The Hausdorff measure of a class of Sierpinski carpets

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1 J. Math. Anal. Appl. 305 (005) The Hausdorff measure of a class of Sierpinski carpets Yahan Xiong, Ji Zhou Department of Mathematics, Sichuan Normal University, Chengdu , PR China Received 9 July 004 Available online 7 January 005 Submitted by J.D.M. Wright Abstract In this paper we obtain the exact value of the Hausdorff measure of a class of Sierpinski carpets with Hausdorff dimension no more than 1 show the fact that the Hausdorff measure of such Sierpinski carpets can be determined by coverings which only consist of basic squares. 004 Elsevier Inc. All rights reserved. Keywords: Hausdorff measure dimension; Sierpinski carpet 1. Introduction One of the most important subjects fractal geometry deals with is the estimation calculation of the Hausdorff measure of fractal sets. But this subject is difficult to tackle the relevant results are not rich even as to seemingly simpler sets. Up to now, the accurate value of Hausdorff measure of some self-similar sets with Hausdorff dimension no more than 1 has been obtained, such as middle-three Cantor set [1]. In [], the Hausdorff measures of certain Sierpinski carpets whose Hausdorff dimension equal to 1 are obtained with the help of the principle of mass distribution. Later, in [3], more general sets are Supported by Applied Fundamental Research Foundation of Sichuan Province. * Corresponding author. address: jzhou1@163.net (J. Zhou) X/$ see front matter 004 Elsevier Inc. All rights reserved. doi: /j.jmaa

2 1 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) considered the corresponding Hausdorff measures are determined by estimating the number of the nth-stage squares that intersect with an open set. In this article, using another way, we obtain that the Hausdorff measures of a class of Sierpinski carpets E, H s (E) = s/, moreover, we show that H s (E) = HF s (E), where H F s (E) refers to the net measure 1 determined by covering that is made up of basic squares s = log λ 4 is the Hausdorff dimension of Sierpinski carpets E. For the net measure, see Section.. Lemmas Take E 0 to be the unit square in R delete all but the four corner squares of side λ to obtain E 1. Continue in this way, at the kth-stage, replacing each square of E k 1 by four corner squares of side λ k to get E k. These retained squares of side λ k are called kth-stage basic squares. We can see that E k consists of 4 k kth-stage basic squares that E 0 E 1 E E n (see Fig. 1). Let E = k=0 E k. Thus E is a compact set that can be considered as a self-similar set generated by four similitudes with the scale factor λ. We call E a Sierpinski carpet. Since it satisfies the open set condition, the Hausdorff dimension 1 of E is s = log λ 4. Let f k be any kth-stage basic square, F k the class of all the kth-stage basic squares, F = k 1 F k the class of all the basic squares f k = λ k the diameter of f k. A net N is a class of sets, with the property that, for any x R ε>0, there exists A N such that x A A ε, where A sts for the diameter of A. Bythis definition, F is a net of sets consisting of basic squares. Let E be a subset of R s be a non-negative number. For δ>0 define Hδ s (E) = inf U i s, i=1 where the infimum is over all (countable) δ-coverings U i of E. The Hausdorff s-dimensional measure of E is defined as H s (E) = lim Hδ s (E). δ 0 The s-dimensional net measure determined by the net N, denoted by HN s (E), is constructed in a similar manner to the s-dimensional Hausdorff measure but using N of covering sets in the definition rather than the class of all sets. To show our main result we need the following two lemmas. Fig. 1.

3 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) Lemma 1. H s F (E) = s/ = E 0 s. Proof. Let G ={U i } be an arbitrary countable δ-covering of E with basic squares. By the definition of HF s (E), we may assume that each element in G cannot contain completely the other. Let k = min{j: U i F j for some U i G}. If all the kth-stage basic squares are in G, it follows that U i s = 4 k( λ k) s = s/. U i G If there are l k (0 <l k < 4 k ) kth-stage basic squares that are in G, then there are (4 k l k ) kth-stage basic squares that are not in G. Now we consider such (k + 1)th-stage basic squares which are contained in such (4 k l k )kth-stage basic squares. If such (k + 1)thstage basic squares all are in G, it is easy to obtain that U i s ( = l k λ k ) s ( k )( l k λ (k+1) ) s = s/. U i G Similarly, if there exists some integer t>0such that t = max{j: U i F j for some U i G}, we have that U i G U i s = s/. If such t does not exist, then there are integers l m, m = k,k + 1,...,with 0 l m < 4 m such that U i s ( = l m λ m ) s. U i G m=k Now denote by n p the number of all the pth-stage basic squares which are not contained in any of qth-stage basic squares in G where q<p. From the above discussion, it follows that for any integer p>k, p 1 ( l m λ m ) s ( + np λ p ) s = s/. m=k We claim that n p ( λ p ) s 0, as p. If not, there exist a γ>0 an infinite subset N of positive integers such that n p ( λ p ) s >γ when p N. For each p N, let W p be the union of n p pth-stage basic squares which are not contained in any of qth-stage basic squares in G where q<p. Obviously, W p is non-empty compact W n W m, m>n. Thus, E p N W p p N W p. From the definition of W p, it follows that ( ) ( ) W p U i =, p N U i G which is contrary to the fact that G is a covering of E. Therefore, we obtain that, for each δ>0, inf U i G U i s = s/, where the infimum is taken over all δ-covering of E.Bythe definition of the net measure, the lemma holds. Lemma. Let λ 1 4 U be an open subset of R with λ k+1 U < λ k for some integer k>0. Let F U be a family of the basic squares f σ (f σ F) which are completely contained in U no one is contained in the other. Then U s

4 14 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) Proof. Since λ 1 4, λk 1 λ k λ k > U, U can intersect with only one kth-stage basic square which cannot be contained completely in U. We may assume that the kthstage basic square which meets the U is A =[0,λ k ] [0,λ k ]. Next we will prove the lemma under the following four cases: Case 1. U intersects with one (k + 1)th-stage basic square in A. It is easy to get, similar to the proof of Lemma 1, that f σ s f k+1 s ( = λ 1+k ) s U s. Case. U intersects with two (k + 1)th-stage basic squares in A. LetI n (respectively I n ) (n > k) refer to the nth-stage basic square on the top right (respectively left) in A J n (respectively J n ) (n > k) the nth-stage basic square on the bottom left (respectively right) in A. Subcase 1. Suppose that U J k+1 U I k+1 (as at the left of Fig. ). It is easy to show that U > f k f k+1 = λ k (1 λ), similar to the proof of Lemma 1, f σ s f k+1 s = 1 ( λ k ) s. Let g(λ) = (1 λ) s. Since g(λ) decreases on (0, 1 4 ] g(1 4 ) = 1, we have, when λ 1 4, that (1 λ)s 1 ( λ k ) s (1 λ) s 1 ( λ k ) s. Therefore, U s Subcase. Suppose that U J k+1 U J k+1 (as at the right of Fig. ). Fig..

5 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) Now let a = min { } ( x: (x, y) U J k+1 0 a λ 1+k ), b = max { x: (x, y) U J k+1} ( λ k λ 1+k <b λ k), a = λ 1+k a b = λ k b. If 0 <a λ 1+k λ +k 0 <b λ k+1 λ k+, then we have U >λ k λ 1+k fσ s 4 f k+ s = 4 ( λ +k) s. Thus, similar to the discussion in Subcase 1, when λ 1 4, we have (λk λ 1+k ) s > 4( λ +k ) s hence U s If a >λ 1+k λ +k 0 <b λ 1+k λ +k, we can get U λ k λ 1+k + a + b >λ k λ 1+k + ( λ 1+k λ +k) fσ s 6 f k+ s = 6 ( λ +k) s. Thus, if λ 1 4,wehave(λk λ 1+k + λ 1+k λ +k ) s 6( λ +k ) s so U s If a >λ 1+k λ +k b >λ 1+k λ +k, we obtain that U λ k λ 1+k + a + b λ k λ +k f σ s 8 f k+ s ( = 8 λ +k ) s. Thus, when λ 1 4,wehave(λk λ +k ) s 8( λ k+ ) s so U s Case 3. U intersects with three (k + 1)th-stage basic squares in A (see Fig. 3). Since U < λ k = f k, U cannot completely contain both I k+1 J k+1. Subcase 1. Suppose that neither I k+1 nor J k+1 is completely contained in U, then there exist positive integers n 1 n, such that U I k+n1, U I k+n1 +1 = ; U J k+n, U J k+n +1 =. Assume that n 1 1 n > 1. Set FU I ={f σ F U : f σ I k+1 } FU J ={f σ F U : f σ J k+1 }. Now draw two lines l 1 l both paralleling the diagonal line such that they intersect with the boundary of U U is located between them. Let the distance between l i A i be a i (i = 1, ) (see Fig. 3). Hence, we have a i f k+1 f k+ni, f σ s < ( 4 n 1 1 )( f k+n1 +1 ) s, f σ F I U f σ s < ( 4 n 1 )( f k+n +1 s). f σ F J U

6 16 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) Fig. 3. Since U f k f k+1 +a 1 + a f σ s f σ s + f σ s + f k+1 s, f σ FU I f σ FU J when λ 1 4, n 1 1, n > 1, similar to the proof of Case, we can get U s Assume that n 1 = n = 1. (λ k+1 λ k+ ), then we have U > If 0 <a 1 (λ k+1 λ k+ ) 0 <a f k f k+1 f σ s 6 f k+ s. Hence, when λ 1 4, we get U s If a 1 > (λ k+1 λ k+ ), 0 <a (λ k+1 λ k+ ), we get that U s > f k f k+1 + a 1 > λ k (1 3 λ 1 λ ) f σ s 8 f k+ s. When λ 1 4,itfollows that U s If a 1 > (λ k+1 λ k+ ) a > (λ k+1 λ k+ ), then U f k f k+1 +a 1 + a f σ s 10 f k+ s hold. When λ 1 4, we obtain that U s Subcase. Suppose that either I k+1 or J k+1 is completely contained in U. Let us say I k+1 U, which implies that a 1 > f k+1 f σ FU I f σ s = f k+1 s. Meanwhile, a f k+1 f k+n f σ FU J f σ s <(4 n 1) f k+n +1 s hold just as in Subcase 1. Hence, we have U f k f k+1 +a 1 + a f σ s + f k+1 s. When λ 1 4, we may get U s f σ s f σ F J U Case 4. U intersects with four (k + 1)th-stage basic squares in A.

7 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) Subcase 1. Suppose that U can completely contain none of I k+1, I k+1, J k+1, J k+1, it follows that there exist positive integers n i (1 i 4), such that U I k+n1, U I k+n1 +1 =, U I k+n, U I k+n +1 =, U J k+n3, U J k+n3 +1 =, U J k+n 4, U J k+n 4 +1 =. (1) Let U 1 = U I k+n1, U = U I k+n, U 3 = U J k+n3, U 4 = U J k+n 4. Denote FU I ={f σ F U : f σ I k+1 }, F I U ={f σ F U : f σ I k+1 }, F U J ={f σ F : f σ J k+1 }, F J U ={f σ F U : f σ J k+1 }. Since each (k + n i)th-stage basic square contains four (k + n i + 1)th-stage basic squares, it can be shown by (1) that U i can intersect with at most three (k + n i + 1)th-stage basic squares. Now draw four lines l i (1 i 4) paralleling the diagonal lines such that they bound U meanwhile all intersect with the boundary of U. Let the distance between l i A i be a i (1 i 4) (see Fig. 4). Suppose that each of U i (1 i 4) intersects with only one (k + n i + 1)th-stage basic square, then we have a i f k+1 f k+ni, U f k f k i=1 a i f σ s 4 i=1 (4 n i 3) f k+ni +1 s. Hence, when λ 1 4, U s f σ s holds. Suppose that U 1 intersects with more than one (k + n 1 + 1)th-stage basic squares U i ( i 4) all intersect with only one (k + n i + 1)th-stage ( i 4) basic square. If there exists l(l ) such that U 1 cannot intersect with two (k + n 1 + l)th-stage basic squares, one of which is on the top right of the (k + n 1 + 1)th-stage basic square that is on the top left of I k+n1, the other is on the top right of the (k + n 1 + 1)th-stage basic square that is on the bottom right of I k+n1.infact,ifl satisfies the above condition, so does l for any l l. We choose l to be the minimum of these l. In this case, we have a 1 f k+1 1 ( fk+n1 f k+n1 +1 ) f k+n1 +l 1 Fig. 4.

8 18 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) f σ F I U f σ s ( 4 n 1 1 ) f k+n1 +1 s f k+n1 +l s. Hence, we get U f k f k f σ s 4 i=1 a i 4 ( 4 n i 3 ) f k+ni +1 s + ( 4 n 1 1 ) f k+n1 +1 s f k+n1 +l s. i= Therefore when λ 1 4, we obtain that U s If such l does not exist, we can get a 1 f k+1 1 ( fk+n1 f k+n1 +1 ) f σ F I U f σ s ( 4 n 1 1 ) f k+n1 +1 s. Thus U f k f k f σ s i= 4 i=1 a i 4 ( 4 n i 3 ) f k+ni +1 s + f σ s f σ F I U hold. Similarly, when λ 1 4, we obtain that U s When any two, three, or four of {U i : 1 i 4} intersect with more than one (k + n i + 1)th-stage (1 i 4) basic squares, the proof is entirely similar to the above discussion. Subcase. Suppose U completely contains I k+1 intersects with I k+1, J k+1, J k+1. In this case we have a 1 > f k+1 f σ FU I f σ s = f k+1 s. The rest of the discussion is the same as Subcase 1 it is easy to get our desired result. Subcase 3. Suppose U completely contains both I k+1 I k+1 intersects with J k+1 J k+1. Under this condition, we may similarly obtain U s The proof of Lemma is complete.

9 Y. Xiong, J. Zhou / J. Math. Anal. Appl. 305 (005) Main results The following theorem is our main result. Theorem 1. If λ 1 4, then H s (E) = H s F (E). Proof. Let {U i } be an arbitrary open δ-covering of E. For any given U i there exists k>0 such that λ k+1 U i < λ k.letf Ui be the class of the basic squares completely contained in U i no one is contained in the other. By Lemma, U i s i f σ s holds provided that λ 1 4. E is compact, so we can make F U i satisfy that i i f σ E. Since i U i s i i f σ s, H s (E) HF s (E) holds. Meanwhile, H s (E) HF s (E) is obvious, we can get H s (E) = HF s (E). By Lemma 1, we can obtain the following result. Corollary 1. If λ 1 4, H s (E) = s/. Acknowledgments The authors appreciate the referees good suggestions for this paper. References [1] K.J. Falconer, Fractal Geometry Mathematical Foundation Applications, Wiley, New York, [] Z. Zhou, M. Wu, The Hausdorff measure of a Sierpinski carpet, Sci. China Ser. A 47 (1999) [3] C. Huang, An elementary proof for the Hausdorff measure of Sierpinski carpet, Acta Math. Sinica 43 (000) (in Chinese).