Linear distortion of Hausdorff dimension and Cantor s function

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1 Collect. Math. 57, 2 (2006), c 2006 Universitat de Barcelona Linear distortion of Hausdorff dimension and Cantor s function O. Dovgoshey and V. Ryazanov Institute of Applied Mathematics and Mechanics, NAS of Ukraine 74 Roze Luxemburg str., Donetsk, 834, UKRAINE aleksdov@mail.ru ryaz@iamm.ac.donetsk.ua O. Martio and M. Vuorinen Department of Mathematics and Statistics, P.O. Box 68 FIN 0004 University of Helsinki, FINLAND martio@cc.helsinki.fi vuorinen@csc.fi Received May 9, 2005 Abstract Let f be a mapping from a metric space X to a metric space Y, and let α be a positive real number. Write dim(e) and H s (E) for the Hausdorff dimension and the s-dimensional Hausdorff measure of a set E. We give sufficient conditions that the equality dim(f(e)) = α dim(e) holds for each E X. The problem is studied also for the Cantor ternary function G. It is shown that there is a subset M of the Cantor ternary set such that H s (M) =, with s = log 2/log 3 and dim(g(e)) = (log 3/log 2) dim(e), for every E M.. Statements of main results Let f be a mapping from a metric space (X, ρ) to a metric space (Y, d). It is a simple fact that if the double inequality c (ρ(x, y)) α d(f(x), f(y)) c 2 (ρ(x, y)) α (.0) Keywords: Hausdorff dimension, Cantor function. MSC2000: 28A78, 26A30. 93

2 94 Dovgoshey, Martio, Ryazanov, and Vuorinen holds for all x, y from X where α (0, ) and c and c 2 are some positive constants, then every set A X satisfies dim(f(a)) = α dim(a). We are interested in necessary and sufficient local conditions under which this equality holds for every A X. The following theorem provides conditions for this. Write ac(a) for the set of all accumulation points of a set A. Let f be a mapping from a metric space (X, ρ) to a metric space (Y, d). If x and a are the points of X and x a we put log(d(f(x), f(a))) if f(x) f(a) log(ρ(x, a)) K f (x, a) := + if f(x) = f(a) Theorem. Let f : (X, ρ) (Y, d) be a homeomorphism. Suppose that the it K f (x, a) = α(a) (0, ) (.) x a exists for every a ac(x). Then the following statements are equivalent. (i) There exists a set X 0 acx such that α(a) = α 0 (.2) for all a in ac(x)\x 0 and either dim(z) = 0 or dim(z) = for every Z X 0. (ii) For every A X the equality holds. Corollary.2 dim(f(a)) = α 0 dim(a) (.3) Let f : X Y be a homeomorphism and let dim(x) <. Suppose that the it (.) exists for every a ac(x). Then if and only if dim(f(a)) = dim(a), A X (.4) for every a ac(x)\x 0 where X 0 X is zero-dimensional. α(a) = (.5)

3 Linear distortion of Hausdorff dimension and Cantor s function 95 Remark.3 Note that, as it follows from the proof of Theorem., the equivalence (i) (ii) is also valid if (.) holds only in X\X where dim(x ) = dim(f(x )) = 0. (.6) See further consequences of Theorem. in the end of Section 2. Note also that, if f : X Y is a continuous bijection and (.) holds, then it does not follow that f is a homeomorphism. On the other hand, if f : (X, ρ) (Y, d) is a homeomorphism and for each a ac(x) we have α(a) =, then there need not exist positive constants α and c such that the inequality d(f(x), f(y)) c(ρ(x, y)) α holds for all x and y in some ball B(a, r) X; see Example 3.3. In the third part, we investigate the following problem: Let C [0, ] be the standard Cantor ternary set and let G be the Cantor function. Characterize the set of points x C such that log G(x) G(y) = log 2 y x log x y log 3. In Theorem.4 these points are characterized in terms of the spacing of 0 s and 2 s in ternary expansions. Let x be a point of the Cantor ternary set C. Then x has a triadic representation x = m= 2α m 3 m where α m {0, }. Define a sequence {R x (n)} by the rule inf{m n : α m α n, m > n} if m > n : α m α n R x (n) := 0 if m > n : α m = α n, (.7) R x (n) = (α n α n+ ), i.e. and so on. R x (n) = 2 (α n = α n+ ) & (α n+ α n+2 ) Theorem.4 Let x be a point of C. Then log G(x) G(y) = log 2 y x log x y log 3 (.8) if and only if R x (n) = 0. (.9) n n

4 96 Dovgoshey, Martio, Ryazanov, and Vuorinen This theorem and Theorem. imply the following result. Theorem.5 There exists a set M C such that H s (M) = with s = log 2/log 3 and dim(g(a)) = log 3 log 2 dim(a), for every A M. Remark.6 It is well-known that the Cantor function G satisfies the inequality G(x) G(y) 2 x y log 2/log 3 (.0) for all x and y in [0, ]. The proof can be found in [4], see also [3]. The Hausdorff dimension of the Cantor ternary set equals log 2/log 3 and, moreover, H s (C) = for s = log 2/log Linear distortion of Hausdorff dimension under mappings of metric spaces We recall the definitions of the Hausdorff dimension and the s-dimensional Hausdorff measure. Let (X, ρ) be a metric space and let diam A := sup {ρ(x, y) : x, y A} be the diameter of A X if A, diam = 0. If A E i with 0 < diame i δ for each index i I, then {E i } i I is called a δ-cover of A. If all δ-covers of A are uncountable, then H s δ(a) := for each s 0 and, in the opposite case, Hδ(A) s := inf (diam E i ) s : {E i } i N i N i I is a countable δ-cover of A for s 0. The s-dimensional Hausdorff measure of A is defined by (2.) H s (A) := δ 0 H s δ(a). (2.2) Note that the it exists because Hδ s (A) is nonincreasing function of δ and that the Hausdorff measure H s is a regular Borel measure, see e.g. [7]. The Hausdorff dimension of A is the number dim(a) such that H s (A) = { + if s < dim(a), 0 if s > dim(a). Definition 2. Let (X, ρ) be a metric space. A family B of closed balls B(a, δ) in (X, ρ) is said to fulfil the condition (V) if B(a, δ) B whenever a X and δ (0, (a)] for some (a) > 0.

5 Linear distortion of Hausdorff dimension and Cantor s function 97 Some results closely related to the next lemma can be found in [2, 2.8]. Lemma 2.2 Let A be a subset of the metric space (X, ρ) such that H s (A) = 0 for some s > 0 and let a family B of closed balls in (X, ρ) fulfil the (V ) condition. Then, for all positive numbers δ and η, there is a countable δ-cover {B i } i N B of the set A with Proof. δ 2 (diam B i ) s η. (2.3) i N It follows from (2.), (2.2) and H s (A) = 0 that for every γ > 0 there is a -cover {E() i } i N of the set A such that (d () i ) s γ 2, (2.4) i N where d () i := diam E () i. Suppose B fulfils the (V ) condition. We shall say that i N is a marked index if there is a point a () i 0 E () i 0 A for which d () i 0 < ( a () i 0 ) where is the function from Definition 2.. Let I () be the set of all marked indices. It is obvious that E () i B(a () i, d () i ) B for all i I (). Using (2.4) and we have diam B(a () i, d () i ) 2d () i, i I () (diam B(a () i, d () i )) s 2 s γ 2. (2.5) It should be observed that {B(a () i, d () i )} i I () is a δ-cover of the set { ( ) } γ /s 2 Really, if a 0 A\( It follows from (2.4) that i I () B(a () a A : (a) >. i, d () i )), then there is E () i 0 a 0 with d () i 0 (a 0 ). ( γ 2 ) /s d () i 0. Hence, ( ) γ /s (a 0 ). 2

6 98 Dovgoshey, Martio, Ryazanov, and Vuorinen Reasoning similarly we can define the sequence {I (n) } such that for each positive integer n: (diam B(a (n) i, d (n) i )) s 2 s γ 2 n, (2.6.) i I (n) B(a (n) i, d (n) i ) B and diam B(a (n) i, d (n) i ) δ for each i I (n), (2.6.2) { ( ) } γ /s a A : (a) > 2 n B(a (n) i, d (n) i ). (2.6.3) i I (n) For this purpose we take a δ 2 -cover {E(n) i } i N of A such that (diam E (n) i ) s γ 2 n. i N Now set γ := 2 s η. Then (2.6.) implies that Hence, by (2.6.2) and (2.6.3), the family (diam B (n) i, d (n) i ) s η. i I (n) { B(a (n) i, d (n) i ) : n =, 2,...; i I (n)} is a desired δ-cover of A. Proposition 2.3 Suppose that (X, ρ) and (Y, d) be metric spaces. Let β (0, ) and let f : X Y be a mapping such that inf K f (x, a) β (2.7) x a for each a ac(x). Then we have for every A X. dim(a) β dim(f(a)) (2.8) Proof. If dim(a) =, then the inequality (2.8) is trivial. Suppose that 0 dim(a) < s <. Then by the definition of the Hausdorff dimension we have H s (A) = 0. For each ε (0, β), define a family B ε of the closed balls B(a, δ) in (X, ρ) by the rule (B(a, δ) B ε ) ( x B(a, δ) : d(f(x), f(a)) (ρ(x, a)) β ε ). (2.9) It follows immediately from (2.7) that B ε fulfils the condition (V ). Hence by Lemma 2.2, for every η > 0, there is δ-cover {B i (a i, δ i )} i N B ε of A such that (diam B(a i, δ i )) s η. (2.0) i N

7 Linear distortion of Hausdorff dimension and Cantor s function 99 It follows from (2.9) that diam (f(b(a i, δ i )) 2(diam (B(a i, δ i ))) β ε. The last inequality and (2.0) imply that {f(b(a i, δ i ))} i N is a 2δ β ε -cover of f(a) and (diam f(b(a i, δ i ))) s/(β ε) 2 s/(β ε) η. Consequently that is i N H s/(β ε) (f(a)) = 0, dim(f(a)) s β ε for all ε (0, β) and every s > dim(a). Letting ε 0 and s dim(a) we have (2.8). Corollary 2.4 Suppose that (X, ρ) and (Y, d) are a metric spaces. Let 0 < β α < and let f : X Y be a homeomorphism such that β inf x a for every a ac(x). Then the inequalities hold for every A X. K f (x, a) sup K f (x, a) α (2.) x a α dim(a) dim(f(a)) dim(a) (2.2) β Proof. By Proposition 2.3 it suffices to prove the first inequality in (2.2). Since f is a homeomorphism, we have ac(y ) = f(ac(x)). Let f be the inverse map of f and let a ac (X). Applying inequality (2.) we obtain α { sup K f (x, a)} = inf K f (y, b) x a y b where b = f(a) ac(y ). Now the desired inequality follows from Proposition Proof of Theorem.. (i) (ii) Suppose that there is X 0 ac(x) such that α(a) = α 0 for every a ac(x)\x 0, and for every Z X 0 we have either dim(z) = 0 or dim(z) =. Let A be a subset of X. Then dim(a) = max{dim(a\x 0 ), dim(a X 0 )} and dim(f(a)) = max{dim(f(a\x 0 )), dim(f(a X 0 ))}.

8 200 Dovgoshey, Martio, Ryazanov, and Vuorinen Thus, by Corollary 2.4 it remains to prove that either dim(a X 0 ) = dim(f(a X 0 )) = 0 or dim(a X 0 ) = dim(f(a X 0 )) = +. To prove this we represent A X 0 in the form A X 0 = A n where A n := {a A X 0 : } n α(a) n for n =, 2,.... From this representation we get the equalities f(a X 0 ) = f(a n ), dim(a X 0 ) = dim(f(a X 0 )) = sup (dim A n ), n< sup dim(f(a n )). n< Hence dim(a X 0 ) = 0 iff n N : dim(a n ) = 0. It follows from the alternative dim(a n ) = 0 or dim(a n ) = that we have dim(a X 0 ) = iff there exists n 0 N such that dim(a n0 ) =. Consequently, the conclusion follows by Corollary 2.4. (ii) (i) Suppose that (.3) holds for every A X. Put X + 0 := {a acx : α(a) > α 0}, X 0 := {a acx : α(a) < α 0}, and X 0 := X + 0 X 0. By definition α(a) = α 0 for each a (acx)\x 0. It remains to prove that for each Z X 0. dim(z) = 0 or dim(z) = +

9 Linear distortion of Hausdorff dimension and Cantor s function 20 Suppose that for some Z X 0 Then we have or Then 0 < dim(z) <. 0 < dim(z X + 0 ) < (2.3) 0 < dim(z X0 ) <. (2.4) Consider the case (2.3) first. Set for each positive integer n X + n := { ( a ac(x) : α(a) α 0 + )} n, α 0 + n. (2.5) X + 0 and (2.3) implies that for some n 0 N = X n + 0 < dim(z X + n 0 ) <. Hence, by Corollary 2.4 and (2.5), we have dim(f(z X + n 0 )) α 0 + dim(z X n + 0 ) < dim(z X n + α 0 ). n 0 0 This contradicts (.3) with A = Z X + n 0. The case (2.4) can be proved analogously Proof of Corollary.2. In order to prove this corollary, it suffices to take α 0 =. It should be observed here that the inequality dim(x) < implies the equality dim(z) = 0 for each Z X 0. Corollary 2.7 Suppose that (X, ρ) and (Y, d) are metric spaces, and X = X X, X X =. Let α (0, ) and let ϕ : X Y be a mapping such that: 2.8. For every a ac(x ) X, dim(x ) = dim(ϕ(x )) = 0. (2.6) K x a ϕ (x, a) = α. x X 2.9. The restriction ϕ X : X ϕ(x ) is a homeomorphism. Then dim(ϕ(a)) = dim(a) (2.7) α

10 202 Dovgoshey, Martio, Ryazanov, and Vuorinen for each A X. Proof. It follows from (2.6) that dim(a) = dim(a X ) and dim(ϕ(a)) = dim(ϕ(a X )) for every A X. Consequently, it suffices to prove (2.7) for A X. For this purpose we can use Theorem. with X = X, Y = ϕ(x ), X 0 = and f equals ϕ X : X ϕ(x ). Corollary 2.0 Suppose that all the conditions of Corollary 2.7 hold with an exception of 2.9. If X can be represented in the form X = Xj such that ϕ X : X j j ϕ(x j ) is a j N homeomorphism for every j N, then equality (2.7) holds for every A X. Proof. Reasoning as in the proof of Corollary 2.7 we can easily show that dim(ϕ(a)) = sup(dim(ϕ(a Xj ))) = j N α sup (dim(a Xj )) = j N α dim(a). Corollary 2. Suppose that all the conditions of Corollary 2.7 hold except 2.9. If X is separable and ϕ X : X ϕ(x ) is a local homeomorphism, then equality (2.7) holds for each A X. Proof. Every separable metric space has a countable base, see e.g. [5, 2, II, Theorem 2]. Hence we can use Corollary 2.0. Let cp(x) denote the set of all condensation points of a metric space X, i.e. points whose neighborhoods are not countable sets. Corollary 2.2 Suppose that X and Y are metric spaces, f : X Y is a local homeomorphism and X is separable. Let α (0, ) and let K f (x, a) = α x a for every a cp(x). Then (2.7) holds for every A X. Proof. The set X\cp (X) is a countable set in every separable metric space X, [5, 23, III]. 3. The Cantor ternary function We recall the definitions of the Cantor ternary set C and ternary Cantor function G. Let x [0, ], then x belongs C if and only if x has a base 3 expansion using only the

11 Linear distortion of Hausdorff dimension and Cantor s function 203 digits 0 and 2, i.e. x = m= 2α m 3 m, α m {0, }. (3.) The Cantor function G may be defined on C by the following rule. If x C has the ternary expansion (3.), then α m G(x) := 2 m. In this section we will denote by C the set of all endpoints of complementary intervals of C. We also write C := C\C. m= The proof of Theorem.4 needs the following lemma. Lemma 3. Let x C be a point with the representation (3.) and let R x (n) be the sequence from (.7). Then R x (n) = R x (n). (3.2) If x is not a right endpoint of a complementary interval of C, then m=n+ α m 2 (m n) 2 (+Rx(n+)), n N. (3.3) Proof. Since x = m= 2( α m) 3 m we have (3.2). It remains to prove (3.3). If α n+ =, then (3.3) is obvious. In the opposite case, 2 (+Rx(n+)) is the first positive element of the series α m 2 (m n). m=n Proof of Theorem.4. Consider first the case where x is not an endpoint of some complementary interval of C. Suppose that x has representation (3.), y tends to x, and 2β m y = 3 m m= where β m = β m (y) {0, }. Let n 0 = n 0 (x, y) be the smallest index m with β m α m 0. Then using the definition of the Cantor function we have log G(x) G(y) y x log x y = y x log (α m β m )2 (m n 0) n 0 log 2 m=n 0 log 2. (α m β m )3 (m n 0) n 0 log 3 m=n 0 It is easy to make sure that 2 (α m β m )3 (m n 0) 3 m=n 0

12 204 Dovgoshey, Martio, Ryazanov, and Vuorinen for all y C and that n 0 (x, y) tends to infinity if y x, y C. Hence, log 3 log 2 log G(x) G(y) y x log x y = y x ( log (α n0 β n0 ) + (α m β m )2 (m n 0) ) m=n 0 + n 0 log 2. Writing z(x, y) := (α n 0 β n0 ) + m=n 0 + we see that the it relation (.8) is equivalent to then Next we obtain bounds for z(x, y). If and hence z(x, y) = + m=n 0 + Consequently, by (3.3) we obtain (α m β m )2 (m n 0) log z(x, y) = 0. (3.4) y x n 0 (x, y) (α n0 β n0 ) =, m=n 0 + (α m β m )2 (m n 0) α m 2 (m n 0) z(x, y) 2. 2 (+Rx(n 0+)) z(x, y) 2. (3.5) If (α n0 β n0 ) =, then z(x, y) = (α m β m )2 (m n0), m=n 0 + and hence α m 2 (m n0) z(x, y) 2. (3.6) m=n 0 + Since α m 2 (m n0) = ( α m )2 (m n0), m=n 0 + m=n 0 + relations (3.6), (3.3) and (3.2) imply that z(x, y) 2 (+R x(n 0 +)) = 2 (+Rx(n0+)).

13 Linear distortion of Hausdorff dimension and Cantor s function 205 Consequently, as in the first case, we have (3.5). Now, (3.4) follows from (3.5) and (.9). Thus, the implication (.9) (.8) follows. Suppose now that (.9) does not hold. Then there is a strictly increasing sequence {n j } j= such that and R x ( + n j ) R x (n) = sup = a (0, ] (3.7) j + n j n n α nj α nj +, n j (3.8) where α nj and α nj + are digits in the representation (3.). Let {α m } m= be a sequence of digits in (3.). Putting α m if n j m n j + R x ( + n j ) β m (j) := otherwise and we claim that α m y j := m= 2β (j) m 3 m, log G(x) G(y j ) = log 2 j log x y j log 3 Indeed, (3.8) and (3.9) imply the equalities ) nj + ( nj ( 2 x y j = 3) 3 ( ) nj ( nj + G(x) G(y j ) = 2 2) Hence, by (3.7), we have ( + a). ( nj +R x(n j +)..., 3) ( nj +R x(n j +).... 2) log G(x) G(y j ) = log 2 j log x y j log 3 n j + R x ( + n j ) = ( + a) log 2 j n j log 3. (3.9) Consider now the case where x C. In this case there is n 0 N such that R x (n) = 0 for every n n 0. It remains only to show that log G(x) G(y) = log 2 y x log x y log 3. Suppose that x is a right endpoint of a complementary interval, (i.e. α n = 0 for all large enough n), y tends to x, and y = m= 2β m (y) 3 m, β m(y) {0, }.

14 206 Dovgoshey, Martio, Ryazanov, and Vuorinen Then there are positive integers m and m 2 = m 2 (y) such that x = m m= 2α m 3 m, β m(y) = α m if m m, m 2 (y) > m, β m2 (y) =, and β m (y) = 0 if m < m < m 2 (y). Hence ( ) log β m 2 m log G(x) G(y) m=m 2 (y) m = ( ) 2 (y)log 2 y x log x y y x = y x m log 2 β m 3 m 2 (y)log 3 = log 2 log 3. m=m 2 (y) The case where x is a left endpoint of a complementary interval is similar. Example 3.3 Here we give an example of a homeomorphism f : (X, ρ) (Y, d) such that: (i) For every a ac(x) x a K f (x, a) =. (ii) For arbitrary positive α, c and for every ball B(a, δ) X there exist x, y B(a, δ) for which d(f(x), f(y)) c(ρ(x, y)) α. The example is constructed with aid of the Cantor ternary set C and the Cantor function G. and Set M := x C : y x log G(x) G(y) log x y = log 2 log 3 I := G(M C ). (3.0) Let F : I M C be a function such that F (G(y)) = y for every y M C. It is easy to see that F is a homeomorphism and by the definition of M we see that K F y a (a, y) = log 3 log 2 (3.) for every a I. Let E be the space of infinite strings from the two-letters alphabet {0, }. We may define a metric d on E by setting d(α, β) = max n< 2 n α n β n if α = {α n }, β = {β n} are elements of E. Evidently, if d(α, β) 0, then there is a positive integer n such that d(α, β) = 2 n.

15 Linear distortion of Hausdorff dimension and Cantor s function 207 The space (E, d) is an ultrametric space and the map C 2ε n 3 n Φ {εn } E is a homeomorphism. (The proof can be found in [, Chapter 2].) We claim that the inequality 2 x y log 2/log 3 d(φ(x), Φ(y)) x y log 2/log 3 (3.2) holds for all x, y C. Indeed, if x and y have the representations x = 2α i 3 i, y = 2β i 3 i, α i, β i {0, }, i= i= and if d(φ(x), Φ(y)) = 2 n, then we get ) x y = (α 2 i β i )3 i 3 ( n 3 i = 2 (d(φ(x), Φ(y)))log 3/log 2. i=n i= A similar argument yields x y 3(d(Φ(x), Φ(y)) log 3/log 2. These inequalities imply (3.2). Let F 2 : M C Φ(M C ) be the restriction of the map Φ on the set M C. Obviously, F 2 is a homeomorphism and it follows from (3.2) that K F z a 2 (a, z) = log 2 log 3 (3.3) for each a (M C ). Now set, for every x I, f(x) := F 2 (F (x)). The function f is a homeomorphism from I to Φ(M C ) and the its (3.), (3.3) imply that x a K f (x, a) = for each a I, i.e., we get (i). Suppose α, c are arbitrary positive constants and O is an open interval in [0, ]. To prove (ii) we can choose x C such that (3.7) holds and α > 2, a (0, ), G(x) O. + a Since M is a dense subset of a perfect set C, there are sequences {x j } j N and {y j } j N in M such that j x j = j y j = x, x j y j j N,

16 208 Dovgoshey, Martio, Ryazanov, and Vuorinen and log F (x j ) F (y j ) j log x j y j = log 3 log 2 ( + a), see the proof of Theorem.4. The last relation and (3.2) imply that Hence there is N 0 N such that log d(f(x j ), f(y j )) = j log x j y j ( + a). d(f(x j ), f(y j )) x j y j 2/(+a) 2 for j N 0. Since +a < α and x j y j = 0, we have (ii). j 0 To prove Theorem.5 we use the following two propositions. Proposition 3.4 Let I be the subset of the unit interval [0, ] from (3.0). Then each number, simply normal to base 2, belongs to I. We recall the definition. Suppose x belongs to [0, ] and has the following base 2 representation a n x = 2 n, a n {0, }. This number x is called a simply normal to base 2 if N N N a n = 2. (3.4) 3.5. Proof of Proposition 3.4. Let x 0 C be a point with the ternary expansion x 0 = 2α n 3 n, α n {0, }. Suppose G(x 0 ) is a number simply normal to base 2. Since G(x 0 ) = α n 2 n, formula (3.4) implies that x 0 C. (If x 0 C, i.e. x is an endpoint of a complementary N N interval of C, then α n = 0 or α n =.) Hence, it suffices to show that N N N N R x0 (n) = 0. (3.5) n n

17 Linear distortion of Hausdorff dimension and Cantor s function 209 Put N 0 := {n : α n = 0} and N := {n : α n = }. Since x 0 C we have card(n 0 ) = card(n ) =. It follows from (3.4) that m+r x0 (m) 2 = α m n m + R m N x0 (m) ( ) m m = m R x0 (m) + α m m + R m N x0 (m) m ( m Rx0 (m) = m m + R m N x0 (m) m + ). 2 Hence we get A similar calculation yields Since N = N 0 N we have (3.5). R (m) x0 m m = 0. m N R (m) x0 m m = 0. m N 0 The proof of the next lemma is well-known. Lemma 3.6 Let m be the Lebesgue measure on R, and let s = log 2/log 3. Then for every A C. m (G(A)) = H s (A) 3.7. Proof of Theorem.5. As in Example 3.3 set M = x C : K y x G (y, x) = log 2 log 3. We claim that for every A M, and dim(g(a)) = log 3 dim(a)) (3.6) log 2 H s (M) = (3.7) for s = log 2/log 3. In order to prove (3.6), we can apply Corollary 2.7 with X = M, X = C, Y = [0, ]. Observe that C is countable and hence we have (2.6). The restriction G M C : M C G(M C )

18 20 Dovgoshey, Martio, Ryazanov, and Vuorinen is strictly increasing and continuous, so that it is a homeomorphism. It remains to verify (3.7). Let N be the set of numbers which are not simply normal to base 2. It is known [6, p.03] that m (N ) = 0. By Proposition 3.4 I is the superset of the set of all simply normal to base 2 numbers. Consequently, m (I ) =, and by Lemma 3.6 we have (3.7). Acknowledgments. Academy of Finland. The first and third authors thank for the support the References. G.A. Edgar, Measure, Topology, and Fractal Geometry, Springer-Verlag, New York, H. Federer, Geometric Measure Theory, Springer-Verlag, New-York, R.E. Gilman, A class of functions continuous but not absolutely continuous, Ann. of Math. (2), 33 (932), E. Hille and J.D. Tamarkin, Remarks on known example of a monotone continuous function, Amer. Math. Monthly 36 (929), K. Kuratowski, Topology, Vol. I, Academic Press, New York-London, I. Niven, Irrational Numbers, John Wiley and Sons, Inc., New York, N.Y., C.A. Rogers, Hausdorff Measures, Cambridge University Press, Cambridge, 970.