Proof. $M_{\mathrm {L}}$ defined in (3.2) is finite, which implies that ${\inf }_{x\in \tilde {M}, n\in \mathbb {Z}}d(x, I^{n}x)\geq \epsilon _{0}'>0$ . Due to the assumption taken on the Poincaré section, $\varphi $ is constant on each $D_{i}\cap T^{-1}D_{j}$ for $i\neq j$ .

Proof. $S_{n}\varphi $ and $S_{n}R$ satisfy the central limit theorem using [Reference Burton and Denker7] and $\int _{0}^{R(x)}\phi (x,u)\,du=\varphi _X(x)$ . We deduce using Theorem 1.1 in [Reference Melbourne and Török11] that $\phi $ satisfies the central limit theorem with variance ${\sigma ^{2}_{\phi }}/{(\int _{X}R\,d\nu _{X})}$ .

Proof. For any $(x,u)\in X_R$ and any $t>0$ , $n_{t+u}(x)$ is the number of visits to X before $t+u$ starting from x (or equivalently the number of visits to X before t starting from $g_u(x)$ ). We have

(3.4) $$ \begin{align} \int_{0}^{t}\phi(\psi_{s}(x,u))\,ds =& \sum_{k=0}^{n_{t+u}(x)-1}\int_{S_{k}R(x)}^{S_{k+1}R(x)}\phi(x,s)\,ds \nonumber\\& \quad+\int_{S_{n_{t+u}(x)}R(x)}^{t+u}\phi(x,s)\,ds-\int_{0}^{u}\frac{\varphi_X (x)}{R(x)}ds. \end{align} $$

Using the definition of $\varphi _X$ and that the points $(x,s+S_{k}R(x))$ and $(T^k x,s)$ are identified, we have

$$ \begin{align*} \int_{S_{k}R(x)}^{S_{k+1}R(x)}\phi(x,s)\,ds & = \int_{0}^{R(T^k x)}\phi(x,s+S_{k}R(x))\,ds \\& = \int_{0}^{R(T^k x)}\phi(T^k x,s)\,ds= \varphi_X(T^k x), \end{align*} $$

from which we get that the second term of (3.4) is equal to $S_{n_{t+u}(x)}\varphi _X(x).$ Now, whenever $S_{n_{t+u}(x)}R(x)\leq s\leq t+u<S_{n_{t+u}(x)+1}R(x)$ , it implies that $0\leq s-S_{n_{t+u}(x)}R(x)<R(T^{n_{t+u}(x)} x)$ . Then we have $\phi (x,s)$ which is identified with $\phi (T^{n_{t+u}(x)}x, s-S_{n_{t+u}(x)}R(x))$ , which is in turn equal to ${(\varphi _{X}(T^{n_{t+u}(x)}(x)))}/{(R(T^{n_{t+u}(x)}(x)))}$ . Thus, we can conclude that the last two terms in (3.4) are bounded by $\Vert \varphi _X\Vert _{\infty }$ . Furthermore,

$$ \begin{align*} \varphi_{t+u}(x)&=\varphi_{(S_{n_{t+u}(x)}R(x))}(x)+\varphi_{(t+u-S_{n_{t+u}(x)}R(x))}(T^{n_{t+u}(x)}(x))\\ &=S_{(n_{t+u}(x))}\varphi_X(x)+\varphi_{(t+u-S_{(n_{t+u}(x))}R(x))}(T^{n_{t+u}(x)}(x)). \end{align*} $$

Setting the constant $N_{\mathrm {L}}=\sup _{x\in M}\sup _{t\leq R(x)}|\varphi _{t}(x)|$ , by combining these estimates together, we get the desired result.

Proof. Setting $\epsilon _{t}:=({2 \Vert \varphi _X\Vert _{\infty }}/{\sqrt {t}})+({N_{\mathrm {L}}}/{\sqrt {t}})$ , and using Lemma 3.3, we get $\vert ({\varphi _{t+u}(x)}/{\sqrt {t}})-({\int _{0}^{t}\phi (\psi _{s}(x,u))\,ds}/{\sqrt {t}})\vert \leq \epsilon _{t}$ , which goes to 0 as t goes to infinity. Therefore, using Lemma 3.5, we deduce the central limit theorem for $\varphi _{t}$ .

Proof. We have

$$ \begin{align*}\sum_{k=0}^{w_{\epsilon}(x)-1}R\circ T^{k}(x)\leq \tau_{\epsilon}(y)+s<\sum_{k=0}^{w_{\epsilon}(x)}R\circ T^{k}(x),\end{align*} $$

from which we get that $\tau _{\epsilon }(y)+s=\sum _{k=0}^{w_{\epsilon }(x)-1}R\circ T^{k}(x)+s',$ where $0\leq s'=s^{\prime }_\epsilon (x)<R\circ T^{w_{\epsilon }(x)}\leq \Vert R\Vert _\infty $ .

Now using the Birkhoff ergodic theorem, for $\nu _{X}$ -almost every $x\in X$ , we have

(4.4) $$ \begin{align} \underset{n\rightarrow\infty}{\lim}\frac{1}{n}\sum_{k=0}^{n-1}R(T^{k}(x))= \int_{X} R\,d\nu_{X}. \end{align} $$

Since the roof function R is bounded, say $R_{\min}\leq R \leq R_{\max}$ , it follows that

$$ \begin{align*} \bigg|\tau_{\epsilon}(y)-\sum_{k=0}^{w_{\epsilon}(x)-1}R\circ T^{k}\bigg|=|s-s'|\leq R_{\max}. \end{align*} $$

Together with (4.4), this yields

$$ \begin{align*} &w_{\epsilon}(x)\int_{X} R\,d\nu_{X}\underset{\epsilon\rightarrow0}{\sim}\sum_{k=0}^{w_{\epsilon}(x)-1}R\circ T^{k}-R_{\max}\leq\tau_{\epsilon}(y)\\ & \quad\leq\sum_{k=0}^{w_{\epsilon}(x)-1}R\circ T^{k}+R_{\max}\underset{\epsilon\rightarrow0}{\sim}w_{\epsilon}(x)\int_{X} R\,d\nu_{X}, \end{align*} $$

since $w_{\epsilon }\rightarrow +\infty \nu _{X}$ -almost surely, and then as $\epsilon \rightarrow 0$ , we get (4.3).

Proof. Let us denote by $\mathcal {C}_{-q,q{'}}$ the set of $(-q,q')$ -cylinders of $\Sigma $ . Let $\delta>0$ be such that $\tfrac 12+(a+b)h_{\nu }> 1+\delta (a+b)$ , and denote by $\mathcal {C}_{-q,q{'}}^{\delta }\subset \mathcal {C}_{-q,q{'}}$ the set of cylinders $C \in \mathcal {C}_{-q,q{'}}$ such that $\nu (C)\in (e^{-(h_{\nu }+\delta )(q+q{'})},e^{-(h_{\nu }-\delta )(q+q{'})})$ . For any ${\omega \in \Sigma _{A}}$ , recall that $C_{-q,q{'}}(\omega )\in \mathcal {C}_{-q,q'}$ is the $(-q,q')$ -cylinder which contains $\omega $ . By the Shannon–McMillan–Breiman theorem, the set $K_{N}^{\delta }=\{\omega \in \Sigma : \text { for all } q,q' \geq N , C_{-q,q'}(\omega )\in \mathcal {C}^{\delta }_{-q,q'}\}$ has a measure $\nu (K_{N}^{\delta })> 1-\delta $ provided N is sufficiently large. According to Proposition 4.3, whenever $n\geq N$ , we have

$$ \begin{align*} \nu(K_{N}^{\delta}\cap G_{n}(q_{n},q_{n}')) &=\nu(\{\omega\in K_{N}^{\delta}: S_{n}\varphi(\omega)=0 \text{ and } \theta^{n}(\omega)\in C_{-q_{n},q_{n}'}(\omega)\})\\ &=\sum_{C\in \mathcal{C}_{-q_{n},q_{n}'}^{\delta}}\nu(C\cap \{S_{n}\varphi=0\}\cap\theta^{-n}(C))\\ &=\sum_{C\in \mathcal{C}^{\delta}_{-q_{n},q_{n}'}}\bigg(\frac{\nu(C)^{2}}{\sqrt{2\pi}\sigma_{\varphi}\sqrt{n}} + O\bigg(\frac{\nu(C)q_{n}\nu(C)}{n} \bigg)\bigg). \end{align*} $$

Hence, it follows that

$$ \begin{align*} \nu(K_{N}^{\delta}\cap G_{n}(q_{n},q_{n}')) =O\bigg(\sum_{C\in \mathcal{C}_{-q_{n},q_{n}'}^{\delta}}\frac{\nu(C)^{2}}{\sqrt{n}}\bigg) =O\bigg(\frac{1}{n^{\frac{1}{2}+(a+b)(h_{\nu}-\delta)}}\bigg), \end{align*} $$

but $\tfrac 12+(a+b)(h_{\nu }-\delta )>1$ then $\sum _{n}\nu (K_{N}^{\delta }\cap G_{n}(q_{n},q_{n}')) < \infty $ . Hence, by the first Borel–Cantelli Lemma, for almost every $\omega\in K_{N}^{\delta}$ , if n is large enough, we get

$$ \begin{align*} \tau_{q_{n},q_{n}'(\omega)}> n. \end{align*} $$

It follows that

$$ \begin{align*} \log\sqrt{\tau_{q_{n},q_{n}'}(\omega)}\geq \frac{1}{2}\log n \quad \text{for all sufficiently large }n. \end{align*} $$

Since $q_{n}+q_{n}'\sim (a+b)\log n$ , we obtain

$$ \begin{align*} \underset{n\rightarrow \infty}{\liminf}\ \frac{\log\sqrt{\tau_{q_{n},q'_{n}}}}{q_{n}+q'_{n}}\geq \frac{1}{2(a+b)} \quad \text{almost everywhere}.\\[-45pt] \end{align*} $$

Proof. We keep the same notation as in the proof of the previous lemma with $\delta>0$ such that $\tfrac 12+(a+b)h_{\nu }< 1+\delta (a+b)$ . For all $l=1,\ldots ,n$ , we define

$$ \begin{align*} A_{l}(q_{n},q_{n}'):=G_{l}(q_{n},q_{n}')\cap \theta^{-l}\{\tau_{q_{n},q_{n}'}>n-l\}. \end{align*} $$

Let us take $L_{n}:=\lceil n^{a'}\rceil $ , with $a'>2(a+b)(h_{\nu }+\delta -\gamma )$ . The sets $A_{l}(q_{n},q_{n}')$ are pairwise disjoint, thus

$$ \begin{align*} 1= \sum_{l=0}^{n}\nu(A_{l}(q_{n},q_{n}'))\geq\sum_{l=L_{n}}^{n}\sum_{C\in \mathcal{C}^{\delta}_{q_{n},q_{n}'}}\nu(C\cap A_{l}(q_{n},q_{n}')). \end{align*} $$

Note that $C\cap \{\tau _{q_{n},q_{n}'}>n-l\}$ is a union of $C_{-q_n,n-l+q^{\prime }_n}$ -cylinders. Then, using Proposition 4.3, for any $C\in \mathcal {C}^{\delta }_{q_{n}}$ provided $q_{n}\geq N$ and $l\geq L_{n}$ , we get

$$ \begin{align*} \nu(C\cap A_{l}(q_{n},q_{n}')) &=\nu(C\cap\{S_{l}\varphi=0\}\cap\theta^{-l}(C\cap\{\tau_{q_{n},q_{n}'}>n-l\}))\\ &=\bigg(\frac{\nu(C)}{\sqrt{2\pi}\sigma_{\varphi}}+O\bigg(\frac{q_{n}\nu(C)}{\sqrt{l-2q_{n}}}\bigg)\bigg)\frac{1}{\sqrt{l-q_{n}}}\nu(C\cap\{\tau_{q_{n},q_{n}'}>n-l\})\\ &\geq c n^{-(a+b)(h_{\nu}+\delta)}\frac{1}{\sqrt{l}}\nu(C\cap\{\tau_{q_{n},q_{n}'}>n-l\}). \end{align*} $$

Note that

$$ \begin{align*} \nu(K_{N}^{\delta}\cap\{\tau_{q_{n},q_{n}'}>n\})\leq \sum_{C\in \mathcal{C}^{\delta}_{q_{n},q_{n}'}}\nu(C\cap\{\tau_{q_{n},q_{n}'}>n\}), \end{align*} $$

but we observe that, for any $C\in \mathcal {C}_{-q_{n},q_{n}'}^{\delta }$ ,

$$ \begin{align*}\sum_{l=L_{n}}^{n}\nu(C\cap A_{l}(q_{n},q_{n}')) \geq cn^{-(a+b)(h_{\nu}+\delta)}\nu(C\cap\{\tau_{q_{n},q_{n}'}>n\})(\sqrt{n}-\sqrt{L_{n}}), \end{align*} $$

so we obtain

$$ \begin{align*} 1 & \geq \sum_{C\in \mathcal{C}^{\delta}_{q_{n},q_{n}'}}\sum_{l=L_{n}}^{n}\nu(C\cap A_{l}(q_{n},q_{n}'))\\ & \geq \sum_{C\in \mathcal{C}^{\delta}_{-q_{n}+q_{n}'}}n^{-(a+b)(h_{\nu}+\delta)}\nu(C\cap\{\tau_{q_{n},q_{n}'}>n\})(\sqrt{n}-\sqrt{L_{n}}), \end{align*} $$

from which one gets

$$ \begin{align*} \sum_{C\in \mathcal{C}^{\delta}_{k_{n}}}\nu(C\cap\{\tau_{q_{n},q_{n}'}>n\})= O\bigg(\frac{1}{n^{\frac{1}{2}-(a+b)(h_{\nu}+\delta)}}\bigg). \end{align*} $$

Now let us take $n_{p}:=p^{-{4}/{(1-2(a+b)(h_{\nu } +\delta ))}}$ . We have $\sum _{p\geq 1}\nu (K_{N}^{\delta }\cap \{\tau _{q_{n_{p}},q_{n_{p}}'}>n_{p}\})<+\infty .$ Hence, by the first Borel–Cantelli lemma, for $\nu $ -almost every $\omega \in K_{N}^{\delta }$ , for every p large enough, $\tau _{q_{n_{p}},q_{n_{p}}'}\leq n_{p}$ . This implies that

$$ \begin{align*} \underset{n\rightarrow +\infty}{\limsup}\frac{\log\sqrt{ \tau_{q_{n},q_{n}'}}}{q_{n}+q_{n}'}\leq \frac{1}{2(a+b)}. \\[-43pt] \end{align*} $$

Proof of Theorem 3.1

Let $y=g_s(x)$ , with x coded by an $\omega \in \Sigma _{A}$ (that is, $x=\chi (\omega )$ ) and such that

$$ \begin{align*} \lim_{n\rightarrow +\infty}\frac{\log a_n^u(x)}{n}= -L^u =\mathbb E_{\nu}[\log \mathbf{a}^u] \quad\mbox{and}\quad \lim_{n\rightarrow +\infty}\frac{\log a_n^s(x)}{n}=L^s=\mathbb E_{\nu}[\log \mathbf{a}^s]\, , \end{align*} $$

and satisfying (4.3). Fix an $\epsilon>0$ and $s>0$ . Recall that, due to (4.1),

$$ \begin{align*}B_D(x,\epsilon/k_D)\subset P_{D}(B(g_{s}(x),\epsilon))\subset B_D(x,\epsilon k_D)\,. \end{align*} $$

Using Proposition 2.7 and the notation C therein, let $q_\epsilon $ and $q^{\prime }_\epsilon $ be the smallest non-negative integers such that

$$ \begin{align*} C\max(a_{q_\varepsilon}^s(x),a_{q^{\prime}_\varepsilon}^u(x))\le \varepsilon/k_D< C\min(a_{q_\varepsilon-1}^s(x),a_{q^{\prime}_\varepsilon-1}^u(x)). \end{align*} $$

Using Lemma 2.15, we fix $p_\epsilon $ and $p^{\prime }_\epsilon $ the largest integers such that

$$ \begin{align*} c_{\mathrm{L}}\epsilon k_D\leq\min( l^{u}_{p^{\prime}_{\epsilon}}(\omega)\times\delta_n, l^{s}_{p_{\epsilon}}(\omega)\times \delta_{m}).\end{align*} $$

Observe that $P_{D}(B(g_{s}(x),\epsilon ))$ is then controlled by

$$ \begin{align*}\chi(C_{-q_{\epsilon},{q'}_{\epsilon}}(\omega)) \subset B_D(x,\epsilon/k_D)\subset P_{D}(B(g_{s}(x),\epsilon))\subset B_D(x,\epsilon k_D)\subset\chi(C_{-p_{\epsilon},{p'}_{\epsilon}}(\omega)).\end{align*} $$

Hence, we have

(4.6) $$ \begin{align} \tau_{p_{\epsilon},p^{\prime}_{\epsilon}} (\omega) < w_{\epsilon}(x) < \tau_{q_{\epsilon}, q^{\prime}_{\epsilon}}(\omega). \end{align} $$

By definition, $\log a_{q_\epsilon }^s(x)\sim \log \epsilon \sim \log a_{q^{\prime }_\epsilon }^u (x)$ and therefore

$$ \begin{align*} \lim_{\epsilon\rightarrow 0}\frac{\log \epsilon}{ q_\epsilon}= \lim_{\epsilon\rightarrow 0}\frac{\log a^s_{q_\epsilon}(x)}{ q_\epsilon}=L^s\quad\mbox{and}\quad\lim_{\epsilon\rightarrow 0}\frac{\log \epsilon}{ q^{\prime}_\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{\log a^u_{q^{\prime}_\epsilon}(x)}{ q^{\prime}_\epsilon}= -L^u. \end{align*} $$

Furthermore, by definition of $p_\epsilon $ and $p^{\prime }_\epsilon $ ,

$$ \begin{align*} l_{1+p_\epsilon}^s(\omega)\times \delta_m<c_{\mathrm{L}}\epsilon k_D<l_{p_\epsilon}^s(\omega)\times\delta_m \quad\mbox{and}\quad l_{1+p^{\prime}_\epsilon}^u(\omega)\times\delta_n<\epsilon k_D<l_{p^{\prime}_\epsilon}^u(\omega)\times\delta_n, \end{align*} $$

and so

$$ \begin{align*} \lim_{\epsilon\rightarrow 0}\frac{\log\epsilon}{p_\epsilon} =\lim_{\epsilon\rightarrow 0}\frac{\log (l_{p_\epsilon}^s(\omega)\times\delta_m)}{p_\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{\log l_{p_\epsilon}^s(\omega)}{p_\epsilon}=L^s\end{align*} $$

and

$$ \begin{align*} \lim_{\epsilon\rightarrow 0}\frac{\log\epsilon}{p^{\prime}_\epsilon} =\lim_{\epsilon\rightarrow 0}\frac{\log (l_{p^{\prime}_\epsilon}^u(\omega)\times\delta_n)}{p^{\prime}_\epsilon}=\lim_{\epsilon\rightarrow 0}\frac{\log l_{p^{\prime}_\epsilon}^u(\omega)}{p^{\prime}_\epsilon}= -L^u \,. \end{align*} $$

Let $\eta \in (0,1)$ and $ \alpha _{q}=({(1-\eta )}/{2h_{\nu }})({1}/{({1}/{L^u})-({1}/{L^s})}).$ For $\epsilon>0$ large enough, we set $ a:={\alpha _{q}}/{|L^s|}$ , $ b:={\alpha _{q}}/{L^u}$ , and $N_{\eta ,\epsilon }:=\lceil \log \epsilon ^{{1}/{\alpha }}\rceil $ . We have $a+b= ({1-\eta }/{2h_{\nu }})<{1}/{(2h_\nu )}$ . Then, using Lemma 4.5 (applied to $a, b, N_{\alpha _{q},\epsilon }$ ), we have almost surely

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\limsup}\frac{\log\sqrt{\tau_{q_{\epsilon},q_{\epsilon}'}}}{q_{\epsilon}+q_{\epsilon}'}\leq\frac{h_{\nu}}{1-\eta}, \end{align*} $$

but $q_{\epsilon }+q^{\prime }_{\epsilon }\leq ({(a+b)|\!\log \epsilon |}/{\alpha })\leq (({1}/{L^u})-({1}/{L^s}))|\!\log \epsilon |$ , which gives almost everywhere

(4.7) $$ \begin{align} \underset{\epsilon\rightarrow0}{\limsup}\frac{\log\sqrt{\tau_{q_{\epsilon},q_{\epsilon}'}}}{|\!\log\epsilon|}\leq \frac{h_{\nu}}{1-\eta} \bigg(\frac{1}{L^u}-\frac{1}{L^s}\bigg). \end{align} $$

On the other hand, for all $\epsilon>0$ large enough and $\eta \in (0,1)$ , let $\alpha _{p}=({1+\eta }/{2h_{\nu }}) ({1}/{({1}/{L^u})-({1}/{L^s})}).$ Set $a:={\alpha _{p}}/{L^s}$ , $b:={\alpha _{p}}/{L^u}$ , and $N_{\eta ,\epsilon }:=\lceil \log \epsilon ^{{1}/{\alpha }}\rceil $ . As ${a+b=({1+\eta }/{2h_{\nu }})>({1}/{2h_{\nu }})}$ , we have almost everywhere, using Lemma 4.4 (applied to $a, b, N_{\alpha _{p},\epsilon }$ ), that

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\liminf}\frac{\log\sqrt{\tau_{p_{\epsilon},p_{\epsilon}'}}}{p_{\epsilon}+p_{\epsilon}'}\geq\frac{h_{\nu}}{1+\eta}. \end{align*} $$

In the same way as previously, $p_{\epsilon }+p^{\prime }_{\epsilon }\geq ({(a+b)|\!\log \epsilon |}/{\alpha })\geq (({1}/{L^u})-({1}/{L^s}))|\!\log \epsilon |$ , from which we get

(4.8) $$ \begin{align} \underset{\epsilon\rightarrow0}{\liminf}\frac{\log\sqrt{\tau_{p_{\epsilon},p_{\epsilon}'}}}{|\!\log\epsilon|}\geq \frac{h_{\nu}}{1+\eta}\bigg(\frac{1}{L^u}-\frac{1}{L^s}\bigg). \end{align} $$

Therefore for all $\eta \in (0,1)$ , due to (4.6), using (4.7), (4.8) and Lemma 4.2, we have the formulas

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\limsup}\frac{\log\sqrt{\tau_{\epsilon}}}{|\!\log\epsilon|} &\leq\underset{\epsilon\rightarrow0}{\limsup}\frac{\log\sqrt{\tau_{q_{\epsilon},q_{\epsilon}'}}} {|\!\log\epsilon|}\leq\frac{h_{\nu}}{1-\eta}\bigg(\frac{1}{L^u}-\frac{1}{L^s}\bigg),\\[3pt]\underset{\epsilon\rightarrow0}{\liminf}\frac{\log\sqrt{\tau_{\epsilon}}}{|\!\log\epsilon|}&\geq\underset{\epsilon\rightarrow0}{\liminf}\frac{\log\sqrt{\tau_{p_{\epsilon},p_{\epsilon}'}}}{|\!\log\epsilon|}\geq\frac{h_{\nu}}{1+\eta}\bigg(\frac{1}{L^u}-\frac{1}{L^s}\bigg), \end{align*} $$

almost everywhere.

Now using the relation between $L^u$ , $L^s$ and the Lyapunov exponents of the flow $(g_t)_t$ (see Proposition 2.9), and Abramov’s formula (2.8), we get

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\lim}\ \frac{\log\sqrt{\tau_{\epsilon}}}{|\!\log\epsilon|}=h_{\nu}\bigg(\frac{1}{L^u}-\frac{1}{L^s}\bigg)= h_{\mu}\bigg(\dfrac{1}{\unicode{x3bb}^u} -\dfrac{1}{ \unicode{x3bb}^s}\bigg) \quad \text{almost everywhere.} \end{align*} $$

It follows from Theorem 6.2 in [Reference Barreira3] that this limit coincides with the Hausdorff dimension $\dim _{\mathrm {H}}\nu $ of the measure $\nu $ , that is, with $\dim _{\mathrm {H}}\mu -1$ .

Proof. Let $\omega ,\omega '\in \Sigma _{A}$ . By (2.27), there exists $c_{\mathrm {L}}>0$ such that

$$ \begin{align*} \text{diam}(\chi(D_{\epsilon,i}))\leq c_{\mathrm{L}}(l^{s}_{n_{2}}(\omega_{\epsilon,i})+l^{u}_{n_{1}}(\omega_{\epsilon,i}))\leq c_{\mathrm{L}}\bigg(\frac{\theta_{\epsilon}}{2c_{g}c_{\mathrm{L}}}+\frac{\theta_{\epsilon}}{2c_{g}c_{\mathrm{L}}}\bigg) <\frac{\theta_{\epsilon}}{c_{g}}, \end{align*} $$

from which we get that

$$ \begin{align*}d(g_{s}(\chi(\omega), g_{s'}(\chi(\omega'))))\leq c_{g}\max(|s-s'|, d(\chi(\omega), \chi(\omega')))<\theta_{\epsilon}. \\[-37pt] \end{align*} $$

Proof. For all $\omega \in \Sigma _{A}$ , set $q=n_{1}(\omega )$ and $q'=n_{2}(\omega )$ . Using the definitions of $n_{1}(\omega )$ and $l^{u}_{q}$ , there exist $\omega '$ and $c_{1}={(\mathbf {a}^{u}(\sigma ^{k}(\omega '))^{-1})}/{(2Cc_{g}c_{\mathrm {L}})}>0$ , such that $l^{u}_{q}(\omega ')\geq c_{1}\theta _{\epsilon }$ . Thus, in turn, there exists $\omega ^{u}\in W^{u}_{q}(\omega ')$ such that

$$ \begin{align*} B^{u}\bigg(\omega^{u}, \frac{c_{1}\theta_{\epsilon}}{3}\bigg)\subset W^{u}_{q}(\omega'). \end{align*} $$

Similarly, using the definitions of $n_{2}(\omega )$ and $l^{s}_{q'}$ , there exist $\omega "\in \Sigma _{A}$ , $c_{2}>0$ such that $l^{s}_{q'}(\omega ")\geq c_{2}\theta _{\epsilon }$ , and $\omega ^{s}\in W^{s}_{q'}(\omega ")$ such that $B^{s}(\omega ^{s},{(c_{2}\theta _{\epsilon })}/{3})\subset W^{s}_{q'}(\omega ).$ Let ${k>{3c_{\mathrm {L}}}/{(\min (c_{1},c_{2}))}}$ . Set $\omega _{i}\in \Sigma _{A}$ such that $\chi (\omega _{i})=\{\omega ^{s},\omega ^{u}\}$ . Then, using Lemma 2.11, we get

$$ \begin{align*} B\bigg(\chi(\omega_{i}),\frac{\theta_{\epsilon}}{k}\bigg)\subset C_{-q,q'}, \end{align*} $$

which finishes the proof.

Proof. Let $P\in \bar {\mathcal {P}}_{\epsilon }$ . Then according to Lemma 5.1, P is contained in the $\theta _{\epsilon }$ -neighborhood of the graph of $r:\Sigma _{A}\rightarrow [0,+\infty ]$ . So

$$ \begin{align*} \bigcup_{P\in \bar{\mathcal{P}}_{\epsilon}} P\subset\{(\omega,s)\in\Delta: r(\omega)-\theta_{\epsilon}<s\leq r(\omega)\}, \end{align*} $$

whose measure (with respect to $\mu _{\Delta }$ ) is bounded by ${\theta _{\epsilon }}/{(\int _{\Sigma _{A}}r\,d\nu )}$ .

Proof. This comes from the transversality of $W^{s}(x)$ and $W^{u}(x)$ and the fact that $\partial A_{\epsilon }(y)$ is $C^{1}$ of order ${1}/{\epsilon }$ .

Proof. We know that the cylinders of $\mathcal {D}_{\epsilon }$ are of diameter less than $\theta _{\epsilon }$ , and the projection map on $\Sigma _{A}$ is Lipschitz with constant $k_{\mathrm {L}}$ . Set $\delta =(k_{\mathrm {L}}+1)\theta _{\epsilon }$ and let $k_{0}$ be the smallest positive number satisfying $k_{\mathrm {L}} \epsilon +\delta + \epsilon \leq k_{0}\epsilon $ . By the coding map $\chi $ , we have on the Poincaré section X that

$$ \begin{align*} A_{\epsilon}^{+}(y)\backslash A_{\epsilon}^{-}(y)&=A_{\epsilon}(y)^{[(k_{\mathrm{L}}+1)\theta_{\epsilon}]}\cap (X\backslash A_{\epsilon}(y))^{[(k_{\mathrm{L}}+1)\theta_{\epsilon}]}\\ &=(\partial A_{\epsilon}(y))^{[\delta]}, \end{align*} $$

thus we need to prove that the measure of the $\delta $ -neighborhood of $A_{\epsilon }(y)$ is negligible with respect to the measure of $A_{\epsilon }(y)$ . Indeed, let $m_{F}>0$ be such that $k_{0}\leq 2^{m_{F}}\alpha _{1}$ . Then, using the Federer property (Remark 5.5) and the inclusion in (5.4), there exists $c_{F}>0$ such that

(5.6) $$ \begin{align} \nu(\chi^{-1}(B(x, k_{0}\epsilon)))\leq c_{F}^{m_{F}}\nu(\chi^{-1}(B(x,\alpha_{1}\epsilon)))\leq c_{F}^{m_{F}}\nu(A_{\epsilon}(x)), \end{align} $$

hence it is sufficient to prove that $\nu ((\partial A_{\epsilon }(y))^{[\delta ]})$ is negligible with respect to $\nu (\chi ^{-1}(B(x, k_{0}\epsilon )))$ .

Now, due to Lemma 5.6, we will work on two parts of $\partial A_{\epsilon }(y)^{[\delta ]}$ .

Let $x_{0}$ be a point in D. Let $x\in W^{s}(x_{0})$ . We set $I^{u}(x):=W^{u}_{\mathrm {loc}}(x)\cap (\partial A_{\epsilon }(y))^{[\delta ]}$ such that, for all $\xi _{x}\in I^{u}(x)$ , there exists $z\in \partial A_{\epsilon }$ as in Lemma 5.6 and where $z\in W^{u}(\xi _{x})$ . On the other hand, we set $I^{s}(x):=W^{s}_{\mathrm {loc}}(x)\cap (\partial A_{\epsilon }(y))^{[\delta ]}$ , such that, for all $ \xi _{x}'\in I^{s}(x)$ , there exists $z'\in \partial A_{\epsilon }$ as in Lemma 5.6 and where $z'\in W^{s}(\xi _{x}')$ .

Hence, we have $\partial A_{\epsilon }(y)^{[\delta ]}=\mathcal {I}^{u}\cup \mathcal {I}^{s}$ , where

$$ \begin{align*} \mathcal{I}^{u}:=\bigcup_{x\in W^{s}(x_{0})}I^{u}(x) \quad\text{and}\quad \mathcal{I}^{s}:=\bigcup_{x\in W^{s}(x_{0})}I^{s}(x). \end{align*} $$

Our aim is to estimate the measure of $\mathcal {I}^{u}$ with respect to that of $B(x, k_{0}\epsilon )$ . We will start working on $\nu ^{u}(\pi ^{+}(\chi ^{-1}(I^{u})))$ and then after integrating over $\Sigma _{A}^{-}\times \Sigma _{A}^{+}$ , we will conclude the estimation of $\nu (\chi ^{-1}(\mathcal {I}^{u}))$ .

Let $q_{i}$ be the greatest integer such that $l^{u}_{q_{i}}(\omega _{i})>3c\delta $ . Then we have

$$ \begin{align*} I^{u}(x)\subset \bigcup_{i=1}^{2}\chi(C_{-\infty,q_{i}}(\omega_{i}))\subset B(x,k_{0}\epsilon), \end{align*} $$

and so $\nu ^{u}(\pi ^{+}(\chi ^{-1}(I^{u})))\leq \sum _{i=1}^{2}\nu ^{u}(C_{0,q_{i}}(\pi ^{+}(\omega _{i})))$ . Knowing that $W^{u}(x)= \chi (C_{-\infty ,0}(\omega )) =\bigcap _{i=1}^{\infty }\chi (C_{-i, 0}(\omega ))$ , let $p_{i}$ be the smallest integer such that

$$ \begin{align*} \prod_{k=0}^{p_i -1}(\mathbf{a}^{u}(\sigma^{k}(\omega_i)))^{-1}> k_{0}\epsilon, \quad \prod_{k=0}^{p_i}(\mathbf{a}^{u}(\sigma^{k}(\omega_i)))^{-1} \leq k_{0}\epsilon. \end{align*} $$

Then using Proposition 2.7, we have

(5.7) $$ \begin{align} \chi(C_{-\infty, p_{i}}(\omega))\subset W^{u}(x)\cap B(x,k_{0}\epsilon), \end{align} $$

but $\nu ^{u}(C_{0,p_{i}}(\pi ^{+}(\omega _{i})))\leq \nu ^{u}(\pi ^{+}(\chi ^{-1}(B(x, k_{0}\epsilon )\cap W^{u}(x))))$ , thus, since $\nu ^u$ is a Gibbs measure (see (2.11) and (2.5)), by Proposition 2.2 there exists a constant $k_{\tilde {h}}\geq 1$ , and knowing that $\max {\tilde {h}}<-\gamma $ for some $\gamma>0$ , we get

$$ \begin{align*} \frac{\nu^{u}(\pi^{+}(\chi^{-1}(I^{u})))}{\nu^{u}(\pi^{+}(\chi^{-1}(B(x, k_{0}\epsilon)\cap W^{u}(x))))}&\leq \sum_{i=1}^{2}\frac{\nu^{u}(C_{0,q_{i}}(\pi^{+}(\omega_{i})))}{\nu^{u}(C_{0,p_{i}}(\pi^{+}(\omega_{i})))}\\ &\leq \sum_{i=1}^{2} k^{2}_{\tilde{h}} \exp\bigg(\sum_{k=p_{i}}^{q_{i}-1}\tilde{h}(\sigma^{k}(\omega_{i}))-(p_i -q_i)P_\sigma(\tilde{h})\bigg)\\ &\leq \sum_{i=1}^{2} k^{2}_{\tilde{h}} \exp((-\gamma+P_{\sigma}(\tilde{h}))(q_{i}-p_{i}))\\ &\leq\eta(\epsilon), \end{align*} $$

where $\eta (\epsilon )\rightarrow 0$ as $\epsilon $ goes to 0, since $q_{i}=o(p_{i})$ , and $p_{i}\rightarrow \infty $ . Set $\tilde {I^{u}}=\{(\omega ^{-}, \omega ^{+}): \text { there exists } \omega =(\omega _{n})_{n\in \mathbb {Z}}\in \Sigma _{A} \mbox { such that } \chi (\omega )\in I^{u}, \omega ^{-}=(\omega _{n})_{n\leq 0} \text { and } \omega ^{+}= (\omega _{n})_{n\geq 0}\}$ , and in the same way we define the set $\tilde {B}(x, k_{0}\epsilon )$ . We have

On the other hand, the measure $\nu $ is absolutely continuous with respect to the product measure $\nu ^{u}\otimes \nu ^{s}$ with bounded density function h (see Proposition 2.3 in [Reference Pesin and Sadovskaya17]). Then there exist $\kappa _{1}$ , $\kappa _{2}>0$ such that, using the previous inequality, we get

where $\tilde {\eta }(\epsilon )=({\kappa _{2}}/{\kappa _{1}})\eta (\epsilon )$ .

Proceeding with the same strategy as above, there exists $\bar {\eta }(\epsilon )$ such that $\nu (\chi ^{-1}(\mathcal {I}^{s}))\leq \bar {\eta }(\epsilon )\nu (\chi ^{-1}(B(x,k_{0}\epsilon )))$ , and this finishes the proof since $\tilde {\eta }(\epsilon )$ and $\bar {\eta }(\epsilon )$ go to 0 as $\epsilon \rightarrow 0$ .

Proof. We use the same method as Dvoretzky and Erdös in [Reference Dvoretzky and Erdös9, Reference Pène and Saussol12, Reference Pène and Saussol13, Reference Pène, Saussol and Zweimüller15]. We make a partition of the cylinder D according to the last passage $r\leq n$ in the time interval $0,\ldots ,n$ of the orbit of $(x,0)$ by the map $\tilde {T}$ into $A_{\epsilon }^{\pm }(y)\times \{0\}$ . This can be seen as follows.

For simplicity, we will write $A_{\epsilon }^{\pm }$ instead of $A_{\epsilon }^{\pm }(y)$ :

(5.8) $$ \begin{align} \nu(D)=\sum_{r=0}^{n}& \nu(x\in D;S_{r}\varphi(x)=0\mbox{, } \sigma^{r}(x)\in A_{\epsilon}^{\pm};\text{ for all }\nonumber\\ & \quad l=r,\ldots,n, S_{l}\varphi(x)\neq0 \text{ or } \sigma^{l}(x)\notin A_{\epsilon}^{\pm}). \end{align} $$

Knowing that $S_{r}\varphi (x)=0$ , then

$$ \begin{align*} \{\text{for all } l=r,\ldots,n, S_{l}\varphi(x)-S_{r}\varphi(x)\neq0 \mbox{ or } \sigma^{l}(x)\notin A_{\epsilon}^{\pm}\} \end{align*} $$

is equal to

$$ \begin{align*} \{\text{for all } k=1,\ldots,n-r, S_{k+r}\varphi(x)-S_{r}\varphi(x)\neq0 \mbox{ or } \sigma^{k+r}(x)\notin A_{\epsilon}^{\pm}\}, \end{align*} $$

but $S_{k+r}\varphi (x)-S_{r}\varphi (x)=\sum _{l=0}^{k-1}\varphi \circ \sigma ^{l+r}(x)=S_{k}\varphi (\sigma ^{r}x)$ . It follows that $\nu (D)$ is equal to

$$ \begin{align*} &\sum_{r=0}^{n} \nu(x\in D;S_{r}(x)=0,\sigma^{r}(x)\in A_{\epsilon}^{\pm}; \sigma^{-r}\{z\in\Sigma_{A}: \text{for all }\\& \qquad k=1,\ldots,n-r, S_{k}\varphi(z)\neq0,\sigma^{k}(z)\notin A_{\epsilon}^{\pm}\}), \end{align*} $$

from which we get

$$ \begin{align*} \nu(D)&=\sum_{r=0}^{n}\nu(D\cap\{S_{r}=0\}\cap \sigma^{-r}(A^{\pm}_{\epsilon}\cap\{w_{A^{\pm}_{\epsilon }}>n-r\}))\\&\geq \nu(D\cap A^{\pm}_{\epsilon}\cap\{w_{A^{\pm}_{\epsilon}}>n\})\\&\quad +\sum_{r=2q+1}^{n}\nu(D\cap\{S_{r}=0\}\cap \sigma^{-r}(A_{\epsilon}^{\pm}\cap\{w_{A_{\epsilon}^{\pm}}(x)>n-r\})). \end{align*} $$

Recall that D is a $(-q, q')$ -cylinder with $q\leq Q_{\epsilon }$ , and $A_{\epsilon }^{\pm }$ is a union of $(-Q_{\epsilon }, q")$ -cylinders for some $q">0$ . Let $E=A_{\epsilon }^{\pm }(y)\cap \{w_{A_{\epsilon }^{\pm }}>n\}$ be the set of points of $A_{\epsilon }^{\pm }(y)$ which do not return to $A_{\epsilon }^{\pm }$ before n, that is, $E=A_{\epsilon }^{\pm }\cap \bigcap _{s=1}^{n}(\{S_{s}\varphi \neq 0\}\cup \sigma ^{-s}(\Sigma _{A}\setminus A_{\epsilon }^{\pm }))$ .

We know that $\varphi $ is constant on $1$ -cylinders, then $\{S_{s}\varphi \neq 0\}$ is a union of s-cylinders, and $\sigma ^{-s}(\Sigma _{A}\setminus A_{\epsilon }^{\pm })=\Sigma _{A}\setminus \sigma ^{-s}(A_{\epsilon }^{\pm })$ is a union of $(-Q_{\epsilon }+s,q"+s)$ -cylinders. Then we get that E is a union of $(-Q_{\epsilon }, q")$ -cylinders, for some $q">0$ . Hence, E can be written as $\sigma ^{Q_{\epsilon }}({\Pi _{+}}^{-1}B)$ , where B is a subset of $\Sigma _{A}^{+}$ , and $\Pi _{+}$ is the canonical projection of $\Sigma _{A}$ on $\Sigma _{A}^{+}$ .

Now let $n=n_{\epsilon }$ . With all these conditions, we apply the local limit theorem in Proposition 4.3. Thus, there is $C_{1}>0$ such that

$$ \begin{align*} \nu(D) &\geq\nu(D\cap\{w_{A_{\epsilon}^{\pm}}(x)>n_{\epsilon}\})+\sum_{r=M_{\epsilon}}^{n_{\epsilon}}\frac{\nu(D)\nu(A_{\epsilon}^{\pm}\cap\{w_{A_{\epsilon}^{\pm}}(x)>n_{\epsilon}-r\})}{\sqrt{2\pi}\sqrt{r-(Q_{\epsilon}+l)}\sigma_{\varphi}}\\&\quad - C_{1}\sum_{r=M_{\epsilon}}^{n_{\epsilon}}\frac{\nu(A_{\epsilon}^{\pm}\cap\{w_{A_{\epsilon}^{\pm}}(x)>n_{\epsilon}-r\})Q_{\epsilon}\nu(D)}{r-2Q_{\epsilon}}. \end{align*} $$

There is $C_{3}>0$ such that the error term is controlled by

(5.9) $$ \begin{align} \sum_{r=M_{\epsilon}}^{n_{\epsilon}}\frac{\nu(A_{\epsilon}^{\pm})q\nu(D)}{r-2Q_{\epsilon}}\leq C_{3}Q_{\epsilon}\log{n_{\epsilon}}\nu(D)\nu(A_{\epsilon}^{\pm}), \end{align} $$

since $Q_{\epsilon }\nu (A_{\epsilon }^{\pm })\log {n_{\epsilon }}=o(1)$ , hence we get

$$ \begin{align*} \nu(D)&\geq \nu(D\cap\{w_{A^{\pm}_{\epsilon}}(x)>n_{\epsilon}\}) +\nu(D)\sum_{r=M_{\epsilon}}^{n_{\epsilon}}\frac{\nu(A_{\epsilon}^{\pm}\cap\{w_{A_{\epsilon}^{\pm}}(x)>n_{\epsilon}-r\}}{\sqrt{2\pi}\sqrt{r-Q_{\epsilon}}\sigma_{\varphi}}\\&\quad -o(\nu(D)), \end{align*} $$

and thus dividing by $\nu (D)$ yields (a) for $D\in \mathcal {D}_{\epsilon }$ such that $D\subset A_{\epsilon }^{\pm }(y)$ .

We prove inequality (b) following the same strategy. For $n_{\epsilon }=({t}/{\nu (A_{\epsilon }^{\pm }(y))})^{2}$ , using the same decomposition as in (5.8), we have

$$ \begin{align*} \nu(D)=\sum_{r=0}^{n_{\epsilon}}\nu(D\cap\{S_{r}=0\}\cap \sigma^{-r}(A_{\epsilon}^{\pm}(y)\cap\{w_{A^{\pm}_{\epsilon}(y)}(x)>n_{\epsilon}-r\})):=\sum_{r=0}^{n_{\epsilon}}G_{r,\epsilon}(x). \end{align*} $$

For $r=0$ , the first term is equal to $\nu (D\cap \{w_{A^{\pm }_{\epsilon }(y)}(x)>n_{\epsilon }\})$ . We then compute the sum of terms between 1 and $M_{\epsilon }$ :

$$ \begin{align*} &\sum\limits_{r=1}^{M_{\epsilon}}\nu(D\cap\{S_{r}=0\}\cap \sigma^{-r}(A_{\epsilon}^{\pm}(y)\cap\{w_{A_{\epsilon}^{\pm}(y)}>n_{\epsilon}-r\}))\\ &\quad= \nu\bigg(\bigcup_{r=1}^{M_{\epsilon}} D\cap\{S_{r}=0\}\cap \sigma^{-r}(A_{\epsilon}^{\pm}(y)\cap\{w_{A_{\epsilon}^{\pm}(y)}>n_{\epsilon}-r\})\bigg)\\&\quad\leq\nu(D\cap\{w_{A_{\epsilon}^{\pm}(y)}\leq M_{\epsilon}\}). \end{align*} $$

Then, using the local limit theorem in Proposition 4.3, there exists $C_{4}>0$ such that we have

$$ \begin{align*} \sum^{n_{\epsilon}}_{r=M_{\epsilon}} G_{r,\epsilon}(x)&\leq \frac{1}{\sqrt{2\pi}\sigma_{\varphi}}\sum^{n_{\epsilon}}_{r=M_{\epsilon}}\frac{\nu(D)\nu(A^{\pm}_{\epsilon}\cap\{w_{A_{\epsilon}^{\pm}}>n_{\epsilon}-r\})}{\sqrt{r-Q_{\epsilon}}}\\& \quad +C_{4}\frac{\nu(A^{\pm}_{\epsilon}\cap\{w_{A_{\epsilon}^{\pm}}>n_{\epsilon}-r\})Q_{\epsilon}\nu(D)}{r-2Q_{\epsilon}}. \end{align*} $$

Similarly as in (5.9), the error term is controlled by $o(\nu (D))$ . Combining these assumptions together, we obtain

(5.10) $$ \begin{align} \nu(D)&\leq o(\nu(D))+\nu(D\cap\{w_{A_{\epsilon}^{\pm}}>n_{\epsilon}\})+\nu(w_{A_{\epsilon}^{\pm}}\leq M_{\epsilon}|D)\nu(D)\nonumber\\&\quad +\frac{1}{\sqrt{2\pi}\sigma_{\varphi}}\nu(D)\nu(A^{\pm}_{\epsilon}\cap\{w_{A_{\epsilon}^{\pm}}>n_{\epsilon}\})\bigg(\sum^{n_{\epsilon}}_{r=M_{\epsilon}}\frac{1}{\sqrt{r-Q_{\epsilon}}}\bigg). \end{align} $$

Dividing by $\nu (D)$ yields the desired inequality.

Proof. Let $t>0$ . Using inequality (a) of Proposition 5.10 with $n_{\epsilon }=({t}/{\nu (A^{\pm }_{\epsilon }(y))})^{2}$ , we have

$$ \begin{align*} 1&\geq\nu(w_{A_{\epsilon}^{\pm}(y)}>n_{\epsilon}|A^{\pm}_{\epsilon}(y))+\frac{1}{\sqrt{2\pi}\sigma_{\varphi}}\sum_{M_{\epsilon}}^{n_{\epsilon}}\frac{\nu(A_{\epsilon}^{\pm}(y)\cap \{w_{A_{\epsilon}^{\pm}(y)}>n_{\epsilon}-r\})}{\sqrt{r-Q_{\epsilon}}}- o(1)\\ &\geq\nu(w_{A_{\epsilon}^{\pm}(y)}>n_{\epsilon}|A^{\pm}_{\epsilon}(y))+\frac{1}{\sqrt{2\pi}\sigma_{\varphi}}\nu(A_{\epsilon}^{\pm}(y)\cap \{w_{A_{\epsilon}^{\pm}(y)}>n_{\epsilon}\})\sum_{M_{\epsilon}}^{n_{\epsilon}}\frac{1}{\sqrt{r-Q_{\epsilon}}}- o(1), \end{align*} $$

from which it follows that, for all $t>0$ ,

(5.11) $$ \begin{align} \nu\bigg(\ w_{A_{\epsilon}^{\pm}(y)}>\bigg(\frac{t}{\nu(A^{\pm}_{\epsilon}(y))}\bigg)^{2}\bigg| A^{\pm}_{\epsilon}(y)\bigg)\leq \frac{1+o(1)}{1+{1}/({\sqrt{2\pi}\sigma_{\varphi}})\nu(A_{\epsilon}^{\pm}(y))\sum^{n_{\epsilon}}_{M_{\epsilon}} {1}/({\sqrt{r-Q_{\epsilon}}})}, \end{align} $$

but ${\sum }_{M_{\epsilon }}^{{n_{\epsilon }}}{1}/{(\sqrt {r-Q_{\epsilon }})}=2(\sqrt {n_{\epsilon }}-\sqrt {Q_{\epsilon }-M_{\epsilon }})\geq \sqrt {n_{\epsilon }}\geq ({t}/{\nu (A_{\epsilon }^{\pm }(y))})$ . Then we get

(5.12) $$ \begin{align} \underset{\epsilon\rightarrow0}{\limsup}\ \nu(\nu(A^{\pm}_{\epsilon}(y))^{2}w_{A_{\epsilon}^{\pm}(y)}>t^{2}\mid A^{\pm}_{\epsilon}(y))\leq\frac{1}{1+t}, \end{align} $$

which proves the tightness.

Proof. Let $\alpha>0, k_{0}>0, m_{F}>0, c_{F}>0$ . Choose some $a\in (0,\alpha )$ and set, for some $\epsilon _{0}>0$ ,

$$ \begin{align*} F_{a}=\{x\in\Sigma_{A}:\text{ for all }\epsilon\leq\epsilon_{0},M_{\epsilon}^{-1}R_{B(x,2k_{0}\epsilon)}(x)>a\}. \end{align*} $$

We will show that $\nu (F_{a})\rightarrow 1$ as $\epsilon _{0}\rightarrow 0$ . To this end, let us prove that for $\nu $ -almost every $x\in \Sigma _{A}$ ,

(5.13) $$ \begin{align} \underset{\epsilon\rightarrow0}{\lim}\ M_{\epsilon}^{-1}R_{B(x,2k_{0}\epsilon)}(x)=+\infty. \end{align} $$

An intermediate step in doing so is to prove that

(5.14) $$ \begin{align} \underset{\epsilon\rightarrow0}{\lim}\ \frac{\log R_{B(x,2k_{0}\epsilon)}}{-\log\epsilon}= \text{dim}_{\mathrm{H}}\mu-1, \quad \nu\mbox{-almost everywhere}. \end{align} $$

Let $r_{\min }>0$ and $r_{\max }>0$ the lower and upper bounds of the height function r. Using (4.3), for $\mu _{\Delta }$ -almost every $y=(x,s)\in \Delta $ , we have

$$ \begin{align*} \bar{\tau_{\epsilon}}(y)\frac{1}{r_{\max}}\leq R_{B(x,2k_{0}\epsilon)}(x)\leq\bar{\tau_{\epsilon}}(y)\frac{1}{r_{\min}}, \end{align*} $$

which gives

$$ \begin{align*} \frac{\log r_{\max}}{\log\epsilon}+\frac{\log\bar{\tau_{\epsilon}}(y)}{-\log\epsilon}\leq\frac{\log R_{B(x,2k_{0}\epsilon)}(x)}{-\log\epsilon}\leq \frac{\log r_{\min}}{\log\epsilon}+\frac{\log\bar{\tau_{\epsilon}}(y)}{-\log\epsilon}, \end{align*} $$

using Theorem 5.14. It follows that $\lim _{\epsilon \rightarrow 0}{(\log R_{B(x,2k_{0}\epsilon )})}/{(-\log \epsilon )}= \dim _{\mathrm {H}}\mu -1$ , $\nu $ -almost everywhere.

We now go back to show (5.13). Let $0<\delta _{0}<\dim _{\mathrm {H}}\mu -1$ , we have

(5.15) $$ \begin{align} \log(M_{\epsilon}^{-1}R_{B(x,2k_{0}\epsilon)}(x))=\log(M_{\epsilon}^{-1}\epsilon^{-\delta_{0}})+\log(\epsilon^{\delta_{0}}R_{B(x,2k_{0}\epsilon)}(x)). \end{align} $$

Note that the first term of (5.15) goes to $+\infty $ as $\epsilon $ goes to 0. This is true using the definition of $M_{\epsilon }$ . There exists $\rho _{0}>0$ such that $M_{\epsilon }<2 ({ |\!\log ({\theta _{\epsilon }}/{2Cc_{g}c_{l}})|}/{|\!\log \rho _{u}|})+1\leq \rho _{0}|\!\log \epsilon |$ , which is negligible with respect to $\epsilon ^{-\delta _{0}}$ .

Now, considering the second term, we observe that

$$ \begin{align*} \frac{\log(\epsilon^{\delta_{0}}R_{B(x,2k_{0}\epsilon)}(x))}{-\log\epsilon}=-\delta_{0}+\frac{\log R_{B(x,2k_{0}\epsilon)}(x)}{-\log\epsilon}, \end{align*} $$

from which we get

$$ \begin{align*}\log(\epsilon^{\delta_{0}}R_{B(x,2k_{0}\epsilon)}(x))\underset{\epsilon\rightarrow0}{\rightarrow} (-\delta_{0}+ {\dim}_{\mathrm{H}}\mu-1)(+\infty).\end{align*} $$

Therefore, (5.13) is proved.

Taking $\epsilon _{0}\rightarrow 0$ , by Egoroff’s theorem, $M_{\epsilon }^{-1}R_{B(x,2k_{0}\epsilon )}$ converges uniformly to $+\infty $ on $F_{a}$ , and $\nu (F_{a}^{C})\rightarrow 0$ .

Let $\epsilon <\epsilon _{0}$ . If $x'\in B(x,k_{0}\epsilon )$ and $R_{B(x,k_{0}\epsilon )}(x')\leq M_{\epsilon }$ , we have $R_{B(x,2k_{0}\epsilon )}(x')\leq M_{\epsilon }$ as well, hence $x'\in F_{a}^{C}$ . Therefore,

$$ \begin{align*} D_{\epsilon}\cap\{x'\in\Sigma_{A},R_{B(x,k_{0}\epsilon)}(x')\leq M_{\epsilon}\}\subset D_{\epsilon}\cap F^{C}_{a}. \end{align*} $$

This means that for any density point x of the set $F_{a}$ relative to the Lebesgue basis given by $(B(\cdot,\delta ))_{\delta }$ , ${(\nu (F_{a}\cap B(x,\delta )))}/{(\nu (B(x,\delta )))}\rightarrow 1$ , as $\delta \rightarrow 0$ , and this is true for almost every point x. Therefore, we get

$$ \begin{align*} \nu(R_{B(x,k_{0}\epsilon)}\leq M_{\epsilon}|D_{\epsilon})&\leq\nu(F_{a}^{C}|D_{\epsilon})\\ &=\frac{\nu(F^{C}_{a}\cap D_{\epsilon})}{\nu(D_{\epsilon})}\\ &\leq\frac{\nu(F^{C}_{a}\cap B(x,\text{diam}(D_{\epsilon})))}{\nu(D_{\epsilon})}\\ &\leq \nu(F^{C}_{a}| B(x,\text{diam}(D_{\epsilon})))\frac{\nu(B(x,\text{diam}(D_{\epsilon})))}{\nu(D_{\epsilon})}, \end{align*} $$

which converges to 0 as $\epsilon \rightarrow 0$ , since $\lim _{\epsilon \rightarrow 0} \nu (F_{a}^{C}|B(x,\text {diam} (D_{\epsilon })))=0$ and ${\nu (B(x,\text {diam}(D_{\epsilon }))}/{\nu (D_{\epsilon })}$ is bounded above by $c_{F}^{m_{F}}$ .

Proof. After proving the tightness in Proposition 5.13, it will be enough to prove that the advertised limit is the only possible accumulation point of the distribution. Let $(\epsilon _{p})_{p\geq 1}$ be a positive sequence with $\lim _{p\rightarrow \infty } \epsilon _{p}=0$ and such that the conditional distribution of the $(\nu (A_{\epsilon _{p}}^{+})\sqrt {w_{A_{\epsilon _{p}}^{+}}}|A^{+}_{\epsilon _{p}})_{p\geq 0}$ converges to the law of some random variable $\mathcal {X}$ . Then, using Lemma 3.5 in [Reference Yassine23], it follows from Lemma 5.15 that $\mathcal {X}$ satisfies the integral equation

(5.16) $$ \begin{align} 1=\mathbb{P}(\mathcal{X}>t)+ t\frac{1}{\sqrt{2\pi}\sigma_{\varphi}}\int_{0}^{1}\frac{\mathbb{P}(\mathcal{X}>t\sqrt{1-u})}{\sqrt{u}}\,du \quad \text{for all } t>0. \end{align} $$

Then the conclusion follows as in Theorem 3.2 in [Reference Yassine23], the distribution of $\mathcal {X}$ coinciding with that of $\sigma _{\varphi }({\mathcal {E}}/{\mid \mathcal {N}\mid })$ .

Proof. By density in $L^1(\mu )$ , it is enough to prove the result for $\mathbb P$ with density h such that $\sum _{k\in \mathbb {Z}} h\circ I^{k}$ is uniformly continuous on $\Lambda $ . Let $\tilde {\mu }=\nu \otimes {\mathrm {Leb}}\otimes \sum _{k\in \mathbb {Z}}\delta _{k}$ be the measure on $\tilde {M}$ . Set $\bar {\mu }:=\nu \otimes {\mathrm {Leb}}\otimes \delta _{0}$ the measure restricted on the 0-cell and let $\mathbb {P}$ be a probability measure absolutely continuous with respect to $\mu $ with a density function h. Set ${H(\cdot):=\sum _{k\in \mathbb {Z}}h(\cdot,k)}$ . By $\mathbb {Z}$ -periodicity, the distribution of $\tau _{\epsilon }$ under $\mathbb {P}$ is the same as under the probability measure absolutely continuous with respect to $\bar {\mu }$ with density H. We have $y=(x,s)\in \bigcup _{P\in \mathcal {P}_{\epsilon }}P$ , and there exist $i,j$ such that $y\in \mathcal {P}_{\epsilon ,i,j}$ . Let $y_{\epsilon ,i,j}\in \mathcal {P}_{\epsilon ,i,j}$ . Then

$$ \begin{align*} H(y_{\epsilon,i,j})-w(H,\theta_{\epsilon})\leq H(y)\leq H(y_{\epsilon,i,j})+\omega(H,\theta_{\epsilon}), \quad y\in\mathcal{P}_{\epsilon,i,j}. \end{align*} $$

Then in turn

(5.17) $$ \begin{align} \underset{y\in\mathcal{P}_{\epsilon,i,j}}{\min}H(y)\geq H(y_{\epsilon,i,j})-\omega(H,\theta_{\epsilon}) \end{align} $$

and

(5.18) $$ \begin{align} \underset{y\in\mathcal{P}_{\epsilon,i,j}}{\max}H(y)\leq H(y_{\epsilon,i,j})-\omega(H,\theta_{\epsilon}). \end{align} $$

We are now in a position to proof our main result. For simplicity we will denote $A^{\pm }_{\epsilon ,i,j}:=A^{\pm }_{\epsilon }(y_{\epsilon ,i,j})$ . Due to (4.3) and Slutzky’s lemma, it is enough to prove that $(\nu (A_{\epsilon })(\cdot))^{2}w_{A_{\epsilon }(\cdot))}\circ P_{D}(\cdot))_{\epsilon }$ converges in distribution with respect to $(H\bar {\mu })$ to $\sigma _{\varphi }^{2}({\mathcal {E}^{2}}/{\mathcal {N}^{2}})$ .

For all $\epsilon>0$ , $\text { for all } t>0$ we set

(5.19)

Then, due to the inclusions in (5.2), for $y=(x,s)\in \mathcal {P}_{\epsilon ,i,j}$ ,

(5.20) $$ \begin{align} \nu(A_{\epsilon,i,j}^{-})^{2}w_{A_{\epsilon,i,j}^{+}}(x)\leq\nu(A_{\epsilon}(x,s))^{2}w_{A_{\epsilon}(x,s)}(x)\leq\nu(A_{\epsilon,i,j}^{+})^{2}w_{A^{-}_{\epsilon,i,j}}(x), \end{align} $$

therefore, using inequality (5.18), we get

For all $y\in \mathcal {P}_{\epsilon ,i,j}$ , we have $A_{\epsilon }^{-}(y)=A_{\epsilon ,i,j}^{-}$ , $A_{\epsilon }^{+}(y)=A_{\epsilon ,i,j}^{+}$ , and $D_{\epsilon ,i}=D_{\epsilon }(x)$ . Then, for all $t>0$ ,

(5.21) $$ \begin{align} E_{\epsilon,t}&\leq\sum_{i,j}H(y_{\epsilon,i,j})\int_{P_{i,j}}\nu(\nu(A_{\epsilon}^{+}(y))^{2}w_{A^{-}_{\epsilon}(y)}>t| D_{\epsilon}(x))\,d\bar{\mu}(y)+\omega(H,\theta_{\epsilon})\nonumber\\&\leq\int_{M}H(y)\nu(\nu(A_{\epsilon}^{+}(y))^{2}w_{A^{-}_{\epsilon}(y)}>t| D_{\epsilon}(x))\,d\bar{\mu}(y)+2\omega(H,\theta_{\epsilon}). \end{align} $$

We now have that $\nu (A^{-}_{\epsilon }(y))$ and $\nu (A_{\epsilon }^{+}(y))$ are equivalent (Lemma 5.7). Then there exists $\alpha (\epsilon )^{(y)}>0$ such that $\alpha (\epsilon )^{(y)}\rightarrow 0$ as $\epsilon \rightarrow 0$ and

(5.22) $$ \begin{align} \nu(A_{\epsilon}^{-}(y))(1-\alpha(\epsilon)^{(y)})\leq\nu(A_{\epsilon}^{+}(y))\leq\nu(A_{\epsilon}^{-}(y))(1+\alpha(\epsilon)^{(y)}), \end{align} $$

and, using Lemma 5.12, we get

$$ \begin{align*} \nu(\nu(A_{\epsilon}^{+}(y))^{2}w_{A^{-}_{\epsilon}(y)}>t| D_{\epsilon}(x)) &\leq\nu((1+\alpha(\epsilon)^{(y)})^{2}\nu(A_{\epsilon}^{-}(y))^{2}w_{A^{-}_{\epsilon}(y)}>t| D_{\epsilon}(x))\\ &\leq\nu((1+\alpha(\epsilon)^{(y)})^{2}\nu(A^{-}_{\epsilon}(y))^{2}w_{A^{-}_{\epsilon}(y)}>t| A^{-}_{\epsilon}(y))\\ &\quad +\ \nu(w_{A_{\epsilon}^{-}(y)}\leq M_{\epsilon}|A_{\epsilon}^{-}(y))\\ &\quad -\nu(w_{A_{\epsilon}^{-}(y)}\leq M_{\epsilon}| D_{\epsilon}(x))+o(1). \end{align*} $$

Moreover, due to Lemma 5.15, applied with $A_{\epsilon }^{-}(y)$ and $D_{\epsilon }(x)$ , we have $\lim _{\epsilon \rightarrow 0} \nu (w_{A_{\epsilon }^{-}(y)} \leq M_{\epsilon }|A_{\epsilon }^{-}(y))=0$ , and $\lim _{\epsilon \rightarrow 0} \nu (w_{A_{\epsilon }^{-}(y)}\leq M_{\epsilon }| D_{\epsilon }(x))=0$ . And since H is uniformly continuous, $\lim {\epsilon \rightarrow 0} \omega (H,\nu _{\epsilon })=0$ . Thus, using the Lebesgue’s dominated convergence theorem in (5.21), we end up with

(5.23) $$ \begin{align} \underset{\epsilon\rightarrow0}{\limsup}\ E_{\epsilon,t}\leq \mathbb{P}\bigg(\bigg(\sigma^{2}_{\varphi}\int R\,d\nu_{X}\bigg)\frac{\mathcal{E}^{2}}{\mathcal{N}^{2}}>t\bigg) \quad \text{for all } t>0. \end{align} $$

We deal with the second inequality analogously; thus, using (5.17) and the first inequality in (5.20), we get

Similarly, due to Lemma 5.12, there exists $\alpha (\epsilon )^{(y)}>0$ such that

$$ \begin{align*} & \nu(\nu(A_{\epsilon}^{-}(y))^{2}w_{A^{+}_{\epsilon}}>t| D_{\epsilon}(x)) \geq\nu\bigg(\frac{\nu(A^{+}_{\epsilon}(y))^{2}}{(1+\alpha(\epsilon)^{(y)})^{2}}w_{A^{+}_{\epsilon}}>t| A^{+}_{\epsilon}(y)\bigg)\\ &\quad+\ \nu(w_{A_{\epsilon}^{+}(y)}\leq M_{\epsilon}|A_{\epsilon}^{+}(y))-\nu(w_{A_{\epsilon}^{+}(y)} \leq M_{\epsilon}| D_{\epsilon}(x))+o(1). \end{align*} $$

Then, using Proposition 5.16, we get

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\lim}\ \nu\bigg(\frac{\nu(A^{+}_{\epsilon})^{2}}{(1+\alpha(\epsilon)^{(y)})^{2}}w_{A^{+}_{\epsilon}(y)}(x)>t| A^{+}_{\epsilon}(y)\bigg)=\mathbb{P}\bigg(\frac{\sigma^{2}_{\varphi}\mathcal{E}^{2}}{\mathcal{N}^{2}}>t\bigg) \quad \text{for all } t>0. \end{align*} $$

By using the same reasoning as demonstrated in the proof of (5.23), we can also establish that

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\liminf}\ E_{\epsilon,t}\geq \mathbb{P}\bigg(\sigma^{2}_{\varphi}\frac{\mathcal{E}^{2}}{\mathcal{N}^{2}}>t\bigg) \quad \text{for all } t>0. \end{align*} $$

Combining these results, we can deduce the convergence in law of $\nu (A_{\epsilon })^{2}w_{A_{\epsilon }(\cdot)}\circ P_{D}:\lim _{\epsilon \rightarrow 0} E_{\epsilon ,t}= \mathbb {P}(\sigma ^{2}_{\varphi }({\mathcal {E}^{2}}/{\mathcal {N}^{2}})>t)$ . Therefore, by applying Slutsky’s lemma, we obtain

$$ \begin{align*} \underset{\epsilon\rightarrow0}{\lim} \ \mathbb{P}(\nu(A_{\epsilon}(\cdot))^{2}\tau_{\epsilon}(\cdot)>t)=\mathbb{P}\bigg(\bigg(\sigma^{2}_{\varphi}\int R\,d\nu_{X}\bigg)\frac{\mathcal{E}^{2}}{\mathcal{N}^{2}}>t\bigg) \quad \text{for all } t>0. \\[-41pt] \end{align*} $$

Proof. Let $\tilde {y}\in \tilde {M}$ . Note that since we are interested in the asymptotic behavior of $\tilde \nu _0(B(\tilde y,\epsilon ))$ when $\epsilon $ tends to $0$ , we can replace $\tilde D_0(\tilde y)$ by a smaller disk if we want.

If the disk $\tilde D_0(\tilde y)$ and $\epsilon $ are small enough, the image measure of $\tilde \mu ({\bigcup }_{-\epsilon <s<\epsilon }\tilde g_s (\tilde D_0(\tilde y)\cap \cdot ))$ by the projection of $\tilde M$ to M is equal to the measure $\mu ({\bigcup }_{-\epsilon <s<\epsilon }g_s(D_0(y) \cap \cdot )$ , where $D_0=D_0(y)$ is the disk around y obtained by the projection of the disk $\tilde D_0=\tilde D_0(\tilde y)$ .

So now we have to show that, for $\mu $ -almost every y in M, the measure $\tilde {\nu }_0^{y}$ defined on $D_0(y)$ by

$$ \begin{align*} \tilde{\nu}_0^{y}(A)=\underset{\epsilon\rightarrow0}{\lim}\frac{1}{2\epsilon}\mu\bigg(\underset{-\epsilon<s<\epsilon}{\bigcup}g_{s}(D_{0}(y)\cap A)\bigg) \quad \text{for all } A\subset D_{0}(y), \end{align*} $$

is well defined and that $\tilde {\nu }_0^{y}(B(y,\epsilon ))$ is equivalent, as $\epsilon $ tends to 0, to $\nu (A_\epsilon (y))/\int _X R\,d\nu _{X}$ .

We assume that $y\in M\setminus X$ and that y is represented by a pair $(x,s)\in X_R$ with $x\not \in \partial X$ . Let us call D the connected component of X containing the point x. For $\epsilon $ small enough, $D_0(y)$ small enough, $\bigcup _{-\epsilon <s<\epsilon }g_s(D_0 (y))$ is contained in $\{(x',s')\in X_R\ :\ x\in D\}$ .

In $X_R$ , the set $D_0 (y)$ is represented by a curve $\{(x',h(x')) : x'\in P_{D}(D_0)\}$ with h of class $C^1$ of uniformly bounded derivative. We have, for every measurable set $A\subset D_0(y)$ ,

$$ \begin{align*} \mu\bigg(\underset{-\epsilon<s<\epsilon}{\bigcup}g_s(D_0(y) \cap A)\bigg) = \frac{2\epsilon \nu(\chi^{-1}( P_{D}(D_0(y)\cap A)))}{\int_X R\ d\nu_X} \end{align*} $$

This ensures the existence of $\nu _0$ and the fact that

$$ \begin{align*} \nu_{0}^{y}(\cdot)=\frac{\nu(\chi^{-1}(P_{D}(\cdot\cap D_{0}(y))))}{\int_{X} R\,d\nu_{X}}. \end{align*} $$

We have to prove that $\nu (A_{\epsilon }(y))$ is equivalent to $\nu (\chi ^{-1}(P_{D}(B(y,\epsilon )\cap D_{0}(y))))$ . We observe that $D_{0}(y)\cap B(y,\epsilon )\subset B(y,\epsilon )$ . Therefore, we have

$$ \begin{align*} A_{\epsilon}(y)=\chi^{-1}(P_{D}(B(y,\epsilon))))\supset \chi^{-1}(\Pi(D_{0}(y)\cap B(y,\epsilon))). \end{align*} $$

We need to prove that $\nu (A_\epsilon (y)) \setminus (\chi ^{-1}(P_{D}(D_0(y)\cap B(y,\epsilon ))))$ is negligible with respect to $\nu (A_\epsilon (y))$ .

We recall that $(x,s)\mapsto g_s(x)$ is a $C^1$ -diffeomorphism of $X_R$ on its image which contains the ball $B(y,\epsilon )$ , assuming $\epsilon $ is sufficiently small.