3.1 Translation surfaces

There are various approaches to defining a translation surface, each tailored to different interests.

One common depiction of a translation surface involves considering a connected topological surface S alongside a discrete subset $ \Sigma \subseteq S$ . Within $ S \setminus \Sigma $ , a maximal atlas called the translation atlas $ \{ \phi _\alpha , U_\alpha \}_{\alpha \in \Lambda }$ is defined, where the transition maps are Euclidean translations.

For the purposes of studying billiard dynamics within polygons and generalized polygons, a more practical definition of a translation surface involves an at-most-countable collection of convex polygons $\{ P_n \} \subset \mathbb {R}^2 $ , where each pair of edges is identified by translations. Denoting the collection of vertices of these polygons as $ V $ , we define the surface as

$$ \begin{align*} S^\circ = \bigg( \bigg(\!\bigcup_{n \in \mathbb{N}} P_n \bigg) \setminus V \bigg) / \sim. \end{align*} $$

If $S^\circ $ is a connected space, we call S, the metric completion of $S^\circ $ , a translation surface.

These two definitions are equivalent (see [Reference Delecroix, Hubert and ValdezDHV24]). A translation surface S is categorized as finite type if the collection of polygons is finite or if the topological surface S on which the translation atlas is defined is homeomorphic to a compact surface X; otherwise, it is termed infinite type.

Now, for any direction $[0,2\pi )$ , we have a well-defined translation flow in $\mathbb {C} = \mathbb {R}^2$ given by

$$ \begin{align*} F_{\theta}^t(z) = z + t e^{i\theta}, \end{align*} $$

which yields a constant vector field

$$ \begin{align*} X_{\theta} = \frac{\partial F_{\theta}^t}{\partial t} \bigg|_{t=0}(z). \end{align*} $$

The definition of a translation surface allows us to pull back this vector field to $S \setminus \Sigma $ . For any $z \in S \setminus \Sigma $ , we can consider a maximal integral curve $\gamma _z: I \to S \setminus \Sigma $ such that $\gamma _z(0) = z$ and we can define a flow on $S \setminus \Sigma $ as

$$ \begin{align*} \phi_{\theta}^t(z) = \gamma_z(t). \end{align*} $$

This is the translation flow in direction $\theta $ . When the direction is not stated, meaning only writing $\phi _t$ , we will mean the translation flow in the vertical direction.

3.2 Ergodic theory

In terms of the Birkhoff theorem, it can be stated as for every measurable function f,

$$ \begin{align*} \lim \frac{1}{n}\sum_{i=0}^n f\circ T^n(x)\to \int_X f\,d\mu \end{align*} $$

for $\mu -$ almost every x.

In the ergodic theory category, we have the following chain of implications:

$$ \begin{align*} \textit{Mixing} \to \textit{Weakly}\ \textit{mixing} \to \textit{Ergodic.} \end{align*} $$

Our aim is to study the continuous version of these properties.

Definition 3.4. We say that the flow $\phi _t$ on a measurable space $(X,\mathcal {B},\mu )$ is mixing if for every $A,B\in \mathcal {B}$ ,

$$ \begin{align*}\lim_{t\to\infty} \mu(\phi_{-t}(A)\cap B)= \mu(A)\mu(B).\end{align*} $$

As we shall see in following sections, the notion of equidistribution plays an important role.

Definition 3.5. Let $\{x_n\}_{n\ge 1}$ be a sequence of points in $[a,b]$ . We say that $\{x_n\}$ is equidistributed in $[a,b]$ if for every subinterval $[c,d] \subset [a,b] $ , one has

$$ \begin{align*} \lim_{n\to\infty} \dfrac{1}{n} \#\{1 \le k \le n : x_k \in [c,d]\} = \frac{d-c}{b-a}. \end{align*} $$

Remark 3.6. By the equidistribution theorem, if q is an irrational number, then the sequence $\{nq\!\! \mod {1}\:\: n\in \mathbb {N}\} $ is equidistributed in the unitary interval.

Definition 3.7. A transformation $T:X\to X$ is said to be uniquely ergodic if there exists only one invariant measure for T.

When a transformation is uniquely ergodic, the unique invariant measure is ergodic and the conclusion of the Birkhoff ergodic theorem applies for every $x\in X$ .

Observe that if T is a continuous uniquely ergodic transformation in an interval $[a,b)$ , then for any x, the sequence $x_n=T^n(x)$ fulfills that for every continuous function f,

$$ \begin{align*} \lim_{n\to \infty} \frac{1}{n}\sum_{i=0}^n f(x_n)\to\frac{1}{b-a} \int_a^b f(x)\,dx, \end{align*} $$

in particular, the sequence $x_n$ equidistributes in the interval $[a,b)$ .

3.3 Cutting and stacking constructions

Cutting and stacking constructions represent a powerful tool extensively used in ergodic theory to construct measurable invertible transformations on the interval. Intuitively, they can be conceptualized as a limit process involving compositions of periodic maps on subintervals. This process entails considering a sequence of transformations $T_1, T_2, \ldots , T_n, \ldots $ satisfying the following properties:

1. (1) Domain $(T_i)\subseteq \textbf {Domain}(T_{i+1})$ ;

2. (2) $T_{n+1|_{\textbf {Domain}(T_n)}}=T_n.$

We can define a map T such that

$$ \begin{align*} \textbf{Domain}(T)=\bigcup_{n\in \mathbb{N}}\textbf{Domain}(T_n) \end{align*} $$

and such that T is the pointwise limit of the sequence $T_n$ .

We are going to pay attention to a class of transformations derived from cutting and stacking constructions, referred to as rank-one transformations. These transformations originate from a geometric framework inspired by Rokhlin’s column theorem [Reference Cornfeld, Fomin and Grigor’evǐc SinaiCFS12]. Their significance was highlighted by Ornstein [Reference OrnsteinOrn72], who employed them to construct examples of interval transformations demonstrating mixing properties without the presence of a square root. Ornstein’s stochastic argument established the almost sure mixing property of these transformations. Subsequent contributions by Adams [Reference AdamsAda98], or Creutz and Silva [Reference Creutz and SilvaCS04, Reference Creutz and SilvaCS06] and others have provided concrete examples and conditions under which these transformations exhibit mixing behavior.

3.3.1 Rank-one transformations

We provide an algorithmic definition of such transformations. Consider an interval $I=[0,b)$ with b possibly infinite, and a subinterval $J=[0,a)$ with a necessarily finite. We denote the length of the interval J as A. The interval J serves as the initial step of the transformation, with the initial transformation $T_0$ being the identity. The interval I acts as the domain of the transformation T.

Now, let $\{r_n\}$ be a sequence of positive integers. First, we divide interval J into $r_1$ subintervals $J_i$ with equal lengths ${A}/{r_1}$ . Then, we introduce another sequence of non-negative integers $\{s_n\}$ . Each interval $[a,a+ ({s_1A}/{r_1}))$ is divided into $s_1$ equal parts and each of these subintervals, referred to as spacers, is placed above one of the parts into which we divided the interval J.

After adding all these spacers above the corresponding subintervals, the set of all intervals stacked above each subinterval $J_i$ of the partition of J is called a column $C_{1,i}$ . We stack all these columns one above the other such that column $C_{1,i}$ lies beneath $C_{1,i+1}$ for each $i\leq r_1-1$ . The height $h_1$ of this first step of the transformation T is given by

$$ \begin{align*} h_1=r_1+ \sum_{i=1}^{r_1}s_{1,i}, \end{align*} $$

where $s_{1,i}$ is the number of spacers added above interval $J_i$ .

We define a transformation $T_1$ on the first $h_1-1$ intervals of this column $\bigcup _i C_{1,i}$ such that

$$ \begin{align*}T_1(x,j)=(x,j+1),\end{align*} $$

where the second entry indicates the interval in the column. Throughout this text, this column will be referred to as the first step of the configuration of the transformation T.

To define the transformation $T_2$ , we partition the first step of the transformation’s configuration into $r_2$ equal parts, resulting in $r_2$ new subcolumns. We further divide the interval $[a+({As_1}/{r_1}),a+({As_1}/{r_1})+ ({As_2}/{r_1r_2}))$ into $s_2$ equal parts, each denoted as spacers. These new spacers are positioned above the $r_2$ new subcolumns. The number of spacers above each subcolumn is denoted as $s_{2,i}$ . Similar to the previous step, we stack the new subcolumns $C_{2,i}$ one above the other.

The height $h_2$ is given by

$$ \begin{align*}h_2=r_2h_1+ \sum_{i=1}^{r_2}s_{2,i}.\end{align*} $$

The transformation $T_2$ is defined on the first $h_2-1$ intervals as $T_2(x,j)=(x,j+1)$ . It is important to note that $T_2$ coincides with $T_1$ in the intervals corresponding to the first step of the configuration. This new stack is referred to as the second step of the configuration of the transformation T.

Continuing this process, suppose we have defined the nth step of the configuration of the transformation T, with height $h_n$ and a well-defined transformation $T_n$ . To define the next step, we partition the column with $h_n$ intervals into $r_{n+1}$ equal parts, resulting in $r_{n+1}$ new subcolumns. We further divide the interval

$$ \begin{align*} \bigg[a+\sum_{i=1}^{n}\frac{s_nA}{\prod_{j=1}^{i}r_j},a+\sum_{i=1}^{n+1}\frac{s_{n+1}A}{\prod_{j=1}^{i}r_j}\bigg) \end{align*} $$

into $s_{n+1}$ equal parts and they are positioned above each of the $r_{n+1}$ new subcolumns according to the provided extra information. The number of intervals attached to the top of each subcolumn is denoted as $s_{n+1,i}$ .

We stack each of these new stacks one above the other, similar to previous cases. The height is given by

$$ \begin{align*} h_{n+1}=r_{n+1}h_n+\sum_{i=1}^{r_{n+1}}s_{n+1,i}. \end{align*} $$

The transformation $T_{n+1}$ is defined such that $T_{n+1}(x,j)=(x,j+1)$ for each $j\leq h_{n+1}-1$ .

Continuing this process to infinity, we obtain a transformation T defined on the interval

$$ \begin{align*} [0,b)=\bigg[0, \lim_{n\to\infty}\frac{h_n A}{r_1\cdots r_n}\bigg). \end{align*} $$

Definition 3.8. The sequences $\{r_n\}$ and $\{s_n\}$ are referred as cutting sequence and spacer sequence, respectively, for the transformation T that comes from the described cutting and stacking construction.

Definition 3.9. We call the interval $J=[0,a)$ the non-spacer interval and the interval $I\setminus {J}$ as the spacer interval.

Under suitable growth assumptions, for example, when

$$ \begin{align*} \frac{r_n^2}{h_n} \longrightarrow 0, \end{align*} $$

the interval $[0,b)$ has finite length; this is the regime of interest here. Such assumptions are referred to as restricted growth conditions. Moreover, by a result of Ryzhikov [Reference RyzhikovRyz92], mixing staircase transformations satisfying restricted growth are n-fold mixing for every $n\in \mathbb {N}$ . We refer to §6 for further discussion.

3.4 Translation surfaces from rank-one transformations

The description we present here is a specific instance of a broader framework outlined in [Reference Lindsey and TreviñoLTn16].

Let us recall the definition of a flow constructed under a function, often referred to as a suspension flow.

Let $ T:I\to I$ be a measurable transformation on the unit interval and let $f:I\to \mathbb {R}^+ $ be a bounded, square-integrable function that is strictly positive. Consider the space

$$ \begin{align*} X=\{(x,y)\mid x\in I, \ y<f(x)\}. \end{align*} $$

We define a suspension flow $\phi _t\hspace{-0.5pt}=\hspace{-0.5pt}(I,T,f,\mu )$ as follows: fix $p,q\hspace{-0.5pt}\in\hspace{-0.5pt} \mathbb {R}^+$ ; for $(x,y)\hspace{-0.5pt}\in\hspace{-0.5pt} X$ , the flow $\phi _t(x,y)=(x,y+t)$ if $y+t<f(x)$ , and $\phi _t(x,y)=(T(x),0)$ if $y+t=f(x)$ . This flow preserves the Lebesgue measure $\mu $ and provides an identification of the interval I with the graph of f.

Now, consider a transformation T derived from a cutting and stacking process, and let $f:I\to \mathbb {R}^+$ be such that for some $c\in I$ point of discontinuity for T, $f(x)=p$ for every $x\in [0,c)$ , and $f(x)=q$ for every $x\in [c,1]$ . Moreover, we require that f is constant on each subinterval of continuity of the transformation T.

Observe that from this construction, we obtain S together with the segments $\{0\} {\kern-1.5pt}\times{\kern-1.5pt} [0,p]$ , $\{c\}\times [ \min (\{p,q\}),\max (\{p,q\})]$ , and $\{1\}\times [0,q]$ as a rectangle or an $L-$ shaped polygon.

Without loss of generality, we may assume that $p>q$ . We do the following identification.

1. (1) Identify $\{0\}\times [0,q]$ with $\{1\}\times [0,q]$ .

2. (2) Identify $\{c\}\times [q,p]$ with $\{0\}\times [q,p]$ .

3. (3) If $f(x)=\epsilon (x)\in \{p,q\}$ , identify $(x,0)$ with $(T(x),\epsilon (x))$ .

See Figure 1. Observe that after these identifications, we obtain a translation surface S, and the suspension flow $\phi _t$ built under f is isomorphic in the measure-theoretical sense to the vertical translation flow in the surface S.

Figure 1

Identify horizontal sides according to as shown in the picture and vertical sides with respect the transformation T (colour online).

Our main result is stated as follows.

4.1 Outline of the proof of Theorem 4.2

Staircase transformations arise from the Rokhlin tower theorem. This theorem implies that for each step in the construction of the transformation, we obtain a finite number of subintervals whose lengths strictly decrease at each step. As the process continues indefinitely, the union of these subintervals will cover the interval $[0,1]$ . These unions at each step provide natural candidates for partial coverings of $[0,1]$ .

For any $\delta> 0$ , there exists an m such that for any subinterval $I_j^{(m)}$ from the m-step of the staircase transformation configuration, the following conditions are met.

1. (1) The measure of $I_j^{(m)}$ satisfies that $\unicode{x3bb} (I_j^{(m)})<\delta $ .

2. (2) The measure of the union of these subintervals satisfies that

   $$ \begin{align*}\unicode{x3bb}\bigg(\!\bigcup_{i=1}^{h_m}I_i^{(m)}\bigg)>1-\delta.\end{align*} $$

Therefore, for that $\delta $ , the partial partition $\eta $ is the union of those intervals. Note that the partial partition depends only on $\delta $ and not $t\in \mathbb {R}_+$ .

It is noteworthy that for any interval $I^{(m)}$ , the application of the flow $\phi _t$ will produce a partition of $ I^{(m)} $ into several horizontal segments. Our goal is to approximate the distribution of these horizontal segments.

To prove that the partitions arising from the staircase construction satisfy the second requirement of Lemma 4.3, we first consider the cases where $ R \subseteq [0, x_0] \times [0, p] $ and $ R \subseteq [x_0, 1] \times [0, q] $ , with $ x_0 $ being the point that divides the spacer and non-spacer intervals. These two rectangles yield a natural partition of the L-shaped polygon, the first is the non-spacer part of the polygon and the second is the spacer part of the polygon. Then, we have to approximate the product of measures $\mu (R) \unicode{x3bb} (I)$ by $\unicode{x3bb} (\phi _t(I) \cap R)$ . For the remainder of the paper, we assume that $(p,q)=(1,q)$ .

First, we prove a lemma inspired by the fact that a staircase transformation $ T $ exhibits mixing properties [Reference AdamsAda98]. Lemma 4.4 states that there exists an $m_0$ such that for every $n\geq m_0$ , any level $ I^{(n)} $ at the nth step of the staircase transformation, and any rectangle $ R $ , the measure

$$ \begin{align*} \unicode{x3bb}(\pi_x(\phi_t(I^{(n)}))\cap \pi_x(R)) \end{align*} $$

converges to $ \unicode{x3bb} (I^{(n)})\unicode{x3bb} (\pi _x(R)) $ .

Next, we establish that for any level at any step of the staircase construction, the action of the flow $ \phi _t $ on this interval produces an equidistributed set of heights. More precisely, given any level $ I^{(n)} $ , we define the $ I^{(n)} $ -cylinder as either $ I^{(n)} \times [0,1) $ or $ I^{(n)} \times [0,q) $ , depending on whether $ I^{(n)} $ is a non-spacer or a spacer interval. We then show that for any $ I^{(m)} $ , the application of $ \phi _t $ generates different heights. In the non-spacer case, we prove that for any vertical subinterval $ [a,b] \subseteq [0,1] $ and any horizontal interval $ I^{(n)} $ , the ratio of the number of heights attained by $ \phi _t(I^{(m)}) $ within $ [a,b] $ (such that $ \phi _t(I^{(m)}) $ remains in the $ I^{(n)} $ -cylinder) to the total number of heights within the $ I^{(n)} $ -cylinder converges to the measure of $ [a,b] $ . Similarly, for any $ [a,b) \subseteq [0,q) $ , the ratio of the number of heights attained by $ \phi _t(I^{(m)}) $ within $ [a,b] $ in the $ I^{(n)} $ -cylinder to the total number of heights in $ I^{(n)} $ converges to $ \unicode{x3bb} ([a,b]) / q $ .

From there, if $R=b(R)\times h(R)$ and

$$ \begin{align*}b(R)=\bigcup_{i\in J}I^{(n_i)},\end{align*} $$

where J is some index and $I^{(n_i)}$ is an interval in the $n_i$ th step of the staircase configuration, then

$$ \begin{align*} \lim_{t\to\infty}\unicode{x3bb}(I^{(m)}\cap \phi_{-t}(R))=\lim_{t\to\infty}\sum_{i\in J}\unicode{x3bb}(\phi_t(I)\cap I^{(n_i)}\times h(R))\geq \mu(R)\unicode{x3bb}(I^{(m)}). \end{align*} $$

For the classical staircase case, we can calculate explicitly the intervals where the roof function is $1$ and q. If the first interval of definition has length A, then the first spacer $s_2$ (remember that $s_1=0$ ) will have length $({A}/{2})$ , and for the third step, we have to add $3$ spacers each of length $({A}/{6})$ . In general, we have that the total length of all the spacers added after k steps is

$$ \begin{align*} \sum_{n=1}^k \frac{A}{n!}\sum_{i=1}^n(i-1)=\sum_{n=1}^{k} \frac{A(n-1)}{2(n-1)!}=\frac{A}{2} + \sum_{n=2}^{k}\frac{A}{2(n-2)!}, \end{align*} $$

which converges to $({A}/{2}) + ({Ae}/{2})$ . Therefore, if we want the interval I to have length $1$ , then we should have that $A={2}/({3+e})$ .

Lemma 4.4. There exists an $m_0$ depending solely on q such that for every $m\geq m_0$ , when $I^{(m)}$ is an m-level of the configuration of the classical staircase T and R is a rectangle under the graph of f, if $\pi _x$ is the horizontal projection and $b(R)=\pi _x(R)$ is the base of the rectangle R, then

$$ \begin{align*} \lim_{t\to \infty} \unicode{x3bb}(\phi_t(I^{(m)}))\cap \pi_x^{-1}(\pi_x(R))=\unicode{x3bb}(I^{(m)})\unicode{x3bb}(b(R)). \end{align*} $$

Proof. For this proof, we use the fact that the classical staircase transformation T is mixing.

First, we need to check that if m is large, then for arbitrary t,

$$ \begin{align*} \unicode{x3bb}(\pi_x(\phi_t(I^{(m)})))=\unicode{x3bb}(I^{(m)}).\end{align*} $$

Suppose that for a level $I^{(m)}$ ,

$$ \begin{align*}\unicode{x3bb}(\pi_x(\phi_t(I^{(m)})))<\unicode{x3bb}(I^{(m)}).\end{align*} $$

This implies that there exist two different sublevels $I^{(n+m)}_{i}$ and $I^{(n+m)}_{j}$ and non-negative real numbers $t,s$ such that

$$ \begin{align*} \phi_{t-s}(I^{(n+m)}_{i})=\phi_t(I^{(n+m)}_{j}). \end{align*} $$

Moreover, $s\geq 0$ must be smaller than $1$ or q.

Figure 3

Part of the orbit of the special flow built under f and the classical staircase T (colour online).

Observe that if we order upwards the sublevels in which $I^{(m)}$ is partitioned in the $(m+n)$ th step of the configuration of the staircase, then $j<i$ . This complication arises because under the intervals positioned beneath $1$ , there may be accumulations of other intervals at varying heights. To illustrate this concept, if an interval lies below q, a single iteration of $\phi _q$ elevates this interval to the subsequent level in the staircase configuration. Conversely, if the interval lies beneath $1$ , there exists a value $N_0>1$ such that $(N_0-1)q<1\leq N_0q$ . This indicates that it takes several iterations (specifically, at most $N_0$ iterations) before the interval advances to the next level in the staircase configuration; therefore, if two sublevels $I^{(n+m)}_i,I^{(n+m)}_j$ are in a distance smaller than $2N_0$ , it follows that under the action of the flow $\phi _t$ , we might encounter several instances $t_l$ in which

$$ \begin{align*}\pi_x(\phi_{t_l}(I_i^{(n+m)}))\cap \pi_x(\phi_{t_l}(I_j^{(n+m)}))\neq\emptyset.\end{align*} $$

We demonstrate that if the initial interval is sufficiently small, we will not find this problem.

Observe that if $I^{(m)}$ is in the mth step of the configuration of the staircase, when we consider the $m+1$ subintervals in which it is divided in the $(m+1)$ th step of the configuration of the staircase, then the distance between any two of those levels will be bounded from below by $h_m$ , the height of the mth step of the configuration. Since $h_m$ is an increasing sequence, then there exists an $m_0$ such that $h_{m_0}\geq M_0= 4N_0$ ; therefore, for any $m\geq m_0$ and for any interval $I^{(m)}$ in the mth step of the configuration of the staircase, it follows that the number of levels between any two subintervals $I^{(m+n)}_i$ and $I^{(m+n)}_j$ for any $n\geq 0$ will be bounded from below by $4N_0$ and, therefore,

$$ \begin{align*}\pi_x(\phi_{t}(I_i^{(n+m)}))\cap \pi_x(\phi_{t}(I_j^{(n+m)}))=\emptyset.\end{align*} $$

Continuing with the proof, notice that for each t, there exists an $n\geq 0$ such that $\phi _t(I^{(m)})$ is scattered in ${n!}/{m!}$ subintervals. For each subinterval $I_j^{(m)}$ with $j=1,\ldots ,{n!}/{m!}$ , there exists $l_j$ such that $\pi _x(\phi _t(I_j^{(m)}))=T^{l_j}(I_j^{(m)})$ . From previous observation,

$$ \begin{align*} \pi_x(\phi_t(I^{(m)}))=\bigcup_{j=1}^{n!/m!}T^{l_j}(I_j^{(m)}). \end{align*} $$

Define $m(t)$ as the minimum of such $l_j$ and $M(t)$ as the maximum. We can assume that for large t, it holds true that

$$ \begin{align*} |\unicode{x3bb}(b(R)\cap T^{M(t)}(I^{(m)}))-\unicode{x3bb}(b(R))\unicode{x3bb}(I^{(m)})| & \leq |\unicode{x3bb}(b(R)\cap \pi_x(\phi_t(I^{(m)})))-\unicode{x3bb}(b(R))\unicode{x3bb}(I^{(m)})| \\ & \leq |\unicode{x3bb}(b(R)\cap T^{m(t)}(I^{(m)}))-\unicode{x3bb}(b(R))\unicode{x3bb}(I^{(m)})|. \end{align*} $$

Since T is mixing, this implies that

$$ \begin{align*} \begin{aligned} \unicode{x3bb}(b(R)\cap \pi_x(\phi_t(I^{(m)})))\to\unicode{x3bb}(b(R))\unicode{x3bb}(I^{(m)}). \end{aligned}\\[-36pt] \end{align*} $$

Remark 4.5. Let $m_0$ be the integer such that for every $m\geq m_0$ and every $I^{(m)}$ , Lemma 4.4 holds. If $I^{(n)}$ is an interval in the nth step of the configuration of the staircase with $n<m_0$ , observe that $I^{(n)}$ is made up by

$$ \begin{align*}I^{(n)}=\bigsqcup_{i=1}^{m!/n!}I^{(m)}_i.\end{align*} $$

Therefore, for any rectangle R, it follows that

$$ \begin{align*} \lim_{t\to\infty}\unicode{x3bb}(\phi_t(I^{(n)})\cap \pi_x^{-1}(b(R)))=\sum_{i=1}^{m!/n!}\unicode{x3bb}(I^{(m)}_i)(\unicode{x3bb}(b(R)))=\unicode{x3bb}(I^{(n)})\unicode{x3bb}(b(R)).\end{align*} $$

It is worth noting that when we consider an interval $I^{(n)}$ and observe how it evolves under the flow $\phi _t$ , over time, this interval will break up into several horizontal segments scattered throughout the polygon. For the second part of the proof, our goal is to approximate the distribution of the heights achieved by these horizontal segments.

Our strategy will be to assume, without loss of generality, that the heights of the polygon are 1 and q. We then consider a discretization of the flow given by the iterates of $\phi _q$ . By doing so, we aim to emulate the behavior of an irrational rotation and demonstrate that the heights of the horizontal segments follow a distribution similar to the set $ \{nq\!\! \mod {1} : n \in \mathbb {N}\} $ .

4.2 Behavior of the first subcolumn

The discussion that follows is crucial for delving into the second part of the proof, with the aim of convincing us that by discretizing the flow $ \phi _t $ through iterations of $ \phi _q $ , applying $ \phi _{mq} $ to any level $ I^{(n)} $ yields an increasing sequence of heights contained in $ \{nq, n \in \mathbb {N}\!\! \mod {1}\} $ .

Notation. We say that $ x \in I $ lies beneath $ q $ if $ f(x) = q $ ; similarly, we say that $x\in I$ lies beneath $1\, f(x) = 1 $ .

Note that if a subinterval $ I^{(m)}$ lies beneath $ q $ at height zero, applying $ \phi _q $ resets its height to zero. Subsequent applications of $ \phi _q $ maintain this zero height as long as the subinterval remains beneath $ q $ , until it eventually reaches a height of zero beneath $ 1 $ . At this point, the application of $ \phi _q $ results in alternating heights: first $ q $ , then $ 2q\!\! \mod {1} $ , and so forth. Due to the construction of the classical staircase, the subinterval passes through the roof whose value is $ 1 $ twice consecutively before going to the spacer side.

Now, consider $ I^{(m)} $ as the last level of the $ m $ th step configuration of the staircase before it breaks into $ m+1 $ new subintervals $ I_j^{(m)} $ with $ j \leq m+1 $ . We explain that the heights attained by $ \phi _{nq}(I^{(m)}) $ with $m+1\leq n\leq h_m$ coincide with the heights attained by $ \phi _{nq}(I^{(m)}_1),\phi _{(n-1)q}(I^{(m)}_1),\ldots ,\phi _{(n-1-m)q}(I^{(m)}_1) $ .

Let $ I^{(m)} $ be as defined in the previous paragraph. Observe that each subinterval $ I^{(m)}_j $ has $ j-1 $ spacers above it, all of which lie beneath $ q $ by construction. Consequently, applying $ \phi _q $ to $ I^{(m)}_j $ up to $ j-1 $ times ensures that $ \phi _{nq}(I^{(m)}_j) $ remains beneath $ q $ and has height zero for any $ n \leq j-1 $ . Moreover, applying $ \phi _q $ exactly $ j $ times to $ I^{(m)}_j $ results in an image that lies beneath $ 1 $ and also has height zero.

Thus, for any $ j \leq m $ , the height of $ \phi _q(I^{(m)}_{j+1}) $ matches that of $ I^{(m)}_{j} $ . This observation implies that the heights attained by successive iterations of $ \phi _{nq} $ on $ I^{(m)}_{j+1} $ are directly related to those attained by iterating $ \phi _{(n-1)q} $ on $ I^{(m)}_{j} $ . Therefore, in the $ (m+1) $ th step configuration of the staircase, the heights of the intervals $ I^{(m)}_j $ are fully determined by the heights attained by $ I^{(m)}_1 $ . We show that this reasoning extends to arbitrarily large iterations. For now, it is clear this holds when $ n \leq h_m $ . See Figures 4, 5, and 6 for a visualization of this concept.

Figure 4

Left: Third step of the staircase configuration, the first two levels are non-spacers, the last three are spacers, the level $I^{(3)}$ is partitioned in the sublevels before it breaks into three levels. Right: Representation of $I^{(3)}$ in the L-shaped polygon (colour online).

Figure 5

After one application of $\phi _q$ , the heights are reset to zero, but the intervals spread in the staircase configuration (colour online).

Figure 6

After two iterations of $\phi _q$ , we obtain a first height different than zero (colour online).

This behavior continues until the level $I^{(m)}_{m+1}$ reaches the final level in the $(m+1)$ th step of the staircase configuration, just before it splits into $m+2$ new sublevels $I^{(m)}_{m+1,l}$ , where $l \leq m+2$ . Using a similar argument, we find that the dynamics at this step of the configuration for the sublevels $I^{(m)}_{m+1,l}$ are governed by those attained by $I^{(m)}_{m+1,1}$ . Furthermore, the heights reached by $I^{(m)}_{m+1,1}$ remain connected to the heights attained by the sublevels $I^{(m)}_1,\ldots ,I_{m}^{(m)}$ , as explained in the previous paragraph. Therefore, to analyze the dynamics in the $(m+2)$ th step of the staircase configuration, it is sufficient to focus on $I^{(m)}_{1,1}$ . Following this reasoning, we conclude that for any interval $I^{(m)}$ , when considering its representation in the $(m+n)$ th step of the staircase configuration, the heights attained after several iterations, say n, coincide with the heights obtained by the set

$$ \begin{align*}{\phi_q(I^{(m)}_{1,\ldots,1}), \ldots, \phi_{nq}(I^{(m)}_{1,\ldots,1})}.\end{align*} $$

From the previous paragraphs, we can draw the following conclusion. Let $I^{(m)}$ represent the final interval in the mth step of the staircase configuration. Consider how this interval is represented in the $(m+n)$ th step of the staircase configuration. The heights reached by applying $\phi _q$ to $I^{(m)}\, n$ times correspond to the heights achieved by the intervals $I^{(m+n)}_1, \phi _q(I^{(m+n)}_1),\ldots , \phi _{nq}(I^{(m+n)}_1)$ , where $I^{(m+n)}_1$ is the lowest level in the representation of $I^{(m)}$ within the $(m+n)$ th step of the staircase configuration, corresponding to $I^{(m)}_{1,\ldots ,1}$ .

More precisely, if k iterations of $\phi _q$ produce that $I^{(m)}$ lies in the $(m+n)$ th step of the staircase configuration, $I^{(m)}$ will be broken into several subintervals $I^{(m)}_J$ , where J is an index of length bounded by $(m+n-1)!/m!$ ; for one of such subintervals $I^{(m)}_{a_1,\ldots ,a_j}$ , it will hold true that the height attained by $\phi _{kq}(I^{(m)}_{a_1,\ldots ,a_j})$ will coincide with the height attained by $\phi _{(K-N)q}(I^{(m)}_{1,\ldots ,1})$ , where

$$ \begin{align*} N=\sum_{i=1}^j|1-a_i|. \end{align*} $$

As $k\to \infty $ , the staircase where $I^{(m)}$ will lie is the $(m+n)$ th step of the staircase configuration with $n\to \infty $ , then the number of different heights also diverges. This result hints that, in the limiting process of iterations, and due to the equidistribution of $\{nq\!\! \mod 1 : n \in \mathbb {N}\}$ , the heights obtained will also equidistribute.

We would like to prove that

$$ \begin{align*} \lim_{t\to\infty} \unicode{x3bb}(\phi_t(I^{(m)})\cap R)\geq\unicode{x3bb}(I^{(m)})\unicode{x3bb}(b(R))\unicode{x3bb}(h(R)). \end{align*} $$

We already have the first part; for the second part, we would like that the concentration of heights contained in the vertical interval spanned by R does not depend on $I^{(m)}$ but only on $h(R)$ . For this purpose, we need a notion of equidistribution.

As noted in §4.2, when examining the images of a subinterval from any m-step configuration of the staircase under the flow $\phi _t$ , the subinterval breaks into further segments. Due to the presence of spacers arranged in a staircase-like pattern, this process results in a sequence of different heights. Our objective is to study the distribution of these heights.

The distribution of heights behaves differently depending on whether the intervals lie beneath $1$ or beneath q. Specifically, if an interval lies under $1$ , every iteration of $\phi _q$ changes its height. In contrast, if the interval lies under q, it is mapped to another level of the staircase while preserving its height. Moreover, blocks of spacer intervals all share the same height. Thus, to analyze the distribution of heights generated by iterations of $\phi _q$ , it is essential to distinguish between intervals lying beneath $1$ and those lying beneath q. For the next lemma, we introduce a convenient notation.

Definition 4.9. Let $t \in \mathbb {R}$ and let $I,J$ be horizontal intervals, while $[a,b] \subseteq [0,1)$ (or $[0,q)$ , depending on whether we are in the non-spacer or spacer part) is a vertical interval. We define the set of heights as follows:

$$ \begin{align*} H_t^{ns}(I,J,[a,b]) &= \{ z \in [a,b] \;:\;\text{there exists } y \in I \\& \qquad \text{ such that } \pi_y(\phi_t(y)) = z \text{ and } \pi_x(\phi_t(y)) \in J \}, \end{align*} $$

when J is contained in the non-spacer part. Similarly, we set

$$ \begin{align*} H_t^{s}(I,J,[a,b]) &= \{ z \in [a,b] \;:\; \text{there exists } y \in I \\& \qquad \text{ such that } \pi_y(\phi_t(y)) = z \text{ and } \pi_x(\phi_t(y)) \in J \}, \end{align*} $$

when J is contained in the spacer part.

Definition 4.10. Let $I^{(m)}$ be a level in the m-step of the staircase configuration. We say that $I^{(m)}$ is H-equidistributed if, for any interval $[a,b] \subseteq [0,1]$ in the vertical direction,

$$ \begin{align*} \lim_{t\to\infty} \frac{\textbf{card}(H_t^{ns}(I^{(m)},I_{ns},[a,b]))}{\textbf{card}((H_t^{ns}(I^{(m)},I_{ns},[0,1]))}=b-a, \end{align*} $$

and if, for any $[a,b] \subseteq [0,q)$ ,

$$ \begin{align*} \lim_{t\to\infty} \frac{\textbf{card}(H_t^{s}(I^{(m)},I_s,[a,b]))}{\textbf{card}((H_t^{ns}(I^{(m)},I_s,[0,q]))}=\frac{b-a}{q}. \end{align*} $$

Achieving $ H $ -equidistribution in the spacer part is not straightforward. Although iterating $ \phi _q $ yields an increasing sequence of heights, understanding the precise distribution of these heights in the spacer part is crucial. While this might initially appear to obstruct $ H $ -equidistribution, the following result shows that the structured nature of these repetitions ensures that $ H $ -equidistribution is still achieved.

Remark 4.11. As noted in §4.2, the heights attained by $\phi _{kq}(I^{(m)})$ are determined by the heights of the iterates

$$ \begin{align*} \phi_{kq}(I^{(m)}_{1,\ldots ,1}), \quad \phi_{(k-1)q}(I^{(m)}_{1,\ldots ,1}), \quad \ldots, \quad \phi_{q}(I^{(m)}_{1,\ldots ,1}), \end{align*} $$

where the length of the sequence $1,\ldots ,1$ reflects the number of times $I^{(m)}$ has passed through the endpoint of different steps in the staircase configuration.

Among these iterations, there exist specific times $n^s_1,\ldots , n^s_i$ at which $\phi _{n^s_l q}(I^{(m)}_{1,\ldots ,1})$ falls into a spacer interval, and times $n_1^{ns},\ldots , n_j^{ns}$ at which $\phi _{n_j^{ns} q}(I^{(m)}_{1,\ldots ,1})$ falls into a non-spacer interval.

From the observations in §4.2, we note that the heights attained by subintervals of $I^{(m)}$ after multiple iterations of $\phi _q$ are determined by those achieved by $I^{(m)}_{1,\ldots ,1}$ . Moreover, applying $\phi _q$ to a spacer interval preserves its height, whereas applying it to a non-spacer interval changes its height. This leads to the following conclusions.

• • For each time $n_l^s$ , there is an associated height $h_l^s$ such that these heights correspond to subintervals of $I^{(m)}$ that, after k iterations of $\phi _q$ , lie in cylinders whose base is a spacer interval.

• • Similarly, for each time $n_l^{ns}$ , there is an associated height $h_l^{ns}$ such that these heights correspond to subintervals of $I^{(m)}$ that, after k iterations of $\phi _q$ , lie in cylinders whose base is a non-spacer interval.

The heights $h_l^s$ and $h_l^{ns}$ correspond to the values attained by $\phi _{n_l^sq}(I^{(m)}_{1,\ldots ,1})$ and $\phi _{n_l^{ns}q}(I^{(m)}_{1,\ldots ,1})$ . Since the heights of the iterates of $\phi _q$ for $I^{(m)}_{1,\ldots ,1}$ remain unchanged while passing through spacer blocks, multiple times $n_l^s$ can correspond to a single height. In contrast, for the times $n_l^{ns}$ , the mapping to heights is injective.

Proof. First, consider the non-spacer case. To establish the claim, we assume that $ I^{(m)} $ is initially at the uppermost level in the $ m $ -step configuration of the staircase transformation before breaking into $ m+1 $ new subintervals. If this is not the case, we can shift the interval using the flow until it reaches this uppermost level. This adjustment preserves the connectedness of the interval, ensuring that during the shifting process, it attains only one height.

Using the ideas developed in §4.2, we observe that the heights attained by $ \phi _{kq}(I^{(m)}) $ coincide with those attained by $ I^{(m)}_{1,\ldots ,1} $ . Since the heights of $ I^{(m)}_{1,\ldots ,1} $ remain unchanged while passing through spacer blocks, it follows that all the heights achieved by $ I^{(m)}_{1,\ldots ,1} $ after multiple iterations of $ \phi _q $ are found in the non-spacer side, while in the spacer, we find only a subsequence of such heights. From the observations in Remark 4.11, we conclude that these heights are attained by subintervals that, after multiple iterations, lie in cylinders whose base is non-spacer. The equidistribution of $ nq\!\! \mod 1 $ then establishes the first part of the proof.

For the second part of the proof, we analyze the heights attained by the spacer subintervals, which correspond to the blocks of spacer levels in the staircase configuration. First, observe that, by Remark 4.11, the heights of the intervals of $ I^{(m)} $ that, after iterations of $ \phi _q $ , lie in cylinders whose base is a spacer interval correspond to the heights $ h^{ns}_l $ . Thus, it suffices to study the distribution of heights achieved by $ I^{(m)}_{1,\ldots ,1} $ after multiple iterations. To achieve this, we construct two sequences as follows.

Let $m_1$ be the smallest integer such that $(m_1 - 1)q < 1 < m_1 q$ . Note that $m_1 q \mod {1} < q$ . Let $k_1'$ be the smallest integer such that

$$ \begin{align*} (k_1' - 1)(m_1 q\!\!\! \mod\kern-1.6pt {1}) < 1 < k_1'(m_1 q\!\!\! \mod\kern-1.6pt {1}) \end{align*} $$

and define $k_1 = k_1' m_1$ . In addition, $k_1 q\!\! \mod 1 < q$ .

Next, let $m_2'$ be the smallest integer such that

$$ \begin{align*}(m_2' - 1)(k_1 q\!\!\! \mod\kern-1.6pt 1) < 1 < m_2' (k_1 q\!\!\! \mod\kern-1.6pt 1)\end{align*} $$

and define $m_2 = m_2' k_1$ .

Then, let $k_2'$ be the smallest integer such that

$$ \begin{align*}(k_2' - 1)(m_2 q\!\!\! \mod\kern-1.6pt 1) < 1 < k_2'(m_2 q\!\!\! \mod\kern-1.6pt 1)\end{align*} $$

and define $k_2 = k_2' m_2$ .

In general, suppose $k_n$ is defined. Let $m_{n+1}'$ be the smallest integer such that

$$ \begin{align*}(m_{n+1}' - 1)(k_n q\!\!\! \mod\kern-1.6pt 1) < 1 < m_{n+1}' (k_n q\!\!\! \mod\kern-1.6pt 1)\end{align*} $$

and define $m_{n+1} = m_{n+1}' k_n$ .

Let $k_{n+1}'$ be the smallest integer such that

$$ \begin{align*}(k_{n+1}' - 1)(m_{n+1} q\!\!\! \mod\kern-1.6pt 1) < 1 < k_{n+1}'(m_{n+1} q\!\!\! \mod\kern-1.6pt 1)\end{align*} $$

and define $k_{n+1} = k_{n+1}' m_{n+1}$ .

It is clear that both sequences satisfy $k_n q\!\! \mod 1 < q$ and $m_n q\!\! \mod 1 < q$ . Moreover, the sequence $k_n q\!\! \mod 1$ describes the heights attained by the spacer intervals under the iterations of $\phi _q$ .

From this, we can convince ourselves that if $a_n$ is the sequence such that $a_n q\!\! \mod 1 < q$ , then since $\{nq\!\! \mod 1\}$ equidistributes in $[0,1)$ , it follows that since $\{a_n q\!\! \mod 1\}$ describes the orbit of $0$ of the first return map of the irrational rotation by q to $[0,q)$ , this first return map g is again an irrational rotation, in particular, all its iterates $g^k$ are uniquely ergodic; therefore, its orbits equidistribute in $[0,q)$ .

From the construction of the sequence $k_i$ , we can observe that $k_i = a_{2i}$ since the odd-indexed terms correspond to $m_i$ . In other words, the sequence $k_nq\!\! \mod 1$ describes the orbit of the second return map to $[0,q)$ , which is $g^2$ in the last paragraph. In particular, it is uniquely ergodic and the orbit of zero will equidistribute. Consequently,

$$ \begin{align*} \lim_{n \to \infty} \frac{\textbf{card} (\{ m \leq n \mid k_m q \mod\kern-1.6pt 1 \in [a,b] \} )}{n} =\frac{b-a}{q} \end{align*} $$

for every $ [a,b] \subseteq [0,q)$ .

As mentioned, the analysis in the spacer case is complicated since multiple intervals can share the same height. According to the last paragraph, the sequence $k_n q\!\! \mod 1$ describes the heights attained by the spacer intervals. In the mth step of the staircase configuration, spacers appear in blocks. Following §4.2, if we apply sufficiently many times $\phi _q$ in such a way that $I^{(m)}_1$ goes its way through one $m(q)$ th step of the staircase configuration, all intervals within a block of spacers share the same height. Rather than considering individual intervals, we focus on these blocks as a whole.

Arrange the blocks of spacers by order of appearance upwards in a staircase configuration. Consider an enumeration $\{d(l)\}_{l\in \mathbb {N}}$ of that appearance such that $d(l)$ describes the number of intervals that the lth spacer block has in order of appearance. For any n, we examine the sequence of blocks containing n intervals. It can be observed that the appearance of these blocks follows a linear pattern. Specifically, for each n, there exists values $a_n$ and $b_n$ such that for any l, the block of spacers $d(b_n+a_n l)$ contains exactly n intervals. Consequently, for each n, the sequence $n_l = k_{b_n+a_n l}$ is such that the set $\{n_l q \ mod 1 : l \in \mathbb {N}\}$ accurately describes the heights of spacer intervals that appear in blocks of n elements.

We can give a description of the sequences $n_j$ . For the case $n=1,2$ , the sequences are given by $1_j=1+3j$ and $2_j=2+3j$ , respectively.

Let n be any other number. Consider $m(n)$ the first step in the staircase configuration such that there exists a block of spacers with n intervals. For $m(n)$ , there will be $m(n)-1$ new numbers $c_i$ such that there exists a block with $c_i$ spacer intervals such that these numbers did not appear in the $(m(n)-1)$ th step of the staircase configuration. Order them from bottom to top. Let $F(n)$ be the first time n appears in the list of blocks. If n is not $c_{m(n)-1}$ , then

$$ \begin{align*} n_j=F(n)+ j\frac{m(n)!}{2} \end{align*} $$

and if n is $c_{m(n)-1}$ , then

$$ \begin{align*} n_j=F(n)+j\frac{(m(n)+1)!}{2}. \end{align*} $$

Continuing with the proof; just as before, using the fact that the first return map of a rotation by $q $ to the subinterval $[0,q)$ is totally ergodic, and observing that for any n, the sequence $n_jq\!\! \mod 1$ is the orbit of $g^{2b_n}(0)$ under the transformation $g^{2a_n}$ , we see that the sequence of heights attained by blocks with n intervals equidistribute in $[0,q)$ .

Define $P_n$ as the probability of finding a block with n elements. Then,

$$ \begin{align*} \lim_{n\to\infty} \frac{\textbf{card}(H_{nq}^{s}(I^{(m)}, I_{s,} [a,b]))}{\textbf{card}(H_{nq}^{s}(I^{(m)}, I_{s},[0,q)])}=\sum_{n\in\mathbb{N}}\bigg(\frac{b-a}{q}\bigg)P_n=\frac{b-a}{q}. \end{align*} $$

Observe that H-equidistribution for $t\to \infty $ follows directly from the H-equidistribution when $nq\to \infty $ , since for large t, the variation when considered, the discrete and continuous version converges to zero. The proposition follows.

Remark 4.13. Observe that H-equidistribution alone is not sufficient to approximate $\unicode{x3bb} (\phi _t(I^{(m)}) \cap R)$ with $\unicode{x3bb} (I^{(m)}),b(R)$ and $h(R)$ . A diagonal concentration phenomenon (see Figure 7) may occur: although the heights of $\phi _t(I^{(m)})$ equidistribute in $[0,1]$ over the non-spacer part and in $[0,q]$ over the spacer part, this only yields equidistribution in the vertical projection $\pi _y(R)$ . It does not imply equidistribution inside the rectangle R itself, in other words, it does not ensure that the vertical coordinates of $\phi _t(I^{(m)}) \cap R$ equidistribute in $[0,1]$ or in $[0,q]$ , respectively.

Figure 7

A diagonal behavior causes the intervals to miss several rectangles (colour online).

To overcome this difficulty, we prove a stronger statement. Given an arbitrary interval $I^{(n)}$ in the staircase configuration, we take $J = I^{(n)}$ in Definition 4.10 and establish equidistribution of the heights within the corresponding cylinders over $I^{(n)}$ . This refined form of equidistribution yields the desired approximation.

Definition 4.14. Let $I^{(m)}$ be a level in the m-step of the staircase configuration. We say that $I^{(m)}$ is $\overline {H}$ -equidistributed if, for any interval $[a,b] \subseteq [0,1]$ in the vertical direction and any interval of the staircase configuration $I^{(n)}$ ,

$$ \begin{align*} \lim_{t\to\infty} \frac{\textbf{card}(H_t^{ns}(I^{(m)},I^{(n)},[a,b]))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n)},[0,1]))}=b-a, \end{align*} $$

and if, for any $[a,b] \subseteq [0,q)$ ,

$$ \begin{align*} \lim_{t\to\infty} \frac{\textbf{card}(H_t^{s}(I^{(m)},I^{(n)},[a,b]))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n)},[0,q]))}=\frac{b-a}{q}. \end{align*} $$

Proof. Using the $ H $ -equidistribution stated in Proposition 4.12 and Lemma 4.4, we establish that for any intervals $ I^{(m)} $ and $ I^{(n)} $ , the heights attained by $ \phi _t(I^{(m)}) $ become equidistributed within the $ I^{(n)} $ -cylinder.

We begin by analyzing the case where $ I^{(n)} $ lies in the non-spacer region. Without loss of generality, we assume $ m \geq n $ . Considering a discretization of the flow given by iterates of $ q $ , the evolution of $ \phi _{kq}(I^{(m)}) $ through the staircase causes it to break into subintervals. Over multiple iterations, some of these subintervals return to the $ I^{(n)} $ -cylinder due to the staircase-like structure. This implies the existence of subintervals $ I^{(m)}_{h_1},\ldots , I^{(m)}_{h_j} $ such that every $ \phi _{kq}(I^{(m)}_{h_i}) $ lies within the $ I^{(n)} $ -cylinder.

These subintervals, in turn, belong to specific subcylinders of $ I^{(n)} $ , denoted by $ I^{(n)}_{g_1},\ldots , I^{(n)}_{g_j} $ . Among the different heights achieved by these subintervals, there exists a leading interval, meaning that if the heights are given by $ n_iq\!\! \mod 1 $ , one interval attains the maximum height $n_{i_0}q\!\! \mod 1$ . More precisely, we can label the subintervals as $ I^{(m)}_1,\ldots , I^{(m)}_j $ such that $ I_1^{(m)} = I^{(m)}_{i_0} $ , and the height of each $ I_l^{(m)} $ coincides with that of $ \phi _{-N_{1,l}q}(I^{(m)}_1) $ , where $ N_{1,l} $ represents the distance between $ l $ and $ 1 $ in this enumeration.

By iterating this process, we observe that if $\phi _{Kq}$ causes $I^{(m)}$ to reach the $(m+d)$ th step of the staircase configuration, an even smaller portion of $ I^{(m)} $ will lie within the $ I^{(n)} $ -cylinder. More precisely, there exist subintervals $ I^{(m)}_{J_1},\ldots , I^{(m)}_{J_k} $ , where each index $ J_i $ has length bounded by $ {(m+d-1)!}/{m!} $ .

Similarly, there are subcylinders over $ I^{(n)}_{J^{\prime }_1},\ldots , I^{(n)}_{J^{\prime }_k} $ , where each $ J^{\prime }_i $ has length bounded by $ {(n+d-1)!}/{n!} $ . For each $ J_i $ , there exist subsubintervals $ I^{(m)}_{J_i,i(1)},\ldots , I^{(m)}_{J_i,i(p)} $ that lie within some subcylinder $ I^{(n)}_{J^{\prime }_i} $ . Among these subsubintervals, one achieves the greatest height within the subcylinder, which we denote as the leading subsubinterval $ I^{(m)}_{J_i,i(j_0)} $ .

Following the reasoning from the previous paragraph, we can enumerate these subsubintervals as $ I^{(m)}_{J_i,1},\ldots , I^{(m)}_{J_i,p} $ so that $ I^{(m)}_{J_i,1} = I^{(m)}_{J_i,i(j_0)} $ . The heights achieved by these subsubintervals satisfy

$$ \begin{align*} \pi_y(\phi_{Kq(} I^{(m)}_{J_i,l})) = \pi_y(\phi_{(K-N_{1,l})q}(I^{(m)}_{J_i,1})), \end{align*} $$

where $ N_{1,l} $ denotes the distance between $ l $ and $ 1 $ in this enumeration.

Next, considering the heights of the leading subsubintervals across different subcylinders $ I^{(m)}_{J_i,1} $ , we apply the previous argument. If multiple subsubintervals attain the maximum height, we order them from left to right according to their appearance in the subcylinders of $ I^{(n)} $ . We then relabel the leftmost interval with the leading height as $ I^{(m)}_{1,1} $ and establish an enumeration such that the height of $ I^{(m)}_{j,1} $ satisfies

$$ \begin{align*} \pi_y(\phi_{Kq}(I^{(m)}_{j,1})) = \pi_y(\phi_{(K-N_{1,j})q}(I^{(m)}_{1,1})), \end{align*} $$

where $ N_{1,j} $ represents the distance between $ j $ and $ 1 $ . Extending this enumeration within each subcylinder, we ensure that for each subsubinterval $ I^{(m)}_{i,j} $ ,

$$ \begin{align*} \pi_y(\phi_{Kq}(I^{(m)}_{i,j})) = \pi_y(\phi_{(K-N_{1,j}+ N_{1,i})q}(I^{(m)}_{1,1})). \end{align*} $$

See Figure 8 for a toy model illustrating this behavior.

Figure 8

Toy model of the behavior of $\phi _{nq}(I^{(m)})$ in the $I^{(n)}$ -cylinder (colour online).

With further iterations of $ \phi _q $ , the measure of the portion of $ I^{(m)} $ that lies within the $ I^{(n)} $ -cylinder converges to $ \unicode{x3bb} (I^{(m)}) \unicode{x3bb} (I^{(n)}) $ , as established in Lemma 4.4. Moreover, these portions become equidistributed among the subcylinders $ I^{(n)}_J $ , where $ J $ is an index of a specified length that depends on the number of iterations of $ \phi _q $ . The heights attained within these subintervals follow the same pattern as before, forming an increasing sequence of the form

$$ \begin{align*} kq, (k+1)q,\ldots , (k+r)q \mod\kern-1.6pt 1. \end{align*} $$

As $ q \to \infty $ , the length of these sequences increases, leading to the equidistribution of heights within the cylinder generated by $ I^{(n)} $ .

If $ I^{(n)} $ is a spacer interval, we observe that, due to the staircase-like pattern, the behavior is analogous to the non-spacer case. After multiple iterations of $ \phi _q $ , the subintervals forming $ I^{(m)} $ distribute across subcylinders of $ I^{(n)} $ , attaining different heights. The only distinction from the non-spacer case is that these heights do not differ by one iteration of $q\!\! \mod 1$ as in the non-spacer case, rather they follow the subsequence described in Proposition 4.12, which, as proved there, ensures their equidistribution within the $ I^{(n)} $ -cylinder.

Now, we can observe that Lemma 4.4 and Lemma 4.15 imply the following theorem.

Proof. According to Lemma 4.3, we need to prove that for any $\epsilon ,\delta>0$ and for every rectangle R, there is a $t_0>0$ such that for every $t\geq t_0$ , there exists a partition $\eta (t)$ such that:

1. (1) $\unicode{x3bb} (\eta (t))>1-\delta $ and $\mathrm{Mesh}(\eta (t))\leq \delta $ ;

2. (2) for every $I\in \eta (t)$ , it follows that $\mu (I\cap \phi _{-t}(R))\geq (1-\epsilon )\mu (R)\unicode{x3bb} (I).$

The partitions we construct depend on $\delta $ , while $t_0$ is determined by $\epsilon $ .

The natural partitions under consideration stem from the m-steps within the staircase configuration. Referring to §4.1, if A denotes the length of $I^{(1)}$ , then for any subinterval $I^{(m)}$ in the m-step, we have

(4.1) $$ \begin{align} \unicode{x3bb}(I^{(m)})=\frac{A}{(m)!}\to0. \end{align} $$

Moreover, since all the intervals in this choice of partition have the same measure, then $\mathrm{Mesh}$ of any of this partition will converge to zero according to (4.1).

However, we observe that the measure of the covers converges to 1 by construction. Therefore, there exists an m such that $\unicode{x3bb} (I^{(m)})<\delta $ and simultaneously $h_m\unicode{x3bb} (I^{(m)})<1-\delta $ and also $m\geq m_0(q)$ , where $h_m$ represents the number of intervals in the m-step of the transformation and $m_0(q)$ is the constant given in Lemma 4.4. This satisfies the initial requirement.

Without loss of generality, we may assume that $b(R)$ is one interval $I^{(n)}$ of the nth step of the staircase configuration for some n. For the second requirement, we need to consider three cases.

(1) The rectangle R is inside the non-spacer part. Then,

$$ \begin{align*} \unicode{x3bb}(\phi_t(I^{(m)})\cap R)\to \unicode{x3bb}(\phi_t(I^{(m)})\cap \pi_x^{-1}(b(R)))\frac{\textbf{card}(H_t^{ns}(I^{(m)},I^{(n)},h(R)))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n)},[0,1]))}.\end{align*} $$

By Lemma 4.4,

$$ \begin{align*}\unicode{x3bb}(\phi_t(I^{(m)})\cap \pi_x^{-1}(b(R)))\to \unicode{x3bb}(I^{(m)})\unicode{x3bb}(b(R)).\end{align*} $$

In addition, by Lemma 4.15, we have that

$$ \begin{align*} \frac{\textbf{card}(H_t^{ns}(I^{(m)},I^{(n)},h(R)))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n)},[0,1]))}\to\unicode{x3bb}(h(R)).\end{align*} $$

This implies, in particular, that

$$ \begin{align*}\unicode{x3bb}(I^{(m)}\cap \phi_{-t}(R))\to \unicode{x3bb}(I^{(m)})\unicode{x3bb}(b(R))\unicode{x3bb}(h(R))=\unicode{x3bb}(I^{(m)})\mu(R).\end{align*} $$

(2) The rectangle R is inside the spacer part of the polygon. Then, by Lemma 4.15,

$$ \begin{align*} \frac{\textbf{card}(H_t^{ns}(I^{(m)},I^{(n)},h(R)))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n)},[0,1]))}\to\frac{\unicode{x3bb}(h(R))}{q}.\end{align*} $$

Then,

$$ \begin{align*}\unicode{x3bb}(I^{(m)}\cap \phi_{-t}(R))\to \frac{\unicode{x3bb}(I^{(m)})\mu(R)}{q}>\unicode{x3bb}(I^{(m)})\mu(R).\end{align*} $$

(3) The rectangle R has a part inside the spacer and non-spacer part. Let $R_1$ be the part inside the non-spacer part and $R_2$ be the part inside the spacer part. Observe that $h(R_1)=h(R_2)=h(R)$ and $b(R_1)+b(R_2)=b(R)$ . Suppose that $b(R_1)=I^{n_1}$ and $b(R_2)=I^{(n_2)}$ . By Lemmas 4.4 and 4.15, we have that

$$ \begin{align*} \unicode{x3bb}(I^{(m)}\cap \phi_{-t}(R))&=\unicode{x3bb}(\phi_t(I^{(m)})\cap \pi_x^{-1}(b(R_1)))\frac{\textbf{card}(H_t^{ns}(I^{(m)},I^{(n_1)},h(R)))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n_1)},[0,1]))} \\ &\quad+ \unicode{x3bb}(\phi_t(I^{(m)})\cap \pi_x^{-1}(b(R_2)))\frac{\textbf{card}(H_t^{ns}(I^{(m)},I^{(n_2)},h(R)))}{\textbf{card}((H_t^{ns}(I^{(m)},I^{(n_2)},[0,1]))}\\ &\quad \to \unicode{x3bb}(I^{(m)})\unicode{x3bb}(b(R_1))\unicode{x3bb}(h(R_1))\\ &\quad +\unicode{x3bb}(I^{(m)})\unicode{x3bb}(b(R_2))\frac{\unicode{x3bb}(h(R_2))}{q}>\unicode{x3bb}(I^{(m)})\mu(R). \end{align*} $$

Observe that in all three cases,

$$ \begin{align*}\lim_{t\to\infty}\unicode{x3bb}(I^{(m)}\cap \phi_{-t}(R))\geq\unicode{x3bb}(I^{(m)})\mu(R).\end{align*} $$

In particular, for $\epsilon>0$ , there exists a $t_0$ such that for any $t\geq t_0$ , we have that

$$ \begin{align*}\unicode{x3bb}(I^{(m)}\cap \phi_{-t}(R))\geq (1-\epsilon)\unicode{x3bb}(I^{(m)})\mu(R).\end{align*} $$

To conclude the proof, we state that the $t_0$ constructed above will be useful, and the covers that serve for this purpose will consist of the intervals for the m-step configuration of the staircase, where the m is the one that is useful for the $\delta $ . With this, and by virtue of Lemma 4.3, we conclude the proof.

In particular, this implies the following scenario.