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VNU JOURNAL OF SCIENCE, Mathematics - Physics T.XXI, N0 - 2005 SINGULARITY OF FRACTAL MEASURE ASSOCIATED WITH THE (0, 1, 7) - PROBLEM Truong Thi Thuy Duong Vinh University, Nghe An Vu Hong Thanh Pedagogical College of Nghe An ∞ −n Xn , where Abstract Let µ be the probability measure induced by S = n=1 X1 , X2 , is a sequence of independent, identically distributed (i.i.d) random variables each taking values 0, 1, a with equal probability 1/3 Let α(s) (resp.α(s), α(s)) denote the local dimension (resp lower, upper local dimension) of s ∈ supp µ, and let α = sup{α(s) : s ∈ supp µ}; α = inf{α(s) : s ∈ supp µ}; E = {α : α(s) = α for some s ∈ supp µ} In the case a = 3k + for k = 1, E = [1 − √ log(1+ 5)−log , 1], log see [10] It is conjectured that in the general case, for a = 3k + ( k ∈ N), the local dimension is of the form as the case k = 1, i.e., E = [1 − log a b log , 1] for a, b depends on k In fact, our result shows that for k = (a = 7), we have α = 1, α = − √ log(1+ 3) log and E = [1 − √ log(1+ 3) log , 1] Introduction Let X be random variable taking values a1 , a2 , , am with probability p1 , p2 , , pm , respectively and let X1 , X2 , be a sequence of independent random variables with the same distribution as X Let S = ∞ ρn Xn , for < ρ < 1, and let µ be the probability n=1 measure induced by S, i.e., µ(A) = Prob{ω : S(ω) ∈ A} It is known that the measure is either purely singular or absolutely continuous An intriguing case when m = 3, ρ = p1 = p2 = p3 = 1/3 and a1 = 0, a2 = 1, a3 = a According to the ”pure theorem” of Lagarias and Wang, in [7], if a ≡ (mod 3) or a ≡ (mod 3) then µ is purely singular Let us recall that for s ∈ supp µ the local dimension α(s) of µ at s is defined by α(s) = lim+ h→0 log µ(Bh (s)) , log h (1) Typeset by AMS-TEX Truong Thi Thuy Duong, Vu Hong Thanh provided that the limit exists, where Bh (s) denotes the ball centered at s with radius h If the limit (1) does not exist, we define the upper and lower local dimension, denoted α(s) and α(s), by taking the upper and lower limits respectively Denote α = sup{α(s) : s ∈ supp µ} ; α = inf{α(s) : s ∈ supp µ}; and let E = {α : α(s) = α for some s ∈ supp µ} be the attainable values of α(s), i.e., the range of function α definning in the supp µ In the case a = 3k + it is conjectured that the local dimension is of the form as k = 1, it means that E = [1 − blog a3 , 1] for a, b depends on k Our aim in this note is to log prove that this conjecture is true for k = In fact, our result is the following: √ √ Main Theorem For a = we have α = 1, α = − log(1+ 3) and E = [1 − log(1+ 3) , 1] log log The paper is organized as follows In Section we establish some auxiliary results used in the proof of the Main Theorem The proof of the Main Theorem will be given in the last section Some Auxiliary Results Let X1 , X2 , be a sequence of i.i.d random variables each taking values 0, 1, ∞ n with equal probability 1/3 Let S = i=1 3−i Xi , Sn = i=1 3−i Xi be the n-partial sum of S, and let µ, µn be the probability measures induced by S, Sn , respectively For any ∞ n s = n=1 3−n xn ∈ supp µ, xn ∈ D: = {0, 1, 7}, let sn = i=1 3−i xi be its n-partial sum It is easy to see that for any sn , sn ∈ supp µn , |sn − sn | = k3−n for some k ∈ N Let n sn = {(x1 , x2 , , xn ) ∈ Dn : i=1 3−i xi = sn } Then we have µn (sn ) = # sn 3−n for every n, (2) where # sn denotes the cardinality of set sn Two sequences (x1 , x2 , , xn ) and (x1 , x2 , , xn ) in Dn are said to be equivalent, denoted by (x1 , x2 , , xn ) ≈ (x1 , x2 , , xn ), if n 3−i xi = n 3−i xi We have i=1 i=1 n 2.1 Claim (i) For any (x1 , x2 , , xn ), (x1 , x2 , , xn ) in Dn and sn = i=1 3−i xi , sn = n 3−i xi If sn − sn = 3k , where k ∈ Z, then xn − xn ≡ k (mod 3) n i=1 (ii) Let sn > sn > sn be three arbitrary consecutive points in supp µn Then either sn − sn or sn − sn is not 31 and either sn − sn or sn − sn is not 32 n n k Proof (i) Since sn − sn = 3n , we have 3n−1 (x1 − x1 ) + 3n−2 (x2 − x2 ) + + 3(xn−1 − xn−1 ) + xn − xn = k, Singularity of Fractal Measure Associated with which implies xn − xn ≡ k (mod 3) The claim (i) is proved (ii) We can write sn = sn−1 + xn xn xn , sn = sn−1 + n and sn = sn−1 + n , n 3 where sn−1 , sn−1 , sn−1 ∈ supp µn−1 and xn , xn , xn ∈ D Assume on the contrary that sn − sn = sn − sn = 31 Then n + xn − xn + xn − xn = , n 3 3n−1 + xn − xn + xn − xn = = n 3 3n−1 sn−1 − sn−1 = sn−1 − sn−1 Since sk − sk = 3tk , t ∈ Z whenever sk , sk ∈ supp µk , we have (1 + xn − xn ) ≡ (mod 3) and (1 + xn − xn ) ≡ (mod 3) Because (1 + xn − xn ) ≡ (mod 3) and xn , xn ∈ D, we obtain xn = Then + xn − xn = + xn ∈ {1, 2, 8} for any xn ∈ D, a contradiction Similarly, we have either sn − sn or sn − sn is not 32 The claim (ii) is proved n 2.2 Claim (i) Let sn+1 ∈ supp µn+1 and sn+1 = sn + # sn+1 = # sn , for every n # sn 3n+1 , sn ∈ supp µn We have (ii) For any sn , sn ∈ supp µn if sn − sn = 31 , then xn = 0, xn = or xn = and n # sn If xn = 1, then sn−1 = sn−1 and if xn = 7, then sn−1 = sn−1 + 3n−1 , x n n where sn = sn−1 + xn , sn = sn−1 + 3n and sn−1 , sn−1 ∈ supp µn−1 , xn , xn ∈ D (iii) For any sn , sn ∈ supp µn , if sn − sn = 32 , then xn = 0, xn = or xn = n Moreover, if xn = 1, then sn−1 − sn−1 = 3n−1 and if xn = 7, then sn−1 − sn−1 = 3n−1 , x n n where sn = sn−1 + xn , sn = sn−1 + 3n and sn−1 , sn−1 ∈ supp µn−1 , xn , xn ∈ D (iv) For any sn , sn , sn ∈ supp µn , if sn − sn = 31 and sn − sn = 32 , then sn = n n sn−1 + 31 = s∗ + 37 and sn = sn−1 + 37 or sn = sn−1 + 31 = sn−1 + 37 and n n n n n n−1 sn = sn−1 + 31 , where sn−1 , sn−1 , s∗ , sn−1 ∈ supp µn−1 n n−1 Proof (i) It follows directly from Claim 2.1 (ii) Since sn − sn = 31 , by Claim 2.1 (i) xn − xn ≡ (mod 3) Then xn = 0, xn = n or xn = Therefore sn = sn−1 + 30 By Claim 2.1 (i) we have # sn = # sn−1 and n sn = sn−1 + 31 = sn−1 + 31 If sn has an other representation sn = s∗ + 37 , then n n n n−1 # sn # sn−1 = # sn If xn = 1, then sn−1 − sn−1 = sn−1 − sn−1 = If xn = 7, then sn = s∗ + 37 , sn = sn−1 + 30 It implies n n n−1 = sn − sn = s∗ − sn−1 + n n−1 n 3 Therefore sn−1 − s∗ = n−1 3n−1 Truong Thi Thuy Duong, Vu Hong Thanh 10 (iii) It is proved similarly to Claim 2.2 (ii) (iv) Since sn − sn = 31 , by Claim 2.2 (ii) we have n sn = sn−1 + , sn = sn−1 + n = s∗ + n n−1 3n 3 On the other hand sn − sn = so if sn = sn−1 + , 3n = sn−1 + n , n 3 then sn−1 − sn−1 = sn−1 − s∗ = n−1 , 3n−1 a contradiction to Claim 2.1 (ii) Therefore sn = sn−1 + assertion 3n Similarly, we get the last Remark 1) By Claim 2.1 (i), it follows that if sn+1 ∈ supp µn+1 and sn+1 = sn + 3n+1 , then it can not be represented in the forms sn+1 = sn + , or sn+1 = sn + n+1 , 3n+1 where sn , sn , sn ∈ supp µn Thus, any sn+1 ∈ supp µn+1 has at most two representations through points in supp µn 2) Assume that sn , sn ∈ supp µn , if sn − sn = 31 or sn − sn = 32 , then sn , sn are n n two consecutive points in supp µn 2.3 Lemma For any two consecutive points sn and sn in supp µn , we have µn (sn ) µn (sn ) n Proof By (2) it is sufficient to show that # sn n We will prove the inequality by # sn induction Clearly the inequality holds for n = Suppose that it is true for all n k Let sk+1 > sk+1 be two arbitrary consecutive points in supp µk+1 Write sk+1 = sk + xk+1 , sk ∈ supp µk , xk+1 ∈ D 3k+1 We consider the following cases for xk+1 Case If xk+1 = then sk+1 = sk + 3k+1 Assume that sk+1 has an other representation xk+1 sk+1 = s∗ + 3k+1 , xk+1 ∈ D Then s∗ > sk , where s∗ ∈ supp µk k k k Let sk ∈ supp µk be the smallest value larger than sk Then sk > sk are two consecutive points in supp µk Singularity of Fractal Measure Associated with a) For sk = sk + 3k 3k sk then sk − sk If sk sk+1 = sk + 3k+1 11 < sk + = 3k+1 3k+1 We have sk , so sk+1 has the unique representation through point sk in supp µk Hence # sk+1 = # sk Since sk+1 > sk+1 are two arbitrary consecutive points in supp µk+1 and sk+1 = sk + = sk + 3k+1 , we have sk+1 = sk + 3k+1 Assume that sk+1 has an other representation 3k+1 sk+1 = sk + 3k+1 Then sk −sk = 3k It implies sk −sk = sk −sk = 31 , which contradicts k to Claim 2.1 (ii) Hence # sk+1 = # sk Therefore # sk+1 # sk = # sk+1 # sk b) For sk = sk + 3k = sk + 3k+1 k < k + We have sk+1 = sk + 3k+1 = sk + 3k+1 It follows that # sk+1 = # sk + # sk and sk+1 = sk + 3k+1 Hence # sk+1 = # sk Therefore # sk+1 # sk + # sk = # sk+1 # sk c) For sk sk + 3k = sk + 3k+1 sk+1 = sk + k + We have 3k+1 < sk + 3k+1 sk , so sk+1 has the unique representation through point sk in supp µk Hence # sk+1 = # sk Since sk+1 > sk+1 are two consecutive points in supp µk+1 and sk+1 = sk + 3k+1 < sk , sk+1 only represents through points not bigger then sk in supp µk Let sk + 3k+1 s∗ < sk be the consecutive point for sk in supp µk We consider following three cases k c1) If sk = s∗ + 31 , then sk+1 = s∗ + 3k+1 is the unique representation through point s∗ k k k k in supp µk It implies # sk+1 = # s∗ Therefore k # sk+1 # sk = # sk+1 # s∗ k c2) If sk = s∗ + k , 3k then sk+1 = s∗ + k 3k+1 = sk + k < k + 1 3k+1 So # sk+1 = # s∗ + # sk k Truong Thi Thuy Duong, Vu Hong Thanh 12 Therefore # sk+1 # s∗ + # sk k = # sk+1 # sk k + 1 c3) If sk s∗ + 33 , then sk+1 = sk + 3k+1 is the unique representation through point sk k k in supp µk Hence # sk+1 = # sk Therefore # sk # sk+1 = = < k + # sk+1 # sk Case If xk+1 = 0, then sk+1 = sk + 3k+1 By Claim 2.2 (i), we have # sk+1 = # sk x∗ k+1 Then sk+1 = s∗ + 3k+1 < sk+1 = sk It implies s∗ < sk k k Let sk ∈ supp µk be the biggest value smaller than sk Then sk < sk are two consecutive points in supp µk a) If sk = sk + 31 , then by Claim 2.2 (ii) k # sk # sk We have sk+1 = sk = sk + Hence sk+1 = sk + 3k+1 3k+1 (3) > sk + sk + 3k+1 3k+1 , where sk > sk are two consecutive points in supp µk ( Because by Claim 2.1, sk −sk Thus, # sk+1 # sk + # sk Therefore # sk+1 # sk+1 b) If sk = sk + # sk + # sk # sk , 3k # sk + # sk # sk ) 3k (k + 1) then sk+1 = sk = sk + = sk + k+1 k 3 sk + + k+1 = sk + k+1 , k 3 3 with sk > sk are two consecutive points in supp µk and sk − sk = 31 or sk − sk k 3k # sk , sk+1 = sk + 3k+1 and it is the unique represenb1) If sk = sk + 3k , then # sk tation of sk+1 through points in supp µk (If it is not the case, sk+1 = s∗ + 3k+1 , then k sk − s∗ = s∗ − sk = 31 , a contradictions to Claim 2.1) Hence # sk+1 = # sk Therefore k k k # sk+1 # sk = # sk+1 # sk To show that # sk+1 # sk+1 # sk # sk k < k + k + 1, we note that sk+1 = sk = sk−1 + xk xk , s = sk−1 + k 3k k Singularity of Fractal Measure Associated with 13 Since sk − sk = 31 , we have xk = and # sk = # sk−1 k Since sk − sk = 33 , we have xk − xk ≡ ( mod 3) By xk = 0, we get xk = It k implies # sk = # sk−1 Moreover, sk−1 −sk−1 = 3k−1 , so sk−1 , sk−1 are two consecutive points in supp µk−1 Therefore, by the inductive hypothesis we have # sk # sk−1 # sk+1 = = # sk+1 # sk # sk−1 b2) If sk sk + 3k k − < k + then sk+1 = sk = sk + 3k+1 sk + 10 > sk + k+1 3k+1 Hence sk+1 = sk + 3k+1 and this is the unique representation of sk+1 through points in supp µk Hence # sk+1 = # sk Therefore # sk+1 # sk = # sk+1 # sk k < k + c) If sk sk + 33 = sk + 3k+1 , then sk+1 = sk + k of sk+1 So # sk+1 = # sk Therefore # sk+1 # sk = # sk+1 # sk 3k+1 is the unique representation k < k + 1 Case If xk+1 = 1, then sk+1 = sk + 3k+1 Note that if sk+1 has an other representation then sk+1 = s∗ + 3k+1 and sk −s∗ = 32 It implies s∗ , sk are two consecutive points in supp k k k k ∗ # sk + # sk Since sk+1 > sk+1 are two arbitrary consecutive µk Clearly # sk+1 points in supp µk+1 , we have sk+1 = sk + 3k+1 Hence # sk+1 = # sk Therefore # sk+1 # sk+1 # sk + # s∗ k # sk k + The lemma is proved The following proposition provides a useful formula for calculating the local dimension and it is proved similarly to the proof of Proposition 2.3 in [10] and using Lemma 2.3 2.4 Proposition For s ∈ supp µ, we have α(s) = lim n→∞ | log µn (sn )| , n log provided that the limit exists Otherwise, by taking the upper and lower limits respectively we get the formulas for α(s) and α(s) Truong Thi Thuy Duong, Vu Hong Thanh 14 For each infinite sequence x = (x1 , x2 , ) ∈ D∞ defines a point s ∈ supp µ by ∞ s = S(x) := 3−n xn n=1 We denote x(k) = {(y1 , , yk ) ∈ Dk : (y1 , , yk ) ≈ (x1 , , xk )}, where x(k) = (x1 , , xk ) It is easy to check that (1, 0, 1) ≈ (0, 1, 7), (0, 7, 0, 1) ≈ (1, 1, 7, 7) and (1, a, 0, 1) ≈ (0, a, 7, 7) for any a ∈ D We call each element in the set {(1, 0, 1), (0, 1, 7), (0, 7, 0, 1), (1, 1, 7, 7), (1, a, 0, 1), (0, a, 7, 7)} a generator 2.5 Claim Let x(3n) = (x1 , x2 , , x3n ) = (1, 0, 1, , 1, 0, 1) y(3n + 1) = (y1 , , y3n+1 ) = (1, x1 , , x3n ) and z(3n + 2) = (z1 , , z3n+2 ) = (1, 1, x1 , , x3n ), (4) where x3k+1 = x3k+3 = 1, x3k+2 = 0, for k = 0, 1, 2, Putting j 3−i xi , sj = i=1 we have (i) # s3 = 2, # s6 = 6, # y(4) = 3, # y(7) = and # z(5) = 4, # z(8) = 10 and (ii) # s3(n+1) = 2# s3n + 2# s3(n−1) , # y(3(n + 1) + 1) = 2# y(3n + 1) + 2# y(3(n − 1) + 1) and # z(3(n + 1) + 2) = 2# z(3n + 2) + 2# z(3(n − 1) + 2) , for n = 1, 2, Proof (i) Claim (i) is clear Singularity of Fractal Measure Associated with 15 (ii) We only prove the case # s3(n+1) = 2# s3n +2# s3(n−1) , the other cases are proved similarly We have s3n+3 = s3n + = s3n + = s3n + = s3n + 33n+1 33n+1 33n+1 + + + 33n+2 33n+2 33n+2 + + + 33n+1 33n+1 33n+1 7 + 3n+2 + 3n+1 , 33n+1 3 (5) where s3n , s3n and s3n ∈ supp µ3n Therefore # s3n+3 = 2# s3n + # s3n + # s3n Using Claim 2.2, we have s3n = s3n−3 + 33n−2 + 33n−1 + 0 , s3n = s3n−3 + 3n+1 + 3n+2 + 3n+1 3n 3 3 So # s3n = # s3n−3 Assume that s3n = s3n−3 + 33n−2 + 33n−1 + a contradiction to x3n−3 = Hence # s3n = # s3n−3 Thus 33n Then x3n−3 = 0, # s3n+3 = 2# s3n + 2# s3n−3 The claim is proved Putting F3n = # s3n , G3n+1 = # y(3n + 1) and H3n+2 = # z(3n + 2) , from Claim 2.5 we have √ √ F3n = √ [(1 + 3)n+1 − (1 − 3)n+1 ], √ √ G3n+1 = √ [(1 + 3)n+2 − (1 − 3)n+2 ] and √ √ H3n+2 = [(1 + 3)n+1 + (1 − 3)n+1 ] 2.6 Claim Let x = (x1 , x2 , ) = (1, 0, 1, , 1, 0, 1, ) or x = (1, 1, 0, 1, , 1, 0, 1, ) ∞ or x = (1, 1, 1, 0, 1, , 1, 0, 1, ) ∈ D∞ and s = i=1 3−i xi ∈ supp µ, we have √ log(1 + 3) α(s) = − log Proof The proof of the claim is similar to the proof of Claim 2.6 in [10] We say that x = (x1 , x2 , , xn ) ∈ Dn is a maximal sequence if # tn where sn = n i=1 3−i xi # sn for any tn ∈ supp µn , Truong Thi Thuy Duong, Vu Hong Thanh 16 The following fact will be used to estimate the greatest lower bound of local dimension 3n+j 2.7 Proposition For every n ∈ N, let t3n+j = i=1 3−i ti be an arbitrary point in supp F3n , # t3n+1 G3n+1 and # t3n+2 H3n+2 µ3n+j , for j = 0, 1, Then # t3n Proof We will prove the proposition by induction It is straightforward to check that the assertion holds for n = 1, 2, Suppose that it is true for all n k(k 3) We show that the proposition is true for n = k + Let t3(k+1) be an arbitrary point in supp µ3k+3 We consider the following cases Case (y3k+1 , y3k+2 , y3k+3 ) is a generator Without loss of generality, we assume that (y3k+1 , y3k+2 , y3k+3 ) = (1, 0, 1) 1.1 If y3k = 0, then t(3k + 3) = (t(3k − 1), 0, 1, 0, 1) ∪ (t(3k − 1), 0, 0, 1, 7) Hence 2H3(k−1)+2 + G3k+1 F3(k+1) # t3k+3 1.2 If y3k = or y3k = 1, then t(3k + 3) = (t(3k), 1, 0, 1) ∪ (t(3k), 0, 1, 7) It implies # t3k+3 F3k + H3k+2 = F3(k+1) Case (y3k+1 , y3k+2 , y3k+3 ) is not a generator 2.1 If y3k+3 = then by Claim 2.2.(i), inductive hypothesis and (6) we have # t3k+3 = # t3k+2 H3k+2 F3(k+1) 2.2.1 Similarity as above, we have if y3k+3 = and y3k+2 = or 7, then # t3k+3 = # t3k+2 H3k+2 F3(k+1) 2.2.2 If y3k+3 = 1, y3k+2 = and (y3k , y3k+1 , 0, 1) is not a generator, then # t3k+3 G3k+1 F3(k+1) 2.2.3 If (y3k , y3k+1 , 0, 1) is a generator, then (y3k , y3k+1 , 0, 1) ∈ {(0, 7, 0, 1), (1, 0, 0, 1), (1, 7, 0, 1)} 2F3k a) If (y3k , y3k+1 , 0, 1) = (0, 7, 0, 1) or (1, 0, 0, 1), then # t3k+3 b) If (y3k , y3k+1 , 0, 1) = (1, 7, 0, 1) We consider two cases b1) If y3k−1 = or y3k−1 = 1, then # t3k+3 2F3k F3(k+1) b2) If y3k−1 = 0, then (0, 1, 7, 0, 1) ≈ (1, 0, 1, 0, 1), hence t(3k + 3) = (y(3k), 1, 0, 1) for y(3k) ∈ D3k According to the Case 1, we have # t(3k + 3) = # (y(3k), 1, 0, 1) F3(k+1) F3(k+1) Singularity of Fractal Measure Associated with 17 2.3 If y3k+3 = By similar argument as above cases, we get # t3k+3 F3(k+1) Thus, by using inductive hypothesis and the formula for calculating for F3n , H3k+2 and G3k+1 , we finished the proof of the first inequality, that means # t3n F3n , for all n Similar argument and using the result # t3n F3n , for all n, we have # t3n+1 G3n+1 for all n By repeating the above argument and using the two above results we have the last of the assertion The proposition is proved Proof of The Main Theorem We call an infinite sequence x = (x1 , x2 , ) ∈ D∞ a prime sequence if # sn = n for every n, where sn = i=1 3−i xi √ 3.1 Claim α = 1, α = − log(1+ 3) log Proof For any prime sequence x = (x1 , x2 , ), for example x = (x1 , x2 , ) = (7, 7, ) or x = (x1 , x2 , ) = (0, 0, ), we have # sn = for every n, where sn = n 3−i xi i=1 Therefore, by Proposition 2.4 we get | log µn (sn )| = 1, n→∞ n log α = α(s) = lim where s = S(x) From Claim 2.6 we have α √ log(1 + 3) 1− log # s3n = For any t ∈ supp µ, by Claim 2.5 and Proposition 2.7, we have # t3n √ n+1 √ n+1 √ ((1 + 3) − (1 − 3) ) for every n Hence √ √ √ | log 2√3 ((1 + 3)n+1 − (1 − 3)n+1 )3−3n | | log µ3n (t3n )| log(1 + 3) lim lim = 1− , n→∞ n→∞ 3n log 3n log 3 log where tn be n - partial sum of t Similar argument as above, we have | log µ3n+1 (t3n+1 )| lim n→∞ (3n + 1) log √ log(1 + 3) | log µ3n+2 (t3n+2 )| 1− , lim n→∞ log (3n + 2) log So we get α Therefore The claim is proved √ log(1 + 3) 1− log √ log(1 + 3) α=1− log √ log(1 + 3) 1− log Truong Thi Thuy Duong, Vu Hong Thanh 18 Now we will show that for any β ∈ (1 − which α(s) = β Let r = 3(1 − log √ β) log(1+3 3) ki = there exists s ∈ supp µ for Clearly < r < For i = 1, 2, , define 3i if i is odd; [ 3i(1−r) ] r if i is even, x Let nj = where [x] denotes the largest integer Ej = {i : i √ log(1+ 3) log , 1) j i=1 j and i is even} ; Oj = {i : i ej = ki ; oj = i∈Ej ki and let j and i is odd}, ki i∈Oj Then nj = oj + ej 3.2 Claim With the above notation we have lim j→∞ j nj−1 oj = ; lim = and lim = r j→∞ nj j→∞ nj nj Proof The proof of the claim is similar to the proof of Claim 3.2 in [10] We define s ∈ supp µ by s = S(x), where x = (1, 0, 0, 0, , 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, , 0, ) k1 =3 k2 k3 =9 (6) k4 Note that, for i ∈ Oj , # ski ⎧ √ ki +1 √ ki +1 ⎨ > = √ [(1 + 3) − (1 − 3) ] ⎩< √ ki +1 3) √ ki +2 √ (1 + 3) √ (1 + (7) Let s ∈ supp µ be defined (6) and let nj−1 n < nj By the multiplication principle, we have # ski # sn # ski i∈Oj−1 i∈Oj Hence, by (7) yield √ oj−1 j−1 j−1 ( √ ) (1 + 3) + 2 # sn √ oj j+1 ( √ ) (1 + 3) +(j+1) , which implies log[( 2√3 ) j−1 (1 + √ oj−1 + j−1 ] 3) nj log log # sn n log log[( 2√3 ) j+1 (1 + √ oj +(j+1) 3) ] nj−1 log Singularity of Fractal Measure Associated with 19 From the latter and Claim 3.1 we get √ r log(1 + 3) log # sn lim = n→∞ n log 3 log Therefore | log # sn 3−n | log # sn = − lim n→∞ n→∞ n log n log √ r log(1 + 3) = β =1− log α(s) = lim The Main Theorem is proved References K J Falconer, Techniques in Fractal Geometry, John Wiley & Sons, 1997 K J Falconer, Fractal Geometry, Mathematical Foundations and Applications,John Wiley & Sons, 1993 A Fan, K S Lau and S M Ngai, Iterated function systems with overlaps, Asian J Math 4(2000), 527 - 552 T Hu, The local dimensions of the Bernoulli convolution associated with the golden number, Trans Amer Math Soc 349(1997), 2917 - 2940 T Hu and N Nguyen, Local dimensions of fractal probability measures associated with equal probability weight, Preprint T Hu, N Nguyen and T Wang, Local dimensions of the probability measure associated with the (0, 1, 3) - problem, Preprint T Hu, Some open questions related to Probability, Fractal, Wavelets, East - West J of Math Vol 2, No 1(2000), 55-71 J C Lagarias and Y Wang, Tiling the line with translates of one tile, Inventions Math 124(1996), 341 - 365 S M Ngai and Y Wang, Hausdorff dimention of the self - similar sets with overlaps,J London Math Soc ( to appear) 10 Le Xuan Son, Pham Quang Trinh and Vu Hong Thanh Local Dimension of Fractal Measure associated with the (0,1,a)-Problem, the case a = 6.Journal of Science, VNU, T.XXI, 1(2005), 31-44  ... that (1, 0, 1) ≈ (0, 1, 7), (0, 7, 0, 1) ≈ (1, 1, 7, 7) and (1, a, 0, 1) ≈ (0, a, 7, 7) for any a ∈ D We call each element in the set { (1, 0, 1), (0, 1, 7), (0, 7, 0, 1), (1, 1, 7, 7), (1, a,... = (1, 0, 1, , 1, 0, 1, ) or x = (1, 1, 0, 1, , 1, 0, 1, ) ∞ or x = (1, 1, 1, 0, 1, , 1, 0, 1, ) ∈ D∞ and s = i=1 3−i xi ∈ supp µ, we have √ log(1 + 3) α(s) = − log Proof The. .. dimensions of fractal probability measures associated with equal probability weight, Preprint T Hu, N Nguyen and T Wang, Local dimensions of the probability measure associated with the (0, 1, 3) - problem,