A type III_1 Bernoulli shift

TL;DR

This paper constructs a specific product measure making the shift on a two-point space a type III_1 transformation, advancing understanding of ergodic theory and measure-preserving systems.

Contribution

It introduces a novel construction of a product measure that results in a type III_1 shift, providing new examples in ergodic theory.

Findings

Demonstrates existence of a type III_1 shift under a specific measure

Advances the classification of ergodic transformations

Provides a new example in the theory of measure-preserving systems

Abstract

We provide a construction of a product measure under which the shift on the two point product space is a type III_1 transformation.

Equations

$$P\circ T(A):=P\left(TA\right)=0.$$

$$P\circ T\left(A\right)=P(A).$$

$$P\circ T^n(A)={\int }_A\left(\frac{dP\circ T^n}{dP}\right)dP.$$

$$\forall A\in {\mathcal{B}}_{+}^{\mu },\forall ϵ>0,\exists n\in \mathbb{N},A\cap T^{-n}A\cap \left[{\left(T^n\right)}^{\prime }\doteq a\pm ϵ\right]\in {\mathcal{B}}_{+}^{\mu }.$$

$${\left(Tw\right)}_i=w_{i+1}.$$

$${\left[b\right]}_k^l=\left\{w\in \mathbb{X}:\forall i=k,...,l,w_i=b_i\right\}.$$

$$P\left({\left[b\right]}_k^l\right)=\prod _{j=k}^l{P}_j\left(\left\{b_j\right\}\right).$$

$$\sum _{k=-\infty }^{\infty }{\left(P_k\left(\left\{0\right\}\right)-P_{k-1}\left(\left\{0\right\}\right)\right)}^2<\infty .$$

$${\left(T^n\right)}^{\prime }(w)=\prod _{k=-\infty }^{\infty }\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}.$$

$$\forall i\ge 0,P_i\left(0\right)=P_i\left(1\right)=\frac{1}{2}.$$

$$N_t:=M_{t-1}+n_t.$$

$$M_t:=N_t+m_t.$$

$${\lambda }_t^{M_{t-1}}<e^{{ϵ}_t}.$$

$$A_{t-1}:=\left\{\prod _{u=1}^{t-1}{\lambda }_u^{x_u}:x_u\in \left[-n_u,n_u\right]\right\}.$$

$${\lambda }_t^{n_t/4}\ge \max \left\{a^2:a\in A_{t-1}\right\}.$$

$$N_t\left(2+2^{3N_t}\right)=m_t.$$

$$P_k\left(\left\{0\right\}\right)=P_k\left(\left\{1\right\}\right)=\frac{1}{2}.$$

$$P_k(\{0\})=1-P_k(\{1\})=\{\begin{array}{ll}\frac{1}{1+{\lambda }_t}& if-N_t<k\le M_{t-1}\\ \frac{1}{2}& \mathrm{for}\mathrm{all}-M_t<k\le -N_t\end{array}$$

$$\sum _{k=-\infty }^{\infty }{\left(P_k\left(\left\{0\right\}\right)-P_{k-1}\left(\left\{0\right\}\right)\right)}^2<\infty .$$

$$P_k\left(\left\{0\right\}\right)-P_{k-1}\left(\left\{0\right\}\right)=\{\begin{array}{ll}\frac{1}{2}-\frac{1}{1+{\lambda }_t}& \mathrm{if}k=-M_{t-1}+1\\ \frac{1}{1+{\lambda }_t}-\frac{1}{2}& \mathrm{if}k=-N_t+1\\ 0& \mathrm{otherwise}\end{array}.$$

$$\sum _{k=-\infty }^{\infty }{\left(P_k\left(\left\{0\right\}\right)-P_{k-1}\left(\left\{0\right\}\right)\right)}^2=2\sum _{t=1}^{\infty }{\left(\frac{{\lambda }_t-1}{2(1+{\lambda }_t)}\right)}^2.$$

$$\underset{n\to \infty }{\lim }\left[{\left(T^n\right)}^{\prime }(w)/\left(\prod _{k=-N_{t(n)}+1}^{n-1}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}\right)\right]=1\mathrm{uniformly}\mathrm{in}w\in \mathbb{X}.$$

$${\left(T^n\right)}^{\prime }(w)=\prod _{k=-\infty }^{\infty }\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}=\left(\prod _{k=-\infty }^{-N_{t(n)}}\times \prod _{k=-N_{t(n)}+1}^{n-1}\times \prod _{k=n}^{\infty }\right)\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}.$$

$$\prod _{k=n}^{\infty }\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}=1.$$

$$\prod _{k=-\infty }^{-N_{t(n)}}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}=\prod _{u=t(n)+1}^{\infty }\left(\prod _{k=-N_u+1}^{-M_{u-1}}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}\times \prod _{k=-M_u+1}^{-N_u}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}\right).$$

$$\prod _{k=-N_u+1}^{-M_{u-1}}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}=\prod _{k=-N_u+1}^{-N_u+n}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}.$$

$$-M_u=-N_u-m_u<k-n\le -N_u.$$

$$\prod _{k=-N_u+1}^{-M_{u-1}}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}=\prod _{k=-N_u+1}^{-N_u+n}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}$$

$$=\prod _{k=-N_u+1}^{-N_u+n}\frac{1/2}{{\lambda }_u^{w_k}/(1+{\lambda }_u)}=\frac{{\lambda }_u^{-{\sum }_{k=-N_u+1}^{N_u+n}{w}_k}}{{\left(2\left(1+{\lambda }_u\right)\right)}^n}.$$

$$\prod _{k=-M_u+1}^{-N_u}\frac{P_{k-n}\left(w_k\right)}{P_k\left(w_k\right)}={\lambda }_u^{{\sum }_{k=-M_u+1}^{-M_u+n}{w}_k}{\left(2\left(1+{\lambda }_u\right)\right)}^n.$$

Full text

This research was was supported by THE ISRAEL SCIENCE FOUNDATION grant No. 114/08.

0.1. Introduction:

A conservative non-singular transformation $T$ of the probability space $\left(X,\mathcal{B},P\right)$ for which there exists no $\sigma$-finite measure which is $P$-equivalent and $T$-invariant is said to be of type ${\rm III}$ [KW, Kri].

The first constructions of type ${\rm III}$ tranformations were given in [Orn, Arn]. Both constructions are odometers and hence don’t have the $K$-property.

In [Ham], Hamachi introduced a class of product measures $P={\displaystyle\prod_{k=-\infty}^{\infty}}P_{k}$ under which the full shift is a type ${\rm III}$ transformation. Thus obtaining the first examples of type ${\rm III}$ transformations which satisfy the $K$-property.

The classification of the type ${\rm III}$ dynamical systems was refined by Krieger [Kri] and Araki-Woods who introduced the ratio set [see below]. Thus the type ${\rm III}$ transformations can be further subdivided into type ${\rm III}_{\lambda}$, with $0\leq\lambda\leq 1$. Krieger showed [Kri, KW] that for each $0<\lambda\leq 1$ there is a unique orbit equivalence class.

In this work we construct a product measure $P={\displaystyle\prod_{k=-\infty}^{\infty}}P_{k}$ under which the full shift is of type ${\rm III}_{1}$.

Acknowledgments. This work is a part of the authors master thesis done under the supervision of Prof. Jon Aaronson. The author would like to thank Prof. Jon Aaronson for his time, patience and helpful suggestions.

0.2. Preliminaries.

Let $\left(X,\mathcal{B},P\right)$ be a probability space. A measurable transformation $T:X\to X$ is said to be non-singular with respect to $P$ if it preserves the $P-$null sets. i.e for every $A\in\mathcal{B},$ if $P(A)=0$ then

[TABLE]

A measure $m$ on $X$ is said to be $T-$invariant if for every* $A\in\mathcal{B}$,*

[TABLE]

The class of non-singular transformations correspond to automorphisms of the measure space* $(X,\mathcal{B},P)$. *

If $T$ is non-singular, then for every $n$, $P\circ T^{n}$ is absolutely continuous with respect to $P$. By the Radon-Nikodym theorem there exist measurable functions $\frac{dP\circ T^{n}}{dP}\in L_{1}(X,P)_{+}$ such that for every $A\in\mathcal{B}$ ,

[TABLE]

For convenience, we will use the notation $\left(T^{n}\right)^{\prime}(x):=\frac{dP\circ T^{n}}{dP}(x)$ .

Notation : Let $\left[f\doteq a\pm\epsilon\right]$ stand for $\left\{x:\ \left|f(x)-a\right|\leq\epsilon\right\}$. Similarly

$x\doteq e^{\pm a}$ will stand for $e^{-a}<x<e^{a}$.

For every set $A\in\mathcal{B}$, $\overline{A}$ will denote the compliment of $A$.

Unless otherwise stated all segments $[a,b]$ will stand for $[a,b]\cap\mathbb{Z}$.

If $\mu$ is a measure on $(X,\mathcal{B})$, denote by $\mathcal{B}_{+}^{\mu}$ the collection of sets with positive $\mu-$measure.

Definition**.**

The Ratio set $R(T,\mu)$: Let $\left(X,\mathcal{B},\mu,T\right)$ be a non-singular dynamical system. Let $a\geq 0$. We shall say that $a\in R(T,\mu)$ if

[TABLE]

The set $R(T,\mu)\cap\mathbb{R}_{+}$ is always a closed multiplicative subgroup of the positive real numbers. It follows from Maharam’s theorem [Mah] that if $T$ is conservative then $1\in R(T,\mu)$.

One can show that a system is of type ${\rm III}$ if and only if $0\in R(T,\mu)$.

Hence for a type ${\rm III}$ conservative system $R(T,\mu)$ is either $\{0,1\}$ ,

$\left\{\lambda^{n}:n\in\mathbb{Z}\right\}\cup\{0\}$ , for some $0<\lambda<1$ , or $[0,\infty)$ . These systems are denoted by ${\rm III}_{0}$, ${\rm III}_{\lambda}$ and ${\rm III}_{1}$ respectively.

0.3. The shift on the product space:

Let $\mathbb{X}=\{0,1\}^{\mathbb{Z}}$ and $T$ be the left shift action on $\mathbb{X}$, that is

[TABLE]

Denote the cylinder sets by

[TABLE]

Let $\mathcal{F}(k,l)$ denote the smallest $\sigma-$algebra which contains the cylinders $\left\{\left[b\right]_{k}^{l}:\ b\in{\displaystyle\prod_{j=l}^{k}}\{0,1\}\right\}$. For convenience write $\mathcal{F}(n)$ for $\mathcal{F}(-n,n)$.

A measure $P={\displaystyle\prod_{k=-\infty}^{\infty}}P_{k}\in\mathcal{P}(\mathbb{X})$ is called a product measure if for every $k<l$, and for every cylinder $[b]_{k}^{l},$

[TABLE]

The following claim is a direct consequence of Kakutani’s Theorem on equivalence of product measures[Kak].

Claim*.*

Let $P={\displaystyle\prod_{k=-\infty}^{\infty}P_{k}}$ be a product measure for which there exists $0<p<1$ such that for every $k\in\mathbb{Z}$, $p<P_{k}\left(\left\{1\right\}\right)<1-p.$ Then the shift is non-singular if and only if

[TABLE]

In addition for every $w\in\mathbb{X}$,

[TABLE]

1. A Type ${\rm III}_{1}$ shift

In this section we construct the product measure.

The product measure will be $P={\displaystyle\prod_{k=-\infty}^{\infty}}P_{k}$ , where

[TABLE]

The definition of $P_{k}-$ for negative $k$’s is more complicated as it involves an inductive procedure.

1.1. The inductive definition of $P_{k}$ for $k<0$:

We will need to define inductively 5 sequences $\left\{\lambda_{t}\right\}_{t=1}^{\infty},\left\{n_{t}\right\}_{t=1}^{\infty}$ , $\left\{m_{t}\right\}_{t=1}^{\infty},\ \left\{M_{t}\right\}_{t=0}^{\infty}$ and $\left\{N_{t}\right\}_{t=1}^{\infty}$ . The sequence $\left\{\lambda_{t}\right\}$ is of real numbers which decreases to $1$. The other four, $\left\{n_{t}\right\}_{t=1}^{\infty}$ , $\left\{m_{t}\right\}_{t=1}^{\infty},\ \left\{M_{t}\right\}_{t=0}^{\infty}$ and $\left\{N_{t}\right\}_{t=1}^{\infty}$ are increasing sequences of integers.

First choose a positive summable sequence $\left\{\epsilon_{t}\right\}_{t=1}^{\infty}$ and set $M_{0}=1$.

Base of the induction: Set $\lambda_{1}=e$ , $n_{1}=2$ , $m_{1}=4$ . Set also $N_{1}=M_{0}+n_{1}=3$ and $M_{1}=N_{1}+m_{1}=7.$

Given $\left\{\lambda_{u},n_{u},N_{u},m_{u},M_{u}\right\}_{u=1}^{t-1}$ , we will choose the next level $\left\{\lambda_{t},n_{t},N_{t},m_{t},M_{t}\right\}$ in the following order. First we choose $\lambda_{t}$ depending on $M_{t-1}$ and $\epsilon_{t}$. Given $\lambda_{t}$ we will choose $n_{t}$ and then $N_{t}$ will be defined by

[TABLE]

Then given $N_{t}$ we will choose $m_{t}$ and finally set

[TABLE]

Choice of $\lambda_{t}$: Set $k_{t}:=\left\lfloor\log_{2}\left(\frac{M_{t-1}}{\epsilon_{t}}\right)\right\rfloor+1$ , where $\left\lfloor x\right\rfloor$ denotes the integral part of $x$. Then set $\lambda_{t}=e^{\frac{1}{2^{k_{t}}}}$. With this choice of $\lambda_{t}$ we have,

[TABLE]

This choice of $\lambda_{t}$ has the property that for every $u<t$, $\lambda_{u}=\lambda_{t}^{2^{k_{t}-k_{u}}}.$

Define

[TABLE]

The set $A_{t-1}$ has the following properties:

1. (1)

$a\in A_{t-1}$ if and only if $\frac{1}{a}\in A_{t-1}$. 2. (2)

By choosing $x_{t}=0$, one can see that for every $t\in\mathbb{N}$, $A_{t-1}\subset A_{t}$. 3. (3)

Since for every $u<t$, $\lambda_{t}=\lambda_{u}^{2^{k_{t}-k_{u}}}$, then $A_{t-1}$ is a subset of

$\lambda_{t}^{\mathbb{Z}}=\left\{\lambda_{t}^{l}:\ l\in\mathbb{Z}\right\}$ . In addition $A_{t-1}$ is a finite set.

Choice of $n_{t}$: Given $\{\lambda_{u},n_{u},m_{u}\}_{u=1}^{t-1}$ and $\lambda_{t}$, the set $A_{t-1}$ is a finite subset of $\lambda_{t}^{\mathbb{Z}}$. Choose $n_{t}$ large enough so it satisfies the following property:

**Property ***: $n_{t}$ is large enough so that for every $a,b\in A_{t-1}$, there exists $p=p(a,b)\in\left(\frac{-n_{t}}{4},\frac{n_{t}}{4}\right)$ for which $\lambda_{t}^{p}=b\cdot a$.

This is possible since $a,b\in A_{t-1}\subset\lambda_{t}^{\mathbb{Z}}$.

Notice that this property is equivalent to

[TABLE]

Choice of $m_{t}$: Now that $n_{t}$ is chosen we set $N_{t}=M_{t-1}+n_{t}$ . Set

[TABLE]

1.2. Definition of the product measure

Let $P={\displaystyle\prod_{k=-\infty}^{\infty}}P_{k}$ , where for $k\geq 0$,

[TABLE]

And for $k<0$ ,

[TABLE]

Define a function $t(\cdot):\mathbb{N}\to\mathbb{N}$ by $t(n)=\min\left\{t\in\mathbb{N}:n<N_{t}\right\}$ .

1.2.1. Statement of the main theorem:

Theorem 1**.**

The shift $T:\left(\mathbb{X},\mathcal{B}(\mathbb{X}),P\right)\circlearrowleft$ is non-singular and of type $III_{1}$.

Remark*.*

By Kakutani’s theorem using the fact that for every

k\in\mathbb{Z},$$\frac{1}{1+e}<P_{k}\left(\{1\}\right)<\frac{e}{1+e} , the non-singularity of the shift under $P$ is equivalent to

[TABLE]

Furthermore it follows from the structure of $P$ that

[TABLE]

Therefore,

[TABLE]

This sum converges or diverges together with $\sum_{t=1}^{\infty}\left(\log\lambda_{t}\right)^{2}$ which by (1.2) is less then $\sum\frac{\epsilon_{t}}{M_{t-1}}<\infty$. Therefore the shift is non-singular with respect to $P$ and we can calculate the Radon-Nikodym derivatives of $T$.

1.3. Radon-Nikodym derivatives of the shift:

Lemma 2**.**

If $P$ is a product measure which is built by the inductive construction then

[TABLE]

Proof.

By Kakutani’s theorem, for every $w\in\mathbb{X}$,

[TABLE]

For every $k>n$ and for every $w_{k}\in\{0,1\}$, $P_{k-n}\left(w_{k}\right)=P_{k}\left(w_{k}\right)=\frac{1}{2}$. Therefore

[TABLE]

Also

[TABLE]

For every $u>t(n)$, the length of the segment $\left(-N_{u},-M_{u-1}\right]$ is $n_{u}$ and the length of the segment $\left(-M_{u},-N_{u}\right]$ is $m_{u}$. Both $n_{u}$ and $m_{u}$ are greater than $N_{t(n)}$ , hence greater than $n$.

Therefore since for every $k\in\left(-N_{u}+n,-M_{u-1}\right]$, $P_{k-n}=P_{k}=\left(\frac{1}{1+\lambda_{u}},\frac{\lambda_{u}}{1+\lambda_{u}}\right)$ then

[TABLE]

Since $n<m_{u}$ then for every $k\in\left(-N_{u},N_{u+n}\right]$,

[TABLE]

So $P_{k-n}=\left(\frac{1}{2},\frac{1}{2}\right)$ and $P_{k}=\left(\frac{1}{1+\lambda_{u}},\frac{\lambda_{u}}{1+\lambda_{u}}\right)$. In conclusion, for every $u>t(n)$,

[TABLE]

[TABLE]

Similarly for every $u>t(n)$,

[TABLE]

Putting together (1.5), (1.6) and (1.7) we have

[TABLE]

[TABLE]

Since for every $u>t(n).\ n<M_{u-1}$ , it follows that

[TABLE]

So $\prod_{k=-\infty}^{-N_{t(n)}}\frac{P_{k-n}\left(w_{k}\right)}{P_{k}\left(w_{k}\right)}\doteq e^{\pm\sum_{u=t(n)+1}^{\infty}\epsilon_{u}}\to 1$ as $n\to\infty$, since $\sum\epsilon_{t}<\infty$.

The conclusion follows. ∎

Definition**.**

For every $u\in\mathbb{N}$ , write

[TABLE]

Set also

[TABLE]

$f_{t}(\cdot)$ is $\mathcal{F}\left(-N_{t},0\right)$ measurable.

The next lemma uses the previous lemma to show that if $n$ belongs to an interval of the form $\left[N_{t-1},M_{t-1}\right)$ then we can obtain a reasonable approximation of $\left(T^{n}\right)^{\prime}$ . This approximation becomes more accurate as $t\to\infty$. Notice that for $n\in\left[N_{t-1},M_{t-1}\right)$, $t=t(n)$.

Lemma 3**.**

For every $N_{t-1}\leq n<M_{t-1}$ and for every $w\in\mathbb{X}$,

[TABLE]

Where $\lim_{n\to\infty}K_{n}(w)=1$ uniformly in $w\in\mathbb{X}$.

Proof.

By lemma 2 it is enough to show that for $N_{t-1}\leq n<m_{t-1}$,

[TABLE]

Where $\lim_{n\to\infty}h_{n}(w)=1$ uniformly in $w\in\mathbb{X}$ .

First notice that because of the product structure

[TABLE]

And

[TABLE]

By (1.1) , $P\left(\left[w\right]_{0}^{n-1}\right)=\frac{1}{2^{n}}$. By (1.4) ,

[TABLE]

[TABLE]

Multiplying all the terms

[TABLE]

Similarly

[TABLE]

[TABLE]

By the definition of $P_{k}$ for negative $k$,

[TABLE]

therefore

[TABLE]

[TABLE]

Combining the estimates (1.8) and (1.9) , we get ,

[TABLE]

The conclusion will follow once we show that for every

$\begin{gathered}N_{t-1}\leq n<M_{t-1}\ ,\ h_{n}(w):=\lambda_{t(n)}^{\Upsilon_{t(n)}\circ T^{n}(w)-\Upsilon_{t(n)}(w)}\to 1\end{gathered}$ uniformly in $w\in\mathbb{X}$ as $n\to\infty$.

Notice that,

[TABLE]

[TABLE]

Therefore

[TABLE]

The conclusion follows. ∎

The following two remarks will be useful in the proof of type ${\rm III}_{1}$.

Remark*.*

Since ${\rm Image}\left(\Upsilon_{u}\circ T^{n}-\Upsilon_{u}\right)=\left[-n_{u},n_{u}\right]$, the set $A_{t}$ is the set of all values of

[TABLE]

Since for every $t_{0}<t$, $\lambda_{t_{0}}\in A_{t-1}$ and also $\frac{1}{f_{t-1}(w)}\in A_{t}$ we can formulate (property *) in the following way.

Remark 4*.*

It follows from property (*) that $n_{t}$ is large enough so that for every $w\in\mathbb{X}$, $t_{0}<t$ and $k\in\mathbb{N}$ there exists $p=p(w)\in\left(-\frac{n_{t}}{4},\frac{n_{t}}{4}\right)$ so that

[TABLE]

1.4. The proof of type ${\rm III}_{1}$

Since $\lambda_{t}\downarrow 1$ , and $\mathcal{R}(T)$ is always a closed multiplicative subgroup of $\mathbb{R}_{+}$, the following lemma will yield that $\left(\mathbb{X},\mathcal{B},P,T\right)$ is of type $III_{1}$.

Lemma 5**.**

For every $t\in\mathbb{N}$, $\lambda_{t}$ belongs to the ratio set of $T$.

The proof of lemma 5 will be a series of lemmas. We will show that $\lambda_{t}$ satisfies an $EVC$ property as in [ALV] , where the generating partitions of $\mathbb{X}$, will be cylinders $\{\left[b\right]_{-n}^{n}:b\in\{-1,1\}^{n}\}$ and the neighborhoods of $\lambda_{t}$ will be $U_{\epsilon}=\left[\lambda_{t}-\epsilon,\lambda_{t}+\epsilon\right]$.

Lemma 6**.**

For every $t_{0}\in\mathbb{N}$ the following holds:

For every cylinder set $B=[b]_{-n}^{n}$ and for every $\epsilon>0$, there exists a $\tau=\mathfrak{\tau}(B,t_{0},\epsilon)\in\mathbb{N}$, for which

[TABLE]

Throughout the proofs and the lemmas the letter $n$ will denote the radius of the cylinder.

Since the proof is delicate and involves many details we give a sketch of the idea of the proof.

Sketch of the proof of Lemma 6: Let $B=[b]_{-n}^{n}$ , $t_{0}\in\mathbb{N}$ and $\epsilon>0$.

We will first choose $\mathfrak{\tau}$ large enough so that for every $l=2,\ldots,\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}$,

[TABLE]

Since $P$ is a product measure and $f_{\mathfrak{\tau}}$ is $\mathcal{F}\left(-N_{\mathfrak{\tau}},0\right)$ measurable, if we restrict our attention to cylinders of the form $C=[c]_{-N_{\mathfrak{\tau}}}^{0}$then

[TABLE]

So we will decompose $B$ to $\left\{B\cap[c]_{-N_{\mathfrak{\tau}}}^{0}:c\in{\displaystyle\prod_{k=-N_{\mathfrak{\tau}-1}}^{0}\{0,1\}}\right\}$ and prove that for the cylinders $C=[c]_{-N_{\mathfrak{\tau}}}^{0}$ which cover most of $\mathbb{X}$,

[TABLE]

Then we make use of the independence of $B\cap C,\left\{T^{-lN_{\mathfrak{\tau}}}B\right\}_{l=2}^{\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}}$ and $\left\{\left[f_{\mathfrak{\tau}}\circ T^{lN_{\mathfrak{\tau}}}=f_{\mathfrak{\tau}}(c)\right]\right\}_{l=2}^{\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}}$ to get

[TABLE]

Afterwards we will finish the proof by summing over $C$.

Fact 7**.**

Since for every $u\in\mathbb{N}$, $1<\lambda_{u}\leq e$, we have that for every $u\in\mathbb{N}$,

[TABLE]

and since $\Upsilon_{u}(w)\sim{\rm Bin}\left(n_{u},\frac{\lambda_{u}}{1+\lambda_{u}}\right)$, the strong law of large numbers implies that

[TABLE]

Write $\mathcal{C}_{t}$ for the collection of cylinders of the form $C=\left[c\right]_{-N_{t}}^{-1}$ such that

[TABLE]

It follows that

[TABLE]

Furthermore since $f_{t}$ is $\mathcal{F}\left(-N_{t},0\right)$ measurable then for all cylinders $C=[c]_{-N_{t}}^{0}\in\mathcal{C}_{t}$,

[TABLE]

Lemma 8**.**

Let $t_{0}\in\mathbb{N}$. For every $t$ such that $t>t_{0}+1$ the following holds:

For every cylinder $C=[c]_{-N_{t}}^{0}\in\mathcal{C}_{t}$ and for every $l=2,\ldots,\frac{m_{t}}{N_{t}}$,

[TABLE]

Proof.

Let $C\in\mathcal{C}_{t}$ and $2\leq l\leq\frac{m_{t}}{N_{t}}$.

We will build a cylinder set $D=D(l,C,t_{0})=\left[d\right]_{\left(l-1\right)N_{t}+1}^{lN_{t}}$ such that

[TABLE]

Notice that

[TABLE]

Set

[TABLE]

It is easy to see that for every $u<t-1$, $\Upsilon_{u}\circ T^{lN_{t}}$ is $\mathcal{F}\left(\left(l-1\right)N_{t}+M_{t-1},lN_{t}\right]$ measurable and therefore for every $w\in D,u<t-1$,

[TABLE]

So for every $w\in D,$

[TABLE]

and

[TABLE]

Since $n_{t}$ satisfies property (*) [ see Remark 4], there exists a $p\in\left(-\frac{n_{t-1}}{4},\frac{n_{t-1}}{4}\right)$ such that

[TABLE]

Also since $C\in\mathcal{C}_{t}$ , $\Upsilon_{t}(c)\in\left(\frac{n_{t}}{4},\frac{3n_{t}}{4}\right)$ and so

[TABLE]

Since $\Upsilon_{t}\circ T^{lN_{t}}(w)$ is $\mathcal{F}\left(\left(l-1\right)N_{t-1}+1,\left(l-1\right)N_{t}+M_{t-1}-1\right)$ measurable this means that there exists

[TABLE]

such that

[TABLE]

By setting $d_{k}=\tilde{w}_{k}$ for every $lN_{t}-N_{t-1}\leq k\leq lN_{t}-M_{t-2}$ we have finished the construction of $D$.

Observe that for every $w\in D$,

[TABLE]

[TABLE]

Therefore

[TABLE]

Since $P(D)={\displaystyle\prod_{k=lN_{\mathfrak{\tau}}}^{\left(l+1\right)N_{\mathfrak{\tau}}}}P_{k}\left(d_{k}\right)=\frac{1}{2^{N_{t}}}$ the lemma follows.

∎

Notation: Given a cylinder $C=[c]_{-N_{t-1}}^{0}$ , $t_{0}\in\mathbb{N}$ and $l=2,\ldots,\frac{m_{t}}{N_{t}}$ , we will abuse notation and write $D(C,t_{0},l)$ for $\left\{w:f_{t}\circ T^{lN_{t}}(w)=f_{t}(c)\cdot\lambda_{t_{0}}\right\}.$

Lemma 9**.**

Let $B=[b]_{-n}^{n}$, and $t_{0}\in\mathbb{N}$. Then for every $t\in\mathbb{N}$ such that $2n<N_{t}$ and $t_{0}<t$ the following holds:

For every $C\in\mathcal{C}_{t}$

[TABLE]

Proof.

Notice that $f_{t}$ is $\mathcal{F}\left(-N_{t-1},0\right)$ measurable and therefore for every $l\leq\frac{m_{t}}{N_{t}}$, $f_{t}\circ T^{lN_{t}}$ is $\mathcal{F}\left(\left(l-1\right)N_{t},lN_{t}\right)$ measurable. By the product structure of $P$ , the random variables $\left\{f_{t}\circ T^{lN_{t}}\right\}_{l=0}^{\frac{m_{t}}{N_{t}}}$ are independent. Therefore the sets $\left\{D\left(C,t_{0},l\right)\right\}_{l=2}^{\frac{m_{t}}{N_{t}}}$ are independent and each one is measurable $\mathcal{F}(N_{t},\infty).$

Since $B\in\mathcal{F}(-n,n)$, then for every $l\geq 2$, $T^{-lN_{t}}B\in\mathcal{F}\left(-n-lN_{t},n-lN_{t}\right)$. Since $2n<N_{t}$ the sets $\left\{T^{-lN_{t}}B\right\}_{l=2}^{\frac{m_{t}}{N_{t}}}$ are independent and independent of $\mathcal{F}\left(-N_{t},\infty\right)$. So the sets $\left\{D\left(C,t_{0},l\right)\right\}_{l=2}^{\frac{m_{t}}{N_{t}}},(B\cap C)$ and $\left\{T^{-lN_{t}}B\right\}_{l=2}^{\frac{m_{t}}{N_{t}}}$ are independent.

Therefore for every $l=2,...,\frac{m_{t}}{N_{t}}$

[TABLE]

Plugging in Lemma 8 we have

[TABLE]

Furthermore, since for every $2\leq l\leq\frac{m_{t}}{N_{t}}-1,$

[TABLE]

the product structure of $P$ (see (1.4)) implies

[TABLE]

Therefore

[TABLE]

Since the sets $\left\{T^{-lN_{t}}B\cap D\left(C,t_{0},l\right)\right\}_{l=2}^{\frac{m_{t}}{N_{t}}},\left(B\cap C\right)$ are independent ( and (1.16)) ,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

∎

Proof.

(Lemma 6): Let $B=[b]_{-n}^{n}$ be fixed, $\epsilon>0$ and $t_{0}\in\mathbb{N}$. We want to choose a $\mathfrak{\tau}\in\mathbb{N}$ such that (1.10) holds.

The choice of $\mathfrak{\tau}$: Choose a $\mathfrak{\tau}\in\mathbb{N}$ which satisfies the following properties:

[TABLE]

(By this condition $\lambda_{t_{0}}\in A_{\mathfrak{\tau}}$ )

[TABLE]

[TABLE]

(so “nice” cylinders exhaust enough of $B$) and

[TABLE]

By Lemma 3 we can, by enlarging $\mathfrak{\tau}$ if needed , demand in addition to (1.17).(1.18) and (1.19) that $\mathfrak{\tau}$ satisfies the following : For every

$N_{\mathfrak{\tau}}\leq k<M_{\mathfrak{\tau}}$,

[TABLE]

This means (derivative approximation),

[TABLE]

We will now prove that this $\mathfrak{\tau}$ satisfies (1.10) with $B,\epsilon$ and $t_{0}$.

First we will show that for every $C=[c]_{-N_{\mathfrak{\tau}}}^{0}\in C_{\mathfrak{\tau}}$,

[TABLE]

Since $\mathfrak{\tau}$ satisfies (1.20) and $\lambda_{t_{0}}<3$ then for every $l=2,\ldots,\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}$,

[TABLE]

Since $f_{\mathfrak{\tau}}$ is $\mathcal{F}\left(-N_{\mathfrak{\tau}},0\right)$ measurable,

[TABLE]

[TABLE]

Therefore $\cup_{l=1}^{\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}}\left\{\left(B\cap C\right)\cap T^{-lN_{\mathfrak{\tau}}}B\cap D\left(C,t_{0},l\right)\right\}$ is a subset of

$\cup_{l=1}^{\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}}\left\{\left(B\cap C\right)\cap T^{-lN_{\mathfrak{\tau}}}B\cap\left[\left(T^{lN_{\mathfrak{\tau}}}\right)^{{}^{\prime}}\doteq\lambda\pm\epsilon\right]\right\}$.

So by Lemma 8,

[TABLE]

[TABLE]

[TABLE]

[TABLE]

The lemma follows by ( the last inequality uses 1.18)

[TABLE]

[TABLE]

∎

Then next lemma is a standard lemma in measure theory. The proof is included here for the sake of completeness,

Lemma 10**.**

For all $\epsilon>0$ and for every $A\in\mathcal{B}_{+}$ there exists a cylinder of the form $B=\left[b_{-n},...,b_{n}\right]_{0}$ for which

[TABLE]

Proof.

Define a partition $\Sigma_{n}=\left\{Cylinders\ of\ the\ form\ \left[b_{-n},...,b_{n}\right]_{0}\right\}$ . Denote by $F_{n}=\sigma\left(\Sigma_{n}\right)$ the sigma algebra generated by $\Sigma_{n}$. Then $\mathcal{F}_{n}\uparrow\mathcal{B}$ therefore for every $A\in\mathcal{B}_{+}$ we have

[TABLE]

This means that if $P(A)>0$ then there exists an $n\in\mathbb{N}$ and an atom of $\Sigma_{n}$ which we will denote by $B_{n}$ for which

[TABLE]

But

[TABLE]

By setting $B=B_{n}$ the lemma is proved.

∎

Proof.

(Lemma 5),

Let $t_{0}\in\mathbb{N}$ such that $\lambda_{t_{0}}<1.4$, we will show that $\lambda_{t_{0}}$ belongs to the ratio set of the shift.

Let $0<\epsilon<0.1$ and $A\in\mathcal{B}_{+}$ . By lemma 10 there exists a cylinder $B=\left[b\right]_{-n}^{n}$ for which

[TABLE]

By Lemma 6 there exists $\mathfrak{\tau}\in\mathbb{N}$ for which,

[TABLE]

Set $\mathcal{D}=\cup_{l=1}^{\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}}\left[B\cap T^{-lN_{\mathfrak{\tau}}}B\cap\left\{\left(T^{lN_{\mathfrak{\tau}}}\right)^{{}^{\prime}}(\cdot)\doteq\lambda_{t_{0}}\pm\epsilon\right\}\right]$ and define a map

$\phi:\mathcal{D}\to\mathbb{N}$ by,

[TABLE]

Define an automorphism $S:\mathcal{D}\to B$ by

[TABLE]

Write $\mathfrak{J}$ for the image $S(\mathcal{D}\cap A)$. We have $\mathcal{D},\mathcal{\mathfrak{J}}\subset B$ , and

[TABLE]

Moreover $0.9<\left(S\right)^{\prime}(w)=\lambda_{t_{0}}\pm\epsilon<1.5,$ and therefore

[TABLE]

Also

[TABLE]

Therefore

[TABLE]

This together with $\left(S\right)^{\prime}(w)<1.5$ means that

[TABLE]

So

[TABLE]

[TABLE]

Whence there is an $2\leq l\leq\frac{m_{\mathfrak{\tau}}}{N_{\mathfrak{\tau}}}$, so that

[TABLE]

Hence $\lambda_{t_{0}}$ belongs to the ratio set of $T$ . ∎