The Upper Densities of Perturbed Cantor Sets

1. Introduction

The fractal dimension and fractal measure constitute two key components in the study of fractal geometry. Significant progress has been made in the calculation of fractal dimensions [1–5]. However, the research on computing fractal measures remains rather limited. To date, only a few results about self-similar sets are known, specifically for the middle-third Cantor set in $\mathbb{R}$ and its variations [6–10], as well as several special cases of the Cartesian product of the symmetric Cantor set with itself in ${\mathbb{R}}^2$ [11,12]. In most cases, only upper and lower bounds can be derived [13–16]. The local geometry of self-similar sets has recently been further investigated by Morán et al. [17]. In that work, the spectrum of asymptotic densities of metric measures was characterized in terms of the packing-centered and Hausdorff-centered measures.

The Hausdorff centered measure [18] is an important fractal measure, closely related to the upper density. Let $0\le s<\infty$ and $\nu$ be a measure on $R^n$, where ${\mathbb{R}}^n:=\{(x_1,x_2,\dots ,x_n)\mid x_i\in \mathbb{R},i=1,2,\dots ,n\}$ is the n-dimensional Euclidean space. The upper s-density of $\nu$ at $x\in R^n$ is defined by

$${\Theta }^{*s}(\nu ,x):=\underset{r\to 0^{+}}{\mathrm{lim\; sup}}{\left(2r\right)}^{-s}\nu \left(B(x,r)\right),$$

(1.1)

where $B(x,r)$ denotes the closed ball with center x and radius r. Over the past 20 years, there have been many works devoted to the calculation of the Hausdorff centered measures of self-similar sets [19–21].

Linear Cantor sets are typical self-similar sets. Research on the upper density and the Hausdorff centered measure of linear Cantor sets has also attracted extensive attention. The first result known may be due to Feng, who determined the exact upper s-density [22]. Let C be the classical middle-third Cantor set, and let ${\mu }_C$ be the restriction of the Hausdorff measure ${\mathcal{H}}^s$ to C. Then ${\Theta }^{*s}({\mu }_C,x)=2/4^s$ for ${\mu }_C$ a.e. $x\in C$. Zhu [23] computed the exact value of the upper s-density of a more general linear Cantor set, namely the symmetric Cantor set $C_{\lambda }$. Ruan [24] and Li [25] further investigated non-symmetric Cantor sets and independently derived a formula for calculating the exact value of the upper s-density.

The symmetric perfect set [26] is a class of important non-self-similar sets. Qu [27] derived an explicit formula for the exact value of the upper s-density of the symmetric perfect set. Let E be a symmetric perfect set with contracting ratios $c_k\le 1/3$ for all $k\ge k_0$, where $k_0$ is a sufficiently large positive integer. Let s denote the Hausdorff dimension of E, and let ${\mu }_E$ be the restriction of the normalized Hausdorff measure ${\left({\mathcal{H}}^s\left(E\right)\right)}^{-1}{\mathcal{H}}^s$ to E. Let $x_k=c_1{c}_2\dots c_k$, and assume that $\underset{k\to \infty }{\lim }{2}^k{x}_k^s$ exists and is finite and positive. Then for ${\mu }_E$-a. e. $x\in E$,

$${\Theta }^{*s}({\mu }_E,x)={\displaystyle \frac{2}{2^s{(2^{\frac{1}{s}}-1)}^s\underset{k\to \infty }{\lim }{2}^k{x}_k^s}}.$$

Garcia [28] further investigated the upper s-density of the symmetric perfect set without assuming that $\underset{k\to \infty }{\lim }{2}^k{x}_k^s$ exists and is finite and positive. Let E be a symmetric perfect set with contracting ratios $c_k\le 1/3$ for all $k\ge k_0$. Define $y_k=x_{k-1}-2x_k$. Assume $0<{\mathcal{H}}^s\left(E\right)<\infty$, and let $\mu ={\left({\mathcal{H}}^s\left(E\right)\right)}^{-1}{\mathcal{H}}^s$. Then for ${\mu }_E$-a. e. $x\in E$,

$${\Theta }^{*s}({\mu }_E,x)={\displaystyle \frac{2}{2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{2}^k{(x_k+y_k)}^s}}.$$

Further calculations show that if $\underset{k\to \infty }{\lim }{2}^k{x}_k^s$ exists, then

$${\Theta }^{*s}({\mu }_E,x)={\displaystyle \frac{2}{2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{2}^k{(x_k+y_k)}^s}}={\displaystyle \frac{2}{2^s{(2^{\frac{1}{s}}-1)}^s\underset{k\to \infty }{\lim }{2}^k{x}_k^s}}.$$

In the present paper, the upper s-density of a perturbed Cantor set has been investigated [29](specifically, a non-symmetric perfect set), which is constructed as follows. Suppose $I=[0,1]$, and let ${\{a_k,b_k\}}_{k\ge 1}$ be a sequence of real numbers satisfying $0<a_k,b_k<1/2$ ($k\ge 1$). For any $k\ge 1$, let $D_k=\{(i_1,i_2,\dots ,i_k):1\le i_j\le 2,1\le j\le k\},D={\bigcup }_{k\ge 0}{D}_k,$ where $D_0=\varnothing .$ If $\sigma =({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_k)\in D_k$, $\tau =({\tau }_1,{\tau }_2,\dots ,{\tau }_m)\in D_m,$ let $\sigma \ast \tau =({\sigma }_1,{\sigma }_2,\dots ,{\sigma }_k,{\tau }_1,{\tau }_2,\dots ,{\tau }_m).$ Let $\mathcal{F}=\{I_{\sigma }:\sigma \in D\}$ be the collection of closed sub-intervals of I satisfying:

i) $I_{\varnothing }=I;$

ii) For any $k\in {\mathbb{N}}^{+}$(the set of positive integers) and $\sigma \in D_{k-1}$, $I_{\sigma \ast i}$ ($1\le i\le 2$) are sub-intervals of $I_{\sigma }.$$I_{\sigma \ast 1}$ and $I_{\sigma }$ have the same left endpoint, while $I_{\sigma \ast 2}$ and $I_{\sigma }$ have the same right endpoint.

iii) For any $k\in {\mathbb{N}}^{+}$ and $\sigma \in D_{k-1}$, the relations

$${\displaystyle \frac{|I_{\sigma \ast 1}|}{|I_{\sigma }|}}=a_k,{\displaystyle \frac{|I_{\sigma \ast 2}|}{|I_{\sigma }|}}=b_k$$

are satisfied.

Let $E_k={\bigcup }_{\sigma \in D_k}{I}_{\sigma }$ and $E={\bigcap }_{k\ge 0}{E}_k$. E is called the perturbed Cantor set determined by the contracting ratios $a_k$ and $b_k$, and

$${\mathcal{F}}_k=\{I_{\sigma }:\sigma \in D_k\}$$

is called the collection of basic intervals of order k of E. The symmetric perfect set is a well-known example of such perturbed sets.

Assume that

(1) There exists $k_0\in N^{+}$ such that for all $k>k_0$

$${\left({\displaystyle \frac{1}{\sqrt{2}+1}}\right)}^2\le \min \{a_k,b_k\}\le {\displaystyle \frac{1}{3}},$$

and

$$1\le {\displaystyle \frac{\max \{a_k,b_k\}}{\min \{a_k,b_k\}}}\le \sqrt{2}+1.$$

(2) $\underset{n\to \infty }{\mathrm{lim\; inf}}{\prod }_{k=1}^n(a_k^s+b_k^s)$ is finite and positive.

In fact, E can be considered as a Moran set [30] on $I=[0,1]\subset \mathbb{R}$. From [30], the Hausdorff dimension is given by

$$s={\dim }_{H}E=\underset{k\to \infty }{\mathrm{lim\; inf}}{s}_k,$$

and $0<{\mathcal{H}}^s\left(E\right)<\infty$, where $s_k$ satisfies the following equation

$${\displaystyle \prod _{i=1}^k}(a_i^{s_k}+b_i^{s_k})=1.$$

The main results of the present paper are:

Theorem 1.1.

Let E be a perturbed Cantor set with contracting ratios $a_k$ and $b_k$. If assumptions (1) and (2) hold, then

$${\Theta }^{*s}({\mu }_E,x)={\displaystyle \frac{1}{2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{(1-c_k)}^s{\displaystyle \prod _{i=1}^{k-1}}(a_i^s+b_i^s)}}\mathrm{for}\mu -\mathrm{a}.\mathrm{e}.x\in E,$$

where $\mu ={\left({\mathcal{H}}^s\left(E\right)\right)}^{-1}{\mathcal{H}}^s$, and $c_k=\max \left\{a_k,b_k\right\}.$

Theorem 1.2.

Let E be a perturbed Cantor set with contracting ratios $a_k$ and $b_k$. If assumptions (1) and (2) hold, then the Hausdorff centered measure of E is given by

$$C^s\left(E\right)=2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{(1-c_k)}^s{\displaystyle \prod _{i=1}^{k-1}}(a_i^s+b_i^s).$$

Remark 1.

If E is the symmetric perfect set, then $a_k=b_k$. If ${\mathcal{H}}^s\left(E\right)$ is finite and positive, Theorems 6 and 7 in [28] are recovered immediately.

Remark 2.

If E is the non-symmetric Cantor set $K_{\lambda }$, then $a_k\equiv a,b_k\equiv b$. In this case, Theorem 4.1 in [24] can be recovered immediately.

The paper is organized as follows. In Section 2, several lemmas required for calculating the upper density are presented. In Section 3, Theorems 1.1 and 1.2 are proved.

2. Notations and Lemmas

Suppose E is a perturbed Cantor set with contracting rations $a_k$ and $b_k$, and assume that conditions of Theorem 1.1 are satisfied.

For any $\sigma ={\sigma }_1{\sigma }_2\dots {\sigma }_m$ and $0<k\le m$, let

$$\sigma |k:={\sigma }_1{\sigma }_2\dots {\sigma }_k\in D_k$$

be the truncation of $\sigma$.

In this paper, the following fact is derived from assumption (1): Let $\sigma \in D_{k-1}$ and $J_{\sigma }$ be the length of the gap between sub-intervals $I_{\sigma \ast 1}$ and $I_{\sigma \ast 2}$. Then, there exists a $k_0\in N^{+}$ such that for any $k>k_0$ and $\sigma \in D_k,$

$$|J_{\sigma }|>\max \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}\left|\right\}.$$

(2.1)

For any $\sigma ,\tau \in D_k$, let $a\left(\sigma \right)$ be the left endpoint of $I_{\sigma }$, and let $b\left(\tau \right)$ be the right endpoint of $I_{\tau }$. Let

$$B=\underset{k\to \infty }{\mathrm{lim\; inf}}{\displaystyle \prod _{i=1}^k}(a_i^s+b_i^s).$$

Assumption (2) implies that $0<B<\infty$. Thus, for any $ϵ>0$, there exists a positive integer $k_0$ such that

$${\displaystyle \prod _{i=1}^k}(a_i^s+b_i^s)>B-ϵ$$

(2.2)

for all $k\ge k_0$.

Let $\mu$ be the restriction of the normalized Hausdorff measure ${\left({\mathcal{H}}^s\left(E\right)\right)}^{-1}{\mathcal{H}}^s$ to the set E. If assumptions (1) and (2) hold, then for $\sigma \in D_k$, one can get

$$\mu \left(I_{\sigma }\right)={\displaystyle \frac{|I_{\sigma }{|}^s}{{\sum }_{\sigma \in D_k}{\left|I_{\sigma }\right|}^s}}={\displaystyle \frac{|I_{\sigma }{|}^s}{{\prod }_{i=1}^k(a_i^s+b_i^s)}}.$$

(2.3)

Lemma 2.1.

Let E be a perturbed Cantor set with contracting rations $a_k$ and $b_k$. If assumptions (1) and (2) hold, then for any $ϵ>0$ there exists a $k_0\in N^{+}$ such that for any σ,$\tau \in D_k(k\ge k_0)$ satisfying $a\left(\sigma \right)<b\left(\tau \right)$ and $\sigma |k_0=\tau |k_0$ , it holds

$$\mu \left([a\left(\sigma \right),b\left(\tau \right)]\right)\le {(B-ϵ)}^{-1}{(b\left(\tau \right)-a\left(\sigma \right))}^s.$$

(2.4)

Proof. 

The proof follows and develops the method introduced in [31]. If $\sigma =\tau \in D_k(k\ge k_0)$, then $b\left(\tau \right)-a\left(\sigma \right)=|I_{\sigma }|$. It follows immediately from (2.2) and (2.3) that

$$\mu \left([a\left(\sigma \right),b\left(\tau \right)]\right)\le {(B-ϵ)}^{-1}{(b\left(\tau \right)-a\left(\sigma \right))}^s.$$

Thus, only the case where $\sigma$,$\tau \in D_k(k\ge k_0)$ and $\sigma \ne \tau$ needs to be considered. In this case, (2.4) will be proved by induction.

(1) For $n=k_0+1$, since $\sigma |k_0=\tau |k_0$ and $\sigma \ne \tau$, we have

$${(b\left(\tau \right)-a\left(\sigma \right))}^s={\left|I_{\sigma |k_0}\right|}^s\ge (B-ϵ)\mu \left([a\left(\sigma \right),b\left(\tau \right)]\right).$$

Therefore

$$\mu \left([a\left(\sigma \right),b\left(\tau \right)]\right)\le {(B-ϵ)}^{-1}{(b\left(\tau \right)-a\left(\sigma \right))}^s.$$

(2) Now suppose that (2.4) holds for $n=k(k>k_0)$.For $n=k+1(k>k_0)$, the following two cases will be dealt with.

1) $\sigma$,$\tau \in D_{k+1}$, $\sigma |k=\tau |k$. In this case, similarly to the proof of (1), (2.4) is obtained.

2) $\sigma$,$\tau \in D_{k+1}$, $\sigma |k\ne \tau |k$.

i) Suppose $a\left(\sigma \right)=a\left(\sigma \right|k)$ and $b\left(\tau \right)\ne b\left(\tau \right|k)$. This implies that $I_{\tau |k}$ lies to the right of $I_{\sigma |k}$. Let $\widehat{\tau }\in D_k$ be such that $I_{\widehat{\tau }}$ is the k-order basic interval immediately to the left of $I_{\tau }$. Define

$$\lambda :=(b\left(\tau \right|k)-b\left(\tau \right))/(b\left(\tau \right|k)-b\left(\widehat{\tau }\right)),0<\lambda <1.$$

Then

$$b\left(\tau \right)-a\left(\sigma \right)=b\left(\tau \right)-a\left(\sigma \right|k)=\lambda (b\left(\widehat{\tau }\right)-a\left(\sigma \right|k))+(1-\lambda )(b\left(\tau \right|k)-a\left(\sigma \right|k)).$$

By the convexity of $x^s$, it follows that

$${(b\left(\tau \right)-a\left(\sigma \right))}^s\ge \lambda {(b\left(\widehat{\tau }\right)-a\left(\sigma \right|k))}^s+(1-\lambda ){(b\left(\tau \right|k)-a\left(\sigma \right|k))}^s.$$

(2.5)

By the inductive assumption and (2.5), it holds that.

$$\begin{array}{ccc}\hfill {(b\left(\tau \right)-a\left(\sigma \right))}^s& \ge & (B-ϵ)(\lambda \mu \left([a\left(\sigma \right|k),b\left(\widehat{\tau }\right)]\right)+(1-\lambda )\mu \left([a\left(\sigma \right|k),b\left(\tau \right|k)]\right))\hfill \\ & =& (B-ϵ)(\mu \left([a\left(\sigma \right|k),b\left(\widehat{\tau }\right)]\right)+(1-\lambda )\mu \left([a\left(\tau \right|k),b\left(\tau \right|k)]\right)).\hfill \end{array}$$

It is now demonstrated that

$$(1-\lambda )\mu \left(\right[a\left(\tau \right|k),b(\tau \left|k\right)\left]\right)\ge \mu \left(\right[a\left(\tau \right),b\left(\tau \right)\left]\right).$$

(2.6)

Recall that

$$1-\lambda ={\displaystyle \frac{b\left(\tau \right)-b\left(\widehat{\tau }\right)}{b\left(\tau \right|k)-b\left(\widehat{\tau }\right)}}={\displaystyle \frac{|I_{\tau }|+(a\left(\tau \right)-b\left(\widehat{\tau }\right))}{|I_{\tau |k}|+(a\left(\tau \right)-b\left(\widehat{\tau }\right))}}.$$

The assumption that $\max \{a_k,b_k\}\le 1/3$ implies $a\left(\tau \right)-b\left(\widehat{\tau }\right)>\left|I_{\tau |k}\right|$. Consequently, the following is obtained

$$1-\lambda \ge {\displaystyle \frac{|I_{\tau }|+|I_{\tau |k}|}{|I_{\tau |k}|+|I_{\tau |k}|}}={\displaystyle \frac{1+a_{k+1}}{2}}.$$

It remains to be shown that

$${\displaystyle \frac{1+a_{k+1}}{2}}-{\displaystyle \frac{a_{k+1}^s}{a_{k+1}^s+b_{k+1}^s}}={\displaystyle \frac{b_{k+1}^s-a_{k+1}^s+a_{k+1}^{s+1}+a_{k+1}{b}_{k+1}^s}{2(a_{k+1}^s+b_{k+1}^s)}}\ge 0.$$

(2.7)

If $a_{k+1}\le b_{k+1}$, it is straightforward to verify that inequality (2.7) holds.

If $a_{k+1}>b_{k+1}$, let $a_{k+1}=m·b_{k+1}$, it follows that

$$\begin{array}{ccc}\hfill b_{k+1}^s-a_{k+1}^s+a_{k+1}^{s+1}+a_{k+1}{b}_{k+1}^s& =& b_{k+1}^s((1-m^s)+m(m^s+1)b_{k+1})\hfill \\ & \ge & b_{k+1}^s((1-m^s)+{(\sqrt{2}-1)}^{2}m(m^s+1)).\hfill \end{array}$$

On the other hand,

$${\displaystyle \frac{m^s-1}{m(m^s+1)}}={\displaystyle \frac{1}{m}}-{\displaystyle \frac{2}{m(m^s+1)}}\le {\displaystyle \frac{1}{m}}-{\displaystyle \frac{2}{m(m+1)}}={\displaystyle \frac{2}{m+1}}-{\displaystyle \frac{1}{m}}.$$

Define the function $f\left(x\right)={\displaystyle \frac{2}{x+1}}-{\displaystyle \frac{1}{x}},1\le x\le \sqrt{2}+1$, then

$$f^{\prime }\left(x\right)={\displaystyle \frac{((\sqrt{2}+1)x+1)(1-(\sqrt{2}-1)x)}{x^2{(x+1)}^2}}\ge 0.$$

Thus, $f\left(x\right)$ attains its maximum at $x=\sqrt{2}+1$ with maximum value $f(\sqrt{2}+1)={(\sqrt{2}-1)}^2$. This implies that

$$b_{k+1}^s-a_{k+1}^s+a_{k+1}^{s+1}+a_{k+1}{b}_{k+1}^s\ge 0,$$

and consequently, (2.7) holds. Therefore, it is obtained that

$$(1-\lambda )\mu \left([a\left(\tau \right|k),b\left(\tau \right|k)]\right)\ge |I_{\tau }{|}^s{\displaystyle \prod _{i=1}^{k+1}}{(a_i^s+b_i^s)}^{-1}=\mu \left([a\left(\tau \right),b\left(\tau \right)]\right).$$

(2.8)

Combining this with the previous result, it follows that

$$\mu \left([a\left(\sigma \right),b\left(\tau \right)]\right)\le {(B-ϵ)}^{-1}{(b\left(\tau \right)-a\left(\sigma \right))}^s.$$

ii) If $a\left(\sigma \right)\ne a\left(\sigma \right|k)$ and $b\left(\tau \right)=b\left(\tau \right|k)$, then $I_{\sigma |k}$ lies to the left of $I_{\tau |k}$. Let $\widehat{\sigma }\in D_k$, and let $I_{\widehat{\sigma }}$ be the k-order basic interval immediately to the right of $I_{\sigma }$. By using the same methods, (2.4) is also obtained.

iii) If $a\left(\sigma \right)\ne a\left(\sigma \right|k)$ and $b\left(\tau \right)\ne b\left(\tau \right|k)$, then $\sigma =\sigma |k\ast 2$ and $\tau =\tau |k\ast 1$. In this case, let $\widehat{\tau }\in D_k$ be such that $I_{\widehat{\tau }}$ is the k-order basic interval immediately to the left of $I_{\tau }$. Define

$$\lambda :=(b\left(\tau \right|k)-b\left(\tau \right))/(b\left(\tau \right|k)-b\left(\widehat{\tau }\right)),0<\lambda <1.$$

Using an argument analogous to case i), it is obtained that

$$b\left(\tau \right)-a\left(\sigma \right)=\lambda (b\left(\widehat{\tau }\right)-a\left(\sigma \right))+(1-\lambda )(b\left(\tau \right|k)-a\left(\sigma \right))$$

and

$${(b\left(\tau \right)-a\left(\sigma \right))}^s\ge \lambda {(b\left(\widehat{\tau }\right)-a\left(\sigma \right))}^s+(1-\lambda ){(b\left(\tau \right|k)-a\left(\sigma \right))}^s.$$

On the other hand, it follows from case ii) that

$$\begin{array}{ccc}\hfill \mu \left([a\left(\sigma \right),b\left(\widehat{\tau }\right)]\right)& \le & {(B-ϵ)}^{-1}{(b\left(\widehat{\tau }\right)-a\left(\sigma \right))}^s,\hfill \\ \hfill \mu \left(\right[a\left(\sigma \right),b\left(\tau \right|k\left)\right])& \le & {(B-ϵ)}^{-1}{(b\left(\tau \right|k)-a\left(\sigma \right))}^s.\hfill \end{array}$$

Therefore, it holds that

$$\begin{array}{ccc}\hfill {(b\left(\tau \right)-a\left(\sigma \right))}^s& \ge & (B-ϵ)(\mu \left([a\left(\sigma \right),b\left(\widehat{\tau }\right)]\right)+(1-\lambda )\mu \left([a\left(\tau \right|k),b\left(\tau \right|k)]\right)).\hfill \end{array}$$

Combining this with (2.8), it follows that

$$\mu \left([a\left(\sigma \right),b\left(\tau \right)]\right)\le {(B-ϵ)}^{-1}{(b\left(\tau \right)-a\left(\sigma \right))}^s.$$

This completes the proof of Lemma 2.1. □

Analogous to the proof of Lemma 2.3 in [27], the following lemma is obtained.

Lemma 2.2.

Let E be a perturbed Cantor set with contracting rations $a_k$ and $b_k$. If assumptions (1) and (2) hold, then for any $ϵ>0$, there exists $\delta >0$ such that for any $a,b\in R$ with $b-a<\delta$, it follows that

$$\mu \left([a,b]\right)\le {(B-ϵ)}^{-1}{(b-a)}^s.$$

(2.9)

3. Proofs of Theorems

First, a lemma is introduced, which plays a crucial role in the subsequent discussion. This lemma provides an upper bound for the upper density of the measure $\mu$ on the set E.

Lemma 3.1.

Let E be a perturbed Cantor set with contracting rations $a_k$ and $b_k$. If assumptions (1) and (2) hold, then

$${\Theta }^{*s}({\mu }_E,x)\le {\displaystyle \frac{1}{2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{(1-c_k)}^s{\prod }_{i=1}^{k-1}(a_i^s+b_i^s)}}forallx\in E.$$

Proof. 

For any $x\in E$, let r be small enough to ensure $r<\min \left\{\right|I_{\sigma }|:\sigma \in D_{k_0}\}$. Then, there exists $k>k_0$ such that the interval $[x-r,x+r]$ contains a k-order basic interval $I_{\sigma }$, but does not contain any $(k-1)$-order basic interval. Assumption (1) implies that $[x-r,x+r]$ only intersects with the $(k-1)$-order basic interval $I_{\sigma \left|\right(k-1)}$.

(1) If $\sigma =\sigma \left|\right(k-1)\ast 1$, it follows that $\max \{a\left(\sigma \right)-x+|I_{\sigma }|,x-a\left(\sigma \right)\}<r\le a\left(\sigma \right)-x+|I_{\sigma \left|\right(k-1)}|$, and

$${\displaystyle \frac{\mu \left(\right[x-r,x+r\left]\right)}{{\left(2r\right)}^s}}={\displaystyle \frac{\mu \left(\right[a\left(\sigma \right),x+r\left]\right)}{{\left(2r\right)}^s}}.$$

If $\max \{a\left(\sigma \right)-x+|I_{\sigma }|,x-a\left(\sigma \right)\}<r\le a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}|$, then

$$\mu \left([a\left(\sigma \right),x+r]\right)={\displaystyle \frac{|I_{\sigma }{|}^s}{{\prod }_{i=1}^k(a_i^s+b_i^s)}}.$$

Thus, the function $\mu \left([x-r,x+r]\right)/{\left(2r\right)}^s$ is strictly decreasing on the interval $\left[\max \right\{a\left(\sigma \right)-x+|I_{\sigma }|,x-a\left(\sigma \right)\},a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}\left|\right]$, and therefore attains its maximum at $r=\max \{a\left(\sigma \right)-x+|I_{\sigma }|,x-a\left(\sigma \right)\}$, with the maximum value

$${\displaystyle \frac{|I_{\sigma }{|}^s}{\left(2\max \right\{a\left(\sigma \right)-x+|I_{\sigma }{|,x-a\left(\sigma \right)\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}.$$

Furthermore, it follows that

$$\max \{a\left(\sigma \right)-x+|I_{\sigma }|,x-a\left(\sigma \right)\}\ge |J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}\left|\right\}.$$

Otherwise, $a\left(\sigma \right)-x+|I_{\sigma }|<|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}\left|\right\}$ and $x-a\left(\sigma \right)<|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}\left|\right\}$, this implies

$$\max \left\{\right|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}\left|\right\}<x-a\left(\sigma \right)<|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}\left|\right\}.$$

which contradicts the assumption that $x\in E$. It holds that

$${\displaystyle \frac{\mu \left(\right[x-r,x+r\left]\right)}{{\left(2r\right)}^s}}\le {\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}.\left(3.1\right)$$

Now, assume $a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}|<r\le a\left(\sigma \right)-x+|I_{\sigma \left|\right(k-1)}|$, then

$$\mu \left([a\left(\sigma \right),x+r]\right)={\displaystyle \frac{|I_{\sigma }{|}^s}{{\prod }_{i=1}^k(a_i^s+b_i^s)}}+\mu \left(\right[a\left(\sigma \right)+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}|,x+r]),$$

and thus, by Lemma 2.2, it follows that

$$\begin{array}{ccc}& & {\displaystyle \frac{\mu \left(\right[x-r,x+r\left]\right)}{{\left(2r\right)}^s}}\hfill \\ & & \le {\displaystyle \frac{|I_{\sigma }{|}^s{\prod }_{i=1}^k{(a_i^s+b_i^s)}^{-1}+\mu \left(\right[a\left(\sigma \right)+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}|,x+r])}{{\left(2r\right)}^s}}\hfill \\ & & \le {\displaystyle \frac{|I_{\sigma }{|}^s{\prod }_{i=1}^k{(a_i^s+b_i^s)}^{-1}+{(B-ϵ)}^{-1}(x-a\left(\sigma \right)+r-|I_{\sigma }|-|J_{\sigma \left|\right(k-1)}{\left|\right)}^s}{{\left(2r\right)}^s}}.\hfill \end{array}$$

Let $t=x-a\left(\sigma \right)+r-|I_{\sigma }|-|J_{\sigma \left|\right(k-1)}|$, and

$$f\left(t\right)={\displaystyle \frac{|I_{\sigma }{|}^s{\prod }_{i=1}^k{(a_i^s+b_i^s)}^{-1}+{(B-ϵ)}^{-1}{t}^s}{(a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}{|+t)}^s}}.$$

Then, it is obtained that $t\le |I_{\sigma \left|\right(k-1)\ast 2}|$ and

$$\begin{array}{ccc}\hfill f^{\prime }\left(t\right)& =& {\displaystyle \frac{{s((B-ϵ)}^{-1}{t}^{s-1}(a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}\left|\right)-|I_{\sigma }{|}^s{\prod }_{i=1}^k{(a_i^s+b_i^s)}^{-1})}{(a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}{|+t)}^{s+1}}}\hfill \\ & \ge & {\displaystyle \frac{s\left({(B-ϵ)}^{-1}{t}^{s-1}\right|J_{\sigma \left|\right(k-1)}|-|I_{\sigma }{|}^s{\prod }_{i=1}^k{(a_i^s+b_i^s)}^{-1})}{(a\left(\sigma \right)-x+|I_{\sigma }|+|J_{\sigma \left|\right(k-1)}{|+t)}^{s+1}}}.\hfill \end{array}$$

Since $t\le |I_{\sigma \left|\right(k-1)\ast 2}|$, $|J_{\sigma \left|\right(k-1)}|\ge \max \{|I_{\sigma \left|\right(k-1)\ast 1}|,|I_{\sigma \left|\right(k-1)\ast 2}\left|\right\}$ and by (2.2), it is obtained that $f^{\prime }\left(t\right)\ge 0$. This implies that $f\left(t\right)$ is strictly increasing. Therefore

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill {\displaystyle \frac{\mu \left(\right[x-r,x+r\left]\right)}{{\left(2r\right)}^s}}& \le {\displaystyle \frac{|I_{\sigma \left|\right(k-1)\ast 1}{|}^s+{\left|I_{\sigma \left|\right(k-1)\ast 2}\right|}^s{\prod }_{i=1}^k(a_i^s+b_i^s){(B-ϵ)}^{-1}}{2^s{\prod }_{i=1}^k(a_i^s+b_i^s)(a\left(\sigma \right)-x+|I_{\sigma \left|\right(k-1)}{\left|\right)}^s}}\hfill \\ & \le {\displaystyle \frac{|I_{\sigma \left|\right(k-1)}{|}^s(a_k^s+b_k^s)+2ϵ{(B-ϵ)}^{-1}{\left|I_{\sigma \left|\right(k-1)\ast 2}\right|}^s}{2^s{\prod }_{i=1}^k(a_i^s+b_i^s)\left(\right|J_{\sigma \left|\right(k-1)}|+|I_{\sigma \left|\right(k-1)\ast 2}{\left|\right)}^s}}\hfill \\ & \le {\displaystyle \frac{|I_{\sigma \left|\right(k-1)}{|}^s}{2^s{\prod }_{i=1}^{k-1}(a_i^s+b_i^s)\left(\right|J_{\sigma \left|\right(k-1)}|+|I_{\sigma \left|\right(k-1)\ast 2}{\left|\right)}^s}}+{\displaystyle \frac{2ϵ}{4^s(B^2-{ϵ}^2)}}\hfill \\ & ={\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}+{\displaystyle \frac{2ϵ}{4^s(B^2-{ϵ}^2)}}.\hfill \end{array}\end{array}$$

(2) If $\sigma =\sigma \left|\right(k-1)\ast 2$, it is also obtained that $\max \{b\left(\sigma \right)-x,|I_{\sigma }|+x-b\left(\sigma \right)\}<r\le x-b\left(\sigma \right)+|I_{\sigma \left|\right(k-1)}|$. Using the same methods, it follows that

$$\begin{array}{c}\hfill {\displaystyle \frac{\mu \left(\right[x-r,x+r\left]\right)}{{\left(2r\right)}^s}}\le {\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}+{\displaystyle \frac{2ϵ}{4^s(B^2-{ϵ}^2)}}.\end{array}$$

Let $c_k:=\max \{a_k,b_k\}$. Combining (1) with (2), it holds that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\mu \left(\right[x-r,x+r\left]\right)}{{\left(2r\right)}^s}}& \le & {\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}+{\displaystyle \frac{2ϵ}{4^s(B^2-{ϵ}^2)}}\hfill \\ & =& {\displaystyle \frac{1}{2^s{(1-c_{k+1})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}+{\displaystyle \frac{2ϵ}{4^s(B^2-{ϵ}^2)}}.\hfill \end{array}$$

Since $ϵ$ is arbitrary, it follows that

$${\Theta }^{*s}({\mu }_E,x)\le {\displaystyle \frac{1}{2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{(1-c_k)}^s{\prod }_{i=1}^{k-1}(a_i^s+b_i^s)}}forallx\in E.$$

This completes the proof of Lemma 3.1. □

Lemma 3.2.

Let E be a perturbed Cantor set with contracting rations $a_k$ and $b_k$. If conditions (1) and (2) hold, then

$${\Theta }^{*s}({\mu }_E,x)\ge {\displaystyle \frac{1}{2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{(1-c_k)}^s{\prod }_{i=1}^{k-1}(a_i^s+b_i^s)}}for\mu -\mathrm{a}.\mathrm{e}.x\in E.$$

Proof. 

Fix $k>0$, and put $\alpha :=\underset{n\to \infty }{\mathrm{lim\; inf}}{(1-c_n)}^s{\prod }_{i=1}^{n-1}(a_i^s+b_i^s)$. There exists a sequence $n_i\to \infty$ with $n_{i+1}-n_i>k$ such that

$$\alpha :=\underset{i\to \infty }{\lim }{(1-c_{n_i+1})}^s{\displaystyle \prod _{j=1}^{n_i}}(a_j^s+b_j^s),$$

which implies that for any $\sigma \in D_{n_i}$,

$$\begin{array}{c}\hfill \underset{i\to \infty }{\lim }{\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{j=1}^{n_i}(a_j^s+b_j^s)}}={\displaystyle \frac{1}{2^s\alpha }}.\end{array}$$

Hence, for any $ϵ>0$, there exists a positive integer $k_0$ such that for any $\sigma \in D_{n_i}$ with $n_i>k_0$, it holds

$${\displaystyle \frac{1}{2^s\alpha }}-ϵ<{\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{j=1}^{n_i}(a_j^s+b_j^s)}}<{\displaystyle \frac{1}{2^s\alpha }}+ϵ.\left(3.2\right)$$

To prove this lemma, it is necessary to construct a full measure set $A\subset E$ satisfying $\mu \left(A\right)=1$. Let

$${\Omega }_{0,p}=\{i\in N:i\ge p,a_{n_i+1}<b_{n_i+1}\},$$

$${\Omega }_{1,p}=\{i\in N:i\ge p,a_{n_i+1}\ge b_{n_i+1}\}.$$

For $k>0$, let $\sigma \left(0\right)=2\underset{k-1}{\underbrace{11\dots 1}}$ and $\sigma \left(1\right)=1\underset{k-1}{\underbrace{22\dots 2}}$. Define

$$\begin{array}{ccc}\hfill A_p^k& =& \left(\bigcup _{l\in {\Omega }_{0,p}}\bigcup _{\sigma \in D_{n_l}}{I}_{\sigma \ast \sigma \left(0\right)}\right)\bigcup \left(\bigcup _{l\in {\Omega }_{1,p}}\bigcup _{\sigma \in D_{n_l}}{I}_{\sigma \ast \sigma \left(1\right)}\right),\hfill \\ \hfill A^k& =& {\displaystyle \bigcap _{p=1}^{\infty }}{A}_p^k,A={\displaystyle \bigcap _{k=1}^{\infty }}{A}^k.\hfill \end{array}$$

For $i=0,1,2,\dots$, define

$$\kappa \left(i\right):=\left\{\begin{array}{cc}1\hfill & \mathrm{if}{a}_{n_i+1}\ge b_{n_i+1}\hfill \\ 0\hfill & \mathrm{if}{a}_{n_i+1}<b_{n_i+1}\hfill \end{array}\right.$$

and

$$\alpha \left(i\right):=a_{n_i+1}{b}_{n_i+2}\dots b_{n_i+k},\beta \left(i\right)=b_{n_i+1}{a}_{n_i+2}\dots a_{n_i+k}.$$

$$\nu \left(i\right):={\displaystyle \frac{\alpha {\left(i\right)}^{\kappa \left(i\right)s}\beta {\left(i\right)}^{(1-\kappa (i\left)\right)s}}{{\prod }_{j=n_i+1}^{n_i+k}(a_j^s+b_j^s)}},\overline{\nu }\left(i\right)=1-\nu \left(i\right).$$

The quantity $\mu \left(A_p^k\right)$ is now computed. It suffices to calculate $\mu \left(A_p^k\right)$ for the case $p=1$, and the remaining cases follow by similar arguments.

Define $D_1^{\prime }=\{\sigma \in D_{n_1}\}$, and

$$\begin{array}{ccc}\hfill D_2^{\prime }& =& \{\sigma \in D_{n_2}:\sigma |n_1\in D_1^{\prime },\sigma |(n_1+k)\notin D_1^{\prime }\ast \sigma \left(\kappa \left(1\right)\right)\},\hfill \\ \hfill D_3^{\prime }& =& \{\sigma \in D_{n_3}:\sigma |n_2\in D_2^{\prime },\sigma |(n_2+k)\notin D_2^{\prime }\ast \sigma \left(\kappa \left(2\right)\right)\},\hfill \\ & & \dots ,\hfill \\ \hfill D_l^{\prime }& =& \{\sigma \in D_{n_l}:\sigma |n_{l-1}\in D_{l-1}^{\prime },\sigma |(n_{l-1}+k)\notin D_{l-1}^{\prime }\ast \sigma \left(\kappa (l-1)\right)\},\hfill \end{array}$$

where $D_{l-1}^{\prime }\ast \sigma \left(\kappa (l-1)\right)=\{\tau \ast \sigma \left(\kappa (l-1)\right):\tau \in D_{l-1}^{\prime }\}$.

Then, it follows that

$$\begin{array}{ccc}\hfill \mu \left(A_1^k\right)& \ge & \mu \left(\left(\bigcup _{l\in {\Omega }_{0,1}}\bigcup _{\sigma \in D_l^{\prime }}{I}_{\sigma \ast \sigma \left(0\right)}\right)\bigcup \left(\bigcup _{l\in {\Omega }_{1,1}}\bigcup _{\sigma \in D_l^{\prime }}{I}_{\sigma \ast \sigma \left(1\right)}\right)\right)\hfill \\ & =& \nu \left(1\right)+\overline{\nu }\left(1\right)\nu \left(2\right)+\overline{\nu }\left(1\right)\overline{\nu }\left(2\right)\nu \left(3\right)+\overline{\nu }\left(1\right)\overline{\nu }\left(2\right)\overline{\nu }\left(3\right)\nu \left(4\right)+\dots \hfill \\ & =& \left(1-\overline{\nu }\left(1\right)\right)+\left(\overline{\nu }\left(1\right)-\overline{\nu }\left(1\right)\overline{\nu }\left(2\right)\right)+\dots \hfill \\ & & +\left(\overline{\nu }\left(1\right)\dots \overline{\nu }\left(2\right)-\overline{\nu }\left(1\right)\dots \overline{\nu }(l+1)\right)+\dots ,\hfill \end{array}$$

where

$$\overline{\nu }\left(i\right)=1-\nu \left(i\right)<1-{\left(2(\sqrt{2}+1)\right)}^{-k}<1,$$

which implies $\mu \left(A_1^k\right)=1$, and $\mu \left(A\right)=1$.

Now, take any $x\in A$. By the definition of A, for all integers $k\ge 1,p\ge 1$, it follows that $x\in {\bigcup }_{l\in {\Omega }_{0,p}}{\bigcup }_{\sigma \in D_{n_l}}{I}_{\sigma \ast \sigma \left(0\right)},$ or $x\in {\bigcup }_{l\in {\Omega }_{1,p}}{\bigcup }_{\sigma \in D_{n_l}}{I}_{\sigma \ast \sigma \left(1\right)}$.

If $x\in {\bigcup }_{l\in {\Omega }_{0,p}}{\bigcup }_{\sigma \in D_{n_l}}{I}_{\sigma \ast \sigma \left(0\right)},$ then there exist infinitely many i such that there exists $\sigma \in D_{n_i}$ with $|x-a(\sigma \ast 2)|<{\displaystyle \frac{1}{3^k}}\left|I_{\sigma }\right|$. Let $r_i=|J_{\sigma }|+|I_{\sigma \ast 1}|+{\left({\displaystyle \frac{1}{3}}\right)}^k\left|I_{\sigma }\right|$. Then $[x-r_i,x+r_i]\supset [a(\sigma \ast 2)-\left(\right|J_{\sigma }|+|I_{\sigma \ast 1}\left|\right),a(\sigma \ast 2)+\left(\right|J_{\sigma }|+|I_{\sigma \ast 1}\left|\right)$ and

$$\begin{array}{ccc}& & {\displaystyle \frac{\mu \left([x-r_i,x+r_i]\right)}{{\left(2r_i\right)}^s}}\hfill \\ & & \ge {\displaystyle \frac{\mu \left(\right[a(\sigma \ast 2)-\left(\right|J_{\sigma }|+|I_{\sigma \ast 1}\left|\right),a(\sigma \ast 2)+\left(\right|J_{\sigma }|+|I_{\sigma \ast 1}\left|\right)\left]\right)}{\left(2\right(|J_{\sigma }|+|I_{\sigma \ast 1}{\left|\right))}^s}}{\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}.\hfill \end{array}$$

From (3.2) and noting that

$$\begin{array}{ccc}& & {\displaystyle \frac{\mu \left(\right[a(\sigma \ast 2)-\left(\right|J_{\sigma }|+|I_{\sigma \ast 1}\left|\right),a(\sigma \ast 2)+\left(\right|J_{\sigma }|+|I_{\sigma \ast 1}\left|\right)\left]\right)}{\left(2\right(|J_{\sigma }|+|I_{\sigma \ast 1}{\left|\right))}^s}}\hfill \\ & \ge & {\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}\ge \left({\displaystyle \frac{1}{2^s\alpha }}-ϵ\right).\hfill \end{array}$$

It follows that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\mu \left([x-r_i,x+r_i]\right)}{{\left(2r_i\right)}^s}}& \ge & {\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}{\left|\right)}^s{\prod }_{j=1}^{n_i}(a_j^s+b_j^s)}}{\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}\hfill \\ & \ge & \left({\displaystyle \frac{1}{2^s\alpha }}-ϵ\right){\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}.\hfill \end{array}$$

If $x\in {\bigcup }_{l\in {\Omega }_{1,p}}{\bigcup }_{\sigma \in D_{n_l}}{I}_{\sigma \ast \sigma \left(1\right)}$, then there exist infinitely many i such that there exists $\sigma \in D_{n_i}$ with $|x-b(\sigma \ast 1)|<{\displaystyle \frac{1}{3^k}}\left|I_{\sigma }\right|$. Let $r_i=|J_{\sigma }|+|I_{\sigma \ast 2}|+{\left({\displaystyle \frac{1}{3}}\right)}^k\left|I_{\sigma }\right|$. Then $[x-r_i,x+r_i]\supset [b(\sigma \ast 1)-\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}\left|\right),b(\sigma \ast 1)+\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}\left|\right)]$, and

$$\begin{array}{ccc}& & {\displaystyle \frac{\mu \left([x-r_i,x+r_i]\right)}{{\left(2r_i\right)}^s}}\hfill \\ & & \ge {\displaystyle \frac{\mu \left(\right[b(\sigma \ast 1)-\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}\left|\right),b(\sigma \ast 1)+\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}\left|\right)\left]\right)}{\left(2\right(|J_{\sigma }|+|I_{\sigma \ast 2}{\left|\right))}^s}}{\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}.\hfill \end{array}$$

From (3.2) and noting that

$$\begin{array}{ccc}& & {\displaystyle \frac{\mu \left(\right[b(\sigma \ast 1)-\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}\left|\right),b(\sigma \ast 1)+\left(\right|J_{\sigma }|+|I_{\sigma \ast 2}\left|\right)\left]\right)}{\left(2\right(|J_{\sigma }|+|I_{\sigma \ast 2}{\left|\right))}^s}}\hfill \\ & \ge & {\displaystyle \frac{|I_{\sigma }{|}^s}{2^s\left(\right|J_{\sigma }|+\min \{|I_{\sigma \ast 1}|,|I_{\sigma \ast 2}{\left|\right\})}^s{\prod }_{i=1}^k(a_i^s+b_i^s)}}\ge \left({\displaystyle \frac{1}{2^s\alpha }}-ϵ\right).\hfill \end{array}$$

In this case, it follows that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\mu \left([x-r_i,x+r_i]\right)}{{\left(2r_i\right)}^s}}& \ge & \left({\displaystyle \frac{1}{2^s\alpha }}-ϵ\right){\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}.\hfill \end{array}$$

Thus, in either of the above cases, it follows that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{\mu \left([x-r_i,x+r_i]\right)}{{\left(2r_i\right)}^s}}& \ge & \left({\displaystyle \frac{1}{2^s\alpha }}-ϵ\right){\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}.\hfill \end{array}$$

Consequently, it is obtained that

$${\Theta }^{*s}({\mu }_E,x)\ge \left({\displaystyle \frac{1}{2^s\alpha }}-ϵ\right){\left(1+{\displaystyle \frac{1}{3^k}}\right)}^{-s}forallx\in A.$$

Note that $ϵ$ is arbitrary. Taking $k\to \infty$, it follows that

$${\Theta }^{*s}({\mu }_E,x)\ge {\displaystyle \frac{1}{2^s\alpha }}forallx\in A.$$

□

Proof of Theorem 1.1. 

Theorem 1.1 follows immediately from Lemmas 3.1 and 3.2. □

Proof of Theorem 1.2. 

By Theorem 1.1 and equation (1.1), it follows that

$$C^s\left(E\right)=2^s\underset{k\to \infty }{\mathrm{lim\; inf}}{(1-c_k)}^s{\displaystyle \prod _{i=1}^{k-1}}(a_i^s+b_i^s).$$

□

4. Discussion

In this paper, the exact upper density and the exact Hausdorff centered measure have been determined for the perturbed Cantor set, which is determined by two sequences of contracting ratios. The perturbed Cantor set is a Moran set on the unit interval and is in general not self-similar; the absence of a self-similarity structure makes its density and measure substantially harder to compute than in the self-similar case. It has been proved that, for almost every point of the set with respect to the normalized Hausdorff measure, the upper density is equal to the reciprocal of an explicit constant depending only on the contracting ratios; consequently, the Hausdorff centered measure of the set is given by an explicit constant of the same type.

By the results of this paper, the exact computation of the upper density and the Hausdorff centered measure is extended from self-similar sets to non-self-similar Moran sets, and from symmetric perfect sets to non-symmetric ones. The following problems remains open and will be pursued in future work. The present results are expected to be extended to more general Moran sets on the real line and to higher-dimensional Moran-type constructions.