A Note on Fractal Measures and Cartesian Product Sets

Abstract

In this paper, we give a new product formula :

$$\begin{aligned} {\mathsf {H}}^t(E) {\mathsf {H}}^s(F)\le c_1 {\mathsf {H}}^{t+s}(E\times F)\le & {} c_2 {\mathsf {H}}^s(E) {\mathsf {P}}^t(F)\\\le & {} c_3 {\mathsf {P}}^{t+s}(E\times F) \le c_4 {\mathsf {P}}^s(E) {\mathsf {P}}^t(F). \end{aligned}$$

where $E\subseteq \mathbb {R}^d$, $F\subseteq \mathbb {R}^l$ , $t,s\ge 0$ and ${\mathsf {H}}^t$ and ${\mathsf {P}}^s$ denote, respectively, the lower and upper Hewitt–Stromberg measures. Using these inequalities, we give lower and upper bounds for the lower and upper Hewitt–Stromberg dimensions ${\mathsf {b}}(E\times F)$ and ${\mathsf {B}}(E\times F)$ in terms of the Hewitt–Strombeg dimensions of E and F.

1 Introduction

In this paper, we are interested on the study of the lower t-dimensional Hewitt–Stromberg measure ${\mathsf {H}}^t$ and the upper t-dimensional Hewitt–Stromberg measure ${\mathsf {P}}^t$. More precisely, we are trying to answer the question : for E and F two subsets of $\mathbb {R}^d$ and $\mathbb {R}^l$, does there exist a constant $\gamma$ such that $\Lambda ^{s+t}(E\times F) =\gamma \Lambda ^s(E) \Lambda ^t(F)$ ? where $\Lambda \in \{{\mathsf {H}}, {\mathsf {P}}\}$. In fact, such results are far from being true. For example, in [8], it is proved that,

$$\begin{aligned} {\mathsf {b}}(E\times F) \ge {\mathsf {b}}(E) + {\mathsf {b}}(F) \end{aligned}$$

(1.1)

where ${\mathsf {b}}$ stands for the lower Hewitt–Stomberg dimension. However, it is possible to construct two subsets E and F such that ${\mathsf {b}}(E\times F) > {\mathsf {b}}(E) + {\mathsf {b}}(F)$. Similar results were proved for s-dimensional Hausdorff measure $\mathcal {H}^s$ and s-dimensional packing measure $\mathcal {P}^s$ [2, 15, 16, 31]. The reader can see also [14, 25, 32, 33] for various results on this problem.

The aim of this paper is to prove Theorem A and C below. This is by constructing a metric outer measure, denoted ${\mathsf {H}}^{*t}$, that is comparable to the Hewitt Stromberg measure ${\mathsf {H}}^{t}$ but significantly easier to analyse.

Theorem A

Let $E\subseteq \mathbb {R}^d$, $F\subseteq \mathbb {R}^l$ and $s,t \ge 0$. There exists a positive constant c such that

$$\begin{aligned} {\mathsf {H}} ^{t+s}(E\times F)\le & {} c\; {\mathsf {H}} ^s(E) {\mathsf {P}} ^t(F) \end{aligned}$$

(1.2)

$$\begin{aligned} {\mathsf {H}} ^s(E) {\mathsf {P}} ^t(F)\le & {} c\; {\mathsf {P}} ^{t+s}(E\times F)\end{aligned}$$

(1.3)

$$\begin{aligned} {\mathsf {P}} ^{t+s}(E\times F)\le & {} c\; {\mathsf {P}} ^s(E) {\mathsf {P}} ^t(F). \end{aligned}$$

(1.4)

To prove (1.1), the authors in [8] prove that

$$\begin{aligned} {\mathsf {H}}^t(E) {\mathsf {H}}^s(F)\le c \, {\mathsf {H}}^{t+s}(E\times F). \end{aligned}$$

The key idea of the proof is to construct a new multifractal measure, on the Euclidean space $\mathbb {R}^d$, equivalent to the lower Hewitt–Stomberg. Therefore, the authors use the class of all half-open dyadic cubes of covering sets in the definition rather than the class of all closed balls. In this paper, we will use the class of all half-open semi-dyadic cubes rather than the dyadic ones to construct two measures similar to ${\mathsf {H}}^t$ and ${\mathsf {P}}^t$. This choice is due to the fact that the semi-dyadic cube $v_n(x)$ is less sensitive to the location of x than the corresponding dyadic cube $u_n(x)$. Semi-dyadic cubes have been used in [15, 26, 35]. A direct consequence of Theorem A is the following result.

Corollary B

For all $E, F\subseteq \mathbb {R}^d$ we have

$$\begin{aligned} {\mathsf {b}} (E) + {\mathsf {b}} (F) \le {\mathsf {b}} (E\times F)\le & {} {\mathsf {b}} (E) + {\mathsf {B}} (F) \end{aligned}$$

(1.5)

$$\begin{aligned}\le & {} {\mathsf {B}} (E\times F) \end{aligned}$$

(1.6)

$$\begin{aligned}\le & {} {\mathsf {B}} (E) + {\mathsf {B}} (F). \end{aligned}$$

(1.7)

where ${\mathsf {B}}$ (resp. ${\mathsf {b}}$)stands for the upper (resp. lower) Hewitt–Stomberg dimension.

The set E is said to be regular if ${\mathsf {b}}(E) = {\mathsf {B}}(E)$. Regular sets are defined with respect to the Hausdorff measure [3, 4, 17, 18], to packing measure [34, 35] or to Hewitt–Stromberg measure [20,21,22]. In particular Tricot et al. [29, 34] managed to show that a subset E of $\mathbb {R}^d$ has an integer Hausdorff and packing dimension if it is strongly regular, i.e., $\mathcal {H}^t(E)=\mathcal {P}^t(E)$, for $t\ge 0$. Now, remark that, as a consequence of (1.5), we have if E or F is regular then

$$\begin{aligned} {\mathsf {b}}(E\times F) = {\mathsf {b}}(E)+ {\mathsf {b}}(F). \end{aligned}$$

(1.8)

This is the case, for example, when $E = {\mathcal {K}}$ the triadic Cantor set. Moreover, if E and F are regular then

$$\begin{aligned} {\mathsf {b}}(E\times F) = {\mathsf {B}}(E\times F). \end{aligned}$$

(1.9)

Thus, ${\mathsf {b}}({\mathcal {K}} \times {\mathcal {K}}) = {\mathsf {B}}({\mathcal {K}} \times {\mathcal {K}}) = 2\log 2/\log 3$. In the following, we give a sufficient condition so that (1.8) and (1.9) hold. Let $\mu$ be a positive measure on $\mathbb {R}^d$ and $\alpha \ge 0$, we define the lower and upper $\alpha$-dimensional densities at $x\in E$ by

$$\begin{aligned} \underline{d}_\mu ^\alpha (x) =\liminf _{r\rightarrow 0}\frac{\mu (E\cap B(x, r))}{(2r)^{\alpha }} \quad \text {and}\quad \overline{d}_\mu ^\alpha (x) =\limsup _{r\rightarrow 0}\frac{\mu (E\cap B(x, r))}{(2r)^{\alpha }} \end{aligned}$$

(1.10)

respectively. We refer to the common value as the $\alpha$-dimensional density at x and denote it by ${d}_\mu ^\alpha (x)$. Let $t,s \ge 0$ and assume that $0< {\mathsf {H}}^s(E)<+\infty$ and $0< {\mathsf {H}}^t(F)<+\infty .$ It is well known [8, Theorem 2] that (1.8) holds provided that, for all $y\in F$, $d^t_{{\mathsf {H}}^t} (y) >0$. In the next Theorem we will improve this result.

Theorem C

Let $E, F\subseteq \mathbb {R}^d$, and $\alpha , \beta >0$.

• (1) Assume that $0<{\mathsf {H}} ^{\alpha }(E)< \infty$ and for all $x\in E$ we have $\underline{d}^\alpha _{{\mathsf {H}} ^{\alpha }}(x)>0$ then

  $$\begin{aligned} {\mathsf {b}} (E\times F) = {\mathsf {b}} (E)+ {\mathsf {b}} (F), \qquad \quad \forall F\subseteq \mathbb {R}^d. \end{aligned}$$

• (2) If, in addition, $0<{\mathsf {H}} ^{\beta }(E)< \infty$ and for all $y\in F$ we have $\underline{d}^\beta _{{\mathsf {H}} ^{\beta }}(y)>0$ then

  $$\begin{aligned} {\mathsf {b}} (E\times F) = {\mathsf {b}} (E)+ {\mathsf {b}} (F) = {\mathsf {B}} (E\times F) = {\mathsf {B}} (E)+ {\mathsf {B}} (F). \end{aligned}$$

Remark that, if E and F are regular then ${\mathsf {B}}(E\times F) = {\mathsf {b}}(E)+{\mathsf {B}}(F)$. This is the case, for example, if we take $E=F$ to be the triadic Cantor set. In the last section of this paper, we will construct a non-standard set F satisfying this equality with E the triadic Cantor set.

2 Construction of Multifractal Measure

Let’s recall that Hausdorff dimension, packing dimension and Hewitt–Stromberg dimension are defined as follows. For $d\ge 1$, $\mathbb {R}^d$ is endowed with the Euclidean distance. If $E\subset \mathbb {R}^d$, $t\ge 0$ and $\delta >0$, then we write

$$\begin{aligned} \mathcal {H}_\delta ^t(E)= \inf \left\{ \sum _i \Big (|E_i|\Big )^t \;\; E\subseteq \bigcup _i E_i,\;\; |E_i|< \delta \right\} , \end{aligned}$$

where |A| is the diameter of the set A defined as $|A| = \sup \big \{ |x - y|, \; x, y \in A \big \}$. We define the t-dimensional Hausdorff measure of E as

$$\begin{aligned} \mathcal {H}^t(E)=\sup _{\delta >0}\mathcal {H}_\delta ^t(E). \end{aligned}$$

The reader can be referred to Rogers’ classical text [30] for a systematic discussion of $\mathcal {H}^t$.

We also define,

$$\begin{aligned} \overline{\mathcal {P}}_\delta ^t(E)= \sup \left\{ \sum _i \Big (2r_i\Big )^t \right\} , \end{aligned}$$

where the supremum is taken over all disjoint closed balls $\Big ( B(x_i, r_i) \Big )_i \; \text {such that}\; r_i \le \delta$ and $x_i \in E$. The t-dimensional packing premeasure, of E is now defined by

$$\begin{aligned} \overline{\mathcal {P}}^t(E)=\sup _{\delta >0}\overline{\mathcal {P}}_\delta ^t(E), \end{aligned}$$

and the packing measure of E by

$$\begin{aligned} {\mathcal {P}}^t(E)=\inf \left\{ \sum _i\overline{\mathcal {P}}^h(E_i)\;\Big |\;\;E\subseteq \bigcup _i E_i\right\} . \end{aligned}$$

While Hausdorff and packing measures are defined using coverings and packings by families of sets with diameters less than a given positive number $\delta$, the Hewitt–Stromberg measures are defined using covering of balls with the same diameter.

Several authors have investigated the Hewitt–Stromberg measures, that were introduced in [13, Exercise (10.51)], explicitly, for example, in Pesin’s monograph [28, 5.3] or implicitly in Mattila’s text [23]. One can also cite, [9,10,11,12, 19, 27, 36] where these measures have been used on the study of the local properties of fractals and products of fractals. In particular, in [5, pp. 32-36], Edgar provided an outstanding and fine introduction to these measures. One of the purposes of this paper is to define and study a class of natural multifractal generalizations of the Hewitt–Stromberg measures. Although packing and Hausdorff measures are defined using packings and coverings by families of sets with diameters less than a given positive number $\varepsilon$, say, the Hewitt–Stromberg measures are defined using packings of balls with a fixed diameter $\varepsilon$. For $t>0$, the Hewitt–Stromberg pre-measures are defined as follows,

$$\begin{aligned} \overline{\mathsf {H}}^t_0(E)=\liminf _{r\rightarrow 0} \overline{\mathsf {H}}^t_r(E)\quad \text {with }\quad \overline{\mathsf {H}}^t_r(E) = N_r(E) \;(2r)^t \end{aligned}$$

and

$$\begin{aligned} \overline{\mathsf {P}}^t(E)=\limsup _{r\rightarrow 0} \overline{\mathsf {P}}^t_r(E)\quad \text {with }\quad \overline{\mathsf {P}}^t_r(E) =M_r(E) \;(2r)^t, \end{aligned}$$

where the covering number $N_r(E)$ of E and the packing number $M_r(E)$ of E are given by

$$\begin{aligned} N_r(E)=\inf \left\{ \sharp \{I\}\;\Big |\; \Big ( B(x_i, r) \Big )_{i\in I} \; \text {is a family of closed balls with}\; x_i \in E \; \text {and}\; E\subseteq \bigcup _i B(x_i, r)\right\} \end{aligned}$$

and

$$\begin{aligned} M_r(E)=\sup \left\{ \sharp \{I\}\; \Big |\; \Big ( B(x_i, r) \Big )_{i\in I} \; \text {is a family of closed balls with}\; x_i \in E \; \text {and}\; d(x_i, x_j)\ge r\; \text {for}\; i\ne j\right\} . \end{aligned}$$

It’s clear that $\overline{\mathsf {P}}^t$ is increasing and $\overline{\mathsf {P}}^t(\emptyset ) =0$. However it’s not $\sigma$-additive. Therefore, we introduce the upper t-dimensional Hewitt–Stromberg measure, which we denote by $\mathcal {\mathsf {P}}^t$, defined by

$$\begin{aligned} {\mathsf {P}}^t(E)=\inf \left\{ \sum _i\overline{\mathsf {P}}^t(E_i)\;\Big |\;\;E\subseteq \bigcup _i E_i, \;\; E_i \;\;\text {is bounded}\right\} . \end{aligned}$$

Since $\overline{\mathsf {H}}^t_0$ is not increasing and not countably subadditive, one needs a standard modification to get an outer measure. Hence, we modify the definition as follows

$$\begin{aligned} \overline{\mathsf {H}}^t(E)=\inf \left\{ \sum _i\overline{\mathsf {H}}^t_0(E_i)\;\Big |\;\;E\subseteq \bigcup _i E_i, \;\; E_i \;\;\text {is bounded}\right\} \end{aligned}$$

and

$$\begin{aligned} {\mathsf {H}}^t(E)=\sup _{F\subseteq E} \overline{\mathsf {H}}^t (F). \end{aligned}$$

We recall some basic inequalities satisfied by the Hewitt–Stromberg, the Hausdorff and the packing measures (see [1, 19, 27])

$$\begin{aligned} {\mathcal {H}}^t(E)\le {\mathsf {H}}^t(E)\le {\mathsf {P}}^t(E)\le {\mathcal {P}}^t(E) \le \overline{\mathcal {P}}^t(E) . \end{aligned}$$

(2.1)

The lower and upper Hewitt–Stromberg dimension ${\mathsf {b}}(E)$ and ${\mathsf {B}}(E)$ are defined by

$$\begin{aligned} {\mathsf {b}}(E)=\inf \Big \{t\ge 0\;\Big |\; {\mathsf {H}}^t(E)=0\Big \}=\sup \Big \{t\ge 0\;\Big |\; {\mathsf {H}}^t(E)=+\infty \Big \} \end{aligned}$$

and

$$\begin{aligned} {\mathsf {B}}(E)=\inf \Big \{t\ge 0\;\big |\; {\mathsf {P}}^t(E)=0\Big \}=\sup \Big \{t\ge 0\;\big |\; {\mathsf {P}}^t(E)=+\infty \Big \}. \end{aligned}$$

As a consequence of (2.1) we get

$$\begin{aligned} \dim _H(E)\le {\mathsf {b}}(E) \le {\mathsf {B}}(E) \le \dim _P(E)\le \overline{\dim }_P(E) \end{aligned}$$

where $\dim _H$ and $\dim _P$ denote respectively Hausdorff and packing dimensions and $\overline{\dim }_P$ is defined as

$$\begin{aligned} \overline{\dim }_P (E)=\inf \Big \{t\ge 0\;\Big |\; \overline{\mathcal {P}}^t(E)=0\Big \}. \end{aligned}$$

Example 1

Consider for $n\ge 1$, the set $A_n = \{ 0\}\bigcup \{ 1/k, \;\; k\le n\}$ and

$$\begin{aligned} K= \bigcup _n A_n = \Big \{ 0 \Big \}\; \bigcup \; \Big \{ 1/n, \;\; n\in \mathbb {N}\Big \}. \end{aligned}$$

K is a countable set and then $\dim _H (K) ={\mathsf {b}}(K) = {\mathsf {B}}(K)=\dim _p(K)=0$. Nevertheless, we have $\overline{\dim }_p(K) =\frac{1}{2}$ ( Corollary 2.5 in [7] ). Moreover, we can construct ( [1]) a set E such that

$$\begin{aligned} \dim _H(E) = {\mathsf {b}}(E) \ne {\mathsf {B}}(E) =\dim _P(E). \end{aligned}$$

Another two widely used notions of dimensions are those of the lower and upper box dimensions, denoted by $\underline{\dim }_{B}$ and $\overline{\dim }_{B}$, respectively, and defined by

$$\begin{aligned} \underline{\dim }_{B}(E)=\liminf _{r\rightarrow 0}\frac{\log N_r(E)}{-\log r}=\liminf _{r\rightarrow 0}\frac{\log M_r(E)}{-\log r} \end{aligned}$$

and

$$\begin{aligned} \overline{\dim }_{B}(E)=\limsup _{r\rightarrow 0}\frac{\log M_r(E)}{-\log r}=\limsup _{r\rightarrow 0}\frac{\log N_r(E)}{-\log r} \end{aligned}$$

for any bounded subset E of $\mathbb {R}^d$. We refer to the common value as the box dimension and denote it by $\dim _B$. These dimensions satisfy the following inequalities,

$$\begin{aligned} \dim _H(E)\le \dim _P(E)\le \overline{\dim }_{B}(E) \end{aligned}$$

and

$$\begin{aligned} \dim _H(E)\le \underline{\dim }_{B}(E)\le \overline{\dim }_{B}(E). \end{aligned}$$

In addition we have, $\underline{\dim }_{B}(E) = \underline{\dim }_{B}(\overline{E})$ and $\overline{\dim }_{B}(E) = \overline{\dim }_{B}(\overline{E})$. An immediate consequence of this is that if E is a dense subset of an open region of $\mathbb {R}^d$ then $\underline{\dim }_{B}(E)=\overline{\dim }_{B}(E)=d$. In particular, countable sets can have nonzero box dimension. Thus, these notions of dimensions are not $\sigma$-stable in general, i.e., it is not true in general that $\dim _B (\bigcup _i E_i) = \sup _i \dim _B (E_i)$. In fact, the lower box dimension is not even finitely stable in general. It is possible, through a slight modification of their definitions, to obtain countably stable notions of dimension called the lower and upper modified box dimensions and defined, for any $E\subseteq \mathbb {R}^d$, by (see [6, 24])

$$\begin{aligned} \underline{\dim }_{MB}(E)= \inf \left\{ \sup _{i} \underline{\dim }_{B}(E_i)\;\Big |\;E\subseteq \bigcup _i E_i, \;\; E_i \;\;\text {is bounded}\;\; \right\} \end{aligned}$$

and

$$\begin{aligned} \overline{\dim }_{MB}(E)= \inf \left\{ \sup _{i} \overline{\dim }_{B}(E_i) \;\Big |\; E\subseteq \bigcup _i E_i,\;\; E_i \;\;\text {is bounded} \;\; \right\} . \end{aligned}$$

These dimensions satisfy the following inequalities (see for example [1, 6])

$$\begin{aligned} \dim _H(E)\le \underline{\dim }_{MB}(E)\le \overline{\dim }_{MB}(E)\le \dim _P(E), \end{aligned}$$

The reader is referred to [6] for an excellent discussion of lower and upper box dimensions.

The pre-measure $\overline{{\mathsf {P}}}^{t}$ assigns in the usual way a multifractal dimension to each subset E of $\mathbb {R}^d$. It is denoted by ${ \Delta }(E)\in [-\infty ,+\infty ]$ such that

$$\begin{aligned} \overline{\mathsf {P}}^{t}(E)=\left\{ \begin{matrix} \infty &{}\text {if}&{} t< { \Delta }(E),\\ \\ 0 &{} \text {if}&{} { \Delta }(E) < t.\end{matrix}\right. \end{aligned}$$

In addition, we have

$$\begin{aligned} {\mathsf {b}}(E) \le {\mathsf {B}}(E) \le { \Delta }(E). \end{aligned}$$

We note also that there exists a unique number ${\Theta }(E)\in [-\infty ,+\infty ]$ such that

$$\begin{aligned} {\overline{{\mathsf {H}}}}^{t}_{0}(E)=\left\{ \begin{matrix} \infty &{}\text {if}&{} t< {\Theta }(E),\\ \\ 0 &{} \text {if}&{} {\Theta }(E) < t.\end{matrix}\right. \end{aligned}$$

Example 2

Consider the set K defined in Example 1. We will prove that $\Theta ( K) = 1/2$. For $n\ge 1$ and $r_k < \frac{1}{n+n^2}$, remark that

$$\begin{aligned} N_{r_k}(A_n) = n+1. \end{aligned}$$

It follows that

$$\begin{aligned} \overline{{\mathsf {H}}}_{r_k}^{1/2}(K)\ge \overline{{\mathsf {H}}}_{r_k}^{1/2}(A_n) = \sqrt{2} \frac{n+1}{\sqrt{n+n^2}}. \end{aligned}$$

Thereby, we have $\overline{{\mathsf {H}}}_{r_k}^{1/2}(A_n) \ge 1$ which implies that

$$\begin{aligned} \Theta ( K) \ge \Theta (A_n) \ge 1/2. \end{aligned}$$

Let $\Big (B(x_i , r )\Big )_{ i\in J}$ be a centred covering of K. Then, using Besicovitch’s covering theorem, we can construct $\xi$ finite sub-families $\Big (B(x_{1j}, r)\Big )_j$, ...,$\Big (B(x_{\xi j}, r )\Big )_j$, such that each $K \subseteq \displaystyle \bigcup _{i=1}^\xi \bigcup _jB(x_{ij}, r)$ and $\Big (B(x_{ij}, r)\Big )_j$ is a packing of K. Therefore,

$$\begin{aligned} \overline{{\mathsf {H}}}_0^t(K) \le \xi \overline{{\mathsf {P}}}^t(K) \le \xi \overline{\mathcal {P}}^t(K). \end{aligned}$$

Which imply that

$$\begin{aligned} \Theta (K) \le \Delta (E) \le \overline{\dim }_p(K)=\frac{1}{2}. \end{aligned}$$

It is well known [1, Propositon 2] that, for any bounded set $E \subseteq \mathbb {R}^d$,

$$\begin{aligned} {\Theta }(E) =\underline{\dim }_{B}(E) \qquad \text {and }\qquad {\Delta }(E) = \overline{\dim }_{B}(E). \end{aligned}$$

It is clear that $\Delta$ and $\Theta$ are monotonic. In addition, for any sets $E, F\subseteq \mathbb {R}^d$, we have $\overline{{\mathsf {P}}}^t(E\cup F)\le \overline{{\mathsf {P}}}^t(E)+\overline{{\mathsf {P}}}^t(F).$ Thus, $\Delta$ is stable, i.e.,

$$\begin{aligned} \Delta (E\cup F) =\max \big (\Delta (E), \Delta (F)\big ). \end{aligned}$$

(2.2)

Let $n\in \mathbb {N}$. A closed interval of the kind $[k 2^{-n}, (k+1) 2^{-n}]$, $k\in \mathbb {Z}$, is called $2^n$-mesh in $\mathbb {R}$. A $2^n$-mesh in $\mathbb {R}^d$ is a closed ball, the projection of which on each axis is a $2^n$-mesh. We denote by $w(2^n, E)$ the number of $2^n$-meshes meeting the bounded set E. There are other definitions of $\Delta$ and $\Theta$ in $\mathbb {R}^d$ which are more convenient to use in this paper [6, Section 3.1]

$$\begin{aligned} {\Delta }(E)= & {} \limsup _{n\rightarrow 0} \frac{\log w(2^n, E) }{n\log 2}, \end{aligned}$$

(2.3)

$$\begin{aligned} {\Theta }(E)= & {} \liminf _{n\rightarrow 0} \frac{\log w(2^n, E) }{n\log 2}. \end{aligned}$$

(2.4)

Example 3

Let E be the set in [0, 1] defined by $E = \displaystyle \bigcap \nolimits _n E_n$, where $E_n$ is the union of $2^n$ non overlapping intervals of length $b_n$, each of them containing two intervals of $E_{n+1}$. Assume that $b_0=1$ and $b_n>2b_{n+1}$. Now, if $2r\in [b_{n+1}, b_n]$, then $2^n\le N_{2r}(E) \le 2^{n+1}$. Hence,

$$\begin{aligned} \Theta (E) = \liminf _{n\rightarrow \infty }\frac{n\log 2}{-\log b_n} \quad \text {and}\quad \Delta (E) = \limsup _{n\rightarrow \infty }\frac{n\log 2}{-\log b_n}. \end{aligned}$$

In the special case when E is the triadic Cantor set $(b_n=3^{-n})$ then

$$\begin{aligned} \dim _H (E) ={\mathsf {b}}(E) = {\mathsf {B}}(E)=\dim _p(E)=\Theta (E) = \Delta (E)= \frac{\log 2}{\log 3}. \end{aligned}$$

3 Proofs of Main Results

For $n\in \mathbb {N}$, we write

$$\begin{aligned} \mathcal {U}_n = \Big \{ \prod _{i=1}^d \Big [\frac{ l_i}{2^n},\frac{ l_i +1}{2^n} \Big [, \;\; l_1,\ldots , l_d\in \mathbb {Z}\Big \} \end{aligned}$$

and

$$\begin{aligned} \mathcal {V}_n = \Big \{ \prod _{i=1}^d \Big [\frac{\frac{1}{2} l_i}{2^n},\frac{\frac{1}{2} l_i +1}{2^n} \Big [, \;\; l_1,\ldots , l_d\in \mathbb {Z}\Big \}. \end{aligned}$$

The family $\mathcal {U}_n$ is the class of half-open dyadic cubes of order n. We denote $u_n(x)$, for $x\in \mathbb {R}^d$, the unique $u\in \mathcal {U}_n$ that contains x. The family $\mathcal {V}_n$ is the class of half-open dyadic semi-cubes of order n. We denote $v_n(x)$, for $x\in \mathbb {R}^d$, the unique $v\in \mathcal {V}_n$ that contains x whose complement is at distance $2^{-n-2}$ from $u_{n+2}(x)$. Let $\mathcal {K}=\{(k_1,\ldots , k_d),\; k_i=0, \frac{1}{2}\}$. For each $\mathbf{k} =(k_1,\ldots , k_d)\in \mathcal {K}$ let

$$\begin{aligned} \mathcal {V}_\mathbf{k , n}=\Big \{ \prod _{i=1}^d \Big [\frac{ k_i+l_i}{2^n},\frac{k_i+ l_i +1}{2^n} \Big [, \;\; l_1,\ldots , l_d\in \mathbb {Z}\Big \}. \end{aligned}$$

Observe that, for $v\ne v'\in \mathcal {V}_\mathbf{k , n}$, we have $v\cap v'=\emptyset$. Furthermore $\Big (\mathcal {V}_\mathbf{k , n}\Big )_\mathbf{k \in \mathcal {K}}$ is a partition of $\mathcal {V}_n$. Moreover, if $v, v'\in \mathcal {V}_\mathbf{k} :=\bigcup _{n\ge 0} \mathcal {V}_\mathbf{k , n}$, then either $v\cap v'=\emptyset$ or one of them is contained in the other. Finally, for $E\subset \mathbb {R}^d$, write

$$\begin{aligned} \mathcal {V}_n(E) = \{v_n(x),\; x\in E\}\quad \text {and} \quad \mathcal {V}_\mathbf{k , n}(E) = \mathcal {V}_n(E)\cap \mathcal {V}_\mathbf{k , n}. \end{aligned}$$

Now, we will construct new measures, on the Euclidean space $\mathbb {R}^d$, in a similar manner to the lower and upper Hewitt–Stromberg measures using the class of all half-open dyadic semi-cubes in the definition rather than the class of all closed balls. More precisely, let $E\subseteq \mathbb {R}^d$ and $t\ge 0$, let

$$\begin{aligned} \overline{\mathsf {H}}^{*t}(E)=\liminf _{n\rightarrow +\infty } N_n^*(E) \; 2^{-nt}, \quad \text {and}\quad \overline{\mathsf {P}}^{*t}(E)=\limsup _{n\rightarrow +\infty } M_n^*(E) \; 2^{-nt}, \end{aligned}$$

where the numbers $N_n^*(E)$ and $M_n^*(E)$ of E are given by

$$\begin{aligned} N_n^{*}(E)=\inf \left\{ \sharp \{I\}\; \Big |\; \Big ( v_i \Big )_{i\in I} \; \; \text {is a family of coverings of}\;\; E\;\; \text {with } \;\; v_i\in \mathcal {V}_n(E) \right\} . \end{aligned}$$

and

$$\begin{aligned} M_n^{*}(E)=\sup \left\{ \sharp \{I\}\; \; v_i \in \mathcal {V}_n(E), \; i\in I, \quad \text {and }\quad \overline{v}_i \cap \overline{v}_j =\emptyset \quad \text {for } i\ne j \right\} . \end{aligned}$$

It’s clear that $\overline{\mathsf {H}}^{*t}$ and $\overline{\mathsf {P}}^{*t}$ are increasing with $\overline{\mathsf {H}}^t(\emptyset ) = \overline{\mathsf {P}}^t(\emptyset ) =0$. However they are not $\sigma$-additive. Therefore, we define

$$\begin{aligned} {\mathsf {H}}^{*t}(E)= & {} \inf \left\{ \sum _i\overline{\mathsf {H}}^{*t}(E_i)\;\Big |\;\;E\subseteq \bigcup _i E_i\;\; E_i \;\;\text {is bounded} \right\} .\\ {\mathsf {P}}^{*t}(E)= & {} \inf \left\{ \sum _i\overline{\mathsf {P}}^{*t}(E_i)\;\Big |\;\;E\subseteq \bigcup _i E_i \;\; E_i \;\;\text {is bounded}\right\} . \end{aligned}$$

Lemma 1

For every set $E\subset \mathbb {R}^d$, for any $t\ge 0$, there exists a constant $c>0$ such that

$$\begin{aligned} c^{-1}{{\mathsf {P}} }^t(E) \le {{\mathsf {P}} ^*}^t(E) \le c {\mathsf {P}} ^t(E) \quad \text {and}\quad c^{-1}{{\mathsf {H}} }^t(E) \le {{\mathsf {H}} ^*}^t(E) \le c {\mathsf {H}} ^t(E) \end{aligned}$$

(3.1)

Proof

It follows from the fact that $B(x, 2^{-n-2}) \subseteq v_n(x) \subseteq B(x, \sqrt{d} 2^{-n})$ $\square$

In a similar manner, we can construct two measures ${\mathsf {H}}^{**t}$ and ${\mathsf {P}}^{**t}$ using the class of all half-open dyadic cubes in the definition rather than the class of all half-open dyadic semi-cubes. Nevertheless, the new pre-measure $\overline{\mathsf {P}}^{**t}$ is not equivalent to the pre-measure $\overline{\mathsf {P}}^{t}$, (see [35, Example 3.5]).

Now, we state a useful proposition which will be applied in the proof of our main results.

Proposition 1

Let $E \subseteq \mathbb {R}^d$. Then,

$$\begin{aligned} {\mathsf {b}} (E) =\sup _{F\subseteq E} \left\{ \inf \left\{ \sup _{i} {\Theta }(F_i) \;\Big |\;\displaystyle F\subseteq \bigcup _i F_i, \;\; F_i \;\;\text {is bounded in} \;\; \mathbb {R}^d \right\} \right\} \end{aligned}$$

and

$$\begin{aligned} {\mathsf {B}} (E) = \inf \left\{ \sup _{i} {\Delta } (E_i) \;\Big |\;\displaystyle E\subseteq \bigcup _i E_i, \;\; E_i \;\;\text {is bounded in} \;\; \mathbb {R}^d\right\} . \end{aligned}$$

(3.2)

Proof

Denote by

$$\begin{aligned} \beta = \inf \left\{ \sup _{i} {\Delta } (E_i) \;\Big |\;\displaystyle E\subseteq \bigcup _i E_i, \;\; E_i \;\;\text {is bounded in} \;\; \mathbb {R}^d\right\} . \end{aligned}$$

Assume that $\beta < {\mathsf {B}}(E)$ and take $\alpha \in (\beta , {\mathsf {B}}(E) )$. Then, there exists $\{E_i\}$ a sequence of subsets of $\mathbb {R}^d$ such that $E\subseteq \cup _i E_i$, and $\sup _i {\Delta }(E_i) < \alpha$. Now observe that $\overline{{\mathsf {P}}}^{ \alpha }(E_i) = 0$ which implies that ${{\mathsf {P}}}^{ \alpha }(E) = 0$. It is a contradiction. Now suppose that ${{\mathsf {B}}}(E) < \beta$, then, for $\alpha \in ( {{\mathsf {B}}}(E), \beta ),$ we have ${{\mathsf {P}}}^{ \alpha }(E) = 0$. Thus, there exists $\{E_i\}$ a sequence of subsets of $\mathbb {R}^d$ such that $E\subseteq \cup _i E_i$, and $\sup _i \overline{{\mathsf {P}}}^{ \alpha } (E_i) < \infty$. We conclude that, $\sup _i {\Delta } (E_i) \le \alpha$. It is also a contradiction. The proof of the second statement is identical to the first one and is therefore omitted. $\square$

3.1 Proof of Inequality (1.2) and (1.5)

Let $s,t\ge 0$, $E\subseteq \mathbb {R}^d$ and $F\subseteq \mathbb {R}^l$. We will prove that, there exists a constant $c>0$ such that

$$\begin{aligned} {\mathsf {H}}^{s+t}(E\times F) \le c {\mathsf {H}}^s(E) {\mathsf {P}}^t(F). \end{aligned}$$

Let $r>0$ and $H\subseteq E\times F$. Let $\{B(x_i, r)\}_i$ be a centred r-covering of E. We denote n to be the integer such that $\sqrt{l} 2^{-n} < r \le \sqrt{l} 2^{-n+1}$. For each i and $v\in \mathcal {V}_n(F)$ with $(B(x_i,r)\times v) \cap H) \ne \emptyset$ choose a point $y_{i,v}\in B(x_i, r)$ and a point $y'_{i,v}\in v$ such that $(y_{i,v}, y'_{i,v}) \in (B(x_i,r)\times v) \cap H)$. Observe that

$$\begin{aligned} H\subseteq & {} \bigcup _i \Big ( \bigcup _{{\mathop {(B(x_i,r)\times v)\cap H\ne \emptyset }\limits ^{v\in \mathcal {V}_n(F)}}} B(x_i, r)\times v \Big ) \subseteq \bigcup _i \Big ( \bigcup _{{\mathop {(B(x_i,r)\times v)\cap H\ne \emptyset }\limits ^{v\in \mathcal {V}_n(F)}}} B(y_{i,v}, 2 r)\times B(y'_{i,v}, 2r) \Big ) \\\subseteq & {} \bigcup _i \Big ( \bigcup _{{\mathop {(B(x_i,r)\times v)\cap H\ne \emptyset }\limits ^{v\in \mathcal {V}_n(F)}}} B((y_{i,v},y'_{i,v}), 2r) \Big ). \end{aligned}$$

Therefore, the family $\big (B((y_{i,v},y'_{i,v}), 2r) \big )_{i\in \mathbb {N}, v\in \mathcal {V}_n(F), B(x_i,r)\times v)\cap H\ne \emptyset }$ is a centred (2r)-covering of H. Moreover, we have $B(y'_{i,v}, \eta _r)\subseteq B(y'_{i,v}, 2^{-n-2})$ for $\eta _r= 2^{-3}\sqrt{l} r$. Hence, for each $\mathbf{k} \in \mathcal {K}$, the family

$$\begin{aligned} \big (B(y'_{i,v}, \eta _r \big ), \;\; i\in \mathbb {N}, v\in \mathcal {V}_\mathbf{k ,n}(F), B(x_i,r)\times v)\cap H\ne \emptyset \end{aligned}$$

is a centred $\eta _r$-packing of F. In follows that

$$\begin{aligned} \overline{{\mathsf {H}}}_{2r}^{s+t}(H)\le & {} \sum _i\Big ( \sum _{{\mathop {(B(x_i,r)\times v)\cap H\ne \emptyset }\limits ^{v\in \mathcal {V}_n(F)}}} (4r)^{s+t}\Big )\\\le & {} 2^{s+t} \left( \frac{8}{\sqrt{l}}\right) ^t \sum _i (2r)^s \Big ( \sum _\mathbf{k \in \mathcal {K}} \sum _{{\mathop {(B(x_i,r)\times v)\cap H\ne \emptyset }\limits ^{v\in \mathcal {V}_\mathbf{k ,n}(F)}}} (2\eta _r)^{t}\Big )\\\le & {} 2^{s+t} \left( \frac{8}{\sqrt{l}}\right) ^t \sum _i (2r)^s \Big ( \sum _\mathbf{k \in \mathcal {K}} \overline{{\mathsf {P}}}_{\eta _r}^{t}(F)\Big )\\\le & {} 2^l 2^{s+t} \overline{{\mathsf {P}}}_{\eta _r}^{t}(F) \left( \frac{8}{\sqrt{l}}\right) ^t \sum _i (2r)^s \end{aligned}$$

Then, by taking the infimum over all centred r-covering of E we obtain

$$\begin{aligned} \overline{{\mathsf {H}}}_{2r}^{s+t}(H)\le 2^l 2^{s+t} \left( \frac{8}{\sqrt{l}}\right) ^t \overline{{\mathsf {H}}}^s_{r}(E) \overline{{\mathsf {P}}}_{\eta _r}^{t}(F) . \end{aligned}$$

Therefore,

$$\begin{aligned} \overline{{\mathsf {H}}}_{0}^{s+t}(H)\le c \liminf _{r\rightarrow 0} \overline{{\mathsf {H}}}^s_{r}(E) \limsup _{r\rightarrow 0} \overline{{\mathsf {P}}}_{\eta _r}^{t}(F)= c \overline{{\mathsf {H}}}^s_{0 }(E) \overline{{\mathsf {P}}}^t (F), \end{aligned}$$

where $c=2^l 2^{s+t} \Big (\frac{8}{\sqrt{l}}\Big )^t$. Now, assume that $E\subseteq \bigcup _i E_i$ and $F\subseteq \bigcup _j F_j$. Then $H\subseteq E\times F \subseteq \bigcup _{i,j} E_i\times F_j$. Therefore

$$\begin{aligned} \overline{{\mathsf {H}}}^{s+t}(H)\le & {} \sum _{i,j} \overline{{\mathsf {H}}}_0^{s+t} (E_i\times F_j)\le c \sum _{i,j} \overline{{\mathsf {H}}}^s_{0 }(E_i) \overline{{\mathsf {P}}}^t (F_j).\\\le & {} c \Big (\sum _{i} \overline{{\mathsf {H}}}^s_{0 }(E_i)\Big )\Big (\sum _j\overline{{\mathsf {P}}}^t (F_j)\Big ). \end{aligned}$$

Since the cover $(E_i)$ of E and the cover $(F_j)$ of F were arbitrarily chosen, we deduce that

$$\begin{aligned} \overline{{\mathsf {H}}}^{s+t}(H) \le c \overline{{\mathsf {H}}}^s(E) {\mathsf {P}}^t(F) \le c {\mathsf {H}}^s(E) {\mathsf {P}}^t(F). \end{aligned}$$

This is true for all $H\subseteq E\times F$, hence

$$\begin{aligned} {\mathsf {H}}^{s+t}(E\times F) \le c {\mathsf {H}}^s(E) {\mathsf {P}}^t(F). \end{aligned}$$

As a direct consequence, we obtain the inequality (1.5) which can be obtained directly without the use of the lower and upper Hewitt–Stomberg measures. More precisely, let $E\subseteq \mathbb {R}^d$ and $F\subseteq \mathbb {R}^l$ and assume, without loss of generality, that E and F are bounded. Suppose that we have :

$$\begin{aligned} {\mathsf {b}}(E\times F) \le {\mathsf {b}}(E) + \Delta (F). \end{aligned}$$

(3.3)

Therefore, if $F=\bigcup _n F_n$, we get

$$\begin{aligned} {\mathsf {b}}(E\times F_n) \le {\mathsf {b}}(E) + \Delta (F_n) \quad \text {and then}\quad {\mathsf {b}}(E\times F) \le {\mathsf {b}}(E) + \sup _n\Delta (F_n), \end{aligned}$$

where we have used the $\sigma$-stability of ${\mathsf {b}}$, i.e., $b(\cup F_n) =\sup _n b(F_n)$. Thus, it is sufficient to apply Proposition 1 to get the inequality (1.5). Now, we will prove (3.3). Let $\alpha > {\mathsf {b}}(E)$ and $\beta >\Delta (F)$. Let $\widetilde{E} \subseteq E$ and let $\widetilde{F} \subseteq F$, using (2.3), we can choose $N\in \mathbb {N}$ such that,

$$\begin{aligned} \forall \; n\ge N, \qquad w(2^n, \widetilde{F})2^{-n\beta }< 1. \end{aligned}$$

(3.4)

In addition $\overline{{\mathsf {H}}}^\alpha (\widetilde{E}) = 0$, then we can find a countable sequence $\{\widetilde{E}_i\}_{i\ge 1}$ such that, for each i, we have $\overline{{\mathsf {H}}}^\alpha _0(\widetilde{E}_i) <1/2^{i}$. Thus, there exists a decreasing sequence $\{r_{im}\}_m$ and $M_i \ge N$ such that

$$\begin{aligned} \forall \, m \ge M_i, \qquad N_{r_{im}}(\widetilde{E}_i) (2 r_{im})^\alpha <1/2^{i}. \end{aligned}$$

For each i and $m\ge M_i$, we can find a family of closed balls ${\mathcal {B}}_{im}:= \{B(x_j, r_{im})\}_{j}$ with $x_j\in \widetilde{E}_i$ and $\widetilde{E}_i \subseteq \bigcup _{j} B(x_j, r_{im})$ such that

$$\begin{aligned} \# \{{\mathcal {B}}_{im}\} (2 r_{im})^\alpha <1/2^{i} \end{aligned}$$

(3.5)

Let $\widetilde{B} \in {\mathcal {B}}_{im}$ and choose $n\in \mathbb {N}$ such that $2^{-n-1}< 2 r_{im}\le 2^{-n}$. Let’s consider the set

$$\begin{aligned} {\mathcal {A}}(\widetilde{B}) = \Big \{ \widetilde{B} \times u,\;\; u \;\;\text {is a}\;2^n\text {-mesh}, u\cap F\ne \emptyset \Big \}. \end{aligned}$$

${\mathcal {A}}(\widetilde{B})$ is covered by $(2l)^l w(2^n, F)$ balls centred in $\widetilde{F}$ of diameter $2^{-n}$. Thus, using (3.4) and (3.5), we have

$$\begin{aligned} N_{2^{-n-1}}({\widetilde{E}}_{i} \times \widetilde{F}) (2^{-n})^{\alpha +\beta }\le & {} \# \{{\mathcal {B}}_{im}\} \; (2l)^l w(2^n, F) 2^{-n(\alpha +\beta )}.\\\le & {} (2l)^l 2^\alpha \; 2^{-i}. \end{aligned}$$

Therefore,

$$\begin{aligned} \overline{{\mathsf {H}}}^{\alpha +\beta }(\widetilde{E}\times \widetilde{F}) \le \sum _i \overline{{\mathsf {H}}}_0^{\alpha +\beta }(\widetilde{E}_i\times \widetilde{F}) \le (2l)^l 2^\alpha . \end{aligned}$$

Since this is true of all $\widetilde{E}\subseteq E$ and $\widetilde{F}\subseteq F$, we obtain :

$$\begin{aligned} {{\mathsf {H}}}^{\alpha +\beta }( E\times F) < \infty \end{aligned}$$

and then (3.3) as required.

Corollary 1

Assume that ${\mathsf {b}} (F) ={\mathsf {B}} (F)$ then

$$\begin{aligned} {\mathsf {b}} (E\times F) = {\mathsf {b}} (E)+{\mathsf {b}} (F). \end{aligned}$$

Example 4

In this example, we will consider the Cantor target set H and we will calculate ${\mathsf {b}}(H).$ Let $t>0$. $S: \mathbb {R}^d\rightarrow \mathbb {R}^l$ is a Lipschitz function if

$$\begin{aligned} |S(x) - S(y) | \le c |x - y|, \qquad \forall x, y\in \mathbb {R}^d, \end{aligned}$$

for some $c>0$ and S is a bi-Lipschitz function if

$$\begin{aligned} c_1 |x - y| \le |S(x) - S(y) | \le c_2 |x - y|, \qquad \forall x, y\in \mathbb {R}^d, \end{aligned}$$

for some $c_1, c_2 >0$. Now, we will prove the following elementary lemma.

Lemma 2

Let $t>0$ and S be a Lipschitz function, Then, we have

$$\begin{aligned} {\mathsf {H}} ^t(S(A)) \le c^t \, {\mathsf {H}} ^t(A), \quad \forall A\subseteq \mathbb {R}^d, \end{aligned}$$

(3.6)

and then ${\mathsf {b}} (S(A)) \le {\mathsf {b}} (A)$. Moreover, if S is a bi-Lipschitz transformation then ${\mathsf {b}} (S(A))={\mathsf {b}} (A)$.

Proof

Let $({\mathcal {B}}_i)_{i\in I}$ be a (r/c)-cover of a bounded set $E\subset \mathbb {R}^d$ by centred closed balls, then there exists $({\mathcal {B}'}_i)_{i\in I}$ an r-cover of S(E) by centred closed balls. Therefore,

$$\begin{aligned} N_{r}(S(E)) (2r)^t \le c^t \; N_{r/c} (E) (2r/c)^t. \end{aligned}$$

Thus, letting r tend to 0, we get

$$\begin{aligned} {\overline{{\mathsf {H}}}}^t_0(S(E)) \le c^t\; {\overline{{\mathsf {H}}}}^t_0(E). \end{aligned}$$

Therefore, for any set $F\subseteq A$ such that $F \subseteq \bigcup _i E_i$ with $E_i$ are bounded, we get

$$\begin{aligned} \overline{\mathsf {H}}^t(S(F)) \le \sum _i\overline{\mathsf {H}}^t_0 (S(E_i)) \le c^t \sum _i\overline{\mathsf {H}}^t_0 (E_i). \end{aligned}$$

Since $\bigcup _i E_i$ is an arbitrary cover of F we obtain

$$\begin{aligned} \overline{\mathsf {H}}^t(S(F)) \le c^t\; \overline{\mathsf {H}}^t(F) \le c^t\; {\mathsf {H}}^t(A). \end{aligned}$$

(3.7)

Since this is true for any subset $F\subseteq A$, we get the desired result. $\square$

Let E be the uniform Cantor set. Then, from Example 3, we have ${\mathsf {b}}(E) ={\mathsf {B}}(E) =\frac{\log 2}{\log 3}$. Therefore for any subset F of $\mathbb {R}$ we have

$$\begin{aligned} {\mathsf {b}}(E\times F) = {\mathsf {b}}(E) + {\mathsf {b}}(F). \end{aligned}$$

(3.8)

Now, we consider the plane set given in polar coordinates by

$$\begin{aligned} H = \Big \{ (r, \theta ), \; r\in E, \; \theta \in [0, 2\pi ]\, \Big \} \end{aligned}$$

We will prove that ${\mathsf {b}}(H) = 1+ \frac{\log 2}{\log 3}$. We consider the function $S : \mathbb {R}^2\rightarrow \mathbb {R}^2$ defined as $S((x, y)) = (x\cos y, x\sin y)$. It is clear that S is a Lipschitz mapping and $H= S(E\times [0, 2\pi ])$. It follows, using Lemma 2 and (3.8), that

$$\begin{aligned} {\mathsf {b}}(H)= & {} {\mathsf {b}}(S(E\times [0, 2\pi ])) \le {\mathsf {b}}(E\times [0, 2\pi ])\\= & {} {\mathsf {b}}(E) + {\mathsf {b}}([0, 2\pi ])= 1+ \frac{\log 2}{\log 3}. \end{aligned}$$

On the other hand, if we restrict S to $[\frac{2}{3}, 1]\times [0 , \pi ]$ then S is bi-Lipschitz function on this domain. Since$S\big ( (E\cap [\frac{2}{3}, 1]) \times [0, \pi ]\big ) \subset H$, we deduce that

$$\begin{aligned} {\mathsf {b}}(H)\ge & {} {\mathsf {b}}\Big ( S\big ( (E\cap [\frac{2}{3}, 1]) \times [0, \pi ]\big )\Big )={\mathsf {b}}\Big ((E\cap [\frac{2}{3}, 1]) \times [0 , \pi ]\Big )\\= & {} {\mathsf {b}}\big ( E\cap [\frac{2}{3}, 1] \big ) + {\mathsf {b}}\big ( [0 , \pi ]\big ) = 1+ \frac{\log 2}{\log 3}. \end{aligned}$$

Corollary 2

If $E_1, E_2, \ldots , E_n$ are subsets of $\mathbb {R}$ then

$$\begin{aligned} {\mathsf {b}}(E_1\times \cdots \times E_n)\le n-1 +\min _{1\le i\le n} {\mathsf {b}}(E_i). \end{aligned}$$

3.2 Proof of Inequalities (1.3) and (1.6)

Let $s,t\ge 0$, $E\subseteq \mathbb {R}^d$ and $F\subseteq \mathbb {R}^l$. We will prove that, there exists a constant $c>0$, such that

$$\begin{aligned} {\mathsf {P}}^{s+t}(E\times F) \ge c {\mathsf {H}}^s(E) {\mathsf {P}}^t(F). \end{aligned}$$

For simplicity, we restrict the result to subsets of the plane, though the work extends to higher dimensions without difficulty. Let $\mathcal {Q}$ be any packing of F containing semi-dyadic intervals and $\mathcal {R}$ be any covering of E containing semi-dyadic intervals. We denote

$$\begin{aligned} \mathcal {R}_1= & {} \Big \{u_i \in \mathcal {R}, u_i \; \text {is dyadic and}\;\; \overline{u}_i \cap \overline{u}_j = \emptyset \;\; \text {for}\;\; i\ne j \;\; \Big \}\\ \mathcal {R}_2= & {} \Big \{u_i \in \mathcal {R}, u_i \; \text {is not dyadic and}\;\; \overline{u}_i \cap \overline{u}_j = \emptyset \;\; \text {for}\;\; i\ne j \;\; \Big \}\\ \mathcal {R}_3= & {} \Big \{u_i \in \mathcal {R}, u_i \; \text {is dyadic} \;\; \Big \}\bigcap \mathcal {R}\backslash \mathcal {R}_1\;\;\text {and}\\&\mathcal {R}_4= \Big \{u_i \in \mathcal {R}, u_i \; \text {is not dyadic} \;\; \Big \}\bigcap \mathcal {R}\backslash \mathcal {R}_2. \end{aligned}$$

Clearly, we have each of $\mathcal {R}_i$ is a packing of E and $\mathcal {R}_i\times \mathcal {Q}$ is a packing of $E\times F$. Therefore,

$$\begin{aligned} 4 M_n^{*} (E\times F) 2^{-n(s+t)}\ge \sum _{\mathcal {Q}} 2^{-nt} \Big ( \sum _{\mathcal {R}_1} 2^{-ns} + \sum _{\mathcal {R}_2} 2^{-ns}+\sum _{\mathcal {R}_3} 2^{-ns}+\sum _{\mathcal {R}_4} 2^{-ns}\Big ). \end{aligned}$$

Since $\mathcal {Q}$ is an arbitrary packing of F and $\mathcal {R}=\bigcup _i \mathcal {R}_i$, we obtain

$$\begin{aligned} 4 M_n^{*} (E\times F) 2^{-n(s+t)}\ge M_n^*(F) 2^{-nt} \sum _{\mathcal {R}} 2^{-ns}\ge M_n^*(F) 2^{-nt} \mathbb {N}_n^*(E) 2^{-ns}. \end{aligned}$$

Thus,

$$\begin{aligned} \overline{{\mathsf {P}}}^{*t+s}(E\times F) \ge \frac{1}{4} \overline{{\mathsf {P}}}^{*t}(F) \overline{{\mathsf {H}}}^{*t}(E) \ge \frac{1}{4} {{\mathsf {P}}}^{*t}(F) {{\mathsf {H}}}^{*t}(E). \end{aligned}$$

Finally, using (3.1), we get the desired result. As a direct consequence, we obtain the inequality (1.6) which can be obtained directly using the definitions. More precisely, remark that for each $n\in \mathbb {N}$, we have

$$\begin{aligned} w(2^n, E\times F) = w(2^n, E) \; w(2^n, F). \end{aligned}$$

(3.9)

Therefore, using Proposition 1, we get

$$\begin{aligned} \Delta (E\times F) \ge \Theta (E) +\Delta (F) \ge {\mathsf {b}}(E) +\Delta (F). \end{aligned}$$

(3.10)

In addition, by Proposition 1 and (2.2), we may assume, without loss of generality, that for $\epsilon >0$, there exists $(E_n) \nearrow E$ and $(F_n)\nearrow F$ such that

$$\begin{aligned} \sup \Delta (E_n\times F_n) \le {\mathsf {B}}(E\times F) + \epsilon . \end{aligned}$$

(3.11)

Thus, by (3.10), we have $\Delta (E_n\times F_n) \ge {\mathsf {b}}(E_n) +\Delta (F_n)$ and using (3.11), we get

$$\begin{aligned} {\mathsf {B}}(E\times F) \ge \sup _n\Big ( {\mathsf {b}}(E_n) +\Delta (F_n)\Big )-\epsilon \ge {\mathsf {b}}(E) +{\mathsf {B}}(F) -\epsilon \end{aligned}$$

as required.

3.3 Proof of Inequalities (1.4) and (1.7)

Let $s,t\ge 0$, $E\subseteq \mathbb {R}^d$ and $F\subseteq \mathbb {R}^l$. We will prove that, there exists a constant $c>0$, such that

$$\begin{aligned} {\mathsf {P}}^{s+t}(E\times F) \le c {\mathsf {P}}^s(E) {\mathsf {P}}^t(F). \end{aligned}$$

For simplicity, we restrict the result to subsets of the plane, though the work extends to higher dimensions without difficulty. Let $\mathcal {B}$ be any packing of $E\times F$ containing semi-dyadic squares, each of which is the product of two semi-dyadic intervals. We denote

$$\begin{aligned} \mathcal {R}=\Big \{u_n(x) : \exists v_n(y) \;\; \text {such that }\;\; w_n(x,y)=u_n(x)\times v_n(y) \in {\mathcal {B}}, \;\; x\in E, y\in F\Big \} \end{aligned}$$

and

$$\begin{aligned} \mathcal {Q}=\Big \{v_n(x) : \exists u_n(y) \;\; \text {such that }\;\; w_n(x,y)=u_n(x)\times v_n(y) \in {\mathcal {B}}, \;\; x\in E, y\in F\Big \}. \end{aligned}$$

Now, we consider the subclasses

$$\begin{aligned} \mathcal {R}_1= & {} \Big \{u_n(x) \in \mathcal {R},\;\; u_n(x) \;\; \text {is dyadic} \Big \} \quad \mathcal {Q}_1 =\Big \{v_n(x) \in \mathcal {Q},\;\; v_n(x) \;\; \text {is dyadic} \Big \}\\ \mathcal {R}_2= & {} \Big \{u_n(x) \in \mathcal {R},\;\; u_n(x) \;\; \text {is not dyadic} \Big \} \quad \mathcal {Q}_2 =\Big \{v_n(x) \in \mathcal {Q},\;\; v_n(x) \;\; \text {is not dyadic} \Big \}. \end{aligned}$$

Clearly, we have each of $\mathcal {R}_1, \mathcal {R}_2$ is a packing of E and each of $\mathcal {Q}_1$, $\mathcal {Q}_2$ is a packing of F. Moreover, each square of the packing ${\mathcal {B}}$ is in the collection $\mathcal {R}_i\times \mathcal {Q}_j$, $i, j\in \{1, 2\}$. Therefore,

$$\begin{aligned} \sum _{\mathcal {B}} 2^{-n(t+s)}\le & {} \Big [ \sum _{\mathcal {R}_1} 2^{-nt} + \sum _{\mathcal {R}_2} 2^{-nt}\Big ]\Big [ \sum _{\mathcal {Q}_1} 2^{-ns} + \sum _{\mathcal {Q}_2} 2^{-ns}\Big ]\\\le & {} 4 M_n^*(E) 2^{-nt} M_n^*(F) 2^{-ns}. \end{aligned}$$

Since, this is true for any packing of $E\times F$, we obtain $M_n^*(E\times F) 2^{-n(t+s)} \le 4 M_n^*(E) 2^{-nt} M_n^*(F) 2^{-ns}$ and then

$$\begin{aligned} \overline{{\mathsf {P}}}^{*t+s} (E\times F) \le 4 \overline{{\mathsf {P}}}^{*t}(E)\; \overline{{\mathsf {P}}}^{*s}(F). \end{aligned}$$

Now, for $E\subseteq \bigcup _i E_i$, and $F\subseteq \bigcup _j F_j$, we have :

$$\begin{aligned} {\mathsf {P}}^{*t+s} (E\times F)\le & {} \sum _{i,j} \overline{{\mathsf {P}}}^{*t+s}(E_i\times F_j)\le 4 \sum _{i,j} \overline{{\mathsf {P}}}^{*t}(E_i)\; \overline{{\mathsf {P}}}^{*s}(F_j).\\\le & {} 4 \Big (\sum _i \overline{{\mathsf {P}}}^{*t}(E_i)\Big ) \; \Big ( \sum _j \overline{{\mathsf {P}}}^{*s}(F_j)\Big ). \end{aligned}$$

Since $(E_i)$ is an arbitrary covering of E and $(F_j)$ is an arbirary covering of F, we get ${\mathsf {P}}^{*t+s} (E\times F) \le 4 {\mathsf {P}}^{*t} (E){\mathsf {P}}$ $^{*s} ( F)$. Finally, using (3.1), we obtain the desired result. As a direct consequence, we obtain the inequality (1.7) which can be obtained directly using the definitions. Indeed, using again (3.9), we obtain

$$\begin{aligned} \Delta (E\times F) \le \Delta (E) +\Delta (F). \end{aligned}$$

Hence, by Proposition 1, we get the inequality (1.7).

Corollary 3

If $E_1, E_2, \ldots , E_n$ are subsets of $\mathbb {R}$ then

$$\begin{aligned} {\mathsf {B}}(E_1\times \cdots \times E_n)\le n-1 +\min _{1\le i\le n} {\mathsf {B}}(E_i). \end{aligned}$$

3.4 Proof of Theorem C

To prove the first assertion, we only have to prove that E is regular. Assume, without loss of generality, that the set E is bounded and consider, for $k, n\ge 1$ the set

$$\begin{aligned} E(n, k) = \Big \{ x\in E; \; r< \frac{1}{k} \implies \frac{(2r)^\alpha }{n} < {\mathsf {H}}^\alpha (E\cap B(x,r)\Big \}. \end{aligned}$$

Therefore, under our assumption, we have $E= \bigcup _{n, k} E(n,k)$. Let $n, k\ge 1$ such that $E(n,k)\ne \emptyset$. Then, for each ball B(x, r) such that $x\in E$ and $r< \frac{1}{k}$ we have $(2r)^\alpha < n \; {\mathsf {H}}^\alpha (E\cap B(x,r))$ and then

$$\begin{aligned} M_r(E(n,k))(2r)^\alpha\le & {} \sum _B (2r)^\alpha \le n \sum _B \; {\mathsf {H}}^\alpha (E\cap B(x,r)) \\\le & {} n \; {\mathsf {H}}^\alpha (E). \end{aligned}$$

It follows that $M_r(E(n,k)) \le n (2r)^{-\alpha } {\mathsf {H}}^\alpha (E)$ and then $\Delta (E(n,k)) \le \alpha$. Now, remark that our assumptions imply that ${\mathsf {b}}(E) =\alpha$, on the other hand, we have

$$\begin{aligned} {\mathsf {B}}(E) = \sup _{n, k} {\mathsf {B}}(E(n,k)) \le \sup _{n, k} \Delta (E(n,k))\le \alpha , \end{aligned}$$

as required. Now, similarly, we have F is regular and the second assertion follows.

4 Example

Let E be the triadic Cantor set and then ${\mathsf {b}}(E) =\Theta (E) =\frac{\log 2}{\log 3}<1.$ In this example, we will construct, for any $\gamma$ such that $\Theta (E)<\gamma <1$, a set $F\subset \mathbb {R}$ such that

$$\begin{aligned} {\mathsf {B}}(F)\ge 1-\gamma \end{aligned}$$

(4.1)

and

$$\begin{aligned} {\mathsf {B}}(E\times F)\ge 1 \end{aligned}$$

(4.2)

Thus, since for any $\epsilon >0$, there exists $\gamma \in ({\mathsf {b}}(E), {\mathsf {b}}(E)+\epsilon )$ and

$$\begin{aligned} \gamma -\epsilon + 1-\gamma \le {\mathsf {b}}(E) + {\mathsf {B}}(F) \le {\mathsf {B}}(E\times F)\le 1 \end{aligned}$$

and then by arbitrariness of $\epsilon$ we get ${\mathsf {b}}(E) + {\mathsf {B}}(F) = {\mathsf {B}}(E\times F).$

First, we will give a sufficient condition to get $\Delta ={\mathsf {B}}$ on a compact set. Let E be a compact set on $\mathbb {R}^d$ and let

$$\begin{aligned} \Delta (E, x) =\lim _{r\rightarrow 0} \Delta (E\cap B_x( r)), \end{aligned}$$

where $B_x( r)$ is the open ball with center x and radius r. $\Delta$ is said to be uniform on E if $\Delta (E, x)$ is constant on E. This constant will be denoted by $\Delta ^*(E).$

Theorem 1

Let E be a compact set on $\mathbb {R}^d$. If $\Delta$ is uniform on E then

$$\begin{aligned} \Delta (E) = {\mathsf {B}}(E). \end{aligned}$$

Proof

We will prove that $\Delta (E)= \Delta ^*(E).$ Let $x\in E$ and $\epsilon >0$. Since $\Delta$ is uniform on E, there exists $r_x>0$ such that $\Delta (E\cap B_x( r_x)) \le \Delta ^*(E)+\epsilon$. Now E is compact then we can extract from $\{B_x(r_x)\}_{x\in E}$ a finite subfamily $\{B_i\}_{i=1}^n$ that cover E. Therefore

$$\begin{aligned} \Delta (E) \le \Delta \Big (\bigcup _{i=1}^nB_i\Big )=\sup _{i} \Delta (B_i)\le \Delta ^*(E) +\epsilon . \end{aligned}$$

Since $\epsilon$ is arbitrarily chosen, we get $\Delta (E)\le \Delta ^*(E)$ and hence

$$\begin{aligned} \Delta (E)= \Delta ^*(E) \end{aligned}$$

since the converse inequality is trivial. This equality implies, for each $x\in E$ and $r>0$, that

$$\begin{aligned} \Delta (E\cap B_x(r))=\Delta (E). \end{aligned}$$

Recall (3.2) and assume that ${\mathsf {B}}(E)< \Delta (E)$, then there exists a sequence $(E_n)$ such that $E = \bigcup E_n$ and $\Delta (E_n)< \Delta (E)$ for all $n\ge 1$. Let $x\in E_n$ and $r>0$, then

$$\begin{aligned} \Delta (E_n)=\Delta (E_n \cap B_x(r)) < \Delta (E)= \Delta (E\cap B_x(r)). \end{aligned}$$

Using the fact that $\Delta (F) =\Delta (\overline{F})$ for every set $F\subseteq \mathbb {R}^d$, we conclude that $E_n$ is nowhere dense (that is contained in a closed set with empty interior) in E and then E is meager (or of first category). Therefore E is not a closed set by Baire’s theorem. That is contradiction, then ${\mathsf {B}}(E) =\Delta (E)$. $\square$

4.1 Construction of the Set F

Since $\Theta (E)<\gamma ,$ then for n big enough and for any decreasing sequence $(\delta _n)_{n\ge _0}$ which converges to 0 we have

$$\begin{aligned} N_{\delta _n}(E) < \delta _n^{-\gamma }. \end{aligned}$$

(4.3)

In addition, we may choose $(\delta _n)$ such that

$$\begin{aligned} \delta _0=1 \quad \text {and} \quad \delta _n = a_n^{1/(\gamma -1)} \delta _{n-1} \end{aligned}$$

with $(a_n)$ is a positive sequence of integers such that $a_n\rightarrow \infty$. Hence, for n big enough, we have

$$\begin{aligned} \Big (\frac{ \delta _{n}}{\delta _{n-1}}\Big )^\gamma < \frac{1}{2}. \end{aligned}$$

(4.4)

Now, we will construct by inductively a sequence of sets $(F_n)_{n\ge 1}$. For $n = 1$, let

$$\begin{aligned} b_1= \Big ( \frac{\delta _{0}}{\delta _1}\Big )^{1-\gamma } \end{aligned}$$

and let

$$\begin{aligned} F_1= \bigcup _{j=1}^{b_1} J_{i_1}, \end{aligned}$$

where $J_{i_1}$, $i_1=1, 2,\ldots , b_1$, are closed subintervals of [0, 1] of length $\delta _1$. We arrange those intervals such that they are equally spaced with gaps $\delta _1$ and they are contained in an interval of length at most $2 b_1 \delta _1 = 2 \delta _1^\gamma <1$ by (4.4).

Let $n\ge 2$ and assume that $F_{n-1}$ has been constructed as the union of $\frac{1}{\delta _{n-1}^{1-\gamma }}$ disjoint closed intervals $J_{i_1,\ldots , i_{n-1}}$ each of them of length $\delta _{n-1}$. Let

$$\begin{aligned} b_n= \Big (\frac{\delta _{n-1}}{\delta _n}\Big )^{1-\gamma } , \end{aligned}$$

in each interval of $F_{n-1}$ we construct $b_n$ closed intervals $J_{i_1,\ldots , i_{n}}$ of length $\delta _n$ in such a way that these intervals are equall with gaps $\delta _n$ and they are contained in an interval of length at most $2 b_n \delta _n = 2 \delta _{n-1}^{1-\gamma } \delta _n^\gamma < \delta _{n-1}$ by (4.4).

Therefore, $F_n$ is the union of $b_1\times \cdots \times b_n$ intervals $J_{i_1,\ldots , i_{n}}$ and then

$$\begin{aligned} F_n= \bigcup _{i_1=1}^{b_1} \cdots \bigcup _{i_n=1}^{b_n} J_{i_1,\ldots , i_{n}}, \end{aligned}$$

where $J_{i_1,\ldots , i_{n}}$, are closed subintervals of [0, 1] of length $\delta _n$. Finally let

$$\begin{aligned}F= \bigcap _{n} F_n.\end{aligned}$$

4.2 Proofs of (4.1) and (4.2)

It is clear that, for $r = \delta _n/2$, we have

$$\begin{aligned} N_r(F) = \frac{1}{\delta _{n}^{1-\gamma }} \end{aligned}$$

and then, by definition of $\Delta$ and Theorem 1, we obtain

$$\begin{aligned} {\mathsf {B}}(F) =\Delta (F) \ge 1-\gamma . \end{aligned}$$

Now, we will prove (4.2). Let $0<r<2^{-1/\gamma }$ and choose n such that $\delta _n \le r< \delta _{n-1}$. If   $2 \delta _{n-1}^{1-\gamma } \delta _n^{\gamma }\le r < \delta _{n-1}$, then, using (4.3), there exists a positive constant C such that

$$\begin{aligned} N_r(E\times F ) \le N_{\delta _{n-1}}(E) C \frac{\delta _{n-1}}{r} \frac{1}{\delta _{n}^{1-\gamma }}\le \frac{C}{r}. \end{aligned}$$

Now, we consider the case where $\delta _n \le r< 2 \delta _{n-1}^{1-\gamma } \delta _n^{\gamma },$ then, using again (4.3) we obtain

$$\begin{aligned} N_r(E\times F)\le N_{\delta _n}(E) \frac{1}{\delta _{n}^{1-\gamma }} C \frac{2 \delta _{n-1}^{1-\gamma } \delta _n^{\gamma }}{r}\le \frac{C}{r}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} \Delta (E\times F) = {\mathsf {B}}(E\times F) \le 1. \end{aligned}$$