2 Setup

Let $\mathcal {A}$ be a finite or countable index set and $\mathcal {P}=\{P_a\}_{a\in \mathcal {A}}$ a collection of subintervals in $[0,1]$ with disjoint interiors covering $\mathcal {X}$ . We say $f:\mathcal {X}\rightarrow [0,1]$ is a piecewise expanding Markov map if:

1. (1) for any $a\in \mathcal A$ , $f_a:=f|_{P_a}$ is continuous, injective and $f(P_a)$ a union of elements in  $\mathcal P$ ;

2. (2) there is a uniform constant $\gamma>1$ such that for all $a\in \mathcal A$ , $|Df_a|\ge \gamma $ .

The repeller of f, denoted by $\Lambda $ , is the collection of points with all their forward iterates contained in $\mathcal P$ , namely

$$ \begin{align*}\Lambda:=\bigg\{x\in\mathcal{X}:f^k(x)\in\bigcup_{a\in\mathcal A}P_a\text{ for all }k\ge0 \bigg\}.\end{align*} $$

We study the dynamics of $f:\Lambda \to \Lambda $ together with an ergodic invariant probability measure $\mu $ supported on $\Lambda $ . There is a shift system associated to f: let M be an $\mathcal {A}\times \mathcal {A}$ matrix such that $M_{ab}=1$ if $f(P_a)\cap P_b\neq \emptyset $ and 0 otherwise. Here, f is topologically transitive if for all $a,b\in \mathcal A$ , there exists k such that $M^k_{ab}>0$ . Let $\Sigma $ denote the space of all infinite admissible words, that is,

$$ \begin{align*}\Sigma:=\{{x}=(x_0,x_1,\ldots)\in\mathcal A^{\mathbb N_0}:M_{x_k,x_{k+1}}=1\;\text{ for all } \,k\ge0\}.\end{align*} $$

A natural choice of metric on $\Sigma $ is $d_s(x,y):=2^{-\inf \{j\ge 0:\,x_j\neq y_j\}}$ , and we define the projection map $\pi :\Sigma \to \Lambda $ by

$$ \begin{align*}{x}=\pi(x_0,x_1,\ldots) \quad\text{if and only if } x\in\bigcap_{i=0}^{\infty}f^{-i}P_{x_i}.\end{align*} $$

The dynamics on $\Sigma $ is the left shift $\sigma :\Sigma \to \Sigma $ given by $\sigma (x_0,x_1,\ldots ,)=(x_1,x_2,\ldots )$ , then $\pi $ defines a semi-conjugacy $f\circ \pi =\pi \circ \sigma $ . The corresponding symbolic measure of $\mu $ is given by , that is, for all Borel-measurable sets $B\in \mathcal B([0,1])$ , .

Denote $\mathcal {P}^n:=\bigvee _{j=0}^{n-1}f^{-j}\mathcal {P}$ , each $P\in \mathcal P^n$ corresponds to an n-cylinder in $\Sigma $ : let ${\Sigma _n\subseteq \mathcal A^n}$ denote all finite words of length n and for any $\textbf {i}\in \Sigma $ , the n-cylinder defined by $\textbf {i}$ is

$$ \begin{align*}[\textbf{i}]=[i_0,\ldots,i_{n-1}]:=\{y\in\Sigma:y_j=i_j,\,j=0,\ldots,n-1\}.\end{align*} $$

Then, $\pi [i_0,i_1,\ldots ,i_{n-1}]=\bigcap _{j=0}^{n-1}f^{-j}P_{i_j}=:P_{\textbf {i}}$ . The depth of a cylinder $[\textbf {i}]$ is the length of $\textbf {i}$ .

Furthermore, $(f,\mu )$ is required to have the following mixing property.

Definition 2.1. Say $\mu $ is exponentially $\psi $ -mixing if there are $C_1,\rho>0$ and a monotone decreasing function $\psi (k)\le C_1e^{-\rho k}$ for all $k\in \mathbb N$ , such that the corresponding symbolic measure satisfies: for all $n,k\in \mathbb N$ , $\textbf {i}\in \Sigma _n$ and $\textbf {j}\in \Sigma ^*=\bigcup _{l\ge 1}\Sigma _l$ ,

3 Examples

Theorem 1.4 is applicable to the following systems.

Example 3.1. Finitely branched Gibbs–Markov maps: let f be a topologically transitive piecewise expanding Markov map with $\mathcal A$ finite. Here, f is said to be Gibbs–Markov if for some potential $\phi :\Sigma \to \mathbb R$ which is locally Hölder with respect to the symbolic metric $d_s$ , there exists $G>0$ and $P\in \mathbb R$ such that for all $n\in \mathbb N$ , all $x=(x_0,x_1,\ldots )\in \Sigma $ ,

For maps of this kind, $|Df|$ is uniformly bounded; thus for each ball at scale r, it is possible to approximate any ball with finitely many cylinders of the same depth (see for example the proof of [Reference Jurga and ToddJT, Lemma 3.2]), and by the Gibbs property of , the asymptotic decay rate converges so ${\mathrm {dim}}_M(\mu )$ exists and is finite. Since Gibbs measures are exponentially $\psi $ -mixing (see [Reference BowenBow, Proposition 1.14]), by Theorem 1.1, we have

$$ \begin{align*}\lim_{r\to0}\frac{\log\tau_r(x)}{-\log r}={\mathrm{dim}}_M(\mu)\end{align*} $$

for $\mu $ -a.e. x in the repeller of f.

In the next example, when $r\to 0$ at a polynomial rate, $M_\mu (r)$ decays exponentially; hence, $\overline {\mathrm {dim}}_M(\mu )$ is infinite and the stretched Minkowski dimensions are needed.

Example 3.2. Similar to [Reference Jurga and ToddJT, Example 7.4], consider the following class of infinitely full-branched maps: pick $\kappa>1$ and set $c=\zeta (\kappa )=\sum _{n\in \mathbb N}(1/{n^{\kappa }})$ . Let $a_0=0$ , $a_j=\sum _{j=1}^n(1/{cj^{\kappa }})$ and define f by

$$ \begin{align*} \text{ for all }\,n\in\mathbb N_0=\mathbb N\cup\{0\},\quad f(x)=cn^\kappa(x-a_{n-1})\text{ for }x\in[a_{n-1},a_{n})=:P_{n}. \end{align*} $$

Then, f is an infinitely full-branched affine map and we can associate this map with a full-shift system on $\mathbb N$ : $x=\pi (i_0,i_1,\ldots )$ if for all $j\ge 1$ , $f^j(x)\in P_{i_j}$ .

Let $\omega>1$ and construct the finite Bernoulli measure by

so the push-forward measure has $\mu (P_n)=\omega ^{-n}$ .

Proposition 3.3. For $(f,\mu)$ defined in the example above, $\overline {{\mathrm {dim}}}_M(\mu )=\infty $ , but ${\mathrm {dim}}_M^s(\mu )=1/({\kappa -1})$ .

Proof. For each $r>0$ , the r-ball of minimum measure is found near $1$ . In particular, along the sequence $r_n=({1}/{2c})\sum _{j\ge n}{j^{-\kappa }}\approx {1}/{2c(\kappa -1)n^{\kappa -1}}$ , the ball that realises $M_\mu (r_n)$ is contained in $\bigcup _{j=n}^\infty P_j$ ; hence,

$$ \begin{align*}\omega^{-n}\le M_\mu(r_n)\le \frac{\omega^{-n}}{1-\omega^{-1}}.\end{align*} $$

Therefore,

$$ \begin{align*}\overline{{\mathrm{dim}}}_M(\mu)\ge\limsup_{n\to\infty}\frac{n\log \omega}{(\kappa-1)\log n} =\infty,\end{align*} $$

whereas for all n,

$$ \begin{align*}\frac{\log n+\log(\log\omega+{\log(1-1/\omega)}/{n})}{(\kappa-1)\log n+\log(2c(\kappa-1))}&\le\frac{\log|\log M_\mu(r_n)|}{-\log r_n}\\&\le \frac{\log n+\log\log \omega}{(\kappa-1)\log n+\log (2c(\kappa-1))}.\end{align*} $$

As for all $r>0$ , there is unique $n\in \mathbb N$ such that $r_{n+1}\le r <r_n$ , while $\lim _{n\to \infty }({\log r_{n+1}}/{\log r_n})=1$ . We can conclude with ${\mathrm { dim}}^{s}_M(\mu )={1}/({\kappa -1})$ .

As in [Reference Jurga and ToddJT, Example 7.4], it is very difficult for the system to cover small neighbourhoods of 1, so Theorem 1.1 says $\limsup _{r\to 0}({\log \tau _r(x)}/{-\log r})\ge \overline {\mathrm {dim}}_M(\mu )=\infty $ , but since is Bernoulli and hence $\psi $ -mixing, Theorem 1.4 asserts that

$$ \begin{align*}\lim_{r\to0}\frac{\log\log\tau_r(x)}{-\log r}=\frac1{\kappa-1}\quad \mu\text{-almost everywhere}.\end{align*} $$

4 Proof of Theorem 1.4

The proofs in this section are adapted from those of [Reference Bárány, Jurga and KolossváryBJK, Propositions 3.1 and 3.2]. We will only demonstrate the proofs for Theorem 1.4, that is, the asymptotics are determined by stretched Minkowski dimensions; the proofs for Theorem 1.1 are obtained by replacing all stretched exponential sequences in the proofs below by some exponential sequence, for example, for a given constant $s\in \mathbb R$ , $e^{\pm n^{s}}$ will be replaced by $2^{\pm ns}$ .

Assuming the inequalities in (1.1), we first prove (1.2), which requires the exponentially $\psi $ -mixing condition.

Remark 4.1. Assuming the conditions of Theorem 1.4, we will prove that the statements hold along the subsequence $r_n=n^{-1}$ such that for each $r>0$ , there is a unique $n\in \mathbb N$ with $r_{n+1}<r\le r_n$ , while $\lim _{n\to \infty }({\log r_{n+1}}/{\log r_n})=1$ (if $\overline {{\mathrm {dim}}}_M(\mu )$ or $\underline {{\mathrm {dim}}}_M(\mu )$ are finite, we choose $r_n=2^{-n}$ instead). Since $\log \tau _{r}(x)$ is increasing as $r\to 0$ ,

$$ \begin{align*}\limsup_{n\to\infty}\frac{\log\log\tau_{r_n}(x)}{-\log r_n}=\limsup_{r\to0}\frac{\log\log\tau_r(x)}{-\log r},\end{align*} $$

and similarly for liminfs.

4.1 Proof of (1.2)

Proposition 4.2. Suppose $(f,\mu )$ is exponentially $\psi $ -mixing and the upper stretched Minkowski dimension $\overline {\mathrm {dim}}_M^s(\mu )$ is finite, then for $\mu $ -a.e. $x\in \Lambda $ ,

$$ \begin{align*}\limsup_{n\rightarrow\infty}\frac{\log \log\tau_{r}(x)}{-\log r}\le \overline{\mathrm{dim}}^s_M(\mu).\end{align*} $$

Proof. Let $\varepsilon>0$ and, for simplicity, denote $\overline \alpha :=\overline {{\mathrm {dim}}}^s_M(\mu )$ .

For any finite k-word $\textbf {i}=x_0,\ldots ,x_{k-1}\in \Sigma _k$ , let $\textbf {i}^-=x_0,\ldots ,x_{k-2}$ , that is, $\textbf {i}$ dropping the last letter. Recall that for each $\textbf {i}\in \Sigma ^*$ , $P_{\textbf {i}}=\pi [\textbf {i}]$ , and we define

$$ \begin{align*}\mathcal W_r:=\{\textbf{i}\in\Sigma^*:\,\text{diam}(P_{\textbf{i}})\le r<\text{diam}(P_{\textbf{i}^-})\}.\end{align*} $$

By expansion, for each $n\in \mathbb N$ , the lengths of the words in $\mathcal W_{n^{-1}}$ are bounded from above; hence, we can define

$$ \begin{align*}L(n):=\frac{\log n}{\log\gamma}+1 \ge\max\{|\textbf{i}|:\textbf{i}\in \mathcal W_{n^{-1}}\}.\end{align*} $$

Given $y\in [0,1]$ and $r>0$ such that $B(y,r)\subset \mathrm {supp}(\mu )$ , define the corresponding symbolic balls by

Note that if for some $x\in P_{\textbf {i}}$ and , $d(x,y)\le r+\text {diam}(P_{\textbf {i}})\le 2r$ , then

Let $\mathcal Q_{n}$ be a cover of $\Lambda $ with balls of radius $r_n=1/2n$ . Denote the collection of their centres by $\mathcal Y_n$ and $\#\mathcal Q_n=\#\mathcal Y_n\le n$ . Let $\tau (\mathcal Q_n,x)$ be the minimum time for the orbit of x to have visited each element of $\mathcal Q_n$ at least once,

$$ \begin{align*}\tau(\mathcal Q_n,x):=\min\{k\in\mathbb N:\text{for all }Q\in\mathcal Q_n,\text{there exists }0\le j\le k: f^j(x)\in Q\}.\end{align*} $$

Then, $\tau _{1/n}(x)\le \tau (\mathcal Q_n,x)$ for all n and all x since for all $y\in \Lambda $ , there is $Q\in \mathcal Q_n$ and $j\le \tau (\mathcal Q_n,x)$ such that $f^j(x)\in Q$ and $y\in Q$ ; hence, $d(f^j(x),y)\le 1/n.$ Let $\varepsilon>0$ be an arbitrary number and for each $k\in \mathbb N$ , set $L'(k)=\lceil L(k)+1/{\rho }(k^{\overline \alpha +\varepsilon }+\log C_1)\rceil $ , where $C_1$ , $\rho $ were given in Definition 2.1 and $\lceil t\rceil $ takes the least integer no smaller than t. We have

(4.1) $$ \begin{align} \begin{aligned} &\mu(x:\tau_{1/n}(x)>e^{n^{\overline\alpha+\varepsilon}}L'(4n))\le \mu(x:\tau(\mathcal Q_n,x)>e^{n^{\overline\alpha+\varepsilon}}L'(4n))\\ &\quad=\mu(x:\text{there exists } y\in\mathcal Y_n:f^j(x)\not\in B(y,{1}/{2n})\;\text{for all } j\le e^{n^{\overline\alpha+\varepsilon}}L'({4n}))\\ &\quad\le \mu(x:\text{there exists } y\in\mathcal Y_n:f^{jL'(4n)}(x)\not \in B(y,1/{2n})\;\text{for all } j\le e^{n^{\overline\alpha+\varepsilon}})\\ &\quad=\mu\bigg(\bigcup_{y\in\mathcal Y_n}\bigcap_{j=1}^{e^{n^{\overline\alpha+\varepsilon}}}(f^{-jL'(4n)}B(y,1/2n))^c\bigg)\le \sum_{y\in\mathcal Y_n}\mu\bigg(\bigcap_{j=1}^{e^{n^{\overline\alpha+\varepsilon}}}(f^{-jL'( 4n)} B(y,1/2n))^c\bigg). \end{aligned} \end{align} $$

A cylinder $[\textbf {i}]$ in has depth at most $L(4n)$ , then by our choice of $L'(4n)$ and the exponentially $\psi $ -mixing property of ,

Similar calculations hold for $\mu (\bigcap _{j=1}^{e^{n^{\overline \alpha +\varepsilon }}}(f^{-jL'( 4n)} B(y,1/2n))^c)$ since the compliment of can be written as a countable union of cylinders of depths no greater than $L(4n)$ .

As for all z and all $r>0$ ,

(4.2)

By definition of $\overline \alpha $ , for all n large such that $(\varepsilon /4)\log n\ge (\overline \alpha +\varepsilon /4)\log 4$ , we have

$$ \begin{align*} \log\bigg(-\log M_\mu\bigg(\frac1{4n}\bigg)\bigg)\le (\overline{\alpha}+\varepsilon/4)(\log 4n)\le (\overline\alpha+\varepsilon/2)\log n. \end{align*} $$

So for all $y\in \mathrm {supp}(\mu )$ and all n large enough,

$$ \begin{align*} \mu\bigg(B\bigg(y,\frac1{4n}\bigg)\bigg)\ge e^{-n^{\overline\alpha+\varepsilon/2}}\ge \frac{e^{n^{\varepsilon/2}}}{e^{n^{\overline\alpha+\varepsilon}}}. \end{align*} $$

As for all $u\in \mathbb R$ and all large k, $(1+u/k)^{k}\approx e^u$ , combining (4.1) and (4.2), for some uniform constant $C_2>0$ ,

$$ \begin{align*} &\mu(x:\tau_{1/n}(x)>e^{n^{\overline\alpha+\varepsilon}}L'(4n))\le (1+e^{-n^{\overline\alpha+\varepsilon}})^{e^{n^{\overline\alpha+\varepsilon}}}\sum_{y\in\mathcal{Y}_{k+1}}(1- e^{-n^{\overline\alpha+\varepsilon/2}})^{e^{n^{\overline\alpha+\varepsilon}}}\\ &\quad\le (1+e^{-n^{\overline\alpha+\varepsilon}})^{e^{n^{\overline\alpha+\varepsilon}}}n\bigg(1-\frac{e^{n^{\varepsilon/2}}}{e^{n^{\overline\alpha+\varepsilon}}}\bigg)^{e^{n^{\overline\alpha+\varepsilon}}}\le C_2\exp(\log n-{e^{n^{\varepsilon/2}}}). \end{align*} $$

The last term is clearly summable over n, then by Borel–Cantelli, for all n large enough, $\tau _{1/n}(x)\le e^{n^{\overline \alpha +\varepsilon }}L'(4n)$ . Since $\log L'(4n)\approx (\overline \alpha +\varepsilon )\log n\ll n^{\overline \alpha +\varepsilon }$ , we have for $\mu $ -a.e. ${x\in \Lambda} $ ,

$$ \begin{align*}\limsup_{n\to\infty}\frac{\log\log\tau_{1/n}(x)}{\log n}\le\limsup_{n\to\infty}\frac{\log\log (e^{n^{\overline\alpha+\varepsilon}}L'(4n))}{\log n}\le \overline{\alpha}+\varepsilon.\end{align*} $$

By Remark 4.1, this upper bound for $\limsup $ holds for all sequences decreasing to 0, and as $\varepsilon>0$ was arbitrary, we can conclude that for $\mu $ -a.e. $x\in \Lambda $ ,

$$ \begin{align*}\limsup_{r\to0}\frac{\log\log\tau_r(x)}{-\log r}=\limsup_{n\to\infty}\frac{\log\log\tau_{1/n}}{\log n}\le \overline{\alpha}.\\[-42pt] \end{align*} $$

Proposition 4.3. Suppose $(f,\mu )$ is exponentially $\psi $ -mixing and the lower stretched Minkowski dimension of $\mu $ , $\underline {\mathrm {dim}}_M^s(\mu )$ , is finite, then for $\mu $ -a.e. $x\in \Lambda $ ,

$$ \begin{align*}\liminf_{r\to0}\frac{\log\log \tau_r(x)}{-\log r}\le \underline{{\mathrm{dim}}}_M^s(\mu).\end{align*} $$

Proof. Again, for simplicity, denote $\underline \alpha :=\underline {{\mathrm {dim}}}^s_M(\mu )$ . Let $\varepsilon>0$ and by definition of liminf, there is a subsequence $\{n_k\}_k\to \infty $ such that for all k,

$$ \begin{align*}\frac{\log (-\log M_\mu(1/n_k))}{\log n_k }\le \underline{\alpha}+\varepsilon,\end{align*} $$

then repeating the proof of Proposition 4.2 by replacing n by $n_k$ everywhere, one gets that for $\mu $ -a.e. x,

$$ \begin{align*}\liminf_{k\to\infty}\frac{\log\log\tau_{1/n_k}(x)}{\log n_k}\le\underline{\alpha}+\varepsilon.\end{align*} $$

Again sending $\varepsilon \to 0$ and using the fact that liminf over the entire sequence is no greater than the liminf along any subsequence, the proposition is proved.

4.2 Proof of the inequalities (1.1)

Proposition 4.4. For $\mu $ -a.e. $x\in \Lambda $ ,

$$ \begin{align*}\liminf_{n\to\infty}\frac{\log\log \tau_{r}(x)}{-\log r}\ge\underline{\mathrm{dim}}^s_M(\mu).\end{align*} $$

Proof. We continue to use the notation $\underline \alpha =\underline {\mathrm {dim}}^s_M(\mu )$ . Let $\varepsilon>0$ be arbitrary and by definition of $\underline \alpha $ for all large n, there exists $y_n\in \mathrm {supp} (\mu )$ such that $\mu (B(y_n,1/n))\le e^{-n^{\underline \alpha -\varepsilon }}$ . Let

$$ \begin{align*}T(x,y,r):=\inf\{j\ge0:f^j(x)\in B(y,r)\},\end{align*} $$

so for all $n\in \mathbb N$ and all x, $\tau _{1/n}(x)\ge T(x,y_n,1/n)$ . Then, by invariance,

$$ \begin{align*} &\mu(x:\tau_{1/n}(x)< e^{n^{\underline\alpha-\varepsilon}}/n^2)\le \mu(x:T(x,y_n,1/n)< e^{n^{\underline\alpha-\varepsilon}}/n^2)\\ &\quad=\mu(x:\text{there exists }\,0\le j< e^{n^{\underline\alpha-\varepsilon}}/n^2:\,f^j(x)\in B(y_n,1/n))\\&\quad\le \sum_{j=0}^{ e^{n^{\underline\alpha-\varepsilon}}/n^2-1 }\mu(x:f^j(x)\in B(y_n,1/n))\\ &\quad=\sum_{j=0}^{e^{n^{\underline\alpha-\varepsilon}}/n^2-1}\mu\bigg(f^{-j}B\bigg(y_n,\frac1n\bigg)\bigg)\le\frac{ e^{n^{\underline\alpha-\varepsilon}}}{n^2} e^{-n^{\overline\alpha-\varepsilon}}=\frac1{n^2}, \end{align*} $$

which is summable. By Borel–Cantelli, since $2\log n\ll n^{\underline \alpha -\varepsilon }$ , for $\mu $ -a.e. x,

$$ \begin{align*}\liminf_{n\to\infty}\frac{\log\log \tau_{1/n}(x)}{\log n}\ge\underline\alpha-\varepsilon.\end{align*} $$

Since $\varepsilon>0$ is arbitrarily small, the proposition is proved.

Similar to Propositions 4.2 and 4.3, we get the following proposition.

Proposition 4.5. For $\mu $ -a.e. $x\in \Lambda $ ,

$$ \begin{align*}\limsup_{r\to0}\frac{\log\log \tau_r(x)}{-\log r}\ge\overline{{\mathrm{dim}}}_M^s(\mu).\end{align*} $$

Proof. Let $\varepsilon>0$ , then by definition of limsup, there exists a subsequence $\{n_k\}_k\to \infty $ such that for all k,

$$ \begin{align*}\frac{\log \log(-M_\mu(1/n_k))}{\log n_k}\ge\overline{\alpha}-\varepsilon.\end{align*} $$

Then, repeating the proof of Proposition 4.4 along $\{n_k\}_k$ , one gets that for $\mu $ -a.e. x:

$$ \begin{align*}\limsup_{k\to\infty}\frac{\log\log\tau_{1/n_k}(x)}{\log n_k}\ge \overline{\alpha}-\varepsilon.\end{align*} $$

As $\varepsilon $ can be arbitrarily small,

$$ \begin{align*}\limsup_{r\to0}\frac{\log\log\tau_r(x)}{-\log r}\ge\limsup_{k\to\infty}\frac{\log\log\tau_{1/n_k}(x)}{\log n_k}\ge\overline{\alpha}.\\[-42pt] \end{align*} $$

5 Irrational rotations

The proof of (1.2) requires an exponentially $\psi $ -mixing rate which is a strong mixing condition, and it is natural to ask if the same asymptotic growth in Theorem 1.4 remains the same under different mixing conditions, for example, exponentially $\phi $ -mixing and $\alpha $ -mixing, or even polynomial $\psi $ -mixing. Although these questions are unresolved, in this section, we will show that the limsup and liminf of the asymptotic growth rate can differ if the system is not mixing at all.

Let $\theta \in (0,1)$ be an irrational number and define $T(x)=T_\theta (x)=x+\theta \text { (mod 1)}$ . Denote the one-dimensional Lebesgue measure on $[0,1)$ by $\mu $ , then $(T,\mu )$ is an ergodic probability preserving system with ${\mathrm {dim}}_M(\mu )=1$ .

Definition 5.1. For a given irrational number $\theta $ , the type of $T_\theta $ is given by the following number:

$$ \begin{align*}\eta=\eta(\theta):=\sup\Big\{\beta:\liminf_{n\to\infty}n^{\beta}\|n\theta\|=0\Big\},\end{align*} $$

where for every $r\in \mathbb R$ , .

For any irrational number $\theta \in (0,1)$ , there is a unique continued fraction expansion

$$ \begin{align*}\theta=[a_1,a_2,\ldots]:=\frac{1}{a_1+\frac{1}{a_2+\cdots}},\end{align*} $$

where $a_i\ge 1$ for all $i\ge 1$ . Set $p_0=0$ and $q_0=1$ , and for $i\ge 1$ , choose $p_i,q_i\in \mathbb N$ coprime such that

$$ \begin{align*}\frac{p_i}{q_i}=[a_1,\ldots,a_i]=\frac1{a_1+\frac1{\cdots\frac1{a_i}}}.\end{align*} $$

Definition 5.3. The $a_i$ terms are called the ith partial quotient and $p_i/q_i$ the ith convergent. In particular (see [Reference KhintchineK]),

$$ \begin{align*}\eta(\theta)=\limsup_{n\to\infty}\frac{\log q_{n+1}}{\log q_n}.\end{align*} $$

Theorem 5.4. For any irrational rotation $T_\theta $ ,

$$ \begin{align*}\liminf_{r\to0}\frac{\log \tau_r(x)}{-\log r}={\mathrm{dim}}_M(\mu)=1\le\eta(\theta)=\limsup_{r\to0}\frac{\log \tau_r(x)}{-\log r} \mu\text{-a.e.}\end{align*} $$

By Remark 5.2, there exists irrational rotations such that the asymptotic cover time does not converge. The proof of this theorem relies on the algebraic properties of $\eta (\theta )$ . For simplicity, we fix $\theta $ and write $\eta $ from now on.

The following propositions use results given in [Reference Kim and SeoKS, Propositions 6 and 10].

Proposition 5.6. For $\mu $ -a.e. x,

(5.1) $$ \begin{align} \limsup_{r\to0}\frac{\log \tau_r(x)}{-\log r}\ge \eta. \end{align} $$

Proof. First, it is easy to see that for all $r>0$ and all $x,y\in [0,1)$ , by the nature of rotation, $\tau _r(x)=\tau _r(y)$ . In particular, $\tau _r(x)=\tau _r(Tx)$ ; hence, the function $x\mapsto \limsup _{r\to 0}({\log \tau _r(x)}/{-\log r})$ is T invariant; therefore, constant $\mu $ -almost everywhere by ergodicity of $\mu $ .

By [Reference Kim and SeoKS, Proposition 10], for almost every $x,y$ ,

$$ \begin{align*}\limsup_{r\to0}\frac{\log W_{B(y,r)}(x)}{-\log r}\ge \eta,\end{align*} $$

where $W_E(x):=\inf \{n\ge 1:T^nx\in E\}$ denotes the waiting time of x before visiting E. Hence, there exists a set of strictly positive measures consisting of points that satisfy

$$ \begin{align*}\limsup_{r\to0}\frac{\log \tau_r(x)}{-\log r}\ge \limsup_{r\to0}\frac{\log W_{B(y,r)}(x)}{-\log r}\ge \eta,\end{align*} $$

since for all $y\in [0,1)$ , $\tau _r(x)\ge W_{B(y,r)}(x)$ . As $\limsup _{r\to 0}({\log \tau _r(x)}/{-\log r})$ is $\mu $ -almost everywhere constant, the inequality above holds for $\mu $ -a.e. x and hence the proposition is proved.

Proposition 5.7. For $\mu $ -a.e. x,

$$ \begin{align*}\limsup_{r\to0}\frac{\log \tau_r(x)}{-\log r}\le \eta.\end{align*} $$

Proof. Let $\mathcal Q_n:=\{[2^{-n}j,2^{-n}(j+1)):j=0,\ldots ,2^n-1\}$ and set $\tau ({\mathcal Q_n},x)$ as the minimum time for x to have visited each element of $\mathcal Q_n$ . Again, we have $\tau _{2^{-n+1}}(x)\le \tau (\mathcal Q_n,x)$ for all x. By Lemma 5.5(a) and (c), $\{\|q_i\theta \|\}_i$ is a decreasing sequence, and for each $n\in \mathbb N$ , there exists a minimal j such that $\|q_{j}\theta \|< 2^{-n}\le \|q_{j-1}\theta \|$ , write $j=j_n$ .

By [Reference Kim and SeoKS, Proposition 6] for all n, there is $\mu (W_{[0,2^{-n})}>q_{j_n}+q_{j_n-1})=0$ . Notice that for all $a,b\in [0,1)$ ,

(5.2) $$ \begin{align} \mu\{W_{[a,a+b)}(x)=k\}=\mu\{\{x:W_{[0,b)}(x)=k\}+a\}=\mu\{W_{[0,b)}(x)=k\}, \end{align} $$

as $\mu =Leb$ is translation invariant. Then, by (5.2),

$$ \begin{align*} \begin{split} &\mu\{\tau({\mathcal Q_n},x)>q_{j_n}+q_{j_n-1}\}=\mu\{x:\text{for all } Q\in\mathcal Q_{n}:\,W_Q(x)>q_{j_n}+q_{j_n-1}\}\\ &\quad=\mu\bigg(x:\bigcup_{Q\in\mathcal Q_{n}}\{W_Q(x)>q_{j_n-1}+q_{j_n}\}\bigg)\le \sum_{Q\in\mathcal Q_{n}}\mu(W_Q>q_{j_n-1}+q_{j_n})\\ &\quad=\sum_{j=0}^{2^n-1}\mu(W_{[2^{-n}j,2^{-n}(j+1))}>q_{j_n}+q_{j_n-1})=\sum_{j=0}^{2^n-1}\mu(W_{[0,2^{-n})}>q_{j_n}+q_{j_n-1})=0. \end{split} \end{align*} $$

Hence, by Borel–Cantelli, for all n large enough, $\tau _{2^{-n+1}}(x)\le (q_{j_n}+q_{j_n-1})$ for $\mu $ -a.e. $x\in [0,1)$ .

Let $\varepsilon>0$ , and by Lemma 5.5(b) and (d), there exists $C_\varepsilon $ such that

$$ \begin{align*} \log(q_{j_n}+q_{j_n-1})\le \log( 2q_{j_n})\le\log\frac2{\|q_{j_n}\theta\|}\le(\eta+\varepsilon)\log q_{j_n}+\log2-\log C_\varepsilon. \end{align*} $$

Again by Lemma 5.5 and our choice of $j_n$ , for $\mu $ -a.e. x and all n large enough,

$$ \begin{align*} \log\tau_{2^{-n+1}}(x)&\le \log(q_{j_n}+q_{j_n-1})\lesssim(\eta+\varepsilon)\log q_{j_n}\\&\le-(\eta+\varepsilon)\log\|q_{j_n-1}\theta\|\le (\eta+\varepsilon)n\log 2, \end{align*} $$

where $a\lesssim b$ means $a\le b$ up to a uniform constant. Hence, $\limsup _{n\to \infty }(({\log \tau _{2^{-n}}(x)})/ {n\log 2})\le \eta +\varepsilon $ for $\mu $ -a.e. x. Again, since for each $r<0$ there is a unique $n\in \mathbb N$ for which $2^{-n}<r\le 2^{-n+1}$ , we can apply the subsequence trick again. As $\varepsilon>0$ is arbitrarily small, the proposition is proved.

Proposition 5.8. For $\mu $ -a.e. $x\in [0,1)$ ,

$$ \begin{align*}\liminf_{r\to0}\frac{\log \tau_r(x)}{-\log r}= 1.\end{align*} $$

Proof. Let $\varepsilon>0$ , and using the same arguments as in the previous proof, that is, the cover time is greater than the hitting time of the ball of smallest measure at scale r, then along the sequence $r_n=2^{-(n+1)}$ , one gets for all $[a-r_n,a+r_n)\subset [0,1)$ ,

$$ \begin{align*} &\sum_{n\ge1}\mu(\tau_{r_n}(x)< 2^{n(1-\varepsilon)})\le \sum_{n\ge1}\mu(W_{[a-2^{-n-1},a+2^{-n-1})}(x)< 2^{n(1-\varepsilon)})\\ &\quad\le \sum_{n\ge1}\sum_{k=0}^{2^{n(1-\varepsilon)}}\mu(T^{-k}[a-2^{-n-1},a+2^{-n-1}))=\sum_{n\ge1}2^{n(1-\varepsilon)}2^{-n}=\sum_{n\ge1}2^{-\varepsilon n}<\infty. \end{align*} $$

Since for each r there is a unique n such that $r_n\kern1.4pt{<}\kern1.4pt r \kern1.4pt{\le}\kern1.4pt r_{n-1}$ while ${\lim _n({\log r_n}/{\log r_{n-1}})\kern1.4pt{=}\kern1.4pt1}$ , so by Borel–Cantelli,

$$ \begin{align*}\liminf_{r\to0}\frac{\log\tau_r(x)}{-\log r}=\liminf_{n\to\infty}\frac{\log \tau_{2^{-n}}(x)}{n\log 2}\ge1-\varepsilon,\end{align*} $$

and as $\varepsilon $ is arbitrarily small, the lower bound is proved.

For the upper bound of liminf, recall that $\tau ({\mathcal Q_n},x)\ge \tau _{2^{-n}}(x)$ . We can repeat the proof of Proposition 5.7, apart from that this time, we choose $\{2^{-n_i}\}_i$ according to $\{q_i\}_{i\in \mathbb N}$ : for each i, choose $n_i\in \mathbb N$ to be the smallest number such that

$$ \begin{align*}\|q_{i+1}\theta\|< 2^{-n_i}\le \|q_{i}\theta\|.\end{align*} $$

Hence, as in Proposition 5.7,

$$ \begin{align*} \mu(\tau({\mathcal Q_{n_i}},x)>q_{i+1}+q_i)\le \sum_{Q\in\mathcal Q_{n_i}}\mu(W_Q>q_{i+1}+q_i)=0. \end{align*} $$

Again by Lemma 5.5(b), $q_{i+1}+q_i\le 2q_{i+1}\le (2/{\|q_i\theta \|})<2^{n_i+1}$ by our choice of $n_i$ , so $\lim _{i\to \infty }({\log (q_i+q_{i+1})}/{n_i\log 2})\le 1$ ; therefore, for $\mu $ -a.e. x,

$$ \begin{align*}\liminf_{r\to0}\frac{\log\tau_r(x)}{-\log r}\le \liminf_{i\to\infty}\frac{\log \tau_{2^{-n_i}}(x)}{n_i\log2}\le \liminf_{i\to\infty}\frac{\log \tau({\mathcal Q_{n_i}},x)}{n_i\log2}\le 1.\\[-42pt] \end{align*} $$

6 Cover time for flows

In this section, we prove results analogous to Theorem 1.1 regarding cover times for a class of flows similar to those discussed in [Reference Rousseau and ToddRT, §4].

Let $\{f_t\}_{t\in \mathbb R}$ be a flow on a metric space $(\mathcal {X},d_{\mathcal {X}})$ preserving an ergodic measure $\nu $ , that is, $\nu (f_t^{-1}A)=\nu (A)$ for every $t\ge 0$ and A measurable. Define the cover time of x at scale r by

$$ \begin{align*}\tau_r(x):=\inf\{T>0:\text{for all } y\in\Omega, \text{ there exists } t\le T:\, d(f_t(x),y)<r\}.\end{align*} $$

We will assume the existence of a Poincaré section $Y\subset \mathcal {X}$ , and let $R_ Y(x)$ denote the first hitting time to Y, that is, $R_Y(x):=\inf \{t>0:f_t(x)\in Y\}$ , with $\overline R:=\int R_Yd\nu <\infty $ . Define the Poincaré map by $(Y,F,\mu )$ , where $F=f_{R_Y}$ and $\mu $ is the induced measure on Y given by $\mu =(1/{\overline {R}})\nu |_Y$ . Additionally, assume the following conditions are satisfied:

1. (H1) ${\mathrm {dim}}_M(\mu )$ exists and is finite for $(F,\mu )$ ;

2. (H2) $(Y,F,\mu )$ is Gibbs–Markov, so Theorem 1.1 is applicable for $\mu $ -a.e. $y\in Y$ ;

3. (H3) $\{f_t\}_t$ has bounded speed: there exists $K>0$ such that for all $t>0$ , $d(f_s(x), f_{s+t} (x))<Kt$ ;

4. (H4) $\{f_t\}_t$ is topologically mixing and there exists $T_1>0$ such that

   (6.1) $$ \begin{align} \bigcup_{0<t\le T_1}f_t(Y)=\mathcal{X}; \end{align} $$

5. (H5) there exists

   $$ \begin{align*}C_f&:=\sup\{\text{diam}(f_t(I))/\text{diam}(I):I\text{ an interval contained in } Y, 0<t\le T_1\}\\&\quad\,\in(0,\infty).\end{align*} $$

Remark 6.1. The last condition is satisfied when condition (H3) holds and the flow is, for example, Lipschitz, that is, there exists $L>0$ such that for all $x,y\in \mathcal {X}$ ,

$$ \begin{align*}d_{\mathcal{X}}(f_t(x),f_t(y))\le L^td_{\mathcal{X}}(x,y).\end{align*} $$

Theorem 6.2. Let $(f_t,\nu )$ be a measure-preserving flow satisfying conditions (H1)–(H5) and $\underline {\mathrm {dim}}_{M}(\nu )>1$ , then for $\nu $ -a.e. $x\in \Omega $ ,

(6.2) $$ \begin{align}\liminf_{r\to0}\frac{\log\tau_r(x)}{-\log r}\ge \underline{{\mathrm{dim}}}_M(\nu)-1.\end{align} $$

Furthermore, if $\overline {\mathrm {dim}}_M(\nu )={\mathrm {dim}}_M(\mu )+1$ ,

(6.3) $$ \begin{align}\limsup_{r\to0}\frac{\log \tau_r(x)}{-\log r}\le\overline{\mathrm{dim}}_M(\mu)\hspace{2mm} \nu\text{-a.e}. \end{align} $$

Proof of (6.2)

This proof is analogous to those of Proposition 4.3 and [Reference Rousseau and ToddRT, Theorem 4.1].

Fix some $y\in \Omega $ and $r>0$ , and consider the random variable

$$ \begin{align*}S_{T,r}(x):=\int_0^T\mathbf{1}_{B(y,r)}(f_t(x))\,dt.\end{align*} $$

Observe that by the bounded speed property, for all $T>r/K$ ,

$$ \begin{align*}\{x:\text{there exists } 0\le t\le T \text{ such that }\,d(f_t(x),y))<r\}\subset\{S_{2T,2r}(x)>r/K\},\end{align*} $$

since if $d(f_s(x),y)<r$ for some s, then for all $t<r/K$ , $d(f_{t+s}(x),y)<2r$ . Also set

$$ \begin{align*}T(x,y,r):=\inf\{t\ge0:f_t(x)\in B(y,r)\},\end{align*} $$

and similarly for all $r>0$ and all $x,z$ , $\tau _{r}(x)\ge T(x,y,r)$ .

Let $\varepsilon>0$ be arbitrary and by definition of $\underline \alpha $ for all large $n\in \mathbb N$ , there exists $y_n\in \Omega $ such that $\nu (B(y_n,2^{-n}))\le 2^{-n(\underline \alpha -\varepsilon )}$ . By Markov’s inequality, for some $\mathcal T_n>0$ to be decided later,

$$ \begin{align*} &\nu(x:\tau_{2^{-n}}(x)< \mathcal T_n)\le \nu(x:T(x,y_n,2^{-n})< \mathcal T_n)\\&\quad=\nu(x:\text{there exists }0\le t< \mathcal T_n:\,f_t(x)\in B(y_n,2^{-n}))\\ &\quad\le \nu(x:S_{2\mathcal T_n,2^{-n+1}}(x)>2^{-n}/K)\le K2^n\int_0^{2\mathcal T_n}\int\mathbf{1}_{B(y_{n-1},2^{-n+1})}(f_t(x))\,d\nu(x)\,dt\\ &\quad\le K2^{n+1}\mathcal T_n\nu(B(y_{n-1},2^{-n+1}))\le 4K\mathcal T_n2^{-(n-1)(\underline{\alpha}-\varepsilon-1)}. \end{align*} $$

Choosing $\mathcal T_n=2^{(n-1)(\underline \alpha -\varepsilon -1)}/n^2$ , the last term above is summable along n; hence, by Borel–Cantelli, for $\nu $ -a.e. x

$$ \begin{align*}\liminf_{r\to0}\frac{\log \tau_r(x)}{-\log r}\ge\liminf_{n\to\infty}\frac{\log\mathcal T_n}{n\log2}=\underline\alpha-1-\varepsilon.\end{align*} $$

Since $\varepsilon>0$ was arbitrarily small, the lower bound is $\underline \alpha -1$ , and by Remark 4.1, the proposition is proved.

Note that the proof of lower bound is independent of the existence or mixing properties of the Poincaré map $(Y,F,\mu )$ . For upper bound, we first prove that the cover time of the Poincaré F in Y is comparable to the cover time of the flow.

Lemma 6.3. Define

$$ \begin{align*}\tau_r^F(x):=\min\{n\in\mathbb N_0:\text{for all } y\in Y,\text{there exists } 0\le j\le n:d(y,F^jx)<r \}.\end{align*} $$

There exists $\unicode{x3bb} =(1/{C_f})$ for $C_f$ defined in condition (H5) such that $\tau _{r}(x)\le T_1+\sum _{j=0}^{\tau _{\unicode{x3bb} r}^F(x)}R_Y(F^jx).$

Proof of (6.3)

Now assume $\overline {\mathrm {dim}}_M(\nu )={\mathrm {dim}}_M(\mu )+1$ . Let $\xi>0$ be arbitrary and define the sets

$$ \begin{align*}U_{\xi,N}:=\{x\in Y:|R_n(x)-n\overline{R}|\le \xi n \text{ for all } n\ge N\},\end{align*} $$

where $R_n(x)=\sum _{j=0}^{n-1}R_Y(F^j(x))$ . By ergodicity, $\lim _N\mu (U_{\xi ,N})=1$ , so for N large, $\nu (U_{\xi ,N})>0$ ; hence, by invariance,

(6.4) $$ \begin{align}\lim_{N\to\infty}\nu\bigg(\bigcup_{t=0}^{\xi N}f_{-t}(U_{\xi,N})\bigg)=1.\end{align} $$

Let $\varepsilon>0$ be arbitrary. By (6.4), one can pick $N^*$ such that for each $\nu $ typical $x\in \mathcal {X}$ , there is some $t^*\le \xi N^*$ such that $f_{t^*}(x)\in Y$ . By Theorem 1.1 applied to the Poincaré map and Lemma 6.3, for all sufficiently small $r>0$ , we have the following two inequalities:

$$ \begin{align*}\frac{\log \tau_{\unicode{x3bb} r}^F(f_{t^*}x)}{-\log\unicode{x3bb} r}\le {\mathrm{dim}}_M(\mu)+\varepsilon,\quad\frac{\log (\tau_{r}(x)-T_1)}{-\log r}\le \frac{\log ((\overline R+\xi)\tau_{\unicode{x3bb} r}^F(f_{t^*}x))}{-\log r}. \end{align*} $$

Then, as $\unicode{x3bb} , \overline R$ are constants and $\varepsilon $ is arbitrary, for $\nu $ -a.e. x,

$$ \begin{align*}\limsup_{r\to0}\frac{\log \tau_r(x)}{-\log r}\le {\mathrm{dim}}_M(\mu)=\overline{\mathrm{dim}}_M(\nu)-1.\\[-42pt] \end{align*} $$

6.1 Example: suspension semi-flows over topological Markov shifts

In this section, we give an example of a flow for which ${\mathrm {dim}}_M(\nu )={\mathrm {dim}}_M(\mu )+1$ is satisfied, so Theorem 6.2 is applicable.

Let $\mathcal {A}$ be a finite alphabet and M an $\mathcal {A}\times \mathcal {A}$ matrix with $\{0,1\}$ entries, we will consider two-sided topological Markov shift systems $(\Sigma ,\sigma ,\phi ,\mu )$ , where

$\sigma $ the usual left shift, $\phi $ a Hölder potential and $\mu $ is the unique Gibbs measure with respect to $\phi $ . We assume that ${\mathrm {dim}}_M(\mu )\in (0,\infty )$ . The natural symbolic metric on $\Sigma $ is $d(x,y)=2^{-x\land y}$ , where

$$ \begin{align*}x\land y=\sup\{k\ge0:x_j=y_j \text{ for all } |j|<k\}.\end{align*} $$

An n-cylinder in this setting is given by $[x_{-(n-1)},\ldots ,x_0,\ldots ,x_{n-1}]:=\{y\in \Sigma , \,y_j=x_j \text {for all } |j|<n\}$ , and it is a well-known fact that balls in $\Sigma $ are precisely the cylinder sets. The left-shift map $\sigma $ is bi-Lipschitz with Lipschitz constant $L=2$ . For a more detailed description of the shift space, see [Reference BowenBow, §1].

Let $\varphi \in L^1(\mu )$ be a positive Lipschitz function, define the space

$$ \begin{align*}Y_\varphi:=\{(x,s)\in \Sigma\times \mathbb R_{\ge0}:0\le s\le \varphi(x)\}{/\sim},\end{align*} $$

where $(x,\varphi (x))\sim (\sigma (x),0)$ for all $x\in I$ . The suspension flow $\Psi $ over $\sigma $ is the function that acts on $Y_\varphi $ by

$$ \begin{align*}\Psi_t(x,s)=(\sigma^k(x),v),\end{align*} $$

where $k,v\ge 0$ are determined by $s+t=v+\sum _{j=0}^{k-1}\varphi (\sigma ^j(x)).$ The invariant measure $\nu $ for the flow $\Psi $ on $Y_\varphi $ satisfies the following: for every $g:Y_\varphi \to \mathbb R$ continuous,

(6.5) $$ \begin{align} \int g\,d\nu=\frac1{\int_\Sigma\varphi \,d\mu}\int_\Sigma\int_0^{\varphi(x)}g(x,s)\,ds\,d\mu(x).\end{align} $$

The metric on $Y_\varphi $ is the Bowen–Walters distance $d_Y$ (see for example [Reference Bowen and WaltersBW]). Define another metric $d_\pi $ on $Y_\varphi $ : for all $(x_i,t_i)_{i=1,2}\in Y_\varphi $ ,

$$ \begin{align*}d_\pi((x_1,t_1),(x_2,t_2)):=\min\left\{\begin{aligned} &d(x,y)+|s-t|,\\ &d(\sigma x,y)+\varphi(x)-s+t,\\ &d(x,\sigma y)+\varphi(y)-t+s \end{aligned}\right\},\end{align*} $$

and the following proposition says $d_\pi $ is comparable to the Bowen–Walters distance.

Proposition 6.4. [Reference Barreira and SaussolBS, Proposition 17]

There exists $c=c_\pi $ such that

$$ \begin{align*}c^{-1}d_\pi((x_1,t_1),(x_2,t_2))\le d_Y((x_1,t_1),(x_2,t_2))\le c\, d_\pi((x_1,t_1),(x_2,t_2)).\end{align*} $$

Then, the Minkowski dimension of the flow-invariant measure $\nu $ is given by the following.

Proposition 6.5. For $(\mu)$ the Gibbs measure with respect to $\phi$ on the two-sided subshift and $\nu$ the flow invariant measure, ${\mathrm {dim}}_M(\nu )={\mathrm {dim}}_M(\mu )+1$ .

Proof. The proof is based on the proof of [Reference Rousseau and ToddRT, Theorem 4.3] for correlation dimensions.

By Proposition 6.4 for all $r>0$ ,

$$ \begin{align*}(B(x,r/2c)\times(s-r/2c,s+r/2c))\cap Y\subset B_Y((x,s),r),\end{align*} $$

where $B_Y$ denotes the ball with respect to the metric $d_Y$ , then for all $(x,s)\in Y_\varphi $ , put $\overline \varphi =\int _\Sigma \varphi \,d\mu $ , then

$$ \begin{align*}\nu(B_Y((x,s),r))\ge \nu\bigg(B(x,r/2c)\times\bigg(s-\frac{r}{2c},s+\frac{r}{2c}\bigg)\bigg),\end{align*} $$

$$ \begin{align*}\frac{\log\nu(B_Y((x,s),r))}{\log r}\le \frac{\log(({r}/{c\overline\varphi})\mu(B(x, r/{2c})))}{\log r}.\end{align*} $$

Hence, $\overline {{\mathrm {dim}}}_M(\nu )\kern1.4pt{=}\kern1.4pt\limsup _{r\to 0}{\log \min _{(x,s)\in \mathrm {supp}(\nu )}\nu (B_Y((x,s),r)}/{\log r})\kern1.4pt{\le}\kern1.4pt {{\mathrm {dim}}_M(\mu )\kern1.4pt{+}1}$ .

For lower bound, define

$$ \begin{align*}B_1:=B(x,c r)\times(s-c r,s+c r),\quad B_2:=B(\sigma x,c r)\times[0,cr),\end{align*} $$

$$ \begin{align*}B_3:=\{(y,t):y\in B(\sigma^{-1}x,2cr)\text{ and } \varphi(y)-cr\le t\le \varphi(y)\}.\end{align*} $$

Then, as in the proof of [Reference Rousseau and ToddRT, Theorem 4.3], $B_Y((x,s),r)\subset (B_1\cup B_2\cup B_3)\cap Y_\varphi $ .

For all $r>0$ and $(x,s)\in Y_\varphi $ by (6.5), and as $\mu $ is $\sigma ,\sigma ^{-1}$ invariant,

$$ \begin{align*} &\nu(B_1\cap Y_\varphi)={2cr}\mu(B(x,cr))/{\overline\varphi},\quad \nu(B_2,Y_\varphi)\le cr\mu(B(x,cr))/\overline\varphi\\ &\nu(B_3\cap Y_\varphi)\le cr\mu(\sigma^{-1}B(x,2cr))/\overline\varphi=cr\mu(B(x,2cr))/\overline\varphi. \end{align*} $$

Therefore,

$$ \begin{align*}\nu(B_Y((x,s),r))\le\frac1{{\overline\varphi}}(3r\mu(B(x,cr))+cr\mu(B(x,2cr))),\end{align*} $$

which is enough to conclude that $\underline {{\mathrm {dim}}}_M(\nu )\ge {\mathrm {dim}}_M(\mu )+1$ .