The isomorphism problem of Hausdorff measures and Hölder restrictions of functions

Transcription

1 The isoorphis proble of Hausdorff easures and Hölder restrictions of functions Doctoral thesis András Máthé PhD School of Matheatics Pure Matheatics Progra School Leader: Prof. Miklós Laczkovich Progra Leader: Prof. András Szűcs Supervisor: Prof. Miklós Laczkovich eber of the Hungarian Acadey of Sciences Eötvös Loránd University Institute of Matheatics 2009

2 Acknowledgeents I would like to express y deepest gratitude to y advisor Miklós Laczkovich for calling y attention to any inspiring probles, especially to the probles fro which this thesis eerged, for any useful suggestions and invaluable discussions, and for his guidance in preparing this anuscript. I a extreely grateful to Taás Keleti for any long and enlightening discussions and for his continuous support and encourageent. I a deeply indebted to Márton Elekes for various discussions and suggestions. It would be hard to overestiate the ipact of his previous results and ideas on this thesis. I a also grateful to Károly Sion for his inspiring talk on rando Cantor sets, which eventually led to the solution of the isoorphis proble of Hausdorff easures. I would also like to thank David Frelin and David Preiss for soe helpful suggestions and rearks.

3 Contents 1 Introduction The questions Restrictions of functions Borel aps and Hausdorff diension Borel aps and Hausdorff easures Notation and definitions Hölder restrictions and restrictions of bounded variation Outline Discrete Hausdorff pre-easure Bounded variation discrete version Overview of the proof Foralizing the discrete proble Solution of the discrete proble Bounded variation the continuous case Liit theores Proof of Theore Hölder restrictions Overview of the proof Preliinaries Proofs Generalizations and open questions Borel aps and Hausdorff diension Outline Preliinaries The rando tree Upper estiate Lower estiate ii

4 3.6 Proof of Theore Generalization of Theore to Euclidean spaces Borel aps and Hausdorff easures The rando tree Upper estiate Lower estiate Estiates of the exponential oent Main results Results on the real line Results in Euclidean spaces Bibliography 71 Suary 72 Magyar nyelvű összefoglalás (Suary in Hungarian 73 iii

5 Chapter 1 Introduction 1.1 The questions It was an unsolved proble for several years whether Hausdorff easures of different diensions can be Borel isoorphic. This proble was attributed to B. Weiss and was popularized by D. Preiss. Let B denote the Borel σ-algebra of R, and let H d denote the d-diensional Hausdorff easure; then the exact question reads as follows. Question 1. Let 0 s, t 1 and s t. (i Can the easure spaces (R, B, H s and (R, B, H t be isoorphic? (ii Does there exist a Borel bijection f : R R such that for every Borel set B, 0 < H s (B < 0 < H t (f(b <? The two parts are not equivalent but it is easy to see that a negative answer to (ii iplies a negative answer to (i. It is iportant to ake the distinction that we are looking for Borel isoorphiss only. M. Elekes [4] proved that if we assue the continuu hypothesis then the easure spaces (R, M H s, H s and (R, M H t, H t are isoorphic whenever s, t (0, 1, where M H d denotes the σ-algebra of the sets which are easurable with respect to H d. Reark In fact, the bijection f : R R that M. Elekes constructed (assuing the continuu hypothesis satisfies that if B R is Borel and H s (B <, then f(b is Borel and H s (B = H t (f(b; if f(b is Borel and H t (f(b <, then B is Borel and H s (B = H t (f(b. However, this ap f is not Borel easurable.

6 1.2 Restrictions of functions 2 M. Elekes, aiing to solve the original proble (Question 1, raised the following question [4]. Question 2. Let 0 < α < 1. Can we find for every Borel (or continuous, or typical continuous function f : [0, 1] R a Borel set B [0, 1] of positive Hausdorff diension such that f restricted to B is Hölder continuous of exponent α? How is this question related to the previous one? Suppose that we have an answer to Question 2 saying that (for soe fixed α for every Borel function f there exists a Borel set B of diension β such that f is Hölder-α on B. As it is wellknown, this iplies that f(b has diension at ost β/α. It is easy to see that this would answer (both parts of Question 1 in the negative for those s and t for which 0 s < β < β/α < t 1 holds. According to a theore of P. Huke and M. Laczkovich [6], a typical continuous function f : [0, 1] R is not onotonic on any set of positive Hausdorff diension. Since every function of bounded variation is the su of two onotonic functions, this theore otivated M. Elekes to raise an analogue of Question 2. Question 3. Can we find for every Borel (or continuous, or typical continuous function f : [0, 1] R a Borel set B [0, 1] of positive Hausdorff diension such that f restricted to B is of bounded variation? Can we even find such a set of diension 1/2? This proble has also been circulated by D. Preiss, and a siilar question was already asked by P. Huke and M. Laczkovich, see also Z. Buczolich [2, 3]. M. Elekes gave partial answers to Question 3 and Question 2 in [4]. He proved that a typical continuous function f : [0, 1] R is not of bounded variation on any set of Hausdorff diension larger than 1/2. Regarding Question 2, he also gave an upper bound for the possible diension by showing that for every 0 < α 1, a typical continuous function is not Hölder-α on any set of diension larger than 1 α. 1.2 Restrictions of functions We answer Question 2 and Question 3 by proving the following theores. (In this chapter all theores are nubered as they will appear in the following chapters. Theore Let f : [0, 1] R be Lebesgue easurable. Then there exists a copact set C [0, 1] of Hausdorff diension 1/2 such that f C is of bounded variation.

7 1.2 Restrictions of functions 3 Theore Let f : [0, 1] R be Lebesgue easurable and let 0 < α < 1. Then there exists a copact set C [0, 1] of Hausdorff diension 1 α such that f C is a Hölder-α function. That is, the diension bounds found by M. Elekes are sharp. These theores will be proved in Chapter 2. To prove Theore 2.1.1, first we will define discrete Hausdorff pre-easures on Z (the integers. Using this notion we will be able to foralize and solve a discrete (quantitative version of the proble. Then suitable liit theores will yield Theore It is possible to prove Theore in exactly the sae way. However, we will present a siple proof instead, using a theore of P. Mattila about Hausdorff diensions of plane sections. We will also ention soe generalizations of Theores and Reark Theores and belong to the faily of restriction theores. The setting of a restriction theore usually is the following. Given soe function f fro soe class X, one tries to find a large set A such that f A belongs to soe other (nice class Y. Here largeness usually eans that A is infinite, uncountable, perfect, not porous, or A is of positive easure or of second category. It is interesting that for the above questions of M. Elekes, the proper notion of largeness is Hausdorff diension. We refer to the survey article of J. B. Brown [1] on restriction theores and to the references therein. Reark Notice that if A R and f : A R is a given function, then there exists a function g : R R extending f (that is, g A = f such that the total variation of g and f are equal and that f is Hölder-α if and only if g is Hölder-α (0 < α 1. (Given f, one can easily define g on the closure of A, and then the linear extension works. This yields that for every function f : [0, 1] R and β [0, 1] the following are equivalent: (i there exists a set A of diension at least β such that f A is of bounded variation; (ii there exists a function g : [0, 1] R of bounded variation such that the set [f = g] (that is, {x : f(x = g(x} has diension at least β. The sae equivalence holds for the Hölder-α property. Thus Questions 3 and 2 and Theores and could have been forulated equivalently corresponding to (ii as well.

8 1.3 Borel aps and Hausdorff diension Borel aps and Hausdorff diension As explained above, Theore gives soe partial results on the isoorphis proble of Hausdorff easures (Question 1. For exaple, applying the theore for α = 2/3 iplies that the Hausdorff easures of diension 1/3 ε and 1/2+ε cannot be Borel isoorphic. However, this approach does not see to answer Question 1 in its full generality. In Chapter 3 we will show the following. Theore Let 0 d 1 and let f : R R be Borel easurable. Then there exists a copact set C of Hausdorff diension d such that f(c is of Hausdorff diension at ost d. Clearly, this theore iediately iplies that the Hausdorff easures of different diensions cannot be Borel isoorphic; that is, we answer (both parts of Question 1 in the negative. The proof of Theore is based on two types of rando constructions. One of the can be used to obtain a rando Cantor set of diension at ost d alost surely; and the other to obtain a rando copact set of diension at least d alost surely. We will also prove the following generalization of Theore Theore Let D R n be a Borel set and let f : D R be Borel easurable. Then for every 0 d 1 there exists a Borel set A D such that di A = d did and di f(a d di f(d. 1.4 Borel aps and Hausdorff easures After we have shown that Hausdorff easures of different diensions are not Borel isoorphic, another question arises. Can it happen that H s and 2 H s are Borel isoorphic? That is, does there exist a Borel bijection f such that 2 H s (B = H s (f(b for every Borel set B? Of course, if we consider R (and R R Borel bijections, then the answer is positive: siilarities of ratio 2 1/s realize the isoorphis. However, if we consider the unit interval [0, 1] only, then the question is highly non-trivial. We will answer this question in Chapter 4 by showing the following. Theore Let f : [0, 1] [0, 1] be Borel easurable. Then for every 0 < s < 1 there exists a copact set C [0, 1] such that H s (C = 1 and H s (f(c 1. The proof of this theore is siilar to the proof of Theore That is, we will use two types of rando constructions. However, to prove that the rando

9 1.5 Notation and definitions 5 set produced by one of the constructions is sufficiently large, requires uch ore effort than in the analogous proof in Chapter 3. This is the reason why we discuss Theore separately, despite the fact that Theore is a stronger stateent. In Chapter 4 we will also prove several related theores. Aong others, we will show that if f : R n R n is a Borel bijection which preserves the s-diensional Hausdorff easure (for soe 0 < s < n, then it also preserves the Lebesgue easure (Theore Notation and definitions We say that the real function f is of bounded variation on the set A if f restricted to A is a function of bounded variation. We say that the real function f is Hölder continuous of exponent α (or briefly Hölder-α on the set A if f A is Hölder-α; that is, there exists a real nuber B > 0 such that for every x, y A, f(x f(y B x y α. Given A R and f : A R, we denote the total variation of f by Varf; that is, Varf = sup { n 1 } f(x i+1 f(x i : n 1, x 1 < x 2 <... < x n, x i A. i=1 Given A R and f : A R, we say that f B-Hölder α if x, y A f(x f(y B x y α. The word typical is used in the Baire category sense. For x R, we denote by x the ceiling of x, that is, the sallest integer not saller than x. We denote by N the set of non-negative integers. As usual in set theory, we identify each n N with the set of its predecessors: n = {0, 1,..., n 1}. The Lebesgue easure on R and on any Euclidean space R n will be denoted by λ. It will be always clear fro the context which Lebesgue easure we are using. We denote the diaeter of a set U R n by dia U. Let H (A s denote the s-diensional Hausdorff pre-easure of the set A R; that is, { } H s (A = inf (dia I i s : A I i. i=1 i=1

10 1.5 Notation and definitions 6 Let H s (A denote the s-diensional Hausdorff easure of A R n ; that is, where i=1 H s (A = li δց0 H s δ(a, { } Hδ s (A = inf (dia U i s : A U i, dia U i δ. The Hausdorff diension of a set A is then defined as i=1 di A = inf{s > 0 : H s (A = 0}. By diension we always ean the Hausdorff diension. We call an interval non-trivial if it has positive length. The diaeter of an interval I will soeties be denoted by I. However, for finite sets S, S will denote the nuber of eleents of S. For ulti-indices i, i will denote the length of i. The intended interpretation will be always clear fro the context. If µ is a Borel easure on R, we will denote its support by supp µ; that is, supp µ = {x R : r > 0 µ((x r, x + r > 0}. We say that µ is supported on K if supp µ K. Probability will be denoted by P. The expected value of a rando variable X will be denoted by EX or E(X. Conditional expectation will be denoted by E(X A (where A is an event of positive probability. We denote the indicator variable by ½; that is, ½ A = 1 if the event A is satisfied and zero otherwise. Let f : D R be an injective ap defined on D R n. For any set A R n, we define f(a to be f(a D. (If we regard f as (f 1 1, this is the generally accepted notation. Let µ be a Borel easure on R n (for exaple λ or H s. Let D R n be a Borel set, and let f : D R n be a Borel apping. We say that f preserves the easure µ, if for every Borel set B f(d we have µ(f 1 (B = µ(b. (Note that this iplies that we have µ(f 1 (B µ(b for every Borel set B R n. We will use any ties in our proofs (without explicitly stating that the iage of a Borel set by a Borel apping is analytic, thus Lebesgue and H s -easurable. We will also use the fact that the iage of a Borel set by an injective Borel apping is Borel. We will also need that if A R n is analytic and H s (A > 1, then there exists a copact set C A such that H s (C = 1. See [10, Theore 8.13] or [5, Theore 471S].

11 Chapter 2 Hölder restrictions and restrictions of bounded variation 2.1 Outline As we stated in the Introduction ( 1.2, our goal here is to answer Question 2 and Question 3 by proving the following theores. Theore Let f : [0, 1] R be Lebesgue easurable. Then there exists a copact set C [0, 1] of Hausdorff diension 1/2 such that f C is of bounded variation. Theore Let f : [0, 1] R be Lebesgue easurable and let 0 < α < 1. Then there exists a copact set C [0, 1] of Hausdorff diension 1 α such that f C is a Hölder-α function. Fro the results of M. Elekes [4] it follows that these theores are sharp. The chapter will be organized as follows. First, in 2.2, we define discrete Hausdorff pre-easures on the integers. Using this notion we foralize and solve a discrete (quantitative version of Theore in 2.3. Then in 2.4 we prove suitable liit theores, and finally deduce Theore In 2.5 we present a siple proof of Theore 2.1.2, using a theore of P. Mattila about Hausdorff diensions of plane sections. Finally, in 2.6, we ention a variant of Theore regarding generalized variations and we extend our theores to Euclidean spaces. We note that Theore can be proved in exactly the sae way as Theore Such a proof of Theore (and the proof of Theore presented here can be found in [9].

12 2.2 Discrete Hausdorff pre-easure Discrete Hausdorff pre-easure We define the discrete Hausdorff pre-easure on the subsets of the set of integers Z. The covering sets will be intervals I Z, and here by interval we ean a set of finitely any consecutive integers. By I we denote the nuber of eleents of I. Let d(x, Y denote the usual distance of X and Y (X, Y R. If X or Y is epty, then we define their distance to be. Definition Let 0 < s 1. The discrete Hausdorff pre-easure of diension s is the function µ s : P(Z [0, ] defined by { µ s (A = in I s : I is a collection of intervals of Z such that A } I. I I It is reasonable to call µ s a pre-easure since it is subadditive. Lea Let 0 < s 1 and A, B Z. Then µ s (A B µ s (A + µ s (B. We also have a lower bound of µ s (A B. Lea Let 0 < s 1 and A, B Z. Then ( µ s (A B in d(a, B s, µ s (A + µ s (B. Proof. Consider a covering of A B by intervals of integers. If there is an interval of size at least d(a, B then the inequality clearly holds. Suppose that every interval is of size at ost d(a, B. Then each interval can intersect either A or B but not both, so we can split the covering into two parts to cover A and to cover B, which corresponds to the case µ s (A B µ s (A + µ s (B. The following stateent connects µ s to the (real Hausdorff pre-easure H. s Lea Let 0 < s 1. For a set A Z, let us define A def = i { [i, i ] : i A}. Then H s (A µ s (A 2 s H s (A.

13 2.3 Bounded variation discrete version 9 Proof. We ay suppose that A is finite (that is, bounded, otherwise H s (A = µ s (A =. We iediately obtain the inequality H s (A µ s (A if we change each covering interval I Z of A to the interval [in I, axi + 1]. Since A is copact, it is enough to consider finite coverings of A with closed intervals to calculate H s (A. Notice that we ay suppose that the covering intervals are disjoint, since (a + b s a s + b s for every a, b 0. Hence we ay also suppose that all the intervals covering A are of the for [n, n + l + 1 ] (of length 2 l + 1 for soe integer n and l N. Hence one can cover A by the corresponding 2 intervals {n, n +1,...,n+l} of size l + 1. Since (l +1 s 2 s (l s for every l N, we obtain the inequality µ s (A 2 s H s (A. Definition Let A Z. We say that a apping ϕ : A Z is non-contractive if ϕ(x ϕ(y x y for every x, y A. In the sequel we will use the following observation any ties. Lea If ϕ : A Z is non-contractive, then µ s (ϕ(a µ s (A. The proof is left to the reader. 2.3 Bounded variation discrete version Overview of the proof Before we start we give an inforal overview of the proof of Theore So let f : [0, 1] R be easurable. Then f is continuous on soe copact set of positive easure. Let us just suppose that f is continuous on the whole interval [0, 1], it will not ake uch difference. We would like to prove that f possesses the property that there exists a set C [0, 1] of large diension such that f C has finite variation. The key observation is that this property (or at least a quantitative version of this property goes through unifor convergence. That is, if soe (not necessarily continuous functions f n converge uniforly to f, and there exist copact sets C n such that H (C s n ε and Varf n Cn B for soe ε > 0 and finite B, then there exists a copact set C such that H s (C ε and Var f C B. (2.1

14 2.3 Bounded variation discrete version 10 (Note that H s (C > 0 iplies that the diension of C is at least s. Hence it is enough to show that this (quantitative property holds for a dense faily of functions. We choose the faily of those functions g : [0, 1] R which are piecewise constant on the intervals [ i, i+1 (i = 0,...,n 1. But for siplicity, we will deal with the n n discrete functions h : n R related to g by h(i = g( i (i = 0,...,n 1. n Now it is not difficult (using Lea to relate the following two stateents: (i there exists a copact set C [0, 1] such that H s (C is large and Varg C is sall; (ii there exists a set A n such that µ s (A is large and Var h A is sall. Thus we only need to show that stateent (ii (after properly forulated holds for all n and all functions h : n R, when s = 1/2. (Unfortunately, we can show this for any fixed s < 1/2 only, but this will be enough to prove Theore In soe sense, the proof will go by induction on n. This stateent is what we will forulate precisely and prove in this section Foralizing the discrete proble In the following definition n is a positive integer, B is a positive real nuber, and s, α (0, 1]. Definition b(n, B, s = in f:n [0,1] { ax µ s (A : Var f A B }. A n Notice that b is onotone increasing both in n and in B. Clearly, b(n, B, s 1 for all n, since the µ s -easure of a single point is 1. The discrete analogue of Theore is the following. Theore For every 0 < s < 1/2 and B > 0, b(n, B, s inf > 0. n 1 n s Note that the denoinator n s is present because, when we exchange a function g : [0, 1] R for the function h : n R (as in the inforal overview at the beginning of this section, there is a scaling by a factor of n, and this changes s-diensional easures by a factor of n s.

15 2.3 Bounded variation discrete version Solution of the discrete proble First we state and prove two ain induction steps, and then what reains, will be just calculations. Lea Fix a positive integer K. Then for every s < 1/2 and B > 0 b(n, KB + K 1, s in (( n s, K b( n, B, s 4K 2K for every n N large enough (depending only on K. Proof. Let us choose K intervals of integers I 1,...,I K inside n = {0, 1,..., n 1} of size n 2K such that the distance of any two of the is at least n ; clearly this can 4K be done if n is sufficiently large (depending on K. Fix any function f : n [0, 1]. We have to find a set A n of large µ s -easure such that Var f A KB + K 1. For each interval I j, consider the function f Ij. Since I j = n n, by the definition of b(, B, s we can find a set A 2K 2K j I j such that Var f Aj B and µ s (A j b( n, B, s. 2K Put A = K j=1a j. Applying Lea inductively to the sets A j we get µ s (A in (( n s 4K, K b( n, B, s. (2.2 2K Since Varf A K j=1 Varf A j + (K 1 KB + K 1, (2.2 instantly gives Lea Lea For each positive integer L, b(n, B, s b( n, BL, s. L Proof. Fix any function f : n [0, 1]. We have to find a set A n such that Varf A B and that µ s (A b( n, BL, s. For each i L let L S i = {x n : i L f(x i+1 L }. There exists an i L such that S i n L ; let S be a subset of this S i of size exactly S = n. Let ϕ : S S be the enueration of S; that is, ϕ is the onotone L increasing bijection fro S to S. Thus ϕ is a non-contractive apping. Define g : S [0, 1] by setting g(x = L (f(ϕ(x i L. (2.3

16 2.3 Bounded variation discrete version 12 By the definition of b( S, BL, s, there exists a set T S such that µ s (T b( S, BL, s and Var g T BL. Using Lea and (2.3, µ s (ϕ(t µ s (T b( S, BL, s and Varf ϕ(t = 1 L Varg T B. Thus A can be chosen as ϕ(t, which proves this Lea. Proof of Theore We consider s < 1/2 to be fixed. Let K be a sufficiently large positive integer, in fact, let K > 2 2s K 2s hold. First we will prove the Theore for B = 2K 1. Let us apply Lea with B = 1. We obtain an N N such that for all n N, b(n, 2K 1, s in (( n s 4K, K b( n, 1, s. 2K Now apply Lea to the right hand side with L = 2K 1. We obtain b(n, 2K 1, s in (( n s, K b( n 2K, 2K 1, s (n 4K 2K 1 N. Since b is onotone increasing in its first coordinate, b(n, 2K 1, s in ( ( n s n, K b(, 2K 1, s (n N. (2.4 4K 2K(2K 1 Fix an arbitrary positive integer n 0, and define the sequence n i+1 = n i 2K(2K 1 Let j be the sallest nonnegative integer for which either b(n j, 2K 1, s ( n j s 4K or n j N holds. Thus fro (2.4 we obtain (i N. (2.5 b(n i, 2K 1, s K b(n i+1, 2K 1, s (0 i < j, (2.6 and b(n j, 2K 1, s ( n j 4KN s, (2.7 since if n j N, then the right hand side is saller than 1, which is a trivial lower

17 2.4 Bounded variation the continuous case 13 bound. Thus fro (2.6 and (2.7 we get Using (2.5 we obtain the lower bound b(n 0, 2K 1, s K j ( n j b(n 0, 2K 1, s K j ( n j 4KN s. 4KN = ( 1 4KN s n s 0 s Kj ( 1 ( 4KN s n s 0 ( 1 js 2K(2K 1 K j ( 1 s n s 2 s K s (2K 1 s 4KN 0 (2.8 K provided that > 1, which clearly holds since K was chosen so that K > 2 s K s (2K 1 s 2 2s K 2s holds. Since n 0 was arbitrary, fro (2.8 we iediately obtain that b(n, 2K 1, s inf ( 1 s n 1 n 4KN > 0. (2.9 s Now let B > 0 be arbitrary. Let L N be so large that BL 2K 1 holds. Using Lea 2.3.4, the fact that b is onotone increasing in its second coordinate, and then (2.9, b(n, B, s inf n 1 n s b( n, BL, s L inf n 1 n s b( n, 2K 1, s L inf n 1 n s inf n 1 b(n, 2K 1, s (n L s > Bounded variation the continuous case Liit theores The inforal overview in contains a precise stateent (2.1 about unifor convergence. We will not prove that stateent for two reasons. On the one hand, it is not sufficient for us, because we also have to deal with functions f : [0, 1] R which are not continuous, just easurable. On the other hand, we do not need the stateent in this generality, since it is ore convenient to prove a siilar theore for soe specific sequence f n only. Let K R be copact, and let f : K R be continuous. Let K n = { [ i n, i+1 n ] : K [ i n, i+1 n ], i Z},

18 2.4 Bounded variation the continuous case 14 and define f n : K n R by setting f n (x = f ( in(k [ i n, i+1 n ] where i is the largest integer for which x [ i n, i+1 n ] K n holds. Thus f n is constant on each of the intervals [ i n, i+1 n. For X R, let B(X, r denote the r-neighborhood of the set X. Lea Let 0 < s 1. Suppose that C n K n are copact sets with H s (C n ε (n N for soe ε > 0. Let C be an accuulation point of (C n in the Hausdorff etric. Then C K and H s (C ε. Proof. Since li sup K n def = N n=n K n = K, C K is trivial. Suppose to the contrary that H s (C < ε. Then there exists an r > 0 such that Hs (B(C, r < ε also holds. There exists an n such that C n B(C, r (since C is an accuulation point, which contradicts the fact that H s (C n ε. Lea Suppose that C n K n are copact sets such that Var f n Cn B for soe B 0. Let C be an accuulation point of (C n in the Hausdorff etric. Then Varf C B also holds. Proof. We know fro the previous proof that C K. Let n j be a sequence of integers such that C nj C in the Hausdorff etric. Let x 1 < x 2 <... < x k be points in C. Let ε > 0 be arbitrary. There exist an n = n j and δ > 0 such that C B(C n, δ and f(x f(y < ε if x y < δ + 1 (x, y K. n Let y i C n be such that x i y i < δ (i = 1,..., k and y 1 y 2 y k. By the definition of f n, there exist z i K such that f n (y i = f(z i and z i y i 1 n (i = 1,...,k. Since z i x i < δ + 1 n, we have f(z i f(x i < ε. Using that Var f n Cn B, we have and thus k 1 k 1 B f n (y i f n (y i+1 = f(z i f(z i+1, i=1 i=1 k 1 f(x i f(x i+1 B + 2kε. i=1 This holds for all ε > 0, therefore the total variation of f C is at ost B. Note that the sets C n in the previous leas are all contained in a copact interval, hence the sequence (C n has an accuulation point in the Hausdorff etric.

19 2.4 Bounded variation the continuous case Proof of Theore Proposition Let K R be a copact set of positive Lebesgue easure, let f : K R be continuous, and let s < 1/2, B > 0. There exists a copact set C K of Hausdorff diension at least s such that Varf C B. Proof. We ay suppose without loss of generality that f(k [0, 1]. Let K n and f n be defined as above. Recall that we denote the Lebesgue easure by λ. Let λ n = n λ(k n ; this is a positive integer. Then K n = [ ϕn (0 n, ϕ ] [ n(0 + 1 ϕn (λ n 1..., ϕ ] n(λ n n n n for soe integers ϕ n (0 < ϕ n (1 <... < ϕ n (λ n 1. Note that ϕ n : λ n Z is a non-contractive apping. Define the function g n : λ n [0, 1] by setting g n (k = f n ( ϕn(k n (k λ n. (2.10 Let us apply Theore to the functions g n. We obtain soe ε > 0 and subsets A n λ n such that µ s (A n λ s n ε λ(ks n s ε and Varg n An B. Let C n = n 1 (ϕ n (A n (see the definition of in Lea 2.2.4, thus C n K n. It is easy to see that we have Varf n Cn = Varg n An B. Fro Lea we obtain H s (C n = n s H s ((ϕ n (A n n s 2 s µ s (ϕ n (A n, and since ϕ n is a non-contractive apping we get fro Lea that H s (C n n s 2 s µ s (A n n s 2 s λ(k s n s ε = λ(k s 2 s ε. Now choose an accuulation point C of (C n. We iediately see fro Lea that C K, H s (C λ(ks 2 s ε > 0, thus the Hausdorff diension of C is at least s; and fro Lea we conclude that Varf C B. Now we are ready to prove our ain theore about bounded variation. Proof of Theore There exists a copact set K [0, 1] of positive Lebesgue easure such that f K is continuous. We ay suppose that every non-epty intersection of K with an open interval has positive Lebesgue easure, since we ay reove those non-epty intersections fro K which are of Lebesgue easure zero (and we need to reove only countably any. Therefore we ay use Proposition not only for K, but for any non-epty portion of K.

20 2.5 Hölder restrictions 16 Let x K, and let x n ց x be a strictly decreasing sequence in K converging fast enough to ensure n=1 sup y [x,x n] K f(y f(x 1. For each positive integer n, let us apply Proposition to the function f restricted to K [x 2n+2, x 2n ] to obtain a copact set C n K [x 2n+2, x 2n ] of diension at least 1/2 1/n such that Var f Cn 2 n. Let C be the closure of n C n (which is n C n {x}. Thus C is of diension at least 1/2 and Var f C 1 + n 2 n = 2. We ay choose a copact subset of C of diension exactly 1/2 (see e.g. [10], which concludes the proof. 2.5 Hölder restrictions Overview of the proof We give an inforal outline of the proof of Theore Let 1 q < p be integers. A theore of Mattila states that if A R p is an analytic set and H s (A > 0, then we can find any q-diensional planes W R p such that di(a W s + q p. Let < n and let f : R n R be Borel easurable. Let A R n R be the graph of f. Then clearly H n (A > 0. Applying the previous theore with p = n + and q = n, we obtain an affine apping ϕ : R n R such that di(a graph ϕ n + n (n + = n. This iplies that f is actually linear (affine restricted to a set of diension n. Thus f is Lipschitz on a set of diension n. We can transfor this result to the case of R R functions. It is possible to ap a large portion of R to a large portion of R n by a Hölder-1/n apping such that its inverse is Hölder-n (see and Clai Now let us pretend that there is a Hölder-1/n apping p n : R R n such that its inverse is Hölder-n, and that there is an analogous ap p : R R. Let g : R R be an arbitrary Borel function. Set f = p g p 1 n, this is an Rn R ap. Apply the previous result to f. Since f is Lipschitz on soe C R n of diension n, the function g = p 1 f p n ust be Hölder-/n on the set p 1 n (C of diension 1 /n. A suitable approxiation of any nuber 0 < α < 1 by fractions /n would give Theore

21 2.5 Hölder restrictions Preliinaries Let us recall soe notions about self-siilar sets. We say that a non-epty copact set F R n is a self-siilar set satisfying the strong separation condition if there exist contractive siilarities ϕ i : R n R n (i = 1, 2,..., k, k 2, such that F = ϕ 1 (F ϕ 2 (F... ϕ k (F, where denotes disjoint union. If the siilarity ratio of ϕ i is denoted by r i, then the Hausdorff diension s of F is deterined by the well-known equation (see e.g. [10] k ri s = 1. i=1 It is also known that H s (F is positive and finite. Setting Ω k = {1, 2,..., k} N (the coding space, we can define the coding ap π : Ω k F by π((i 0, i 1,... = li ϕ i 0 ϕ i1 ϕ i (0. Now let 0 < ε < 1, and let n 1 be an integer. Let r be such that 2 n r 1 ε = 1 (2.11 holds. Then r < 1/2 n. For i = 1,..., 2 n we define siilarities ϕ i : R R by setting ϕ i (x = rx + 1 r 2 n 1 (i 1 (i = 1,..., 2n. Then ϕ i ([0, 1] are disjoint copact intervals in [0, 1]. Let E n,ε be the self-siilar set generated by the siilarities ϕ 1, ϕ 2,..., ϕ 2 n; that is, E n,ε = ϕ 1 (E n,ε ϕ 2 (E n,ε... ϕ 2 n(e n,ε [0, 1]. Fro equation (2.11 we conclude that the Hausdorff diension of E n,ε is 1 ε. We also define a self-siilar set in R n. Taking a direct product of the self-siilar set E 1,ε [0, 1], let F n,ε = n E 1,ε [0, 1] n i=1 Then F n,ε is also self-siilar, and its Hausdorff diension is n(1 ε. Clearly F n,ε can be generated by 2 n siilarities of scaling ratio r 1/n.

22 2.5 Hölder restrictions 18 Clai There exists a bijection p n : E n,ε F n,ε such that p n is Hölder-1/n and its inverse is Hölder-n. Proof. The self-siilar sets E n,ε and F n,ε have the sae coding space Ω 2 n. It is easy to verify that p n = π F π 1 E satisfies the requireents where π F and π E are the coding aps of F n,ε and E n,ε, respectively. Clai There exists a onotonic continuous surjective function g : E n,ε [0, 1] such that g is Hölder-(1 ε. Proof. We leave the proof to the reader Proofs Recall that we say that f B-Hölder α (where B > 0 if f(x f(y B x y α for every x and y in the doain of f. Proposition Let K [0, 1] be a copact set of positive Lebesgue easure, and let f : K R be continuous. Let 0 < α < 1, and s < 1 α. There exists a copact set C K of Hausdorff diension at least s such that f C 1-Hölder α. Proof. There exists a copact set K K of positive Lebesgue easure such that f(k [y, y + 1] for soe y R. Therefore we ay suppose without loss of generality that f(k [0, 1]. Choose positive integers 1 < n and soe 0 < ε < 1 such that (1 ε/n α and 1 ε /n s. (2.12 Let c > 0 be sufficiently sall (depending on n,, and ε. Let µ be the Borel easure obtained by restricting H 1 ε to the self-siilar set E n,ε. Then 0 < µ(r <. Let G = {(x, y R 2 : x + y K}. Since λ(k > 0, an application of Fubini s theore yields that µ(k + t dλ(t = λ µ(g = λ(k + t dµ(t > 0. R R Therefore there exists soe t R such that µ(k +t > 0. We ay suppose without loss of generality that in fact t = 0; that is, µ(k > 0. Then H 1 ε (K E n,ε > 0. Let g : E,ε [0, 1] be the function given by Clai Let h : [0, 1] E,ε be a Borel function such that g h is the identity. Then h f K En,ε : K E n,ε E,ε is a Borel apping. Since H 1 ε (K E n,ε > 0, by Luzin s theore [12, Theore 2.24]

23 2.5 Hölder restrictions 19 there exists a copact set D K E n,ε of positive H 1 ε easure such that h f D is continuous. Let p n : E n,ε F n,ε and p : E,ε F,ε be the appings given by Clai Let D n = p n (D F n,ε ; this is a copact set. We define a continuous function ψ : D n F,ε R by setting ψ = p h f D p 1 n D n. (2.13 Since the apping p n is Hölder-1/n and its inverse is Hölder-n, fro H 1 ε (D > 0 we obtain H n(1 ε (D n > 0. Define the copact set A = graphψ = {(x, ψ(x R n+ : x D n }. As the projection of A to R n is D n, we clearly have H n(1 ε (A > 0. A theore of Mattila [10, Theore 10.10] iplies that for alost every n-plane in R n+ there exists a parallel n-plane W such that A W has Hausdorff diension at least n(1 ε + n (n + = n(1 ε. In fact, there exists an affine ap ϕ : R n R with Lipschitz constant at ost c such that W = graph ϕ satisfies that di A W n(1 ε. Let C n = {x D n : ψ(x = ϕ(x} (a copact set. Since A graph ϕ is a Lipschitz (affine iage of C n, we clearly have di C n n(1 ε. Clearly ψ Cn = ϕ Cn is a Lipschitz function with constant at ost c. Fro (2.13 it is easy to deduce that f D = g (h f D = g p 1 ψ p n D. Since p n is Hölder-1/n, the ap p 1 is Hölder-, the function g is Hölder-(1 ε, and ψ Cn is Lipschitz, we obtain that f D is a Hölder function with exponent (1 ε/n on the copact set C def = p 1 n (C n D. That is, f C is a Hölder function with exponent (1 ε/n. Moreover, the Hölder constant for this exponent is clearly at ost 1 if c is sufficiently sall (copared to the Hölder constant of p n, p 1 and g for the appropriate exponents. As dic n n(1 ε, we have di C 1 ε /n. Therefore, using (2.12, we obtain that f C 1-Hölder α and di C s. Proof of Theore There exists a copact set K [0, 1] of positive Lebesgue easure such that f K is continuous. We ay suppose that every non-epty intersection of K with an open interval has positive Lebesgue easure, since we ay reove those non-epty intersections fro K which are of Lebesgue easure zero (and we need to reove only countably any. Therefore we ay use Proposition not

24 2.6 Generalizations and open questions 20 only for K, but for any non-epty portion of K. Let us apply Proposition for soe 0 < s < 1 α to obtain a copact set C K of diension at least s such that f C 1-Hölder α. Choose a strictly decreasing sequence (x n in C. Thus f is also Hölder-α with constant at ost 1 on the sequence (x n. For each positive integer n, let ε n > 0 be very sall. Now for each positive integer n apply Proposition to f restricted to K [x n ε n, x n + ε n ]. We obtain copact sets C n K [x n ε n, x n +ε n ] of diension at least 1 α 1/n such that f Cn 1-Hölder α. Let C be the closure of n C n. Thus C is of diension at least 1 α. It is clear that if the nubers ε n are chosen to be sall enough, then C = n C n {li x n }, and fro the continuity of f, that f C 2-Hölder α. We ay choose a copact subset of C of diension exactly 1 α, which concludes the proof. 2.6 Generalizations and open questions Definition The β-variation of a function f : A R (or f : A R is defined as { n 1 } sup f(x i+1 f(x i β : x 1 < x 2 <... < x n, x i A. i=1 Closely following the ethods used in the proofs of Theore and Theore 2.1.1, one can generalize these theores to bounded β-variations instead of bounded 1-variation. Theore Let f : [0, 1] R be Lebesgue easurable, β > 0. There exists a copact set C [0, 1] of Hausdorff diension β 1+β on C. such that f has finite β-variation This result is sharp. Indeed, the ethods of M. Elekes used in [4] can also be generalized to show that a typical continuous function has infinite β-variation on any set of diension larger than β 1+β. Using standard techniques it is straightforward to generalize Theore and Theore to higher diensional Euclidean spaces. (Naely, we can exploit the fact that it is possible to ap a large portion of R to a large portion of R n by a Hölder-1/n apping such that its inverse is Hölder-n; see Clai Theore Let f : R R be Lebesgue easurable, β > 0. There exists a copact set C R of Hausdorff diension on C. β +β such that f has finite β-variation

25 2.6 Generalizations and open questions 21 Theore Let f : R n R be Lebesgue easurable and let 0 < α < n. There exists a copact set C R n of Hausdorff diension n α such that f is Hölder-α on C. These theores (that is, the stated diensions are again sharp for all β, and n. We have shown that every R R Borel function is of bounded variation on soe copact set of Hausdorff diension 1/2. However, we do not know anything about the possible (1/2-diensional Hausdorff easure of such sets. Question 4. Can we find for every Borel function f : [0, 1] R a Borel set B of positive 1/2-diensional Hausdorff easure such that f restricted to B is of bounded variation? A slightly ore general variant is the following. Question 5. Does there exist a Borel function f : [0, 1] R such that if f is of bounded variation on soe Borel set B, then B has zero/finite/σ-finite 1/2- diensional Hausdorff easure? The analogous questions for the Hölder-α property are also open.

26 Chapter 3 Borel aps and Hausdorff diension 3.1 Outline Our ain ai in this chapter is to solve a proble of D. Preiss and B. Weiss (Question 1 as discussed in the Introduction ( 1.3. Theore For every 0 s < t 1 the easure spaces (R, B, H s and (R, B, H t are not isoorphic. Moreover, there does not exist a Borel bijection f : R R such that for any Borel set B R, 0 < H s (B < 0 < H t (f(b <. (3.1 Later we will also prove the analogous result in R n (see Theore Theore Let f : R R be Borel (or Lebesgue easurable. For every 0 d 1 there exists a copact set A R such that di A = d and di f(a d. Theore clearly iplies Theore Indeed, let f be Borel easurable and choose a d for which s < d < t. By applying Theore we obtain a copact set A of diension d with di f(a d. Since s < d, there exists a Borel subset B of A for which 0 < H s (B < (see e.g. [10]. Now f(b f(a, so it has diension at ost d, which iplies that H t (f(b = 0. So f cannot be an isoorphis of the easure spaces (R, B, H s and (R, B, H t, and cannot satisfy (3.1 either. To prove Theore it is clearly enough to show the following. Theore Suppose that K is a copact set of positive Lebesgue easure, and f : K R is continuous. For every 0 d 1 there exists a copact set A K of Hausdorff diension d such that f(a has Hausdorff diension at ost d.

27 3.2 Preliinaries 23 In fact, we will prove the following stronger theore. Theore Suppose that K is a copact set of positive Lebesgue easure and f : K R is continuous. For every 0 d 1 there exists a copact set A K of Hausdorff diension d such that f(a has Hausdorff diension at ost d dif(k. We outline the proof of this theore. First, in 3.2 we prove soe auxiliary leas. In 3.3 we define a rando subtree of a specific rooted tree. Using this rando subtree we construct rando Cantor sets in 3.4 and prove an upper estiate for their diension. In 3.5 we construct rando copact sets and prove a lower estiate for their diension using energy integrals. Then in 3.6 we prove Theore as follows: for the given K and continuous function f : K R, we apply the forer construction in the range space and the latter in the doain space in such a way that they produce a set A with the desired property. Then, using straightforward techniques, we will deduce the following generalization of Theore fro Theore in 3.7. Theore Let D R n be a Borel set and let f : D R be Borel easurable. Then for every 0 d 1 there exists a Borel set A D such that di A = d did and di f(a d di f(d. Reark We can state Theore (alost equivalently in the following for. If f : R R is Borel easurable and 0 d 1, then there exists a copact set B R such that di B = d and di f 1 (B d. We ight ask if there is a set B R such that di B = d and di f 1 (B d. However, this is far fro true. In Chapter 4 we present a continuous real function f such that f 1 (x is of diension 1 for every x R, see Clai All results of this chapter were published in [8] with essentially the sae proof. However, in [8] first we give an easier upper estiate (siilar to 4.2 which is sufficient to prove Theore but not Theore We also have a strong upper estiate in [8] which is sufficient to prove Theore (and thus Theore as well. In fact, it turned out that this arguent contained soe (inor error. Here we only give this stronger upper estiate correcting also the error ade in [8]. 3.2 Preliinaries Notation. For a Borel easure µ on R, let I t (µ denote the t-diensional energy of µ; that is, I t (µ = x y t dµ(x dµ(y. For Borel easures µ k (k N and µ, µ k µ denotes that µ k weakly converges to µ.

28 3.2 Preliinaries 24 The following stateents are probably well-known. Lea Suppose that µ and µ k (k N are probability easures on R such that µ k µ. Then µ k µ k µ µ. Proof. We have to show that for every copactly supported continuous function h : R 2 R, h d(µ R 2 k µ k h d(µ µ. Clearly it is enough to show this for R 2 a dense subset of the copactly supported continuous functions. It is well known that functions of the for n f i (xg i (y (f, g : R R continuous functions with copact support i=1 are dense, so it is enough to check that f(xg(y d(µ k µ k f(xg(yd(µ µ. R 2 R 2 By Fubini, f(xg(y d(µ k µ k = R 2 R f(x dµ k (x g(y dµ k (y R which tends to R f(x dµ(x g(y dµ(y = f(xg(y d(µ µ R R as k, using µ k µ and Fubini again. Lea Suppose that µ k (k N are probability easures on R supported on [ R, R] for soe R > 0. If µ k µ then I t (µ li inf I t (µ k. Proof. Let ϕ be a copactly supported continuous function on the plane which equals 1 on the square [ R, R] 2 and for which 0 ϕ(x, y 1 everywhere. For each positive integer i define h i : R 2 R by setting h i (x, y = ϕ(x, y in( x y t, i. Using Lea we have h i (x, y dµ dµ = li k li inf k h i (x, y dµ k dµ k x y t dµ k dµ k = li inf k I t(µ k.

29 3.3 The rando tree 25 The support of µ µ is in [ R, R] 2 since µ k is supported on [ R, R] for all k, so we have li i h i (x, y dµ(x dµ(y = Thus I t (µ li inf k I t (µ k. x y t dµ(x dµ(y = I t (µ. Lea Let 0 < t < 1, let A be a copact set in R and let I = [0, λ(a]. Then A A x y t dλ(x dλ(y where c t is a constant depending only on t. Proof. Define the function ϕ : A [0, λ(a] by setting I I x y t dλ(xdλ(y = c t λ(a 2 t ϕ(x = λ ( (, x] A. Using first the fact that ϕ is 1-Lipschitz and then that it is a easure preserving transforation between λ A and λ I, we obtain x y t dλ(x dλ(y ϕ(x ϕ(y t dλ(x dλ(y A A A A = x y t dλ(x dλ(y = λ(ax λ(ay t λ(a 2 dλ(x dλ(y I I [0,1] [0,1] = λ(a 2 t x y t dλ(x dλ(y = c t λ(a 2 t where c t is finite if t < 1. [0,1] [0,1] 3.3 The rando tree Let M 3 and be integers with 2 M 1. Let M <ω = {(i 0, i 1,...,i n 1 : n N, i τ {0, 1,..., M 1} = M}. We will consider M <ω as a set of ulti-indices and also as the M-adic tree with root, where every node has M children. For an i M <ω let i denote the length of the ulti-index; that is, the level of the node i. For i, j M <ω we write i j if j is a descendant of i in the tree. We say that i and j are incoparable if i j and j i. If i = (i 0, i 1,...,i n 1 M <ω and r M, we adopt the notation ir = (i 0, i 1,...,i n 1, r.

30 3.4 Upper estiate 26 We will use the notation i j for the node which is the nearest coon ancestor of i and j; that is, i j is the longest ulti-index for which i j i and i j j hold. Now we choose a rando -adic subtree S of M <ω in the following way. Let X i (i M <ω be independent rando variables with unifor distribution over the set of -eleent subsets of M. That is, for each set T {0, 1,..., M 1} of eleents, Define the rando subtree as P(X i = T = 1 ( M. S = {(i 0, i 1,...,i n 1 M <ω : i τ X (i0,i 1,...,i τ 1 for every 0 τ n 1}. So S, and for each i S exactly children of i are in S. It is easy to see that {i S : i = n} = n for every n N, and ( i P(i S =. (3.2 M Set S n = {i S : i = n} (n N. 3.4 Upper estiate Let E R be a copact set. Suppose that for each i M <ω a copact non-trivial interval U i R is given satisfying the following conditions: (1 the endpoints of U i lie in E; (2 for every node i and its child ir (r M we have U ir U i ; and (3 for every node i and for every two distinct r, r M, the intervals U ir and U ir can have at ost one point in coon. In fact (3 iplies that if i and j are incoparable nodes, then U i and U j can have at ost one point in coon. Using our rando subtree S M <ω (see 3.3, we define rando copact sets C n as C n = {U i i S n }.

31 3.4 Upper estiate 27 Then U = C 0 C 1 C 2. Put C = n C n. Lea Alost surely C E. Proof. It is easy to check that li n ax i Sn U i = 0 iplies that C E. Fro the properties of the intervals U i we know that U i dia E i =n for every n. Therefore for each fixed n, the nuber of those intervals U i for which i = n and U i δ is at ost (dia E/δ. Hence the probability that there exists an i S n for which U i δ holds, tends to zero as n. That is, the probability that ax i Sn U i δ holds, tends to zero as n. Keeping in ind that ax i Sn U i is onotone decreasing, this iplies that li n ax i Sn U i δ alost surely. Since this holds for every δ > 0, the proof is finished. Proposition The rando copact set C defined above has Hausdorff diension at ost log di E alost surely. log M Proof. Let t > di E be arbitrary and let 0 < ε < U t be sufficiently sall. Then H t (E = 0 (this akes sense even if t > 1. Therefore we can choose a finite collection of open intervals I covering the copact set E such that I I I t < ε. Clearly, we ay suppose that I is inial in the sense that no interval is covered by the others. Then we can split I into two collections I and I such that both of the contains disjoint intervals only. Let and I 2 = I \ I 1. I 1 = {I I i M <ω U i I} Let I I 1 be arbitrary. Let j(i be a sallest ulti-index for which U j(i I holds; that is, U i I if i < j(i. Since ε < U t, we cannot have U I, thus j(i 1. If for three ulti-indices i, i, i of length j(i 1 all the corresponding intervals U i, U i, U i intersected I, then at least one of the would be contained in I, which is ipossible. Therefore at ost two intervals U i (with i = j(i 1 can intersect I, hence by (3.2 ( j(i 1 P(C I 2. M

32 3.4 Upper estiate 28 Now let I I 2. Let N be a (large positive integer. Then there can be at ost two intervals U i for which i = N and U i I. Using (3.2 this iplies that ( N P(C I 2. M Since this holds for all N, we obtain P(C I = 0. Since the intervals I I are disjoint, the nodes j(i (when I I 1 I for an anti-chain in M <ω ; that is, none of the is an ancestor of any other. Thus 1 1. M j(i I I 1 I The sae inequality holds for the intervals in I 1 I, therefore I I 1 1 M j(i 2. (3.3 Let s = log log M t, hence s < t and t/s = M. By Lea 3.4.1, C E alost surely. Therefore, alost surely, C can be covered by those intervals I I which intersect C. Fro the previous arguents we obtain ( E(H (C s E I s ½ C I = I I I I P(C I I s = I I 1 P(C I I s ( j(i 2 I s M I I 1 2c I I 1 ( j(i t/s I t s/t cm j(i, where we choose c so that 1 I I 1 = 1 holds, hence c 2 by (3.3. Applying cm j(i Jensen s inequality to the concave function x x s/t and using t/s = M we obtain E(H s (C 2c I I 1 ( j(i t/s I t s/t cm j(i 2c ( I I 1 s/t j(i t/s I t cm j(i

33 3.5 Lower estiate 29 ( s/t ( = 2c I t = 2c 1 s/t I t c I I 1 I I 1 ( s/t 4 I t 4ε s/t. I I 1 Because ε can be chosen arbitrarily sall, we obtain E(H s (C = 0 and thus H (C s = 0 alost surely. Therefore dic s alost surely. This holds for every s > log di E, since t > di E was arbitrary. So the diension of C is at ost log log M log M di E alost surely. s/t 3.5 Lower estiate Suppose that for each i M <ω a copact set P i R is given satisfying the following conditions: (1 λ(p i = M i ; (2 for every node i and its child ir (r M we have P ir P i ; (3 for every node i and for every two distinct r, r M the intersection of P ir and P ir is of Lebesgue easure zero. In fact (3 iplies that if i and j are incoparable then λ(p i P j = 0. Using our rando subtree S M <ω (see 3.3, we define rando copact sets D n as D n = {P i i S n }. Then P = D 0 D 1 D 2. Put D = n D n. Theore The rando copact set D defined above has Hausdorff diension at least log log M alost surely. Proof. We define rando Borel easures µ k on R by setting µ k (A = λ(a D k ( k M,