Proof. We will check that the hypothesis of Theorem 5.1 is met for ${\mathbb P}$ -a.e. $\omega $ with ${T_n = T_{\sigma ^{n-1} \omega }}$ , $\mu = \nu ^\omega $ . Recall that $c_n=c_\alpha (\epsilon ) = 0$ when $\alpha \in (0,1)$ . Using [Reference Kounias and WengKW69, Theorem 1] (see Theorem 5.6) and the equivariance of the family of measures $\{ \nu ^\omega \}_{\omega \in \Omega }$ , we have

which shows that the condition in equation (5.3) implies the condition in equation (5.1) for all $\delta> 0$ and $\ell \ge 1$ .

Proof. As in the proof of Proposition 5.4, we check the hypothesis of Theorem 5.1 with $T_n = T_{\sigma ^{n-1} \omega }$ , $\mu = \nu ^\omega $ for ${\mathbb P}$ -a.e. $\omega \in \Omega $ . We will see that equation (5.1) follows from equations (5.4) and (5.5).

Using the equivariance of $\{ \nu ^\omega \}_{\omega \in \Omega }$ , we see that the condition in equation (5.1) is implied by the equation (5.4) and

(5.6)

We next show that the condition in equation (5.5) implies equation (5.6).

Since

we obtain that, using again the equivariance, for ${\mathbb P}$ -a.e. $\omega \in \Omega $ ,

Thus, the condition in equation (5.5) implies equation (5.6), which concludes the proof.

Proof. We check the conditions of Theorem 5.3.

The proof for $\alpha \in (0,1)$ is similar to the proof of Proposition 5.4, the proof of the case $\alpha \in [1,2)$ is similar to the proof of Proposition 5.7.

Proof. By condition (Dec), and as all the operators $P_\omega $ are Markov with respect to m, we have

(5.10)

which proves that is a Cauchy sequence in BV converging to a unique limit $h_\omega \in BV$ satisfying $P_\omega h_\omega = h_{\sigma \omega }$ for all $\omega $ . The lower bound in equation (5.8) follows from the condition (Min), while the upper bound is a consequence of the uniform Lasota–Yorke inequality of condition (LY), as actually the family $\{h_\omega \}_{\omega \in \Omega }$ is bounded in BV. To prove the Hölder continuity of $\omega \mapsto h_\omega $ with respect to the distance $d_\theta $ , we remark that if $\omega $ and $\omega '$ agree in coordinates $|k| \le n$ , then

$$ \begin{align*} \| h_\omega - h_{\omega'}\|_{\mathrm{BV}} = \|P_{\sigma^{-k}\omega}^k(h_{\sigma^{-k}\omega} - h_{\sigma^{-k} \omega'})\|_{\mathrm{BV}} \le C \theta^n \le C d_\theta(\omega, \omega').\\[-37pt] \end{align*} $$

Proof. We obtain the second claim by taking the difference between two values of $\ell $ in the first claim.

Fix $\ell \ge 1$ . For $\delta>0$ , let

$$ \begin{align*} U_k^n(\delta) = \bigg\{ \omega \in \Omega : \bigg| \sum_{j=0}^{kn -1} ({\mathbb E}_{\nu^{\sigma^j \omega}}(\chi_n) - {\mathbb E}_\nu(\chi_n) ) \bigg| \ge \delta \bigg\} \end{align*} $$

and

$$ \begin{align*} B^n(\delta) = \bigg\{ \omega \in \Omega : \sup_{0 \le k \le \ell} \bigg| \sum_{j=0}^{kn -1} ({\mathbb E}_{\nu^{\sigma^j \omega}}(\chi_n) - {\mathbb E}_\nu(\chi_n) ) \bigg| \ge \delta \bigg\}. \end{align*} $$

Note that

$$ \begin{align*} B^n(\delta) = \bigcup_{k=0}^\ell U_k^n(\delta). \end{align*} $$

We define $f_n(\omega ) = {\mathbb E}_{\nu ^\omega }(\chi _n)$ and $\overline {f}_n = {\mathbb E}_{{\mathbb P}}(f_n)$ . We claim that $f_n:\Omega \to \mathbb {R}$ is Hölder with norm $\|f_n \|_\theta = {\mathcal O}(n^{-1}\widetilde {L}(n))$ . Indeed, for $\omega \in \Omega $ , we have

$$ \begin{align*} |f_n(\omega)| = \bigg| \int_Y \chi_n(x) \,d \nu^\omega(x) \bigg| \le \|h_\omega\|_{L^\infty_m} \|\chi_n\|_{L^1_m} \le \frac{C}{n}\widetilde{L}(n), \end{align*} $$

and for $\omega , \omega ' \in \Omega $ , we have

$$ \begin{align*} |f_n(\omega) - f_n(\omega')| &= \bigg|\int_Y \chi_n(x) \,d \nu^\omega(x) - \int_Y \chi_n(x) \,d\nu^{\omega'}(x) \bigg| \\ & \le \int_Y | \chi_n(x) | \cdot |h_\omega(x) - h_{\omega'}(x)| \,dm(x) \\ & \le \| h_\omega - h_{\omega'}\|_{L^\infty_m} \| \chi_n \|_{L^1_m} \\ & \le \frac C n \widetilde{L}(n) d_\theta(\omega, \omega'), \end{align*} $$

since $\omega \in \Omega \mapsto h_\omega \in L^\infty (m)$ is Hölder continuous. In particular, we also have that ${\overline {f}_n = {\mathcal O}(n^{-1}\widetilde {L}(n))}$ .

We have, using Chebyshev’s inequality,

$$ \begin{align*} {\mathbb P}(U_k^n(\delta)) &= {\mathbb P}\bigg( \bigg\{ \omega \in \Omega : \bigg|\sum_{j=0}^{kn -1} ( f_n \circ \sigma^j - \overline{f}_n ) \bigg| \ge \delta \bigg\}\bigg) \\ & \le \frac{1}{\delta^2} {\mathbb E}_{{\mathbb P}}\bigg(\bigg(\sum_{j=0}^{kn -1} ( f_n \circ \sigma^j - \overline{f}_n )\bigg)^2 \bigg) \\ & \le \frac{1}{\delta^2} \bigg[ \sum_{j=0}^{kn - 1} ({\mathbb E}_{{\mathbb P}} |f_n \circ \sigma^j - \overline{f}_n|^2\\ &\quad+ 2 \sum_{0 \le i < j \le kn - 1} {\mathbb E}_{{\mathbb P}}((f_n \circ \sigma^i - \overline{f}_n)(f_n \circ \sigma^j - \overline{f}_n)) \bigg]. \end{align*} $$

By the $\sigma $ -invariance of ${\mathbb P}$ , we have

$$ \begin{align*} {\mathbb E}_{{\mathbb P}} |f_n \circ \sigma^j - \overline{f}_n|^2 = {\mathbb E}_{{\mathbb P}} |f_n - \overline{f}_n|^2, \end{align*} $$

and, since $(\Omega , {\mathbb P}, \sigma )$ admits exponential decay of correlations for Hölder observables, there exist $\unicode{x3bb} \in (0,1)$ and $C>0$ such that

$$ \begin{align*} {\mathbb E}_{{\mathbb P}}((f_n \circ \sigma^i - \overline{f}_n)(f_n \circ \sigma^j - \overline{f}_n)) &= {\mathbb E}_{{\mathbb P}}((f_n - \overline{f}_n)(f_n \circ \sigma^{j-i}- \overline{f}_n)) \\ & \le C \unicode{x3bb}^{j-i} \| f_n - \overline{f}_n\|_\theta^2. \end{align*} $$

We then obtain that

$$ \begin{align*} {\mathbb P}(U_k^n(\delta)) &\le \frac{C}{\delta^2} \bigg[kn \|f_n - \overline{f}_n\|_{L^2_m}^2 + 2 \sum_{0 \le i < j \le kn - 1} \unicode{x3bb}^{j-i} \| f_n - \overline{f}_n\|_\theta^2 \bigg]\\ & \le C \frac{nk}{\delta^2} \|f_n\|_\theta^2 \\ &\le C \frac{k}{n \delta^2}(\widetilde{L}(n))^2, \end{align*} $$

which implies that

$$ \begin{align*} {\mathbb P}(B^n(\delta)) \le C \frac{\ell^2}{n \delta^2} (\widetilde{L}(n))^2. \end{align*} $$

Let $\eta>0$ . By the Borel–Cantelli lemma, it follows that for ${\mathbb P}$ -a.e. $\omega \in \Omega $ , there exists $N(\omega , \delta ) \ge 1$ such that $\omega \notin B^{\lfloor p^{1+\eta }\rfloor } (\delta )$ for all $p \ge N(\omega , \delta )$ .

Let now $P:= \lfloor p^{1+\eta } \rfloor < n \le P' = \lfloor (p+1)^{1+\eta } \rfloor $ for p large enough. Let $0 \le k \le \ell $ . Then, since $\|f_n\|_\infty = {\mathcal O}(n^{-1}\widetilde {L}(n))$ ,

$$ \begin{align*} \bigg| \sum_{j=0}^{kP -1 } (f_n(\sigma^j \omega) - \overline{f}_n ) - \sum_{j=0}^{kn - 1}(f_n(\sigma^j \omega) - \overline{f}_n )\bigg| &\le \sum_{j=kP}^{kn - 1} |f_n(\sigma^j \omega) - \overline{f}_n | \\ & \le C \frac{P' - P}{P} \widetilde{L}(n) \le C \frac{\widetilde{L}(p^{1 + \eta})}{p}, \end{align*} $$

because on the one hand,

$$ \begin{align*} \frac{P' - P}{P} = \frac{\lfloor (p+1)^{1+\eta} \rfloor- \lfloor p^{1+\eta}\rfloor }{\lfloor p^{1+\eta} \rfloor} = {\mathcal O} \bigg( \frac 1 p \bigg), \end{align*} $$

and on the other hand, by Potter’s bounds, for $\tau> 0$ ,

$$ \begin{align*} \widetilde{L}(n) \le C \widetilde{L}(P) \bigg( \frac{n}{P} \bigg)^\tau \le C \widetilde{L}(P) \bigg( \frac{P'}{P} \bigg)^\tau \le C\widetilde{L}(P). \end{align*} $$

Since

$$ \begin{align*} \bigg| \sum_{j=0}^{kP -1} (f_n(\sigma^j \omega) - \overline{f}_n ) \bigg| < \delta \end{align*} $$

for all $0 \le k \le \ell $ , it follows that for ${\mathbb P}$ -a.e. $\omega $ , there exists $N(\omega , \delta )$ such that $\omega \notin B^n(2 \delta )$ for all $n \ge N(\omega , \delta )$ , which concludes the proof.

Proof. In the setting of intermittent maps, we must modify the argument of Lemma 7.1 slightly as $h_\omega $ is not a Hölder function of $\omega $ . Instead, we consider and use that, by equation (5.11),

(7.1) $$ \begin{align} \|h^i_{\omega}-h_{\omega}\|_{L^{1}(m)} \le C i^{1-{1}/{\gamma_{\mathrm{max}}}}\quad \text{(leaving out the log term).} \end{align} $$

Note that $h^i_{\omega }$ is the ith approximate to $h_{\omega }$ in the pullback construction of $h_{\omega }$ . Let $\nu _\omega ^i$ be the measure such that ${d \nu _\omega ^i}/{dm} = h_\omega ^i$ .

Consider

$$ \begin{align*} f_n^i(\omega) = {\mathbb E}_{\nu_\omega^i}(\chi_n)\:,\quad f_n(\omega) = {\mathbb E}_{\nu^\omega}(\chi_n), \end{align*} $$

$$ \begin{align*} \overline{f}_n^i = {\mathbb E}_{\mathbb P}(f_n^i)\:,\quad \overline{f}_n = {\mathbb E}_{\mathbb P}(f_n). \end{align*} $$

By equation (5.12), on the set $[\delta , 1]$ , the densities involved ( $h_\omega ^k, h_\omega , h=d \nu /d m$ ) are uniformly bounded above and away from zero. Thus, $\|f_n^i \|_\infty = {\mathcal O}(n^{-1})$ .

Pick $0<a<1$ such that $\beta :=(({1}/{\gamma _{\mathrm {max}}})-1)a -1>0$ .

For a given n, take $i=i_n=n^a$ . By equation (7.1), for all $\omega $ , n, and $i=n^a$ ,

$$ \begin{align*} |f_{n}^i (\omega) - f_{n} (\omega)| \le \|h_\omega^i -h_\omega\|_{L^1(m)} \| \chi_n \|_{L^\infty(m)} = {\mathcal O}(n^{-(\beta+1)}). \end{align*} $$

Then

$$ \begin{align*} |\overline{f}_n^i -\overline{f}_n| = {\mathcal O}(n^{-(\beta+1)}) \end{align*} $$

and

$$ \begin{align*} \bigg|\sum_{r=0}^{kn - 1} [f_{n}^i (\sigma^r \omega)-f_n (\sigma^r \omega)] \bigg|\le C \ell n^{-\beta}. \end{align*} $$

Given $\epsilon $ , choose n large enough that for all $0 \le k \le \ell $ ,

$$ \begin{align*} \bigg\{ \omega \in \Omega : \bigg| \sum_{r=0}^{kn -1} (f_n (\sigma^r \omega)- \overline{f}_n) \bigg|>\epsilon\bigg\} \subset \bigg\{ \omega \in \Omega : \bigg| \sum_{r=0}^{kn-1} (f^i_n (\sigma^r \omega)-\overline{f}^i_n) \bigg| >\frac{\epsilon}{2}\bigg\}. \end{align*} $$

By Chebyshev,

$$ \begin{align*} &{\mathbb P} \bigg(\ \bigg| \sum_{r=0}^{kn-1} (f^i_n \circ \sigma^r-\overline{f}^i_n)\bigg|>\frac{\epsilon}{2}\bigg) \le \frac{4}{\epsilon^2} \sum_{r=0}^{kn-1} {\mathbb E}_{{\mathbb P}} \bigg( [f_n^i \circ \sigma^r -\overline{f}^i_n]^2 \bigg)\\ &\quad +\frac{4}{\epsilon^2} \bigg[2 \sum_{r=0}^{kn-1} \sum_{u=r+1}^{kn-1} | {\mathbb E}_{{\mathbb P}}[(f^i_n \circ \sigma^r -\overline{f}^i_n)(f_n^i \circ\sigma^u-\overline{f}^i_n)]|\bigg]. \end{align*} $$

We bound

$$ \begin{align*} \sum_{r=0}^{kn-1} {\mathbb E}_{{\mathbb P}} ( [f_n^i -\overline{f}^i_n]^2 ) \le C \sum_{r=0}^{kn-1} \|f_n^i\|_\infty^2 \le \frac{C \ell}{ n} \end{align*} $$

and note that if $|r-u|>n^{a}$ , then by independence,

$$ \begin{align*} {\mathbb E}_{{\mathbb P}} [(f^i_n \circ\sigma^r -\overline{f}^i_n)(f_n \circ\sigma^u -\overline{f}^i_n)]= {\mathbb E}_{{\mathbb P}}[f^i_n \circ\sigma^r - \overline{f}^i_n] {\mathbb E}_{{\mathbb P}}[f^i_n \circ\sigma^u - \overline{f}^i_n]=0 \end{align*} $$

and hence, we may bound

$$ \begin{align*} \sum_{r=0}^{kn-1} \sum_{u=r+1}^{kn-1} | {\mathbb E}_{{\mathbb P}}[(f^i_n \circ \sigma^r -\overline{f}^i_n)(f_n^i \circ\sigma^u-\overline{f}^i_n)]| \le \frac{C \ell}{n^{1-a}}. \end{align*} $$

Thus, for n large enough,

$$ \begin{align*} {\mathbb P} \bigg( \bigg\{ \omega \in \Omega : \bigg| \sum_{r=0}^{kn-1} [f_n (\sigma^r \omega)-\overline{f}_n]\bigg|>\epsilon \bigg\}\bigg) \le \frac{C\ell}{n^{1-a}\epsilon^2}. \end{align*} $$

The rest of the argument proceeds as in the case of Lemma 7.1 using a speedup along a sequence $n=p^{1+\eta }$ , where $\eta>{a}/({1-a})$ , since $\|f_n\|_\infty = {\mathcal O}(n^{-1})$ still holds.

Proof of Theorem 7.3

When $\alpha \in (0,1)$ , we check the hypothesis of Proposition 5.4. Using equation (5.8), we have

Using Proposition 3.2, we see that the condition in equation (5.3) is satisfied since $\alpha <1$ , thus proving the theorem in this case.

When $\alpha \in [1,2)$ , we consider instead Proposition 5.7. First, we remark that the condition in equation (5.5) is implied by equations (7.2) and (5.8). It remains to check the condition in equation (5.4), which constitutes the rest of the proof.

If $\alpha \in (1,2)$ , we have

(7.3)

with

$$ \begin{align*} A_n^\omega(t) = \bigg| \frac{1}{b_n} \bigg[ \sum_{j=0}^{\lfloor nt \rfloor - 1} {\mathbb E}_{\nu^{\sigma^j \omega}}(\phi) - t c_n \bigg] \bigg|, \end{align*} $$

and

Since $\phi $ is regularity varying with index $\alpha>1$ , it is integrable and the function ${\omega \mapsto {\mathbb E}_{\nu ^\omega }(\phi )}$ is Hölder. Hence, it satisfies the law of the iterated logarithm, and we have for ${\mathbb P}$ -a.e. $\omega \in \Omega $ ,

$$ \begin{align*} \bigg|\frac{1}{k} \sum_{j=0}^{k-1} {\mathbb E}_{\nu^{\sigma^j \omega}}(\phi) - {\mathbb E}_\nu(\phi) \bigg| = {\mathcal O}\bigg( \frac{\sqrt{\log \log k}}{\sqrt k}\bigg). \end{align*} $$

Thus, we have

$$ \begin{align*} \sup_{0 \le t \le \ell} A_n^\omega(t) = \mathcal{O} \bigg( \frac{\sqrt{n \ell} \sqrt{\log \log (n \ell)}}{b_n} \bigg). \end{align*} $$

As a consequence, we can deduce that $\lim _{n \to \infty } \sup _{0 \le t \le \ell } A_n^\omega (t) = 0 $ since $b_n = n^{\frac {1}{\alpha }} \widetilde {L}(n)$ for a slowly varying function $\widetilde {L}$ , with $\alpha <2$ .

By Proposition 3.2, we also have

In particular, we have

$$ \begin{align*} \lim_{n \to \infty} \sup_{0 \le t \le \ell} C_{n, \epsilon}^\omega(t) = 0. \end{align*} $$

This also implies that ${\mathbb E}_m(| \chi _n| ) = {\mathcal O}(n^{-1})$ if we define . From Lemma 7.1, it follows that $\lim _{n \to \infty } \sup _{0 \le t \le \ell } B_{n, \epsilon }^\omega (t) = 0$ .

Putting all these estimates together concludes the proof when $\alpha \in (1,2)$ .

When $\alpha = 1$ , we estimate the right-hand side of equation (7.3) by $A_{n, \epsilon }^\omega (t) + B_{n, \epsilon }^\omega (t)$ with

and

We define . By Proposition 3.2, we have ${\mathbb E}_m(| \chi _n |) = {\mathcal O}(n^{-1} \widetilde {L}(n))$ for some slowly varying function $\widetilde {L}$ , and so by Lemma 7.1,

$$ \begin{align*} \lim_{n \to \infty} \sup_{0 \le t \le \ell} A_{n, \epsilon}^\omega(t) = 0. \end{align*} $$

However, by Proposition 3.2, we have

and so $\lim _{n \to \infty } \sup _{0 \le t \le \ell } B_{n, \epsilon }^\omega (t) = 0$ which completes the proof.

Proof of Theorem 6.1

Due to rounding errors when taking the integer parts, we have

$$ \begin{align*} &| \nu^{\sigma^{\lfloor ns \rfloor}\omega} (R_{A_n}(\sigma^{\lfloor ns \rfloor}\omega)> \lfloor n(t-s) \rfloor) - \nu^{\sigma^{\lfloor ns \rfloor}\omega} (R_{A_n}(\sigma^{\lfloor ns \rfloor}\omega) > \lfloor nt \rfloor - \lfloor ns \rfloor )| \\ &\quad\le \nu^{\sigma^{\lfloor nt \rfloor}\omega}(A_n) \le C m(A_n) \to 0, \end{align*} $$

and it is thus enough to prove the convergence of $\nu ^{\sigma ^{\lfloor ns \rfloor }\omega } (R_{A_n}(\sigma ^{\lfloor ns \rfloor }\omega )> \lfloor nt \rfloor - \lfloor ns \rfloor )$ .

By Lemmas 8.4 and 8.5, for all $N \ge 1$ , we have

(8.4) $$ \begin{align} \bigg| \nu^{\sigma^{\lfloor ns \rfloor}\omega} (R_{A_n}(\sigma^{\lfloor ns \rfloor}\omega)> \lfloor nt \rfloor - \lfloor ns \rfloor ) - \prod_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor} (1 - \nu^{\sigma^j \omega}(A_n) )\bigg| \le \mathrm{(I)} + \mathrm{(II)} + \mathrm{(III)}, \end{align} $$

with

$$ \begin{align*} \mathrm{(I)} = \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor}\nu^{\sigma^j \omega}(A_n \cap \{R_{A_n}(\sigma^j \omega) \le N \}), \end{align*} $$

$$ \begin{align*} \mathrm{(II)} = \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor} \nu^{\sigma^j \omega}(A_n) \nu^{\sigma^j \omega}(R_{A_n}(\sigma^j \omega) \le N), \end{align*} $$

and

To estimate (I), we choose $\epsilon> 0$ such that $J \subset \{ |x|> \epsilon \}$ and we introduce ${V_n = \{ | \phi |> \epsilon b_n \}}$ . For a measurable subset $V \subset Y$ , we also define the shortest return to V by

$$ \begin{align*} r_\omega(V) = \inf_{x \in V} R_V(\omega)(x), \end{align*} $$

and we set

$$ \begin{align*} r(V) = \inf_{\omega \in \Omega} r_\omega(V). \end{align*} $$

We have

It follows from condition (Dec) that

as BV is a Banach algebra, and both and $\|h_{\sigma ^j \omega } \|_{\mathrm { BV}}$ are uniformly bounded. (Recall that, from the definition of $\phi $ , it follows that $V_n$ is an open interval, and thus has a uniformly bounded BV norm.)

Consequently,

$$ \begin{align*} \mathrm{(I)} & \le \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor} \sum_{i=r_{\sigma^j \omega}(V_n)}^N [ \nu^{\sigma^j \omega}(V_n) \nu^{\sigma^{i+j} \omega}(V_n) + {\mathcal O}( \theta^i m(V_n))] \\ & \le C (m(V_n)^2 n N + m(V_n) n \theta^{r(V_n)} ). \end{align*} $$

However, we have by equation (8.2),

$$ \begin{align*} \mathrm{(II)} & \le \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor} \nu^{\sigma^j \omega}(A_n) \sum_{i=1}^N \nu^{\sigma^{i+j} \omega}(A_n) \\ & \le C n N m(A_n)^2, \end{align*} $$

and it follows from condition (Dec) that

since $\{h_\omega \}_{\omega \in \Omega }$ is a bounded family in BV, $A_n$ is the union of at most two intervals, and thus is uniformly bounded. We can thus bound equation (8.4) by

$$ \begin{align*} C ( m(V_n)^2 n N + m(V_n) n \theta^{r(V_n)} + m(A_n)^2 n N + n \theta^N ) \le C ( n^{-1} N + \theta^{r(V_n)} + n \theta^N ), \end{align*} $$

and, assuming for the moment that $r(V_n) \to + \infty $ , we obtain the conclusion by choosing $N = N(n) = 2 \log n$ and letting $n \to \infty $ .

It thus remains to show that $r(V_n) \to + \infty $ . Recall that $V_n$ is the ball of center $x_0$ and radius $b^{-1} \epsilon ^{-\alpha } n^{-1}$ . Let $R \ge 1$ be a positive integer. Since $x_0$ is assumed to be non-recurrent, and that the collection of maps $T_\omega ^j$ for $\omega \in \Omega $ and $0 \le j < R$ is finite, we have that

$$ \begin{align*} \delta_R := \inf_{\omega \in \Omega} \inf_{0 \le j < R} |T_\omega^j(x_0) - x_0|>0 \end{align*} $$

is positive. Since all the maps $T_\omega ^j$ are continuous at $x_0$ by assumption, there exists $n_R \ge 1$ such that for all $n \ge n_R$ , $j < R$ , and $\omega \in \Omega $ ,

$$ \begin{align*} x \in V_n \Longrightarrow | T_\omega^j(x) - T_\omega^j(x_0)| < \frac{\delta_R}{2}. \end{align*} $$

Increasing $n_R$ if necessary, we can assume that $b^{-1} \epsilon ^{-\alpha } n^{-1} < ({\delta _R}/{2})$ for all $n \ge n_R$ .

Then, for all $n \ge n_R$ , $\omega \in \Omega $ , $j < R$ , and $x \in V_n$ , we have

$$ \begin{align*} |T_\omega^j(x) - x_0| \ge |T_\omega^j(x_0) - x_0| - |T_\omega^j(x) - T_\omega^j(x_0)|> \frac{\delta_R}{2} > b^{-1} \epsilon^{-\alpha} n^{-1}, \end{align*} $$

and thus $T_\omega ^j(x) \notin V_n$ .

This implies that $r(V_n)> R$ for all $n \ge n_R$ , which concludes the proof as R is arbitrary.

Proof. Let $N = \lfloor n (\log n)^{-1} \rfloor $ an $V_n = B_{c n^{-1}}(x_0)$ . First, we remark that for m-a.e. $x_0$ and ${\mathbb P}$ -a.e. $\omega $ ,

(8.5) $$ \begin{align} m (V_n \cap \{R_{V_n}(\omega) \le N \} ) = o(n^{-1}). \end{align} $$

This is a consequence of [Reference Haydn, Rousseau and YangHRY20, Theorem 7.2]. Their result is stated for two intermittent LSV maps both with $\gamma <\tfrac 13$ but generalizes immediately to a finite collection of maps with a uniform bound of $\gamma _{\mathrm {max}}<\tfrac 13$ . The exponential law for return times to nested balls implies that for a fixed t, for m-a.e $x_0$ , and ${\mathbb P}$ -a.e. $\omega $ ,

$$ \begin{align*} \lim_{n \to \infty} \frac{1}{\nu^\omega(V_n)} \nu^\omega (V_n \cap \{R_{V_n}(\omega) \le nt \} ) = 1-e^{-t}, \end{align*} $$

which shows in particular, since $\{R_{V_n}(\omega ) \le N \} \subset \{R_{V_n}(\omega ) \le nt \}$ for all n large enough, that for all $t>0$ , m-a.e $x_0$ , and ${\mathbb P}$ -a.e. $\omega $ ,

(8.6) $$ \begin{align} \limsup_{n \to \infty} \frac{1}{\nu^\omega(V_n)} \nu^\omega (V_n \cap \{R_{V_n}(\omega) \le N \} ) \le 1 - e^{-t}. \end{align} $$

Using equation (5.12), taking the limit $t \to 0$ proves equation (8.5). Note that, even though the set of full measure of $x_0$ and $\omega $ such that equation (8.6) holds may depend on t, it is enough to consider only a sequence $t_k \to 0$ .

Now, for $k \ge 0$ and $n_0 \ge 1$ , we introduce the set

$$ \begin{align*} \Omega_k^{n_0} = \bigg\{ \omega \in \Omega : m(V_n \cap \{R_{V_n}(\omega) \le N \} ) \le \frac{2^{-k}}{n} \text{ for all } n \ge n_0 \bigg\}. \end{align*} $$

According to equation (8.5), we have for all $k \ge 0$ ,

$$ \begin{align*} \lim_{n_0 \to \infty} {\mathbb P}(\Omega_k^{n_0}) = {\mathbb P}\bigg( \bigcup_{n_0 \ge 1} \Omega_k^{n_0} \bigg) = 1. \end{align*} $$

By the Birkhoff ergodic theorem, for all $k \ge 0$ , $n_0 \ge 1$ , and ${\mathbb P}$ -a.e. $\omega $ ,

which implies that for all $0 \le s < t$ ,

Let $n_0 = n_0(\omega , k)$ such that ${\mathbb P}(\Omega _k^{n_0}) \ge 1 - 2^{-k}$ , and for all $n \ge n_0$ ,

Then, for all $n \ge n_0(\omega , k)$ , we have

Consequently,

$$ \begin{align*} \sum_{\lfloor ns \rfloor +1}^{\lfloor nt \rfloor} m(V_n \cap \{R_{V_n}(\omega) \le N \} ) \le (\lfloor nt \rfloor - \lfloor ns \rfloor ) \frac{2^{-k}}{n} + (\lfloor nt \rfloor - \lfloor ns \rfloor ) 2^{-(k-1)} m(V_n). \end{align*} $$

This proves that

$$ \begin{align*} \limsup_{n \to \infty} \sum_{\lfloor ns \rfloor +1}^{\lfloor nt \rfloor} m(V_n \cap \{R_{V_n}(\omega) \le N \} ) \le C \, 2^{-k} \end{align*} $$

and the result follows by taking the limit $k \to \infty $ .

Note that the set of $x_0$ and $\omega $ for which the lemma holds depends a priori on $c>0$ , but it is enough to consider a countable and dense set of c, since for $c < c'$ ,

$$ \begin{align*} \{ B_{c n^{-1}}(x_0) \cap \{ R_{B_{c n^{-1}}(x_0)}^\omega \le N \} \} \subset \{ B_{c' n^{-1}}(x_0) \cap \{ R_{B_{c' n^{-1}}(x_0)}^\omega \le N \} \}.\\[-42pt] \end{align*} $$

Proof of Theorem 6.2

We consider the three terms in equation (8.4) with ${N = \lfloor n (\log n)^{-1} \rfloor }$ .

Let $V_n = \{ | \phi |> \epsilon b_n\}$ , where $\epsilon>0$ is such that $A_n \subset V_n$ for all $n\ge 1$ . Since $V_n$ is a ball of center $x_0$ and radius $b^{-1} \epsilon ^{-\alpha } n^{-1}$ , and since $V_n \subset [\delta , 1]$ , the term

$$ \begin{align*} \mathrm{(I)} = \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor}\nu^{\sigma^j \omega}(A_n \cap \{R_{A_n}(\sigma^j \omega) \le N \}) \le C \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor} m(V_n \cap \{R_{V_n}(\sigma^j \omega) \le N \}) \end{align*} $$

tends to zero by Lemma 8.7 for m-a.e $x_0$ .

The term

$$ \begin{align*} \mathrm{(II)} = \sum_{j = \lfloor ns \rfloor +1}^{\lfloor nt \rfloor} \nu^{\sigma^j \omega}(A_n) \nu^{\sigma^j \omega}(R_{A_n}(\sigma^j \omega) \le N) \le CnNm(A_n)^2 \end{align*} $$

also tends to zero since $N=o(n)$ . Lastly, we consider

We approximate by a $C^1$ function g such that $\|g\|_{C^1}\le n^{\tau }$ , on $A_n$ , and (recall $A_n$ is two intervals of length roughly ${1}/{n}$ so a simple smoothing at the endpoints of the intervals allows us to find such a function g). Later we will specify $\tau>1$ as needed. By [Reference Nicol, Pereira and TörökNPT21, Lemma 3.4] with $h = h_\omega $ and ${\varphi = g - m(g h_\omega )}$ , for all $\omega $ ,

$$ \begin{align*} \| P_{\omega}^N ([g-m(g h_\omega)] h_{\omega})\|_{L^1}&\le C n^\tau N^{1-{1}/{\gamma_{\mathrm{max}}}} (\log N)^{{1}/{\gamma_{\mathrm{max}}}} \\ & \le C n^{\tau +1 - {1}/{\gamma_{\mathrm{max}}}} (\log n)^{({2}/{\gamma_{\mathrm{max}}}) - 1}. \end{align*} $$

Using the decomposition , we estimate, leaving out the log term,

$$ \begin{align*} \mathrm{(III)} \le C [n^{1-\tau}+n^{\tau +2-{1}/{\gamma_{\mathrm{max}}}}], \end{align*} $$

where the value of C may change line to line. Taking $\gamma _{\mathrm {max}}<\tfrac 13$ and $1<\tau <({1}/{\gamma _{\mathrm {max}}})-2$ suffices.

Proof. We proceed in four steps.

Step 1. We define

$$ \begin{align*} g_\omega^n = \prod_{j=0}^n f_j \circ T_\omega^j, \end{align*} $$

where we have set . We observe that for all $n \ge 0$ , there exists $C_n> 0$ such that for all $\omega \in \Omega $ ,

(9.1) $$ \begin{align} \|g_\omega^n \|_{L^\infty(m)} \le \Big( \sup_{j \ge 1} \| f_j \|_{L^\infty(m)} \Big)^{n+1} \le 1 \: \text{ and } \: \|g_\omega^n\|_{\mathrm{BV}} \le C_n \Big( \sup_{j \ge 1} \|f_j\|_{\mathrm{BV}} \Big). \end{align} $$

The first estimate is immediate, and the second follows, because

$$ \begin{align*} \mathrm{Var}(g_\omega^{n+1}) &\le \mathrm{Var}(g_\omega^n) \|f_{n+1} \circ T_\omega^{n+1}\|_{L^\infty(m)} + \|g_\omega^n\|_{L^\infty(m)} \mathrm{Var}(f_{n+1}\circ T_\omega^{n+1}) \\ & \le \mathrm{Var}(g_\omega^n) + \mathrm{Var}(f_{n+1}\circ T_\omega^{n+1}) \\ & = \mathrm{Var}(g_\omega^n) + \sum_{I \in {\mathcal A}_\omega^{n+1}} \mathrm{Var}_I(f_{n+1}\circ T_\omega^{n+1}) \\ & = \mathrm{Var}(g_\omega^n) + \sum_{I \in {\mathcal A}_\omega^{n+1}} \mathrm{Var}_{T_\omega^{n+1}(I)}(f_{n+1}) \\ & \le \mathrm{Var}(g_\omega^n) + (\# {\mathcal A}_\omega^{n+1}) \mathrm{Var}(f_{n+1}), \end{align*} $$

and so we can define by induction $C_{n+1} = C_n + \sup _{\omega \in \Omega } \# {\mathcal A}_\omega ^{n+1}$ which is finite, as there are only finitely many maps in ${\mathcal S}$ .

Step 2. We first prove the lemma in the case where $r=1$ in the condition (LY). Before, we claim that for $f \in \mathrm {BV}$ and sequences $(f_j) \subset \mathrm {BV}$ as in the statement, we have

(9.2) $$ \begin{align} \mathrm{Var}(P_\omega^n(f g_\omega^n)) &\le \sum_{j=0}^n \rho^j \| P_\omega^{n-j}(f g_\omega^{n-j-1}) \|_{L^\infty(m)} \|f_{n-j}\|_{\mathrm{BV}} \nonumber\\ &\quad + D \sum_{j=0}^{n-1} \rho^j \|P_\omega^{n-1-j}(f g_{\omega}^{n-1-j})\|_{L^1(m)} \|f_{n-j}\|_{L^\infty(m)}. \end{align} $$

This implies the lemma when $r=1$ , since

$$ \begin{align*} \| P_\omega^{n-j}(f g_\omega^{n-j-1}) \|_{L^\infty(m)} \le \|g_\omega^{n-j-1}\|_{L^\infty(m)} \|P_\omega^{n-j} |f| \|_{L^\infty(m)} \le C \|f\|_{\mathrm{BV}} \end{align*} $$

and

$$ \begin{align*} \|P_\omega^{n-j}(f g_{\omega}^{n-j})\|_{L^1(m)} \le \|f g_{\omega}^{n-j} \|_{L^1(m)}\le \|f\|_{L^\infty(m)} \|g_\omega^{n-j}\|_{L^1(m)} \le \|f\|_{\mathrm{BV}}. \end{align*} $$

We prove the claim by induction on $n \ge 0$ . It is immediate for $n=0$ , and for the induction step, we have, using condition (LY),

$$ \begin{align*} &\mathrm{Var} (P_\omega^{n+1} (f g_\omega^{n+1})) \\ &\!\!\quad = \mathrm{Var}(P_\omega^{n+1}(f g_\omega^n f_{n+1} \circ T_\omega^{n+1})) = \mathrm{Var}(P_\omega^{n+1}(f g_\omega^n) f_{n+1}) \\ &\!\!\quad \le \mathrm{Var}(P_\omega^{n+1}(f g_\omega^n)) \|f_{n+1}\|_{L^\infty(m)} + \|P_\omega^{n+1} (f g_\omega^n) \|_{L^\infty(m)} \mathrm{Var}(f_{n+1}) \\ &\!\!\quad \le ( \rho \mathrm{Var}(P_\omega^{n}(f g_\omega^n)) + D \|P_\omega^n (f g_\omega^n) \|_{L^1(m)} ) \|f_{n+1}\|_{L^\infty(m)}\\ &\!\!\qquad + \|P_\omega^{n+1} (f g_\omega^n) \|_{L^\infty(m)} \mathrm{Var}(f_{n+1}) \\ &\!\!\quad \le \rho \mathrm{Var}(P_\omega^{n}(f g_\omega^n)) \kern1.3pt{+}\kern1.3pt D \|P_\omega^n (f g_\omega^n) \|_{L^1(m)} \|f_{n+1}\|_{L^\infty(m)} \kern1.3pt{+}\kern1.3pt \|P_\omega^{n+1} (f g_\omega^n) \|_{L^\infty(m)} \|f_{n+1}\|_{\mathrm{BV}}, \end{align*} $$

which proves equation (9.2) for $n+1$ , assuming it holds for n.

Step 3. Now, we consider the general case $r \ge 1$ and we assume that n is of the particular form $n = pr$ , with $p \ge 0$ . We note that the random system defined with ${\mathcal T} = \{ T_\omega ^r \}_{\omega \in \Omega }$ satisfies the condition (LY) with $r = 1$ . Consequently, by the second step and equation (9.1), we have

$$ \begin{align*} \| P_\omega^n (f g_\omega^n)\|_{\mathrm{BV}} & = \| P_{\sigma^{r-1} \omega}^r \circ \cdots \circ P_\omega^r \bigg(f \prod_{j=1}^p g_{\sigma^{jr}\omega}^r \circ T_\omega^{jr}\bigg ) \|_{\mathrm{BV}} \\ & \le C \|f\|_{\mathrm{BV}} \Big( \sup_{j \ge 1} \| g_{\sigma^{jr} \omega}^r \|_{\mathrm{BV}}\Big) \le C C_r \|f\|_{\mathrm{BV}} \Big(\sup_{j \ge 1} \|f_j\|_{\mathrm{BV}} \Big). \end{align*} $$

Step 4. Finally, if $n = pr + q$ , with $p \ge 0$ and $q \in \{0, \ldots , r-1\}$ , as an immediate consequence of condition (LY), we obtain

$$ \begin{align*} \|P_\omega^n(f g_\omega^n) \|_{\mathrm{BV}} & = \|P_{\sigma^{pr}\omega}^q P_\omega^{pr}(f g_\omega^{pr} g_{\sigma^{pr}\omega}^q \circ T_\omega^{pr})\|_{\mathrm{BV}} \\ & = \|P_{\sigma^{pr}\omega}^q (P_\omega^{pr}(f g_\omega^{pr})g_{\sigma^{pr} \omega}^q) \|_{\mathrm{BV}} \le C \| P_\omega^{pr}(f g_\omega^{pr})g_{\sigma^{pr} \omega}^q\|_{\mathrm{BV}}. \end{align*} $$

However, from Step 3, we have

$$ \begin{align*} \| P_\omega^{pr}(f g_\omega^{pr})g_{\sigma^{pr} \omega}^q \|_{L^1(m)} & \le \| g_{\sigma^{pr} \omega}^q \|_{L^\infty(m)} \| P_\omega^{pr} (f g_\omega^{pr})\|_{L^1(m)} \\ & \le \| P_\omega^{pr} (f g_\omega^{pr})\|_{L^1(m)} \le C \|f\|_{\mathrm{BV}} \Big(\sup_{j \ge 1} \|f_j \|_{\mathrm{BV}}\Big), \end{align*} $$

and, using equation (9.1),

$$ \begin{align*} \mathrm{Var}( P_\omega^{pr}(f g_\omega^{pr})g_{\sigma^{pr} \omega}^q) & \le \|P_\omega^{pr}(f g_\omega^{pr})\|_{L^\infty(m)} \mathrm{Var}(g_{\sigma^{pr}\omega}^q)\\ &\quad+ \mathrm{Var}(P_\omega^{pr}(f g_\omega^{pr})) \| g_{\sigma^{pr}\omega}^q \|_{L^\infty(m)} \\ & \le [ C_q \|g_\omega^{pr}\|_{L^\infty(m)} \|P_\omega^{pr} |f|\|_{L^\infty(m)} + C \|f\|_{\mathrm{BV}}] \Big( \sup_{j \ge 1 } \| f_j\|_{\mathrm{BV}}\Big) \\ & \le C \Big( 1 + \max_{q = 0, \ldots, r-1} C_q \Big) \|f\|_{\mathrm{BV}} \Big( \sup_{j \ge 1 } \| f_j\|_{\mathrm{ BV}}\Big), \end{align*} $$

which concludes the proof of the lemma.

Proof of Theorem 6.3

We denote by ${\mathcal R}$ the family of finite unions of rectangles R of the form $R = (s,t] \times J$ with $J \in {\mathcal J}$ . By Kallenberg’s theorem, see [Reference KallenbergKal76, Theorem 4.7] or [Reference ResnickRes87, Proposition 3.22], $N_n^\omega \, {\stackrel {d}{\to } \,} N_{(\alpha )}$ if for any $R \in {\mathcal R}$ ,

$$ \begin{align*}\mathrm{(a)}~\lim_{n \to \infty} \nu^\omega(N_n^\omega(R) = 0) = {\mathbb P}(N_{(\alpha)}(R) = 0)\end{align*} $$

and

$$ \begin{align*}\mathrm{(b)}~\lim_{n \to \infty} {\mathbb E}_{\nu^\omega}N_n^\omega(R) = {\mathbb E} N_{(\alpha)}(R).\end{align*} $$

We first prove equation (b). We write

$$ \begin{align*} R = \bigcup_{i=1}^k R_i, \end{align*} $$

with $R_i = (s_i, t_i] \times J_i$ disjoint.

Then

$$ \begin{align*} {\mathbb E} N_{(\alpha)}(R) = \sum_{i=1}^k (t_i - s_i) \Pi_\alpha(J_i) \end{align*} $$

and

By Lemma 8.1, for ${\mathbb P}$ -a.e. $\omega \in \Omega $ , we have

$$ \begin{align*} \lim_{n \to \infty} \sum_{i=1}^k \sum_{j=\lfloor ns_i \rfloor}^{\lfloor nt_i \rfloor - 1} \nu^{\sigma^j \omega}(\phi_{x_0}^{-1}(b_n J_i)) = (t_i - s_i) \Pi_\alpha(J_i), \end{align*} $$

which proves equation (b).

We next establish equation (a). We will use induction on the number of ‘time’ intervals $(s_i,t_i] \subset (0,\infty ]$ . Let $R=(s_1,t_1]\times J_1$ , where $J_1\in \mathcal {J}$ . Define

$$ \begin{align*} A_n =\phi_{x_0}^{-1} (b_n J_1). \end{align*} $$

Since

$$ \begin{align*} \{N_n^{\omega} (R)=0\} & = \{ x: T_{\omega}^j (x)\not \in A_n, ns_1<j+1 \le nt_1 \} \\ &= \{ 1_{A_n^c}\circ T_{\omega}^{\lfloor ns_1 \rfloor} \cdot 1_{A_n^c}\circ T_{\omega}^{\lfloor ns_1 \rfloor+1} \cdot \cdots \cdot 1_{A_n^c}\circ T_{\omega}^{\lfloor nt_1 \rfloor-1} \not= 0\} \\ & = \bigg\{ x: \bigg(\prod_{j=0}^{\lfloor n t_1 \rfloor -1 - \lfloor n s_1 \rfloor}1_{A_n^c}\circ T_{\sigma^{\lfloor n s_1 \rfloor}\omega}^{j}\bigg) \circ T_{\omega}^{\lfloor n s_1 \rfloor}(x) \not= 0\bigg\}, \end{align*} $$

we have that

(9.3) $$ \begin{align} |\nu^{\omega} (N_n^{\omega} (R)&=0) - \nu^{\sigma^{\lfloor ns_1 \rfloor} \omega} (R_{A_n}(\sigma^{\lfloor ns_1 \rfloor} \omega)> \lfloor n(t_1-s_1) \rfloor )|\notag \\ &\ \ \le \nu^{\sigma^{\lfloor n s_1 \rfloor} \omega}(R_{A_n}(\sigma^{\lfloor n s_1 \rfloor}\omega) = 0) = \nu^{\sigma^{\lfloor n s_1 \rfloor}\omega}(A_n) \le C m(A_n)\to 0, \end{align} $$

because, due to rounding when taking integer parts, $\lfloor n t_1 \rfloor - \lfloor n s_1 \rfloor -1$ is either equal to $\lfloor n( t_1 - s_1) \rfloor -1$ or to $\lfloor n(t_1 - s_1) \rfloor $ . By Theorem 6.1,

$$ \begin{align*} \nu^{\sigma^{\lfloor ns_1\rfloor} \omega} (R_{A_n}(\sigma^{\lfloor ns_1 \rfloor} \omega)> \lfloor n(t_1-s_1) \rfloor)\to e^{-(t_1-s_1) \Pi_{\alpha} (J)} \end{align*} $$

as desired.

Now let $R=\bigcup _{j=1}^k (s_i, t_i]\times J_i$ with $0\le s_1 < t_1<\cdots < s_k < t_k$ and $J_i \in \mathcal {J}$ . Furthermore, define $s_i'=s_i-s_1$ and $t_i'=t_i-s_1$ .

Observe that, accounting for the rounding errors when taking integer parts as for equation (9.3), we get

(9.4) $$ \begin{align} &\bigg|\nu^{\omega} \bigg(N_n^{\omega} \bigg(\bigcup_{i=1}^k (s_i,t_i]\times J_i\bigg)=0\bigg) -\nu^{\sigma^{\lfloor n s_1 \rfloor}\omega} \bigg(N_n^{\sigma^{\lfloor n s_1 \rfloor}\omega} \bigg(\bigcup_{i=1}^k (s'_i,t'_i]\times J_i\bigg)=0\bigg)\bigg| \nonumber\\ &\quad\le 2C \sum_{i=1}^k m (\phi_{x_0}^{-1}(b_n J_i)) \to 0 \end{align} $$

so, after replacing $\omega $ by $\sigma ^{\lfloor n s_1 \rfloor }\omega $ , we can assume that $s_1=0$ . Let

$$ \begin{align*} &R_1=(0,t_1] \times J_1, \\ &R_2=\bigcup_{i=2}^k (s_i,t_i]\times J_i,\\ &R'_2=\bigcup_{i=2}^k (s_i-s_2,t_i-s_2]\times J_i. \end{align*} $$

Then, with $A_n =\phi _{x_0}^{-1} (b_n J_1)$ ,

(9.5) $$ \begin{align} |\nu^{\eta} (N_n^{\eta} (R_1\cup R_2)=0)-\nu^{\eta} [\{R_{A_n}(\eta)> \lfloor nt_1 \rfloor\}\cap T_{\eta}^{-\lfloor ns_2\rfloor} (N_n^{\sigma^{ \lfloor ns_2\rfloor}\eta}(R'_2)=0)]|\to 0 \end{align} $$

as $n\to \infty $ , uniformly in $\eta \in \Omega $ , as in equation (9.4). Moreover, as we check below,

(9.6) $$ \begin{align} &\vert\nu^{\eta} [\{R_{A_n}(\eta)> \lfloor nt_1 \rfloor\} \cap T_{\eta}^{-\lfloor ns_2 \rfloor} (N_n^{\sigma^{\lfloor ns_2\rfloor}\eta}(R'_2)=0)] \nonumber\\ &\quad- \nu^{\eta} (R_{A_n}(\eta)> \lfloor nt_1 \rfloor)\cdot \nu^{\eta} (N_n^{\eta}(R_2)=0) \vert\to 0 \end{align} $$

as $n\to \infty $ , uniformly in $\eta \in \Omega $ . Therefore, setting $\eta = \sigma ^{\lfloor n s_2 \rfloor } \omega $ in equations (9.5) and (9.6), we have, by Theorem 6.1,

$$ \begin{align*} \lim_{n\to\infty} | \nu^{\sigma^{\lfloor n s_2 \rfloor} \omega} (N_n^{\sigma^{\lfloor n s_2 \rfloor} \omega}(R_1 \cup R_2)=0) - e^{-t_1 \Pi_{\alpha} (J_1)}\nu^{\sigma^{\lfloor n s_2 \rfloor} \omega} (N_n^{\sigma^{\lfloor n s_2 \rfloor} \omega}(R_2)=0) | = 0, \end{align*} $$

which gives the induction step in the proof of equation (a).

We prove now equation (9.6). Our proof uses the spectral gap for $P^{n}_{\omega }$ and breaks down for random intermittent maps.

Similarly to equation (9.4),

$$ \begin{align*} | \nu^\eta(N_n^{\eta} (R_2)=0) - \nu^\eta(T_{\eta}^{-\lfloor ns_2 \rfloor} (N_n^{\sigma^{\lfloor ns_2 \rfloor}\eta}(R'_2)=0)) | \to 0 \quad\text{as } n\to\infty, \text{ uniformly in } \eta. \end{align*} $$

We have, using the notation

$$ \begin{align*} U=\{R_{A_n}(\eta)> \lfloor nt_1 \rfloor\}, \quad V=\{N_n^{\sigma^{\lfloor ns_2 \rfloor}\eta}(R'_2)=0\}, \end{align*} $$

that

where the last inequality follows from the decay, uniform in $\eta $ , of $\{P_{\eta }^k\}_{k}$ in BV (condition (Dec)).

However,

(9.7)

which proves equation (9.6). This follows from Lemma 9.1 applied to $f = h_\eta $ and , because

and both $\|h_\eta \|_{\mathrm {BV}}$ and are uniformly bounded. Note that for the stationary case, the estimate in equation (9.7) is used in the proof of [Reference Tyran-KamińskaTK10b, Theorem 4.4], which refers to [Reference Aaronson, Denker, Sarig and ZweimüllerADSZ04, Proposition 4].

Proof of Theorem 6.4

We will show that for ${\mathbb P}$ -a.e. $\omega \in \Omega $ , the assumptions of Kallenberg’s theorem [Reference KallenbergKal76, Theorem 4.7] hold.

Recall that $\mathcal {J}$ denotes the set of all finite unions of intervals of the form $(x,y]$ , where $x <y$ and $0\not \in [x,y]$ .

By Kallenberg’s theorem [Reference KallenbergKal76, Theorem 4.7], $N^{\omega }_n[(0,1]\times \cdot ) \to ^{d} N_{(\alpha )} ((0,1] \times \cdot )$ if for all $J \in \mathcal {J}$ ,

$$ \begin{align*} \mathrm{(a)} \lim_{n\to \infty} \nu^{\omega} (N_n^{\omega} ((0,1] \times J)=0)={\mathbb P} (N_{(\alpha)} ((0,1]\times J)=0) \end{align*} $$

and

$$ \begin{align*} \mathrm{(b)} \lim_{n\to \infty} {\mathbb E}_{\nu^{\omega}} N^{\omega}_n ((0,1]\times J)= {\mathbb E}[N_{(\alpha)} ((0,1]\times J)]. \end{align*} $$

We prove first equation (b) following [Reference Tyran-KamińskaTK10b, p. 12]. Write

$$ \begin{align*} J=\bigcup_{i=1}^k J_i \end{align*} $$

with $J_i=(x_i,y_i]$ disjoint.

Then

$$ \begin{align*} {\mathbb E} N_{(\alpha)} ((0,1]\times J)=\sum_{i=1}^k \Pi_{\alpha} (J_i)=\Pi_{\alpha} (J) \end{align*} $$

and

We check that

for $J=\bigcup _{i=1}^k J_i$ .

Write $A_n:=\phi _{x_0}^{-1} (b_n J)$ . Then

hence

by Lemma 7.2.

Now we prove equation (a), that is,

$$ \begin{align*} \lim_{n\to \infty} \nu^{\omega} (N_n^{\omega} ((0,1] \times J)=0)=P(N_{(\alpha)} ((0,1]\times J)=0) \end{align*} $$

for all $J \in \mathcal {J}$ .

Let $J \in \mathcal {J}$ and denote as above $A_n := \phi _{x_0}^{-1} (b_n J)\subset X =[0,1]$ . Then

$$ \begin{align*} \{N_{n}^{\omega} ((0,1]\times J)=0\}=\{ x: T_{\omega}^{j} (x) \not \in A_n, 0 < j+1 \le n\} = \{R_{A_n}(\omega)> n-1\}\cap A^c_n. \end{align*} $$

Hence,

$$ \begin{align*} | \nu^{\omega} (N_n^\omega ((0,1]\times J)=0) -\nu^{\omega} (R_{A_n}(\omega)> n)| \le C m(A_n) \to 0 \end{align*} $$

and by Theorem 6.2, for m-a.e. $x_0$

$$ \begin{align*} \nu^{ \omega} (R_{A_n}(\omega)> n) \to e^{- \Pi_{\alpha} (J)}. \end{align*} $$

This proves equation (a).

Proof. We follow the proof of [Reference Holland, Nicol and TörökHNT12, Lemma 3.4], conveniently adapted to our setting of random non-Markov maps. Recall that ${\mathcal A}_\omega ^n$ is the partition of monotonicity associated to the map $T_\omega ^n$ . Consider $I \in {\mathcal A}_\omega ^n$ . Since $\inf _I |(T_\omega ^n)'| \ge \unicode{x3bb} ^n> 1$ , there exists at most one solution $x_I^\pm \in I$ to the equation

(10.1) $$ \begin{align} T_\omega^n(x_I^\pm) = x_I^\pm \pm \epsilon, \end{align} $$

and since there is no sign change of $(T_\omega ^n)'$ on I, we have

(10.2) $$ \begin{align} {\mathcal E}_n^\omega(\epsilon) \cap I \subset [x_I^-, x_I^+]. \end{align} $$

We have

$$ \begin{align*} T_\omega^n(x_I^+) - T_\omega^n(x_I^-) = x_I^+ - x_I^- + 2 \epsilon, \end{align*} $$

and by the mean value theorem,

$$ \begin{align*} | T_\omega^n(x_I^+) - T_\omega^n(x_I^-) | = | (T_\omega^n)'(c) | \, | x_I^+ - x_I^- | \quad \text{for some } c \in I. \end{align*} $$

Consequently,

(10.3) $$ \begin{align} | x_I^+ - x_I^- | \le \bigg( \sup_I \frac{1}{|(T_\omega^n)'|} \bigg) [ | x_I^+ - x_I^-| + 2 \epsilon ] \le \unicode{x3bb}^{-n} | x_I^+ - x_I^-| + 2 \epsilon \sup_I \frac{1}{|(T_\omega^n)'|}. \end{align} $$

Note that if there is no solutions to equation (10.1), then the estimate in equation (10.3) is actually improved. Rearranging equation (10.3) and summing over $I\in {\mathcal A}_\omega ^n$ , we obtain, thanks to equation (10.2),

$$ \begin{align*} m({\mathcal E}_n^\omega(\epsilon)) \le \sum_{I \in {\mathcal A}_\omega^n} | x_I^+ - x_I^-| \le \frac{2 \epsilon}{1 - \unicode{x3bb}^{-n}} \sum_{I \in {\mathcal A}_\omega^n} \sup_I \frac{1}{|(T_\omega^n)'|} \le C \epsilon. \end{align*} $$

The fact that

(10.4) $$ \begin{align} \sum_{I \in {\mathcal A}_\omega^n} \sup_I \frac{1}{|(T_\omega^n)'|} \le C \end{align} $$

for a constant $C>0$ independent from $\omega $ and n follows from a standard distortion argument for one-dimensional maps that can be found in the proof of [Reference Aimino, Nicol and VaientiANV15, Lemma 8.5(3)] (see also [Reference Aimino and RousseauAR16, Lemma 7]), where finitely many piecewise $C^2$ uniformly expanding maps with finitely many discontinuities are also considered. Since it follows from condition (LY) that $\| P_\omega ^n f \|_{\mathrm {BV}} \le C \|f\|_{\mathrm {BV}}$ for some uniform $C>0$ , we do not have to average equation (10.4) over $\omega $ as in [Reference Aimino, Nicol and VaientiANV15], but instead, we can simply have an estimate that holds uniformly in $\omega $ .

Proof. Let

$$ \begin{align*} E_n^\omega = \{ x \in [0,1] : |T_\omega^j(x) - x | \le 2 n^{- \psi} \text{ for some } 0 < j \le \lfloor a \log n \rfloor \}. \end{align*} $$

Since $ B_{n^{-\psi }}(x_0) \cap \{R_{ B_{n^{-\psi }}(x_0)}^{\sigma ^i \omega } \le \lfloor a \log n \rfloor \} \subset B_{n^{-\psi }}(x_0) \cap E_n^{\sigma ^i \omega }$ , it is enough to consider

$$ \begin{align*} (\log n) \sum_{i=0}^{n-1} m( B_{n^{-\psi}}(x_0) \cap E_n^{\sigma^i \omega}). \end{align*} $$

According to Lemma 10.1, we have

$$ \begin{align*} m(E_n^\omega) \le \sum_{j=1}^{\lfloor a \log n \rfloor } m( {\mathcal E}_j^\omega(2n^{-\psi})) \le C \frac{\log n}{n^\psi}. \end{align*} $$

We introduce the maximal function

By [Reference RudinRud87, Equation (5), p. 138], for all $\unicode{x3bb}> 0$ , we have

(10.5)

Let $\rho>0$ and $\xi>0$ to be determined later. We define

$$ \begin{align*} F_n^\omega = \{ x_0 \in [0,1] : m( B_{n^{-\psi}}(x_0) \cap E_n^\omega ) \ge 2n^{-\psi(1+\rho)} \}, \end{align*} $$

so that we have

By definition of the maximal function $M_n^\omega $ , this implies that

from which it follows, by equation (10.5) with $\unicode{x3bb} = ( \log n) n^{\xi - \psi \rho }$ ,

$$ \begin{align*} m (A_n^\omega ) \le m ( M_n^\omega> (\log n) n^{\xi - \psi \rho} ) \le C n^{-(\xi + (1 - \rho)\psi -1)} =: \gamma_1(n), \end{align*} $$

where

If $x_0 \notin A_n^\omega $ , then

Since $\tfrac 23 < \psi < 1$ and $0 < \kappa < 3 \psi - 2$ , it is possible to choose $\rho>0$ and $\xi>0$ such that $\kappa = \xi + (1-\rho )\psi - 1$ , $\psi> \xi $ , and $\psi (1+\rho )>1$ (for instance, take $\xi = \psi - \delta $ and $\rho = \psi ^{-1} - 1 + \delta \psi ^{-1}$ with $\delta = ({3 \psi - 2 - \kappa })/{2}$ ), which concludes the proof.

Proof. Let $0 < \kappa < 3 \psi - 2$ to be determined later. Consider the sets $(A_n^\omega )_{n \ge 1}$ given by Lemma 10.2, with $m(A_n^\omega ) \le \gamma _1(n) = {\mathcal O}(n^{- \kappa })$ . Since $\kappa < 1$ , we need to consider a subsequence $(n_k)_{k \ge 1}$ such that $\sum _{k \ge 1} \gamma _1(n_k) < \infty $ . For such a subsequence, by the Borel–Cantelli lemma, for m-a.e. $x_0$ , there exists $K = K(x_0, \omega )$ such that for all $k \ge K$ , $x_0 \notin A_{n_k}^\omega $ . Since $\lim _{k \to \infty } \gamma _2(n_k) = 0$ , this implies

$$ \begin{align*} \lim_{k \to \infty} (\log n_k) \sum_{i=0}^{n_k - 1} m( B_{n_k^{-\psi}}(x_0) \cap \{R_{ B_{n_k^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n_k \rfloor \}) = 0. \end{align*} $$

We take $n_k = \lfloor k^\zeta \rfloor $ for some $\zeta>0$ to be determined later. To have $\sum _{k \ge 1} \gamma _1(n_k) < \infty $ , we need to require that $\kappa \zeta> 1$ . Set $U_n^\omega (x_0) = B_{n^{-\psi }}(x_0) \cap \{R_{ B_{n^{-\psi }}(x_0)}^\omega \le \lfloor a \log n \rfloor \}$ . To obtain the convergence to $0$ of the whole sequence, we need to prove that

(10.6) $$ \begin{align} \lim_{k \to \infty} \sup_{n_k \le n < n_{k+1}} \bigg| (\log n) \sum_{i=0}^{n-1} m(U_n^{\sigma^i \omega}(x_0)) - (\log n_k ) \sum_{i=0}^{n_k - 1} m(U_{n_k}^{\sigma^i \omega}(x_0)) \bigg| = 0. \end{align} $$

For this purpose, we estimate

$$ \begin{align*} &\bigg| (\log n) \sum_{i=0}^{n-1} m(U_n^{\sigma^i \omega}(x_0)) - (\log n_k ) \sum_{i=0}^{n_k - 1} m(U_{n_k}^{\sigma^i \omega}(x_0)) \bigg|\\ &\quad\le \mathrm{(I)} + \mathrm{(II)} + \mathrm{(III)} + \mathrm{ (IV)} + \mathrm{(V)}, \end{align*} $$

where

$$ \begin{align*} \mathrm{(I)} = | \log n - \log n_k| \sum_{i=0}^{n-1} m(U_n^{\sigma^i \omega}(x_0)), \: \: \mathrm{(II)} = (\log n_k) \sum_{i=n_k}^{n-1} m(U_n^{\sigma^i \omega}(x_0)), \end{align*} $$

$$ \begin{align*} \mathrm{(III)} &= (\log n_k) \sum_{i=0}^{n_k - 1} | m(B_{n^{-\psi}}(x_0) \cap \{R_{ B_{n^{-\psi}}(x_0)}^{\sigma^i \omega} \le\lfloor a \log n \rfloor \})\\ &\quad- m(B_{n_k^{-\psi}}(x_0) \cap \{R_{ B_{n^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n \rfloor \})|, \end{align*} $$

$$ \begin{align*} \mathrm{(IV)} &= (\log n_k) \sum_{i=0}^{n_k - 1} |m(B_{n_k^{-\psi}}(x_0) \cap \{R_{ B_{n^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n \rfloor \})\\ &\quad- m(B_{n_k^{-\psi}}(x_0) \cap \{R_{ B_{n_k^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n \rfloor \}) |, \end{align*} $$

$$ \begin{align*} \mathrm{(V)} &= (\log n_k) \sum_{i=0}^{n_k - 1} | m( B_{n_k^{-\psi}}(x_0) \cap \{R_{ B_{n_k^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n \rfloor \} )\\ &\quad - m( B_{n_k^{-\psi}}(x_0) \cap \{R_{ B_{n_k^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n_k \rfloor \} ) |. \end{align*} $$

Before proceeding to estimate each term, we note that $|n_{k+1} - n_k| = {\mathcal O}(k^{-(1- \zeta )})$ , $|n_{k+1}^{- \psi } - n_k^{-\psi }| = {\mathcal O}(k^{-(1 + \zeta \psi )})$ , $| \log n_{k+1} - \log n_k | = {\mathcal O}(k^{-1})$ and $m(U_n^\omega (x_0)) \le m(B_{n^{-\psi }}(x_0)) = {\mathcal O}(k^{-\zeta \psi })$ .

From these observations, it follows

$$ \begin{align*} \mathrm{(I)} \le C | \log n_{k+1} - \log n_k | n_{k+1} k^{- \zeta \psi} \le C k^{-(1-(1 - \psi) \zeta)}, \end{align*} $$

$$ \begin{align*} \mathrm{(II)} \le C (\log n_k) |n_{k+1} - n_k| k^{-\zeta \psi} \le C (\log k) k^{-(1-(1 - \psi) \zeta)}, \end{align*} $$

$$ \begin{align*} \mathrm{(III)} &\le C (\log n_k) n_k m(B_{n_k^{-\psi}}(x_0) \setminus B_{n^{-\psi}}(x_0))\\ &\le C (\log n_k) n_k | n_{k+1}^{-\psi} - n_k^{-\psi}| \le C (\log k) k^{-(1 - (1 - \psi) \zeta)}, \end{align*} $$

$$ \begin{align*} \mathrm{(IV)} & \le C (\log n_k) \sum_{i=0}^{n_k - 1} m (B_{n_k^{-\psi}}(x_0) \cap \{ R_{B_{n_k^{-\psi}}(x_0) \setminus B_{n^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n \rfloor \} ) \\ & \le C (\log n_k) \sum_{i=0}^{n_k - 1} a (\log n) m(B_{n_k^{-\psi}}(x_0) \setminus B_{n^{-\psi}}(x_0)) \\ & \le C (\log k)^2 k^{-(1 - (1 - \psi) \zeta)}, \end{align*} $$

and

$$ \begin{align*} \mathrm{(V)} & \le C (\log n_k) \sum_{i=0}^{n_k -1} m( B_{n_k^{-\psi}}(x_0) \cap \{ \lfloor a \log n_k \rfloor < R_{ B_{n_k^{-\psi}}(x_0)}^{\sigma^i \omega} \le \lfloor a \log n \rfloor \}) \\ & \le C (\log n_k) \sum_{i=0}^{n_k-1 } a | \log n_{k+1} - \log n_k | m(B_{n_k^{-\psi}}(x_0)) \\ & \le C (\log k) k^{-(1 - (1 - \psi) \zeta)}. \end{align*} $$

To obtain equation (10.6), it is thus sufficient to choose $\kappa>0$ and $\zeta>0$ such that $\kappa < 3 \psi - 2$ , $\kappa \zeta> 1$ , and $(1 - \psi ) \zeta < 1$ , which is possible if $\psi> \frac 3 4$ .

Proof of Theorem 2.4

We apply Theorem 7.3. By Theorem 6.3, we have $N_n^\omega \, {\stackrel {d}{\to } \,} N_{(\alpha )}$ under the probability $\nu ^\omega $ for ${\mathbb P}$ -a.e. $\omega \in \Omega $ . It thus remains to check that equation (7.2) holds for m-a.e. $x_0$ when $\alpha \in [1,2)$ to complete the proof. For this purpose, we will use a reverse martingale argument from [Reference Nicol, Török and VaientiNTV18] (see also [Reference Aimino and RousseauAR16, Proposition 13]). Because of equation (5.8), it is enough to work on the probability space $([0,1], \nu ^\omega )$ for ${\mathbb P}$ -a.e. $\omega \in \Omega $ . Let $\mathcal {B}$ denote the $\sigma $ -algebra of Borel sets on $[0,1]$ and

$$ \begin{align*} \mathcal{B}_{\omega, k} =(T_{\omega}^k)^{-1}(\mathcal{B}). \end{align*} $$

To simplify notation a bit, let

From equation (5.8), it follows that ${\mathbb E}_m(|f_{\omega , j, n}|) \le C \epsilon b_n$ , and from the explicit definition of $\phi $ , we can estimate the total variation of $f_{\omega ,j,n}$ and obtain the existence of $C> 0$ , independent of $\omega $ , $\epsilon $ , n, and j, such that

(10.7) $$ \begin{align} \|f_{\omega, j,n}\|_{\mathrm{BV}} \le C \epsilon b_n. \end{align} $$

We define

$$ \begin{align*} S_{\omega, k, n}:=\sum_{j=0}^{k-1} f_{\omega, j, n}\circ T_{\omega}^j \end{align*} $$

and

(10.8) $$ \begin{align} H_{\omega, k, n} \circ T_{\omega}^{n} :={\mathbb E}_{\nu^\omega} (S_{\omega, k, n}| \mathcal{B}_{\omega, k}). \end{align} $$

Hence, $H_{\omega , 1, n}=0$ and an explicit formula for $H_{\omega , k, n}$ is

$$ \begin{align*} H_{\omega, k, n}=\frac{1}{ h_{\sigma^k \omega}} \sum_{j=0}^{k-1} P_{\sigma^j \omega}^{k-j}(f_{\omega, j, n} h_{\sigma^j \omega}). \end{align*} $$

From the explicit formula, the exponential decay in the BV norm of $P_{\sigma ^j \omega }^{n-j}$ from condition (Dec), equations (5.8) and (10.7), we see that $\|H_{\omega , k, n}\|_{\mathrm {BV}}\le C \epsilon b_n$ , where the constant C may be taken as constant over $\omega \in \Omega $ . If we define

$$ \begin{align*} M_{\omega, k, n}=S_{\omega, k, n}-H_{\omega, k, n}\circ T_{\omega}^{k}, \end{align*} $$

then the sequence $\{ M_{\omega , k, n}\}_{k \ge 1}$ is a reverse martingale difference for the decreasing filtration $\mathcal {B}_{\omega , k}=(T_{\omega }^n)^{-1}( \mathcal {B})$ as

$$ \begin{align*} {\mathbb E}_{\nu^\omega} (M_{\omega, k, n} |\mathcal{B}_{\omega, k})=0. \end{align*} $$

The martingale reverse differences are

$$ \begin{align*} M_{\omega, k+1, n}-M_{\omega, k, n}=\psi_{\omega, k, n} \circ T_{\omega}^k, \end{align*} $$

where

$$ \begin{align*} \psi_{\omega, k, n}:=f_{\omega, k, n} +H_{\omega, k, n}-H_{\omega, k+1, n}\circ T_{\sigma^{k+1} \omega}. \end{align*} $$

We see from the $L^{\infty }$ bounds on $\| H_{\omega , k, n}\|_{\infty } \le C b_n \epsilon $ and the telescoping sum that

(10.9) $$ \begin{align} \bigg|\sum_{j=0}^{k-1} \psi_{\omega, j, n}\circ T_{\omega}^j -\sum_{j=0}^{k-1} f_{\omega, j, n}\circ T_{\omega}^j \bigg|\le C\epsilon b_n. \end{align} $$

By Doob’s martingale maximal inequality,

$$ \begin{align*} \nu^\omega \bigg\{ \max_{1\le k \le n} \bigg| \sum_{j=0}^{k-1} \psi_{\omega, j, n}\circ T_{\omega}^j \bigg|\ge b_n \delta \bigg\} \le \frac{1}{b_n^2 \delta^2} {\mathbb E}_{\nu^\omega} \bigg|\sum_{j=0}^{n-1} \psi_{\omega, j, n} \circ T_{\omega}^j \bigg|^2. \end{align*} $$

Note that

$$ \begin{align*} \sum_{j=0}^{n-1} {\mathbb E}_{\nu^\omega} [\psi^2_{\omega, j, n} \circ T_{\omega}^j]={\mathbb E}_{\nu^\omega} \bigg[\sum_{j=0}^{n-1} \psi_{\omega, j, n} \circ T_{\omega}^j\bigg]^{2} \end{align*} $$

by pairwise orthogonality of martingale reverse differences.

As in [Reference Haydn, Nicol, Török and VaientiHNTV17, Lemma 6],

$$ \begin{align*} {\mathbb E}_{\nu^\omega} [(S_{\omega, n, n})^2]=\sum_{j=0}^{n-1} {\mathbb E}_{\nu^\omega} [\psi^2_{\omega, j, n} \circ T_{\omega}^j]+ {\mathbb E}_{\nu^\omega}[H_{\omega, 1, n}^2]- {\mathbb E}_{\nu^\omega}[H_{\omega, n, n}^2\circ T_{\omega}^{n}]. \end{align*} $$

So we see that

(10.10) $$ \begin{align} \nu^\omega \bigg\{ \max_{1\le k \le n} \bigg| \sum_{j=0}^{k-1} \psi_{\omega, j, n}\circ T_{\omega}^j \bigg|\ge b_n \delta \bigg\} \le \frac{1}{b_n^2 \delta^2} {\mathbb E}_{\nu^\omega} [(S_{\omega, n, n})^2] +2 \frac{C^2 \epsilon^2 }{\delta^2}, \end{align} $$

where we have used $\| H^2_{\omega , j, n}\|_{\infty } \le C^2 b_n^2 \epsilon ^2$ .

Now we estimate

(10.11) $$ \begin{align} {\mathbb E}_{\nu^\omega} [(S_{\omega, n, n})^2]\le \sum_{j=0}^{n-1} {\mathbb E}_{\nu^\omega}[f^2_{\omega, j, n} \circ T_{\omega}^j]+2\sum_{i=0}^{n-1} \sum_{i < j} {\mathbb E}_{ \nu^\omega} [f_{\omega, j, n} \circ T_{\omega}^j \cdot f_{\omega, i, n} \circ T_{\omega}^i]. \end{align} $$

Using the equivariance of the measures $\{ \nu ^\omega \}_{\omega \in \Omega }$ and equation (5.8), we have

(10.12)

by Proposition 3.2 and that

$$ \begin{align*} \lim_{n \to \infty} n \, \nu( | \phi_{x_0} |> \unicode{x3bb} b_n) = \unicode{x3bb}^{-\alpha} \quad \text{ for } \unicode{x3bb} >0, \end{align*} $$

since $\phi _{x_0}$ is regularly varying.

However, we are going to show that for m-a.e. $x_0$ ,

(10.13) $$ \begin{align} \lim_{\epsilon \to 0} \limsup_{n \to \infty} \frac{1}{b_n^2}\sum_{i=0}^{n-1} \sum_{i< j} {\mathbb E}_{\nu^\omega} [f_{\omega, j, n} \circ T_{\omega}^j \cdot f_{\omega, i, n} \circ T^i_{\omega}] = 0. \end{align} $$

The first observation is that, due to condition (Dec),

$$ \begin{align*}{\mathbb E}_{\nu^\omega} [f_{\omega, j, n} \circ T_{\omega}^j \cdot f_{\omega, i, n} \circ T^i_{\omega}]\le C \theta^{j-i} \|f_{\omega, i, n}\|_{\mathrm{BV}} \|f_{\omega, j, n}\|_{L^1_m} \le C \epsilon^2 b_n^2 \theta^{j-i},\end{align*} $$

where $\theta <1$ . Hence, there exists $a> 0$ independently of n and $\epsilon $ such that

$$ \begin{align*}\sum_{j-i> \lfloor a \log n \rfloor}{\mathbb E}_{\nu^\omega} [f_{\omega, j, n} \circ T_{\omega}^j \cdot f_{\omega, i, n} \circ T^i_{\omega}]\le C \epsilon^2 n^{-2} b_n^2\end{align*} $$

and it is enough to prove that for $\epsilon> 0$ ,

$$ \begin{align*} \sum_{i=0}^{n-1} \sum_{j = i+1}^{i + \lfloor a \log n \rfloor} {\mathbb E}_{\nu^\omega} [f_{\omega, j, n} \circ T_{\omega}^j \cdot f_{\omega, i, n} \circ T^i_{\omega}] = o(b_n^2) = o(n^{ 2/ \alpha}). \end{align*} $$

By construction, the term ${\mathbb E}_{\nu ^\omega }[f_{\omega , i, n} \circ T_\omega ^i \cdot f_{\omega , j, n} \circ T_\omega ^j]$ is a covariance, and since $\phi $ is positive, we can bound this quantity by ${\mathbb E}_{\nu ^\omega }[f \circ T_\omega ^i \cdot f \circ T_\omega ^j] = {\mathbb E}_{\nu ^{\sigma ^i \omega }}[f_n \cdot f_n \circ T_{\sigma ^i \omega }^{j-i}]$ , where . Then, since the densities are uniformly bounded by equation (5.8), we are left to estimate

(10.14) $$ \begin{align} \sum_{i=0}^{n-1} \sum_{j= i+1}^{i + \lfloor a \log n \rfloor} {\mathbb E}_m[f_n \cdot f_n \circ T_{\sigma^i \omega}^{j-i}]. \end{align} $$

Let $\tfrac 3 4 < \psi < 1$ and $U_n = B_{n^{-\psi }}(x_0)$ . We bound equation (10.14) by $\mathrm {(I)} + \mathrm {(II)} + \mathrm {(III)}$ , where

$$ \begin{align*} \mathrm{(I)} = \sum_{i=0}^{n-1} \sum_{j= i+1}^{i + \lfloor a \log n \rfloor} \int_{U_n \cap (T_{\sigma^i \omega}^{j-i})^{-1}(U_n)} f_n \cdot f_n \circ T_{\sigma^i \omega}^{j-i} \,dm, \end{align*} $$

$$ \begin{align*} \mathrm{(II)} = \sum_{i=0}^{n-1} \sum_{j= i+1}^{i + \lfloor a \log n \rfloor} \int_{U_n \cap (T_{\sigma^i \omega}^{j-i})^{-1}(U_n^c)} f_n \cdot f_n \circ T_{\sigma^i \omega}^{j-i} \,dm, \end{align*} $$

and

$$ \begin{align*} \mathrm{(III)} = \sum_{i=0}^{n-1} \sum_{j= i+1}^{i + \lfloor a \log n \rfloor} \int_{U_n^c} f_n \cdot f_n \circ T_{\sigma^i \omega}^{j-i} \,dm. \end{align*} $$

Since $\|f_n\|_\infty \le \epsilon b_n$ , it follows that

$$ \begin{align*} \mathrm{(I)} &\le \epsilon^2 b_n^2 \sum_{i=0}^{n-1} \sum_{j=i +1}^{i + \lfloor a \log n \rfloor }m(U_n \cap (T_{\sigma^i \omega}^{j-i})^{-1}(U_n)) \\ & \le a \epsilon^2 b_n^2 (\log n) \sum_{i=0}^{n-1} m(U_n \cap \{ R_{U_n}^{\sigma^i \omega} \le a \log n \} ), \end{align*} $$

which by Lemma 10.3 is a $o(b_n^2)$ as $n \to \infty $ for m-a.e. $x_0$ .

To estimate terms (II) and (III), we will use Hölder’s inequality. We first observe by a direct computation that

(10.15) $$ \begin{align} \int_{U_n^c} \phi_{x_0}^2 \,dm = {\mathcal O}(n^{\psi ({2}/{\alpha} - 1 )}). \end{align} $$

We consider term (III) first. Let $ A = U_n^c$ . We have

(10.16) $$ \begin{align} \int_{U_n^c} f_n \cdot f_n \circ T_{\sigma^i \omega}^{j-i} \,dm &\le \int_A \phi_{x_0} \cdot f_n \circ T_{\sigma^i \omega}^{j-i} \,dm\nonumber\\ &\le \bigg( \int_A \phi_{x_0}^2 \,dm \bigg)^{ 1/ 2} \bigg( \int f_n^2 \circ T_{\sigma^i \omega}^{j-i} \,dm \bigg)^{ 1/ 2} \end{align} $$

(10.17) $$ \begin{align} &\qquad\qquad\ \le C \bigg( \int_A \phi_{x_0}^2 \,dm \bigg)^{1/ 2} \bigg( \int f_n^2 \,dm \bigg)^{ 1/ 2}. \end{align} $$

By equation (10.15), $( \int _A \phi _{x_0}^2\,dm )^{ 1/ 2} \le C n^{{\psi }/{2} ({2}/{\alpha } - 1 )}$ and by Proposition 3.2, $( \int f_n^2\,dm )^{ 1/ 2} \le C n^{{1}/{\alpha } - 1 /2}$ . Hence, we may bound equation (10.16) by $C n^{(1+\psi )({1}/{\alpha } - 1 /2)}$ .

To bound term (II), let $B= U_n \cap (T_{\sigma ^i \omega }^{j-i})^{-1}(U_n^c)$ . Then,

(10.18) $$ \begin{align} \int_{U_n \cap (T_{\sigma^i \omega}^{j-i})^{-1}(U_n^c)} f_n \cdot f_n \circ T_{\sigma^i \omega}^{j-i} \,dm &\le \int_B f_n \cdot \phi_{x_0} \circ T_{\sigma^i \omega}^{j-i} \,dm\nonumber\\ & \le \bigg( \int f_n^2 \,dm \bigg)^{ 1/ 2} \bigg( \int_B \phi_{x_0}^2 \circ T_{\sigma^i \omega}^{j-i} \,dm \bigg)^{ 1/ 2}. \end{align} $$

As before, $( \int f_n^2 dm )^{ 1 /2} \le C n^{{1}/{\alpha } - 1 /2}$ and

by equation (10.15), and so equation (10.18) is bounded by $C n^{(1+\psi )({1}/{\alpha } - 1 /2)}$ .

It follows that $\mathrm {(II)} + \mathrm {(III)} \le C (\log n) n^{1 + (1+\psi )({1}/{\alpha } - 1/ 2)} = o(n^{ 2/ \alpha })$ , since $ \psi < 1$ . This proves that equation (10.14) is $o(b_n^2)$ and concludes the proof of equation (10.13).

Finally, from equations (10.11), (10.12), and (10.13), we obtain

(10.19) $$ \begin{align} \lim_{\epsilon \to 0} \limsup_{n \to \infty} \frac{1}{b_n^2} {\mathbb E}_{\nu^\omega}[(S_{\omega, n, n})^2] = 0, \end{align} $$

which gives the result by taking the limit first in n and then in $\epsilon $ in equation (10.10).

Proof of Theorem 2.6

We apply Proposition 5.8. By Theorem 6.4, it remains to prove equation (5.7), since $\alpha \in (0,1)$ . We will need an estimate for which is independent of $\omega $ . For this purpose, we introduce the absolutely continuous probability measure $\nu _{\mathrm {\max }}$ whose density is given by $h_{\mathrm {\max }}(x) = \kappa x^{- \gamma _{\mathrm {\max }}}$ . Since all densities $h_\omega $ belong to the cone L, we have that $h_\omega \le ({a}/{\kappa }) h_{\mathrm {\max }}$ for all $\omega $ . Thus,

We can easily verify that $\phi _{x_0}$ is regularly varying of index $\alpha $ with respect to $\nu _{\mathrm {\max }}$ , with scaling sequence equal to $(b_n)_{n \ge 1}$ up to a multiplicative constant factor. Consequently, by Proposition 3.2, we have that, for some constant $c>0$ ,

which implies equation (5.7).