Proof. Take $\unicode{x3bb} \in \mathbb {R}^{\mathcal {A}}_{>0}$ and consider the surface $(\pi ,\unicode{x3bb} ,\tilde {\tau })$ . Due to the rational independence of $\tilde {\tau }$ , this surface has no horizontal saddle connections. Fix $\alpha \in \mathcal {A}$ and $x_0\in I_{\alpha }$ . Let $J:=\bigcup _{t\in [0,h_{\alpha })} \phi _t(x_0)$ be a vertical segment inside the zippered rectangle $D_{\alpha }=\bigcup _{t\in [0,h_{\alpha })} \phi _t(I_{\alpha })$ . Since J is a transversal interval for the horizontal flow $\psi $ , we may consider a first return map $S:J\to J$ via $\psi $ . Note that the endpoints of J lie on the outgoing separatrix of the singularity at $0$ , hence, due to [Reference Yoccoz29, Corollary 5.5] and the fact that $(\pi ,\unicode{x3bb} ,\tilde {\tau })$ has no horizontal connections, S is an IET of $d = \#\mathcal {A}$ intervals.

The general strategy of the proof is as follows. The lengths of the intervals exchanged by S vary continuously depending on $\tau $ , without changing the combinatorial data of S, as long as the order of the first hitting points of the outgoing and incoming separatrices is retained. Moreover, as long as the perturbation of the parameter $\tau $ is small enough, the number and order of zippered rectangles visited by the points from J via the horizontal flow do not change. Since there are only finitely many exchanged intervals, by taking C as the maximal number of zippered rectangles visited this way, we obtain the desired result. Below, for the sake of clarity and readability, we provide a detailed proof in a particular case, but let us mention that the general case follows along similar lines.

From now on, for simplicity we assume that $\pi _0(\alpha )=1$ and $\pi _1(\alpha )=d$ . Moreover, we assume that

(4.1) $$ \begin{align} \tilde{\tau}_\alpha>0\quad\text{and}\quad\tilde{\tau}_\beta<0 \quad \text{for every }\beta\in \mathcal{A}\setminus\{\alpha\} \end{align} $$

(see Figure 2). Note that, due to the rational independence of $\tilde {\tau }$ , we also have $\sum _{\alpha \in \mathcal {A}} \tilde {\tau }_{\alpha }\neq 0$ . Hence, we can also assume that

(4.2) $$ \begin{align} \sum_{\beta\in\mathcal{A}}\tilde{\tau}_\beta<0. \end{align} $$

Figure 2

The polygonal representation of the surface $(\pi ,\unicode{x3bb} ,\tau )$ with indicated zippered rectangles. The picture shows three different orbits, corresponding to three of the cases described in the proof of Lemma 4.2 (colour online).

Let us describe the parameters of S. First, we have

(4.3) $$ \begin{align} |J|=h_{\alpha}=-(\Omega_{\pi}\tilde\tau)_{\alpha}=-\sum_{\beta\in\mathcal{A}\setminus\{\alpha\}}\tilde\tau_{\beta}. \end{align} $$

Moreover, we know that the discontinuities of S are given by the first backward intersections of J with incoming separatrices for $\psi $ . More precisely, if $\sigma \in \Sigma $ is a singularity and $(\psi _{-t}(\sigma ))_{t\in (0,\infty )}$ is one of its incoming horizontal separatrices, then a point $y\in J$ is a discontinuity of S obtained from this separatrix either if $y=\sigma $ , or there exists $t(y)\in (0,\infty )$ such that

$$ \begin{align*} y=\psi_{-t(y)}(\sigma)\quad\text{and}\quad(\psi_{-t}(\sigma))_{t\in(0,t(y))}\cap J=\emptyset. \end{align*} $$

By choice of J, the only possible case for $y=\sigma $ to hold is if $y=x_0=\partial I_{\alpha }$ . By shifting the interval J slightly via $\psi $ , we may assume that $y\notin \Sigma $ . By shifting it even more, but still inside $D_{\alpha }$ , since $\tau _{\pi _0^{-1}(2)}<0$ , we may assume that if $z_0$ is the intersection of J with the side connecting $0$ and $\unicode{x3bb} _{\alpha }+i\tilde \tau _{\alpha }$ , then

(4.4) $$ \begin{align} \operatorname{Im}(z_0)>\tilde\tau_{\pi_0^{-1}(1)}+\tilde\tau_{\pi_0^{-1}(2)}. \end{align} $$

The initial segments of horizontal incoming separatrices correspond to the horizontal segments in the polygonal representation of $(\pi ,\unicode{x3bb} ,\tilde \tau )$ whose right-hand-side endpoints are in the vertices of the polygon. Due to (4.1), the right-hand-side endpoints of these segments are the vertices

$$ \begin{align*} \bigg\{\!\sum_{\pi_0(\beta)\le k}(\unicode{x3bb}_{\beta}+i\tilde\tau_\beta) \,\bigg| \, k=2,\ldots,d \bigg\}. \end{align*} $$

By condition (4.2), the vertex $\sum _{ \beta \in \mathcal {A} }(\unicode{x3bb} _{\beta }+i\tilde \tau _\beta )$ , that is, the one corresponding to $k=d$ , is below the x-axis. Moreover, due to the fact that $\tilde \tau _{\pi _1^{-1}(d)}=\tilde \tau _{\alpha }>0$ , the subinterval of J given by $\bigcup _{t\in [h_{\alpha }+\sum _{\beta \in \mathcal {A}}\tilde {\tau }_\beta ,h_{\alpha })} \phi _t(x_0)$ is fully below the x-axis. Hence, since $h_{\alpha }+\sum _{\beta \in \mathcal {A}}\tilde {\tau }_\beta =\tilde \tau _{\alpha }=\tilde \tau _{\pi _0^{-1}(1)}$ , the discontinuity of S corresponding to $k=d$ is the point $y_{d}$ given by

(4.5) $$ \begin{align} y_{d}:=\phi_{\tilde\tau_{\pi_0^{-1}(1)}}(x_0). \end{align} $$

On the other hand, for every $k=2,\ldots ,d-1$ , the initial segment of the corresponding incoming separatrix lies above the x-axis, and, due to (4.1) (and by (4.4)), the first backward hit of the incoming separatrix via $\psi $ is the point

(4.6) $$ \begin{align} y_k:=\phi_{t_k}(x_0) \quad \text{where } t_k:= \sum_{\pi_0(\beta)\le k}\tilde\tau_{\beta}. \end{align} $$

Notice that we can use the above notation also for $k=1$ , that is, $t_1=\tilde \tau _{\pi _0^{-1}(1)}$ and $y_1=\phi _{t_1}(x_0)=y_d$ . Note that none of the formulas (4.3), (4.5), or (4.6) depend on $\unicode{x3bb} $ .

It remains to describe the zippered rectangles crossed by the horizontal orbit segment of any point $x\in J$ . Note that due to (4.1) and (4.2) the order of the points $y_k$ on J, counting upwards, is $y_{d-1},y_{d-2},\ldots ,y_{3},y_{2},y_{d}$ . Assume first that $x\in (y_{k+1},y_k)\subseteq J$ for some $k\in \{2,\ldots ,d-2\}$ . Then the orbit segment starting at x given by the forward action of $\psi $ leaves the zippered rectangle $D_{\alpha }$ then goes through the cylinders $D_{\pi _0^{-1}(\ell )}$ for $2\leq \ell \leq k+1$ , and inside the zippered rectangle $D_{\pi _0^{-1}(k+1)}$ it crosses the edge given by the vertices $\sum _{ \pi _0(\beta )\le k}(\unicode{x3bb} _{\beta }+i\tilde \tau _\beta )$ and $\sum _{ \pi _0(\beta )\le k+1}(\unicode{x3bb} _{\beta }+i\tilde \tau _\beta )$ . This describes the part of the segment in the part of the polygonal representation above the x-axis. After crossing the side of the polygon, it goes through zippered rectangles $D_{\pi _1^{-1}(\ell )}$ , with $\pi _1(\pi _0^{-1}(k+1))\leq \ell \leq d-1$ in the lower part of the polygon. After that, the orbit goes to $D_{\pi _1^{-1}(d)}=D_{\alpha }$ and returns directly to J. Hence, in the case when $x\in (y_{k+1},y_k)$ for some $k\in \{2,\ldots ,d-2\}$ , the orbit crosses at most $2d-2$ edges of zippered rectangles different from $D_\alpha $ .

To describe the horizontal orbit segment for $x\in (y_{2},y_{d})=(y_{2},y_{1})$ it suffices to take $k=1$ in the above description of the passage of the orbit through zippered rectangles.

If $x\in (x_0,y_{d-1})$ then the orbit segment starting at x obtained by acting forward via $\psi $ passes all zippered rectangles in the upper part of the polygonal representation of $(\pi ,\unicode{x3bb} ,\tilde {\tau })$ , before crossing the edge given by $\sum _{ \pi _0(\beta ) < d}(\unicode{x3bb} _{\beta }+i\tilde \tau _\beta )$ and $\sum _{\beta \in \mathcal {A}}(\unicode{x3bb} _{\beta }+i\tilde \tau _\beta )$ . Then, before returning to the zippered rectangle $D_{\alpha }$ it passes through the zippered rectangles $D_{\pi _1^{-1}(\ell )}$ , with $\pi _1(\pi _0^{-1}(d))\leq \ell \leq d-1$ . After that, the orbit goes to $D_{\pi _1^{-1}(d)}=D_{\alpha }$ and returns directly to J. Hence, it also crosses no more than $2d-2$ edges of zippered rectangles different from $D_\alpha $ .

Finally, if $x\in ({y_d}, \phi _{h_{\alpha }}(x_0))$ , then the orbit of x does not leave the lower part of the polygonal representation of $(\pi ,\unicode{x3bb} ,{\tilde \tau })$ before returning to $D_{\alpha }$ . After leaving the zippered rectangle $D_{\alpha }$ , it goes through zippered rectangles $D_{\pi _1^{-1}(\ell )}$ , with $\pi _1(\pi _0^{-1}(d))+1\leq \ell \leq d-1$ (note that here, if $\pi _1(\pi _0^{-1}(d))=d-1$ , then the set of the indices $\ell '$ is empty, that is, the horizontal orbit of x does not leave $D_{\alpha }$ before the first return to J). After that, the orbit goes to $D_{\pi _1^{-1}(d)}=D_{\alpha }$ and returns directly to J. Hence, in this case, the horizontal orbit of x does not cross more than $d-1$ edges of zippered rectangles different from $D_\alpha $ . Thus the thesis of the lemma holds for $(\pi ,\unicode{x3bb} ,\tilde {\tau })$ with $C=2d-2$ .

Since the values in (4.3), (4.5) and (4.6) change continuously with respect to $\tilde \tau $ , we can find an $\epsilon>0$ such that the inequalities (4.1) and (4.2) hold for any $\tau \in B_{\Theta _\pi }(\tilde \tau , \epsilon )$ and, in particular, the order in which the discontinuities coming from specific vertices appear on the vertical section remains unchanged. Hence, the order of visited zippered rectangles stays the same, and the thesis holds with $C=2d-2$ .

Proof of Proposition 4.1

Fix $\tilde \tau \in \Theta _\pi $ and let $\epsilon =\epsilon (\tilde \tau )$ and $C=C(\tilde \tau ,\epsilon )$ be given by Lemma 4.2. Fix $\alpha \in \mathcal {A}$ and $n\in \mathbb {N}$ such that $\mathcal R^{n}(\pi ,\unicode{x3bb} ,\tau )\in \{\pi \}\times \mathbb {R}_{>0}^{\mathcal {A}}\times B_{\Theta _{\pi }}(\tilde \tau ,\epsilon )$ . Let $(\pi ,\unicode{x3bb} ^{(n)},\tau ^{(n)})=R^{n}(\pi ,\unicode{x3bb} ,\tau )$ . Then $\mathcal R^{n}(\pi ,\unicode{x3bb} ,\tau )=(\pi ,e^s\unicode{x3bb} ^{(n)},e^{-s}\tau ^{(n)})$ with $s= -{\log}\, |\unicode{x3bb} ^{(n)}|$ . Since ${R}^n(\pi , \unicode{x3bb} , \tau )$ and $(\pi , \unicode{x3bb} , \tau )$ describe the same surface, we can decompose the surface $(\pi , \unicode{x3bb} , \tau )$ into zippered rectangles $\{D^{(n)}_\beta \,|\ \beta \in \mathcal {A}\}$ according to numerical data $(\pi ,\unicode{x3bb} ^{(n)},\tau ^{(n)})$ , that is, $D^{(n)}_\beta =\bigcup _{0\leq t<h^{(n)}_\beta }\phi _t I^{(n)}_\beta $ , where $h^{(n)}=(h^{(n)}_\beta )_{\beta \in \mathcal {A}}=-\Omega _\pi \tau ^{(n)}$ . Then $T^jI^{(n)}_\beta =\phi _{t^\beta _j} I^{(n)}_\beta $ for $0\leq j\leq q^{(n)}_\beta $ , where $0=t^\beta _0<t^\beta _1<\cdots <t^\beta _{q^{(n)}_\beta }=h^{(n)}_{\beta }$ are consecutive recurrence times of $I^{(n)}_\beta $ to the interval I.

Let $x\in T^m(I^{(n)}_{\alpha })$ for some $0\leq m< q^{(n)}_\alpha $ . Consider $T^m(I^{(n)}_{\alpha })$ as a subset of the surface $(\pi ,\unicode{x3bb} ,\tau )$ and let J be a vertical segment of length $h_{\alpha }^{(n)}$ inside $D_{\alpha }^{(n)}$ , which starts at $I_{\alpha }^{(n)}$ (at $x_0=T^{-m}x$ ), which contains x. As $R^n(\pi , \unicode{x3bb} , \tau )$ is obtained from $\mathcal {R}^n(\pi , \unicode{x3bb} , \tau )$ by vertical and horizontal rescaling, they have the same (up to rescaling) decompositions into zippered rectangles. As $\mathcal R^{n}(\pi ,\unicode{x3bb} ,\tau )\in \{\pi \}\times \mathbb {R}_{>0}^{\mathcal {A}}\times B_{\Theta _{\pi }}(\tilde \tau ,\epsilon )$ , in view of Lemma 4.2 and denoting by r the first return time of x to J via the horizontal flow $\psi $ , the orbit $(\psi _{t}(x))_{t\in (0,r)}$ crosses at most C edges of the zippered rectangles $D^{(n)}_\beta $ , $\beta \in \mathcal {A}\setminus \{\alpha \}$ before returning to $J\subset D^{(n)}_\alpha $ . In particular, $\psi _{r}(x)\in J\subset D^{(n)}_\alpha $ .

There are two possibilities: either $(\psi _{t}(x))_{t\in (0,r)}$ is not included in I, which is equivalent to the fact that $T^m(I^{(n)}_{\alpha })$ is the rightmost level of the tower $\bigcup _{j=0}^{q^{(n)}_\alpha -1}T^j(I^{(n)}_{\alpha })$ with respect to the linear order on I, or $(\psi _{t}(x))_{t\in (0,r)}$ is a subinterval of I. Indeed, if x does not belong to the rightmost level of the tower $\bigcup _{j=0}^{q^{(n)}_\alpha -1}T^j(I^{(n)}_{\alpha })$ , then denote by $T^{m'}(I^{(n)}_{\alpha })$ the next level traveling horizontally. Then $\psi _r(x)\in T^{m'}(I^{(n)}_{\alpha })\subset I$ . It follows that the horizontal interval $(\psi _{t}(x))_{t\in (0,r)}$ is a subinterval of I.

Let us assume first that $(\psi _{t}(x))_{t\in (0,r)} \subseteq I$ . Recall that for any $\beta \in \mathcal {A}$ , the tower $\bigcup _{j=0}^{q^{(n)}_\beta -1}{T^j(I^{(n)}_\beta )}$ is fully included inside the zippered rectangle $D_{\beta }^{(n)}$ . Hence, the fact that $(\psi _{t}(x))_{t\in (0,r)}$ crosses at most C times the edges of zippered rectangles is equivalent to the fact that it crosses at most C times the endpoints of levels of towers corresponding to other symbols than $\alpha $ . Moreover, since $\psi _r(x)\in T^{m'}(I^{(n)}_\alpha )$ (this is the next level of the tower traveling horizontally), the interval between $T^{m}(I^{(n)}_\alpha )$ and $T^{m'}(I^{(n)}_\alpha )$ contains at most C levels of towers other than $\bigcup _{j=0}^{q^{(n)}_\alpha -1}T^j(I^{(n)}_{\alpha })$ .

If, on the other hand, we have $(\psi _{t}(x))_{t\in (0,r)} \not \subseteq I$ , then $(\psi _{t}(x))_{t\in (0,r)}$ contains the right-hand-side endpoint of I and hence contains the rightmost interval in $I\setminus \bigsqcup _{j=0}^{q_{\alpha }^{(n)}-1}T^jI^{(n)}_\alpha $ . As $(\psi _{t}(x))_{t\in (0,r)}$ crosses at most C edges of the zippered rectangles $D^{(n)}_\beta $ , $\beta \in \mathcal {A}\setminus \{\alpha \}$ , the argument used in the previous paragraph shows that the rightmost interval in $I\setminus \bigsqcup _{j=0}^{q_{\alpha }^{(n)}-1}T^jI^{(n)}_\alpha $ contains at most C levels of towers other than $\bigcup _{j=0}^{q^{(n)}_\alpha -1}T^j(I^{(n)}_{\alpha })$ .

It remains to consider the leftmost component of $I\setminus \bigsqcup _{j=0}^{q_{\alpha }^{(n)}-1}T^jI^{(n)}_\alpha $ , with respect to the linear order on I. This is done by choosing a point x in the intersection of J and the leftmost level of $\bigsqcup _{j=0}^{q_{\alpha }^{(n)}-1}T^jI^{(n)}_\alpha $ and taking $r_0\geq 0$ such that $\psi _{-r_0}(x)=0$ is the left endpoint of I. Then the orbit $(\psi _{-t}(x))_{t\in (0,r_0)}$ contains the leftmost component of $I\setminus \bigsqcup _{j=0}^{q_{\alpha }^{(n)}-1}T^jI^{(n)}_\alpha $ . Next, choose $x_\delta =\phi _\delta x\in J$ for small enough $\delta>0$ . Let $y=\psi _{-r}x_\delta \in J$ be the first backward return point for the horizontal flow to J. For all $\delta>0$ small enough, we have $r>r_0$ . In view of Lemma 4.2, the orbit $(\psi _{t}(y))_{t\in (0,r)}=(\psi _{-t}(x_\delta ))_{t\in (0,r)}$ crosses at most C edges of the zippered rectangles $D^{(n)}_\beta $ , $\beta \in \mathcal {A}\setminus \{\alpha \}$ . As $r>r_0$ , the orbit $(\psi _{-t}(x_\delta ))_{t\in (0,r_0)}$ , and hence the $\delta $ -close orbit $(\psi _{-t}(x))_{t\in (0,r_0)}$ , crosses at most C edges of the zippered rectangles $D^{(n)}_\beta $ , $\beta \in \mathcal {A}\setminus \{\alpha \}$ . Finally, the argument used in the previous paragraphs shows that the leftmost interval in $I\setminus \bigsqcup _{j=0}^{q_{\alpha }^{(n)}-1}T^jI^{(n)}_\alpha $ contains at most C levels of towers other than $\bigcup _{j=0}^{q^{(n)}_\alpha -1}T^j(I^{(n)}_{\alpha })$ . This completes the proof of the proposition.

Proof. Fix $\pi \in S^{\mathcal {A}}$ and let $\tau \in \Theta _{\pi }$ be rationally independent. We will choose an appropriate subset on which to induce the Zorich induction $\mathcal Z$ . We will define $\Delta^n$ as a properly chosen subinterval of the base of the tower $\bigcup _{j=0}^{q_\alpha^{(n)}-1}T^j(I^{(n)}_{\alpha })$ , for some $\alpha \in \mathcal {A}$ .

Let $\mathcal D_1\subseteq \Lambda ^{\mathcal {A}}$ be a set of length parameters such that

(5.7) $$ \begin{align} \mathcal D_1:=\{\unicode{x3bb}\in \Lambda^{\mathcal{A}}\mid \unicode{x3bb}_{\pi_0^{-1}(d)}\ge \tfrac{7}{8} \text{ and } \tfrac{1}{16}<|w_{\pi_0^{-1}(d)}|< \tfrac{1}{10} \}, \end{align} $$

where $w = \Omega _\pi \unicode{x3bb} $ is the translation vector associated with $(\pi , \unicode{x3bb} )$ .

Note that since $\unicode{x3bb} _{\pi _0^{-1}(d)}\ge \tfrac {7}{8}$ for $\unicode{x3bb} \in \mathcal D_1$ , the IET $(\pi ,\unicode{x3bb} )$ has top type for the Rauzy–Veech induction. Moreover, whenever $\mathcal Z^n(\pi ,\unicode{x3bb} , \tau )\in \{\pi \}\times \mathcal D_1\times \Theta _{\pi }$ , we have that $T^{q^{(n)}_{\pi _0^{-1}(d)}}$ acts on $I_{\pi _0^{-1}(d)}^{(n)}$ via translation by a number between $\tfrac {1}{16}|I_{\pi _0^{-1}(d)}^{(n)}|$ and $\tfrac {1}{8}|I_{\pi _0^{-1}(d)}^{(n)}|$ . Thus, the interval $I_{\pi _0^{-1}(d)}^{(n)}$ satisfies (5.6).

Now let $\tilde \tau \in \Theta _{\pi }$ be such that $(\pi ,\unicode{x3bb} ,\tilde \tau )$ is of bottom backward type and $\tilde \tau $ is rationally independent. Let $\epsilon =\epsilon (\tilde \tau )>0$ and $C_1=C(\tilde \tau ,\epsilon )>0$ be given by Proposition 4.1. Denote

$$ \begin{align*} \mathcal E:= \mathbb{R}_{>0}B_{\Theta_\pi}(\tilde\tau,\epsilon). \end{align*} $$

In view of Proposition 4.1, whenever $\mathcal Z^n(\pi ,\unicode{x3bb} ,\tau )\in \{\pi \}\times \mathcal D_1\times \mathcal E$ , the tower $\bigcup _{j=0}^{q^{(n)}_{\pi _0^{-1}(d)}-1} T^j {I_{\pi _0^{-1}(d)}^{(n)}}$ satisfies (5.5) with $C=C_1$ . Moreover, since the cone $\mathcal E$ is relatively compact (with respect to the Hilbert pseudometric) in $\Theta _{\pi }$ and by considering smaller $\epsilon $ if necessary, we may assume that all $\tau \in \mathcal E$ are of bottom backward type and

$$ \begin{align*} \sup_{\tau\in\mathcal E}\max_{\alpha,\beta\in \mathcal{A}}\frac{h_{\alpha}}{h_{\beta}} \leq C_2, \end{align*} $$

for some $C_2>0$ , where $h=h(\tau )=-\Omega _{\pi }\tau $ is the height vector. It follows that if $\mathcal Z^n(\pi ,\unicode{x3bb} ,\tau )\in \{\pi \}\times \mathcal D_1\times \mathcal E$ , then

$$ \begin{align*} \max_{\alpha,\beta\in \mathcal{A}}\frac{h^{(n)}_{\alpha}}{h^{(n)}_{\beta}} \leq C_2. \end{align*} $$

For every $\tau \in \Theta _\pi $ , let $M(\tau ):=\max _{\alpha \in \mathcal {A}}h_\alpha (\tau )$ , $m(\tau ):=\min _{\alpha \in \mathcal {A}}h_\alpha (\tau )$ and $\delta (\tau ):={M(\tau )}/{m(\tau )}$ . As the zippered rectangle $D_\alpha ^{(n)}$ consists of $q_\alpha ^{(n)}$ rectangles of heights $h_\beta $ for $\beta \in \mathcal {A}$ , we have

$$ \begin{align*}m(\tau)\leq \frac{h^{(n)}_{\alpha}}{q^{(n)}_{\alpha}}\leq M(\tau)\quad\text{for any }\alpha\in \mathcal{A}.\end{align*} $$

It follows that if $\mathcal Z^n(\pi ,\unicode{x3bb} ,\tau )\in \{\pi \}\times \mathcal D_1\times \mathcal E$ , then ${q^{(n)}_{\beta }}/{q^{(n)}_{\alpha }} \leq C_2\delta (\tau )$ for all $\alpha , \beta \in \mathcal {A}$ , and, letting $\alpha =\pi _0^{-1}(d)$ , we have

$$ \begin{align*} 1\hspace{-1pt}=\hspace{-1pt}|I|=\!\sum_{\beta\in\mathcal{A}}q^{(n)}_\beta|I^{(n)}_\beta|\hspace{-1pt}\leq\hspace{-1pt} C_2\delta(\tau)q^{(n)}_\alpha\!\sum_{\beta\in\mathcal{A}}|I^{(n)}_\beta|=C_2\delta(\tau)q^{(n)}_\alpha|I^{(n)}|\leq \frac{8}{7}C_2\delta(\tau)q^{(n)}_\alpha|I^{(n)}_\alpha|. \end{align*} $$

By taking $C:=\max \{C_1,\tfrac 87C_2\delta (\tau )\}$ , both conditions (5.3) and (5.5) are satisfied.

By [Reference Marmi, Moussa and Yoccoz18] (see §1.2.4), for almost every $\unicode{x3bb} \in \mathcal {D}_1$ , there exists $k=k(\pi ,\unicode{x3bb} )\in \mathbb {N}$ such that the Kontsevich–Zorich matrix $Q(k)(\pi ,\unicode{x3bb} )$ is positive (note that this matrix does not depend on $\tau $ ). Using the same argument for successive iterations of Zorich induction and multiplying several matrices with positive (natural) coefficients, for almost every $\unicode{x3bb} \in \mathcal {D}_1$ we can find $N=N(\pi ,\unicode{x3bb} )\in \mathbb {N}$ such that all coefficients of the Kontsevich–Zorich matrix $Q(N)(\pi ,\unicode{x3bb} )$ are greater than $320C^2$ . Since the Zorich induction $\mathcal Z$ and the Kontsevich–Zorich cocycle are locally continuous, there exists a subset $\mathcal D_2\subseteq \mathcal D_1$ of positive measure such that the matrix $Q(N)$ is constant on $\{\pi \}\times \mathcal D_2$ and all its coefficients are greater than $320C^2$ . Next, by the ergodicity of $\mathcal Z$ on $\mathcal M^1$ , there exists a subset $\mathcal D_3\subseteq \mathcal D_2$ of positive measure such that $\mathcal Z^{j}(\{\pi \}\times \mathcal D_3\times \Theta _\pi )$ are pairwise disjoint for $0\leq j\leq N$ . Thus, if $\mathcal Z^{n_k}(\pi ,\unicode{x3bb} ,\tau ),\mathcal Z^{n_{k+1}}(\pi ,\unicode{x3bb} ,\tau )\in \{\pi \}\times \mathcal D_3\times \mathcal E$ are two consecutive visits $\{\pi \}\times \mathcal D_3\times \mathcal E$ , then $q^{(k_{n+1})}_\alpha \geq 320C^2q^{(k_{n})}_\alpha $ for any $\alpha \in \mathcal {A}$ ensuring that condition (5.2) is met. Summarizing, by taking

$$ \begin{align*} \Delta^n:=I_{\pi_0^{-1}(d)}^{(k_n)}\quad\text{and}\quad q_n:=q^{(k_n)}_{\pi_0^{-1}(d)}, \end{align*} $$

where $k_n:=k_n(\pi ,\unicode{x3bb} ,\tau )$ is such that $k_n$ is the nth visit of $(\pi ,\unicode{x3bb} ,\tau )$ in $\mathfrak L:=\{\pi \}\times \mathcal D_3\times \mathcal E$ , we obtain that (5.2)–(5.6) are satisfied and the proof is complete.

Proof. To prove this result the idea is to show that for every $n> k$ a large proportion of the set $A_n$ ‘fills’ the holes left by the sets $A_k,\ldots ,A_{n-1}$ (more precisely, by sets $\tilde A_j$ and $T^{q_j} \tilde A_j$ for $k\leq j<n$ , which are properly chosen subsets of $A_j$ ). Later in the proof, we provide the construction of $\tilde A_n$ .

Note that, for every $n\ge k$ , the set $A_n^c$ is a union of $q_n+1$ disjoint intervals. This follows from the fact that each interval $J^n_i$ belongs to a different level of the tower $\bigcup _{0\leq i<q_n} T^i(\Delta ^n)$ and is separated from its left and right endpoints. In particular, the set $\bigcap _{j=k}^{n-1} A_j^c$ consists of at most $q_{n-1}+q_{n-2}+\cdots +q_k+1$ disjoint intervals. We refer to each of these intervals as a hole. Passing to $\bigcap _{j=k}^{n-1} \hat {A}_j^c$ , where

$$ \begin{align*}\hat{A}_j:=\tilde{A}_j\cup T^{q_j}\tilde{A}_j, \end{align*} $$

we extend the holes such that their lengths are at least ${10C}/{q_n}$ . Roughly speaking, all intervals from $A_n$ that are contained in these holes are used in the construction of $\tilde {A}_n$ .

Preliminary arguments. Assume that $(a,b)\subseteq [0,1)$ is a hole (interval) such that $b-a\geq {10C}/{q_n}$ . By (5.3) and (5.5), we have that the maximal distance between consecutive intervals forming $A_{n}$ is at most ${3C}/{2q_{n}}$ . In view of (6.7), the hole $(a,b)$ intersects l intervals of the form $J^{n}_i$ with $l \in \mathbb {N}$ satisfying

(6.10) $$ \begin{align} b-a\leq l\frac{D_2}{q_{n}\theta(q_{n})}+(l+1)\frac{3C}{2q_{n}}. \end{align} $$

Hence,

$$ \begin{align*} 10\leq l\frac{D_2}{C\theta(q_{n})}+(l+1)\leq l\frac{1}{32}+\frac{3}{2}(l+1), \end{align*} $$

so $l \geq 6$ . In particular, we get that $2(l-2)\geq l+1$ and thus by (6.10) we obtain

(6.11) $$ \begin{align} 2(l-2)\ge \dfrac{b-a}{({D_2}/{q_{n}\theta(q_{n})})+({3C}/{2q_{n}})}. \end{align} $$

Denote by $\mathcal I^n_{(a,b)}$ the set of indices $0\leq i<q_n$ for which $J^{n}_i\cap (a,b)\neq \emptyset $ after throwing out the indices corresponding to the leftmost and rightmost intervals in $A_n$ intersecting $(a, b)$ . Then $\# \mathcal I^n_{(a,b)} = l - 2\geq 4$ and

$$ \begin{align*}(a,b)\setminus \bigcup_{i\in \mathcal I^n_{(a,b)}}J^n_i\end{align*} $$

consists of $l-1$ holes. In view of (6.6), the length of every such hole is at least $\tfrac 12|\Delta ^n|$ and hence, since by (5.6) $T^{q_n}$ translates the interval $J^n_i$ by at most $\tfrac 18|\Delta ^n|$ , we obtain

(6.12) $$ \begin{align} T^{q_n}J^n_i \subseteq (a,b) \quad \text{for every }i\in\mathcal{I}^n_{(a,b)}. \end{align} $$

Note that, by (5.6), we have that

(6.13) $$ \begin{align} \{ J_i^n, T^{q_n}J_i^n \mid 0\leq i<q_n\} \text{ is a collection of pairwise disjoint sets.} \end{align} $$

Indeed, as $T^{q_n}\Delta ^n(\tfrac {1}{4}) \subseteq \Delta ^n(\tfrac {1}{8}) \subseteq \Delta ^n$ , for every $0\leq i<q_n$ , we have $J_i^n,\,T^{q_n}J_i^n \subseteq T^{i}\Delta ^n$ and for $i_1\neq i_2$ the intervals $T^{i_1}\Delta ^n$ and $T^{i_2}\Delta ^n$ are two different levels of the Rokhlin tower. It follows that $J^n_{i_1}$ and $T^{q_n}J^n_{i_1}$ are disjoint from $J^n_{i_2}$ and $T^{q_n}J^n_{i_2}$ and each pair of intervals is distant by at least $\tfrac 14|\Delta ^n|$ . Moreover, by (5.6), $T^{q_n}J_i^n$ is a translation of $J_i^n$ by at least $\tfrac {1}{16}|\Delta ^n|$ . In view of (6.7) and (5.3), $|J^n_i|\leq ({D_2}/{q_n\theta ( q_n)})\leq \tfrac {1}{32}|\Delta ^n|$ . It follows that $J_i^n$ and $T^{q_n}J_i^n$ are also disjoint and they are distant by at least $\tfrac {1}{32}|\Delta ^n|$ . In summary,

(6.14) $$ \begin{align} (a,b)\setminus \bigcup_{i\in \mathcal I^n_{(a,b)}}(J^n_i\cup T^{q_n}J^n_i)\text{ consists of } 2(l-2)+1 \text{ holes of length at least }\frac{1}{32}|\Delta^n|. \end{align} $$

Moreover, by (5.2) and (5.3),

(6.15) $$ \begin{align} \frac{1}{32}|\Delta^n|\geq \frac{1}{32Cq_n}\geq \frac{10C}{q_{n+1}}. \end{align} $$

On the other hand, by (6.7) and (6.11),

(6.16) $$ \begin{align} \mathrm{Leb}\bigg(\bigcup_{i\in \mathcal I^n_{(a,b)}}J^n_i\bigg) &= \mathrm{Leb}\bigg(\bigcup_{i\in \mathcal I^n_{(a,b)}}T^{q_n}J^n_i\bigg) \nonumber\\ &\ge (l-2)\cdot\dfrac{D_1}{q_{n}\theta(q_{n})}\geq \dfrac{b-a}{({2D_2}/{q_{n}\theta (q_{n})})+({3C}/{q_{n}})}\cdot\frac{D_1}{q_{n}\theta(q_{n})} \nonumber\\ &\geq \frac{D_1(b-a)}{(3C+2D_2)\theta( q_{n})}=\frac{c}{\theta( q_n)}(b-a), \end{align} $$

with $c:=D_1/(3C+2D_2)$ .

Inductive construction. Fix any interval $(r,s) \subseteq [0,1)$ and $k\in \mathbb {N}$ such that ${10C}/{q_{k}}\leq (s-r)$ and $\theta ( q_k)\geq 32 CD_2$ . We will define inductively a sequence $\{\tilde {A}_n\}_{n\geq k}$ of subsets of $(r,s)$ such that, for every $n\ge k$ , the set $\tilde A_n \subseteq A_n$ is the union of intervals $J_i^n$ with $0\leq i<q_n$ and

(6.17) $$ \begin{align} \text{all holes of }(r,s)\,\setminus\bigcup_{k\leq j\leq n}(\tilde A_j\cup T^{q_j}\tilde A_j)\text{ have length at least }\frac{10C}{q_{n+1}}, \end{align} $$

(6.18) $$ \begin{align} \mathrm{Leb}(\hat A_n)\geq \frac{2c}{\theta( q_n)}\mathrm{Leb}\bigg((r,s)\setminus\bigcup_{k\leq j< n}\hat A_j\bigg) \quad \text{where }\hat A_j=\tilde A_j\cup T^{q_j}\tilde A_j. \end{align} $$

Moreover, we define this sequence so that the collection $\{ \tilde A_n$ , $T^{q_n} \tilde A_n \mid n \geq k\}$ consists of pairwise disjoint subsets of $(r,s)$ .

First step. Let us define the set $\tilde A_k$ . As $ s-r\geq {10C}/{q_k}$ , using the preliminary arguments at the beginning of the proof, we can define $\tilde A_k$ as

$$ \begin{align*} \tilde A_k:=\bigcup_{i\in \mathcal I^k_{(r,s)}}J^k_i. \end{align*} $$

By (6.13), the sets $\tilde A_k$ and $T^{q_k}\tilde A_k$ are disjoint and, by (6.12), $T^{q_k} \tilde A_k$ is a subset of $(r,s)$ . Moreover, (6.17) follows directly from (6.14) and (6.15), and (6.18) from (6.16).

Inductive step. Suppose that the sets $\tilde A_k,\ldots , \tilde A_n$ are already defined, and let us define the set $\tilde A_{n + 1}$ . Denote by $\mathcal H_n$ the set of all holes in $(r,s)\setminus \bigcup _{k\leq j\leq n}(\tilde A_j\cup T^{q_j}\tilde A_j)$ . By assumption, for any $(a,b)\in \mathcal H_n$ , we have $(b-a)\geq {10C}/{q_{n+1}}$ . Let

$$ \begin{align*}\tilde A_{n+1}:=\bigcup_{(a,b)\in \mathcal H_n}\bigcup_{i\in \mathcal I^{n+1}_{(a,b)}}J^{n+1}_i.\end{align*} $$

By (6.13) the sets $\tilde A_{n+1}$ and $T^{q_{n+1}}\tilde A_{n+1}$ are disjoint, and by (6.12) we have that $T^{q_{n+1}}(\tilde A_{n+1}\cap (a,b)) $ is a subset of $(a,b)$ for any $(a,b)\in \mathcal {H}_n$ . That is, we have

$$ \begin{align*}\tilde A_{n+1},\ T^{q_{n+1}}\tilde A_{n+1} \subseteq \bigcup_{(a,b)\in \mathcal H_n}(a,b)=(r,s)\setminus\bigcup_{k\leq j\leq n}(\tilde A_j\cup T^{q_j}\tilde A_j),\end{align*} $$

and hence, we also have the disjointness of $\tilde A_{n+1}$ and $T^{q_{n+1}}\tilde A_{n+1}$ from $\tilde A_{j}$ and $T^{q_{j}}\tilde A_{j}$ for all $k\leq j\leq n$ . Moreover, (6.17) follows directly from (6.14) and (6.15). In view of (6.16) for any $(a,b)\in \mathcal H_n$ we have

$$ \begin{align*}\mathrm{Leb}\bigg(\bigcup_{i\in \mathcal I^{n+1}_{(a,b)}}(J^{n+1}_i\cup T^{q_{n+1}}J^{n+1}_i)\bigg)\geq\frac{2c}{\theta( q_{n+1})}(b-a).\end{align*} $$

Summing up along $(a,b)\in \mathcal H_n$ , we obtain

$$ \begin{align*} \mathrm{Leb}(\hat A_{n+1})\geq \frac{2c}{\theta( q_{n+1})}\bigcup_{(a,b)\in \mathcal{H}_n}(b-a)= \frac{2c}{\theta (q_{n+1})}\mathrm{ Leb}\bigg((r,s)\setminus\bigcup_{k\leq j\leq n}\hat A_j\bigg), \end{align*} $$

so (6.18) holds.

Final arguments. Using the construction above, we define a collection $\{ \tilde A_n$ , $T^{q_n} \tilde A_n \mid n \geq k\}$ of pairwise disjoint subsets of $(r,s)$ satisfying (6.17) and (6.18). Let $n \geq k$ . Since $\hat A_{n} \subseteq (r,s)$ and is disjoint from $\hat A_{k},\ldots ,\hat A_{n-1}$ , by (6.18), we have

$$ \begin{align*} & \mathrm{Leb}\bigg((r,s)\setminus\bigcup_{k\leq j\leq n}\hat A_j\bigg)\\ & \quad= \mathrm{Leb}\bigg((r,s)\setminus\bigcup_{k\leq j< n}\hat A_j\bigg)-\mathrm{Leb}(\hat A_{n}) \leq \bigg(1-\frac{2c}{\theta( q_{n})}\bigg)\mathrm{Leb}\bigg((r,s)\setminus\bigcup_{k\leq j< n}\hat A_j\bigg). \end{align*} $$

It follows that

$$ \begin{align*} \mathrm{Leb}\bigg((r,s)\setminus\bigcup_{k\leq j\leq n}\hat A_j\bigg)\leq (s-r)\prod_{k\leq j\leq n}\bigg(1-\frac{2c}{\theta( q_{j})}\bigg). \end{align*} $$

As $\sum _{n\geq k}({1}/{\theta ( q_n)})=+\infty $ , this gives

$$ \begin{align*}\mathrm{Leb}\bigg((r,s)\setminus\bigcup_{n\geq k}\hat A_n\bigg)=\lim_{n\to\infty}\mathrm{ Leb}\bigg((r,s)\setminus\bigcup_{k\leq j\leq n}\hat A_j\bigg)=0.\end{align*} $$

Therefore,

$$ \begin{align*}2\mathrm{Leb}\bigg(\bigcup_{n\geq k}\tilde A_n\bigg)=\mathrm{Leb}\bigg(\bigcup_{n\geq k}\tilde A_n\bigg)+\mathrm{ Leb}\bigg(\bigcup_{n\geq k}T^{q_n}\tilde A_n\bigg)=\mathrm{Leb}\bigg(\bigcup_{n\geq k}\hat A_n\bigg)=(s-r),\end{align*} $$

which completes the proof.

Proof of Theorem 6.3

In view of (5.3), the IET T is ergodic.

We will show that every number $a\in (v-D_1(4E)^{-1},v+D_1(4E)^{-1})$ is an essential value of $T_f$ . Notice that, by Proposition 6.2, this is enough to conclude the ergodicity of $T_f$ .

Fix any $a\in (v-D_1(4E)^{-1},v+D_1(4E)^{-1})$ and $0<\epsilon <D_1(4E)^{-1}-|v-a|$ . Note that

$$ \begin{align*}(a-\epsilon,a+\epsilon)\subseteq[v-D_1(4E)^{-1},v+D_1(4E)^{-1}] \subseteq S_{q_n}(f)(\Delta_i^n),\end{align*} $$

for all $n\in \mathbb {N}$ and $0\leq i<q_n$ . Let $J_i^n$ be a subinterval of $\Delta _i^n$ such that $S_{q_n}(f)(J_i^n)= [a-\epsilon ,a+\epsilon ]$ , for any $n\in \mathbb {N}$ and $0\leq i<q_n$ . By (6.5), each of these intervals satisfies ${2\epsilon E^{-1}}/({q_n\theta ( q_n)})\le |J_i^n|\le {2\epsilon E}/({q_n\theta ( q_n)})$ . In particular, the family of intervals $J_i^n$ for $n\in \mathbb {N}$ and $0\leq i<q_n$ satisfies the assumptions of Proposition 6.4.

Let $\Omega \subseteq [0,1)$ be any measurable set of positive Lebesgue measure. By the Lebesgue density theorem, there exists an interval $(r,s)\subseteq [0,1)$ such that

(6.19) $$ \begin{align} \mathrm{Leb}(\Omega\cap (r,s))>\tfrac{9}{10}(s-r). \end{align} $$

By Proposition 6.4, there exists $k\in \mathbb {N}$ such that for every $n\geq k$ there exists a set $\tilde A_n$ consisting of some intervals of the form $J_i^n$ such that $\{ \tilde A_n,\, T^{q_n} (\tilde A_n) \,|\, n \geq k\}$ is a collection of pairwise disjoint subsets of $(r,s)$ and

(6.20) $$ \begin{align} \mathrm{Leb}\bigg(\bigcup_{n\geq k}\tilde A_n\bigg)=\frac{1}{2}(s-r). \end{align} $$

We claim that there exist $n\geq k$ and $0\leq i<q_n$ such that

(6.21) $$ \begin{align} \mathrm{Leb}(\{x\in J_i^n \mid x, T^{q_n}x\in \Omega\cap(r,s)\})>0. \end{align} $$

Before proving this, notice that this implies that a is an essential value. Indeed, since we chose the interval $J_i^n$ so that $S_{q_n}f$ restricted to the interior of $J_i^n$ attains values in $(a-\epsilon ,a+\epsilon )$ , this would verify the condition required to check that a is an essential value.

Therefore, it suffices to show that (6.21) holds for some $n\geq k$ and $0\leq i<q_n$ . For the sake of contradiction, let us assume that (6.21) is never satisfied. Then, for any $n\geq k$ , any $0\leq i<q_n$ , and for almost every $x\in J_i^n$ , we have $x\notin \Omega \cap (r,s)$ or $T^{q_n}x\notin \Omega \cap (r,s)$ , so

$$ \begin{align*} \mathrm{Leb}(J_i^n)&\leq \mathrm{Leb}(J^n_i\setminus (\Omega\cap(r,s)))+\mathrm{Leb}(J^n_i\setminus T^{-q_n}(\Omega\cap(r,s)))\\ &=\mathrm{Leb}(J^n_i\setminus (\Omega\cap(r,s)))+\mathrm{Leb}(T^{q_n}J^n_i\setminus (\Omega\cap(r,s))) \\ & =\mathrm{Leb}((J^n_i\cup T^{q_n}J^n_i)\setminus (\Omega\cap(r,s))). \end{align*} $$

For any $n\geq k$ , by summing up over all $0\leq i<q_n$ with $J^n_i \subseteq \tilde {A}_n$ , we get

$$ \begin{align*} \mathrm{Leb}((\tilde A_n\cup T^{q_n} \tilde A_n)\setminus (\Omega\cap(r,s)))\geq \mathrm{Leb}(\tilde A_n). \end{align*} $$

Then, by summing up over $n\ge k$ and using (6.20), we get

$$ \begin{align*} \mathrm{Leb}((r,s)\setminus (\Omega\cap (r,s)))& \geq \mathrm{Leb}\bigg(\bigcup_{n\geq k}(\tilde A_n\cup T^{q_n} \tilde A_n )\setminus(\Omega\cap(r,s))\bigg)\\ & \geq \mathrm{Leb}\bigg(\bigcup_{n\geq k}\tilde A_n\bigg)=\frac{1}{2}(s-r). \end{align*} $$

Hence,

$$ \begin{align*} \mathrm{Leb}(\Omega\cap(r,s))\le \tfrac{1}{2}(s-r), \end{align*} $$

which is in contradiction to (6.19).

Proof. By the anti-symmetry of f, for any $x\in [0,1)$ and any $n \in \mathbb {N}$ , we have

$$ \begin{align*} \begin{aligned} S_nf(x)&=\sum_{0\leq j<n}f(T^jx)=-\sum_{0\leq j< n}f(T^{-1}\circ \mathcal{I}\circ T^jx)\\ &=-\sum_{0\leq j< n}f( T^{n-j-1}\circ T^{-n}\circ \mathcal{I} x)=-S_nf(T^{-n}\circ \mathcal{I} x)=-S_nf(\mathcal{I}\circ T^{n} x), \end{aligned} \end{align*} $$

which gives (7.1).

Let $-q<i<q$ . Notice that $T^{\pm q}(m_\alpha ) = m_\alpha \pm \delta $ and, more generally,

$$ \begin{align*} T^{\pm q}(x) = x \pm \delta \quad \text{for any }x\in T^i[m_\alpha-2\varepsilon/3,m_\alpha+2\varepsilon/3]. \end{align*} $$

Hence, $T^{\pm q}[m_\alpha -2\varepsilon /3,m_\alpha +2\varepsilon /3] \subseteq \Delta _\alpha $ , and the continuity of $S_qf$ on $T^i[m_\alpha -2\varepsilon /3,m_\alpha +2\varepsilon /3]$ follows directly from our assumptions.

Without loss of generality of reasoning, let us assume that f is piecewise increasing and that $0< \delta <\varepsilon /3$ . In all other cases, the reasoning goes the same way. We now consider the cases $\alpha \in \mathcal {A}$ and $\alpha = I$ separately.

Case 1: $\alpha \in \mathcal {A}$ . By the anti-symmetry of f, for any $x\in [0,1)$ , we have

$$ \begin{align*} S_qf(x)&=\sum_{0\leq j<q}f(T^jx)=f(x)-f(T^qx)+\sum_{0< j\leq q}f(T^jx)\\ &=f(x)-f(T^qx)-\sum_{0< j\leq q}f(T^{-1}\circ \mathcal{I} \circ T^jx)\\ &=f(x)-f(T^qx)-\sum_{0< j\leq q}f( T^{-j-1}\circ \mathcal{I} x)\\ &=f(x)-f(T^qx)-S_qf(T^{-q}\circ T^{-1}\circ \mathcal{I} x). \end{align*} $$

As $T^q(m_\alpha -\delta /2)=m_\alpha +\delta /2$ and $T^{-1}\circ \mathcal {I}(m_\alpha -\delta /{2})=m_\alpha +\delta /{2}$ , it follows that

$$ \begin{align*} S_qf\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)&=f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)- f\bigg(T^q\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)\bigg)-S_qf\bigg(T^{-q}\circ T^{-1}\circ \mathcal{I}\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)\bigg)\\ &=f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)-f\bigg(m_\alpha+\dfrac{\delta}{2}\bigg)- S_qf\bigg(T^{-q}\bigg(m_\alpha+\dfrac{\delta}{2}\bigg)\bigg)\\ &=2f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)-S_qf\bigg(m_\alpha-\dfrac{\delta}{2}\bigg). \end{align*} $$

Therefore,

$$ \begin{align*}S_qf\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)=f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg).\end{align*} $$

Note that, if $0\leq i\leq q$ , for any $x\in [m_{\alpha} -2\varepsilon/3,m_{\alpha} +2\varepsilon/3]$ , we have

$$ \begin{align*} S_qf(T^ix)=\sum_{i\leq j<q}f(T^jx)+\sum_{0\leq j<i}f(T^q T^jx)=\sum_{i\leq j<q}f(T^jx)+\sum_{0\leq j<i}f(T^jx+\delta). \end{align*} $$

As f is piecewise increasing, this yields

$$ \begin{align*}S_qf(x)\leq S_qf(T^ix)\leq S_qf(x+\delta).\end{align*} $$

It follows that

$$ \begin{align*}S_qf\bigg(T^i\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)\bigg)\geq S_qf\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)=f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)\end{align*} $$

and

$$ \begin{align*} S_qf\bigg(T^i\bigg(m_\alpha-\dfrac{3\delta}{2}\bigg)\bigg)\leq S_qf\bigg(m_\alpha-\dfrac{\delta}{2}\bigg)=f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg). \end{align*} $$

By continuity, there exists

$$ \begin{align*} \xi_i\in T^i\bigg[m_\alpha-\dfrac{3\delta}{2},m_\alpha-\dfrac{\delta}{2}\bigg]\quad \text{ such that }\ S_qf(\xi_i)=f\bigg(m_\alpha-\dfrac{\delta}{2}\bigg). \end{align*} $$

If $-q\leq i\leq 0$ , then we take $\xi _i:=\xi _{q+i}$ . Then

$$ \begin{align*} \xi_i=\xi_{q+i}\in T^{q+i}\bigg[m_\alpha-\dfrac{3\delta}{2},m_\alpha-\dfrac{\delta}{2}\bigg] =T^{i}\bigg[m_\alpha-\dfrac{\delta}{2},m_\alpha+\dfrac{\delta}{2}\bigg]. \end{align*} $$

Finally, we have $\xi _i\in T^{i}[m_\alpha -\varepsilon /2,m_\alpha +\varepsilon /2]$ in both cases.

Case 2: $\alpha = I$ . As $T^{-q}(m_I-\delta /2)=m_I+\delta /2$ and $\mathcal {I}(m_I+\delta /2)=m_I-\delta /2$ , by (7.1) it follows that

$$ \begin{align*} S_qf\bigg(m_I-\dfrac{\delta}{2}\bigg)=-S_qf\bigg(m_I-\dfrac{\delta}{2}\bigg)\quad \text{so } S_qf\bigg(m_I-\dfrac{\delta}{2}\bigg)=0. \end{align*} $$

As f is piecewise increasing, if $0 < i \leq q$ , the arguments used in the first case show that

$$ \begin{align*} S_qf(x)\leq S_qf(T^ix)\leq S_qf(x+\delta) \quad \text{for any } x\in [m_I-2\varepsilon/3,m_I+2\varepsilon/3]. \end{align*} $$

It follows that

$$ \begin{align*} S_qf\bigg(T^i\bigg(\hspace{-.5pt}m_I-\dfrac{\delta}{2}\bigg)\bigg)&\geq S_qf\bigg(\hspace{-.5pt}m_I-\dfrac{\delta}{2}\bigg)=0\quad \text{and}\\ S_qf\bigg(T^i\bigg(\hspace{-.5pt}m_I-\dfrac{3\delta}{2}\bigg)\bigg)&\leq S_qf\bigg(\hspace{-.5pt}m_I-\dfrac{\delta}{2}\bigg)=0. \end{align*} $$

By continuity, there exists

$$ \begin{align*} \xi_i\in T^i\bigg[m_I-\dfrac{3\delta}{2},m_I-\dfrac{\delta}{2}\bigg]\quad \text{ such that } S_qf(\xi_i)=0. \end{align*} $$

If $-q\leq i\leq 0$ , then we again take $\xi _i:=\xi _{q+i}$ .

Proof. Note that

$$ \begin{align*} \int_{I_\alpha(\delta)}f(x)\,dx&=\int_{l_\alpha+\delta|I_\alpha|}^{m_\alpha}f(x)\,dx+\int_{m_\alpha}^{r_\alpha-\delta|I_\alpha|}f(x)\,dx\\ &=\int_{l_\alpha+\delta|I_\alpha|}^{m_\alpha}f(x)\,dx-\int_{m_\alpha}^{r_\alpha-\delta|I_\alpha|}f(\mathcal{I}\circ T^{-1}x)\,dx\\ &=\int_{l_\alpha+\delta|I_\alpha|}^{m_\alpha}f(x)\,dx-\int_{l_\alpha+\delta|I_\alpha|}^{m_\alpha}f(x)\,dx=0. \end{align*} $$

As $\mathcal {I}\circ T=T^{-1}\circ \mathcal {I}$ , we have $\mathcal {I}\circ T^n=T^{-n}\circ \mathcal {I}$ for any $n\in \mathbb {Z}$ . Hence, for any $x\in J$ , we have $T^{r_J(x)}\circ \mathcal {I}\circ T_Jx=\mathcal {I} x\in J$ . It follows that $r_J(\mathcal {I}(T_Jx))=r_J(x)$ and $\mathcal {I}\circ T_J=T_J^{-1}\circ \mathcal {I}$ . By assumption, the permutation of $T_J$ belongs to the Rauzy class of $\pi $ , so it is irreducible and non-degenerated. Moreover, since $\unicode{x3bb} $ is rationally independent, the lengths of the intervals exchanged by $T_J$ are different. Therefore, $T_J$ is also symmetric. Moreover, by (7.1),

$$ \begin{align*}f_J(\mathcal{I}(T_Jx))=S_{r_J(\mathcal{I}(T_Jx))}f(\mathcal{I}(T_Jx))=S_{r_J(x)}f(\mathcal{I}(T^{r_J(x)}x))=-S_{r_J(x)}f(x)=-f_J(x).\end{align*} $$

By the first part of the lemma, this gives $m(f_J, J_\alpha (\delta ))=0$ for every $\alpha \in \mathcal {A}$ .

Proof. By (5.3) and (5.5), there exists $\bar {C}>1$ such that for any $x\in T^i\Delta ^n(\tfrac {1}{4})$ with $0\leq i<q_n$ , we have that the distance between two consecutive elements of the set

$$ \begin{align*}\{T^jx:0\leq j<q_n\}\cup \mathrm{End}(T),\end{align*} $$

where $\mathrm {End}(T)$ denotes the set of endpoints of the exchanged intervals of T, is always between $\bar {C}^{-1}/q_n$ and $\bar {C}/q_n$ . As $1/\tau $ is decreasing, this observation and (7.2) together yield

$$ \begin{align*} |S_{q_n}f'(x)|&\leq \sum_{0\leq j<q_n}|f'(T^jx)|\leq 2dZ_\theta(f)\sum_{1\leq j\leq q_n}\frac{1}{\tau(({1}/{\bar{C}q_n})j)}\\ & \leq 4dZ_\theta(f)\bar{C}q_n\int_{{1}/{(2\bar{C}q_n)}}^1\frac{ds}{\tau(s)}. \end{align*} $$

Moreover, by the definition of $\tau $ ,

$$ \begin{align*} \int_{{1}/{(2\bar{C}q_n)}}^1\frac{ds}{\tau(s)}= \int_{{1}/{(2\bar{C}q_n)}}^1 \frac{\theta'(1/s)ds}{s^2} = &\, -\int_{{1}/{(2\bar{C}q_n)}}^1 \frac{d}{ds}\theta(1/s)ds\\ = &\, \theta(2\bar{C}q_n)-\theta(1)\leq \theta(2\bar{C}q_n), \end{align*} $$

which gives (7.4).

Suppose now that f is piecewise monotonic and that $z_\theta (f)$ given by (7.5) is positive. Let us assume without loss of generality that f is increasing and that, for some $\alpha _0 \in \mathcal {A}$ , $z_\theta (f)=\inf _{x\in (l_{\alpha _0},m_{\alpha _0}]}f'(x)\tau (x-l_{\alpha _0})>0$ . In all other cases, the arguments proceed in the same way.

Then, for any $x\in T^i\Delta ^n(\tfrac {1}{4})$ with $0\leq i<q_n$ , we have

$$ \begin{align*} |S_{q_n}f'(x)|&=\sum_{0\leq j<q_n}f'(T^jx)\geq \sum_{\substack{0\leq j<q_n\\ T^jx\in(l_{\alpha_0},m_{\alpha_0}]}}f'(T^jx) \geq z_\theta(f)\sum_{1\leq j\leq ({\unicode{x3bb}_{\alpha_0}}/{2\bar{C}})q_n}\frac{1}{\tau(({\bar{C}}/{q_n})j)}\\ &\geq \frac{z_\theta(f)}{\bar{C}}q_n\int_{{\bar{C}}/{q_n}}^{{\unicode{x3bb}_{\alpha_0}}} \frac{ds}{\tau(s)}= \dfrac{z_\theta(f)}{\bar{C}}q_n \bigg(\theta\bigg(\dfrac{q_n}{\bar{C}}\bigg)-\theta \bigg(\dfrac{1}{\unicode{x3bb}_{\alpha_0}}\bigg)\bigg), \end{align*} $$

which gives (7.6).

Proof of Theorem 1.1

By Theorem 5.1, for almost every T there exist an increasing sequence $\{q_n\}_{n\in \mathbb {N}}$ of natural numbers and a sequence $\{\{T^i(\Delta ^n) \mid 0\leq i<q_n\}\}_{n\in \mathbb {N}}$ of Rokhlin towers satisfying (5.1)–(5.6). Moreover, it follows from the proof of Theorem 5.1 that we can take $\Delta ^n=I_{\pi _0^{-1}(d)}^{(k_n)}$ and $q_n=q^{(k_n)}_{\pi _0^{-1}(d)}$ for some increasing sequence $\{k_n\}_{n\in \mathbb {N}}$ . In view of [Reference Berk, Trujillo and Ulcigrai3, Lemma 3.12], there exist $\alpha \in \mathcal {A}\cup \{I\}$ and $0\leq s_n< q_n$ such that $T^{-s_n}m_{\alpha }=m_n$ is the center of $I_{\pi _0^{-1}(d)}^{(k_n)}$ .

As $T^{q_n}m_n=m_n+\delta _n$ , for some $\tfrac {1}{16}|\Delta ^n|<|\delta _n|<\tfrac {1}{8}|\Delta ^n|$ , the interval $[m_\alpha -\varepsilon _n,m_\alpha +\varepsilon _n]=T^{s_n}[m_m-\varepsilon _n,m_n+\varepsilon _n]$ with $\varepsilon _n=\tfrac {3}{8}|\Delta ^n|$ satisfies the assumptions of Lemma 7.1. Hence, for any $0\leq i<q_n$ , there exists

$$ \begin{align*} \xi^n_i\in T^{i-s_n}[m_\alpha-\varepsilon_n,m_\alpha+\varepsilon_n]= T^i[m_n-\varepsilon_n/2,m_n+\varepsilon_n/2]=T^i\Delta^n(\tfrac{5}{16}) \end{align*} $$

such that

$$ \begin{align*} S_{q_n}f(\xi^n_i) = v_n := \begin{cases} f(m_\alpha-\delta_n/2) & \text{ if } \alpha\in\mathcal{A}, \\ 0 & \text{ if } \alpha = I. \end{cases} \end{align*} $$

By the anti-symmetry of f, we have, $v_n\to f(m_\alpha ) = 0$ as $n\to \infty $ .

As $f\in \Upsilon _\theta (\bigsqcup _{\alpha \in \mathcal {A}}I_{\alpha })$ and $\theta $ is slowly varying, by Proposition 7.3, there exists $E>1$ such that

$$ \begin{align*} E^{-1}q_n\theta(q_n)\leq|S_{q_n}(f')(x)|\leq Eq_n\theta(q_n)\quad\text{for any }x\in T^i\Delta^n(\tfrac{1}{4})\text{ and }0\leq i<q_n. \end{align*} $$

Notice that $[\xi ^n_i-\tfrac {1}{16}|\Delta ^n|,\xi ^n_i+\tfrac {1}{16}|\Delta ^n|]\subseteq \Delta ^n(\tfrac {1}{4})$ . Let $0<D<C^{-1}/8$ , where C is the constant appearing in (5.1)–(5.6) and let $\Delta ^n_i$ be the interval centered at $\xi ^n_i$ such that $|\Delta ^n_i|={D}/({q_n\theta (q_n)})<\tfrac {1}{8}|\Delta^n|$ . Then $\Delta ^n_i\subseteq \Delta ^n(\tfrac {1}{4})$ and $|S_{q_n}f(\xi ^n_i)|=v_n\leq {D}/{4E}$ , for all n large enough. In view of Theorem 6.3, this gives the ergodicity of $T_f$ .

Proof. Note that, for almost every symmetric IET $T = (\pi , \unicode{x3bb} )$ there exist $0<\kappa <1$ and a sequence of renormalizations $\{(\pi ^{(k)},\unicode{x3bb} ^{(k)})\}_{k\geq 1}$ such that:

1. (i) $\pi ^{(k)}=\pi $ for $k\geq 0$ ;

2. (ii) $\unicode{x3bb} ^{(k)}_\alpha>\kappa |I^{(k)}|$ for all $\alpha \in \mathcal {A}$ ;

3. (iii) T satisfies the UDC for the sequence of renormalizations $\{(\pi ^{(k)},\unicode{x3bb} ^{(k)})\}_{k\geq 1}$ ;

4. (iv) $|I^{(k)}|p^{(k)}\geq \kappa>0$ , where $p^{(k)}$ is the maximal number for which $T^j(\mathrm {Int}I^{(k)})$ , $0\leq j<p^{(k)}$ , avoid the discontinuities of T;

5. (v) for every $\alpha \in \mathcal {A}\cup \{I\}$ there exist $\beta _\alpha \in \mathcal {A}\cup \{I\}$ and $0\leq s^{(k)}_\alpha < q_\alpha ^{(k)}$ such that $T^{s^{(k)}_\alpha }m^{(k)}_\alpha =m_{\beta _\alpha }$ , where $q_I^{(k)}= p^{(k)}$ .

Let $J^{(k)}:=I^{(k)}_{\alpha _0}$ if $\beta _{\alpha _0}=I$ , and $J^{(k)}:=I^{(k)}$ if $\beta _I=I$ . We require an additional condition for T:

1. (vi) the lengths of all intervals exchanged by $T_{J^{(k)}}$ are greater than $\kappa |J^{(k)}|$ .

To show that the almost every symmetric IET T satisfies the above conditions, one must construct a set $Y\subset \mathcal {M}$ of positive measure such that, for any $(\pi ',\unicode{x3bb} ,\tau )\in Y$ , we have:

• • $\pi '=\pi $ and $\unicode{x3bb} _\alpha \geq \kappa |\unicode{x3bb} |$ for $\alpha \in \mathcal {A}$ (this is responsible for (i) and (ii));

• • $(\pi ,\unicode{x3bb} ,\tau )$ satisfies (5.8) and (5.9) (in view of Remark 5.2, this is responsible for (v));

• • $(\pi ,\unicode{x3bb} ,\tau )$ satisfies conditions (i) and (ii) in the proof of Lemma 3.6 (this is responsible for (iv));

• • for any $\alpha \in \mathcal A$ the IET induced from $(\pi ,\unicode{x3bb} )$ on $I_\alpha $ is such that the lengths of all its exchanged intervals are greater than $\kappa \unicode{x3bb} _\alpha $ (this is responsible for (vi)).

Finally, we apply the proof of Theorem 3.8 in [Reference Frączek and Ulcigrai12] (to the set $Y\subset \mathcal {M}$ ) to obtain a renormalization sequence (given by recurrence to the set $K\subset Y$ ) for which the UDC is also met.

For every $\alpha \in \mathcal {A}$ , let $r^{(k)}_\alpha \in \mathbb {N}$ be the first return time of $J^{(k)}_{\alpha }$ for $T^{(k)}$ to $J^{(k)}$ . By assumption, $r^{(k)}_\alpha \leq \kappa ^{-2}$ , for all $k\geq 1$ and any $\alpha \in \mathcal {A}$ . For any $0\leq j<r_\alpha ^{(k)}$ , let $\gamma ^{(k)}(\alpha ,j)\in \mathcal {A}$ be such that $(T^{(k)})^j J^{(k)}_{\alpha } \subseteq I^{(k)}_{\gamma ^{(k)}(\alpha ,j)}$ . As the sequences $\{r^{(k)}_\alpha \}_{k\geq 1}$ are bounded, we may pass to a subsequence and assume that $r^{(k)}_\alpha $ and $\gamma ^{(k)}(\alpha ,j)$ do not depend on k, so we will write $r_\alpha $ and $\gamma (\alpha ,j)$ instead. Let us consider the appearance matrix $D\in SL(d,\mathbb {Z})$ given by $D_{\alpha \beta }=\#\{0\leq j<r_\alpha :\gamma (\alpha ,j)=\beta \}$ .