Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 24, 1999, 155–163

REMARKS ON THE NONEXISTENCE OF DOUBLING MEASURES

Eero Saksman

University of Helsinki, Department of Mathematics, P.O. Box 4 (Yliopistonkatu 5) FIN-00014 University of Helsinki, Finland; eero.saksman@helsinki.fi

Abstract. We establish that there are bounded Jordan domains Ω ⊂ R^{n} (n ≥ 2) that do not carry a (nontrivial) doubling measure with respect to the Euclidean distance. More generally, it is shown that every nonempty metric space (X, d) without isolated points has an open and dense subset A such that (A, d) does not carry a doubling measure.

1. Introduction

The fundamental result on the existence of doubling measures, due to Vol’berg and Konyagin [11], states that every doubling and compact metric space (X, d) carries a nontrivial doubling measure. This remains true also if instead of compactness one only requires that X is complete, see [4]. In particular, every closed subset of R^{n} carries a nontrivial doubling measure. The existence question gains special interest from the fact that, in order to extend classical results of analysis to a setting of a metric space X, one is often forced to postulate a doubling measure on X (see e.g. [9, Chapter I]).

In the present note we show that one cannot replace compactness by local compactness in the theorem of Vol’berg and Konyagin, thus answering negatively a question posed in [4]. In fact, counterexamples contain remarkably ‘nice’ spaces that appear in analysis: Theorem 3 yields a bounded Jordan domain Ω ⊂ R^{n} (n ≥ 2) that carries no doubling measure with respect to the Euclidean distance. We also prove (Theorem 5) a more general result of nonexistence: for every nonempty metric space X which does not have isolated points there is an open and dense subset Ω ⊂ X such that Ω does not carry a doubling measure. Further observations are contained in Remarks 1 and 2 below.

We refer to [11], [3], [9, Chapter I, Section 8], [10], [12], [13], and [2] for additional results on doubling measures. In addition, the work of Staples and Ward [8] is directly related to Theorem 5, see Remark 3 below (we are grateful to J. Heinonen who noticed the connection between our results and [8]).

The research for this note was carried out during my visit to the University of Michigan, Ann Arbor. I thank F. Gehring and J. Heinonen for hospitality and for creating a stimulating mathematical atmosphere. Moreover, I thank J. Luukkainen for the careful reading of the manuscript and his valuable comments on it.

1991 Mathematics Subject Classification: Primary 28A12; Secondary 28A75.

Supported by the Academy of Finland.

2. Results

We first recall that if (X, d) is a metric space, then a Borel measure μ on X is doubling if there exists a constant C ≥ 1 so that the inequality

(1)  0 < μ(B(x, 2r)) ≤ Cμ(B(x, r)) < ∞

holds for all x ∈ X and r > 0 (in this case we also say that μ is C -doubling). Here B(x, r) = {y ∈ X | d(x, y) < r}. If B is a ball in X with radius r, then for k > 0 the abbreviation kB denotes the ball with the same center and with radius kr. Moreover, we write B_{c}(x, r) = {y ∈ X | d(x, y) ≤ r}.

A subset A′ of a metric space A is ε-dense in A for ε > 0 if A = ⋃_{y∈A′}B(y, ε). A metric space (X, d) is doubling (in the metric sense) if there is a constant K > 0 such that for every set A ⊂ B(x, 2r) with x ∈ X, r > 0 that satisfies d(y, z) ≥ r for distinct y, z ∈ A, the number of points in A is bounded from above by K. If X carries a doubling measure, then X is known to be doubling. The family of Borel sets of a metric space X is denoted by ℬ(X).

The following simple lemma is essentially known, but for completeness we include a proof.

Lemma 1. Let A be a dense subset of a metric space X. Then every C - doubling measure μ on A extends to a C -doubling measure μ̃ on X for which μ̃(S) = μ(S ∩ A) for S ∈ ℬ(X).

Proof. Note that A need not be a Borel subset of X, so that one needs to verify that μ̃ is a well-defined measure. For that end it is enough to note that {S ∈ ℬ(X) | S ∩ A ∈ ℬ(A)} = ℬ(X), and this follows by observing that the left-hand side is a σ -field that contains all open subsets of X. Let then x ∈ X and r > 0. Choose a sequence of points x_{k} ∈ A so that d(x_{k}, x) < \frac{1}{2}r and d_{k} = d(x_{k}, x) decreases to 0 as k → ∞. The inequality (1) follows by letting k → ∞ in the estimate

\begin{aligned}μ̃(B(x, 2r − 3d_{k})) & ≤ μ(B(x_{k}, 2r − 2d_{k}) ∩ A) \\ & ≤ Cμ(B(x_{k}, r − d_{k}) ∩ A) ≤ Cμ̃(B(x, r)). \end{aligned}

Our second lemma is a geometrically refined variant of known estimates (see e.g. [12, Lemma 1] for doubling measures. In what follows, the word rectangle (or square) refers to the closure of a rectangular domain.

Lemma 2. Let S = [0, a]^{2} ⊂ R^{2} and R = [0, b] × [0, a], where 0 < b < \frac{1}{2}a. Let μ be a C -doubling measure on a set U ⊂ R^{2} so that U ∩ S is (b/16)-dense in S. Then

\frac{μ(R ∩ U)}{μ(S ∩ U)} ≤ (1 − C^{−4})^{log_{2}(a/b)−2}.

Proof. The claim clearly holds for b ∈ [\frac{1}{4}a, \frac{1}{2}a). By induction it is then enough to prove it for b ∈ (0, \frac{1}{4}a) assuming that it is true for 2b. Assume hence that b ∈ (0, \frac{1}{4}a) and denote R′ = [0, 2b] × [0, a]. By the density assumption it is easily verified that there are disjoint balls B_{1}, …, B_{m} with centers in U so that ⋃_{i=1}^{m} B_{i} ⊂ R′ \ R and R′ ⊂ ⋃_{i=1}^{m} 2^{4}B_{i}. The doubling property of μ implies

μ(U ∩ R′) ≤ C^{4}μ(U ∩ ( ⋃_{i=1}^{m} B_{i}))

and hence

μ(U ∩ R) ≤ μ(U ∩ R′) − μ(U ∩ ( ⋃_{i=1}^{m} B_{i})) ≤ (1 − C^{−4})μ(U ∩ R′).

The lemma now follows by applying the induction hypothesis on the rectangle R′ since the needed density assumption is satisfied.

Theorem 3. For each n ≥ 2 there is a bounded Jordan domain Ω ⊂ R^{n} (even the image of B^{n} under a homeomorphism of R^{n}) which does not carry a doubling measure.

Proof. For reasons of clarity we first give a detailed proof in the case n = 2. Let us start by constructing a standard Cantor set F ⊂ R corresponding to a sequence (λ_{j})_{j=1}^{∞} of scalars, such that λ_{j} ∈ (0, \frac{1}{8}) and lim_{j→∞} λ_{j} = 0. Thus F = ⋂_{j=0}^{∞} F_{j}, where F_{0} = [0, 1] and for each j ≥ 1 the set F_{j} is obtained from F_{j−1} by dissecting a middle interval of length λ_{j}|I| from each of the 2^{j−1} closed intervals I that comprise F_{j−1} . Finally we choose λ_{j} = 2^{−5j} for j ≥ 1 (however, we keep writing λ_{j} instead of 2^{−5j} until to the very end of the argument, since we will later make another choice for (λ_{j}) in Remark 2).

Set K = F × F ⊂ R^{2} . Since K is a Cantor set, it is well known that there is a planar Jordan curve J such that K ⊂ J, see e.g. [7, Theorems 12.8 and 13.7]. Let Ω be the bounded component of R^{2} \J. We claim that Ω carries no doubling measure. Towards the proof of the claim denote by U the closure of Ω and assume that μ is a C -doubling measure on U. Lemma 1 implies that it is enough to show that μ cannot be supported on Ω, and this follows if we establish that

(2)  μ(K) > 0.
We may write K = ⋂_{j=0}^{∞} K_{j} with K_{j} = F_{j} × F_{j} and, moreover,

K_{j} = ⋃_{i=1}^{4^{j}} S_{j}^{i},

where each S_{j}^{i} (i = 1, …, 4^{j}) is a square.

Choose j_{0} = j_{0}(C) ≥ 1 so that b_{j} > \frac{1}{2} for j ≥ j_{0} , where
(3)  b_{j} = 1 − C^{4}(1 − C^{−4})^{−3+log_{2}((1−λ_{j})/λ_{j})}.

Let then j ≥ j_{0} and i_{0} ∈ {1, …, 4^{j}} be arbitrary and let S_{j+1}^{i_{r}}, r = 1, 2, 3, 4, be the four squares that comprise S_{j}^{i_{0}} ∩ K_{j+1} . We show that

(4)  ∑_{r=1}^{4} μ(S_{j+1}^{i_{r}} ∩ U) ≥ b_{j+1}μ(S_{j}^{i_{0}} ∩ U).

Before proving (4) we show how (2) follows from it. Note that (4) yields by induction the inequality μ(K_{j} ∩ U) ≥ μ(K_{j_{0}} ∩ U) ∏_{i=j_{0}+1}^{j} b_{i}, and letting j → ∞ we obtain

μ(K) = μ(K ∩ U) ≥ ( ∏_{j=j_{0}+1}^{∞} b_{j})μ(K_{j_{0}} ∩ U).

We observe that evidently μ(K_{j_{0}} ∩ U) > 0. Moreover, the elementary inequality b ≥ e^{−2(1−b)} for b ∈ (\frac{1}{2}, 1) implies that

∏_{j=j_{0}+1}^{∞} b_{j} ≥ exp(−2C^{4} ∑_{j=j_{0}+1}^{∞}(1−C^{−4})^{−3+log_{2}((1−λ_{j})/λ_{j})}) ≥ exp(−C′ ∑_{j=j_{0}+1}^{∞} λ_{j}^{d}),

where the positive constants d and C′ depend only on C. Hence ∏_{j=j_{0}+1}^{∞} b_{j} > 0 (independently of the value of C) provided that the sequence (λ_{j}) satisfies

(5)  ∑_{j=1}^{∞} λ_{j}^{δ} < ∞ for all δ > 0.

We obtain (2) since (5) obviously holds for the choice λ_{j} = 2^{−5j}.

It remains to prove (4). Let a_{j} = 2^{−j} ∏_{i=1}^{j}(1 − λ_{i}) be the sidelength of the square S_{j}^{i_{0}} , and write S_{j}^{i_{0}} in an obvious manner as a union of nine sets:

(6)  S_{j}^{i_{0}} = S′ ∪ ( ⋃_{r=1}^{4} R_{r}) ∪ ( ⋃_{r=1}^{4} S_{j+1}^{i_{r}}),

where each R_{r} (r = 1, 2, 3, 4) is a rectangle with sides λ_{j+1}a_{j} and \frac{1}{2}(1 − λ_{j+1})a_{j}, and S′ is a square with side λ_{j+1}a_{j}. Fix r ∈ {1, 2, 3, 4}. With a possible (and harmless) relabeling of the rectangles we may denote by R′_{r} the closed rectangle of the same size as R_{r}, having a common side with it, and contained in S_{j+1}^{i_{r}}. We now apply Lemma 2 and deduce that

μ(R′_{r} ∩ U) ≤ (1 − C^{−4})^{−3+log_{2}((1−λ_{j+1})/λ_{j+1})}μ(S_{j+1}^{i_{r}} ∩ U),

since the density hypothesis of Lemma 2 is satisfied owing to the construction of the Cantor set F and the fact that λ_{j+2} = λ_{j+1}/32. Because of the density we may also choose disjoint balls B_{1}, …, B_{m} with centers in U so that ⋃_{i=1}^{m} B_{i} ⊂ R′_{r} and S′ ∪ R_{r} ⊂ ⋃_{i=1}^{m} 2^{4}B_{i}, which implies that

μ((S′ ∪ R_{r}) ∩ U) ≤ C^{4}μ(R′_{r} ∩ U).

Combining these observations we deduce that

μ((S_{j}^{i_{0}} \( ⋃_{r=1}^{4} S_{j+1}^{i_{r}}))∩U) ≤ C^{4}(1−C^{−4})^{−3+log_{2}((1−λ_{j+1})/λ_{j+1})}μ(( ⋃_{r=1}^{4} S_{j+1}^{i_{r}})∩U),

which yields (4).

Finally, in the case n ≥ 3 we choose Ω = Ω_{2} ×(0, 1)^{n−2} , where Ω_{2} is the twodimensional domain constructed above. The proof of the theorem and a higher dimensional version of Lemma 2 remains the same almost verbatim if one replaces the two-dimensional sets A figuring in the proof by sets of the form A × [0, 1]^{n−2} . There are only minor differences: the power C^{4} has to be replaced by a power C^{k}, where k depends on the dimension n, and the density condition of Lemma 2 together with the choice of λ_{j} must be adjusted accordingly.

Remark 1. Under some additional conditions on the domain one can ensure the existence of a doubling measure on Ω with respect to the Euclidean distance. We give a simple example: every doubling measure on \overline{Ω} is supported on Ω if the domain Ω satisfies the following condition:

(A) There is a constant k > 1 with the following property: For every x ∈ ∂Ω and for every ε > 0 there is y ∈ Ω so that d(y, ∂Ω) < ε and x ∈ kB(y, d(y, ∂Ω)).

The existence of a doubling measure on Ω satisfying (A) follows from the Vol’berg–Konyagin theorem applied to \overline{Ω} (or from [4] if Ω is unbounded). Notice that (A) is satisfied e.g. by John domains, but also by some bounded domains that are not John (for the definition of a John domain, see [5, 2.1]). In order to prove our claim, we assume for simplicity that Ω is bounded. Denote A_{ε} = {x ∈ Ω | d(x, ∂Ω) ≤ ε}. Given ε > 0, assumption (A) and an application of a standard covering theorem (see e.g. [6, Theorem 2.1]) to the ball family ℬ = {B(y, d(y, ∂Ω)) | y ∈ A_{ε}} yield a constant k_{0} (that does not depend on ε) and disjoint balls B_{1}, B_{2}, … from the family ℬ so that ⋃_{i=1}^{∞} B_{i} ⊂ A_{2ε} and ∂Ω ⊂ ⋃_{i=1}^{∞} 2^{k_{0}}B_{i}. Let μ be C -doubling on \overline{Ω}. It follows that μ(∂Ω) ≤ C^{k_{0}}μ(A_{2ε}), where the right-hand side can be made arbitrarily small since ⋂_{j=1}^{∞} A_{2^{−j}} = ∅.

Remark 2. There are domains that are ‘nearly John’ but which do not carry a doubling measure. More precisely, there is a bounded domain Ω ⊂ R^{2} which does not carry a doubling measure but whose boundary is accessible in the following sense:

(B) Let ε > 0. For every boundary point x ∈ ∂Ω there is a path γ: [0, 1] → \overline{Ω} with finite length, and such that γ(0) = x and γ(1) = x_{0} , where x_{0} ∈ Ω is fixed. Moreover, for all t ∈ (0, 1] it holds that

B(γ(t), C_{ε}s^{1+ε}) ⊂ Ω,

where s is the length of the subarc γ([0, t]).

(Compare with the previous remark.) In fact, we may choose Ω = B(0, 3)\K, where K is the planar Cantor set constructed in the proof of Theorem 3 with the choice λ_{j} = 2^{−j/ log(j+2)} . The proof for our claim is practically equal to the proof of Theorem 3, now only the density conditions are automatically satisfied (independently of the choice of (λ_{j})) so that it remains to verify (5), which is immediate. Finally, it is not difficult to verify that Ω fulfils (B) with the choice x_{0} = (\frac{1}{2}, \frac{1}{2}).

Question. Suppose that Ω is a bounded domain in R^{n} and ∂Ω is piecewise given by the graph of a continuous function. Does Ω support a doubling measure?

We next turn to a general observation about nonexistence.

Lemma 4. Let X be a metric space without isolated points and assume that the set I = {C ≥ 1 | there is a C -doubling measure on X} is nonempty (notice that I is a half-line). Let y ∈ X. Then there are functions f_{1}, f_{2}: (0, 1) × I → (0, ∞) with the following properties:

(i) f_{1} is decreasing with respect to the second variable and f_{2} is increasing with respect to both variables. Moreover, lim_{r→0^{+}} f_{2}(r, C) = 0 for every C ∈ I.

(ii) Every C -doubling measure μ on X satisfies

f_{1}(r, C)μ(B(y, 1)) ≤ μ(B(x, r)) ≤ f_{2}(r, C)μ(B(y, 1))

for x ∈ B(y, 1) and r ∈ (0, 1).

Proof. We first conclude that X is doubling, since I ≠ ∅. Assume that μ is a C -doubling measure on X. The existence of the lower bound is obtained by a standard argument: given r ∈ (0, 1) and x ∈ B(y, 1), choose k_{0} = [log_{2}(1/r)] + 2 and notice that μ(B(y, 1)) ≤ μ(B(x, 2^{k_{0}}r)) ≤ C^{k_{0}}μ(B(x, r)). Hence an appropriate choice for f_{1} is f_{1}(r, C) = C^{−2}r^{log_{2} C}, which is clearly decreasing which respect to C since r < 1.

Towards the upper bound, we first consider the case where X is complete. Then, since metric doubling clearly implies total boundedness for bounded sets, we see that every closed ball of X is compact. For r ∈ (0, 1) and C ∈ I define

f_{2}(r, C) = sup_{x∈B(y,1)}{μ(B(x, r)) | μ is C-doubling on X with μ(B(y, 1)) = 1}.

Clearly f_{2} is well-defined since the supremum is bounded from above by C, and it is increasing with respect to both variables. We show that lim_{r→0^{+}} f_{2}(r, C) = 0, where C ∈ I is fixed. Assuming the contrary we deduce the existence of a sequence (x_{j}) with x_{j} ∈ B(y, 1), a sequence (r_{j}) of positive radii with r_{j} < 1/j, and a sequence of C -doubling measures μ_{j} on X with μ_{j}(B(y, 1)) = 1, and such that μ_{j}(B(x_{j}, r_{j})) ≥ c_{0} > 0 for all j ≥ 1. Note that 1 ≤ μ_{j}(B_{c}(y, 3)) ≤ C^{2} . We may extract a subsequence (j_{k}) such that μ_{j_{k}} → μ in the weak^{∗} -topology of Borel measures on the compact ball B_{c}(y, 3) and x_{j_{k}} → a as k → ∞. A simple reasoning (compare the proof of [4, Theorem 1]) shows that μ is nonzero and satisfies the doubling property (1) (with possibly a larger doubling constant) for x ∈ B_{c}(y, 1) and r ∈ (0, 1). Next, let ε ∈ (0, 1). There are arbitrarily large k such that B(x_{j_{k}}, r_{j_{k}}) ⊂ B(a, \frac{1}{2}ε). It follows that μ(B(a, ε)) ≥ c_{0} and hence μ({a}) > 0. However, this is clearly impossible since μ is doubling (for small radii) on a neighborhood of the nonisolated point a.

Consider finally the general case where X may possibly be noncomplete. Let Y be the completion of X. Then Y is doubling and perfect so that the above reasoning yields an appropriate f_{2} for Y and y ∈ X ⊂ Y . If μ is a C -doubling measure on X, then Lemma 1 yields a C -doubling extension μ̃ on Y . Then μ(B^{X}(x, r)) ≤ f_{2}(r, C)μ(B^{X}(y, 1)) for x ∈ B^{X}(y, 1) and r ∈ (0, 1), by the definition of the extension μ̃. Hence f_{2} (or a suitable restriction of it) will do the job for X.

Theorem 5. Let X be a nonempty metric space without isolated points. Then there is a dense open set A ⊂ X such that A does not carry a doubling measure.

Proof. We may assume that X carries a C_{0} -doubling measure for some C_{0} ≥ 1 since otherwise we may choose A = X. Fix y ∈ X and select functions f_{1} and f_{2} that satisfy the conditions stated in Lemma 4. Choose a dense sequence (y_{k})_{k=1}^{∞} of points in X (because X is doubling it is separable) so that y_{k} ≠ y for k ≥ 1. The set A will be constructed in the form A = ⋃_{k=1}^{∞} B(x_{k}, r_{k}), where the positive radii r_{k} ∈ (0, \frac{1}{8}) and the points x_{k} ∈ X are chosen inductively in such a way that the following conditions are satisfied for each n ≥ 1:

(i) x_{n} = y_{k_{n}}, where k_{n} = inf\{k | y_{k} ∉ \overline{⋃_{j=1}^{n−1} B(x_{j}, r_{j})} \},

(ii) d(x_{n}, \overline{⋃_{j=1}^{n−1} B(x_{j}, r_{j})} )≥ 3r_{n} if n > 1, (iii) y ∉ \overline{⋃_{j=1}^{n} B(x_{j}, r_{j})},

(iv) f_{2}(r_{n}, n) ≤ \frac{1}{2n} min\{f_{1}(r_{n−1}, n − 1), f_{2}(r_{n−1}, n − 1)\} if n ≥ C_{0} + 1,

(v) {z ∈ X | d(z, x_{n}) = 2r_{n}} ≠ ∅. In order to start the induction set x_{1} = y_{1} , and since x_{1} is not an isolated

point we may choose r_{1} > 0 satisfying (v) and with r_{1} < min\{\frac{1}{2}d(y_{1}, y), \frac{1}{8}\}. Assume then that n ≥ 2 and that x_{1}, …, x_{n−1} together with r_{1}, …, r_{n−1} have been chosen so that conditions (i)–(v) hold for the respective indexes. We next choose x_{n} according to (i). This is possible according to the induction hypothesis since (iii) implies that \{y_{k}|k ≥ 1\} \ \overline{⋃_{j=1}^{n−1} B(x_{j}, r_{j})} ≠ ∅. Then (ii)–(iv) hold once r_{n} is chosen small enough, and also (v) may be satisfied since x_{n} is not isolated in X. The induction argument is complete. Note that (iv) implies for n ≥ C_{0} the estimate

(iv)  f_{1}(r_{n}, n) ≥ 2nf_{2}(r_{n+1}, n + 1) ≥ n ∑_{k=n+1}^{∞}f_{2}(r_{k}, k),

since the infinite series is bounded from above by a geometric majorant which is obtained from the observation

f_{2}(r_{k}, k) ≤ f_{2}(r_{k−1}, k − 1)/2k ≤ f_{2}(r_{k−1}, k − 1)/2.

Assume then that μ is a C -doubling measure on A. Note that (i) implies the density of A in X, and hence Lemma 1 extends μ to a C -doubling measure μ̃ on X. We may assume that μ̃(B(y, 1)) = 1 and C ≥ C_{0} . By the fact \overline{A} = X and (iii) we may choose n ≥ 2C^{2} so that x_{n} ∈ B(y, \frac{1}{2}). According to (v) there is z ∈ X satisfying d(x_{n}, z) = 2r_{n}. Write B_{1} = B(x_{n}, r_{n}) and B_{2} = B(z, r_{n}). Then B_{1} ∩ B_{2} = ∅ and (ii) implies that also B_{2} ∩ B(x_{k}, r_{k}) = ∅ for all k < n. Hence we may apply Lemma 4 and (vi) in order to deduce that

\begin{aligned}μ̃(B_{2}) & ≤ μ(  ⋃_{\{k≥n+1|x_{k}∈B(y,1)\}} B(x_{k}, r_{k})) ≤ ∑_{\{k≥n+1|x_{k}∈B(y,1)\}} f_{2}(r_{k}, C) \\ & ≤ ∑_{k=n+1}^{∞} f_{2}(r_{k}, k) ≤ \frac{1}{n}f_{1}(r_{n}, n) ≤ \frac{1}{2C^{2}} f_{1}(r_{n}, C) ≤ \frac{1}{2C^{2}} μ̃(B_{1}). \end{aligned}

This contradicts the facts that μ̃ is C -doubling and B_{1} ⊂ 4B_{2} . Hence A carries no nontrivial doubling measure.

Remark 3. In the case where X = [0, 1], the existence of a dense open subset of X without a doubling measure may in fact be deduced as a simple consequence of a result due to Staples and Ward [8, Theorem 1.2]. Namely, in [8] a subset K ⊂ [0, 1] is called quasisymmetrically thick if there is no quasisymmetric map ϕ from [0, 1] onto [0, 1] such that |ϕ(K)| = 0 (for the definitions and properties of quasisymmetric maps on the real line we refer to [1]). Here | · | refers to the Lebesgue measure. Choose a closed and quasisymmetrically thick subset K ⊂ [0, 1] with dense complement K^{c} (such sets are provided by [8, Theorem 1.2] and, not surprisingly, our reasoning for Theorem 5 partly resembles their proof). Assume that K^{c} carries a normalized doubling measure μ. Let μ̃ be the extension of μ onto [0, 1] provided by Lemma 1. For x ∈ [0, 1] define ϕ(x) = μ̃([0, x]). Then ϕ is a quasisymmetric map of [0, 1] onto itself such that |ϕ(K)| = 0, which is impossible. Hence K^{c} carries no doubling measures.

Remarks on the nonexistence of doubling measures 163

References

[1] Beurling, A., and L. Ahlfors: The boundary correspondence under quasiconformal mappings. - Acta Math. 96, 1956, 125–142.

[2] Buckley, S.M., B. Hansen, and P. MacManus: Doubling conditions for general sets. - In preparation.

[3] Kaufman, R., and J.-M. Wu: Two problems on doubling measures. - Rev. Math. Iberoamericana 11, 1995, 527–545.

[4] Luukkainen, J., and E. Saksman: Every complete doubling metric space carries a doubling measure. - Proc. Amer. Math. Soc. (to appear).

[5] Martio, O., and J. Sarvas: Injectivity theorems in plane and space. - Ann. Acad. Sci. Fenn. Ser. A I Math. 4, 1979, 383–401.

[6] Mattila, P.: Geometry of Sets and Measures in Euclidean Spaces. - Cambridge University Press, 1995.

[7] Moise, E.M.: Geometric Topology in Dimensions 2 and 3. - Springer-Verlag, 1977.

[8] Staples, S.G., and L.A. Ward: Quasisymmetrically thick sets. - Ann. Acad. Sci. Fenn. Math. (to appear).

[9] Stein, E.M.: Harmonic Analysis. - Princeton University Press, 1993.

[10] Tukia, P.: Hausdorff dimension and quasisymmetric mappings. - Math. Scand. 65, 1989, 152–160.

[11] Vol’berg, A.L., and S.V. Konyagin: On measures with the doubling condition. - Math. USSR-Izv. 30, 1988, 629–638.

[12] Wu, J.-M.: Null sets for doubling and dyadic doubling measures. - Ann. Acad. Sci. Fenn. Ser. A I Math. 18, 1993, 77–91.

[13] Wu, J.-M.: Hausdorff dimension and doubling measures on metric spaces. - Proc. Amer. Math. Soc. (to appear).

Received July 1997