A selection of fractals from the book - and a fractal egg, reproduced with the permission of Kevin Van Aelst.

Transcription

1 A selection of fractals from the book - and a fractal egg, reproduced with the permission of Kevin Van Aelst.

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3 Fractals in Probability and Analysis Christopher J. Bishop Stony Brook University, Stony Brook, NY and Yuval Peres Microsoft Research

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5 Contents Preface page 1 Minkowski and Hausdorff dimensions Minkowski dimension Hausdorff dimension and the Mass Distribution Principle Sets defined by digit restrictions Billingsley s Lemma and the dimension of measures Sets defined by digit frequency Slices Intersecting translates of Cantor sets * Notes Exercises 36 2 Self-similarity and packing dimension Self-similar sets The open set condition is sufficient Homogeneous sets Microsets Poincaré sets * Alternative definitions of Minkowski dimension Packing measures and dimension When do packing and Minkowski dimension agree? Notes Exercises 78 3 Frostman s theory and capacity Frostman s Lemma The dimension of product sets Generalized Marstrand Slicing Theorem Capacity and dimension 93 vii

6 viii Contents 3.5 Marstrand s Projection Theorem Mapping a tree to Euclidean space preserves capacity Dimension of random Cantor sets Notes Exercises Self-affine sets Construction and Minkowski dimension The Hausdorff dimension of self-affine sets A dichotomy for Hausdorff measure The Hausdorff measure is infinite * Notes Exercises Graphs of continuous functions Hölder continuous functions The Weierstrass function is nowhere differentiable Lower Hölder estimates Notes Exercises Brownian motion, Part I Gaussian random variables Lévy s construction of Brownian motion Basic properties of Brownian motion Hausdorff dimension of the Brownian path and graph Nowhere differentiability is prevalent Strong Markov property and the reflection principle Local extrema of Brownian motion Area of planar Brownian motion General Markov processes Zeros of Brownian motion Harris inequality and its consequences Points of increase Notes Exercises Brownian motion, Part II Dimension doubling The Law of the Iterated Logarithm Skorokhod s Representation Donsker s Invariance Principle Harmonic functions and Brownian motion in R d 221

7 Contents ix 7.6 The maximum principle for harmonic functions The Dirichlet problem Polar points and recurrence Conformal invariance * Capacity and harmonic functions Notes Exercises Random walks, Markov chains and capacity Frostman s theory for discrete sets Markov chains and capacity Intersection equivalence and return times Lyons Theorem on percolation on trees Dimension of random Cantor sets (again) Brownian motion and Martin capacity Notes Exercises Besicovitch Kakeya sets Existence and dimension Splitting triangles Fefferman s Disk Multiplier Theorem * Random Besicovitch sets Projections of self-similar Cantor sets The open set condition is necessary * Notes Exercises The Traveling Salesman Theorem Lines and length The β -numbers Counting with dyadic squares β and μ are equivalent β -sums estimate minimal paths Notes Exercises 337 Appendix A Banach s Fixed-Point Theorem 343 Appendix B Frostman s Lemma for analytic sets 353 Appendix C Hints and solutions to selected exercises 360 References 379 Index 396

8 Preface The aim of this book is to acquaint readers with some fractal sets that arise naturally in probability and analysis, and the methods used to study them. The book is based on courses taught by the authors at Yale, Stony Brook University, the Hebrew University and UC Berkeley. We owe a great debt to our advisors, Peter Jones and Hillel Furstenberg; thus the book conveys some of their perspectives on the subject, as well as our own. We have made an effort to keep the book self-contained. The only prerequisite is familiarity with measure theory and probability at the level acquired in a first graduate course. The book contains many exercises of varying difficulty. We have indicated with a those for which a solution, or a hint, is given in Appendix C. A few sections are technically challenging and not needed for subsequent sections, so could be skipped in the presentation of a given chapter. We mark these with a * in the section title. Acknowledgments: We are very grateful to Tonći Antunović, Subhroshekhar Ghosh and Liat Kessler for helpful comments and crucial editorial work. We also thank Ilgar Eroglu, Hrant Hakobyan, Michael Hochman, Nina Holden, Pertti Mattila, Elchanan Mossel, Boris Solomyak, Perla Sousi, Ryokichi Tanaka, Tatiana Toro, Bálint Virág, Samuel S. Watson, Yimin Xiao and Alex Zhai for useful comments. Richárd Balka carefully read the entire manuscript and provided hundreds of detailed corrections and suggestions. Many thanks to David Tranah and Sam Harrison at Cambridge University Press for numerous helpful suggestions. Finally, we dedicate this book to our families: Cheryl, David and Emily Bishop, and Deborah, Alon and Noam Peres; without their support and understanding, it would have taken even longer to write.

9 1 Minkowski and Hausdorff dimensions In this chapter we will define the Minkowski and Hausdorff dimensions of a set and will compute each in a few basic examples. We will then prove Billingsley s Lemma and the Law of Large Numbers. These allow us to deal with more sophisticated examples: sets defined in terms of digit frequencies, random slices of the Sierpiński gasket, and intersections of random translates of the middle thirds Cantor set with itself. Both Minkowski and Hausdorff dimensions measure how efficiently a set K can be covered by balls. Minkowski dimension requires that the covering be by balls all of the same radius. This makes it easy to compute, but it lacks certain desirable properties. In the definition of Hausdorff dimension we will allow coverings by balls of different radii. This gives a better behaved notion of dimension, but (as we shall see) it is usually more difficult to compute. 1.1 Minkowski dimension A subset K of a metric space is called totally bounded if for any ε > 0, it can be covered by a finite number of balls of diameter ε. For Euclidean space, this is the same as being a bounded set. For a totally bounded set K, let N(K,ε) denote the minimal number of sets of diameter at most ε needed to cover K. We define the upper Minkowski dimension as dim M (K)=limsup ε 0 and the lower Minkowski dimension dim M (K)=liminf ε 0 1 logn(k, ε), log1/ε logn(k, ε). log1/ε

10 2 Minkowski and Hausdorff dimensions If the two values agree, the common value is simply called the Minkowski dimension of K and denoted by dim M (K). When the Minkowski dimension of a set K exists, the number of sets of diameter ε needed to cover K grows like ε dim M (K)+o(1) as ε 0. We get the same values of dim M (K) and dim M (K) if we replace N(K,ε) by N B (K,ε), which is the number of closed balls of radius ε needed to cover K. This is because N B (K,ε) N(K,ε) N(K,ε/2) (any set is contained in a ball of at most twice the diameter and any ball of radius ε/2 has diameter at most ε; strict inequality could hold in a metric space). For subsets of Euclidean space we can also count the number of axis-parallel squares of side length ε needed to cover K, or the number of such squares taken from a grid. Both possibilities give the same values for upper and lower Minkowski dimension, and for this reason Minkowski dimension is sometimes called the box counting dimension. It is also easy to see that a bounded set A and its closure A satisfy dim M (A)=dim M (A) and dim M (A)=dim M (A). If X is a set and x,y X implies x y ε, wesayx is ε-separated. Let N sep (K,ε) be the number of elements in a maximal ε-separated subset X of K. Clearly, any set of diameter ε/2 can contain at most one point of an ε- separated set X, son sep (K,ε) N(K,ε/2). On the other hand, every point of K is within ε of a maximal ε-separated subset X (otherwise add that point to X). Thus N(K,ε) N sep (K,ε). Therefore replacing N(K,ε) by N sep (K,ε) in the definition of upper and lower Minkowski dimension gives the same values (and it is often easier to give a lower bound in terms of separated sets). Example Suppose that K is a finite set. Then N(K,ε) is bounded and dim M (K) exists and equals 0. Example Suppose K =[0,1]. Then at least 1/ε intervals of length ε are needed to cover K and clearly ε 1 +1 suffice. Thus dim M (K) exists and equals 1. Similarly, any bounded set in R d with interior has Minkowski dimension d. Example Let C be the usual middle thirds Cantor set obtained as follows. Let C 0 =[0,1] and define C 1 =[0, 1 3 ] [ 2 3,1] C0 by removing the central interval of length 1 3. In general, Cn is a union of 2 n intervals of length 3 n and C n+1 is obtained by removing the central third of each. This gives a decreasing nested sequence of compact sets whose intersection is the desired set C. The construction gives a covering of C that uses 2 n intervals of length 3 n. Thus for 3 n ε < 3 n+1 we have N(C,ε) 2 n,

11 1.1 Minkowski dimension 3 Figure The Cantor middle thirds construction. and hence dim M (C) log2 log3. Conversely, the centers of the nth generation intervals form a 3 n -separated set of size 2 n,son sep (C,3 n ) 2 n. Thus dim M (C) log2 log3 = log 3 2. Therefore the Minkowski dimension exists and equals this common value. If at each stage we remove the middle α (0 < α < 1) we get a Cantor set C α with Minkowski dimension log2/(log2 + log 1 1 α ). Example Consider K = {0} {1, 1 2, 1 3, 1 4,...}. Observe that 1 n 1 1 n = 1 n(n 1) > 1 n 2. So, for ε > 0, if we choose n so that (n + 1) 2 < ε n 2, then n ε 1/2 and n distinct intervals of length ε are needed to cover the points 1, 1 2,..., 1 n. The 1 interval [0, n+1 ] can be covered by n + 1 additional intervals of length ε. Thus Hence dim M (K)=1/2. ε 1/2 N(K,ε) 2ε 1/ This example illustrates a drawback of Minkowski dimension: finite sets have dimension zero, but countable sets can have positive dimension. In particular, it is not true that dim M ( n E n )=sup n dim M (E n ), a useful property for a dimension to have. In the next section, we will introduce Hausdorff dimension, which does have this property (Exercise 1.6). In the next chapter, we will introduce packing dimension, which is a version of upper Minkowski dimension forced to have this property.

12 4 Minkowski and Hausdorff dimensions 1.2 Hausdorff dimension and the Mass Distribution Principle Given any set K in a metric space, we define the α-dimensional Hausdorff content as { H α (K)=inf U i α : K U i }, i i where {U i } is a countable cover of K by any sets and E denotes the diameter of a set E. Definition The Hausdorff dimension of K is defined to be dim(k)=inf{α : H α (K)=0}. More generally we define { Hε α (K)=inf U i α : K } U i, U i < ε, i i where eachu i is now required to have diameter less than ε. The α-dimensional Hausdorff measure of K is defined as H α (K)=lim Hε α (K). ε 0 This is an outer measure; an outer measure on a non-empty set X is a function μ from the family of subsets of X to [0, ] that satisfies μ (/0)=0, μ (A) μ (B) if A B, μ ( j=1 A j ) j=1 μ (A j ). For background on real analysis see Folland (1999). The α-dimensional Hausdorff measure is even a Borel measure in R d ; see Theorem below. When α = d N, then H α is a constant multiple of L d, d-dimensional Lebesgue measure. If we admit only open sets in the covers of K, then the value of H ε α(k) does not change. This is also true if we only use closed sets or only use convex sets. Using only balls might increase Hε α by at most a factor of 2 α, since any set K is contained in a ball of at most twice the diameter. Still, the values for which H α (K)=0 are the same whether we allow covers by arbitrary sets or only covers by balls. Definition Let μ be an outer measure on X. A set K in X is μ - measurable, if for Every set A X we have H α (A)=μ (A K)+μ (A K c ).

13 1.2 Hausdorff dimension and the Mass Distribution Principle 5 Definition Let (Ω,d) be a metric space. An outer measure μ on Ω is called a metric outer measure if dist(a,b) > 0 = μ(a B)=μ(A)+μ(B), where A and B are two subsets of Ω. Since Hausdorff measure H α is clearly a metric outer measure, the following theorem shows that all Borel sets are H α -measurable. This implies that H α is a Borel measure (see Folland (1999)). Theorem μ-measurable. Let μ be a metric outer measure. Then all Borel sets are Proof It suffices to show any closed set K is μ-measurable, since the measurable sets form a σ-algebra. So, let K be a closed set. We must show for any set A Ω with μ(a) <, μ(a) μ(a K)+μ(A K c ). (1.2.1) Let B 0 = /0 and for n 1 define B n = {x A : dist(x,k) > 1 n }, so that B n = A K c n=1 (since K is closed). Since μ is a metric outer measure and B n A \ K, μ(a) μ[(a K) B n ]=μ(a K)+μ(B n ). (1.2.2) For all m N, the sets D n = B n \ B n 1 satisfy ( ) m m μ(d 2 j )=μ D 2 j μ(a), m, j=1 j=1 since if x B n, and y D n+2, then dist(x,y) dist(x,k) dist(y,k) 1 n 1 n + 1. Similarly m j=1 μ(d 2 j 1) μ(a). So j=1 μ(d j ) <. The inequality μ(b n ) μ(a K c ) μ(b n )+ μ(d j ) j=n+1 implies that μ(b n ) μ(a K c ) as n. Thus letting n in (1.2.2) gives (1.2.1).

14 6 Minkowski and Hausdorff dimensions The construction of Hausdorff measure can be made a little more general by considering a positive, increasing function ϕ on [0, ) with ϕ(0) =0. This is called a gauge function and we may associate to it the Hausdorff content { H ϕ (K)=inf ϕ( U i ) : K U i }; i i then H ϕ ε (K), and H ϕ (K)=lim ε 0 H ϕ ε (K) are defined as before. The case ϕ(t)=t α is just the case considered above. We will not use other gauge functions in the first few chapters, but they are important in many applications, e.g., see Exercise 1.59 and the Notes for Chapter 6. Lemma If H α (K) < then H β (K)=0 for any β > α. Proof It follows from the definition of Hε α that H β ε (K) ε β α H α ε (K), which gives the desired result as ε 0. Thus if we think of H α (K) as a function of α, the graph of H α (K) versus α shows that there is a critical value of α where H α (K) jumps from to 0. This critical value is equal to the Hausdorff dimension of the set. More generally we have: Proposition For every metric space E we have H α (E)=0 H α (E)=0 and therefore dime = inf{α : H α (E)=0} = inf{α : H α (E) < } = sup{α : H α (E) > 0} = sup{α : H α (E)= }. Proof Since H α (E) H α (E), it suffices to prove. If H α (E) =0, then for every δ > 0 there is a covering of E by sets {E k } with k=1 E k α < δ. These sets have diameter less than δ 1/α, hence H α δ 1/α (E) < δ. Letting δ 0 yields H α (E)=0, proving the claimed equivalence. The equivalence readily implies that dime = inf{α : H α (E)=0} = sup{α : H α (E) > 0}. The other conclusions follow from Lemma The following relationship to Minkowski dimension is clear dim(k) dim M (K) dim M (K). (1.2.3)

15 1.2 Hausdorff dimension and the Mass Distribution Principle 7 Indeed, if B i = B(x i,ε/2) are N(K,ε) balls of radius ε/2 and centers x i in K that cover K, then consider the sum N(K,ε) S ε = i=1 B i α = N(K,ε)ε α = ε α R ε, where R ε = logn(k,ε) log(1/ε).ifα > liminf ε 0 R ε = dim M (K) then inf ε>0 S ε = 0. Strict inequalities in (1.2.3) are possible. Example Example showed that K = {0} n{ 1 n } has Minkowski dimension 1 2. However, any countable set has Hausdorff dimension 0, for if we enumerate the points {x 1,x 2,...} and cover the nth point by a ball of diameter δ n = ε2 n we can make n δn α as small as we wish for any α > 0. Thus K is a compact set for which the Minkowski dimension exists, but is different from the Hausdorff dimension. Lemma (Mass Distribution Principle) Borel measure μ that satisfies If E supports a strictly positive μ(b(x,r)) Cr α, for some constant 0 < C < and for every ball B(x,r), then H α (E) H α (E) μ(e)/c. In particular, dim(e) α. Proof Let {U i } beacoverofe. For{r i }, where r i > U i, we look at the following cover: choose x i in each U i, and take open balls B(x i,r i ). By assumption, μ(u i ) μ(b(x i,r i )) Cr i α. We deduce that μ(u i ) C U i α, whence i Thus H α (E) H α (E) μ(e)/c. U i α μ(u i ) i C μ(e) C. We note that upper bounds for Hausdorff dimension usually come from finding explicit coverings of the set, but lower bounds are proven by constructing an appropriate measure supported on the set. Later in this chapter we will generalize the Mass Distribution Principle by proving Billingsley s Lemma (Theorem 1.4.1) and will generalize it even further in later chapters. As a special case of the Mass Distribution Principle, note that if A R d has positive Lebesgue d-measure then dim(a)=d.

16 8 Minkowski and Hausdorff dimensions Example Consider the Cantor set E obtained by replacing the unit square in the plane by four congruent sub-squares of side length α < 1/2 and continuing similarly. See Figure We can cover the set by 4 n squares of diameter 2 α n. Thus log4 n dim M (E) lim n log( 2α n ) = log4 logα. On the other hand, it is also easy to check that at least 4 n sets of diameter α n are needed, so dim M (E) log4 logα. Thus the Minkowski dimension of this set equals β = log4/logα. Figure Four generations of a Cantor set. We automatically get dim(e) β and we will prove the equality using Lemma Let μ be the probability measure defined on E that gives each nth generation square the same mass (namely 4 n ). We claim that μ(b(x,r)) Cr β, for all disks and some 0 < C <. To prove this, suppose B = B(x,r) is some disk hitting E and choose n so that α n+1 r < α n. Then B can hit at most 4 of the nth generation squares and so, since α β = 1/4, μ(b E) 4 4 n = 4α nβ 16r β. Example Another simple set for which the two dimensions agree and are easy to compute is the von Koch snowflake. To construct this we start with an equilateral triangle. At each stage we add to each edge an equilateral triangle pointing outward of side length 1/3 the size of the current edges and centered on the edge. See Figure for the first four iterations of this process. The boundary of this region is a curve with dimension log4/ log3 (see Theorem 2.2.2). We can also think of this as a replacement construction, in which at

17 1.3 Sets defined by digit restrictions 9 each stage, a line segment is replaced by an appropriately scaled copy of a polygonal curve. Figure Four generations of the von Koch snowflake. Even for some relatively simple sets the Hausdorff dimension is still unknown. Consider the Weierstrass function (Figure 1.2.3) f α,b (x)= n=1 b nα cos(b n x), where b > 1 is real and 0 < α < 1. It is conjectured that the Hausdorff dimension of its graph is 2 α, and this has been proven when b is an integer; see the discussion in Example On the other hand, some sets that are more difficult to define, such as the graph of Brownian motion (Figure 1.2.4), will turn out to have easier dimensions to compute (3/2 by Theorem 6.4.3). Figure The Weierstrass function with b = 2, α = 1/2. This graph has Minkowski dimension 3/2 and is conjectured to have the same Hausdorff dimension. 1.3 Sets defined by digit restrictions In this section we will consider some more complicated sets for which the Minkowski dimension is easy to compute, but the Hausdorff dimension is not

18 10 Minkowski and Hausdorff dimensions Figure dimensional Brownian motion. This graph has dimension 3/2 almost surely. so obvious, and will be left to later sections. These subsets of [0,1] will be defined by restricting which digits can occur at a certain position of a number s b-ary expansion. In a later section we will consider sets defined by the asymptotic distribution of the digits. We start by adapting Hausdorff measures to b-adic grids. Let b 2 be an integer and consider b-adic expansions of real numbers, i.e., to each sequence {x n } {0,1,...,b 1} N we associate the real number x = n=1 x n b n [0,1]. b-adic expansions give rise to Cantor sets by restricting the digits we are allowed to use. For example, if we set b = 3 and require x n {0,2} for all n we get the middle thirds Cantor set C. For each integer n let I n (x) denote the unique half-open interval of the form [ k 1 b n, k b n ) containing x. Such intervals are called b-adic intervals of generation n (dyadic if b = 2). It has been observed (by Frostman (1935) and Besicovitch (1952)) that we can restrict the infimum in the definition of Hausdorff measure to coverings of the set that involve only b-adic intervals and only change the value by a bounded factor. The advantage of dealing with these intervals is that they are nested, i.e., two such intervals either are disjoint or one is contained in the other. In particular, any covering by b-adic intervals always contains a subcover by disjoint intervals (just take the maximal intervals). Furthermore, the b-adic intervals can be given the structure of a tree, an observation that we will use extensively in later chapters.

19 1.3 Sets defined by digit restrictions 11 We define the grid Hausdorff content by H α (A)=inf{ J i α,a J i }, (1.3.1) i i and the grid Hausdorff measures by H ε α (A)=inf{ J i α,a J i, J i < ε}, (1.3.2) i i H α (A)=lim H α (A), (1.3.3) ε 0 where the infimums are over all coverings of A R by collections {J i } of b- adic intervals. The grid measure depends on b, but we omit it from the notation; usually the value of b is clear from context. Clearly H α (A) H α (A), since we are taking the infimum over a smaller set of coverings. However, the two sides are almost of the same size, i.e., H α (A) (b + 1)H α (A), since any interval I can be covered by at most (b + 1) shorter b-adic intervals; just take the smallest n so that b n I and take all b-adic intervals of length b n that hit I. Thus H α (A)=0 if and only if H α (A)=0. Similarly, we can define Ñ(K, ε) to be the minimal number of closed b-adic intervals of length ε needed to cover K. We get N(K,ε) Ñ(K,ε) (b + 1)N(K,ε). Hence, the Hausdorff and the Minkowski dimensions are not changed. One can define grid Hausdorff content H ϕ (A) and grid Hausdorff measures H ε ϕ (A) and H ϕ (A) with respect to any gauge function ϕ by replacing J i α with ϕ( J i ) in (1.3.1) and (1.3.2). Furthermore, the definitions can be extended to subsets of R n by replacing the b-adic intervals J i in (1.3.1) and (1.3.2) with b-adic cubes (products of b-adic intervals of equal length). Definition For S N the upper density of S is d(s)=lim sup N ε #(S {1,...,N}). N Here and later #(E) denotes the number of elements in E, i.e., # is counting measure. The lower density is d(s)=lim inf N #(S {1,...,N}). N

20 12 Minkowski and Hausdorff dimensions If d(s)=d(s), then the limit exists and is called d(s), the density of S. Example Suppose S N, and define { } A S = x = x k 2 k : x k {0,1}. k S The set A S is covered by exactly 2 n k=1 1 S(k) = 2 #(S {1,...,n}) closed dyadic intervals of generation n, where 1 S is the characteristic function of S, i.e., 1 S (n)=1 for n S, and 1 S (n)=0 for n S.So Thus which implies log 2 Ñ(A S,2 n )= logñ(a S,2 n ) log2 n = 1 n n k=1 n k=1 dim M (A S )=d(s), 1 S (k). 1 S (k), dim M (A S )=d(s). It is easy to construct sets S where the liminf and limsup differ and hence we get compact sets where the Minkowski dimension does not exist (see Exercise 1.17). The Hausdorff dimension of A S is equal to the lower Minkowski dimension. We shall prove this in the next section as an application of Billingsley s Lemma. We can also construct A S as follows. Start with the interval [0,1] and subdivide it into two equal length subintervals [0, 1/2] and [1/2, 1]. If 1 S then keep both intervals and if 1 S then keep only the leftmost, [0,1/2]. Cut each of the remaining intervals in half, keeping both subintervals if 2 S and only keeping the left interval otherwise. In general, at the nth step we have a set A S n AS n 1 that is a finite union of intervals of length 2 n (some may be adjoining). We cut each of the intervals in half, keeping both subintervals if n S and throwing away the right-hand one if n S. The limiting set is A S = n=1 A S n. In Figure we have drawn the first generations of the construction corresponding to S = {1,3,4,6,8,10,...}. Example The shifts of finite type are defined by restricting which

21 1.3 Sets defined by digit restrictions 13 Figure First 8 generations of A S for S = {1,3,4,6,8,10,...}. digits can follow other digits. Let A =(A ij ),0 i, j < b, beab b matrix of 0s and 1s, and define { } X A = x n b n : A xn x n+1 = 1 for all n 1. n=1 We will also assume that if the jth column of A is all zeros, so is the jth row; this implies every finite sequence {x n } n 1 satisfying the condition above can be extended to at least one infinite sequence satisfying the condition. Thus such finite strings correspond to b-adic closed dyadic intervals that intersect X A. Notice that a b-ary rational number r will belong to X A if either of its two possible b-ary expansions does (one is eventually all 0s, the other eventually all (b 1)s). The condition on {x n } described in the example is clearly shift invariant and is the intersection of countably many closed conditions, so the set X A is compact and invariant under the map T b, where T b (x)=(bx) mod 1. For example, if ( ) 1 1 A =, 1 0 then x X A if and only if the binary representation of x has no two consecutive 1s. The first ten generations in the construction of the Cantor set X A are shown in Figure Any such matrix can be represented by a directed graph where the vertices represent the numbers {0,...,b 1} and a directed edge connects i to j if A ij = 1. For example, the 2 2 matrix A above is represented by Figure An element of x = n x n b n X A corresponds in a natural way to an infinite path {x 1,x 2,...} on this graph and a b-adic interval hitting X A corresponds to a finite path. Conversely, any finite path of length n corresponds to a b-adic interval of length b n hitting X A. Thus N n (X A ) Ñ(X A,b n ) is the number of

22 14 Minkowski and Hausdorff dimensions 0 1 Figure The directed graph representing A. Figure Approximations of X A. distinct paths of length n in the graph (this uses our assumption about the rows and columns of A). By the definition of matrix multiplication, the number of paths of length n from i to j is (A n ) ij (i.e., the (ith, jth) entry in the matrix A n ). (See Exercise 1.23.) Thus the total number of length n paths is N n (X A )= A n, where the norm of a matrix is defined as B = i, j B ij. (However, since any two norms on a finite-dimensional space are comparable, the precise norm won t matter.) Thus where logn n (X A ) log(n n (X A )) 1/n dim M (X A )= lim = lim = logρ(a) n nlogb n logb logb, ρ(a)= lim n A n 1/n (1.3.4) is, by definition, the spectral radius of the matrix A, and is equal to the absolute value of the largest eigenvalue. That this limit exists is a standard fact about matrices and is left to the reader (Exercise 1.24). For the matrix ( ) 1 1 A =, 1 0 the spectral radius is (1+ 5)/2 (Exercise 1.25), so the Minkowski dimension

23 of X A is 1.3 Sets defined by digit restrictions 15 logρ(a) log2 = log(1 + 5) log2. log2 We shall see in Example 2.3.3, that for shifts of finite type the Hausdorff and Minkowski dimensions agree. Example Now we consider some sets in the plane. Suppose A is a b b matrix of 0s and 1s with rows labeled by 0 to b 1 from bottom to top (the unusual labeling ensures the matrix corresponds to the picture of the set) and columns by 0 to b 1 from left to right. Let Y A = {(x,y) : A yn x n = 1 for all n}, where {x n },{y n } are the b-ary expansions of x and y. For example, if ( ) 1 0 A =, 1 1 then x n = 0 implies that y n can be either 0 or 1, but if x n = 1 then y n must be 0. This matrix A gives a Sierpiński gasket. See Figure Figure The Sierpiński gasket (8th generation) and the Sierpiński carpet (4th generation). If A = 0 0 0, then we get C 1/3 C 1/3, the product of the middle thirds Cantor set with itself.

24 16 Minkowski and Hausdorff dimensions For A = 1 0 1, we get the Sierpiński carpet. See the right side of Figure If the matrix A has r 1s in it, then we can cover Y A by r n squares of side b n and it is easy to see this many are needed. Thus dim M (Y A )=logr/logb = log b r. Defining a measure on Y A which gives equal mass to each of the nth generational squares and using the Mass Distribution Principle, we get that the Hausdorff dimension of the set Y A is also log b r. The Sierpiński gasket (as well as the other examples so far) are self-similar in the sense that they are invariant under a certain collection of maps that are constructed from isometries and contractive dilations. Self-similar sets are discussed in Section 2.1. If we replace the square matrices of the previous example by rectangular ones (i.e., use different base expansions for x and y) we get a much more difficult class of sets that are invariant under affine maps. These sets will be studied in Chapter 4, but we give an example here. Example Self-affine sets. We modify the previous example by taking a non-square matrix. Suppose m < n and suppose A is an m n matrix of 0s and 1s with rows and columns labeled as in the previous example. Let Y A = {(x,y) : A yk x k = 1 for all k}, where {x k } is the n-ary expansion of x and {y k } is the m-ary expansion of y. For example, if ( ) A =, we obtain the McMullen set in Figure To construct the set, start with a rectangle Q and divide it into 6 sub-rectangles by making one horizontal and 2 vertical subdivisions. Choose one of the rectangles on the top row and two rectangles from the bottom row according to the pattern A. Now subdivide the chosen rectangles in the same way and select the corresponding sub-rectangles. Figure illustrates the resulting set after a few iterations. 3 The Minkowski dimension of this set exists and is equal to 1 + log (Theorem 4.1.1). We will prove later in this chapter that the Hausdorff dimension of the McMullen set is strictly larger than 1 (see Example

25 1.4 Billingsley s Lemma and the dimension of measures 17 Figure Four generations of the McMullen set ). We shall eventually prove C. McMullen s (1984) result that the Hausdorff dimension of this set is log 2 (2 log ) , which is strictly less than the Minkowski dimension (Theorem 4.2.1). 1.4 Billingsley s Lemma and the dimension of measures In this section we prove a refinement of the Mass Distribution Principle. The measure in this version need not satisfy uniform estimates on all balls as in the Mass Distribution Principle, but only estimates in a neighborhood of each point, where the size of that neighborhood may vary from point to point. An even more general result was proved by Rogers and Taylor (1959), see Proposition Let b 2 be an integer and for x [0,1] let I n (x) be the nth generation, half-open b-adic interval of the form [ j 1 b n, j b n ) containing x. The following is due to Billingsley (1961). Lemma (Billingsley s Lemma) Let A [0,1] be Borel and let μ be a finite Borel measure on [0,1]. Suppose μ(a) > 0.If logμ(i n (x)) α 1 liminf β 1, (1.4.1) n log I n (x) for all x A, then α 1 dim(a) β 1. Proof and Let α < α 1 < β 1 < β. The inequalities (1.4.1) yield that for all x A, for all x A, limsup n limsup n μ(i n (x)) 1, (1.4.2) I n (x) β μ(i n (x)) 1. (1.4.3) I n (x) α We will show that (1.4.2) implies H β (A) μ([0,1]) and that (1.4.3) implies H α (A) μ(a). The lemma follows from these two claims.

26 18 Minkowski and Hausdorff dimensions We start with the first assertion. Assume (1.4.2) holds. For 0 < c < 1, fix ε > 0. For every x A we can find integers n as large as we wish satisfying μ(i n (x)) I n (x) β > c. Take n(x) to be the minimal integer satisfying this condition and such that b n(x) < ε. Now {I n(x) (x) : x A} isacoverofa, and suppose {J k } is a subcover by disjoint intervals. This covering of A has the property that J k < ε for all k (by our choice of n(x)) and This implies k J k β c 1 μ(j k ) c 1 μ([0,1]). (1.4.4) k H ε β (A) c 1 μ([0,1]). Taking c 1 and ε 0 gives the first assertion. Therefore H β (A) μ([0,1]). Next we prove the second assertion. Assume (1.4.3) holds. For a fixed C > 1 and a positive integer m, let A m = {x A : μ(i n (x)) < C I n (x) α for all n > m}. Since A = m A m and A m+1 A m,wehaveμ(a)=lim m μ(a m ). Fix ε < b m and consider any cover of A by b-adic intervals {J k } such that J k < ε. Then J k α J k α C 1 μ(j k ) C 1 μ(a m ). k k:j k A m /0 k:j k A m /0 This shows H ε α (A) C 1 μ(a m ). Taking ε 0, m and C 1 gives the desired result. The above proof of Billingsley s Lemma can be generalized in several ways: (a) It is clear that the proof generalizes directly to subsets A [0,1] d.in the proof one needs to replace b-adic intervals with cubes that are products of intervals of the form d i=1 [ j i 1 b n, j i b n ). (b) A covering S = {S i } i of A is called a Vitali covering if for every point x A and all δ > 0 there is a set S i such that x S i and such that 0 < S i < δ. We say S has the bounded subcover property if there exists a constant C such that whenever a set E is covered by a subcover S E S then there is a further subcover S E S E of E such that D S E 1 D C. For example, by Besicovitch s covering theorem (see Mattila, 1995, Theorem 2.6) the family of open balls in R d enjoys the bounded subcover property. Replacing

27 1.4 Billingsley s Lemma and the dimension of measures 19 the family of dyadic intervals by a Vitali covering with bounded subcover property in (1.3.2) and (1.3.3) and in the statement of Billingsley s Lemma one can proceed as in the above proof to conclude that H α (A) μ(a) and H β (A) Cμ([0,1]). In particular (1.4.4) is replaced by D β c 1 μ(d)=c 1 1 D dμ c 1 Cμ([0,1]). D S A D S A D S A However, in general, H γ (A) will not be comparable to H γ (A). To replace H α (A) and H β (A) by H α (A) and H β (A) and prove the corresponding version of Billingsley s Lemma, one needs additional assumptions. For example, it suffices to assume that there exists a constant C such that any ball B can be covered by no more than C elements of the cover of diameter less than C B. An example of a covering satisfying all the above assumptions are approximate squares as in Definition 4.2.2; these will be used to compute dimensions of self-affine sets in Chapter 4. (c) The assumptions can be further weakened. For example, instead of the last assumption we can assume that there is a function Ψ: R + R + satisfying logψ(r) lim r 0 log1/r = 0 and such that any ball of radius r can be covered by Ψ(r) elements of the cover of diameter at most Ψ(r)r. Example For S N, recall { } A S = x = x k 2 k : x k {0,1}. (1.4.5) k S We computed the upper and lower Minkowski dimensions of this set in Example and claimed that dim(a S )=liminf N #(S {1,...,N}). N To prove this, let μ be the probability measure on A S that gives equal measure to the nth generation covering intervals. This measure makes the digits {x k } k S in (1.4.5) independent identically distributed (i.i.d.) uniform random bits. For any x A S, logμ(i n (x)) log I n (x) = log2 #(S {1,...,n}) log2 n = #(S {1,...,n}). n Thus the liminf of the left-hand side is the liminf of the right-hand side. By Billingsley s Lemma, this proves the claim. Definition If μ is a Borel measure on R n we define dim(μ)=inf{dim(a) : μ(a c )=0, A R n is Borel }.

28 20 Minkowski and Hausdorff dimensions Observe that the infimum is really a minimum, because if dim(a n ) dim(μ) and μ(a c n)=0 for all n then A = n A n satisfies dim(a)=dim(μ). An equivalent definition is to write dim(μ)=inf{α : μ H α } where μ ν means the two measures are mutually singular, i.e., there is a set A R n such that μ(a c )=ν(a)=0. Lemma Let b be a positive integer. Given x [0,1], let I n (x) denote the b-adic interval of the form [ j 1 b n, j b n ) containing x. Let { logμ(i n (x)) } α μ = esssup lim inf, n log I n (x) where esssup( f )=min{α : μ({x : f (x) > α})=0}. Then dim(μ)=α μ. Proof First take α > α μ, and set { A = x : liminf n log μ(i n (x)) log I n (x) } α. By definition of essential supremum μ(a c )=0. Hence, dim(μ) dim(a).by Billingsley s Lemma, dim(a) α. Taking α α μ gives dim(μ) α μ. To prove the other direction, let α < α μ and consider { B = x : liminf n log μ(i n (x)) log I n (x) } α. By the definition of α μ,wehaveμ(b) > 0. If μ(e c )=0, then we also have 0 < μ(b)=μ(e B) and log μ(i n (x)) lim inf α n log I n (x) on E B. Billingsley s Lemma shows dim(e) dim(e B) α. Therefore dim(μ) α for all α < α μ. We deduce dim(μ)=α μ. Example Consider the measure μ on the middle thirds Cantor set C that gives equal mass to each nth generation interval in the construction. This is called the Cantor singular measure. If we consider a 3-adic interval I of length 3 n then μ(i)=2 n = I log 3 2, if I o (the interior of I) hits C and is 0 otherwise. Thus log μ(i n (x)) lim = log n log I n (x) 3 2,

29 1.5 Sets defined by digit frequency 21 for all x C, and hence for μ almost every x. Therefore Lemma implies dim(μ)=log Sets defined by digit frequency We previously considered sets with restrictions on what the nth digit of the b- ary expansion could be. In this section we do not restrict particular digits, but will require that each digit occurs with a certain frequency. The resulting sets are dense in [0,1], so we need only consider their Hausdorff dimension. Example A p = {x = n=1 x n2 n : x n {0,1},lim j 1 j j k=1 x k = p}. Thus A p is the set of real numbers in [0,1] in which a 1 occurs in the binary expansion with asymptotic frequency p. For typical real numbers we expect a 1 to occur about half the time, and indeed, A 1/2 is a set of full Lebesgue measure in [0,1] (see below). In general we will show dim(a p )=h 2 (p)= plog 2 p (1 p)log 2 (1 p). The quantity h 2 is called the entropy of p and is strictly less than 1 except for p = 1/2. It represents the uncertainty associated to the probability (if p = 0 or 1 the entropy is 0; it is maximized when p = 1/2). See Cover and Thomas (1991). In addition to Billingsley s Lemma we will need Theorem (Strong Law of Large Numbers) Let (X, dν) be a probability space and { f n },n= 1,2... a sequence of orthogonal functions in L 2 (X,dν). Suppose E( fn 2)= f n 2 dν 1, for all n. Then a.e. (with respect to ν)asn. 1 n S n = 1 n n k=1 f k 0, Proof We begin with the simple observation that if {g n } is a sequence of functions on a probability space (X,dν) such that g n 2 dν <, n then n g n 2 < ν-a.e. and hence g n 0 ν-a.e. Using this, it is easy to verify the Strong Law of Large Numbers (LLN) for

30 22 Minkowski and Hausdorff dimensions n along the sequence of squares. Specifically, since the functions { f n } are orthogonal, ( ) 1 2 n S n dν = 1 n 2 S n 2 dν = 1 n n 2 f k 2 dν 1 k=1 n. Thus if we set g n = 1 S n 2 n 2,wehave g 2 n dν 1 n 2. Since the right-hand side is summable, the observation made above implies that g n = n 2 S n 2 0 ν-a.e. To handle the limit over all positive integers, suppose that m 2 n < (m + 1) 2. Then 1 m 2 S n 1 m 2 S m 2 2 dν = 1 m 4 n k=m 2 +1 f k 2 dν = 1 n m 4 f k 2 dν k=m m 3, since the sum has at most 2m terms, each of size at most 1. Set m(n)= n and h n = S n m(n) 2 Sm(n)2 m(n) 2. Now each integer m equals m(n) for at most 2m + 1 different choices of n. Therefore, h n 2 2 dμ n=1 n=1 m(n) 3 (2m + 1) 2 m m 3 <, so by the initial observation, h n 0 a.e. with respect to ν. This yields that 1 m(n) 2 S n 0 a.e., that, in turn, implies that 1 n S n 0 a.e., as claimed. Theorem is called a Strong Law of Large Numbers because it gives a.e. convergence, as opposed to the Weak Law of Large Numbers, that refers to convergence in measure. We frequently make use of the following Corollary If {X k } are independent identically distributed (i.i.d.) random variables and E[Xk 2] < then lim n 1 n n k=1 X k = E[X 1 ].

31 1.5 Sets defined by digit frequency Figure A histogram of μ 1/3 applied to dyadic intervals of length 2 7. Proof Note that {X k E[X k ]} are orthogonal and apply Theorem a.e. Next, define a measure μ p on [0,1] as follows. By Caratheodory s Extension Theorem it suffices to define μ p on dyadic intervals. Let ( μ p [ j 2 n, j + 1 ) 2 n ) = p k( j) (1 p) n k( j), where k( j) is the number of 1s in the binary expression of j. We can also write μ p as μ p (I n (x)) = p n k=1 x k (1 p) n n k=1 x k, where x = k=1 x k2 k. An alternative way of describing μ p is that it gives the whole interval [0,1] mass 1, and gives the two subintervals [0,1/2], [1/2,1] mass 1 p and p respectively. In general, if a dyadic interval I has measure μ p (I), then the left half has measure (1 p)μ p (I) and the right half has measure pμ p (I). Observe that if p = 1/2 then this is exactly Lebesgue measure on [0, 1]. In Figure we have graphed the density of the measure μ p for p = 1/3 on all dyadic intervals of length 2 7 in [0,1]. Lemma dim(a p )=dim(μ p )=h 2 (p) Proof We start by proving that μ p (A p )=1.

32 24 Minkowski and Hausdorff dimensions To see this, let f n (x)=(x n p) (where k x k 2 k is the binary expansion of x). If S n = n k=1 f k, then unwinding definitions shows that A p is exactly the set where 1 n S n 0, so we will be done if we can apply the Law of Large Numbers to { f n } and the measure μ p. Clearly, fn 2 dμ p 1. It is easy to check orthogonality, f n f m dμ p = 0 for n m. Thus μ p (A p )=1. Now, we show that dim(μ p )=h 2 (p). By Lemma 1.4.4, { dim(μ p )=esssup Note that log μ p (I n (x)) = 1 (( 1 log I n (x) log2 n n k=1 lim inf n ) x k log 1 ( 1 p + n Since μ p -almost every point is in A p, we see that log μ(i n (x)) log I n (x) n k=1 }. 1 )) (1 x k )log. 1 p for μ p -a.e. x. Therefore, 1 lim n n n k=1 x k = p, log μ p (I n (x)) log I n (x) plog 1 1 p +(1 p)log 1 p log2 = h 2 (p), for a.e. x with respect to μ p. Thus dim(μ p )=h 2 (p). By Definition of dim(μ p ), we deduce from μ p (A p )=1that dim(a p ) dim(μ p )=h 2 (p).however, since { A p x [0,1] : liminf n log μ p (I n (x)) log I n (x) } = h 2 (p), and μ p (A p )=1 > 0, Billingsley s Lemma implies dim(a p ) h 2 (p), hence equality. Example Consider the related sets { 1 Ã p = x [0,1] : limsup n n { 1 Â p = x [0,1] : liminf n n n k=1 n k=1 } x k p, } x k p.

33 1.5 Sets defined by digit frequency 25 Clearly, A p à p  p. Since dim(a 1/2 )=1, we have dim(ã p )=dim(â p )=1 if p 1/2. If p < 1/2, then by Lemma 1.5.4, h 2 (p)=dim(a p ) dim(ã p ) dim(â p ). On the other hand, we saw before that logμ p (I n (x)) = 1 ( 1 ( log log I n (x) log2 1 p + log 1 p ) 1 p n We have log 1 p p > 0 since p < 1/2, so for x  p, we get logμ p (I n (x)) lim inf h 2 (p). n log I n (x) Therefore, for p < 1/2, Billingsley s Lemma implies that dim(a p )=dim(ã p )=dim(â p )=h 2 (p). n k=1 x k ). Example The same argument used in Lemma to compute the dimension of A p works in a more general setting. Suppose p =(p 0,...,p b 1 ) is a probability vector, i.e., b 1 k=0 p k = 1 and define a measure μ p by μ p (I n (x)) = n j=1 where {x n } is the b-ary expansion of x. Then repeating the proof of the lemma shows b 1 1 dim(μ p )=h b (p)= p k log b. p k k=0 p x j, Similarly, the set of x s in [0,1] such that #({n [0,N] : x n = k}) lim = p k N N for each k = 0,...,b 1 has dimension h b (p). The following is a variant we will need later in the proof of Proposition about intersections of random translates of Cantor sets. Fix an integer b 2 and a set E {0,1,...,b 1}. Let {x n } be the b-ary expansion of the real number x. Lemma The set X E p = { x [0,1) : lim N 1 N N n=1 } 1 E (x n )=p

34 26 Minkowski and Hausdorff dimensions has Hausdorff dimension ( p ) ( 1 p ) plog b (1 p)log #(E) b. b #(E) Proof Let I n (x) denote the b-ary interval of generation n containing x. Define a Borel measure on [0,1) by ( p ) k n 1 E (x k ) ( 1 p ) k n (1 1 E (x k )) μ(i n (x)) =. #(E) b #(E) The proof of Lemma shows that dim(x E p )=dim(μ)= plog b ( p #(E) ) (1 p)log b ( 1 p b #(E) ). 1.6 Slices If A R 2 has dimension α, what should we expect the dimension of A L to be, where L is a typical line? For the present, let us consider only vertical lines and set A x = {y : (x,y) A}. Theorem (Marstrand Slicing Theorem) Let A R 2 and suppose that dim(a) 1. Then dim(a x ) dim(a) 1, for (Lebesgue) almost every x. If dim(a) < 1, then the slice A x is empty for almost every x (in fact, it is empty except for a set of exceptional x of dimension at most dim(a); see Exercise 1.9). On the other hand, it is possible that dim(a x )=dim(a) for some values of x, e.g., if A is a vertical line segment. The inequality can be strict for every x, e.g., there are real-valued functions on R whose graphs have dimension strictly larger than one, even though all the vertical slices are singletons. See Chapter 5. For example, we will prove that the graph of a 1-dimensional Brownian motion has dimension 3/2 almost surely (see Theorem 6.4.3). We shall give several other examples of strict inequality later in this section. Proof We start with the following claim: for 1 α 2 H α (A) c 2 H α 1 (A x )dx. R

35 1.6 Slices 27 (Measurability of the integrand follows from a monotone class argument that we omit.) In fact, we can take c 2 = 1. (But in higher dimensions the analogous constant is c n < 1.) It suffices to prove this claim because for α > dim(a), 0 = H α (A) H α 1 (A x )dx implies H α 1 (A x )=0 for Lebesgue almost every x. Thus dim(a x ) α 1 for Lebesgue almost every x. To prove the claim fix ε > 0 and δ > 0 and let {D j } beacoverofa with D j < ε and D j α < Hε α (A)+δ. j Enclose each D j in a square Q j with sides parallel to the axes and with side length s j D j. Let I j be the projection of Q j onto the x-axis. For each x, the slices {(Q j ) x } form a cover of A x and have length { s j, x I j (Q j ) x =. 0, x I j (In the higher-dimensional case, this is the point where the constant c n < 1 appears. For example, if n = 3, then a slice of a cube Q j has diameter either 2s j or 0.) Wenowhaveanε-cover of A x and Therefore Taking δ 0gives H α 1 ε R R (A x ) (Q j ) x α 1 = s α 1 j. j j:x I j Hε α 1 (A x )dx ( s α 1 j )dx R j:x I j = s j α j Hε α (A)+δ. R Hε α 1 (A x )dx Hε α (A). As ε 0, Hε α 1 (A x ) H α 1 (A x ), so the Monotone Convergence Theorem implies H α 1 (A x )dx H α (A). R

36 28 Minkowski and Hausdorff dimensions There is a generalization of the Slicing Theorem to higher dimensions. Theorem Let A R n be a Borel set with dim(a) > n m and let E m be an m-dimensional subspace. Then for almost every x E m (or equivalently, almost every x R n ) and for α > n m, dim(a (E m + x)) dim(a) (n m), H α (A) c n E m H α (n m) (A (E m + x))dx. The proof is almost the same as of Theorem 1.6.1, so we omit it. As before, the equality can be strict. There are also generalizations that replace the almost everywhere with respect to Lebesgue measure on R n with a more general measure. See Theorem Example Let A = C C be the product of the middle thirds Cantor set with itself. We saw in Example that dim(a)=log 3 4. The vertical slices of A are empty almost surely, so dim(a x )=0 < (log 3 4) 1 almost surely. This gives strict inequality in Theorem for almost all slices. Example A set with more interesting slices is the Sierpiński gasket G (see Example 1.3.4). We claim that (Lebesgue) almost every vertical slice G x of G has dimension 1/2 (note that this is strictly smaller than the estimate in the slicing theorem, dim(g) 1 = log = ). Proof For each x [0,1], let S(x)={n : x n = 0}, where {x n } is the binary expansion of x. Then it is easy to see G x = A S(x), where A S is the set defined in Example and discussed in Example The Law of Large Numbers says that in almost every x the digits 0 and 1 occur with equal frequency in its binary expansion. We deduce dim(g x )=1/2 for almost every x. More generally, dim(g x )=pif and only if limsup n k=1 x k n = 1 p. It follows from the proof of Lemma and Example that the set of such xs has dimension h 2 (p). Thus dim({x : dim(g x )=p})=h 2 (p).

37 1.7 Intersecting translates of Cantor sets * 29 Example Consider the self-affine sets of Example More precisely, consider the McMullen set X obtained by taking ( ) A = The intersection of X with any vertical line is a single point. On the other hand the intersection of the set with a horizontal line of height y = n y n 2 n is a set described as follows: at the nth generation replace each interval by its middle third if y n = 1 but remove the middle third if y n = 0. By Lemma the dimension of such a set is (log 3 2)liminf N 1 N N n=1 (1 y n ). Since for almost every y [0,1] the limit exists and equals 1 2, we see that almost every horizontal cross-section of the McMullen set has dimension 1 2 log 3 2. See Exercise 1.42 for a refinement. While it is easy to compute the dimension of the cross-sections, it is more challenging to compute the dimension of the McMullen set itself. Marstrand s Slicing Theorem implies that its dimension is at least log (we shall use the Law of Large Numbers in Chapter 4 to prove that its dimension is exactly log 2 (1 + 2 log 3 2 ) ). 1.7 Intersecting translates of Cantor sets * In the previous section we considered intersecting a set in the plane with random lines. In this section we will intersect a set in the line with random translates of itself. It turns out that this is easiest when the set has a particular arithmetic structure, such as the middle thirds Cantor set. We include the discussion here, because this special case behaves very much like the digit restriction sets considered earlier. This section also serves to illustrate a principle that will be more apparent later in the book: it is often easier to compute the expected dimension of a random family of sets than to compute the dimension of any particular member of the family. J. Hawkes (1975) proved the following: Theorem Let C be the middle thirds Cantor set. Then dim((c + t) C)= 1 log2 3 log3 for Lebesgue a.e. t [ 1, 1].