) ≤ C_{2}λ μ(R_{k}).

Therefore,

μ \{_{∑j}|φ_{s_{j}}| > 2C_{2}λ\} ≤ \frac{μ(R_{k})}{2}.

So we set

A_{k} = R_{k} ∩ {_{∑j}|φ_{s_{j}}| ≤ 2C_{2}λ} ,

and then μ(A_{k}) ≥ μ(R_{k})/2.

The constant α_{k} is chosen so that for φ_{k} = α_{k} χ_{A_{k}} we have ∫ φ_{k} dμ = ∫_{Q_{k}} f w_{k} dμ. Then we obtain

|α_{k}| ≤ \frac{1}{μ(A_{k})} ∫_{Q_{k}} |f| dμ ≤ \frac{2}{μ(R_{k})} ∫_{\frac{1}{2} R_{k}} |f| dμ ≤ C_{3}λ

(this calculation also applies to k = 1). Thus,

|φ_{k}| + ∑_{j} |φ_{s_{j}}| ≤ (2C_{2} + C_{3}) λ.

If we choose B = 2C_{2} + C_{3}, (2.5) follows.

Now it is easy to check that (2.6) also holds. Indeed we have

∥φ_{i}∥_{L^{∞}(μ)} μ(R_{i}) ≤ C |α_{i}| μ(A_{i}) = C |∫_{Q_{i}} f w_{i} dμ| ≤ C ∫_{Q_{i}} |f| dμ.

Suppose now that the collection of cubes {Q_{i}}_{i} is not finite. For each fixed N we consider the family of cubes {Q_{i}}_{1≤i≤N}. Then, as above, we construct functions φ_{1}^{N}, …, φ_{N}^{N} with supp(φ_{i}^{N}) ⊂ R_{i} satisfying

∫ φ_{i}^{N} dμ = ∫_{Q_{i}} f w_{i} dμ,

∑_{i=1}^{N} |φ_{i}^{N}| ≤ B λ

and

∥φ_{i}^{N}∥_{L^{∞}(μ)} μ(R_{i}) ≤ C ∫_{Q_{i}} |f| dμ.

Notice that the sign of φ_{i}^{N} equals the sign of ∫ f w_{i} dμ and so it does not depend on N.

Then there is a subsequence {φ_{1}^{k}}_{k∈I_{1}} which is convergent in the weak ∗ topology of L^{∞}(μ) to some function φ_{1} ∈ L^{∞}(μ). Now we can consider a subsequence {φ_{2}^{k}}_{k∈I_{2}} with I_{2} ⊂ I_{1} which is also convergent in the weak ∗ topology of L^{∞}(μ) to some function φ_{2} ∈ L^{∞}(μ). In general, for each j we consider a subsequence {φ_{j}^{k}}_{k∈I_{j}} with I_{j} ⊂ I_{j−1} that converges in the weak ∗ topology of L^{∞}(μ) to some function φ_{j} ∈ L^{∞}(μ). It is easily checked that the functions φ_{j} satisfy the required properties. □

Remark 3.1. Recall that Besicovich’s covering theorem asserts that if Ω ⊂ ℝ^{d} is a bounded set and for each x ∈ Ω there is a cube Q_{x} centered at x, then there exists a family of cubes {Q_{x_{i}}}_{i} with finite overlap, that is ∑_{i} χ_{Q_{i}} ≤ C, which covers Ω.

In (a) of the preceeding proof we have applied Besicovich’s covering theorem to Ω = {x : |f(x)| > λ}. However this set may be unbounded, and the boundedness property is a necessary assumption in Besicovich’s theorem (example: take Ω = [0, +∞) ⊂ ℝ and consider Q_{x} = [0, 2x] for all x ∈ Ω).

We can solve this problem using different arguments. One possibility is to consider for each r > 0 the set Ω_{r} = {x : |x| ≤ r, |f(x)| > λ} and to apply Besicovich’s covering theorem to Ω_{r}. With the same arguments as above, we can decompose f = g + b, with |g| ≤ λ only on Ω_{r} and b as above. Then the proof of Theorem 1.1 can be modified to show that for any fixed constants λ, R > 0 one has

μ\{x ∈ B(0, R) : |T_{ε}f(x)| > λ\} ≤ C \frac{∥f∥_{L^{1}(μ)}}{λ} .

However we prefer the following solution. We are interested in showing that the Calderón-Zygmund decomposition of Lemma 2.4 works also without assuming Ω = {x : |f(x)| > λ} bounded. Let us sketch the argument. Consider a cube Q_{0} centered at 0 big enough so that

2^{d+1} ∥f∥_{L^{1}(μ)}/μ(Q_{0}) < λ.

So for any cube Q containing Q_{0} we will have

(3.2)  2^{d+1} ∥f∥_{L^{1}(μ)}/μ(Q) < λ.

For m ≥ 0 we set Q_{m} := (\frac{5}{4})^{m} Q_{0}. For each m we can apply Besicovich’s covering theorem to the annulus Q_{m} \ Q_{m−1} (we take Q_{−1} := ∅), with cubes Q_{x} centered at x ∈ supp(μ) ∩ (Q_{m} \ Q_{m−1}) as in (a) of the proof above, satisfying (3.1).

In this argument we have to be careful with the overlapping among the cubes belonging to coverings of different annuli. Indeed, there exist some fixed constants N and N′ such that if m ≥ N′, for x ∈ supp(μ)∩(Q_{m}\Q_{m−1}) we have

(3.3)  Q_{x} ⊂ Q_{m+N} \ Q_{m−N}.

Otherwise, it is easily seen that ℓ(Q_{x}) > \frac{3}{4}ℓ(Q_{m}), choosing N big enough. It follows that Q_{0} ⊂ 2Q_{x} since ℓ(Q_{0}) ≪ ℓ(Q_{m}) for N′ big enough too. This cannot happen because then 2Q_{x} satisfies (3.2), which contradicts (3.1).

Because of (3.3), the covering made up of squares belonging to the Besicovich coverings of different annuli Q_{m} \ Q_{m−1}, m ≥ 0, will have finite overlap.

Notice that in this argument, it is essential the fact that in (3.1) we are not dividing by μ(Q_{x}), but by μ(2Q_{x}).

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Department of Mathematics, Chalmers, 412 96 Göteborg, Sweden Current address: Département de Mathématique, Université de Paris-Sud, Bât. 425,

91405 Orsay Cedex, France

E-mail address: xavier.tolsa@math.u-psud.fr