# random $\beta$-transformation on fat Sierpinski gasket

Tingyu Zhang1, Karma Dajani2 and Wenxia Li3

###### Abstract.

We consider the iterated function system (IFS)

$f_{\vec{q}}(\vec{z})=\frac{\vec{z}+\vec{q}}{\beta},\vec{q}\in\\{(0,0),(1,0),(0,1)\\}.$

As is well known, for $\beta=2$ the attractor, $S_{\beta}$, is a fractal
called the Sierpiński gasket(or sieve) and for $\beta>2$ it is also a fractal.
Our goal is to study greedy, lazy and random $\beta$-transformations on the
attractor for this IFS with $1<\beta<2$. For $1<\beta\leq 3/2$, $S_{\beta}$ is
a triangle and it is shown that the greedy transformation $T_{\beta}$ and the
lazy transformation $L_{\beta}$ are isomorphic and they both admit an
absolutely continuous invariant measure. We show that all $\beta$-expansions
of a point $\vec{z}$ in $S_{\beta}$ can be generated by a random map
$K_{\beta}$ defined on
$\\{0,1\\}^{\mathbb{N}}\times\\{0,1,2\\}^{\mathbb{N}}\times S_{\beta}$ and
$K_{\beta}$ has a unique invariant measure of maximal entropy when
$1<\beta\leq\beta_{*}$, where $\beta_{*}\approx 1.4656$ is the root of
$x^{3}-x^{2}-1=0$. We also show existence of a $K_{\beta}$-invariant
probability measure, absolutely continuous with respect to $m_{1}\otimes
m_{2}\otimes\lambda_{2}$, where $m_{1},m_{2}$ are product measures on
$\\{0,1\\}^{\mathbb{N}},\\{0,1,2\\}^{\mathbb{N}}$, respectively, and
$\lambda_{2}$ is the normalized Lebesgue measure on $S_{\beta}$. For
$3/2<\beta\leq\beta^{*}$, where $\beta^{*}\approx 1.5437$ is the root of
$x^{3}-2x^{2}+2x=2$, there are radial holes in $S_{\beta}$. In this case,
$K_{\beta}$ is defined on $\\{0,1\\}^{\mathbb{N}}\times S_{\beta}$. We also
show that it has a unique invariant measure of maximal entropy.

MSC2020: 37H12, 37A44, 37A05, 11K55
Keywords: fat Sierpinski gasket, random $\beta$-transformation, absolutely
continuous invariant measures, measures of maximal entropy.

1,3: Department of Mathematics, Shanghai Key Laboratory of PMMP, East China
Normal University, Shanghai 200241, People’s Republic of China.

2: Department of Mathematics, Utrecht University, Fac Wiskunde en informatica
and MRI, Budapestlaan 6, P.O. Box 80.000, 3508 TA Utrecht, the Netherlands.

Email addresses: tingyuzhangecnu@163.com(T. Zhang), k.dajani1@uu.nl(K.
Dajani),
wxli@math.ecnu.edu.cn(W. Li).

##  1\. Introduction

Let $\beta>1$ and consider the $iterated\ function\ system$(IFS):

$f_{\vec{q}_{0}}(\vec{z})=\frac{\vec{z}+\vec{q}_{0}}{\beta},f_{\vec{q}_{1}}(\vec{z})=\frac{\vec{z}+\vec{q}_{1}}{\beta},f_{\vec{q}_{2}}(\vec{z})=\frac{\vec{z}+\vec{q}_{2}}{\beta},$

where the coordinates of the three points
$\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}$ are $(0,0),(1,0),(0,1)$, respectively.
It is well known that there exists a unique non-empty compact set
$S_{\beta}\subset\mathbb{R}^{2}$ such that
$S_{\beta}=\cup_{i=0}^{2}f_{\vec{q}_{i}}(S_{\beta})$; see [12] for further
details. The $attractor$ for the IFS, $S_{\beta}$, is a Sierpinski gasket.
Denote the convex hull of $S_{\beta}$ by $\Delta$ which is a triangle with
vertices at $(0,0),(\frac{1}{\beta-1},0)$ and $(0,\frac{1}{\beta-1})$. For
every point $\vec{z}\in S_{\beta}$, there exists a sequence
$(a_{i})_{i=1}^{\infty}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}$
such that

$\vec{z}=\lim_{n\rightarrow\infty}f_{a_{1}}\circ\cdots\circ
f_{a_{n}}(\vec{q}_{0})=\sum_{i=1}^{\infty}\frac{a_{i}}{\beta^{i}}.$

We call $(a_{i})_{i=1}^{\infty}$ a $coding$ of $\vec{z}$ and
$\sum_{i=1}^{\infty}a_{i}\beta^{-i}$ a $representation$ of $\vec{z}$ in base
$\beta$.

Let $i\in\\{0,1,2\\}$. For $\beta>2$, the images $f_{\vec{q}_{i}}(\Delta)$ are
disjoint. In the case the IFS $\\{f_{\vec{q}_{i}}\\}$ satisfies the strong
separation condition and each point in $S_{\beta}$ has a unique coding. For
$\beta=2$, the sets $f_{\vec{q}_{i}}(\Delta)$ overlap only at the vertices.
Therefore only countably many points in $S_{\beta}$ have two codings, and all
other points have a unique coding.

When $\beta\in(1,2)$, we call $S_{\beta}$ a $fat\ Sierpinski\ gasket$. For
$1<\beta\leq 3/2$, we have a non-empty triple overlap
$C_{012}:=f_{\vec{q}_{0}}(\Delta)\cap f_{\vec{q}_{1}}(\Delta)\cap
f_{\vec{q}_{1}}(\Delta)$(see Figure 1), and $S_{\beta}=\Delta$. In this case
Lebesgue almost every point in $S_{\beta}$ has a continuum of codings (see
[19, Theorem3.5]). For $3/2<\beta<2$, there are holes in $S_{\beta}$ as well
as overlaps (see Figure 3 for example), which make its structure more complex.
In [6], Broomhead et al. described two special types of structures: those in
which holes are radially distributed and those that are totally self-similar.
We are more interested in $S_{\beta}$ with the former characteristics. For
more results of the Hausdorff dimension of the attractors, see [16, 15, 18,
17, 14].

In this article, we shall consider the random transformation on the two-
dimensional Sierpinski gasket $S_{\beta}$. The main motivation of this
consideration is from the random $\beta$-expansion in an interval. In [9],
Dajani and Kraaikamp introduced a random $\beta$-transformation $K_{\beta}$ on
$\\{0,1\\}^{\mathbb{N}}\times[0,[\beta]/(\beta-1)]$ associated with the
‘greedy’ map and the ‘lazy’ map. They showed that the greedy expansion is
isomorphic to the lazy expansion(see [8]). Dajani and de Vries[10] showed that
$K_{\beta}$ has a unique measure of maximal entropy. In [11] they also proved
the existence and uniqueness of a $K_{\beta}$-invariant measure, absolutely
continuous with respect to $m_{p}\otimes\lambda$, where $m_{p}$ is the
Bernoulli measure on $\\{0,1\\}^{\mathbb{N}}$ and $\lambda$ is the normalized
Lebesgue measure on $[0,[\beta]/(\beta-1)]$. Inspired by these results, we
consider transformations which are defined on fat Sierpinski triangles with
overlapping structures.

The rest of the article is organized as follows. We suppose $\beta\in(1,3/2]$
in sections 3, 4 and 5. In section 2, we focus on some basic concepts and
notations needed in the rest of the paper. In section 3, we introduce the
greedy and lazy transformations on $S_{\beta}$ and prove that they are
isomorphic. We also prove the existence of invariant measures that are
absolutely continuous with respect to Lebesgue measure (acim) for these two
transformations. In section 4, we give the definition of the random
transformation $K_{\beta}$ on
$\\{0,1\\}^{\mathbb{N}}\times\\{0,1,2\\}^{\mathbb{N}}\times S_{\beta}$ and
prove that it has a unique invariant measure of maximal entropy for
$1<\beta\leq\beta_{*}$, where $\beta_{*}\approx 1.4656$ is the root of
$x^{3}-x^{2}-1=0$. In section 5, we give a position dependent random map $R$
on $S_{\beta}$. With two skew product transformations, we establish a
connection between $R$ and $K_{\beta}$, and finally prove that $K_{\beta}$ has
an invariant measure of the form $m_{1}\otimes m_{2}\otimes\mu_{\beta}$, where
$m_{1}$ is the product measure on $\\{0,1\\}^{\mathbb{N}}$ with weights
$\\{p,1-p\\}$, $m_{2}$ is the product measure on $\\{0,1,2\\}^{\mathbb{N}}$
with weights $\\{s,t,1-s-t\\}$, and $\mu_{\beta}$ is $R$-invariant and
absolutely continuous with respct to $\lambda_{2}$, the normalized Lebesgue
measure on $S_{\beta}$. In section 6, we modify the definition of the random
transformation $K_{\beta}$ for $3/2<\beta\leq\beta^{*}$ and prove that it also
has a unique invariant measure of maximal entropy, where $\beta^{*}\approx
1.5437$ is the root of $x^{3}-2x^{2}+2x=2$.

##  2\. preliminary

Given $1<\beta<2$, recall that the fat Sierpinski gasket $S_{\beta}$ is the
self-similar set in $\mathbb{R}^{2}$ generated by the IFS

(2.1)
$f_{\vec{q}_{0}}(\vec{z})=\frac{\vec{z}+\vec{q}_{0}}{\beta},f_{\vec{q}_{1}}(\vec{z})=\frac{\vec{z}+\vec{q}_{1}}{\beta},f_{\vec{q}_{2}}(\vec{z})=\frac{\vec{z}+\vec{q}_{2}}{\beta},$

where $\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}$ are $(0,0),(1,0),(0,1)$,
respectively. For every point $\vec{z}\in S_{\beta}$, there exists a sequence
$(a_{i})_{i=1}^{\infty}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}$
such that

$\vec{z}=\sum_{i=1}^{\infty}\frac{a_{i}}{\beta^{i}}.$

Recall that $\Delta$ is the convex hull of $S_{\beta}$ which is an isosceles
right triangle.

Consider the ordering of points on the plane. We write
$(x_{1},y_{1})<(x_{2},y_{2})$ if $x_{1}+y_{1}<x_{2}+y_{2}$, or
$x_{1}+y_{1}=x_{2}+y_{2}$ and $y_{1}<y_{2}$. Notice that
$\vec{q}_{0}<\vec{q}_{1}<\vec{q}_{2}$.

Let $\Omega=\\{0,1\\}^{\mathbb{N}}$ with the product $\sigma$-algebra
$\mathcal{A}$ and $\Upsilon=\\{0,1,2\\}^{\mathbb{N}}$ with the product
$\sigma$-algebra $\mathcal{B}$. Define metrics $d$ and $\rho$ on $\Omega$ and
$\Upsilon$ respectively by

$d((\omega_{i}),(\omega^{\prime}_{i}))=2^{-min\\{k:\omega_{k}\neq\omega^{\prime}_{k}\\}}$

and

$\rho((\upsilon_{i}),(\upsilon^{\prime}_{i}))=3^{-min\\{k:\upsilon_{k}\neq\upsilon^{\prime}_{k}\\}}.$

Throughout the article, the lexicographical ordering on $\Omega$, $\Upsilon$
and $\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}$ are all denoted
by $\prec$ and $\preceq$. More precisely, for two sequences $(c_{i}),(d_{i})$
we write $(c_{i})\prec(d_{i})$ if $c_{1}<d_{1}$, or there exists $k\geq 2$
such that $c_{i}=d_{i}$ for all $1\leq i<k$ and $c_{k}<d_{k}$. Similarly, we
write $(c_{i})\preceq(d_{i})$ if $(c_{i})\prec(d_{i})$ or $(c_{i})=(d_{i})$.

We will use the following concepts and properties.

######  Definition 2.1.

[20] A topological space X is called $separable$ if it has a countable dense
set. A topological space is called $completely\ metrizable$ if its topology is
induced by a complete metric. A $Polish\ space$ is a separable, completely
metrizable topological space.

There are some elementary observations.

  * (i) The real line $\mathbb{R}$ with the usual topology is Polish.

  * (ii) Any countable discrete space is Polish. In particular, $\\{0,1,2\\}$ with the discrete topology is Polish.

  * (iii) The product of countably many Polish spaces is Polish. In particular, $\Omega=\\{0,1\\}^{\mathbb{N}}$ and $\Upsilon=\\{0,1,2\\}^{\mathbb{N}}$ are Polish.

######  Theorem 2.2.

[20] Let $X,Y$ be Polish spaces, $A$ a Borel subset of $X$, and
$f:A\rightarrow Y$ a one-to-one Borel map. Then $f(A)$ is Borel.

######  Definition 2.3.

[7] $A\ dynamical\ system$ is a quadruple $(X,\mathcal{F},\mu,Q)$, where $X$
is a non-empty set, $\mathcal{F}$ is a $\sigma$-algebra on $X$, $\mu$ is a
probability measure on $(X,\mathcal{F}$) and $Q:X\rightarrow X$ is a
surjective $\mu$-measure preserving transformation.

If $(X,\mathcal{F},\mu,Q)$ is a dynamical system, and $x\in X$, we call the
sequence

$x,Qx,\ldots,Q^{n}x,\ldots$

the $Q$-orbit of $x$.

######  Definition 2.4.

[7] Two dynamical systems $(X,\mathcal{F},\mu,Q)$ and $(Y,\mathcal{G},\nu,U)$
are isomorphic if there exist measurable sets $N\subset X$ and $M\subset Y$
with $\mu(N)=0=\nu(M)$ and $Q(X\setminus N)\subset X\setminus N,U(Y\setminus
M)\subset Y\setminus M,$ and finally if there exists a measurable map
$\psi:X\setminus N\rightarrow Y\setminus M$ such that (i)–(iv) are satisfied
for the systems restricted to $X\setminus N$ and $Y\setminus M$:
(i) $\psi$ is one-to-one and onto,
(ii) $\psi$ is measurable, i.e., $\psi^{-1}(G)\in\mathcal{F}$ for all
$G\in\mathcal{G}$,
(iii) $\psi$ preserves the measures, i.e., $\nu(G)=\mu(\psi^{-1}(G))$,
(iv) $\psi$ preserves the dynamics of $Q$ and $U$, i.e., $\psi\circ
Q=U\circ\psi$,
The map $\psi$ is called an $isomorphism$.

######  Proposition 2.5.

[7] Entropy is isomorphism invariant.

##  3\. greedy and lazy transformation for $1<\beta\leq 3/2$

Let $1<\beta\leq 3/2$, then $S_{\beta}=\Delta$. Divide $S_{\beta}$ into the
following sets according to the overlapping structure of
$f_{\vec{q}_{i}}(S_{\beta})$(see Figure 1).

(3.1) $\begin{split}&E_{0}=[0,\frac{1}{\beta})\times[0,\frac{1}{\beta}),\\\
&E_{1}=\\{(x,y):0\leq
y<\frac{1}{\beta},\frac{1}{\beta(\beta-1)}<x+y\leq\frac{1}{\beta-1}\\},\\\
&E_{2}=\\{(x,y):0\leq
x<\frac{1}{\beta},\frac{1}{\beta(\beta-1)}<x+y\leq\frac{1}{\beta-1}\\},\\\
&C_{01}=\\{(x,y):x\geq\frac{1}{\beta},0\leq
y<\frac{1}{\beta},x+y\leq\frac{1}{\beta(\beta-1)}\\},\\\
&C_{12}=\\{(x,y):x\geq\frac{1}{\beta},y\geq\frac{1}{\beta},\frac{1}{\beta(\beta-1)}<x+y\leq\frac{1}{\beta-1}\\},\\\
&C_{02}=\\{(x,y):0\leq
x<\frac{1}{\beta},y\geq\frac{1}{\beta},x+y\leq\frac{1}{\beta(\beta-1)}\\},\\\
&C_{012}=\\{(x,y):x\geq\frac{1}{\beta},y\geq\frac{1}{\beta},x+y\leq\frac{1}{\beta(\beta-1)}\\}.\end{split}$

Notice that $C_{012}$ is a single point set if $\beta=3/2$. Let

$C=C_{01}\cup C_{12}\cup C_{02},\ \ E=\cup_{i=0}^{2}E_{i}.$

######  Definition 3.1.

The greedy transformation $T_{\beta}$ from $S_{\beta}$ into $S_{\beta}$ is
given by

(3.2) $T_{\beta}(\vec{z})=\begin{cases}\beta\vec{z},&\mbox{if }\vec{z}\in
E_{0},\\\ \beta\vec{z}-\vec{q}_{1},&\mbox{if }\vec{z}\in C_{01}\cup E_{1},\\\
\beta\vec{z}-\vec{q}_{2},&\mbox{if }\vec{z}\in C_{012}\cup C_{12}\cup
C_{02}\cup E_{2}.\end{cases}$

The lazy transformation $L_{\beta}$ from $S_{\beta}$ into $S_{\beta}$ is given
by

(3.3) $L_{\beta}(\vec{z})=\begin{cases}\beta\vec{z},&\mbox{if }\vec{z}\in
C_{012}\cup C_{01}\cup C_{02}\cup E_{0},\\\ \beta\vec{z}-\vec{q}_{1},&\mbox{if
}\vec{z}\in C_{12}\cup E_{1},\\\ \beta\vec{z}-\vec{q}_{2},&\mbox{if
}\vec{z}\in E_{2}.\end{cases}$

$\frac{1}{\beta(\beta-1)}$$\frac{1}{\beta(\beta-1)}$$\frac{1}{\beta}$$\frac{1}{\beta-1}$$\frac{1}{\beta}$$\frac{1}{\beta-1}$$x$$y$$E_{0}$$C_{01}$$E_{1}$$C_{02}$$C_{12}$$E_{2}$$C_{012}$
Figure 1. $S_{\beta}$ for $1<\beta\leq\frac{3}{2}$

Notice that

$\begin{split}&T_{\beta}|_{E_{i}}=L_{\beta}|_{E_{i}}=f_{\vec{q}_{i}}^{-1}\text{
for }i=\\{0,1,2\\},\\\
&T_{\beta}|_{C_{ij}}=f_{\vec{q}_{j}}^{-1},L_{\beta}|_{C_{ij}}=f_{\vec{q}_{i}}^{-1}\text{
for }ij\in\\{01,12,02\\},\\\
&T_{\beta}|_{C_{012}}=f_{\vec{q}_{2}}^{-1},L_{\beta}|_{C_{012}}=f_{\vec{q}_{0}}^{-1},\end{split}$

and

$\begin{split}&f_{\vec{q}_{0}}(S_{\beta})=E_{0}\cup C_{01}\cup C_{02}\cup
C_{012},\\\ &f_{\vec{q}_{1}}(S_{\beta})=E_{1}\cup C_{01}\cup C_{12}\cup
C_{012},\\\ &f_{\vec{q}_{2}}(S_{\beta})=E_{2}\cup C_{12}\cup C_{02}\cup
C_{012}.\end{split}$

Then it is easy to verify that $T_{\beta}$ and $L_{\beta}$ are well defined.

Denote the Borel $\sigma$-algebra on $S_{\beta}$ by $\mathcal{S}$. Let $\mu$
be an arbitrary $T_{\beta}$-invariant probability measure on
$(S_{\beta},\mathcal{S})$. Let $\psi:S_{\beta}\rightarrow S_{\beta}$ be given
by

$\psi(\vec{z})=\begin{bmatrix}1&0\\\ -1&-1\\\
\end{bmatrix}\vec{z}+\begin{bmatrix}0\\\ \frac{1}{\beta-1}\\\ \end{bmatrix}.$

######  Theorem 3.2.

For $\beta\in(1,3/2]$, the systems $(S_{\beta},\mathcal{S},\mu,T_{\beta})$ and
$(S_{\beta},\mathcal{S},\nu,L_{\beta})$ are isomorphic, where
$\nu=\mu\circ\psi^{-1}.$

###### Proof.

Denote $\begin{bmatrix}x^{\prime}\\\ y^{\prime}\\\
\end{bmatrix}=\psi(\begin{bmatrix}x\\\ y\\\ \end{bmatrix})$. Then

$x^{\prime}=x,\ y^{\prime}=\frac{1}{\beta-1}-(x+y),\
x^{\prime}+y^{\prime}=\frac{1}{\beta-1}-y.$

Therefore, we can obtain

$\displaystyle\psi(E_{0})=$ $\displaystyle\psi\\{(x,y):0\leq
x<\frac{1}{\beta},0\leq y<\frac{1}{\beta}\\}$ $\displaystyle=$
$\displaystyle\\{(x^{\prime},y^{\prime}):0\leq
x^{\prime}<\frac{1}{\beta},\frac{1}{\beta-1}-\frac{1}{\beta}<x^{\prime}+y^{\prime}\leq\frac{1}{\beta-1}\\}$
$\displaystyle=$ $\displaystyle\\{(x^{\prime},y^{\prime}):0\leq
x^{\prime}<\frac{1}{\beta},\frac{1}{\beta(\beta-1)}<x^{\prime}+y^{\prime}\leq\frac{1}{\beta-1}\\}$
$\displaystyle=$ $\displaystyle E_{2}.$

By doing the same calculation, we can also obtain

$\psi(E_{1})=E_{1},\psi(C_{01})=C_{12},\psi(C_{02})=C_{02},\psi(C_{012})=C_{012}.$

Notice that $\psi^{-1}=\psi$ and $\psi(S_{\beta})=S_{\beta}$, then we can get
$\psi(E_{2})=E_{0}$ and $\psi(C_{12})=C_{01}$. Now we show $\psi\circ
T_{\beta}=L_{\beta}\circ\psi$. For $(x,y)\in E_{0}$,

$\psi\circ T_{\beta}(\begin{bmatrix}x\\\ y\\\
\end{bmatrix})=\psi(\begin{bmatrix}\beta x\\\ \beta y\\\
\end{bmatrix})=\begin{bmatrix}\beta x\\\ \frac{1}{\beta-1}-\beta x-\beta y\\\
\end{bmatrix}.$

Since $\psi((x,y))\in E_{2}$, then

$L_{\beta}\circ\psi(\begin{bmatrix}x\\\ y\\\
\end{bmatrix})=\beta\begin{bmatrix}x\\\ \frac{1}{\beta-1}-x-y\\\
\end{bmatrix}-\begin{bmatrix}0\\\ 1\\\ \end{bmatrix}=\begin{bmatrix}\beta x\\\
\frac{\beta}{\beta-1}-\beta x-\beta y-1\\\ \end{bmatrix}=\psi\circ
T_{\beta}(\begin{bmatrix}x\\\ y\\\ \end{bmatrix}).$

Similarly, we have

$\displaystyle\psi\circ T_{\beta}(\begin{bmatrix}x\\\ y\\\ \end{bmatrix})$
$\displaystyle=\begin{bmatrix}\beta x-1\\\ \frac{\beta}{\beta-1}-\beta x-\beta
y\\\ \end{bmatrix}=L_{\beta}\circ\psi(\begin{bmatrix}x\\\ y\\\
\end{bmatrix}),\text{ for }(x,y)\in C_{01}\cup E_{1},$ $\displaystyle\psi\circ
T_{\beta}(\begin{bmatrix}x\\\ y\\\ \end{bmatrix})$
$\displaystyle=\begin{bmatrix}\beta x\\\ \frac{\beta}{\beta-1}-\beta x-\beta
y\\\ \end{bmatrix}=L_{\beta}\circ\psi(\begin{bmatrix}x\\\ y\\\
\end{bmatrix}),\text{ for }(x,y)\in C_{012}\cup C_{12}\cup C_{02}\cup E_{2}.$

Since $\psi:S_{\beta}\rightarrow S_{\beta}$ is a bijection, then it follows
from $\nu=\mu\circ\psi^{-1}$ that $\psi$ is an isomorphism. ∎

We end this section by recalling some definitions and results in [4] due to
Boyarsky, A. and Góra, P., but rephrased to our setting. Let $S$ be a bounded
region in $\mathbb{R}^{N}$($N\geq 2$ is an integer) and let
$\mathcal{P}=\\{S_{1},S_{2},\ldots,S_{m}\\}$ be a partition of $S$, where
$m>0$ is an integer. Let $\tau$ be a transformation from $S$ into $S$. We say
$\tau$ is $piecewise\ C^{2}\ and\ expanding$ with respect to $\mathcal{P}$ if:

(a) each $S_{i}$ is a bounded closed domain having a piecewise $C^{2}$
boundary of finite $(N-1)$-dimensional measure;

(b) $\tau_{i}=\tau|_{S_{i}}$ is a $C^{2}$, 1-1 transformation from
int($S_{i}$) onto its image and can be extended as a $C^{2}$ transformation
onto $S_{i},i=1,2,\ldots,m$;

(c) there exists $0<c<1$ such that for any $i=1,2,\ldots,m$,

$\|D\tau_{i}^{-1}\|<c,$

where $D\tau_{i}^{-1}$ is the derivative matrix of $\tau_{i}^{-1}$ and
$\|\cdot\|$ is the euclidean matrix norm, i.e.,
$\|A\|=(\sum_{i=1}^{m}\sum_{j=1}^{n}a_{ij}^{2})^{\frac{1}{2}}$ for $m\times n$
order matrix $A=(a_{ij})$.

Assume that the faces of $\partial S_{i}$ meet at angles bounded uniformly
away from 0. Fix $1\leq i\leq m$. Let $F$ denote the set of singular points of
$\partial S_{i}$. At any $x\in F$ we construct the largest cone having a
vertex at $x$ and which lies completely in $S_{i}$. Let $\theta(x)$ denote the
angle subtended at the vertex of this cone. Then define

(3.4) $\gamma(S_{i})=\min_{x\in F}\theta(x).$

Since the faces of $\partial S_{i}$ meet at angles bounded uniformly away from
0, $\gamma(S_{i})>0.$ Let $\alpha(S_{i})=\pi/2+\gamma(S_{i})$ and

(3.5) $a(S_{i})=|\cos(\alpha(S_{i}))|.$

Assume $a=\min\\{a(S_{i}):i=1,\ldots,m\\}$.

######  Corollary 3.3.

[4, Corollary 1] Let $\tau:S\rightarrow S,S\subset\mathbb{R}^{N}$, be
piecewise $C^{2}$ and such that some iterate $\tau^{k}$ satisfies
$c(1+1/a)<1$($c$ and $a$ corresponds to $\tau^{k}$), then $\tau$ admits an
absolutely continuous invariant measure (acim).

Using the above corollary, we can prove the following theorem.

######  Theorem 3.4.

Let $T_{\beta}$ be defined as in (3.2). Then $T_{\beta}$ admits an acim. And
so does $L_{\beta}$ which is defined as in (3.3).

###### Proof.

Since $\beta>1$, $\lim_{n\rightarrow\infty}\beta^{n}=\infty$. We can choose an
integer $k$ such that

$\beta^{k}>2(2+\sqrt{2}).$

Use the partition $\mathcal{P}:\\{S_{1},S_{2},S_{3}\\}$, where

$S_{1}=E_{0},S_{2}=C_{01}\cup E_{1},S_{3}=C_{012}\cup C_{12}\cup C_{02}\cup
E_{2}.$

Consider the iterate $T_{\beta}^{k}$ and the corresponding partition
$\vee_{i=0}^{k-1}T_{\beta}^{-i}(\mathcal{P})$. For
$P_{i}\in\vee_{i=0}^{k-1}T_{\beta}^{-i}(\mathcal{P})$, let
$T^{k}_{\beta,i}=T_{\beta}^{k}|_{P_{i}}$. Since the derivative matrix of
$(T_{\beta,i}^{k})^{-1}$ is $\begin{bmatrix}\frac{1}{\beta^{k}}&0\\\
0&\frac{1}{\beta^{k}}\\\ \end{bmatrix},$ then the Euclidean matrix norm,

$\|D(T_{\beta,i}^{k})^{-1}\|=\frac{\sqrt{2}}{\beta^{k}}<\frac{2\sqrt{2}}{\beta^{k}}:=c<1.$

For the partition $\vee_{i=0}^{k-1}T_{\beta}^{-i}(\mathcal{P})$, we have
$a=\sqrt{2}/2,$ which implies

$c(1+\frac{1}{a})=\frac{2\sqrt{2}}{\beta^{k}}(1+\sqrt{2})<\frac{2\sqrt{2}}{2(2+\sqrt{2})}(1+\sqrt{2})=1.$

By Corollary 3.3, $T_{\beta}$ admits an acim.

For $L_{\beta}$, we use the partition $\mathcal{Q}=\\{C_{012}\cup C_{01}\cup
C_{02}\cup E_{0},C_{12}\cup E_{1},E_{2}\\}$. In a similar way we can also
prove that $L_{\beta}$ has an acim. ∎

##  4\. Random transformation for $1<\beta\leq 3/2$

###  4.1. $K_{\beta}$ and some properties

From Definition 3.1, one can see that on $E_{i}$ both the greedy and lazy maps
are identical which means that they assign the same digits. On $C_{ij}$, the
greedy map assigns the digit $\vec{q}_{j}$, while the lazy map assigns the
digit $\vec{q}_{i}$. On $C_{012}$, the digit $\vec{q}_{1}$ cannot be assigned
by the two maps. Now, in order to define a random transformation, we need to
randomly select the map used in the switch regions $C_{ij}$ and $C_{012}$.
When $\vec{z}$ belongs to $C_{ij}$ we flip a 2-sided coin to decide which map
will be applied to $\vec{z}$, while we flip a 3-sided ‘coin’ when $\vec{z}$
belongs to $C_{012}$.

Recall $\Omega=\\{0,1\\}^{\mathbb{N}}$ and
$\Upsilon=\\{0,1,2\\}^{\mathbb{N}}$. Let $\sigma:\Omega\rightarrow\Omega$ and
$\sigma^{\prime}:\Upsilon\rightarrow\Upsilon$ be the left shifts. Define
$K_{\beta}:\Omega\times\Upsilon\times
S_{\beta}\rightarrow\Omega\times\Upsilon\times S_{\beta}$ by

$K_{\beta}(\omega,\upsilon,\vec{z})=\begin{cases}(\omega,\upsilon,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
$\vec{z}\in E_{i},i=0,1,2$},\\\
(\sigma\omega,\upsilon,\beta\vec{z}-\vec{q}_{i}),&\mbox{if $\omega_{1}=0$ and
$\vec{z}\in C_{ij},\ ij\in\\{01,12,02\\}$},\\\
(\sigma\omega,\upsilon,\beta\vec{z}-\vec{q}_{j}),&\mbox{if $\omega_{1}=1$ and
$\vec{z}\in C_{ij},\ ij\in\\{01,12,02\\}$},\\\
(\omega,\sigma^{\prime}\upsilon,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
$\vec{z}\in C_{012}$ and $\upsilon_{1}=i\in\\{0,1,2\\}$}.\end{cases}$

The digits are given by

$d_{1}=d_{1}(\omega,\upsilon,\vec{z})=\begin{cases}\vec{q}_{i},&\mbox{if
$\vec{z}\in E_{i},i=0,1,2$},\\\ \ &\mbox{or
$(\omega,\upsilon,\vec{z})\in\Omega\times\\{\upsilon_{1}=i\\}\times
C_{012}$},\\\ \ &\mbox{or
$(\omega,\upsilon,\vec{z})\in\\{\omega_{1}=0\\}\times\Upsilon\times C_{ij},\
ij\in\\{01,12,02\\}$},\\\ \vec{q}_{j},&\mbox{if
$(\omega,\upsilon,\vec{z})\in\\{\omega_{1}=1\\}\times\Upsilon\times C_{ij},\
ij\in\\{01,12,02\\}$}.\end{cases}$

Then

$K_{\beta}(\omega,\upsilon,\vec{z})=\begin{cases}(\omega,\upsilon,\beta\vec{x}-d_{1}),&\mbox{if
$\vec{z}\in E_{0}\cup E_{1}\cup E_{2}$},\\\
(\sigma\omega,\upsilon,\beta\vec{z}-d_{1}),&\mbox{if $\vec{z}\in C_{01}\cup
C_{12}\cup C_{02}$},\\\
(\omega,\sigma^{\prime}\upsilon,\beta\vec{z}-d_{1}),&\mbox{if $\vec{z}\in
C_{012}$}.\end{cases}$

Set
$d_{n}=d_{n}(\omega,\upsilon,\vec{z})=d_{1}(K_{\beta}^{n-1}(\omega,\upsilon,\vec{z}))$,
and $\pi_{3}:\Omega\times\Upsilon\times S_{\beta}\rightarrow S_{\beta}$ be the
canonical projection onto the third coordinate. Then

$\pi_{3}(K_{\beta}^{n}(\omega,\upsilon,\vec{z}))=\beta^{n}\vec{z}-\beta^{n-1}d_{1}-\cdots-\beta
d_{n-1}-d_{n},$

and rewriting yiels

$\vec{z}=\frac{d_{1}}{\beta}+\frac{d_{2}}{\beta^{2}}+\cdots+\frac{d_{n}}{\beta^{n}}+\frac{\pi_{3}(K_{\beta}^{n}(\omega,\upsilon,\vec{z}))}{\beta^{n}}.$

Since $\pi_{3}(K_{\beta}^{n}(\omega,\upsilon,\vec{z}))\in S_{\beta}$ and
$S_{\beta}$ is a bounded set in $\mathbb{R}^{2}$, it follows that

$\|\vec{z}-\sum_{i=1}^{n}\frac{d_{i}}{\beta^{i}}\|_{1}=\frac{\|\pi_{3}(K_{\beta}^{n}(\omega,\upsilon,\vec{z}))\|_{1}}{\beta^{n}}\rightarrow
0,$

where $\parallel\cdot\parallel_{1}$ denotes the $L_{1}$ norm, i.e. the sum of
the absolute values of the vector elements. This shows that for all
$\omega\in\Omega$ , $\upsilon\in\Upsilon$ and for all $\vec{z}\in S_{\beta}$
one has

$\vec{z}=\sum_{i=1}^{\infty}\frac{d_{i}}{\beta^{i}}=\sum_{i=1}^{\infty}\frac{d_{i}(\omega,\upsilon,\vec{z})}{\beta^{i}}.$

For each point $\vec{z}\in S_{\beta}$, consider the set

$D_{\vec{z}}=\\{(d_{1}(\omega,\upsilon,\vec{z}),d_{2}(\omega,\upsilon,\vec{z}),\ldots):\omega\in\Omega,\upsilon\in\Upsilon\\}.$

Recall that $\prec$ and $\preceq$ are the lexicographical ordering on
$\Omega$, $\Upsilon$ and
$\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}$. The following
theorem shows how the lexicographical ordering on $\Omega$ and $\Upsilon$
affect the ordering of the elements in $D_{\vec{z}}$.

######  Theorem 4.1.

Suppose
$\omega,\omega^{\prime}\in\Omega,\upsilon,\upsilon^{\prime}\in\Upsilon$ are
such that $\omega\prec\omega^{\prime}$ and $\upsilon\prec\upsilon^{\prime}$.
Then for $\vec{z}\in S_{\beta}$,

$(d_{1}(\omega,\upsilon,\vec{z}),d_{2}(\omega,\upsilon,\vec{z}),\ldots)\preceq(d_{1}(\omega^{\prime},\upsilon^{\prime},\vec{z}),d_{2}(\omega^{\prime},\upsilon^{\prime},\vec{z}),\ldots).$

###### Proof.

Let $m:=\inf\\{i:\omega_{i}<\omega^{\prime}_{i}\\},\
n:=\inf\\{i:\upsilon_{i}<\upsilon^{\prime}_{i}\\}$. Then we have
$\omega_{m}<\omega^{\prime}_{m}$ and $\upsilon_{n}<\upsilon^{\prime}_{n}$.
Denote by $t_{i}$ the time of the $i^{\text{th}}$ visit to the region
$\Omega\times\Upsilon\times C$ of the orbit of $(\omega,\upsilon,\vec{z})$
under $K_{\beta}$. Denote by $s_{j}$ the time of the $j^{\text{th}}$ visit to
the region $\Omega\times\Upsilon\times C_{012}$ of the orbit of
$(\omega,\upsilon,\vec{z})$ under $K_{\beta}$. One can see that
$K_{\beta}^{t_{i}}(\omega,\upsilon,\vec{z})$ hits $\Omega\times\Upsilon\times
C$ for the $i^{\text{th}}$ time and
$K_{\beta}^{s_{j}}(\omega,\upsilon,\vec{z})$ hits $\Omega\times\Upsilon\times
C_{012}$ for the $j^{\text{th}}$ time.

Let $l=\min\\{t_{m},s_{n}\\}$. Then
$\pi_{3}(K_{\beta}^{i}(\omega,\upsilon,\vec{z}))=\pi_{3}(K_{\beta}^{i}(\omega^{\prime},\upsilon^{\prime},\vec{z}))$
for $i=0,\ldots,l$. It follows that
$d_{i}(\omega,\upsilon,\vec{z})=d_{i}(\omega^{\prime},\upsilon^{\prime},\vec{z})$
for $i=0,\ldots,l$.

If $l=\infty$, then
$d_{i}(\omega,\upsilon,\vec{z})=d_{i}(\omega^{\prime},\upsilon^{\prime},\vec{z})$
for all $i$. If $l<+\infty$, then
$K_{\beta}^{l}(\omega,\upsilon,\vec{z})=K_{\beta}^{l}(\omega^{\prime},\upsilon^{\prime},\vec{z})$
hits $\Omega\times\Upsilon\times C$ for the $m^{\text{th}}$ time or
$\Omega\times\Upsilon\times C_{012}$ for the $n^{\text{th}}$ time. Since
$\omega_{m}<\omega^{\prime}_{m}$ and $\upsilon_{n}<\upsilon^{\prime}_{n}$,
then

$d_{l+1}(\omega,\upsilon,\vec{z})=d_{1}(K_{\beta}^{l}(\omega,\upsilon,\vec{z}))<d_{1}(K_{\beta}^{l}(\omega^{\prime},\upsilon^{\prime},\vec{z}))=d_{l+1}(\omega^{\prime},\upsilon^{\prime},\vec{z}).$

∎

Now we show that any representation of $\vec{z}$ can be generated from the map
$K_{\beta}$ by choosing appropriate $\omega\in\Omega$ and
$\upsilon\in\Upsilon$. We need the following lemma.

######  Lemma 4.2.

Let $\beta\in(1,3/2]$. Let $(x,y)\in S_{\beta}$ and
$(x,y)=\sum_{i=1}^{\infty}a_{i}\beta^{-i}$ with
$a_{i}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}$ be a representation of
$(x,y)$ in base $\beta$. One has,

  * (i) If $(x,y)\in E_{i}$ for some $i\in\\{0,1,2\\}$, then $a_{1}=\vec{q}_{i}$;

  * (ii) If $(x,y)\in C_{ij}$ for some $ij\in\\{01,12,02\\}$, then $a_{1}\in\\{\vec{q}_{i},\vec{q}_{j}\\}$;

  * (iii) If $(x,y)\in C_{012}$, then $a_{1}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}$.

###### Proof.

(i) Suppose $a_{1}\neq\vec{q}_{0}$. From
$a_{1}\in\\{\vec{q}_{1},\vec{q}_{2}\\}$ and
$(x,y)=\sum_{i=1}^{\infty}a_{i}\beta^{-i}$ we have

$x\geq\frac{1}{\beta}\ \text{or}\ y\geq\frac{1}{\beta}.$

Then $(x,y)\notin E_{0}$.

Suppose $a_{1}\neq\vec{q}_{1}$. If $a_{1}=\vec{q}_{0}$, then

$x+y=\sum_{i=2}^{\infty}\frac{\parallel
a_{i}\parallel_{1}}{\beta^{i}}\leq\frac{1}{\beta(\beta-1)}.$

If $a_{1}=\vec{q}_{2}$, then $y\geq\frac{1}{\beta}.$ In both cases,
$(x,y)\notin E_{1}$.

Suppose $a_{1}\neq\vec{q}_{2}$. By a similar proof we have $(x,y)\notin
E_{2}$.

(ii) Suppose $a_{1}\notin\\{\vec{q}_{i},\vec{q}_{j}\\}$. If
$a_{1}=\vec{q}_{2}$, then $y\geq\frac{1}{\beta}$, which implies $(x,y)\notin
C_{01}$. If $a_{1}=\vec{q}_{0}$, then $x+y\leq\frac{1}{\beta(\beta-1)}$, which
implies $(x,y)\notin C_{12}$. If $a_{1}=\vec{q}_{1}$, then
$x\geq\frac{1}{\beta}$, which implies $(x,y)\notin C_{02}$.

(iii) holds trivially. ∎

######  Theorem 4.3.

For $\beta\in(1,3/2]$, let $\vec{z}\in S_{\beta}$ and
$\vec{z}=\sum_{i=1}^{\infty}a_{i}\beta^{-i}$ with
$a_{i}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}$ be a representation of
$\vec{z}$ in base $\beta.$ Then there exists an $\omega\in\Omega$ and an
$\upsilon\in\Upsilon$ such that $a_{i}=d_{i}(\omega,\upsilon,\vec{z})$.

###### Proof.

Let $\vec{z}_{n}=\sum_{i=1}^{\infty}\frac{a_{i+n-1}}{\beta^{i}}.$ Notice that
$\vec{z}_{1}=\vec{z}.$ Denote by $k_{n}(\vec{z})$ and $l_{n}(\vec{z})$ the
number of times we flip a 2-sided coin and a 3-sided coin, respectively, which
are

$k_{n}(\vec{z})=\sum_{i=1}^{n}\textbf{1}_{C}(\vec{z}_{n}),\ \
l_{n}(\vec{z})=\sum_{i=1}^{n}\textbf{1}_{C_{012}}(\vec{z}_{n}).$

  * -

If $\vec{z}\in E_{i}$, then $k_{1}(\vec{z})=l_{1}(\vec{z})=0$. It follows from
Lemma 4.2 that $a_{1}=\vec{q}_{i}$. Let
$\Omega_{1}=\Omega,\Upsilon_{1}=\Upsilon.$

  * -

If $\vec{z}\in C_{ij},ij\in\\{01,12,02\\}$, then $k_{1}(\vec{z})=1,\
l_{1}(\vec{z})=0$. Let $\Upsilon_{1}=\Upsilon.$ It follows from Lemma 4.2 that
$a_{1}\in\\{\vec{q}_{i},\vec{q}_{j}\\}$.

    * –

If $a_{1}=\vec{q}_{i}$, we let
$\Omega_{1}=\\{\omega\in\Omega:\omega_{1}=0\\}$.

    * –

If $a_{1}=\vec{q}_{j}$, we let
$\Omega_{1}=\\{\omega\in\Omega:\omega_{1}=1\\}$.

  * -

If $\vec{z}\in C_{012}$, then $k_{1}(\vec{z})=0,l_{1}(\vec{z})=1$ and by Lemma
4.2, $a_{1}=\vec{q}_{i},\ i\in\\{0,1,2\\}$. We let $\Omega_{1}=\Omega,\ \
\Upsilon_{1}=\\{\upsilon\in\Upsilon:\upsilon_{1}=i\\}.$

Obviously $\Omega_{1}$ and $\Upsilon_{1}$ are cylinders of length
$k_{1}(\vec{z})$ and $l_{1}(\vec{z})$ respectively, and
$d_{1}(\omega,\upsilon,\vec{z})=a_{1}$ for all $\omega\in\Omega_{1},\
\upsilon\in\Upsilon_{1}$. By a cylinder of length $0$ we mean the whole space.

Now suppose we have obtained cylinders
$\Omega_{n}\subseteq\cdots\subseteq\Omega_{1}\subseteq\Omega$ and
$\Upsilon_{n}\subseteq\cdots\subseteq\Upsilon_{1}\subseteq\Upsilon$.
$\Omega_{n}$ and $\Upsilon_{n}$ are cylinders of length $k_{n}(\vec{z})$ and
$l_{n}(\vec{z})$ respectively. For all $\omega\in\Omega_{n},\
\upsilon\in\Upsilon_{n}$, we have
$d_{1}(\omega,\upsilon,\vec{z})=a_{1},\cdots,d_{n}(\omega,\upsilon,\vec{z})=a_{n}$
and $\vec{z}_{n+1}=\pi_{3}(K_{\beta}^{n}(\omega,\upsilon,\vec{z})$.

  * -

If $\vec{z}_{n+1}\in E_{i}$, then $k_{n+1}(\vec{z})=k_{n}(\vec{z}),\
l_{n+1}(\vec{z})=l_{n}(\vec{z})$ and by Lemma 4.2, $a_{n+1}=\vec{q}_{i}$. We
let $\Omega_{n+1}=\Omega_{n},\Upsilon_{n+1}=\Upsilon_{n}.$

  * -

If $\vec{z}_{n+1}\in C_{ij},ij\in\\{01,12,02\\}$, then
$k_{n+1}(\vec{z})=k_{n}(\vec{z})+1,\ l_{n+1}(\vec{z})=l_{n}(\vec{z})$. We let
$\Upsilon_{n+1}=\Upsilon_{n}.$ By Lemma 4.2,
$a_{n+1}\in\\{\vec{q}_{i},\vec{q}_{j}\\}$.

    * –

If $a_{n+1}=\vec{q}_{i}$, we let
$\Omega_{n+1}=\\{\omega\in\Omega_{n}:\omega_{k_{n+1}}=0\\}$.

    * –

If $a_{n+1}=\vec{q}_{j}$, we let
$\Omega_{n+1}=\\{\omega\in\Omega_{n}:\omega_{k_{n+1}}=1\\}$.

  * -

If $\vec{z}_{n+1}\in C_{012}$, then $k_{n+1}(\vec{z})=k_{n}(\vec{z}),\
l_{n+1}(\vec{z})=l_{n}(\vec{z})+1$. By Lemma 4.2, $a_{n+1}=\vec{q}_{i},\
i\in\\{0,1,2\\}$. Let $\Omega_{n+1}=\Omega_{n}$ and
$\Upsilon_{n+1}=\\{\upsilon\in\Upsilon_{n}:\upsilon_{l_{n+1}}=i\\}.$

We can see that $\Omega_{n+1}$ is a cylinder of length $k_{n+1}(\vec{z})$ and
$\Upsilon_{n+1}$ is a cylinder of length $l_{n+1}(\vec{z})$. For all
$\omega\in\Omega_{n+1},\ \upsilon\in\Upsilon_{n+1}$,

$d_{1}(\omega,\upsilon,\vec{z})=a_{1},\cdots,d_{n+1}(\omega,\upsilon,\vec{z})=a_{n+1}.$

If the map $K_{\beta}$ hits $\Omega\times\Upsilon\times C$ infinitely many
times, then $k_{n}(\vec{z})\rightarrow\infty$ and $\cap\Omega_{n}$ consists of
a single point. If this happens only finitely many times, then the set
$\\{k_{n}(\vec{z}):n\in\mathbb{N}\\}$ is finite and $\cap\Omega_{n}$ is a
cylinder set. For $\cap\Upsilon_{n}$, it has the same property according to
how many times $K_{\beta}$ hits $\Omega\times\Upsilon\times C_{012}$.

In all cases $\cap\Omega_{n}$ and $\cap\Upsilon_{n}$ are non-empty, and for
$\omega\in\cap\Omega_{n},\upsilon\in\cap\Upsilon_{n}$, we have
$d_{j}(\omega,\upsilon,\vec{z})=a_{j}$ for all $j\geq 1$. ∎

###  4.2. Unique $K_{\beta}$-invariant measure of maximal entropy for
$\beta\in(1,\beta_{*})$

Equip $\Upsilon$ with the uniform product measure $\mathbb{P}$ and recall that
$\sigma^{\prime}$ is the left shift on $\Upsilon$. On the set
$\Omega\times\Upsilon\times S_{\beta}$ we consider the product
$\sigma$-algebra $\mathcal{A\times B\times S}$. Define the function
$\rho_{1}:\Omega\times\Upsilon\times
S_{\beta}\rightarrow\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}$ by

$\rho_{1}(\omega,\upsilon,\vec{z})=(d_{1}(\omega,\upsilon,\vec{z}),d_{2}(\omega,\upsilon,\vec{z}),\ldots).$

Define the function
$\rho_{2}:\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}\rightarrow\Upsilon$
by

$\rho_{2}(\vec{q}_{b_{1}},\vec{q}_{b_{2}},\vec{q}_{b_{3}},\ldots)=(b_{1},b_{2},b_{3},\ldots).$

Denote by $\varphi=\rho_{2}\circ\rho_{1}$ which is the function from
$\Omega\times\Upsilon\times S_{\beta}$ to $\Upsilon$. Then $\varphi\circ
K_{\beta}=\sigma^{\prime}\circ\varphi$, and $\varphi$ is surjective from
Theorem 4.3.

It is easily seen that $\varphi$ is measurable. In fact, the inverse image of
the cylinder set with the first digit fixed is measurable in
$\Omega\times\Upsilon\times S_{\beta}$:

$\displaystyle\varphi^{-1}(\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{1}=0\\})=$
$\displaystyle(\Omega\times\Upsilon\times
E_{0})\cup(\\{\omega\in\Omega:\omega_{1}=0\\}\times\Upsilon\times(C_{01}\cup
C_{02}))$
$\displaystyle\cup(\Omega\times\\{\upsilon\in\Upsilon:\upsilon_{1}=0\\}\times
C_{012}),$
$\displaystyle\varphi^{-1}(\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{1}=1\\})=$
$\displaystyle(\Omega\times\Upsilon\times
E_{1})\cup(\\{\omega\in\Omega:\omega_{1}=1\\}\times\Upsilon\times C_{01})$
$\displaystyle\cup(\\{\omega\in\Omega:\omega_{1}=0\\}\times\Upsilon\times
C_{12})\cup(\Omega\times\\{\upsilon\in\Upsilon:\upsilon_{1}=1\\}\times
C_{012}),$
$\displaystyle\varphi^{-1}(\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{1}=2\\})=$
$\displaystyle(\Omega\times\Upsilon\times
E_{2})\cup(\\{\omega\in\Omega:\omega_{1}=1\\}\times\Upsilon\times(C_{02}\cup
C_{12}))$
$\displaystyle\cup(\Omega\times\\{\upsilon\in\Upsilon:\upsilon_{1}=2\\}\times
C_{012}).$

To show that $\varphi$ is an isomorphism, let

$\displaystyle Z_{1}$
$\displaystyle=\\{(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
S_{\beta}:K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C\text{ infinitely often}\\},$ $\displaystyle Z_{2}$
$\displaystyle=\\{(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
S_{\beta}:K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C_{012}\text{ infinitely often}\\},$ $\displaystyle D_{1}$
$\displaystyle=\\{(b_{1},b_{2},\ldots)\in\Upsilon:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in
C\text{ for infinitely many }j\text{'s}\\},$ $\displaystyle D_{2}$
$\displaystyle=\\{(b_{1},b_{2},\ldots)\in\Upsilon:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in
C_{012}\text{ for infinitely many }j\text{'s}\\}.$

Notice that

$Z_{1}=\cap_{n=1}^{\infty}\cup_{m=n}^{\infty}K_{\beta}^{-m}(\Omega\times\Upsilon\times
C)$

and

$Z_{2}=\cap_{n=1}^{\infty}\cup_{m=n}^{\infty}K_{\beta}^{-m}(\Omega\times\Upsilon\times
C_{012}),$

which imply that $Z_{1}$ and $Z_{2}$ are Borel sets in
$\Omega\times\Upsilon\times S_{\beta}$. Let $Z=Z_{1}\cap Z_{2},\ D=D_{1}\cap
D_{2}$, then we have $K_{\beta}^{-1}(Z)=Z,\ (\sigma^{\prime})^{-1}(D)=D$ and
$\varphi(Z)=D$. Let $\varphi^{\prime}=\varphi|_{Z}$.

######  Lemma 4.4.

The map $\varphi^{\prime}:Z\rightarrow D$ is a bimeasurable bijection.

###### Proof.

For any sequence $(b_{1},b_{2},\ldots)\in D$, we can obtain a point
$\vec{z}=\sum_{i=1}^{\infty}\vec{q}_{b_{i}}\beta^{-i}$. To determine $\omega$
and $\upsilon$, we could define

$\displaystyle r_{1}=\min\\{j\geq
1:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in C\\},\ \
r_{k}=\min\\{j>r_{k-1}:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in
C\\},$ $\displaystyle s_{1}=\min\\{j\geq
1:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in C_{012}\\},\ \
s_{k}=\min\\{j>s_{k-1}:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in
C_{012}\\}.$

  * -

If $\sum_{i=1}^{\infty}\vec{q}_{b_{r_{k}+i-1}}\beta^{-i}\in
C_{ij},ij\in\\{01,12,02\\}$, then $b_{r_{k}}\in\\{i,j\\}$ by Lemma 4.2.

    * –

If $b_{r_{k}}=i$, we let $\omega_{k}=0$.

    * –

If $b_{r_{k}}=j$, we let $\omega_{k}=1$.

  * -

If $\sum_{i=1}^{\infty}\vec{q}_{b_{s_{k}+i-1}}\beta^{-i}\in C_{012}$, then
$b_{s_{k}}\in\\{0,1,2\\}$ by Lemma 4.2. If $b_{s_{k}}=i$, we let
$\upsilon_{k}=i,i\in\\{0,1,2\\}$.

Notice that for any $N>0$, there exsit $r,s>N$ such that

$\sum_{i=1}^{\infty}\vec{q}_{b_{r+i-1}}\beta^{-i}\in C\text{ and
}\sum_{i=1}^{\infty}\vec{q}_{b_{s+i-1}}\beta^{-i}\in C_{012},$

which implies that $K_{\beta}^{n}(\omega,\upsilon,\vec{z})$ hits both
$\Omega\times\Upsilon\times C$ and $\Omega\times\Upsilon\times C_{012}$
infinitely often. Then the infinite sequences
$\omega=(\omega_{1},\omega_{2},\omega_{3},\ldots)\in\Omega$ and
$\upsilon=(\upsilon_{1},\upsilon_{2},\upsilon_{3},\ldots)\in\Upsilon$ can be
uniquely determined. Therefore, we can define the inverse of
$\varphi^{\prime}$. Let $(\varphi^{\prime})^{-1}:D\rightarrow Z$ be

$(\varphi^{\prime})^{-1}((b_{1},b_{2},\ldots))=(\omega,\upsilon,\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{i}}}{\beta^{i}}).$

If
$(\omega,\upsilon,\vec{z})=(\omega^{\prime},\upsilon^{\prime},\vec{z^{\prime}})$
then
$\varphi^{\prime}(\omega,\upsilon,\vec{z})=\varphi^{\prime}(\omega^{\prime},\upsilon^{\prime},\vec{z^{\prime}})$.
Since $Z=Z_{1}\cap Z_{2}$ is a Borel set in $\Omega\times\Upsilon\times
S_{\beta}$, then we have that $(\varphi^{\prime})^{-1}$ is measurable by
Theorem 2.2. Hence $\varphi^{\prime}$ is a bimeasurable bijection. ∎

Let $\beta_{*}\approx 1.4656$ be the root of $x^{3}-x^{2}-1=0$.

######  Lemma 4.5.

If $1<\beta<\beta_{*}$, then $\mathbb{P}(D)=1$.

###### Proof.

Let us first prove $\mathbb{P}(D_{2})=1$. For any sequence
$(b_{1},b_{2},\ldots)\in\Upsilon$ and $n\geq 4$, define

$\vec{z}_{n}=(x_{n},y_{n})=\frac{\vec{q}_{1}}{\beta}+\frac{\vec{q}_{2}}{\beta^{2}}+\frac{\vec{q}_{2}}{\beta^{3}}+\frac{\vec{q}_{0}}{\beta^{4}}+\cdots+\frac{\vec{q}_{0}}{\beta^{n}}+\frac{\vec{q}_{b_{1}}}{\beta^{n+1}}+\frac{\vec{q}_{b_{2}}}{\beta^{n+2}}+\cdots.$

Then we have $x_{n}\geq\frac{1}{\beta}$, and

$\displaystyle y_{n}$
$\displaystyle\geq\frac{1}{\beta^{2}}+\frac{1}{\beta^{3}},$ $\displaystyle
x_{n}+y_{n}$
$\displaystyle\leq\sum_{i=1}^{3}\frac{1}{\beta^{i}}+\sum_{i=1}^{\infty}\frac{1}{\beta^{n+i}}.$

If $1<\beta<\beta_{*}$, then

$\frac{1}{\beta^{2}}+\frac{1}{\beta^{3}}\geq\frac{1}{\beta},$

and

$\frac{1}{\beta}+\frac{1}{\beta^{2}}+\frac{1}{\beta^{3}}<\frac{1}{\beta(\beta-1)}.$

Since
$\sum_{i=1}^{\infty}\frac{1}{\beta^{n+i}}=\frac{1}{\beta^{n}(\beta-1)}\rightarrow
0$ as $n\rightarrow\infty$, there exists an integer $N>3$ such that for any
$n\geq N$,

$\displaystyle y_{n}$ $\displaystyle\geq\frac{1}{\beta},$ $\displaystyle
x_{n}+y_{n}$ $\displaystyle\leq\frac{1}{\beta(\beta-1)}.$

It follows that $(x_{n},y_{n})\in C_{012}$ for any $n\geq N$.

Let

$\bm{b}=122\underbrace{00\ldots 0}_{N-3\text{ times}}.$

Let

$\displaystyle D^{\prime}$
$\displaystyle=\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{j}b_{j+1}\ldots
b_{j+N-1}=\bm{b}\text{ for infinitely many }j\text{'s}\\}$

then $D^{\prime}\subset D_{2}$. Now we show that $\mathbb{P}(D^{\prime})=1$.
Notice that $D^{\prime}=\cap_{n=1}^{\infty}\cup_{m=n}^{\infty}D_{m},$ where
$D_{m}=\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{m}b_{m+1}\ldots
b_{m+N-1}=\bm{b}\\}.$ Let $B_{n}=\cup_{m=n}^{\infty}D_{m}$. If
$(b_{1},b_{2},\ldots)\in\Upsilon\setminus B_{n}$, then we have that for any
$j\geq n$, $b_{j}b_{j+1}\ldots b_{j+N-1}\neq\bm{b}.$ Clearly,

$\Upsilon\setminus B_{n}\subseteq
B^{\prime}:=\\{(b_{1},b_{2},\ldots):b_{n+kN}\ldots
b_{n+(k+1)N-1}\neq\bm{b},k=0,1,2,\ldots\\}.$

Since $\mathbb{P}(\Upsilon\setminus
B_{n})\leq\mathbb{P}(B^{\prime})=\lim_{k\rightarrow\infty}(1-1/3^{N})^{k}=0$,
then $\mathbb{P}(B_{n})=1$. It follows from
$D^{\prime}=\cap_{n=1}^{\infty}B_{n}$ and $B_{1}\supseteq
B_{2}\supseteq\cdots$ that
$\mathbb{P}(D^{\prime})=\lim_{n\rightarrow\infty}B_{n}=1$. Then we get
$\mathbb{P}(D_{2})=1.$

Similarly, define

$\vec{z}_{l}=(x_{l},y_{l})=\frac{\vec{q}_{1}}{\beta}+\frac{\vec{q}_{0}}{\beta^{2}}+\cdots+\frac{\vec{q}_{0}}{\beta^{l}}+\frac{\vec{q}_{b_{1}}}{\beta^{l+1}}+\frac{\vec{q}_{b_{2}}}{\beta^{l+2}}+\cdots.$

Then we have $x_{l}\geq\frac{1}{\beta}$, and

$\displaystyle 0$ $\displaystyle\leq
y_{l}\leq\sum_{i=1}^{\infty}\frac{1}{\beta^{l+i}}=\frac{1}{\beta^{l}(\beta-1)},$
$\displaystyle\frac{1}{\beta}$ $\displaystyle\leq
x_{l}+y_{l}\leq\frac{1}{\beta}+\sum_{i=1}^{\infty}\frac{1}{\beta^{l+i}}=\frac{1}{\beta}+\frac{1}{\beta^{l}(\beta-1)}.$

Since $\lim_{l\rightarrow\infty}\frac{1}{\beta^{l}(\beta-1)}=0$, then there
exists $L>0$ such that for any $l\geq L$,

$0\leq y_{l}<\frac{1}{\beta},x_{l}+y_{l}\leq\frac{1}{\beta(\beta-1)}.$

It follows that $(x_{l},y_{l})\in C_{01}$ for any $l\geq L$. Let

(4.1) $D^{\prime\prime}=\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{j}b_{j+1}\ldots
b_{j+L-1}=1\underbrace{00\ldots 0}_{L-1\text{ times }}\text{ for infinitely
many }j\text{'s}\\},$

then $D^{\prime\prime}\subset D_{1}$. Since $\mathbb{P}(D^{\prime\prime})=1$,
we have $\mathbb{P}(D_{1})=1$. Therefore, $\mathbb{P}(D)=1$. ∎

The following theorem can be obtained from Lemmas 4.4 and 4.5.

######  Theorem 4.6.

For $\beta\in(1,\beta_{*}]$, the dynamical systems
$(\Omega\times\Upsilon\times S_{\beta},\mathcal{A\times B\times
S},\nu_{\beta},K_{\beta})$ and
$(\Upsilon,\mathcal{B},\mathbb{P},\sigma^{\prime})$ are isomorphic, where
$\nu_{\beta}(A)=\mathbb{P}(\varphi(Z\cap A))$.

######  Remark 4.7.

(i) Notice that if we change $\mathbb{P}$ to the general product measure
$m_{2}$ on $\Upsilon$, Lemma 4.5 and Theorem 4.6 remain true.
(ii) Since $\mathbb{P}$ is the unique measure of maximal entropy on
$\Upsilon$, the above theorem implies that any other $K_{\beta}$-invariant
measure with support $Z$ has entropy strictly less than $\log 3$. We now
investigate the entropy of $K_{\beta}$-invariant measure $\mu$ for which
$\mu(Z^{c})>0$.

Divide $Z^{c}$ into three Borel sets as follows:

$\displaystyle Z^{c}$ $\displaystyle=(Z_{1}\cap Z_{2})^{c}$
$\displaystyle=Z_{1}^{c}\cup Z_{2}^{c}$ $\displaystyle=(Z_{1}^{c}\setminus
Z_{2}^{c})\cup(Z_{2}^{c}\setminus Z_{1}^{c})\cup(Z_{1}^{c}\cap Z_{2}^{c})$
$\displaystyle:=Z_{3}\cup Z_{4}\cup Z_{5},$

where

$\displaystyle Z_{3}=$ $\displaystyle(Z_{1}^{c}\setminus Z_{2}^{c})$
$\displaystyle=$
$\displaystyle\\{(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
S_{\beta}:K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C\text{ for finitely many }n\text{'s}$ $\displaystyle\text{and
}K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C_{012}\text{ infinitely often}\\},$ $\displaystyle Z_{4}=$
$\displaystyle(Z_{2}^{c}\setminus Z_{1}^{c})$ $\displaystyle=$
$\displaystyle\\{(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
S_{\beta}:K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C_{012}\text{ for finitely many }n\text{'s},$ $\displaystyle\text{and
}K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times C\text{
infinitely often}\\},$ $\displaystyle Z_{5}=$ $\displaystyle(Z_{1}^{c}\cap
Z_{2}^{c})$ $\displaystyle=$
$\displaystyle\\{(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
S_{\beta}:K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C\text{ for finitely many }n\text{'s},$ $\displaystyle\text{ and
}K_{\beta}^{n}(\omega,\upsilon,\vec{z})\in\Omega\times\Upsilon\times
C_{012}\text{ for finitely many }n\text{'s}\\}.$

We first prove the following lemma.

######  Lemma 4.8.

Let $\beta\in(1,\beta_{*})$. Let $\mu_{3}$ be a $K_{\beta}$-invariant measure
for which $\mu_{3}(Z_{3})=1$. Then $h_{\mu_{3}}(K_{\beta})<\log 3$. Similarly,
let $\mu_{4}$ and $\mu_{5}$ be $K_{\beta}$-invariant measures for which
$\mu_{4}(Z_{4})=\mu_{5}(Z_{5})=1$. Then
$h_{\mu_{4}}(K_{\beta}),h_{\mu_{5}}(K_{\beta})<\log 3$ also holds.

###### Proof.

Let

$\displaystyle
H_{3}=\\{\vec{z}=\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{i}}}{\beta^{i}}\in
S_{\beta}:$
$\displaystyle\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\text{
belongs to }C_{012}\text{ for infinitely many }j\text{'s,}$
$\displaystyle\text{ and never belongs to }C\\}.$

Then $\Omega\times\Upsilon\times H_{3}\subseteq
K_{\beta}^{-1}(\Omega\times\Upsilon\times H_{3})$ and
$\cup_{i=0}^{\infty}K_{\beta}^{-i}(\Omega\times\Upsilon\times H_{3})=Z_{3}$.
It follows that
$\mu_{3}(Z_{3})=\lim_{i\rightarrow\infty}\mu_{3}(K_{\beta}^{-i}(\Omega\times\Upsilon\times
H_{3}))=1.$ Since $\mu_{3}$ is $K_{\beta}$-invariant, then

$\mu_{3}(\Omega\times\Upsilon\times
H_{3})=\mu_{3}(K_{\beta}^{-1}(\Omega\times\Upsilon\times
H_{3}))=\mu_{3}(K_{\beta}^{-2}(\Omega\times\Upsilon\times H_{3}))=\cdots=1.$

Thus it is enough to study the entropy with respect to $\mu_{3}$ of the map
$K_{\beta}$ restricted to $\Omega\times\Upsilon\times H_{3}$. Let
$\pi_{1},\pi_{2},\pi_{3}$ be the canonical projection onto the three
coordinates respectively. Notice that the action of the transformation
$K_{\beta}$ on the first coordinate is an identity, which implies that
$K_{\beta}$ is essentially a product transformation $I_{\Omega}\times
K_{\beta}^{\prime}$, where $K_{\beta}^{\prime}=(\pi_{2}\circ
K_{\beta})\times(\pi_{3}\circ K_{\beta})$ on $\Upsilon\times H_{3}$ and
$I_{\Omega}$ is the identity on $\Omega$. Since
$(\upsilon,\vec{z})\in\Upsilon\times S_{\beta}$ and $\omega\in\Omega$ are
independent, it follows from $h_{\mu}(I_{\Omega})=0$ for any measure $\mu$ on
$(\Omega,\mathcal{A})$ and that

$h_{\mu_{3}}(K_{\beta})=h_{\mu_{3}^{\prime}}(K_{\beta}^{\prime}),$

where $\mu_{3}^{\prime}(B\times H)=\mu_{3}(\Omega\times B\times H)$ for
$B\in\mathcal{B},H\in(H_{3}\cap\mathcal{S})$.

Let

$\displaystyle D_{3}=\\{(b_{1},b_{2},\ldots)\in\Upsilon:$
$\displaystyle\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\text{
belongs to }C_{012}\text{ for infinitely many }j\text{'s,}$
$\displaystyle\text{ and never belongs to }C\\}.$

Define a map $\phi$ from $(\Upsilon\times
H_{3},\mathcal{B}\times(H_{3}\cap\mathcal{S}),\mu_{3}^{\prime},K_{\beta}^{\prime})$
to
$(D_{3},D_{3}\cap\mathcal{B},\mu_{3}^{\prime}\circ\phi^{-1},\sigma^{\prime})$
as

$\phi(\upsilon,\vec{z})=\rho_{2}(\rho_{1}(0^{\infty},\upsilon,\vec{z})).$

Since $\vec{z}\in H_{3}$, then $\phi$ is well defined and bijective. $\phi$ is
measurable and the inverse is also measurable by Theorem 2.2. Finally, $\phi$
preserves the measure and $\phi\circ
K_{\beta}^{\prime}=\sigma^{\prime}\circ\phi$. Then $\phi$ is an isomorphism
and it follows that

$h_{\mu_{3}}(K_{\beta})=h_{\mu_{3}^{\prime}}(K_{\beta}^{\prime})=h_{\mu_{3}^{\prime}\circ\phi^{-1}}(\sigma^{\prime})\leq
h_{\mathbb{P}}(\sigma^{\prime})=\log 3.$

Since $\mathbb{P}$ is the unique measure of maximal entropy on $D_{3}$, to
show $h_{\mu_{3}}(K_{\beta})<\log 3$, it is enough to prove that
$\mu_{3}^{\prime}\circ\phi^{-1}\neq\mathbb{P}$. This is done by contradiction.
If $\mu_{3}^{\prime}\circ\phi^{-1}=\mathbb{P}$, then

$\mathbb{P}(D_{3})=\mu_{3}^{\prime}(\Upsilon\times
H_{3})=\mu_{3}(\Omega\times\Upsilon\times H_{3})=1.$

Since $D_{3}\subset(D^{\prime\prime})^{c}$, where $D^{\prime\prime}$ is
defined as in (4.1), then $\mathbb{P}((D^{\prime\prime})^{c})=1$, which is a
contradiction to $\mathbb{P}(D^{\prime\prime})=1$. Therefore,

$h_{\mu_{3}}(K_{\beta})<\log 3.$

Let

$\displaystyle
H_{4}=\\{\vec{z}=\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{i}}}{\beta^{i}}\in
S_{\beta}:$
$\displaystyle\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\text{
belongs to }C\text{ for infinitely many }j\text{'s,}$ $\displaystyle\text{ and
never belongs to }C_{012}\\},$ $\displaystyle
H_{5}=\\{\vec{z}=\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{i}}}{\beta^{i}}\in
S_{\beta}:$
$\displaystyle\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\text{
never belongs to }C$ $\displaystyle\text{ and never belongs to }C_{012}\\},$
$\displaystyle=\\{\vec{z}\in S_{\beta}:\vec{z}\text{ has a u}$
$\displaystyle\text{nique }\beta\text{-expansion}\\}.$

Then it follows that

$\mu_{4}(\Omega\times\Upsilon\times H_{4})=\mu_{5}(\Omega\times\Upsilon\times
H_{5})=1.$

We can also obtain that $h_{\mu_{4}}(K_{\beta}),h_{\mu_{5}}(K_{\beta})<\log 3$
using the similar method. ∎

From the above lemma we can obtain the upper bound of the entropy of
$K_{\beta}$-invariant measure for which $Z^{c}$ has positive measure.

######  Lemma 4.9.

Let $\beta\in(1,3/2]$. Let $\mu$ be a $K_{\beta}$-invariant measure for which
$\mu(Z^{c})>0$. Then $h_{\mu}(K_{\beta})<\log 3.$

###### Proof.

Notice that $Z,Z_{3},Z_{4}$ and $Z_{5}$ are pairwise disjoint and the union is
$\Omega\times\Upsilon\times S_{\beta}$. Since $Z,Z_{3},Z_{4}$ and $Z_{5}$ are
$K_{\beta}$-invariant, then there exist $K_{\beta}$-invariant probability
measures $\mu_{12},\mu_{3},\mu_{4}$ and $\mu_{5}$ concentrated on
$Z,Z_{3},Z_{4}$ and $Z_{5}$, respectively, such that

$\mu=(1-\alpha_{3}-\alpha_{4}-\alpha_{5})\mu_{12}+\alpha_{3}\mu_{3}+\alpha_{4}\mu_{4}+\alpha_{5}\mu_{5},$

where $0\leq\alpha_{3},\alpha_{4},\alpha_{5}\leq 1$ and
$0<\alpha_{3}+\alpha_{4}+\alpha_{5}\leq 1$. Then

$h_{\mu}(K_{\beta})=(1-\alpha_{3}-\alpha_{4}-\alpha_{5})h_{\mu_{12}}(K_{\beta})+\alpha_{3}h_{\mu_{3}}(K_{\beta})+\alpha_{4}h_{\mu_{4}}(K_{\beta})+\alpha_{5}h_{\mu_{5}}(K_{\beta}).$

Since $h_{\mu_{12}}(K_{\beta})\leq\log 3$ by Remark 4.7 and
$h_{\mu_{3}}(K_{\beta}),h_{\mu_{4}}(K_{\beta}),h_{\mu_{5}}(K_{\beta})<\log 3$
by Lemma 4.8, then the result follows. ∎

Now we obtain the main result in this section.

######  Theorem 4.10.

Let $\beta\in(1,\beta_{*})$. The measure
$\nu_{\beta}(A)=\mathbb{P}(\varphi(Z\cap A))$ is the unique
$K_{\beta}$-invariant measure of maximal entropy.

##  5\. An absolutely continuous invariant measure for $K_{\beta}$ when
$1<\beta\leq\frac{3}{2}$

The analysis is similar in the two cases: $\beta\in(1,3/2)$ and $\beta=3/2$.
Without special statement we assume $\beta\in(1,3/2)$ in this section. A brief
description of the case of $\beta=3/2$ is given in the remarks. Recall the
product measure $m_{1}$ on $\Omega$ with weights $\\{p,1-p\\}$ and $m_{2}$ on
$\Upsilon$ with weights $\\{s,t,1-s-t\\}$. Consider the measure space
$(S_{\beta},\mathcal{S},\lambda_{2})$, where $\lambda_{2}$ is the normalized
Lebesgue measure. In this section we will prove that $K_{\beta}$ has an
invariant measure of the form $m_{1}\otimes m_{2}\otimes\mu_{\beta}$, where
$\mu_{\beta}$ is absolutely continuous with respect to $\lambda_{2}$. We will
show the result by several steps.

Step 1: a position dependent random transformation $R$.

Bahsoun and Góra [3], gave a sufficient condition for the existence of an
absolutely continuous invariant measure for a random map with position
dependent probabilities on a bounded domain of $\mathbb{R}^{N}$. We take some
of their results a little further.

For $k=1,\ldots,K$, let $\tau_{k}:S_{\beta}\rightarrow S_{\beta}$ be piecewise
one-to-one and $C^{2}$, non-singular transformations on a common partition
$\mathcal{P}$ of $S_{\beta}:\mathcal{P}=\\{S_{1},\ldots,S_{q}\\}$ and
$\tau_{k,i}=\tau_{k}|_{S_{i}},i=1,\ldots,q.$ Let
$p_{k}:S_{\beta}\rightarrow[0,1]$ be piecewise $C^{1}$ functions such that
$\sum_{k=1}^{K}p_{k}=1.$ Denote by
$R=\\{\tau_{1},\ldots,\tau_{K};p_{1}(\vec{z}),\ldots,p_{K}(\vec{z})\\}$ the
position dependent random map, i.e., $R(\vec{z})=\tau_{k}(\vec{z})$ with
probability $p_{k}(\vec{z})$. Define the transition function for $R$ as
follows:

$\mathbf{P}(\vec{z},A)=\sum_{k=1}^{K}p_{k}(\vec{z})\mathbbm{1}_{A}(\tau_{k}(\vec{z})),$

where $A$ is any measurable set and $\mathbbm{1}_{A}$ denotes the indicator
function of the set $A$.

The iteration of $R$ is denoted by $R^{n}:=\\{\tau_{k_{1}k_{2}\cdots
k_{n}};p_{k_{1}k_{2}\cdots k_{n}}\\},k_{1}k_{2}\cdots
k_{n}\in\\{1,2,\ldots,K\\}^{n}$, where

$\tau_{k_{1}k_{2}\cdots
k_{n}}(\vec{z})=\tau_{k_{n}}\circ\tau_{k_{n-1}}\circ\cdots\circ\tau_{k_{1}}(\vec{z})$

and

$p_{k_{1}k_{2}\cdots
k_{n}}(\vec{z})=p_{k_{n}}(\tau_{k_{n-1}}\circ\cdots\circ\tau_{k_{1}}(\vec{z}))\cdot
p_{k_{n-1}}(\tau_{k_{n-2}}\circ\cdots\circ\tau_{k_{1}}(\vec{z}))\cdots
p_{k_{1}}(\vec{z}).$

The transition function $\mathbf{P}$ induces an operator $\mathbf{P}_{*}$ on
the set of probability measure on $(S_{\beta},\mathcal{S})$ defined by

$\mathbf{P}_{*}\mu(A)=\int\mathbf{P}(\vec{z},A)d\mu(\vec{z})=\sum_{k=1}^{K}\int_{\tau_{k}^{-1}(A)}p_{k}(\vec{z})d\mu(\vec{z})=\sum_{k=1}^{K}\sum_{i=1}^{q}\int_{\tau_{k,i}^{-1}(A)}p_{k}(\vec{z})d\mu(\vec{z}).$

We say that the measure $\mu$ is $R$-invariant iff $\mathbf{P}_{*}\mu=\mu$.

If $\mu$ has density $f$ with respect to $\lambda_{2}$, then
$\mathbf{P}_{*}\mu$ has also a density which we denote by $P_{R}f$, i.e.,

$\int_{A}P_{R}f(\vec{z})d\lambda_{2}(\vec{z})=\sum_{k=1}^{K}\sum_{i=1}^{q}\int_{\tau_{k,i}^{-1}(A)}p_{k}(\vec{z})f(\vec{z})d\lambda_{2}(\vec{z}).$

We call $P_{R}$ the Perron-Frobenius operator of the random map $R$ and it has
very useful properties[3]:

  * (i) $P_{R}$ is linear;

  * (ii) $P_{R}$ is non-negative;

  * (iii) $P_{R}f=f\iff\mu=f\cdot\lambda_{2}$ is $R$-invariant;

  * (iv) $\|P_{R}f\|_{1}\leq\|f\|_{1}$, where $\|\cdot\|_{1}$ denotes the $L^{1}$ norm;

  * (v) $P_{R\circ T}=P_{R}\circ P_{T}.$ In particular, $P_{R}^{N}=P_{R^{N}}.$

Let each $S_{i}$ be a bounded closed domain having a piecewise $C^{2}$
boundary of finite $1$-dimensional measure. Assume that the faces of $\partial
S_{i}$ meet at angles bounded uniformly away from $0$ and the probabilities
$p_{k}(\vec{z})$ are piecewise $C^{1}$ functions on the partition
$\mathcal{P}$. We assume:

Condition(A):

$\max_{1\leq i\leq
q}\sum_{k=1}^{K}p_{k}(\vec{z})\|D\tau_{k,i}^{-1}(\tau_{k,i}(\vec{z}))\|<c<1,$

where $D\tau_{k,i}^{-1}(\vec{z})$ is the derivative matrix of
$\tau_{k,i}^{-1}$ at $\vec{z}$.

Using the multidimensional notion of variation [13]:

$V(f)=\int_{\mathbb{R}^{N}}\|Df\|d\lambda_{N}=\sup\left\\{\int_{\mathbb{R}^{N}}f\text{div}(g)d\lambda_{N}:g=(g_{1},\ldots,g_{N})\in
C_{0}^{1}(\mathbb{R}^{N},\mathbb{R}^{N})\right\\},$

where $f\in L_{1}(\mathbb{R}^{N})$ has bounded support, $Df$ denotes the
gradient of $f$ in the distributional sense, $\text{div}(g)=\nabla\cdot
g=\frac{\partial g_{1}}{\partial x_{1}}+\frac{\partial g_{2}}{\partial
x_{2}}+\cdots+\frac{\partial g_{N}}{\partial x_{N}}$ is the divergence
operator, and $C_{0}^{1}(\mathbb{R}^{N},\mathbb{R}^{N})$ is the space of
continuously differentiable functions from $\mathbb{R}^{N}$ into
$\mathbb{R}^{N}$ having compact support. Consider the Banach space [13, Remark
1.12],

$BV(S_{\beta})=\\{f\in L_{1}(S_{\beta}):V(f)<+\infty\\},$

with the norm $\|f\|_{BV}=\|f\|_{L_{1}}+V(f)$.

Let $a(S_{i})$ be defined as in (3.5) and $\gamma(S_{i})$ defined as in (3.4).
Now we start at points $y\in F$, where the minimal angle $\gamma(S_{i})$ is
attained, defining $L_{y}$ to be central rays of the largest regular cones
contained in $S_{i}$. Then we extend this field of segments to $C^{1}$ field,
making $L_{y}$ short enough to avoid overlapping. Let $\delta(y)$ be the
length of $L_{y},y\in\partial S_{i}$. By the compactness of $\partial S_{i}$
we have

$\delta(S_{i})=\inf_{y\in\partial(S_{i})}\delta(y)>0.$

Let $\vec{z}$ be a point in $\partial S_{i}$ and $J_{k,i}$ the Jacobian of
$\tau_{k|S_{i}}$ at $\vec{z}$.

We recall the following theorem.

######  Theorem 5.1.

[3, Theorem 6.3] If $R$ is a random map which satisfies Condition (A), then

$V(P_{R}f)\leq c(1+1/a)V(f)+(M+\frac{c}{a\delta})\|f\|_{1},$

where
$a=\min\\{a(S_{i}):i=1,\ldots,q\\}>0,\delta=\min\\{\delta(S_{i}):i=1\ldots,q\\}>0,M_{k,i}=\sup_{\vec{z}\in
S_{i}}(Dp_{k}(\vec{z})-\frac{DJ_{k,i}}{J_{k,i}}p_{k}(\vec{z}))$ and
$M=\sum_{k=1}^{K}\max_{1\leq i\leq q}M_{k,i}.$

Now, let $R$ be a random map which is given by
$\\{\tau_{1},\ldots,\tau_{6};p_{1}(\vec{z}),\ldots,p_{6}(\vec{z})\\}$ where

$\displaystyle\tau_{1}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\}\\\ \beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\}\\\ \beta\vec{z},&\mbox{if }\vec{z}\in
C_{012},\end{cases}$ $\displaystyle\tau_{2}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\}\\\ \beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\}\\\ \beta\vec{z}-\vec{q}_{1},&\mbox{if }\vec{z}\in
C_{012},\end{cases}$ $\displaystyle\tau_{3}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\}\\\ \beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\}\\\ \beta\vec{z}-\vec{q}_{2},&\mbox{if }\vec{z}\in
C_{012},\end{cases}$ $\displaystyle\tau_{4}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\}\\\ \beta\vec{z}-\vec{q}_{j},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\}\\\ \beta\vec{z},&\mbox{if }\vec{z}\in
C_{012},\end{cases}$ $\displaystyle\tau_{5}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\}\\\ \beta\vec{z}-\vec{q}_{j},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\}\\\ \beta\vec{z}-\vec{q}_{1},&\mbox{if }\vec{z}\in
C_{012},\end{cases}$ $\displaystyle\tau_{6}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\}\\\ \beta\vec{z}-\vec{q}_{j},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\}\\\ \beta\vec{z}-\vec{q}_{2},&\mbox{if }\vec{z}\in
C_{012},\end{cases}$

The probabilities are defined as follows.

$\displaystyle p_{1}(\vec{z})=p\cdot s,$ $\displaystyle
p_{4}(\vec{z})=(1-p)\cdot s,$ (5.1) $\displaystyle p_{2}(\vec{z})=p\cdot t,$
$\displaystyle p_{5}(\vec{z})=(1-p)\cdot t,$ $\displaystyle
p_{3}(\vec{z})=p\cdot(1-s-t),$ $\displaystyle
p_{6}(\vec{z})=(1-p)\cdot(1-s-t).$

We have the following lemma.

######  Lemma 5.2.

For any $\vec{z}\in S_{\beta}$, $\sum_{k_{1}k_{2}\cdots
k_{n}\in\\{1,\ldots,6\\}^{n}}p_{k_{1}k_{2}\ldots
k_{n}}(\vec{z})=1,n\in\mathbb{Z}$.

###### Proof.

We prove this lemma by induction. For $n=1$,
$p_{1}(\vec{z})+\dots+p_{6}(\vec{z})=1$. Assume it is true for $n=m$, i.e. for
any $\vec{z}\in S_{\beta}$,

$\sum_{k_{1}k_{2}\cdots k_{m}\in\\{1,\ldots,6\\}^{n}}p_{k_{1}k_{2}\ldots
k_{m}}(\vec{z})=1.$

For $n=m+1$,

$\displaystyle\sum_{k_{1}k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m+1}}p_{k_{1}k_{2}\ldots k_{m+1}}(\vec{z})$
$\displaystyle=$ $\displaystyle\sum_{k_{1}k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m+1}}p_{k_{m+1}}(\tau_{k_{m}}\circ\cdots\circ\tau_{k_{1}}(\vec{z}))\cdot
p_{k_{m}}(\tau_{k_{m-1}}\circ\cdots\circ\tau_{k_{1}}(\vec{z}))\cdots
p_{k_{1}}(\vec{z})$ $\displaystyle=$ $\displaystyle\sum_{1k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m+1}}p_{k_{m+1}}(\tau_{k_{m}}\circ\cdots\circ\tau_{k_{2}}(\tau_{1}(\vec{z})))\cdots
p_{k_{2}}(\tau_{1}(\vec{z}))\cdot p_{1}(\vec{z})$
$\displaystyle+\sum_{2k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m+1}}p_{k_{m+1}}(\tau_{k_{m}}\circ\cdots\circ\tau_{k_{2}}(\tau_{2}(\vec{z})))\cdots
p_{k_{2}}(\tau_{2}(\vec{z}))\cdot p_{2}(\vec{z})$
$\displaystyle+\cdots+\sum_{6k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m+1}}p_{k_{m+1}}(\tau_{k_{m}}\circ\cdots\circ\tau_{k_{2}}(\tau_{6}(\vec{z})))\cdots
p_{k_{2}}(\tau_{6}(\vec{z}))\cdot p_{6}(\vec{z})$ $\displaystyle=$
$\displaystyle p_{1}(\vec{z})\sum_{k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m}}p_{k_{2}\ldots
k_{m+1}}(\tau_{1}(\vec{z}))+p_{2}(\vec{z})\sum_{k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m}}p_{k_{2}\ldots k_{m+1}}(\tau_{2}(\vec{z}))$
$\displaystyle+\cdots+p_{6}(\vec{z})\sum_{k_{2}\cdots
k_{m+1}\in\\{1,\ldots,6\\}^{m}}p_{k_{2}\ldots k_{m+1}}(\tau_{6}(\vec{z}))$
$\displaystyle=$ $\displaystyle p_{1}(\vec{z})+\dots+p_{6}(\vec{z})$
$\displaystyle=$ $\displaystyle 1.$

∎

Now we can prove the existence of an acim for $R$.

######  Theorem 5.3.

Let $R=\\{\tau_{1},\ldots,\tau_{6};p_{1}(\vec{z}),\ldots,p_{6}(\vec{z})\\}$,
then $R$ admits an acim.

###### Proof.

Denote the partition (3.1) by $\mathcal{P}$ with

$S_{1}=E_{0},S_{2}=E_{1},S_{3}=E_{2},S_{4}=C_{01},S_{5}=C_{12},S_{6}=C_{02},S_{7}=C_{012}.$

Consider the iteration of the random map, $R^{n}$, the corresponding partition
is $\vee_{i=0}^{n-1}R^{-i}\mathcal{P}$, where

$R^{-i}\mathcal{P}=\vee_{k_{1}k_{2}\cdots
k_{i}\in\\{1,\ldots,6\\}^{i}}\tau_{k_{1}k_{2}\ldots k_{i}}^{-1}\mathcal{P}.$

For a set $P_{i}\in\vee_{i=0}^{n-1}R^{-i}\mathcal{P}$ and a sequence
$k_{1}\ldots k_{n}\in\\{1,\ldots,6\\}^{n}$, let $\tau_{k_{1}\ldots
k_{n},i}=\tau_{k_{1}\ldots k_{n}}|_{P_{i}}$ and $M_{k_{1}\ldots
k_{n},i}=\sup_{\vec{z}\in P_{i}}(Dp_{k_{1}\ldots
k_{n}}(\vec{z})-\frac{DJ_{k_{1}\ldots k_{n},i}}{J_{k_{1}\ldots
k_{n},i}}p_{k_{1}\ldots k_{n},i}(\vec{z}))$, where $J_{k_{1}\ldots k_{n},i}$
is the Jacobian of $\tau_{k_{1}\ldots k_{n},i}$. Let

$M_{n}=\sum_{k_{1}\ldots
k_{n}\in\\{0,1,2\\}^{n}}\max_{P_{i}\in\vee_{i=0}^{n-1}R^{-i}\mathcal{P}}M_{k_{1}\ldots
k_{n},i}\ \ \text{ and }\ \
\delta_{n}=\min_{P_{i}\in\vee_{i=0}^{n-1}R^{-i}\mathcal{P}}\delta(P_{i}).$

For any set $P_{i}\in\vee_{i=0}^{n-1}R^{-i}\mathcal{P}$, the derivative matrix
of $\tau_{k_{1}k_{2}\ldots k_{n}}^{-1}$ is equal to

$\begin{bmatrix}\frac{1}{\beta^{n}}&0\\\ 0&\frac{1}{\beta^{n}}\\\
\end{bmatrix}.$

Using Lemma 5.2 we have

$\displaystyle\max_{P_{i}\in\vee_{i=0}^{n-1}R^{-i}\mathcal{P}}\sum_{k_{1}k_{2}\cdots
k_{n}\in\\{1,\ldots,6\\}^{n}}p_{k_{1}k_{2}\ldots
k_{n}}(\vec{z})\|D(\tau_{k_{1}k_{2}\ldots
k_{n}}|_{P_{i}})^{-1}\|=\frac{\sqrt{2}}{\beta^{n}}<\frac{2\sqrt{2}}{\beta^{n}}:=c_{n}.$

For the partition $\vee_{i=0}^{n-1}R^{-i}\mathcal{P}$, we have
$a_{n}=\sqrt{2}/2$. Let

$r_{n}=c_{n}(1+\frac{1}{a_{n}})=\frac{2\sqrt{2}+4}{\beta^{n}},\ \ \
R_{n}=M_{n}+\frac{c_{n}}{a_{n}\delta_{n}}.$

We can find $l>\log(2\sqrt{2}+4)/\log\beta$ such that $r_{l}<1$. Fix this $l$
and let
$C_{1}=\max\\{r_{1},r_{2},\ldots,r_{l-1}\\},C_{2}=\max\\{R_{1},R_{2},\ldots,R_{l-1}\\}$.
For any integer $n$, we have $n=jl+i$, where $0\leq i\leq l-1.$ Notice that
$P_{R^{n}}=(P_{R^{l}})^{j}P_{R^{i}}$. Apply Theorem 5.1 on $R^{l}$, then we
get

$\displaystyle V(P_{R^{n}}f)$ $\displaystyle=VP_{R^{l}}^{j}(P_{R^{i}}f)$
$\displaystyle\leq r_{l}\cdot VP_{R^{l}}^{j-1}(P_{R^{i}}f)+R_{l}\|f\|_{1}$
$\displaystyle\leq r_{l}\cdot(r_{l}\cdot
VP_{R^{l}}^{j-2}(P_{R^{i}}f)+R_{l}\|f\|_{1})+R_{l}\|f\|_{1}$
$\displaystyle\cdots$ $\displaystyle\leq
r_{l}^{j}V(P_{R^{i}}f)+(r_{l}^{j-1}+r_{l}^{j-2}+\cdots+r_{l}+1)R_{l}\|f\|_{1}$
$\displaystyle\leq
r_{l}^{j}(C_{1}V(f)+C_{2}\|f\|_{1})+(r_{l}^{j-1}+r_{l}^{j-2}+\cdots+r_{l}+1)R_{l}\|f\|_{1}$
$\displaystyle=C_{1}r_{l}^{j}V(f)+(C_{2}r_{l}^{j}+r_{l}^{j-1}+r_{l}^{j-2}+\cdots+r_{l}+1)R_{l}\|f\|_{1}$
$\displaystyle\leq
C_{1}r_{l}^{j}V(f)+(C_{2}+\frac{1}{1-r_{l}})R_{l}\|f\|_{1}.$

By definition of the norm $\|\cdot\|_{BV}$,

$\displaystyle\|P_{R^{n}}f\|_{BV}$
$\displaystyle=\|P_{R^{n}}f\|_{1}+V(P_{R^{n}}f)$
$\displaystyle\leq\|f\|_{1}+C_{1}r_{l}^{j}V(f)+(C_{2}+\frac{1}{1-r_{l}})R_{l}\|f\|_{1}.$

Then the result follows by the technique in [4, Theorem 1]. We write some
details for completeness. From the above inequality it follows that the set
$\\{P_{R}^{n}\mathbf{1}\\}_{n\geq l}$ is uniformly bounded, where $\mathbf{1}$
is the constant function equal to $1$ on $S_{\beta}$. Hence $P_{R}$ has a
nontrivial fixed point $\mathbf{1}^{*}$ which is the density of an acim by the
Kakutani-Yoshida Theorem. ∎

Step 2: for the skew product transformation $R^{\prime}$ on
$S_{\beta}\times[0,1)$.

Let $(I,\mathcal{B}(I),\lambda_{1})$ be the unit interval $I=[0,1)$, with
$\mathcal{B}(I)$ the Borel $\sigma$-algebra on $I$ and $\lambda_{1}$ being
Lebesgue measure on $(I,\mathcal{B}(I))$. Let $Y=S_{\beta}\times I$ and the
set $J_{k}$ be given by $J_{k}=\\{(\vec{z},w):\sum_{i<k}p_{i}(\vec{z})\leq
w<\sum_{i\leq k}p_{i}(\vec{z})\\}$. Define maps $\varphi_{k}:J_{k}\rightarrow
I$ by

$\varphi_{k}(\vec{z},w)=\frac{1}{p_{k}(\vec{z})}w-\frac{\sum_{r=1}^{k-1}p_{r}(\vec{z})}{p_{k}(\vec{z})}.$

Define the skew product transformation $R^{\prime}:S_{\beta}\times
I\rightarrow S_{\beta}\times I$ by

$R^{\prime}(\vec{z},w)=(\tau_{k}(x),\varphi_{k}(\vec{z},w))$

for $(\vec{z},w)\in J_{k}$.

Since $p_{k}(\vec{z})$ is defined as in (5.1), then we have

$\displaystyle\varphi_{1}(\vec{z},w)=\frac{w}{ps},$
$\displaystyle\varphi_{4}(\vec{z},w)=\frac{w-p}{(1-p)s},$
$\displaystyle\varphi_{2}(\vec{z},w)=\frac{w-ps}{pt},$
$\displaystyle\varphi_{5}(\vec{z},w)=\frac{w-p-(1-p)s}{(1-p)t},$
$\displaystyle\varphi_{3}(\vec{z},w)=\frac{w-ps-pt}{p(1-s-t)},$
$\displaystyle\varphi_{6}(\vec{z},w)=\frac{w-p-(1-p)s-(1-p)t}{(1-p)(1-s-t)}.$

We denote $p_{k}(\vec{z})$ and $\varphi_{k}(\vec{z},w)$ by $p_{k}$ and
$\varphi_{k}(w)$, respectively, since each $p_{k}(\vec{z})$ is a constant.
Therefore,

$R^{\prime}(\vec{z},w)=\begin{cases}(\tau_{1}(\vec{z}),\varphi_{1}(w)),&\mbox{if
}w\in[0,ps),\\\ (\tau_{2}(\vec{z}),\varphi_{2}(w)),&\mbox{if
}w\in[ps,ps+pt),\\\ (\tau_{3}(\vec{z}),\varphi_{3}(w)),&\mbox{if
}w\in[ps+pt,p),\\\ (\tau_{4}(\vec{z}),\varphi_{4}(w)),&\mbox{if
}w\in[p,p+(1-p)s),\\\ (\tau_{5}(\vec{z}),\varphi_{5}(w)),&\mbox{if
}w\in[p+(1-p)s,p+(1-p)s+(1-p)t),\\\
(\tau_{6}(\vec{z}),\varphi_{6}(w)),&\mbox{if }w\in[p+(1-p)s+(1-p)t,1).\\\
\end{cases}$

Denote by $\mu_{\beta}$ an acim for the position dependent random
transformation $R=\\{\tau_{1},\ldots,\tau_{6};p_{1},\ldots,p_{6}\\}$, which
means $\mu_{\beta}$ is $R$-invariant and absolutely continuous with respect to
Lebesgue measure $\lambda_{2}$ in $\mathbb{R}^{2}$. We start by recalling
Lemma 3.2 in [1].

######  Lemma 5.4.

$\mu_{\beta}$ is invariant for the random map $R$ if and only if
$\mu_{\beta}\otimes\lambda_{1}$ is invariant for the skew product
$R^{\prime}$.

Step 3: for the skew product transformation $R_{\beta}$ on
$\Omega\times\Upsilon\times S_{\beta}$ .

Define the $skew\ product\ transformation\ R_{\beta}$ on
$\Omega\times\Upsilon\times S_{\beta}$ as follows:

$R_{\beta}(\omega,\upsilon,\vec{z})=\begin{cases}(\sigma\omega,\sigma^{\prime}\upsilon,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
}\vec{z}\in E_{i},i\in\\{0,1,2\\}\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
}\vec{z}\in C_{ij},ij\in\\{01,02,12\\}\mbox{ and }\omega_{1}=0\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\beta\vec{z}-\vec{q}_{j}),&\mbox{if
}\vec{z}\in C_{ij},ij\in\\{01,02,12\\}\mbox{ and }\omega_{1}=1\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
}\vec{z}\in C_{012},\upsilon_{1}=i,i\in\\{0,1,2\\}.\end{cases}$

######  Lemma 5.5.

$(S_{\beta}\times I,\mathcal{S\times
B}(I),\mu_{\beta}\otimes\lambda_{1},R^{\prime})$ and
$(\Omega\times\Upsilon\times S_{\beta},\mathcal{A\times B\times
S},m_{1}\otimes m_{2}\otimes\mu_{\beta},R_{\beta})$ are isomorphic.

###### Proof.

Let $\pi_{2}:S_{\beta}\times I\rightarrow I$ be the canonical projection onto
the second coordinate. Consider the map $\varphi=\pi_{2}\circ R^{\prime}$ on
$(I,\mathcal{B}(I),\lambda_{1})$. One can see that $\varphi(w)=\varphi_{k}(w)$
for $w\in I_{k}$, where $I_{1}=[0,p_{1})$ and
$I_{k}=\Big{[}\sum_{i=1}^{k-1}p_{i},\sum_{i=1}^{k}p_{i}\Big{)}$ for $2\leq
k\leq 6$. Define

$l(w)=\frac{1}{p_{k}}\quad\text{and}\quad
h(w)=\frac{\sum_{i=1}^{k-1}p_{i}}{p_{k}}$

for $w\in I_{k}$. It follows that $\varphi(w)=l(w)\cdot w-h(w)$. Let

$l_{n}=l_{n}(w):=l(\varphi^{n-1}(w))\quad\text{and}\quad
h_{n}=h_{n}(w):=h(\varphi^{n-1}(w)).$

For $w\in[0,1)$, we can write the generalized Lüroth series (GLS) of $w$,
which is

$w=\frac{h_{1}}{l_{1}}+\frac{h_{2}}{l_{1}l_{2}}+\cdots+\frac{h_{n}}{l_{1}\cdots
l_{n}}+\cdots.$

Consider the system
$\\{\\{0,1,2,3,4,5\\}^{\mathbb{N}},\mathcal{C},m,\sigma^{\prime\prime}\\}$,
where $\mathcal{C}$ is the product $\sigma$-algebra, $\sigma^{\prime\prime}$
is the left shift and $m$ is the product measure with weights
$\\{p_{1},\ldots,p_{6}\\}$ as (5.1). Let
$\phi_{1}:I\rightarrow\\{0,1,2,3,4,5\\}^{\mathbb{N}}$ be given by

$\phi_{1}:w=\sum_{n=1}^{\infty}\frac{h_{i}}{l_{1}l_{2}\cdots
l_{i}}\mapsto(\gamma_{1},\gamma_{2},\ldots,),$

where $\gamma_{n}=\gamma_{n}(w),n\geq 1$ is defines as follows:

$\gamma_{n}:=\gamma_{n}(w)=k-1\iff\varphi^{n-1}(w)\in I_{k},$

for $k\in\\{1,2,3,4,5,6\\}.$ It is known that $\varphi$ preserves the Lebesgue
measure $\lambda_{1}$ and $\phi_{1}$ is an isomorphism between the two
dynamical systems $(I,\mathcal{B}(I),\lambda_{1},\varphi\\}$ and
$\\{\\{0,1,2,3,4,5\\}^{\mathbb{N}},\mathcal{C},m,\sigma^{\prime\prime}\\}$.
See [2] for more details.

Next we give a map $\phi_{2}$ from
$\\{\\{0,1,2,3,4,5\\}^{\mathbb{N}},\mathcal{C},m,\sigma^{\prime\prime}\\}$ to
$\\{\Omega\times\Upsilon,\mathcal{A\times B},m_{1}\otimes
m_{2},\sigma\times\sigma^{\prime}\\}$. Let
$h_{1}:\\{0,1,2,3,4,5\\}\rightarrow\\{0,1\\}$ and
$h_{2}:\\{0,1,2,3,4,5\\}\rightarrow\\{0,1,2\\}$ be given by

$h_{1}(x)=\begin{cases}0,&\mbox{if }x=0,1,2,\\\ 1,&\mbox{if }x=3,4,5,\\\
\end{cases},\quad h_{2}(x)=\begin{cases}0,&\mbox{if }x=0,3,\\\ 1,&\mbox{if
}x=1,4,\\\ 2,&\mbox{if }x=2,5.\\\ \end{cases}$

Define
$\phi_{2}:\\{0,1,2,3,4,5\\}^{\mathbb{N}}\rightarrow\Omega\times\Upsilon$ by
$\phi_{2}(\gamma)=(\omega,\upsilon),$ where

$\displaystyle\omega=(h_{1}(\gamma_{1}),h_{1}(\gamma_{2}),h_{1}(\gamma_{3}),\ldots):=\tilde{h}_{1}(\gamma),$
$\displaystyle\upsilon=(h_{2}(\gamma_{1}),h_{2}(\gamma_{2}),h_{2}(\gamma_{3}),\ldots):=\tilde{h}_{2}(\gamma).$

One can see that $\phi_{2}$ maps a cylinder of rank $n$ in
$\\{0,1,2,3,4,5\\}^{\mathbb{N}}$ to the product of two cylinders of the same
rank $n$ in $\Omega\times\Upsilon$. It follows that $\phi_{2}$ is a
bimeasurable bijection. From the definition of the product measure, we can get
the measure preservingness on cylinders. Finally, it is easy to see that
$\phi_{2}\circ\sigma^{\prime\prime}=(\sigma\times\sigma^{\prime})\circ\phi_{2}$.
Therefore, $\phi_{2}$ is an isomorphism.

Now let $\phi:S_{\beta}\times I\rightarrow\Omega\times\Upsilon\times
S_{\beta}$ be given by

$\phi(\vec{z},w)=(\tilde{h}_{1}(\phi_{1}(w)),\tilde{h}_{2}(\phi_{1}(w)),\vec{z}).$

In fact, $\phi=\iota\circ(I_{S_{\beta}}\times(\phi_{2}\circ\phi_{1}))$, where
$I_{S_{\beta}}$ is the identity map on $S_{\beta}$ and
$\iota(\vec{z},\omega,\upsilon)=(\omega,\upsilon,\vec{z})$ is a transformation
that only changes the order of coordinates. Since $\phi_{2}\circ\phi_{1}$
preserves the dynamics of $\pi_{2}\circ R$ and $\sigma\times\sigma^{\prime}$,
i.e.,

$(\phi_{2}\circ\phi_{1})\circ(\pi_{2}\circ
R)=(\sigma\times\sigma^{\prime})\circ(\phi_{2}\circ\phi_{1}),$

we have that $\phi\circ R^{\prime}=R_{\beta}\circ\phi$. Therefore, the result
follows. ∎

Step 4: for the random transformation $K_{\beta}$ on
$\Omega\times\Upsilon\times S_{\beta}$.

Define a skew product transformation $R_{\beta}$ as follows:

$R_{\beta}(\omega,\upsilon,\vec{z})=\begin{cases}(\sigma\omega,\sigma^{\prime}\upsilon,\tau_{1}(\vec{z})),&\mbox{if
}\omega_{1}=0,\upsilon_{1}=0,\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\tau_{2}(\vec{z})),&\mbox{if
}\omega_{1}=0,\upsilon_{1}=1,\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\tau_{3}(\vec{z})),&\mbox{if
}\omega_{1}=0,\upsilon_{1}=2,\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\tau_{4}(\vec{z})),&\mbox{if
}\omega_{1}=1,\upsilon_{1}=0,\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\tau_{5}(\vec{z})),&\mbox{if
}\omega_{1}=1,\upsilon_{1}=1,\\\
(\sigma\omega,\sigma^{\prime}\upsilon,\tau_{6}(\vec{z})),&\mbox{if
}\omega_{1}=1,\upsilon_{1}=2,\\\ \end{cases}$

Let $\mu$ be an arbitrary probability measure on $S_{\beta}$. We will show
that any product measure of the form $m_{1}\otimes m_{2}\otimes\mu$ is
$K_{\beta}$-invariant if and only if it is $R_{\beta}$-invariant.

######  Lemma 5.6.

$m_{1}\otimes m_{2}\otimes\mu\circ K_{\beta}^{-1}=m_{1}\otimes
m_{2}\otimes\mu\circ R_{\beta}^{-1}=m_{1}\otimes m_{2}\otimes\nu$, where

$\displaystyle\nu=$ $\displaystyle
ps\cdot\mu\circ\tau_{1}^{-1}+pt\cdot\mu\circ\tau_{2}^{-1}+p(1-s-t)\cdot\mu\circ\tau_{3}^{-1}$
$\displaystyle+(1-p)\cdot s\cdot\mu\circ\tau_{4}^{-1}+(1-p)\cdot
t\cdot\mu\circ\tau_{5}^{-1}+(1-p)\cdot(1-s-t)\cdot\mu\circ\tau_{6}^{-1}.$

###### Proof.

Denote by $C_{1}$ and $C_{2}$ arbitrary cylinders in $\Omega$ and $\Upsilon$,
respectively. Let $S$ be a closed set in $S_{\beta}$. It suffices to verify
that the measures coincide on sets of the form $C_{1}\times C_{2}\times S$,
because the collection of these sets forms a generating $\pi$-system. Let
$[i,C_{1}]=\\{\omega_{1}=i\\}\cap\sigma^{-1}(C_{1})$ for $i=0,1$ and
$[i,C_{2}]=\\{\upsilon_{1}=i\\}\cap(\sigma^{\prime})^{-1}(C_{2})$ for
$i=0,1,2$. Notice that

$\displaystyle\tau_{1}(S)\cap E=\tau_{2}(S)\cap E=\tau_{3}(S)\cap
E=\tau_{4}(S)\cap E=\tau_{5}(S)\cap E=\tau_{6}(S)\cap E,$
$\displaystyle\tau_{1}(S)\cap C=\tau_{2}(S)\cap C=\tau_{3}(S)\cap C,$
$\displaystyle\tau_{4}(S)\cap C=\tau_{5}(S)\cap C=\tau_{6}(S)\cap C,$
$\displaystyle\tau_{1}(S)\cap C_{012}=\tau_{4}(S)\cap C_{012},$
$\displaystyle\tau_{2}(S)\cap C_{012}=\tau_{5}(S)\cap C_{012},$
$\displaystyle\tau_{3}(S)\cap C_{012}=\tau_{6}(S)\cap C_{012}.$

We can get

$\displaystyle K_{\beta}^{-1}(C_{1}\times C_{2}\times S)=$ $\displaystyle
C_{1}\times C_{2}\times(\tau_{1}(S)\cap E)$ $\displaystyle\cup[0,C_{1}]\times
C_{2}\times(\tau_{1}(S)\cap C)\cup[1,C_{1}]\times C_{2}\times(\tau_{4}(S)\cap
C)$ $\displaystyle\cup C_{1}\times[0,C_{2}]\times(\tau_{1}(S)\cap C_{012})\cup
C_{1}\times[1,C_{2}]\times(\tau_{2}(S)\cap C_{012})$ $\displaystyle\cup
C_{1}\times[2,C_{2}]\times(\tau_{3}(S)\cap C_{012})$

Hence,

$\displaystyle m_{1}\otimes m_{2}\otimes\mu\circ K_{\beta}^{-1}(C_{1}\times
C_{2}\times S)$ $\displaystyle=$ $\displaystyle m_{1}(C_{1})\cdot
m_{2}(C_{2})\cdot\mu(\tau_{1}(S)\cap E)$ $\displaystyle+p\cdot
m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{1}(S)\cap C)+(1-p)\cdot
m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{4}(S)\cap C)$
$\displaystyle+s\cdot m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{1}(S)\cap
C_{012})+t\cdot m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{2}(S)\cap
C_{012})$ $\displaystyle+(1-s-t)\cdot m_{1}(C_{1})\cdot
m_{2}(C_{2})\cdot\mu(\tau_{3}(S)\cap C_{012}$ $\displaystyle=$ $\displaystyle
ps\cdot m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{1}(S))+pt\cdot
m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{2}(S))$
$\displaystyle+p(1-s-t)\cdot m_{1}(C_{1})\cdot
m_{2}(C_{2})\cdot\mu(\tau_{3}(S))+(1-p)s\cdot m_{1}(C_{1})\cdot
m_{2}(C_{2})\cdot\mu(\tau_{4}(S))$ $\displaystyle+(1-p)t\cdot
m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{5}(S))+(1-p)(1-s-t)\cdot
m_{1}(C_{1})\cdot m_{2}(C_{2})\cdot\mu(\tau_{6}(S))$ $\displaystyle=$
$\displaystyle m_{1}\otimes m_{2}\otimes\nu(C_{1}\times C_{2}\times S).$

On the other hand,

$\displaystyle R_{\beta}^{-1}(C_{1}\times C_{2}\times S)$ $\displaystyle=$
$\displaystyle[0,C_{1}]\times[0,C_{2}]\times\tau_{1}(S)\cup[0,C_{1}]\times[1,C_{2}]\times\tau_{2}(S)$
$\displaystyle\cup[0,C_{1}]\times[2,C_{2}]\times\tau_{3}(S)\cup[1,C_{1}]\times[0,C_{2}]\times\tau_{4}(S)$
$\displaystyle\cup[1,C_{1}]\times[1,C_{2}]\times\tau_{5}(S)\cup[1,C_{1}]\times[2,C_{2}]\times\tau_{6}(S).$

Therefore, we complete the proof. ∎

Now we give the main result in this section.

######  Theorem 5.7.

Let $\beta\in(1,3/2)$. Then $K_{\beta}$ has an invariant measure of the form
$m_{1}\otimes m_{2}\otimes\mu_{\beta}$, where $\mu_{\beta}$ is absolutely
continuous with respect to $\lambda_{2}$.

###### Proof.

By Theorem 5.3, Lemma 5.4, Lemma 5.5, and Lemma 5.6, we complete the proof. ∎

######  Remark 5.8.

When $\beta=3/2$, $C_{012}=\\{(\frac{2}{3},\frac{2}{3})\\}$ is a point. We
modify the definition of $R,R^{\prime},R_{\beta}$, and give relevant
conclusions.
(i) Let $R=\\{\tau_{1},\tau_{2};p_{1}(\vec{z}),p_{2}(\vec{z})\\}$ be a
position dependent random transformation on $S_{\beta}$, where

$\displaystyle\tau_{1}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\},\\\ \beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\},\\\ \beta\vec{z},&\mbox{if
}\vec{z}=(\frac{2}{3},\frac{2}{3}),\end{cases}$
$\displaystyle\tau_{2}(\vec{z})$
$\displaystyle=\begin{cases}\beta\vec{z}-\vec{q}_{i},&\mbox{if }\vec{z}\in
E_{i},i\in\\{0,1,2\\},\\\ \beta\vec{z}-\vec{q}_{j},&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\},\\\ \beta\vec{z},&\mbox{if
}\vec{z}=(\frac{2}{3},\frac{2}{3}),\end{cases}$

and $p_{1}(\vec{z})=p_{2}(\vec{z})=1/2$ for $\vec{z}\in S_{\beta}$. Similar to
Theorem 5.3, it is not difficult to prove that $R$ has an acim $\mu_{\beta}$.
(ii) By Lemma 3.2 in [1], $\mu_{\beta}\otimes\lambda_{1}$ is invariant for the
skew product $R^{\prime}$, where

$R^{\prime}(\vec{z},w)=\begin{cases}(\tau_{1}(\vec{z}),\frac{w}{p}),&\mbox{if
}w\in[0,p),\\\ (\tau_{2}(\vec{z}),\frac{w-p}{1-p}),&\mbox{if }w\in[p,1).\\\
\end{cases}$

(iii) Define the $skew\ product\ transformation\ R_{\beta}$ on $\Omega\times
S_{\beta}$ as follows:

$R_{\beta}(\omega,\vec{z})=\begin{cases}(\sigma\omega,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
}\vec{z}\in E_{i},i\in\\{0,1,2\\},\\\
(\sigma\omega,\beta\vec{z}-\vec{q}_{i}),&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\},\omega_{1}=0,\\\
(\sigma\omega,\beta\vec{z}-\vec{q}_{j}),&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\},\omega_{1}=1,\\\
(\sigma\omega,\beta\vec{z}),&\mbox{if
}\vec{z}=(\frac{2}{3},\frac{2}{3}).\end{cases}$

Then we have that the dynamical systems $(S_{\beta}\times I,\mathcal{S\times
B}(I),\mu_{\beta}\otimes\lambda_{1},R^{\prime})$ and $(\Omega\times
S_{\beta},\mathcal{A\times S},m_{1}\otimes\mu_{\beta},R_{\beta})$ are
isomorphic. The proof is similar and easier than that of Lemma 5.5.
(iv) Let $\mu$ be an arbitrary probability measure on $S_{\beta}$ and let
$\tilde{K}_{\beta}:\Omega\times S_{\beta}\rightarrow\Omega\times S_{\beta}$ be
given by

$\tilde{K}_{\beta}(\omega,\vec{z})=\begin{cases}(\omega,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
}\vec{z}\in E_{i},i\in\\{0,1,2\\}\\\
(\sigma\omega,\beta\vec{z}-\vec{q}_{i}),&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\},\omega_{1}=0\\\
(\sigma\omega,\beta\vec{z}-\vec{q}_{j}),&\mbox{if }\vec{z}\in
C_{ij},ij\in\\{01,02,12\\},\omega_{1}=1\\\ (\omega,\beta\vec{z}),&\mbox{if
}\vec{z}=(\frac{2}{3},\frac{2}{3}).\end{cases}$

It is easy to check that

$m_{1}\otimes\mu\circ\tilde{K}_{\beta}^{-1}=m_{1}\otimes\mu\circ
R_{\beta}^{-1}=m_{1}\otimes\nu,$

where

$\nu=p\cdot\mu\circ\tau_{1}^{-1}+(1-p)\cdot\mu\circ\tau_{2}^{-1}$

by using the same method of calculation in Lemma 5.6. Therefore, it follows
from (i-iv) that $\tilde{K}_{\beta}$ has an invariant measure of the form
$m_{1}\otimes\mu_{\beta}$.

Next we give another method to prove that $R_{\beta}$ has an acim by using
Corollary 3.3. Denote
$I_{0}=[0,p),I_{1}=[p,1],J_{0}=[0,s),J_{1}=[s,s+t),J_{2}=[s+t,1].$ Let

$\displaystyle\tilde{R}(\vec{z},w,v)$
$\displaystyle=\begin{cases}(\beta\vec{z}-\vec{q}_{i},\frac{w}{p},\frac{v}{s})&\mbox{if
}\vec{z}\in E_{i}\times[0,p)\times[0,s),i\in\\{0,1,2\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w-p}{1-p},\frac{v}{s})&\mbox{if }\vec{z}\in
E_{i}\times[p,1]\times[0,s),i\in\\{0,1,2\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w}{p},\frac{v-s}{t})&\mbox{if }\vec{z}\in
E_{i}\times[0,p)\times[s,s+t),i\in\\{0,1,2\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w-p}{1-p},\frac{v-s}{t})&\mbox{if }\vec{z}\in
E_{i}\times[p,1]\times[s,s+t),i\in\\{0,1,2\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w}{p},\frac{v-s-t}{1-s-t})&\mbox{if
}\vec{z}\in E_{i}\times[0,p)\times[s+t,1],i\in\\{0,1,2\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w-p}{1-p},\frac{v-s-t}{1-s-t})&\mbox{if
}\vec{z}\in E_{i}\times[p,1]\times[s+t,1],i\in\\{0,1,2\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w}{p},\frac{v}{s})&\mbox{if }\vec{z}\in
C_{ij}\times[0,p)\times[0,s),ij\in\\{01,02,12\\},\\\
(\beta\vec{z}-\vec{q}_{j},\frac{w-p}{1-p},\frac{v}{s})&\mbox{if }\vec{z}\in
C_{ij}\times[p,1]\times[0,s),ij\in\\{01,02,12\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w}{p},\frac{v-s}{t})&\mbox{if }\vec{z}\in
C_{ij}\times[0,p)\times[s,s+t),ij\in\\{01,02,12\\},\\\
(\beta\vec{z}-\vec{q}_{j},\frac{w-p}{1-p},\frac{v-s}{t})&\mbox{if }\vec{z}\in
C_{ij}\times[p,1]\times[s,s+t),ij\in\\{01,02,12\\},\\\
(\beta\vec{z}-\vec{q}_{i},\frac{w}{p},\frac{v-s-t}{1-s-t})&\mbox{if
}\vec{z}\in C_{ij}\times[0,p)\times[s+t,1],ij\in\\{01,02,12\\},\\\
(\beta\vec{z}-\vec{q}_{j},\frac{w-p}{1-p},\frac{v-s-t}{1-s-t})&\mbox{if
}\vec{z}\in C_{ij}\times[p,1]\times[s+t,1],ij\in\\{01,02,12\\},\\\
(\beta\vec{z},\frac{w}{p},\frac{v}{s})&\mbox{if }\vec{z}\in
C_{012}\times[0,p)\times[0,s),\\\
(\beta\vec{z},\frac{w-p}{1-p},\frac{v}{s})&\mbox{if }\vec{z}\in
C_{012}\times[p,1]\times[0,s),\\\
(\beta\vec{z},\frac{w}{p},\frac{v-s}{t})&\mbox{if }\vec{z}\in
C_{012}\times[0,p)\times[s,s+t),\\\
(\beta\vec{z},\frac{w-p}{1-p},\frac{v-s}{t})&\mbox{if }\vec{z}\in
C_{012}\times[p,1]\times[s,s+t),\\\
(\beta\vec{z},\frac{w}{p},\frac{v-s-t}{1-s-t})&\mbox{if }\vec{z}\in
C_{012}\times[0,p)\times[s+t,1],\\\
(\beta\vec{z},\frac{w-p}{1-p},\frac{v-s-t}{1-s-t})&\mbox{if }\vec{z}\in
C_{012}\times[p,1]\times[s+t,1].\end{cases}$

We can get the following theorem.

######  Lemma 5.9.

Let $\beta\in(1,3/2)$, then $\tilde{R}$ admits an acim.

###### Proof.

Use the partition $\mathcal{P}:\\{E_{i}\times I_{m}\times J_{n},C_{ij}\times
I_{m}\times J_{n},C_{012}\times I_{m}\times J_{n},\\},i=0,1,2;m=0,1;n=0,1,2$.

Consider the iterate $\tilde{R}^{k}$ and the corresponding partition
$\vee_{i=0}^{k-1}\tilde{R}^{-i}(\mathcal{P})$. For
$P_{i}\in\vee_{i=0}^{k-1}\tilde{R}^{-i}(\mathcal{P})$, let
$\tilde{R}^{k}_{i}=\tilde{R}^{k}|_{P_{i}}$. Since the derivative matrix of
$(\tilde{R}_{i}^{k})^{-1}$ is

$\begin{bmatrix}\frac{1}{\beta^{k}}&0&0&0\\\ 0&\frac{1}{\beta^{k}}&0&0\\\
0&0&p^{l}(1-p)^{k-l}&0\\\ 0&0&0&s^{i}t^{j}(1-s-t)^{k-i-j}\\\ \end{bmatrix},$

where $l,i,j\in\\{0,1,\ldots,k\\}$ and $0\leq i+j\leq k$, then the Euclidean
matrix norm,

$\|D(\tilde{R}_{i}^{k})^{-1}\|=\sqrt{\frac{2}{\beta^{2k}}+p^{2l}(1-p)^{2(k-l)}+s^{2i}t^{2j}(1-s-t)^{2(k-i-j)}}.$

Since $\beta>1$ and $0<p,s,t,s+t<1$, then
$\lim_{n\rightarrow\infty}\|D(\tilde{R}_{i}^{k})^{-1}\|=0$ for any
$P_{i}\in\vee_{i=0}^{k-1}\tilde{R}^{-i}(\mathcal{P})$. It means that we can
choose an integer $k$ such that

$\|D(\tilde{R}_{i}^{k})^{-1}\|<c$

for any $0<c<1$. For the partition
$\vee_{i=0}^{k-1}\tilde{R}^{-i}(\mathcal{P})$, $a$ is a positive constant
which is not larger than 1 since any set in the partition is a polyhedra in
$\mathbb{R}^{4}$. Let $c<\frac{a}{a+1}$, then we have

$c(1+\frac{1}{a})<1.$

By Corollary 3.3, $\tilde{R}$ admits an acim. ∎

######  Theorem 5.10.

$R_{\beta}$ admits an acim.

###### Proof.

Set

$l^{\prime}(x)=\begin{cases}\frac{1}{p},&\text{ if }x\in[0,p),\\\
\frac{1}{1-p},&\text{ if }x\in[p,1],\\\ \end{cases}\quad
h^{\prime}(x)=\begin{cases}0,&\text{ if }x\in[0,p),\\\ \frac{p}{1-p},&\text{
if }x\in[p,1].\\\ \end{cases}$

Let $\varphi^{\prime}(x)=l^{\prime}(x)\cdot x-h^{\prime}(x)$ and

$l^{\prime}_{n}=l^{\prime}_{n}(x):=l^{\prime}(\varphi^{\prime
n-1}(x)),\quad\quad
h^{\prime}_{n}=h^{\prime}_{n}(x):=h^{\prime}(\varphi^{\prime n-1}(x)).$

Define the map $\phi_{1}:[0,1]\rightarrow\Omega$ as follows:

$\phi_{1}:x=\sum_{n=1}^{\infty}\frac{h^{\prime}_{i}}{l^{\prime}_{1}l^{\prime}_{2}\cdots
l^{\prime}_{i}}\mapsto(\omega_{1},\omega_{2},\ldots,),$

where $\omega_{n}=\omega_{n}(x)=k-1\iff\varphi^{n-1}(x)\in I_{k}$ for
$I_{1}=[0,p)$ and $I_{2}=[p,1]$.
Similarly, set

$l^{\prime\prime}(x)=\begin{cases}\frac{1}{s},&\text{ if }x\in[0,s),\\\
\frac{1}{t},&\text{ if }x\in[s,s+t),\\\ \frac{1}{1-s-t},&\text{ if
}x\in[s+t,1],\\\ \end{cases}\quad h^{\prime\prime}(x)=\begin{cases}0,&\text{
if }x\in[0,s),\\\ \frac{s}{t},&\text{ if }x\in[s,s+t),\\\
\frac{s+t}{1-s-t},&\text{ if }x\in[s+t,1].\\\ \end{cases}$

and

$l^{\prime\prime}_{n}=l^{\prime\prime}_{n}(x):=l^{\prime\prime}(\varphi^{\prime\prime
n-1}(x)),\quad\quad
h^{\prime\prime}_{n}=h^{\prime\prime}_{n}(w):=h^{\prime\prime}(\varphi^{\prime\prime
n-1}(x)),$

where $\varphi^{\prime\prime}(x)=l^{\prime\prime}(x)\cdot
x-h^{\prime\prime}(x)$. Let $\phi_{2}:[0,1]\rightarrow\Upsilon$ be given by

$\phi_{2}:x=\sum_{n=1}^{\infty}\frac{h^{\prime\prime}_{i}}{l^{\prime\prime}_{1}l^{\prime\prime}_{2}\cdots
l^{\prime\prime}_{i}}\mapsto(\upsilon_{1},\upsilon_{2},\ldots,),$

where $\upsilon_{n}=\upsilon_{n}(x)=k-1\iff\varphi^{n-1}(x)\in J_{k}$ for
$J_{1}=[0,s),J_{2}=[s,s+t)$ and $J_{3}=[s+t,1]$.
Consider the map $\phi=\iota\circ(I_{S_{\beta}}\times\phi_{1}\times\phi_{2})$
from $(S_{\beta}\times[0,1]\times[0,1],\mathcal{S\times
B}[0,1]\times\mathcal{B}[0,1],\nu,\tilde{R})$ to $(\Omega\times\Upsilon\times
S_{\beta},\mathcal{A\times B\times S},\mu,R_{\beta})$, where
$\iota(\vec{z},\omega,\upsilon)=(\omega,\upsilon,\vec{z})$ and $I_{S_{\beta}}$
is the identity map on $S_{\beta}$. It is easy to check that $\phi$ is an
isomorphism by the definition, and the result follows from Lemma 5.9. ∎

##  6\. random transformation $K_{\beta}$ for $\frac{3}{2}<\beta\leq\beta^{*}$

###  6.1. A special class of $S_{\beta}$

For $3/2<\beta<2$, we should notice that there are holes in the attractor.
Broomhead et al.[6] discussed a special structure of those holes: they are all
centered on three radial lines originating from the center of the attractor
and extending to the three vertices. Since the Sierpinski triangle they
discussed can be converted to $S_{\beta}$ through affine transformation, we
can obtain the corresponding range of $\beta$. Their set-up is as follows. Let
$S_{\lambda}$ be the attractor of an IFS

$\\{g_{i}(\vec{z})=\lambda\vec{z}+(1-\lambda)\vec{p}_{i}\\},\ i=0,1,2,$

where $\vec{p}_{i}=\frac{2}{3}(\cos\frac{2\pi i}{3},\sin\frac{2\pi i}{3})$,
which are the vertices of an equilateral triangle $\Delta^{\prime}$. Denote
the central hole by $H^{\prime}$, i.e.
$H^{\prime}=\Delta^{\prime}\setminus\cup_{i=0}^{2}g_{i}(\Delta^{\prime})$.
Recall the following result in [6, Proposition 3.7].

######  Proposition 6.1.

Let $\lambda_{*}\approx 0.6478$ be the appropriate root of

$x^{3}-x^{2}+x=\frac{1}{2}.$

Then $S_{\lambda}$ has a non-empty interior if
$\lambda\in[\lambda_{*},\frac{2}{3})$, and moreover, each hole has the form
$g_{i}^{n}(H^{\prime})$, see Figure 2.

Figure 2. $S_{\lambda}$ for $\lambda_{*}\leq\lambda<\frac{2}{3}$

Notice that the IFS given by (2.1) can also be writen as

$f_{\vec{q}_{i}}(\vec{z})=\frac{1}{\beta}\vec{z}+(1-\frac{1}{\beta})\frac{\vec{q}_{i}}{\beta-1},i=0,1,2.$

Since there is an invertible affine transformation $l$ from $S_{\lambda}$ to
$S_{\beta}$, then the two attractors are affinely equivalent when
$\lambda=1/\beta$. Therefore we have the following proposition.

######  Proposition 6.2.

Let $\beta^{*}\approx 1.5437$ be the root of

$x^{3}-2x^{2}+2x=2.$

Then $S_{\beta}$ has a non-empty interior if
$\beta\in(\frac{3}{2},\beta^{*}]$, and each hole has the form
$f_{\vec{q}_{i}}^{n}(H)$, see Figure 3.

Figure 3. $S_{\beta}$ for $\frac{3}{2}<\beta\leq\beta^{*}$

###### Proof.

Let

$l(\vec{z})=\frac{1}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{z}+\frac{1}{\beta-1}\begin{bmatrix}1/3\\\ 1/3\\\
\end{bmatrix}.$

Then we have $l(\vec{p}_{i})=\frac{1}{\beta-1}\vec{q}_{i},i\in\\{0,1,2\\}$,
which implies

(6.1) $l(\Delta^{\prime})=\Delta.$

Since

$\displaystyle f_{\vec{q}_{i}}\circ l(\vec{z})$
$\displaystyle=\frac{1}{\beta}(\frac{1}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{z}+\frac{1}{\beta-1}\begin{bmatrix}1/3\\\ 1/3\\\
\end{bmatrix})+(1-\frac{1}{\beta})\frac{\vec{q}_{i}}{\beta-1}$
$\displaystyle=\frac{1}{\beta(\beta-1)}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{z}+\frac{1}{\beta(\beta-1)}\begin{bmatrix}1/3\\\ 1/3\\\
\end{bmatrix}+(1-\frac{1}{\beta})l(\vec{p}_{i}),$

and

$\displaystyle l\circ g_{i}(\vec{z})$
$\displaystyle=\frac{1}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}(\lambda\vec{z}+(1-\lambda)\vec{p}_{i})+\frac{1}{\beta-1}\begin{bmatrix}1/3\\\
1/3\\\ \end{bmatrix}$
$\displaystyle=\frac{\lambda}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{z}+\frac{1-\lambda}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{p}_{i}+\frac{1}{\beta-1}\begin{bmatrix}1/3\\\ 1/3\\\
\end{bmatrix}$
$\displaystyle=\frac{\lambda}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{z}+(1-\lambda)(l(p_{i})-\frac{1}{\beta-1}\begin{bmatrix}1/3\\\
1/3\\\ \end{bmatrix})+\frac{1}{\beta-1}\begin{bmatrix}1/3\\\ 1/3\\\
\end{bmatrix}$
$\displaystyle=\frac{\lambda}{\beta-1}\begin{bmatrix}-1/2&\sqrt{3}/2\\\
-1/2&-\sqrt{3}/2\\\
\end{bmatrix}\vec{z}+(1-\lambda)l(p_{i})+\frac{\lambda}{\beta-1}\begin{bmatrix}1/3\\\
1/3\\\ \end{bmatrix},$

then $f_{\vec{q}_{i}}\circ l=l\circ g_{i}$ by $\lambda=1/\beta$. From this, we
can get

(6.2)
$l(H^{\prime})=l(\Delta^{\prime})\setminus\cup_{i=0}^{2}l(g_{i}(\Delta^{\prime}))=\Delta\setminus\cup_{i=0}^{2}f_{\vec{q}_{i}}(\Delta)=H.$

If $\beta\in(\frac{3}{2},\beta^{*}]$, then
$\lambda\in[\lambda_{*},\frac{2}{3})$. By Proposition 6.1(see [6, Proposition
3.7] for details), we have that for $n\geq 2$,

  * (1) $g_{i}g_{j}^{n-1}(H^{\prime})\subset g_{j}(\Delta^{\prime})$;

  * (2) $g_{i}g_{j}^{n-1}(H^{\prime})\cap g_{j}g_{i}^{n-1}(H^{\prime})=\emptyset$.

Then it follows from (6.1) and (6.2) that

$\displaystyle(1)f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n-1}(H)$
$\displaystyle=f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n-1}(l(H^{\prime}))=f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n-2}(l\circ
g_{i}(H^{\prime}))=l(g_{i}g_{j}^{n-1}(H^{\prime}))$ $\displaystyle\subset
l\circ g_{j}(\Delta^{\prime})=f_{\vec{q}_{j}}\circ
l(\Delta^{\prime})=f_{\vec{q}_{j}}(\Delta),$
$\displaystyle(2)f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n-1}(H)$ $\displaystyle\cap
f_{\vec{q}_{j}}f_{\vec{q}_{i}}^{n-1}(H)=l(g_{i}g_{j}^{n-1}(H^{\prime})\cap
g_{j}g_{i}^{n-1}(H^{\prime}))=\emptyset.$

∎

###  6.2. Definition of $K_{\beta}$ for $\beta\in(3/2,\beta^{*}]$

Because of the disappearance of the triple overlap, we only need one two-sided
coin to be tossed to determine the transformation on the point. However, the
partition of $S_{\beta}$ is complicated by the presence of holes.

First we partition $\Delta_{1}$ into $\tilde{E}_{i}$ and $\tilde{C}_{ij}$ (see
Figure 4), where

$\displaystyle\tilde{E}_{0}$
$\displaystyle=([0,\frac{1}{\beta})\times[0,\frac{1}{\beta}))\backslash H,$
$\displaystyle\tilde{E}_{1}$ $\displaystyle=\\{(x,y):0\leq
y<\frac{1}{\beta},\frac{1}{\beta(\beta-1)}<x+y\leq\frac{1}{\beta-1}\\}\backslash
H,$ $\displaystyle\tilde{E}_{2}$ $\displaystyle=\\{(x,y):0\leq
x<\frac{1}{\beta},\frac{1}{\beta(\beta-1)}<x+y\leq\frac{1}{\beta-1}\\}\backslash
H,$ $\displaystyle\tilde{C}_{01}$
$\displaystyle=\\{(x,y):x\geq\frac{1}{\beta},y\geq
0,x+y\leq\frac{1}{\beta(\beta-1)}\\},$ $\displaystyle\tilde{C}_{12}$
$\displaystyle=\\{(x,y):x\geq\frac{1}{\beta},y\geq\frac{1}{\beta},x+y\leq\frac{1}{\beta-1}\\},$
$\displaystyle\tilde{C}_{02}$ $\displaystyle=\\{(x,y):x\geq
0,y\geq\frac{1}{\beta},x+y\leq\frac{1}{\beta(\beta-1)}\\}.$

Here
$H=\\{(x,y):x<\frac{1}{\beta},y<\frac{1}{\beta},x+y>\frac{1}{\beta(\beta-1)}\\}$.

Then we have the following lemma.

$\frac{1}{\beta(\beta-1)}$$\frac{1}{\beta(\beta-1)}$$\frac{1}{\beta}$$\frac{1}{\beta-1}$$\frac{1}{\beta}$$\frac{1}{\beta-1}$$x$$y$$\tilde{E}_{0}$$\tilde{C}_{01}$$\tilde{E}_{1}$$\tilde{C}_{02}$$\tilde{C}_{12}$$\tilde{E}_{2}$$H$
Figure 4. $\Delta_{1}$ for $\frac{3}{2}<\beta\leq\beta^{*}$

######  Lemma 6.3.

For $\beta\in(3/2,\beta^{*}]$. Let $(x,y)\in S_{\beta}$ and
$(x,y)=\sum_{i=1}^{\infty}\frac{a_{i}}{\beta^{i}}$ with
$a_{i}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}$ be a representation of
$(x,y)$ in base $\beta$. one has
(i) If $(x,y)\in\tilde{E}_{i}$ for some $i\in\\{0,1,2\\}$, then
$a_{1}=\vec{q}_{i}$,
(ii) If $(x,y)\in\tilde{C}_{ij}$ for some $ij\in\\{01,12,02\\}$, then
$a_{1}\in\\{\vec{q}_{i},\vec{q}_{j}\\}$.

###### Proof.

The proof is similar with Lemma 4.2. ∎

From the discussion in Proposition 6.2, we know that all holes
$f_{\vec{q}_{i}}^{n}(H)$ in the attractor $S_{\beta}$ are in
$\tilde{E}_{i}$(see Figure 3). And the ‘holes’
$f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n}(H)(i\neq j)$ that should exist in
$\tilde{C}_{i}$ are covered by $f_{\vec{q}_{j}}(S_{\beta})$, i.e.
$f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n}(H)(i\neq j)\subset S_{\beta}$, which
leaves only one choice $\vec{q}_{j}$ for the first digit of the expansions of
the points in $f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n}(H)$. Now we can partition
the Sierpinski carpet $S_{\beta}$ into equality regions $E_{i}$ and switch
regions $C_{ij}$.

$\displaystyle E_{0}$
$\displaystyle=\tilde{E}_{0}\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{1}}f_{\vec{q}_{0}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{2}}f_{\vec{q}_{0}}^{n}(H)\setminus\cup_{n=0}^{\infty}f_{\vec{q}_{0}}^{n}H,$
$\displaystyle E_{1}$
$\displaystyle=\tilde{E}_{1}\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{0}}f_{\vec{q}_{1}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{2}}f_{\vec{q}_{1}}^{n}(H)\setminus\cup_{n=0}^{\infty}f_{\vec{q}_{1}}^{n}H,$
$\displaystyle E_{2}$
$\displaystyle=\tilde{E}_{2}\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{1}}f_{\vec{q}_{2}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{0}}f_{\vec{q}_{2}}^{n}(H)\setminus\cup_{n=0}^{\infty}f_{\vec{q}_{2}}^{n}H,$
$\displaystyle C_{01}$
$\displaystyle=\tilde{C}_{01}\setminus(\cup_{n=1}^{\infty}f_{\vec{q}_{0}}f_{\vec{q}_{1}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{1}}f_{\vec{q}_{0}}^{n}(H)),$
$\displaystyle C_{12}$
$\displaystyle=\tilde{C}_{12}\setminus(\cup_{n=1}^{\infty}f_{\vec{q}_{2}}f_{\vec{q}_{1}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{1}}f_{\vec{q}_{2}}^{n}(H)),$
$\displaystyle C_{02}$
$\displaystyle=\tilde{C}_{02}\setminus(\cup_{n=1}^{\infty}f_{\vec{q}_{0}}f_{\vec{q}_{2}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{2}}f_{\vec{q}_{0}}^{n}(H)).$

The selection of the digits in the expansion becomes more accurate.

######  Lemma 6.4.

For $\beta\in(3/2,\beta^{*}]$. Let $\vec{z}\in S_{\beta}$ and
$\vec{z}=\sum_{i=1}^{\infty}\frac{a_{i}}{\beta^{i}}$ with
$a_{i}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}$ be a representation of
$\vec{z}$ in base $\beta$. one has
(i) If $\vec{z}\in E_{i}$ for some $i\in\\{0,1,2\\}$, then
$a_{1}=\vec{q}_{i}$,
(ii) If $\vec{z}\in C_{ij}$ for some $ij\in\\{01,12,02\\}$, then
$a_{1}\in\\{\vec{q}_{i},\vec{q}_{j}\\}$.

###### Proof.

It is enough to prove that $a_{1}=\vec{q}_{j}$ for $\vec{z}\in
f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n}(H)$.

Suppose $a_{1}\neq\vec{q}_{j}$. Since
$f_{\vec{q}_{i}}f_{\vec{q}_{j}}^{n}(H)\subset\tilde{C}_{ij}$, it follows from
Lemma 6.3 that $a_{1}=\vec{q}_{i}$. Then

$f_{\vec{q}_{i}}^{-1}(\vec{z})=\beta\vec{z}-\vec{q}_{i}=\sum_{i=1}^{\infty}\frac{a_{i+1}}{\beta^{i}}\in
S_{\beta},$

which is a contradiction to the fact that $f_{\vec{q}_{i}}^{-1}(\vec{z})\in
f_{\vec{q}_{j}}^{n}(H)$ which is a hole. ∎

Let

$C=C_{01}\cup C_{12}\cup C_{02},\ \ E=E_{0}\cup E_{1}\cup E_{2}.$

Recall $\Omega=\\{0,1\\}^{\mathbb{N}}$ with the product $\sigma$-algebra
$\mathcal{A}$ and $\sigma:\Omega\rightarrow\Omega$ is the left shift, and
define $K_{\beta}:\Omega\times S_{\beta}\rightarrow\Omega\times S_{\beta}$ by

$K_{\beta}(\omega,\vec{z})=\begin{cases}(\omega,\beta\vec{z}-\vec{q}_{i}),&\mbox{if
$\vec{z}\in E_{i},i=0,1,2$},\\\
(\sigma\omega,\beta\vec{z}-\vec{q}_{i}),&\mbox{if $\omega_{1}=0$ and
$\vec{z}\in C_{ij},\ ij\in\\{01,12,02\\}$,}\\\
(\sigma\omega,\beta\vec{z}-\vec{q}_{j}),&\mbox{if $\omega_{1}=1$ and
$\vec{z}\in C_{ij},\ ij\in\\{01,12,02\\}$.}\end{cases}$

The digits are given by

$d_{1}=d_{1}(w,\vec{z})=\begin{cases}\vec{q}_{i},&\mbox{if $\vec{z}\in
E_{i},i=0,1,2$},\\\ \ &\mbox{or $(\omega,\vec{z})\in\\{\omega_{1}=0\\}\times
C_{ij},\ ij\in\\{01,12,02\\}$,}\\\ \vec{q}_{j},&\mbox{if
$(\omega,\vec{z})\in\\{\omega_{1}=1\\}\times C_{ij},\
ij\in\\{01,12,02\\}$.}\end{cases}$

Then

$K_{\beta}(\omega,\vec{z})=\begin{cases}(\omega,\beta\vec{z}-d_{1}),&\mbox{if
$\vec{z}\in E$},\\\ (\sigma\omega,\beta\vec{z}-d_{1}),&\mbox{if $\vec{z}\in
C$}.\end{cases}$

Set $d_{n}=d_{n}(\omega,\vec{z})=d_{1}(K_{\beta}^{n-1}(\omega,\vec{z}))$, and
$\pi_{2}:\Omega\times S_{\beta}\rightarrow S_{\beta}$ be the canonical
projection onto the second coordinate, then

$\|\vec{z}-\sum_{i=1}^{n}\frac{d_{i}}{\beta^{i}}\|_{1}=\frac{\|\pi_{2}(K_{\beta}^{n}(\omega,\vec{z}))\|_{1}}{\beta_{n}}.$

Since $\pi_{2}(K_{\beta}^{n}(w,\vec{z}))\in\Delta$ and $\Delta$ is a bounded
set in $\mathbb{R}^{2}$, it follows that for all $\omega\in\Omega$ and for all
$\vec{z}\in S_{\beta}$ one has

$\vec{z}=\sum_{i=1}^{\infty}\frac{d_{i}}{\beta^{i}}=\sum_{i=1}^{\infty}\frac{d_{i}(\omega,\vec{z})}{\beta^{i}}.$

One can see that the random map $K_{\beta}$ for $\beta\in(3/2,\beta^{*}]$
satisfies similar properties as that for $\beta\in(1,3/2]$. For each point
$\vec{z}\in S_{\beta}$, consider the set

$D_{\vec{z}}=\\{(d_{1}(\omega,\vec{z}),d_{2}(\omega,\vec{z}),\ldots):\omega\in\Omega\\}.$

######  Theorem 6.5.

Suppose $\omega,\omega^{\prime}\in\Omega$ are such that
$\omega\prec\omega^{\prime}$. Then

$(d_{1}(\omega,\vec{z}),d_{2}(\omega,\vec{z}),\ldots)\preceq(d_{1}(\omega^{\prime},\vec{z}),d_{2}(\omega^{\prime},\vec{z}),\ldots).$

###### Proof.

This result is easier to prove than Theorem 4.1 using the same method. ∎

######  Theorem 6.6.

For $\beta\in(3/2,\beta^{*}]$. Let $\vec{z}\in S_{\beta}$ and
$\vec{z}=\sum_{i=1}^{\infty}\frac{a_{i}}{\beta^{i}}$ with
$a_{i}\in\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}$ be a representation of
$\vec{z}$ in base $\beta.$ Then there exists an $\omega\in\Omega$ such that
$a_{i}=d_{i}(\omega,\vec{z})$.

###### Proof.

Using Lemma 6.4, the proof is similar with Theorem 4.3.

∎

###  6.3. Unique $K_{\beta}$-invariant measure of maximal entropy.

On the set $\Omega\times S_{\beta}$ we consider the product $\sigma$-algebra
$\mathcal{A\times S}$. Recall the function
$\rho_{2}:\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}\rightarrow\Upsilon$
which is given by

$\rho_{2}(\vec{q}_{b_{1}},\vec{q}_{b_{2}},\vec{q}_{b_{3}},\ldots)=(b_{1},b_{2},b_{3},\ldots).$

Define the function $\rho:\Omega\times
S_{\beta}\rightarrow\\{\vec{q}_{0},\vec{q}_{1},\vec{q}_{2}\\}^{\mathbb{N}}$ by

$\rho_{1}(\omega,\vec{z})=(d_{1}(\omega,\vec{z}),d_{2}(\omega,\vec{z}),\ldots).$

Denote by $\varphi=\rho_{2}\circ\rho_{1}$ the function from $\Omega\times
S_{\beta}$ to $\Upsilon$. It is easy to check that $\varphi$ is measurable and
$\varphi\circ K_{\beta}=\sigma^{\prime}\circ\varphi$. It follows from Theorem
6.6 that $\varphi$ is surjective.

Let

$\displaystyle Z$ $\displaystyle=\\{(\omega,\vec{z})\in\Omega\times
S_{\beta}:K_{\beta}^{n}(\omega,\vec{z})\in\Omega\times C\text{ infinitely
often}\\},$ $\displaystyle D$
$\displaystyle=\\{(b_{1},b_{2},\ldots)\in\Upsilon:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in
C\text{ for infinitely many }j\text{'s}\\}.$

Then $\varphi(Z)=D,K_{\beta}^{-1}(Z)=Z$ and
$(\sigma^{\prime})^{-1}(\Upsilon)=\Upsilon.$ Let $\varphi^{\prime}$ be the
restriction of the map $\varphi$ on $Z$.

######  Lemma 6.7.

The map $\varphi^{\prime}:Z\rightarrow D$ is a bimeasurable bijection.

###### Proof.

For any sequence $(b_{1},b_{2},\ldots)\in D$, we can get
$\vec{z}=\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{i}}}{\beta^{i}}$. To get
$\omega$, define

$r_{1}=\min\\{j\geq
1:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in C\\},\ \
r_{k}=\min\\{j>r_{k-1}:\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{j+i-1}}}{\beta^{i}}\in
C\\}.$

If $b_{r_{k}}\in\\{0,2\\}$, let $\omega_{k}=b_{r_{k}}/2$. If $b_{r_{k}}=1$ and
$\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{r_{k}+i-1}}}{\beta^{i}}\in C_{0}$, let
$\omega_{k}=1$. If $b_{r_{k}}=1$ and
$\sum_{i=1}^{\infty}\frac{\vec{q}_{b_{r_{k}+i-1}}}{\beta^{i}}\in C_{1}$, let
$\omega_{k}=0$. Define $(\varphi^{\prime})^{-1}:\Upsilon^{\prime}\rightarrow
Z$ by

$(\varphi^{\prime})^{-1}((b_{1},b_{2},\ldots))=(w,\vec{z}).$

$(\varphi^{\prime})^{-1}$ is measurable by Theorem 2.2, and is the inverse of
$\varphi^{\prime}$. ∎

######  Lemma 6.8.

$\mathbb{P}(D)=1$, where $\mathbb{P}$ is the uniform product measure.

###### Proof.

The proof is similar with Theorem 4.5. For any sequence
$(b_{1},b_{2},\ldots)\in\Upsilon$ and $l\geq 3$, define

$\vec{z}_{l}=(x_{l},y_{l})=\frac{\vec{q}_{1}}{\beta}+\frac{\vec{q}_{2}}{\beta^{3}}+\frac{\vec{q}_{b_{1}}}{\beta^{l+1}}+\frac{\vec{q}_{b_{2}}}{\beta^{l+2}}+\cdots.$

Since $\beta\in[3/2,\beta^{*})$ and $\lim_{n\rightarrow
0}\frac{1}{\beta^{n}(\beta-1)}=0$, there exsits $L>0$ such that for any $l\geq
L$,

$\displaystyle x_{l}\geq\frac{1}{\beta},\ \ \ \ y_{l}\geq\frac{1}{\beta^{3}},$
$\displaystyle
x_{l}+y_{l}\leq\frac{1}{\beta}+\frac{1}{\beta^{3}}+\frac{1}{\beta^{n}(\beta-1)}\leq\frac{1}{\beta(\beta-1)}.$

It follows that $(x_{l},y_{l})\in\tilde{C}_{01}$. Notice that for any point
$(x,y)\in\cup_{n=1}^{\infty}f_{\vec{q}_{0}}f_{\vec{q}_{1}}^{n}(H)\bigcup\cup_{n=1}^{\infty}f_{\vec{q}_{1}}f_{\vec{q}_{0}}^{n}(H)$,
we have $y<1/\beta^{3}$, which implies that $(x_{l},y_{l})\in C_{01}\subset C$
for any $l\geq L$. Let

$D^{\prime}=\\{(b_{1},b_{2},\ldots)\in\Upsilon:b_{j}b_{j+1}\ldots
b_{j+L-1}=1\underbrace{00\ldots 0}_{N-1\text{ times }}\text{ for infinitely
many }j\text{'s}\\},$

then $D^{\prime}\subset D$. Since $\mathbb{P}(D^{\prime})=1$, we have
$\mathbb{P}(D)=1$. ∎

Now, consider the $K_{\beta}$-invariant measure $\nu_{\beta}$ defined on
$\mathcal{A}\times\mathcal{S}$ by $\nu_{\beta}(A)=\mathbb{P}(\varphi(Z\cap
A))$. We have the following theorem.

######  Theorem 6.9.

Let $3/2<\beta\leq\beta^{*}$. The dynamical systems $(\Omega\times
S_{\beta},\mathcal{A}\times\mathcal{S},\nu_{\beta},K_{\beta})$ and
$(\Upsilon,\mathcal{B},\mathbb{P},\sigma^{\prime})$ are isomorphic. Moreover,
the measure $\nu_{\beta}$ is the unique $K_{\beta}$-invariant measure of
maximal entropy.

###### Proof.

It follows from Lemmas 6.7 and 6.8 that $\varphi:(\Omega\times
S_{\beta},\mathcal{A}\times\mathcal{S},\nu_{\beta},K_{\beta})\rightarrow(\Upsilon,\mathcal{B},\mathbb{P},\sigma^{\prime})$
is an isomorphism, which implies that
$h_{\nu_{\beta}}(K_{\beta})=h_{\mathbb{P}}(\sigma^{\prime})=\log 3$ and any
other $K_{\beta}$-invariant measure with support $Z$ has entropy strictly less
than $\log 3$. Let $\mu$ be a $K_{\beta}$-invariant measure for which
$\mu(Z^{c})>0$. By a similar discussion with that of Lemma 4.9, we have
$h_{\mu}(K_{\beta})<\log 3.$ ∎

## Acknowledgements

The first and third author were supported by National Natural Science
Foundation of China (NSFC) No. 12071148 and Science and Technology Commission
of Shanghai Municipality (STCSM) No. 18dz2271000. The first author was also
supported by China Scholarship Council No.202006140156 during her visit to
Utrecht University.

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