Assumption 2.1. Assume $f\colon X \to X$ is a Markov map with respect to a partition ${\mathcal P}$ .

Assumption 2.2. There exists a union of partition elements Y so that the first return map $F\colon Y \to Y$ given by

where

is a topologically mixing $C^2$ uniformly expanding Gibbs–Markov map with respect to a refinement ${\mathcal Q}$ of ${\mathcal P}$ . That is, F is Markov with respect to ${\mathcal Q}$ , topologically mixing, and the following additional properties hold.

• • (Finite images) $\mathrm {Card} \{ F (a) \colon a \in {\mathcal Q} \} < \infty $ , that is, there exist $Y_1,\ldots , Y_L$ so that each $Y_i$ is a union of elements of ${\mathcal Q}$ and $\{F(a)\colon a\in {\mathcal Q}\} = \{ Y_1, \ldots , Y_L\}$ .

• • (Bounded distortion) There exist $\theta \in (0,1)$ and $C> 0$ so that the function $\log ({d m}/{dm \circ F})$ is $d_\theta $ -Lipschitz on elements of ${\mathcal Q}$ , where $d_{\theta }$ is the metric and

  

• • ( $C^{ 2 }$ branches) For every $a\in {\mathcal Q}$ , the map $F|_{\mathrm {int}(a)}\colon \mathrm {int}(a) \to F(\mathrm {int}(a))$ is a $C^{ 2 }$ diffeomorphism. (Notice that we do not necessarily assume that F has $C^{ 2 }$ extension to the closure of a.)

• • (Uniform expansion) There exists $\unicode{x3bb}> 1$ such that $\inf _{ x\in Y} |F'(x)|> \unicode{x3bb} $ .

Proof. Let $\{ y_1 < y_2 \cdots < y_{N - 1} < y_{N}\}$ be the collection of end points of the intervals $Y_i$ ordered so that $y_i < y_{i + 1}$ and let ${\mathcal Y}$ denote the partition of Y given by all intervals of the form $[y_{i}, y_{ i + 1 }]$ that intersect Y. As each $Y_i$ is a union of elements of ${\mathcal P}$ , it follows that each element in ${\mathcal Y}$ is also a union of elements of ${\mathcal P}$ .

As each $Y_i$ is a finite union of elements in ${\mathcal Y}$ , f is a bijection on elements of ${\mathcal P}$ and we can refine ${\mathcal P}$ to form a new partition $\widetilde {{\mathcal Q}}$ so that $f(a) \in {\mathcal Y}$ for all $a\in \widetilde {{\mathcal Q}}$ .

Assumption 2.6. There exist constants $d \in {\mathbb N}$ , $\alpha \in (0,1)$ , and $\gamma _1, \ldots , \gamma _d>0$ , and a partition (mod $\mathrm {Leb}$ ) of $X \setminus Y$ into open sets $X_1,\ldots ,X_d$ such that for every $i,j\in \{1,\ldots ,d\}$ :

1. (1) there is a single fixed point $\xi _j \in X_j$ ;

2. (2) for $i \neq j$ , orbits cannot pass from $X_i$ to $X_j$ without first passing through Y, that is, if $f^{ n } (x) \in X_{ i }$ and $f^{ n + m }(x) \in X_{ j }$ for some $i \neq j$ and $m\in {\mathbb N}$ , then there is $k \in \{1,\ldots ,m-1\}$ such that $f^{ n+k }( x) \in Y$ ,

3. (3) $\mu _Y ( \tau ^{(j)}> n ) \sim \gamma _j n^{-\alpha }$ , where

   

Proof. First, notice that

$$ \begin{align*} \frac{ {\overline{\tau}}_{ n + k } }{ \tau_{ n + k } } (ab) = \frac{ {\overline{\tau}}_{ n } (a) + {\overline{\tau}}_{ k } (b) }{ \tau_{ n } (a) + \tau_{ k }(b) }. \end{align*} $$

The inequality in item (3) then follows from (4.7). Also note that $|{\overline {\tau }}_j/\tau _{ j }| \le 1$ . Hence, (4.8) implies

$$ \begin{align*} \bigg|\frac{{\overline{\tau}}_{ n + k } }{ \tau_{ n + k } } (ab) - \frac{ {\overline{\tau}}_n }{ \tau_n } (a)\bigg| = \bigg| \bigg(\frac{{\overline{\tau}}_{ k } (b) }{\tau_{k}(b)} - \frac{{\overline{\tau}}_{n} (a)}{\tau_{n} (a)} \bigg) \frac{\tau_{k}(b)}{\tau_{n} (a) + \tau_{ k }(b)}\bigg| \le 2\frac{\tau_{k}(b)}{\tau_{n} (a) + \tau_{k}(b)} \end{align*} $$

together with the analogous lower bound. This proves item (1). The proof of item (2) is analogous.

Proof. Let us first prove that there exists a random variable $\boldsymbol {Z}$ whose distribution $\boldsymbol Z_{ * } {\mathbb P}$ on ${\mathcal S}$ is equivalent to $\mathrm {Leb}$ and such that the random variables

$$ \begin{align*} x \mapsto ( e_{ n } (x) ( X_{ 1 } ) , \ldots, e_{ n } (x) (X_{ d })) \end{align*} $$

converge in distribution to $\boldsymbol {Z}$ as $n\to \infty $ , that is,

$$ \begin{align*} & \mu\{x\colon ( e_{ n } (x) ( X_{ 1 } ) , \ldots, e_{ n } (x) (X_{ d })) \in A\}\\ & \quad \to {\mathbb P}(\boldsymbol Z\in A) \quad \text{for every }A \text{ so that } {\mathbb P}(\boldsymbol Z\in\partial A)=0. \end{align*} $$

This follows from [Reference SeraSer20, Corollary 4.2 and Theorem 3.3], provided one can check [Reference SeraSer20, Assumptions 2.1, 2.2, and 2.6]. Assumption 2.6 yields [Reference SeraSer20, Assumption 2.1], and then Assumption 2.6(3) together with [Reference SeraSer20, Lemma 2.4] gives [Reference SeraSer20, Assumption 2.2]. Finally, as $F $ is topologically mixing, it is exponentially continued fraction mixing (see, for example, [Reference AaronsonAar97, §4]), which yields [Reference SeraSer20, Assumption 2.6].

Recall that, as in the proof of Theorem 3.1, we have that for all $x\in X$ ,

$$ \begin{align*} \frac{ {\overline{\tau}}_k }{ \tau_k } (x) = ( e_{\tau_k}(x) (X_1), \ldots, e_{\tau_k}(x) (X_d)). \end{align*} $$

Thus, as the distribution of $\boldsymbol Z_\ast {\mathbb P}$ of $\boldsymbol Z$ is equivalent to Lebesgue on ${\mathcal S}$ , we can take $A = B_{\varepsilon / 2 } ( {\overline {p}} )$ in the above to obtain

$$ \begin{align*} \lim_{n \to \infty} \mu \{x \colon {\overline{\tau}}_{ n } (x) / \tau_{ n } (x) \in B_{ \varepsilon/2 } ({\overline{p}}) \} = {\mathbb P} (\boldsymbol Z \in B_{\varepsilon/2} ({\overline{p}}) )> 0. \end{align*} $$

This implies the conclusion of the lemma for the measure $\mu $ . Finally, as (see Remark 2.5) $\mu $ is equivalent to $\mathrm {Leb}$ , this implies the assertion.

Proof. Fix $i,j \in \{ 1, \ldots , L \}$ and fix $a_i \in {\mathcal Q}$ with $a_i \subset Y_i$ . As F is topologically mixing, we can choose $k_{ij} \in {\mathbb N}$ to be such that, for all $n \ge k_{ij}$ ,

$$ \begin{align*} F^n(a_i)\cap Y_j \ne \emptyset. \end{align*} $$

Let . Hence, there is $a_{ij} \in {\mathcal Q}^{k_0} \cap a_i$ such that $F^{k_0}( a_{ij}) \cap Y_j \neq \emptyset $ . However, we know that $F^{k_0}( a_{ij}) \in \{ Y_1, \ldots , Y_L \}$ , Hence, we get $F^{k_0}(a_{ij}) = Y_j$ .

Proof of Proposition 5.2

Given $n \in {\mathbb N}$ , let

and let ${\mathcal B}(n)^\prime \subset {\mathcal B}(n)$ be a finite sub-collection so that $\mathrm {Leb} ( {\mathcal B}(n)^\prime ) \ge \tfrac 12 \mathrm {Leb}( {\mathcal B}(n) )$ . Thus, by Lemma 5.3, there exists some constant $C^\prime> 0$ so that

(5.3) $$ \begin{align} \mathrm{Leb} \,{\mathcal B}(n)^\prime \ge C^\prime\quad \text{for all } n \text{ sufficiently large.} \end{align} $$

Let $k_0$ and $\{ a_{ij} \}_{i,j = 1}^{L}$ be as given by Lemma 5.4. As F is assumed to be a topologically mixing Gibbs–Markov map, each element of ${\mathcal Y}$ is a union of elements of ${\mathcal Q}$ , and the union of all elements in ${\mathcal Y}$ equals Y. So, for any cylinder $b \in {\mathcal Q}^n$ , we know from (5.1) that there exist $p(b),s(b) \in \{ 1, \ldots , L\}$ so that $b \subset Y_{p(b)}$ and $F^n (b) = Y_{s(b)}$ . Thus, for all $i,j\in \{1,\ldots , L\}$ , the cylinder $a_{i p(b)} b a_{s(b) j} \in {\mathcal Q}^{n + 2 k_0}$ is always non-empty and satisfies

$$ \begin{align*} a_{i p(b)} b a_{s(b)j} \subset Y_{i} \quad\text{and} \quad F^{n+2k_0}(a_{ip(b)}ba_{s(b)j}) = Y_j. \end{align*} $$

To choose now our collection, given $n> 2k _0$ , we set

By construction, ${\mathcal A}(n)$ satisfies item (3) of the proposition.

Letting $i = 1, \ldots , L$ ,

$$ \begin{align*} \mathrm{Leb} ( {\mathcal A}(n) \cap Y_i ) & = \sum_{j = 1 }^L \sum_{ b \in {\mathcal B}(n - 2k_0)^\prime} |a_{i p(b)} b a_{s(b)j} | \\ \text{(by}\ (4.5)) \quad & \ge D^{-2} \sum_{j = 1 }^L \sum_{ b \in {\mathcal B}(n - 2k_0)^\prime} |a_{ip(b)}| |b| | a_{s(b)j} | \\ & \ge D^{-2} \bigg(\min_{\ell,j = 1 ,\ldots, L } |a_{\ell,j}|\bigg)^2 \mathrm{Leb} ( {\mathcal B}( n - 2 k_0)^\prime). \end{align*} $$

So, for all n large enough that (5.3) holds, there is a $C> 0$ such that $\mathrm {Leb}({\mathcal A}(n)\cap Y_i )> C$ . This concludes item (2).

Now, consider an arbitrary element $a_{ip} b a_{sj} \in {\mathcal A}(n)$ for some $n \ge 1$ . From Lemma 4.6(2) and the fact that $\tau _n \ge n$ , we know that

(5.4) $$ \begin{align} \bigg|\dfrac{{\overline{\tau}}_{n}}{\tau_{n}} (a_{ip} b a_{sj})-\dfrac{ {\overline{\tau}}_{ n - k_0 } }{ \tau_{ n - k _0 } }( b a_{sj} )\bigg| < 2\dfrac{\tau_{k_0}(a_{ip})}{\tau_{ n - k_0 }( b a_{sj})+\tau_{k_0}(a_{ip})}\le \dfrac{ 2 M }{n}, \end{align} $$

where

Similarly, by Lemma 4.6(1),

(5.5) $$ \begin{align} \bigg|\dfrac{{\overline{\tau}}_{ n - k _0} }{ \tau_{ n - k_0} } (b a_{sj} )- \dfrac{{\overline{\tau}}_{ n - 2 k_0 } }{ \tau_{ n - 2k_0 } }( b )\bigg| < 2\dfrac{\tau_{k_0}(a_{sj})}{\tau_{n - 2k_0}(b)+\tau_{k_0}(a_{sj})}\le \dfrac{ 2 M }{n - k_0}. \end{align} $$

So, as $b \in {\mathcal B}(n - 2 k_0)^\prime \subset {\mathcal B}(n - 2 k_0)$ , we have

(5.6) $$ \begin{align} \frac{ {\overline{\tau}}_n}{ \tau_n } (b) \in B_{\varepsilon/2} ({\overline{p}}). \end{align} $$

Thus, combining (5.4), (5.5), and (5.6), we find that

$$ \begin{align*} {\mathcal A}(n) \subset \{ a \in {\mathcal Q}^n \colon ({\overline{\tau}}_n/\tau_n) (a) \in B_{3\varepsilon/4} ({\overline{p}}) \} \end{align*} $$

provided that n is sufficiently large. This yields item (1) and completes the proof.

Proof. We first claim that

(5.8) $$ \begin{align} \bigg|\frac{{\overline{\tau}}_{ kn }}{ \tau_{ kn }} (a) - {\overline{p}}\bigg| < \frac{3\varepsilon}{4} \quad\text{for every } a\in{\mathcal Q}_{\Lambda({\mathcal A}(n))}^{kn} \text{ and every } k \ge 1. \end{align} $$

Notice that item (1) of Proposition 5.2 implies (5.8) for $k=1$ . Proceeding inductively on k, let $a \in {\mathcal Q}^{(k+1)n}_{\Lambda (n)}$ . By Remark 4.1, we have $a = a_1 \cdots a_ka_{k+1}$ , for $a_i \in {\mathcal A}(n)$ . It follows from Lemma 4.6(3) that

$$ \begin{align*} \dfrac{ {\overline{\tau}}_{(k + 1)n} } { \tau_{ ( k + 1 ) n } }(a) = \dfrac{ {\overline{\tau}}_{kn+n} } { \tau_{ kn+ n } }(a) \le \max\bigg\{ \dfrac{{\overline{\tau}}_{k n} (a_1 \cdots a_{k })}{ \tau_{ k n } ( a_1 \cdots a_{k} ) }, \dfrac{ {\overline{\tau}}_n (a_{k + 1 })}{\tau_n (a_{k + 1} )} \bigg\} \end{align*} $$

together with the analogous lower bound. Hence, from (5.8) and our inductive hypothesis,

$$ \begin{align*} \dfrac{ {\overline{\tau}}_{(k + 1)n} } { \tau_{ ( k + 1 ) n } }(a) \in B_{ 3 \varepsilon / 4 }({\overline{p}}\,), \end{align*} $$

yielding (5.8).

We finally conclude (5.7). As ${\mathcal A}(n)$ only contains finitely many cylinders, the value

is finite. Let $b \in {\mathcal Q}^{ s } \cap {\mathcal A}(n)$ with $s \in \{1,\ldots , n\}$ . It follows from Lemma 4.6(1) that

(5.9) $$ \begin{align} \bigg|\dfrac{ {\overline{\tau}}_{ k n + s } }{ \tau_{ k n + s } } (bc) -\dfrac{{\overline{\tau}}_{ k n }}{ \tau_{ k n }} (b)\bigg| \le \dfrac{2{\overline{\tau}}_s(c)}{ {\overline{\tau}}_{ k n } (b) + {\overline{\tau}}_s(c) } \le 2 \dfrac{M}{ k n + s }. \end{align} $$

Choose so that $2 M / N_0 < \varepsilon / 4$ . For every $\ell> N_0$ and $a \in {\mathcal Q}^\ell _{\Lambda (n)}$ , relation (5.9) together with (5.8) implies that $( {\overline {\tau }}_{ \ell } / \tau _{ \ell } )(a) \in B_{ \varepsilon } ( {\overline {p}} )$ . This proves the lemma.

Proof. Consider the function $h \colon [0,1] \to {\mathbb R}$ given by

Let so that, by (4.2), $h ( s(n) ) = 1$ . It follows from item (2) in Proposition 5.2 that

(5.10) $$ \begin{align} h (0) = \sum_{ a \in {\mathcal A}(n) } |a| = \mathrm{Leb} ( {\mathcal A}(n) )> C, \end{align} $$

and this property holds for all n sufficiently large. We get

$$ \begin{align*} h ' (t) &= - \sum_{ a \in {\mathcal A}(n) } \log |a| \cdot |a|^{ 1 - t }\\ &\ge n \log \unicode{x3bb} \cdot \sum_{ a \in {\mathcal A} (n) } |a|^{ 1 - t } \ge n \log \unicode{x3bb} \cdot \sum_{ a \in {\mathcal A} (n) } |a|\quad\text{ by}\ (4.4)\\ &\ge C n \log \unicode{x3bb}\quad\text{by}\ (5.10) \end{align*} $$

giving that $h'(t) \ge Cn \log \unicode{x3bb} $ . It follows that

$$ \begin{align*} s (n)\le \frac{ 1 - h (0) }{ \| h' \|_{ \infty } } \le \frac{ 1 - C }{ C n \log \unicode{x3bb} }. \end{align*} $$

Using Lemma 4.3, with the above, we get

$$ \begin{align*}\begin{aligned} \dim_{\text{H }} \Lambda (n) &\ge \mathrm{vdim} \Lambda (n) - \frac{D}{n\log\unicode{x3bb}-D} = 1-s(n) - \frac{D}{n\log\unicode{x3bb}-D}\\ & \ge 1- \frac{ 1 - C }{ C n \log \unicode{x3bb} } - \frac{D}{n\log\unicode{x3bb}-D}. \end{aligned}\end{align*} $$

Letting , the assertion follows.

Proof. As m is geometric for $F^n|_{\Lambda (n)}$ , by (4.6), there exists an $E(n)> 0$ so that

(5.11) $$ \begin{align} \dim_{\text{H }} \Lambda (n) + \frac{ E(n) }{ \log |a| } \le \frac{ \log m (a) }{ \log |a| } \le \dim_{\text{H }} \Lambda (n) - \frac{ E(n) }{ \log |a| }. \end{align} $$

Using (4.4), we can choose $N_1 = N_1 ({\mathcal A}(n)) \in {\mathbb N}$ so that for $\ell \ge N_1$ ,

(5.12) $$ \begin{align} \sup_{a\in{\mathcal Q}^\ell}\frac{E(n)}{-\log|a|} \le \frac{E(n)}{\ell \log\unicode{x3bb}} <\varepsilon/2. \end{align} $$

If $n> 2C^\prime / \varepsilon $ , then Lemma 5.6 ensures $\dim _{\text {H }} \Lambda (n) \ge 1 - \varepsilon /2$ . Combining this with (5.11) and (5.12) concludes the proof.

Proof. It is enough to check that for every index $i\ge 0$ for every $a\in {\mathcal A}_{i+1}$ , there exists $a'\in {\mathcal A}_i$ such that $F^{n_i}(a')\supset a$ . This follows from the fact that Proposition 5.1 ensures $F^{n_i} ( {\mathcal A}_i ) = {\mathcal Y}$ . So, every ${\mathcal B}_i$ is non-empty and the intersection in (6.3) is the intersection of a nested sequence of compact sets. This implies the assertion.

Proof. Note that the collection $\{ b_i (x) \colon x \in \Gamma \}$ is finite. We also know that the quantities on the tops of the fractions in (6.13) and (6.14) depend on $k_0,\ldots ,k_{i-1}$ , but not $k_i$ . So, there exists some constant $L_i < +\infty $ which is independent of $k_i$ (but may depend on $k_0,\ldots ,k_{i-1}$ ) such that

$$ \begin{align*} \sup_{x\in\Gamma} |\!\log m ( b_i (x) )| + \widetilde C_i < L_i \quad\text{and} \quad \sup_{x\in\Gamma}|\!\log |b_i (x) || + |\!\log D | < L_i. \end{align*} $$

We know from Remark 4.2 that $|\!\log |a_i(x)|| \ge k_{i} n_i \log \unicode{x3bb} $ . We also know from (6.17) that $|\!\log m_{i} (a_i)| / \log |a_{i} | \in B_{\varepsilon _i} (1)$ . In particular, we must have that

$$ \begin{align*} |\!\log m_{i} (a_i)| \ge (1 - \varepsilon_i ) k_in_i \log \unicode{x3bb}. \end{align*} $$

So, we can choose $k_i$ recursively to grow fast enough so that (6.13) and (6.14) hold.

Proof. Let $x\in \Gamma $ and consider the associated sequence $(a_i)_i$ and $(b_i)_i$ in (6.8) and (6.9). Using the definition of m in (6.4), the bounded distortion of F in (4.4), we get

$$ \begin{align*} \frac{ \log m ( b_{ i + 1} ) }{ \log |b_{ i + 1}| }& = \frac{ \log m ( b_{ i } a_i ) }{ \log |b_{ i } a_i| } \ge \frac{ \log m ( b_i ) + \log m_i (a_i) + \log C_i (b_i) } { \log |b_i| + \log |a_i| - \log D } \\ &\ge\frac{\log m_i (a_i)}{\log |a_i|}\cdot \bigg(\dfrac{1 + ({\log m ( b_i ) + \widetilde{C}_i })/({\log m_i (a_i) })}{1+ ({\log |b_i| - \log D})/({\log |a_i|})}\bigg)\quad\text{(by}\ (6.7)). \end{align*} $$

Using (6.17), together with (6.13) and (6.14), we find the lower bound

$$ \begin{align*} \frac{ \log m ( b_{ i + 1} ) }{ \log |b_{ i + 1}| } \ge (1 - \varepsilon_i)\cdot \frac{ 1 - \varepsilon_i }{ 1 + \varepsilon_i }. \end{align*} $$

Analogously, we obtain the upper bound

$$ \begin{align*} \frac{ \log m ( b_{i + 1} )}{\log |b_{i + 1 }| } \le (1 + \varepsilon_i)\cdot \frac{ 1 + \varepsilon_i }{ 1 - \varepsilon_i }. \end{align*} $$

Taking the limit as $i\to \infty $ concludes the proof.

Proof. Suppose that $(k_i)_i$ is chosen to grow fast enough so that (6.13) and (6.14) hold by Lemma 6.5. The top of the fractions in (6.15) and (6.16) do not depend on the choice of sequence $(k_i)_i$ . However, we know that

$$ \begin{align*} |\!\log |b_i (x)|| \ge k_{i - 1 }n_{ i - 1 }\log \unicode{x3bb}. \end{align*} $$

Moreover, Lemma 6.6 tells us that then,

$$ \begin{align*} |\!\log |b_i (x)|| \ge ( 1 - \gamma \varepsilon_{ i - 1 } ) k_{i - 1 }n_{ i - 1 }\log \unicode{x3bb}. \end{align*} $$

Choosing $(k_i)_{i}$ to grow sufficiently quickly ensures that (6.15) and (6.16) hold.

Proof of Proposition 6.4

Conditions (6.10) and (6.12) are satisfied whenever $k_i$ is chosen to be large enough. Moreover, the top of the fraction in (6.11) depends only on $k_0,\ldots ,k_{i-1}$ and not on $k_i$ . These observations together with Lemmas 6.5 and 6.7 ensure that $(k_i)_i$ can be chosen such that (6.10)–(6.16) hold. This completes the proof.

Proof. Let $x \in \Gamma $ . By Corollary 3.2, it is enough to show that ${\mathcal T}(x) = {\overline {p}}$ .

Recall the definition of the associated cylinders $a_{i,s} \in {\mathcal Q}_{\Lambda _i}^{s}$ , $a_{i} \in {\mathcal Q}_{\Lambda _i}^{k_in_i}$ , and $b_i \in {\mathcal B}_i$ in (6.8) and (6.9). Recall from Proposition 5.1 that

(6.18) $$ \begin{align} | ( {\overline{\tau}}_{ s } / \tau_{ s } )(a_{i,s}) - {\overline{p}} |<\varepsilon_i \quad\text{for all } N_i\le s \le k_in_i. \end{align} $$

We also make use of the following inequalities which come from the definition (6.6) and the fact that $\tau \ge 1$ :

(6.19) $$ \begin{align} t_i \le \tau_{t_i} ( b_i ) \le \sum_{ j = 0 }^{ i - 1 } M_j k_jn_j \quad\text{and} \quad s \le \tau_s ( a_{ i, s} ) \le s M_i \end{align} $$

for all $s = 1,\ldots , k_in_i $ .

Recalling that $b_i = a_0 \cdots a_{i - 1}$ , we can write $b_{ i + 1 } = b_i a_i$ . Then, from Lemma 4.6(2) together with (6.19) and (6.11),

(6.20) $$ \begin{align} \bigg|\dfrac{{\overline{\tau}}_{t_{i + 1}} }{ \tau_{ t_{ i + 1 } } } (b_{ i + 1 }) - \dfrac{ {\overline{\tau}}_{ k_{i} n_{i } } }{\tau_{ k_{i}n_{i } } } (a_{i} )\bigg| \le \dfrac{2 \tau_{t_i}(b_i)}{\tau_{t_i}(b_i) + \tau_{n_{i}k_{i}} ( a_{i} )} &< \dfrac{ 2\sum_{ j = 0 }^{ i - 1 } M_{ j }k_{ j }n_{j}}{k_{ i }n_{i}}< 2 \varepsilon_{i}. \end{align} $$

So, using (6.18) with $s = k_{ i }n_{i}$ , we obtain

(6.21) $$ \begin{align} ({\overline{\tau}}_{ t_{ i + 1 } } / \tau_{ t_{ i + 1 } } ) (b_{ i + 1 }) \in B_{3\varepsilon_{ i }}({\overline{p}}) \quad \text{for all } i \ge 0. \end{align} $$

We now fix $i \ge 1$ and consider the intermediate times $t_{i} + s$ where $s= 1 ,\ldots , k_in_{ i } - 1$ . Let us first consider the range $s = 1 , \ldots , N_{i} - 1$ . By Lemma 4.6(1) together with (6.19) and (6.12), we have that

(6.22) $$ \begin{align} \bigg|\dfrac{ {\overline{\tau}}_{t_{i} + s}}{\tau_{t_{i} + s}}(b_i a_{i, s}) - \dfrac{{\overline{\tau}}_{ t_{ i } } }{ \tau_{ t_{ i } } }(b_i)\bigg| \le 2\dfrac{ \tau_s ( a_{i,s} ) }{ \tau_{t_i}( b_i )+ \tau_s ( a_{i,s })} &\le \dfrac{ 2N_{ i } M_{ i } }{ t_{ i } } < 2 \varepsilon_{ i - 1 }. \end{align} $$

So, using (6.21), we know that

$$ \begin{align*} ( {\overline{\tau}}_{ t_i + s } / \tau_{ t_i + s } ) ( b_i a_{ i , s } ) \in B_{ 5 \varepsilon_{ i- 1} } ( {\overline{p}} ). \end{align*} $$

Now, let us consider the range $s = N_{ i }, \ldots , k_{i}n_{i} - 1$ . By Lemma 4.6(3), we know that

$$ \begin{align*} \min\bigg\{\frac{ {\overline{\tau}}_{ t_{ i } }}{ \tau_{ t_{ i } } }(b_i), \dfrac{ {\overline{\tau}}_{ s } }{ \tau_{ s } }( a_{ i, s } )\bigg\} \le \dfrac{ {\overline{\tau}}_{ t_{ i } +s } }{ \tau_{ t_{ i } +s} }(b_i a_{i,s})\le\max\bigg\{\dfrac{{\overline{\tau}}_{ t_{ i } } }{ \tau_{ t_{ i } } }(b_i), \dfrac{{\overline{\tau}}_{ s } }{ \tau_{ s } }( a_{ i, s } )\bigg\}. \end{align*} $$

Thus, by (5.7) and (6.21), we find that

As we assume that $\varepsilon _i\to 0$ , this proves ${\mathcal T}(x) = {\overline {p}}$ .

Proof. Let $x\in \Gamma $ . Recall the definition of the associated cylinders $a_{i,s} \in {\mathcal Q}_{\Lambda _i}^{s}$ , $a_{i}\in {\mathcal Q}_{\Lambda _i}^{k_in_i}$ , and $b_i\in {\mathcal B}_i$ in (6.8) and (6.9). We will show that

(6.23) $$ \begin{align} \lim_{i\to\infty} \inf_{s = 0 ,\ldots, k_in_i - 1} \frac{ \log m ( b_i a_{i,s} ) }{ \log |b_i a_{i,s} |} = \lim_{ i \to \infty }\sup_{s = 0 ,\ldots, k_in_i - 1} \frac{ \log m ( b_i a_{i,s} ) }{ \log |b_i a_{i,s} |} = 1. \end{align} $$

Set

and recall from Lemma 6.6 that $\lim _{i\to \infty } \ell _i = 1$ . Consider first the range $s = 0, \ldots , N_i$ . We proceed as in the proof of Lemma 6.6. Using the definition (6.4), the bounded distortion (see Remark 4.2)

(6.24) $$ \begin{align} \nonumber \dfrac{\log m ( b_i a_{i,s})} {\log |b_i a_{i,s}|} &\ge \dfrac{ \log m ( b_i ) + \log m_i (a_{i,s}) + \log C_i (b_i)}{\log |b_{i}| + \log |a_{i,s}| - \log D } \\ &\ge \dfrac{ \log m (b_i) }{ \log |b_i| } \bigg(\dfrac{ 1 + ({ \log m ( a_{i,s} ) + \widetilde C_{i} })/({ \log m (b_i) })}{ 1+ ({ \log |a_{i,s}| - \log D })/({ \log |b_i|})}\bigg) \nonumber\\ &\ge \ell_i \dfrac{ 1 - \varepsilon_{ i - 1} }{ 1 + \varepsilon_{ i - 1 } }, \end{align} $$

where in the final inequality, we have used (6.15) and (6.16).

Now, let us consider the range $s = N_i, \ldots ,k_in_{i}$ . We begin as before, but this time, we make use of (4.7) to obtain

(6.25) $$ \begin{align} \nonumber \frac{\log m ( b_i a_{i,s})} {\log |b_i a_{i,s}|} &\ge\frac{ \log m ( b_i ) + \log m_i (a_{i,s}) + \widetilde C_i }{\log |b_{i}| + \log |a_{i,s}| - \log D } \\ &\ge \min\bigg\{\frac{ \log m ( b_i ) + \widetilde C_i }{ \log |b_{i}| - \log D }, \frac{ \log m_i (a_{i,s}) }{\log |a_{i,s}| }\bigg\}. \end{align} $$

Then, we notice, using (6.15) and (6.16), that

(6.26) $$ \begin{align} \dfrac{ \log m ( b_i ) + \widetilde C_i } { \log |b_{i}| - \log D } = \dfrac{ \log m ( b_i )} { \log |b_{i}| }\bigg(\dfrac{ 1 +({\widetilde C_i})/({\log m ( b_i )}) }{ 1 + ({\log D })/({\log |b_{i}|})}\bigg) \ge \ell_i \dfrac{ 1 - \varepsilon_{ i - 1 } }{ 1 + \varepsilon_{ i - 1 } }. \end{align} $$

Recalling (6.17), we have $\log m_i (a_{i,s}) / \log |a_{i,s}| \ge 1 - \varepsilon _i$ . Combining this with (6.24), (6.25), and (6.26), we obtain

Proceeding in the same way as above, one can find analogous upper bounds for (6.24), (6.25), and (6.26), which yield

As $\lim _{i \to \infty } \underline {\delta }_i = \lim _{ i \to \infty } \overline {\delta }_i = 1$ , we obtain (6.23).

Denote by ${\mathcal Q}^\ell (x)$ the element of the partition ${\mathcal Q}^\ell $ which contains x. Recall that, by the above, for every $x\in \Gamma $ , for $\ell =t_i=k_0n_0+k_1n_1+\cdots +k_{i-1}n_{i-1}$ associated with x, there is a sequence $b_i(x)=a_0(x)a_1(x)\cdots a_{i-1}(x)$ , where $a_j(x)\in {\mathcal Q}^{k_jn_j}_{\Lambda _j}$ and

$$ \begin{align*}\begin{aligned} {\mathcal Q}^\ell(x) &= {\mathcal Q}^{t_i}(x) = {\mathcal Q}^{k_0n_0+k_1n_1+\cdots+k_{i-1}n_{i-1}}(x)\\ &= a_0(x) \cap F^{-k_0n_0}(a_1(x)) \cap\cdots \cap F^{-(k_0n_0+k_1n_1+\cdots+k_{i-1}n_{i-1})}(a_{i-1}(x)). \end{aligned}\end{align*} $$

Likewise, for $\ell =t_i+s$ , with $s=1,\ldots ,k_in_i$ ,

$$ \begin{align*} {\mathcal Q}^\ell(x) = {\mathcal Q}^{t_i}(x) \cap F^{-(t_i+s)}(a_{i,s}(x)). \end{align*} $$

From the above, we get

$$ \begin{align*} \lim_{\ell \to \infty}\frac{ \log m ( {\mathcal Q}^{\ell}(x) ) }{ \log|{\mathcal Q}^{\ell}(x)|} = 1. \end{align*} $$

As the refinements of the partition ${\mathcal Q}$ separate points, the assertion of the lemma then follows from the mass distribution principle (compare [Reference FalconerFal97, Ch. 2]).

Proof. Let $x \in \Gamma $ . From Theorem 3.1, it is enough to show that ${\mathcal T}(x) = {\mathcal C}$ and (3.1) holds.

Recall the definition of the cylinders $a_{i,s}(x) \in {\mathcal Q}_{\Lambda _i}^{s}$ , $a_{i}(x) \in {\mathcal Q}_{\Lambda _i}^{k_in_i}(x)$ , and $b_i(x) \in {\mathcal B}_i$ in (6.8) and (6.9) (see also Figure 3). We suppress the dependence of $a_{i,s},a_i$ , and $b_i$ on x to ease notation. Recall from item (1) of Proposition 5.1 that

(7.1) $$ \begin{align} | ( {\overline{\tau}}_{ s } / \tau_{ s } )(a_{i,s}) - {\overline{p}}_i |<\varepsilon_i \quad\text{for all } N_i\le s \le k_in_i. \end{align} $$

Writing $b_{i+1} = b_i a_{i}$ , it follows from Lemma 4.6(2) together with (6.19) and (6.11) that

(7.2) $$ \begin{align} \bigg| \dfrac{ {\overline{\tau}}_{t_{i+1}} }{ \tau_{ t_{ i+1 } } } (b_{i+1}) - \dfrac{ {\overline{\tau}}_{k_{i} n_{i } } }{\tau_{ k_{i}n_{i}}} (a_{i} ) \bigg| &\le \dfrac{ 2 \tau_{t_i}(b_{i})}{\tau_{t_i}(b_{i}) + \tau_{k_{i}n_{i}} ( a_{i} )} < \dfrac{ 2\sum_{ j = 0 }^{ i - 1 } k_{ j } n_{ j } M_{ j } }{ k_{ i }n_{ i } } < 2 \varepsilon_{ i }. \end{align} $$

So, using (7.1) with $s = k_{ i }n_{i}$ , we obtain

(7.3) $$ \begin{align} ({\overline{\tau}}_{ t_{ i + 1 }} / \tau_{ t_{ i + 1} } )(b_{ i + 1 }) \in B_{3\varepsilon_{ i }}({\overline{p}}_i) \quad \text{for all } i \ge 0. \end{align} $$

This gives the conclusion of the lemma along the subsequence $t_i$ , in the sense that

$$ \begin{align*} \{ \text{limit points of } ({\overline{\tau}}_{t_i}(x)/\tau_{t_i}(x))_{i \in{\mathbb N}}\} = \{ \text{limit points of } ({\overline{p}}_i)_{i\in{\mathbb N}}\} = {\mathcal C}. \end{align*} $$

We now fix $i \ge 1$ and consider the intermediate times $t_{i} + s$ , where $s= 1 ,\ldots , k_{i}n_{i} - 1$ . Let us first consider the range $s = 1 , \ldots , N_{i} - 1$ . By Lemma 4.6(1) together with (6.19) and (6.12), we have that

(7.4) $$ \begin{align} \bigg| \dfrac{ {\overline{\tau}}_{ t_{ i } + s } }{ \tau_{ t_{ i } + s } } ( b_i a_{i, s} ) - \dfrac{ {\overline{\tau}}_{ t_{ i } } }{ \tau_{ t_{ i } } }(b_i) \bigg| \le 2\dfrac{ \tau_s ( a_{i,s} ) }{ \tau_{t_i}( b_i )+ \tau_s ( a_{i,s })} &\le \dfrac{ 2N_{ i } M_{ i } }{ t_{ i } } < 2 \varepsilon_{ i - 1 }. \end{align} $$

So, using (6.21), we know that

$$ \begin{align*} \dfrac{ {\overline{\tau}}_{ t_i + s } }{ \tau_{ t_i + s }} ( b_i a_{ i , s } ) \in B_{ 5 \varepsilon_{i - 1} } ( {\overline{p}}_{i-1} ). \end{align*} $$

Now, let us consider the range $s = N_{ i}, \ldots , k_{i}n_{i}-1, k_{i}n_{i}$ . By Lemma 4.6(3), we know that

$$ \begin{align*} \min\bigg\{\dfrac{{\overline{\tau}}_{ t_{ i } } }{ \tau_{ t_{ i } }} (b_i) ,\dfrac{ {\overline{\tau}}_{ s }}{ \tau_{ s }} ( a_{ i, s } )\bigg\} \le \dfrac{ {\overline{\tau}}_{ t_{ i } +s }}{ \tau_{ t_{ i } +s}} (b_i a_{i,s}) \le\max\bigg\{\dfrac{ {\overline{\tau}}_{ t_{ i } } }{ \tau_{ t_{ i } } }(b_i) ,\dfrac{ {\overline{\tau}}_{ s } }{ \tau_{ s }} ( a_{ i, s } )\bigg\}, \end{align*} $$

where the inequality holds coordinate-wise. Let $\delta _i = \max \{ 3\varepsilon _{i-1}, 3\varepsilon _{i} \}$ , and recall from (7.3) and (7.1) that $({\overline {\tau }}_{ t_{ i } } / \tau _{ t_{ i } } ) (b_i) \in B_{\delta _i}({\overline {p}}_{i-1})$ and $({\overline {\tau }}_{ s } / \tau _{ s } ) (a_{i,s}) \in B_{\delta _i}({\overline {p}}_{i})$ . Thus,

(7.5) $$ \begin{align} \dfrac{ {\overline{\tau}}_{ t_{ i } +s } }{ \tau_{ t_{ i } +s} }(b_i a_{i,s}) \in R_{i}^{\delta_i}, \end{align} $$

where $R_i^{\delta _i}$ is the $\delta _i$ -neighborhood of the d-dimensional rectangle $R_i$ that has ${\overline {p}}_i$ and ${\overline {p}}_{i+1}$ as vertices and is given by

As $\lim _{i \to \infty } | {\overline {p}}_i - {\overline {p}}_{i+1} | = 0$ , and as $\lim _{i \to \infty } \delta _i = 0$ , we know that the rectangles $R^{\delta _i}_{i}$ accumulate precisely at the limit points ${\mathcal C}$ of $({\overline {p}}_i)_i$ . This concludes that ${\mathcal T}(x) = {\mathcal C}$ .

To check that (3.1) holds, let $k\in {\mathbb N}$ and choose i such that $t_i\le k< t_{i+1}$ . Hence, it follows from (7.5) that

$$ \begin{align*} \bigg| \frac{{\overline{\tau}}_{k}}{\tau_{k }}(x) - \frac{ {\overline{\tau}}_{ k +1} }{ \tau_{ k +1}}(x)\bigg|\le2\,\mathrm{diam}(R_i^{\delta_i}). \end{align*} $$

As $k\to \infty $ , we get $i\to \infty $ and hence, the latter tends to $0$ . This implies that (3.1) holds and hence proves the lemma.