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Introduction Hausdorff measure and packing measure are two of the most important fractal measures used in studying fractal sets see [1–5].. They also yield Hausdorff dimension dim and pack

Volume 2007, Article ID 26249, 17 pages

doi:10.1155/2007/26249

Research Article

New Inequalities on Fractal Analysis and Their Applications

Der-Chen Chang and Yong Xu

Received 26 September 2006; Revised 22 November 2006; Accepted 23 November 2006 Recommended by Wing-Sum Cheung

Two new fractal measuresM ∗ sandM s

∗are constructed from Minkowski contentsM ∗ s

andM s

∗ The properties of these two new measures are studied We show that the fractal dimensions Dim andδ can be derived from M ∗ sandM s

∗, respectively Moreover, some inequalities about the dimension of product sets and product measures are obtained Copyright © 2007 D.-C Chang and Y Xu This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited

1 Introduction

Hausdorff measure and packing measure are two of the most important fractal measures used in studying fractal sets (see [1–5]) They also yield Hausdorff dimension dim and packing dimension Dim, whose main properties are the following

Property 1.1 (monotonicity) E1⊂ E2⇒dim (E1)≤dim(E2), Dim(E1)≤Dim(E2)

Property 1.2 (σ-stability) dim(

n En)≤supndim(En), Dim(

n En)≤supnDim(En) Not all dimension indices areσ-stable For example, upper box dimension Δ and lower

box dimensionδ are not σ-stable These two indices can be yielded from the upper and

lower Minkowski contentsM ∗ sandM s

∗ We know that the Minkowski contents are not outer measures as they are not countably subadditive It is known that the modified upper box dimensionΔ and the modified lower box dimension δ are dimension indices which satisfy Properties1.1and1.2 However, until now no measures have been constructed that yieldΔ and δ In the first part of this paper, we construct two Borel regular measures

ᏹ∗ sandᏹs

∗ The properties of these two new measures, many of which mirror those of packing measure, are studied We show that they yieldΔ and δ, respectively.

The first result about the Hausdorff dimension of the Cartesian product of sets in Euclidean space was obtained by Besicovitch and Moran [6] Readers can also consult the book of Falconer [2] for a good survey In [5], Tricot gives a complete description of Hausdorff and packing dimensions as follows:

dim(E) + dim(F) ≤dim(E × F) ≤dim(E) + Dim(F)

≤Dim(E × F) ≤Dim(E) + Dim(F). (1.1)

Connecting toδ, Xiao [ 7] proves the following result:



δ(E) + Dim(F) ≤Dim(E × F). (1.2)

In this paper, we first prove the following inequality:



δ(E) + δ(F) ≤  δ(E × F) ≤  δ(E) + Dim(F). (1.3)

As a consequence, we have the following inequality:



δ(E) + δ(F) ≤  δ(E × F) ≤  δ(E) + Dim(F) ≤Dim(E × F) ≤Dim(E) + Dim(F). (1.4)

We also show that the inequality dim(E × F) ≤dim(E) + δ(F) does not hold.

On the other hand, Haase [8] studies the dimension of product measures and obtains the following result:

dim(μ) + dim(ν) ≤dim(μ × ν) ≤dim(μ) + Dim(ν)

≤Dim(μ × ν) ≤Dim(μ) + Dim(ν). (1.5)

Using the properties ofᏹs

∗, here we prove a new inequality as follows:



δ(μ) + δ(ν) ≤  δ(μ × ν) ≤  δ(μ) + Dim(ν) ≤Dim(μ × ν) ≤Dim(μ) + Dim(ν). (1.6)

2 Background

Let us first recall some basic properties of Hausdorff measure, Hausdorff dimension, packing measure, packing dimension, Minkowski contents, box dimensions, and mod-ified box dimensions

LetU be a nonempty subset ofRn As usual, one may define the diameter ofU as

| U | =sup

| x − y |:x, y ∈ U

LetE be a subset ofRnands > 0 For δ > 0, define

Ᏼs

δ(E) =inf

∞

i =1

Eis

:E ⊂ i

Ei,Ei  ≤ δ (2.2)

It is easy to check thatᏴs

δis an outer measure onRn

We define thes-dimensional Hausdorff measure of E by

Ᏼs(E) =lim

δ →0Ᏼs

It is known thatᏴsis a Borel regular measure (see Stein and Shakarchi [9, Chapter 7]) The Hausdorff dimension of E can be defined as

dim(E) =inf

s > 0 : Ᏼ s(E) =0

=sup

s > 0 : Ᏼ s(E) > 0

Define

P δ s(E) =sup

∞



i =1

2ris:B

xi,ri

s are pairwisely disjoint, xi ∈ E, ri < δ , (2.5)

whereB(x,r) is the closed ball centered at x with radius r Then the premeasure P s(E) of

E is defined as (see Tricot [5])

P s(E) =lim

δ →0P s

It is known thatP s(E) is not an outer measure since it fails to be countably subadditive.

However, thes-dimensional packing measure of E, which is a Borel regular measure, can

be defined as

ᏼs(E) =inf

∞

i =1

P s

E i :E ⊂ i

The packing dimension ofE is defined by

Dim(E) =inf

s > 0 : ᏼ s(E) =0

=sup

s > 0 : ᏼ s(E) > 0

IfE is a bounded subset inRn, forε > 0, denote

E(ε) =x ∈ R n:d(x,E) ≤ ε

which is called a closedε-neighborhood of E Associating to ε, one may also define the

covering number

N(E,ε) =min



k : E ⊂ k

i =1

B

xi,ε

and the packing number

P(E,ε) =max

k : there are disjoint balls B

x i,ε ,i =1, ,k, x i ∈ E

. (2.11)

Thes-dimensional upper and lower Minkowski contents of bounded set E are defined

by

M ∗ s(E) =lim sup

ε ↓0

 (2ε) s − nᏸn

E(ε) ,

M s

∗(E) =lim inf

ε ↓0

 (2ε) s − nᏸn

E(ε) ,

(2.12)

whereε ↓0 andᏸnare the Lebesgue measures onRn

Thus we can define the upper and lower box dimensions by

Δ(E) =inf

s : M ∗ s(E) =0

=sup

s : M ∗ s(E) > 0

,

δ(E) =inf

s : M s ∗(E) =0

=sup

s : M s ∗(E) > 0

It is known that Minkowski contents are not outer measures as they are not countable subadditive, and the indicesΔ,δ are not σ-stable (see, e.g., Tricot [5], Falconer [1]) We can obtainσ-stable indicesΔ and δ, which are called the modified upper and lower box

dimensions, by letting



Δ(E) =inf

 sup

i ΔE i

:E ⊂ i

E i,E i s are bounded ,



δ(E) =inf

 sup

i δ

Ei :E ⊂ i

Ei,Eis are bounded

(2.14)

In [5], Tricot proves that Dim= Δ, and Falconer [1] shows that for any setE ⊂ R n,

0≤dim(E) ≤  δ(E) ≤  Δ(E) =Dim(E) ≤ n. (2.15)

In order to prove the results in this paper, the following two auxiliary lemmas are needed, which can be found by Mattila in [3, Lemmas 5.4 and 5.5]

Lemma 2.1 N(E,2ε) ≤ P(E,ε) ≤ N(E,ε/2) for any subset E ofRn

Lemma 2.2 P(E,ε)anε n ≤ᏸn(E(ε)) ≤ N(E,ε)an(2ε) n , where an =ᏸn(B(0,1)).

The following lemma is from [1, Example 7.8]

Lemma 2.3 There exist sets E,F ⊂ R with δ(E) = δ(F) = 0 and dim( E × F) ≥ 1.

For reader’s convenience, we give the example as follows

Let 0= m0< m1< ··· be a rapidly increasing sequence of integers satisfying a con-dition to be specified below LetE be a set of real numbers in [0,1] with zero in the rth

decimal place whenevermk+ 1≤ r ≤ mk+1 withk =2,  ∈ Z+ Similarly, letF be a set

of real numbers with zero in therth decimal place if mk+ 1≤ r ≤ mk+1withk =2 + 1,

 ∈ Z+ Looking at the firstm k+1decimal places for evenk, there is an obvious cover of E

by 10j kintervals of length 10− m k+1, where

j k =m2− m1

 +

m4− m3

 +···+

m k − m k −1

Then log 10j k / −log 10− m k+1 = jk/mk+1which tends to 0 ask → ∞provided that themkare chosen to increase sufficiently rapidly So we have δ(E)=0 Similarly,δ(F) =0

If 0< w < 1, then we can write w = x + y, where x ∈ E and y ∈ F; just take the rth

decimal digit ofw from E if mk+ 1≤ r ≤ mk+1andk is odd and from F if k is even The

mapping f :R 2→ Rgiven byf (x, y) = x + y is easily seen to be Lipschitz, so

dim(E × F) ≥dimf (E × F) ≥dim

(0, 1)

by [1, Corollary 2.4(a)]

The following lemma summarizes some of the basic properties of Minkowski contents

Lemma 2.4 Let M s be one of M ∗ s and M s

∗ , then for bounded sets E, F, { Ei } ,

(i)M s(∅)= 0;

(ii)M s is monotone: E1⊂ E2⇒ M s(E1)≤ M s(E2);

(iii)M s(E) = M s(E);

(iv) assume that s < t If M s(E) < ∞ , then M t(E) = 0 Moreover, if M t(E) > 0, then

M s(E) = ∞ ;

(v)M ∗ s(E ∪ F) ≤ M ∗ s(E) + M ∗ s(F), M s

∗(

i Ei)≥ i M s

∗(Ei ) for d(Ei,Ej)> c > 0, i j;

(vi) if E = { x } , then M0(E) =2− n an,M s(E) =0,s > 0;

(vii) if 0 < ᏸ n(E) < ∞ , then M n(E) =ᏸn(E), M s(E) = ∞,s < n.

Proof (i), (ii) are trivial (iii) follows from E(ε) = E(ε) (iv) derives from the equality

(2ε) s − nᏸn

E(ε)

=(2ε) s − t(2ε) t − nᏸn

E(ε)

(v) The first inequality is obvious

We have d(Ei(ε),Ej(ε)) > 0 for i j when 0 < 2ε < c, thus

M ∗ s

i Ei



=lim inf

ε ↓0

 (2ε) s − nᏸn

i Ei

 (ε)

=lim inf

ε ↓0

 (2ε) s − n

i

ᏸn

E i(ε)

≥ i

lim inf

ε ↓0

 (2ε) s − nᏸn

Ei(ε)

= i

M ∗ s

Ei

.

(2.19)

(vi) Follows from (2ε) s − nᏸn(x(ε)) = an(2ε) s − n ε n =2s − n anε s

(vii) Holds since

lim

ε ↓0(2ε) n − nᏸn

E(ε)

=ᏸn(E),

lim

ε ↓0

 (2ε) s − nᏸn

E(ε)

≥lim

ε ↓0(2ε) s − nᏸn(E) = ∞ fors < n. (2.20)

3 The dimensions of product sets

In this section, we give a formula about dimensions of product sets First let us state a lemma from Bishop and Peres [10, Lemma 2.1]

Lemma 3.1 Let E be a subset of a separable metric space, with δ(E) > α (or Dim(E) >

α) Then there is a (relatively closed) nonempty subset F of E, such that δ(F ∩ V) > α (or Dim(F ∩ V) > α) for any open set V which intersects F.

Theorem 3.2 For any subsets E, F ofRn ,



δ(E) + δ(F) ≤  δ(E × F) ≤  δ(E) + Dim(F) ≤Dim(E × F) ≤Dim(E) + Dim(F). (3.1)

Proof (i) First we prove the first inequality Here we modify the proof of Theorem 4.1 in

[7], whereE is Borel set and F is compact.

It suffices to show that



for anyα < δ(E), β < δ(F).

ByLemma 3.1, there exist closed setsEα ⊂ E, Fβ ⊂ F such that



δ

Eα ∩ V

> α, δFβ ∩ W> β (3.3)

for any open setsV, W, where V ∩ Eα ,W ∩ Fβ

For anyε > 0, we may find bounded { G n }withE α × F β ⊂n G n, and for anyn,

δ

Gn

≤  δ

Eα × Fβ

+ε ≤  δ(E × F) + ε. (3.4)

Sinceδ(G n)= δ(G n), we may takeG n to be closed andG n ∩(E α × F β) By Baire’s category theorem, we know that there existn and an open set U which intersects Eα × Fβ

such thatU ∩(Eα × Fβ)⊂ Gn Therefore, we may find open setsV,W such that V × W ⊂

U and (V × W) ∩(E α × F β) , then we have

Eα ∩ V

×Fβ ∩ W

hence

α + β ≤  δ

Eα ∩ V

+δFβ ∩ W

≤ δ

Eα ∩ V

+δ

Fβ ∩ W

≤ δ

Eα ∩ V

×Fβ ∩ W

≤ δ

Gn

≤  δ(E × F) + ε,

(3.6)

the third inequality follows from the definitions of the upper and lower box dimensions Sinceε is arbitrary, (3.2) follows immediately

(ii) Now let us turn to the second inequality SupposeE ⊂i Ei,F ⊂i Fi,Eis andFis are bounded, thenE × F ⊂i, j(Ei × Fi), thus



δ(E × F) = inf

E × F ⊂ l V l

 sup

l

δ

Vl :E × F ⊂

l

Vl,Vls are bounded

≤ inf

E × F ⊂ i, j

E i × F j

 sup

i, j δ

Ei × F j

:E × F ⊂

i, j

Ei × F j

≤inf

 sup

i, j

δ

E i +ΔF j

:E ⊂ i

E i,F ⊂

j

F j

≤inf

 sup

i

δ

E i :E ⊂ i

E i + inf

 sup

j ΔF j

:F ⊂ j

F j

=  δ(E) + Δ(F),

(3.7)

the second inequality above follows from the definitions of the upper and lower box di-mensions

(iii) The proof of the third inequality is similar to (i)

(iv) The last one can be referred to Tricot [5, Theorem 3] 

Remark 3.3 (a) One may ask whether dim(E × F) ≤dim(E) + δ(F) holds or not By Lemma 2.3, we know that there exist sets E,F ⊂ R withδ(E) = δ(F) =0 and dim(E × F) ≥1 Hence,

dim(E × F) > δ(E) + δ(F) ≥dim(E) + δ(F) ≥dim(E) + δ(F). (3.8) (b) As a consequence of (2.15) andTheorem 3.2, one has

dim(E) + dim(F) ≤dim(E × F) ≤  δ(E × F) ≤  δ(E) + Dim(F)

≤Dim(E × F) ≤Dim(E) + Dim(F). (3.9)

4.ᏹ∗ s,ᏹs

∗, and their dimensionsD, d

It is known that the Minkowski contents are not outer measures since they fail to be countably subadditive In fact, we may derive this assertion directly fromLemma 2.4 Considers =1 andE = Q ∩[0, 1], the set of rational numbers in [0, 1] ByLemma 2.4, we know thatM1(E) = M1([0, 1])=1 andM1({ q })=0 for anyq ∈ E, thus q ∈ E M1({ q })=0

We use a standard procedure and define

ᏹ∗ s(E) =inf

∞

i =1

M ∗ s

E i :E = i

E i,E i s are bounded ,

ᏹs

∗(E) =inf

∞

i =1

M s

∗

E i :E = i

E i,E i s are bounded

(4.1)

Theorem 4.1 Letᏹs be one ofᏹ∗ s andᏹs

∗ , then

(i)ᏹs is an outer measure;

(ii)ᏹs is metric: d(E,F) > 0 ⇒ᏹs(E ∪ F) =ᏹs(E) + ᏹ s(F);

(iii)ᏹs is a Borel measure;

(iv)ᏹs is Borel regular: for all E ⊂ R n , there is a Borel set B ⊃ E such that ᏹ s(B) =

ᏹs(E);

(v)ᏹs(E) ≤ M s(E) for bounded set E;

(vi)ᏹs(En)→ᏹs(E) for any sequence of sets En ↑ E;

(vii) if E is ᏹ s -measurable, 0 < ᏹ s(E) < ∞ , and ε > 0, there exists a closed set F ⊂ E such thatᏹs(F) > ᏹ s(E) − ε;

(viii) for any E,

ᏹ∗ s(E) =inf

lim

n →∞ M ∗ s

E n :E n ↑ E, E n s are bounded

Proof Let M sbe one ofM ∗ sandM s

∗ (i)ᏹs(∅)=0 and thatᏹs is monotone are obvious, so it suffices to verify that ᏹs

is countably subadditive Suppose thatE =i Ei, for anyε > 0, there exist bounded sets { Ei j }such thatEi =j Ei j, j M s(Ei j)< ᏹ s(Ei) +ε/2 i, thus

E = i

Ei = i

j

Ei j,

ᏹs(E) ≤

i



j

M s

Ei j

≤ i



ᏹs

Ei + ε

2i



= i

ᏹs

Ei +ε.

(4.3)

So we haveᏹs(E) ≤ iᏹs(Ei) by the arbitrariness ofε.

(ii) Assume thatE ∪ F = i A i,A is are bounded, then



i

M s

Ai

E ∩ A i

M s

Ai

F ∩ A i

M s

Ai

thus

inf

i

M s

Ai

≥inf 

E ∩ A i

M s

Ai + inf 

F ∩ A i

M s

Ai

so we have

ᏹs(E ∪ F) ≥ᏹs(E) + ᏹ s(F), (4.6) the opposite inequality holds sinceᏹsis an outer measure by (i)

(iii) Follows from (ii) by Falconer [2, Theorem 1.5]

(iv) We haveM s(E) = M s(E) by (iii) ofLemma 2.4, thus

ᏹs(E) =inf

∞

i =1

M s

B i :E ⊂ i

B i,B i s are closed and bounded (4.7)

Fori =1, 2, , choose closed sets Bi1,Bi2, , such that

E ⊂ j

Bi j,

∞



j =1

M s

Bi j

≤ᏹs(E) +1

ThenB =ij Bi jis a Borel set such thatE ⊂ B and ᏹ s(E) =ᏹs(B).

(v) Is obvious by the definition ofᏹs

(vi) SinceE n ↑ E, we know that limᏹ s(E n) exists and is≤ᏹs(E) by the monotonicity

ofᏹs By (iv), there exists Borel setFi ⊃ Eiwithᏹs(Fi)=ᏹs(Ei), that is,ᏹs(Fi \ Ei)=0 Let

Bn = n

i =1

Fi, B =

n

thenBns are Borel sets withBn ↑ B, En ⊂ Bn Furthermore, we have

ᏹs

Bn

=ᏹs

 n

i =1

Fi



=ᏹs

Fn +ᏹs

n −1

i =1

Fi



Fn



≤ᏹs

En +

n−1

i =1

ᏹs

Fi En

≤ᏹs

E n +

n−1

i =1

ᏹs

F i \ E i

=ᏹs

E n ,

(4.10)

hence

ᏹs(E) ≥lim

n →∞ᏹs

E n

=lim

n →∞ᏹs

B n

=ᏹs(B) ≥ᏹs(E) (4.11)

by the fact that

E = n

En ⊂ n

(vii) LetE be ᏹ s-measurable, then there exists a Borel setB ⊃ E with ᏹ s(B) =ᏹs(E),

that is,ᏹs(B \ E) =0 We can find a Borel setB1⊃(B \ E) with ᏹ s(B1)=0, thenB2= B \ B1

is Borel,B2⊂ E, and ᏹ s(B2)=ᏹs(E) By [3, Theorem 1.9 and Corollary 1.11], we know thatᏹs | B2, the restriction of measureᏹstoB2, is a Radon measure, thus is an inner regular measure since 0< ᏹ s(E) =ᏹs(B2)< ∞, so there exists a closed setF ⊂ B2such thatᏹs | B2(F) > ᏹ s | B2(B2)− ε which gives ᏹ s(F) > ᏹ s(B2)− ε =ᏹs(E) − ε.

(viii) The proof is the same as that of [4, Lemma 5.1(vii)] 

Corollary 4.2 For any subset E ofRn ,

ᏹs(E) =inf

∞

i =1

M s

Ei :E ⊂ i

Ei,Eis are bounded Borel sets (4.13)

Proof We denote the right-hand side of the above equality by μ(E), then ᏹ s(E) ≤ μ(E)

follows from the definition ofᏹs(E) and ᏹ s(E) ≥ μ(E) follows from (4.7) 

Corollary 4.3 Let B be Borel set ofRn , then

ᏹs(B) =inf

∞

i =1

M s

Bi :B = i

Bi,Bis are disjoint bounded Borel sets (4.14)

Proof From (4.7), we have

ᏹs(B) =inf

∞

i =1

M s

Fi :B ⊂ i

Fi,Fis are closed and bounded , (4.15) thenEi = Fi ∩ B is a bounded Borel set and B =i Ei Take

B1= E1,B2= E2\ B1, ,Bn = En

n −1

i =1

Bi

 , , (4.16) then{ Bi }are disjoint bounded Borel sets andB =i Bi, so we have

ᏹs(B) ≥inf

∞

i =1

M s

Bi :B = i

Bi,Bis are disjoint bounded Borel sets (4.17)

by the fact thatBi ⊂ Fi

The opposite inequality holds by the definition ofᏹs 

Theorem 4.4 For any subset E ofRn , the following inequality holds:

2− s − n a nᏴs(E) ≤ᏹs

∗(E) ≤ᏹ∗ s(E) ≤2s a nᏼs(E). (4.18)

Proof The assertionᏹs

∗(E) ≤ᏹ∗ s(E) is trivial We first prove the right-hand inequality,

by Lemmas2.1and2.2, for all bounded setB ⊂ R n,

M ∗ s(B) =lim sup

ε ↓0

 (2ε) s − nᏸn

B(ε)

≤lim sup

ε ↓0

 (2ε) s − n N(B,ε)a n(2ε) n

≤lim sup

ε ↓0



2s a n P



B, ε

2



ε s

≤2s anlim sup

ε ↓0 P s(B) ≤2s anP s(B),

(4.19)

thus

ᏹ∗ s(E) =inf

∞

i =1

M ∗ s

Ei :E = i

Ei,Eis are bounded

≤inf

∞

i =1

2s a n P s

E i :E = i

E i,E i s are bounded

=2s a nᏼs(B).

(4.20)

. (2.11)

Thes-dimensional upper and lower Minkowski contents of bounded set E are defined

by... definitions of the upper and lower box dimensions Sinceε is arbitrary, (3.2) follows immediately

(ii)...

Theorem 4.1 Letᏹs be one ofᏹ∗ s and< /i>ᏹs

∗