A new ergodic proof of a theorem of W. Veech

TL;DR

This paper presents a novel ergodic proof of a classical theorem by W. Veech, utilizing methods developed in the authors' prior research to offer new insights into the result.

Contribution

The paper introduces a new ergodic proof of Veech's theorem, expanding the methodological toolkit for ergodic theory.

Findings

Provides a new ergodic proof of Veech's theorem

Builds upon previous methods in ergodic theory

Enhances understanding of Veech's result

Abstract

Our goal in the present paper is to give a new ergodic proof of a well-known Veech's result, build upon our previous works.

Equations

$$w_n:={\Phi }_{r_n}({\Phi }_{r_{n-1}}(\dots ({\Phi }_{r_2}({\Phi }_{r_1}(x)))\dots ))$$

$$\Psi :X\times Y\to X\times Y\text{defined by}\Psi (x,r):=({\Phi }_{r_1}(x),T(r))$$

$${\Phi }_{\infty }\equiv I{d}_X(I{d}_X\text{the identity on}X).$$

$$d(\sigma ,\nu ):=\sum _{n=1}^{\infty }\frac{|\int f_{n}d\sigma -\int f_{n}d\nu |}{2^n\Vert f_n\Vert }$$

$$\phi :M(X)\to M(X)\text{given by}(\phi \sigma )(B)=\sigma ({\Phi }^{-1}(B))$$

$${\mu }_m^{\nu }:=\frac{1}{m_G(F_m)}{\int }_{F_m}\phi (\nu )d{m}_G(\Phi )$$

$${\int }_{X}f(x)d{\mu }_m^{\nu }(x):=\frac{1}{m_G(F_m)}{\int }_{F_m}{\int }_{X}f(\Phi (x))d\nu (x)d{m}_G(\Phi )$$

$$d({\mu }_{m_n}^{{\nu }_n},\mu )>\epsilon .$$

$${\int }_{X}f(x)d{\mu }_{m_n}^{{\nu }_n}(x):=\frac{1}{m_G(F_{m_n})}{\int }_{F_{m_n}}{\int }_{X}f(\Phi (x))d{\nu }_n(x)d{m}_G(\Phi )$$

$${\int }_{X}f(x)dh({\mu }_{m_n}^{{\nu }_n}(x)):$$

$$=\frac{1}{m_G(F_{m_n})}{\int }_{H{F}_{m_n}}{\int }_{X}f(\Phi (x))d{\nu }_n(x)d{m}_G(\Phi ).$$

$$\displaystyle\bigg{|}\int_{X}f(x)\,d\mu^{\nu_{n}}_{m_{n}}(x)$$

$$\le \frac{1}{m_G(F_{m_n})}{\int }_{F_{m_n}△H{F}_{m_n}}{\int }_X|f(\Phi (x)|d{\nu }_n(x)d{m}_G(\Phi )$$

$$\le \frac{m_G(F_{m_n}△H{F}_{m_n})}{m_G(F_{m_n})}\Vert f{\Vert }_{\infty }\to 0\text{for}n\to \infty .$$

$$Q_{k}:=(X\backslash E_{k})\Big{\backslash}\bigcup_{\varPhi\in G}\varPhi^{-1}(E_{k})$$

$$x\in X\big{\backslash}\bigcap^{\infty}_{k=1}E_{k}.$$

$$d({\rho }_m(\sigma ),\mu )\to 0\text{uniformly for}\sigma \in M(X).$$

$${\int }_{X}fd{\mu }_m^{\sigma }:=\frac{1}{m_G(F_m)}{\int }_{F_m}{\int }_{X}f(\Phi (x))d\sigma d{m}_G(\Phi )\text{for}f\in C(X)$$

$$d({\mu }_m^{\sigma },\mu )<{\epsilon }_m\text{for}m=1,2,\dots \text{and}\sigma \in M(X).$$

$$g_n^i:G\to \mathbb{R},\text{where}{g}_n^i(\Phi )={\int }_X{f}_n\circ \Phi (x)d{\delta }_{x_i}.$$

$${\int }_X{f}_{n}d{\mu }_m^{{\delta }_{x_i}}=\frac{1}{m_G(F_m)}{\int }_{F_m}{g}_n^i(\Phi )d{m}_G.$$

$$\sum _{k=1}^{k_m}{\lambda }_k{\delta }_{{\Phi }_{{\ell }_k}},{\Phi }_{{\ell }_k}\in B,k=1,\dots ,k_m$$

$$\bigg{|}\frac{1}{m_{G}(F_{m})}\int_{F_{m}}g^{i}_{n}(\varPhi)dm_{G}-\sum^{k_{m}}_{k=1}\lambda_{k}g^{i}_{n}(\varPhi_{\ell_{k}})\bigg{|}\leq\varepsilon_{m}\cdot\|f_{n}\|.$$

$$\displaystyle\bigg{|}\int_{X}f_{n}d\mu^{\delta_{x_{i}}}_{m}-\sum^{k_{m}}_{k=1}\lambda_{k}\int_{X}f_{n}\circ\varPhi_{\ell_{k}}(y)d\delta_{x_{i}}\bigg{|}\leq\varepsilon_{m}\cdot\|f_{n}\|.$$

$$\bigg{|}\int_{X}f_{n}d\mu^{\delta_{x_{i}}}_{m}-\int_{X}f_{n}d\rho_{m}(\delta_{x_{i}})\bigg{|}\leq\varepsilon_{m}\cdot\|f_{n}\|.$$

$$d({\mu }_m^{{\delta }_{x_i}},{\rho }_m({\delta }_{x_i}))$$

$$<{\epsilon }_m+\frac{1}{2^{m-1}}.$$

$$d({\rho }_m({\delta }_{x_i}),\mu )<2{\epsilon }_m+\frac{1}{2^{m-1}}.$$

$$\frac{1}{2^{m-1}}<\epsilon \text{and}{\epsilon }_m<\epsilon \text{for}m>m_0.$$

$$|f_n(x)-f_n(x^{\prime })|<\epsilon \cdot \Vert f_n\Vert \text{for}n=1,\dots ,m_0$$

$$v({\Phi }_{\ell }(y),{\Phi }_{\ell }(y^{\prime }))<\delta \text{for every}{\Phi }_{\ell }\in B.$$

$$\bigg{|}\sum^{k_{m}}_{k=1}\lambda_{k}\int_{X}f_{n}\circ\varPhi_{\ell_{k}}(y)d\delta_{x_{i_{\ast}}}-\sum^{k_{m}}_{k=1}\lambda_{k}\int_{X}f_{n}\circ\varPhi_{\ell_{k}}(y)d\delta_{x}\bigg{|}<\varepsilon\cdot\|f_{n}\|$$

Full text

A new ergodic proof of a theorem of W. Veech ††footnotetext: P. Georgopoulos††footnotetext: Department of Mathematics, University of Athens, 15784, Athens, Greece††footnotetext: email: pangeorgopoul@gmail.com & pgeorgop@math.uoa.gr

Panagiotis Georgopoulos

Abstract. Our goal in the present paper is to give a new ergodic proof of a well-known Veech’s result, build upon our previous works [4,5].

Keywords: Invariant measure $\cdot$ Skew product $\cdot$ Uniformly distributed sequence $\cdot$ Uniquely ergodic and non-sensitive action $\cdot$ amenable group $\cdot$ Bernoulli shift.

Mathematics Subject Classification (2010) Primary 28D15, 37B05, 43A07; Secondary 11K06.

1 Introduction

W. Veech in his remarkable paper [11, Theorem 3] (see also [7, p. 235] and [8, Commentary of Problem 116, p. 203]), proved the following:

“Almost all” sequences $(r_{1},\ldots,r_{n},\ldots)$ of positive integers have the following “universal” property: Whenever $G$ is a compact separable group and $z_{1},z_{2},\ldots,z_{n},\ldots$ a sequence of elements of $G$ that generates a dense subgroup of $G$, then the sequence $y_{1},y_{2},\ldots,y_{n},\ldots$, where $y_{n}:=z_{r_{1}}\cdot z_{r_{2}}\ldots z_{r_{n}}$ is uniformly distributed for the Haar measure on G. Veech called such sequences, “uniformly distributed sequence generators”.

In [5] we prove that:

“Almost all” sequences $(r_{1},\ldots,r_{n},\ldots)$ of positive integers have the following “universal” property: Whenever $(X,\mu)$ is a Borel probability measure, compact metric space and $\varPhi_{1},\varPhi_{2},\ldots,\varPhi_{n},\ldots$ a sequence of continuous, measure preserving maps on $(X,\mu)$, such that the action (by composition) on $(X,\mu)$ of the semigroup with generators $\varPhi_{1},\ldots,\varPhi_{n},\ldots$ is amenable (as discrete), uniquely ergodic and non-sensitive on $\text{supp}\mu$, then for every $x\in X$ the sequence $w_{1},w_{2},\ldots,w_{n},\ldots$ where

[TABLE]

is uniformly distributed for $\mu$.

In the present paper we prove the next most special, albeit not direct, corollary of [5].

“Almost all” sequences $(r_{1},\ldots,r_{n},\ldots)$ of positive integers have the following “universal” property: Whenever $G$ is a locally compact, amenable, separable group acting (continuously) on $(X,\mu)$ (a Borel probability measure compact metric space), by measure preserving homeomorphisms, such that the action is uniquely ergodic for $\mu$ and non-sensitive on $\mbox{supp}\mu$ (it turns out that such an action is necessarily equicontinuous) and if $\varPhi_{n}$, $n\in\mathbb{N}$ is a sequence in $G$ that generates (by composition) a dense semigroup in $G$ and $x\in X$, then the sequence $w_{n}:=\varPhi_{r_{n}}(\varPhi_{r_{n-1}}(\ldots(\varPhi_{r_{2}}(\varPhi_{r_{1}}(x)))\ldots))$, $n\in\mathbb{N}$ is uniformly distributed for $\mu$.

This completes investigation of [4,5] and gives Veech’s theorem, at least for metrizable groups.

The new element in the present paper is Proposition 4.1 that allows us to use a combination of the methods of [4,5]. In fact, in many aspects, most parts of the arguments of [4,5] are much simpler.

Next, let us explain how Veech’s theorem falls in the frame of the above result.

Clearly, $G$ acts on $G$ (uniformly equicontinuously) by multiplication, i.e. for $g\in G$, $x\in G$, $(x,g)\mapsto x\cdot g$, $G$ is amenable (as compact) and the Haar measure $m_{G}$ is the unique invariant measure for this action. Also, the assumption that $z_{1},z_{2},\ldots,z_{n},\ldots$ generate a dense subgroup of $G$, implies that the action of this subgroup on $G$ (by right translations) is uniquely ergodic for $m_{G}$.

On the other hand, the assumption that $z_{1},z_{2},\ldots,z_{n},\ldots$ generate a dense subgroup of $G$, is equivalent to the assumption that $z_{1},z_{2},\ldots,z_{n},\ldots$ generate a dense semigroup in $G$ (see [6, Theorem 9.16]).

Under these circumstances for $G$ metrizable, in view of our result (in particular for $x=e$) the sequence $y_{n}:=z_{r_{1}}\cdot z_{r_{1}}\ldots z_{r_{n}}$, $n\in\mathbb{N}$ is uniformly distributed for $G$.

And a final remark: The general case, where the group $G$ is not necessarily metrizable, can be treated by similar methods, since the topology of $G$ is defined by a family of pseudometrics (see [3, Chapter IX, Section 11]).

2 The main results

Throughout this paper $(p_{1},\ldots,p_{n},\ldots)$ is a probability sequence with non-zero entries (i.e. $p_{n}>0$ for each $n$ and $\sum\limits^{\infty}_{n=1}p_{n}=1$). We consider now the set of natural numbers $\mathbb{N}=\{1,2,\ldots\}$ endowed with the discrete topology. Then, we take the one-point compactification of $\mathbb{N}$ and we get the compact space $\widetilde{\mathbb{N}}:=\mathbb{N}\cup\{\infty\}$. Let $(\widetilde{\mathbb{N}},m)$ be the measure space, where $m$ is a probability measure on $\widetilde{\mathbb{N}}$, defined by $m(\{n\})\!\!=\!\!p_{n}$, for every point $n$ on $\mathbb{N}$ and $m(\{\infty\})\!=\!0$. On the space $Y:=\widetilde{\mathbb{N}}^{\mathbb{Z}}$, $\mathbb{Z}$ the integers, we consider the product measure $\lambda:=\!\prod\limits^{+\infty}_{-\infty}\!m$ and the two-sided Bernoulli shift $T\!:\!Y\!\rightarrow\!Y$, with $T(\{x_{n}\})\!=\!\{y_{n}\}$, where $y_{n}=x_{n+1}$, for every $n\in\mathbb{Z}$.

Also, throughout this paper, $G$ is an amenable, locally compact separable group acting (continuously) on a Borel probability measure, compact metric space $(X,\mu)$ and the action is uniquely ergodic for $\mu$ and non-sensitive on $\mbox{supp}\mu$. It turns out (see Corollary 4.1), that such an action is necessarily equicontinuous.

Next, let $\varPhi_{1},\ldots,\varPhi_{n},\ldots$ be a sequence in $G$, that generates a dense semigroup in $G$. (Note that the action of this semigroup in $(X,\mu)$ is also uniquely ergodic).

We set up the skew product

[TABLE]

where $r:=(\ldots,r_{-n},\ldots,r_{-1},r_{0},r_{1},\ldots,r_{n},\ldots)$, conventionally we set

[TABLE]

Clearly ${\varPsi}$ is Borel measurable and $\mu\times\lambda$ is invariant under ${\varPsi}$.

Theorem 2.1**.**

If $\tau$ is a Borel probability measure on $X\times Y$, invariant for ${\varPsi}$, such that the projection of $\tau$ on $Y$ equals $\lambda$, then $\tau$ coincides with $\mu\times\lambda$.

From the above theorem, taking $r=(\ldots,r_{-n},\ldots,r_{-1},r_{0},r_{1},\ldots,\linebreak r_{n},\ldots)\in\mathbb{N}^{\mathbb{Z}}$ a generic point for $T$, it is easily seen, using some standard results (see [5, pp. 193-194]), that $(r_{1},\ldots,r_{n},\ldots)$ has the property mentioned in the abstract.

3 Invariant measures for continuous maps

The space $M(X)$ of all Borel probability measures on $X$ is metrizable in the weak∗ topology. If \big{\{}f_{n}\big{\}}^{\infty}_{n=1} is a dense subset of $C(X)$ (the space of continuous functions on $X$), then

[TABLE]

is a metric on $M(X)$ giving the weak∗ topology. Also, $M(X)$ is compact in this topology.

For $\varPhi:X\rightarrow X$ continuous, hence Borel measurable, we have the continuous affine map

[TABLE]

for $B$ a Borel set.

We have

Theorem 3.1**.**

Let $F_{m}$, $m\in\mathbb{N}$ be a Fölner sequence in $G$. For $\nu\in M(X)$ and $m\in\mathbb{N}$ we consider the measures

[TABLE]

(where $m_{G}$ is the Haar measure on $G$), or more concretely

[TABLE]

*for every $f\in C(X)$ and every $m\in\mathbb{N}$.

Then, $d(\mu^{\nu}_{m},\mu)\rightarrow 0$ for $m\rightarrow\infty$ uniformly for $\nu\in M(X)$.*

Proof. Suppose that the conclusion of the theorem does not hold. Then, there exist an $\varepsilon>0$, a subsequence $F_{m_{n}}$, $n\in\mathbb{N}$ of $F_{m}$, $m\in\mathbb{N}$ and a sequence $\nu_{n}$, $n\in\mathbb{N}$ in $M(X)$ such that

[TABLE]

For $f\in C(X)$ we have

[TABLE]

and for $H\in G$ $(h:M(X)\rightarrow M(X)$ the induced map),

[TABLE]

So

[TABLE]

Hence, every $w^{\ast}$-limit of the sequence $\mu^{\nu_{n}}_{m_{n}}$, $n\in\mathbb{N}$ is invariant under the action of $G$, so equals $\mu$ contradicting (1). $\square$

4 Some results on amenable, non-sensitive actions

We recall the following

Definition 4.1**.**

(See also [1, p. 23]) A continuous action of a group $G$, on a compact metric space $(X,\upsilon)$ ($\upsilon$ denotes the metric on $X$), is called sensitive on a subset $X^{\prime}\subset X$, if there exists a $\beta>0$, such that for every $x\in X^{\prime}$ and $\delta>0$, there exist a $y\in X$ with $\upsilon(x,y)<\delta$ and an $h\in G$, such that $\upsilon(h(x),h(y))\geq\beta$. Otherwise the action is called non-sensitive on $X^{\prime}\subset X$.

We set for $k\in\mathbb{N}$

$E_{k}:=\{x\in X:$ there exists an open neighborhood $U$ of $x$ such that

                               $x_{1},x_{2}\in U\Rightarrow\upsilon(\varPhi(x_{1}),\varPhi(x_{2}))<\dfrac{1}{k}$, for all $\varPhi\in G\}$.

Clearly, $E_{k}$ is open and since the action of $G$ is non-sensitive on $\text{supp}\mu$, $E_{k}\cap\text{supp}\mu\neq\emptyset$, for every $k\in\mathbb{N}$.

Note that a $x\in X$ is an equicontinuity point for $G$, if for every $\varepsilon>0$ there exists a $\delta>0$ such that $\upsilon(x,y)<\delta$ implies $\upsilon(\varPhi(x),\varPhi(y))<\varepsilon$, for every $\varPhi\in G$. Clearly, $\displaystyle\bigcap^{\infty}_{k=1}E_{k}$ is the set of equicontinuity points for $G$.

Lemma 4.1**.**

Let $k\in\mathbb{N}$. Then for every $x\in X\backslash E_{k}$ there exists a $\varPhi_{i_{x}}\in G$, such that $\varPhi_{i_{x}}(x)\in E_{k}$.

Proof. For $k\in\mathbb{N}$, the set

[TABLE]

is compact and forward invariant under the elements of $G$.

In case that $Q_{k}\neq\emptyset$, by an application of Day’s fixed point theorem [2, Theorem 1], there exists a Borel probability measure $\tau$ supported on $Q_{k}$ and invariant under $G$, so $\tau=\mu$. But this contradicts the fact that $E_{k}\cap\text{supp}\mu\neq\emptyset$, for every $k$. So, $Q_{k}=\emptyset$ and the conclusion of the lemma followsimmediately. $\square$

Corollary 4.1**.**

The group $G$ acts on $X$ equicontinuously.

Proof. Since the maps $\varPhi:X\rightarrow X$, $\varPhi\in G$ are open (as homeomorphisms), it is easily seen that $\varPhi(E_{k})\subseteq E_{k}$ for every $k\in\mathbb{N}$ and $\varPhi\in G$.

Let $x\in X$. Suppose, if possible, that $x$ is not an equicontinuity point for the action of $G$ in $X$. Then

[TABLE]

So, there exists a $k_{0}\in\mathbb{N}$ such that $x\notin E_{k_{0}}$. By the previous lemma, there exists a $\varPhi_{i_{x}}\in G$ such that $\varPhi_{i_{x}}(x)\in E_{k_{0}}$. Since $\varPhi(E_{k_{0}})\subseteq E_{k_{0}}$, for every $\varPhi\in G$, clearly we have $\varPhi^{-1}_{i_{x}}\circ\varPhi_{i_{x}}(x)=x\in E_{k_{0}}$, a contradiction. $\square$

We set $Seq:=\bigcup\limits^{\infty}_{n=1}\mathbb{N}^{n}$ the set of finite sequences of positive integers, and for $r=(r_{1},\ldots,r_{n})\in Seq$, $\varPhi_{r}:=\varPhi_{r_{n}}\circ\cdots\circ\varPhi_{r_{1}}$, $\varphi_{r}:=\varphi_{r_{n}}\circ\cdots\circ\varphi_{r_{1}}$ and ${\mit\Theta}:=\{\varphi_{r}:r\in Seq\}$.

Under the above setting we have the following proposition, which is the new element that gives the possibility to use a combination of the methods of [4,5] in the present situation (see [5, Proposition 3.1]).

Proposition 4.1**.**

There exists a sequence $\rho_{m}$, $m\in\mathbb{N}$ in $conv({\mit\Theta})$ (the convex hull of ${\mit\Theta}$) such that

[TABLE]

Proof. By Theorem 3.1, we can assume that there exist a Fölner sequence $F_{m}$, $m\in\mathbb{N}$ in $G$, and $\varepsilon_{m}>0$, $m\in\mathbb{N}$ with $\varepsilon_{m}\rightarrow 0$ for $m\rightarrow\infty$ such that setting, for $\sigma\in M(X)$, $\mu^{\sigma}_{m}\in M(X)$ with

[TABLE]

we have

[TABLE]

Let $D\!\subseteq\!X$ be denumerable, with $\overline{D}\!=\!X$. We enumerate $D=\{x_{i}:i\in\mathbb{N}\}$ and set $A:=\{\delta_{x_{i}}:x_{i}\in D$, $i\in\mathbb{N}$ and $\delta_{x_{i}}$ is the Dirac measure on $x_{i}\}$ $(\subseteq M(X))$.

Also, let $\{f_{n}:n\in\mathbb{N}\}(\subseteq C(X))$ be dense in $C(X)$ (clearly $\{f_{n}:n\in\mathbb{N}\}$ defines the metric on $M(X)$, see above).

Let $m\in\mathbb{N}$. For $n=1,\ldots,m$, $i=1,\ldots,m$ we set

[TABLE]

It is easily seen, that the above $g^{i}_{n}$ are continuous.

Clearly, for $m\in\mathbb{N}$ and $n=1,\ldots,m$, $i=1,\ldots,m$ we have

[TABLE]

We set $B:=\{\varPhi_{\ell}:\ell\in Seq\}$. By assumption we have $\overline{B}=G$.

By [9, Chapter II, Theorem 6.3], for $m\in\mathbb{N}$ there exists a convex combination

[TABLE]

of Dirac measures on $M(G)$, such that for $i=1,\ldots,m$ and $n=1,\ldots,m$

[TABLE]

So, in view of (3) and the definition of the $g^{i}_{n}$’s, for $m\in\mathbb{N}$, $i=1,\ldots,m$ and $n=1,\ldots,m$

[TABLE]

Setting $\rho_{m}:=\sum\limits^{k_{m}}_{k=1}\lambda_{k}\varphi_{\ell_{k}}$, we have for $m\in\mathbb{N}$, $i=1,\ldots,m$ and $n=1,\ldots,m$

[TABLE]

So, for $m\in\mathbb{N}$ and $i=1,\ldots,m$

[TABLE]

Combining (2) and (5), it follows that for $m\in\mathbb{N}$ and $i=1,\ldots,m$

[TABLE]

Claim 1. $\rho_{m}(\delta_{x})\rightarrow\mu$ uniformly for $x\in X$.

Let $\varepsilon>0$. There exists an $m_{0}\in\mathbb{N}$ such that

[TABLE]

Let $f_{1},\ldots,f_{m_{0}}$. For the given $\varepsilon>0$ there exists $\delta>0$, such that for $x,x^{\prime}\in X$ with $v(x,x^{\prime})<\delta$

[TABLE]

(where $v$ denotes the metric on $X$).

Since $B:=\{\varPhi_{\ell}:\ell\in Seq\}$ is equicontinuous, for the above $\delta>0$ there exists $\theta>0$ such that for $y,y^{\prime}\in X$ with $v(y,y^{\prime})<\theta$

[TABLE]

Since $\overline{D}=X$, there exists an $m_{\ast}>m_{0}$ such that for every $x\in X$, there exists a $x_{i_{\ast}}\in D$, $i_{\ast}\in\{1,\ldots,m_{\ast}\}$ with $v(x_{i_{\ast}},x)<\theta$.

So, for every $x\in X$, $m>m_{\ast}$ and $n=1,2,\ldots,m_{0}$ we have

[TABLE]

and in view of (4), since $i_{\ast}\in\{1,\ldots,m_{\ast}\}$, we have for every $x\in X$, $m>m_{\ast}$ and $n=1,2,\ldots,m_{0}$

[TABLE]

(note that $\varepsilon_{m}<\varepsilon$ for $m>m_{\ast}>m_{0}$).

So, for every $x\in X$, $m>m_{\ast}$ we have

[TABLE]

Finally, by (6) we have that for every $x\in X$ and $m>m_{\ast}$

[TABLE]

(note that for $m>m_{\ast}>m_{0}$, $\varepsilon_{m}<\varepsilon$ and $\dfrac{1}{2^{m-1}}<\varepsilon$).

Claim 2. $\rho_{m}(\sigma)\rightarrow\mu$ uniformly for \sigma\in\Big{\{}\sum\limits^{s}_{k=1}\lambda_{k}\delta_{x_{k}}:\sum\limits^{s}_{k=1}\lambda_{k}=1, x_{k}\in D\Big{\}}.

Indeed, the claim holds from Claim 1, since $\rho_{m}(\sigma)$ is a convex combination of measures of the form $\rho_{m}(\delta_{x})$, $x\in X$.

Finally, $\rho_{m}(\sigma)\!\rightarrow\!\mu$ uniformly for every $\sigma\in M(X)$, since the set \Big{\{}\sum\limits^{s}_{k=1}\lambda_{k}\delta_{x_{k}}\!:\sum\limits^{s}_{k=1}\lambda_{k}=1, x_{k}\in D\Big{\}} is dense in $M(X)$ by [9, Chapter II, Theorem 6.3]. $\square$

The following lemma is a simplification of [5, Lemma 4.4].

Lemma 4.2**.**

Let $\rho_{m}$, $m\in\mathbb{N}$ a sequence in $\text{conv}({\mit\Theta})$ as in Proposition 4.1, $\nu_{m}$, $m\in\mathbb{N}$, $h_{m}$, $m\in\mathbb{N}$ sequences in $M(X)$ and $Seq$ respectively and $f\in C(X)$. Then

[TABLE]

for some subsequence $m_{\ell}$, $\ell\in\mathbb{N}$, of $m\in\mathbb{N}$.

Proof. Since the action of $G$ on $X$ is equicontinuous, the sequence $\varPhi_{h_{m}}$, $m\in\mathbb{N}$ is equicontinuous for every sequence $h_{m}$, $m\in\mathbb{N}$ in Seq. Then $f\circ\varPhi_{h_{m}}$, $m\in\mathbb{N}$ is equicontinuous, so by Arzela-Ascoli theorem it has a uniformly convergent subsequence

[TABLE]

Then for $\varepsilon>0$ there exists an $\ell_{1}\in\mathbb{N}$ such that

[TABLE]

So

[TABLE]

On the other hand, by Proposition 4.1 there exists an $\ell_{2}\in\mathbb{N}$ such that

[TABLE]

By (7) and (8) there exists an $\ell_{0}\in\mathbb{N}$ so that

[TABLE]

Hence

[TABLE]

Now it suffices to show that $\displaystyle\int_{X}fd\mu=\displaystyle\int_{X}\widetilde{f}d\mu$.

Indeed, $\displaystyle\int_{X}f\circ\varPhi_{h_{m_{\ell}}}d\mu=\displaystyle\int_{X}fd\mu$, since the $\varPhi_{r}$’s, $r\in Seq$ preserve the measure $\mu$ and $f\circ\varPhi_{h_{m_{\ell}}}\overset{u}{\longrightarrow}\widetilde{f}$, so $\displaystyle\int_{X}fd\mu=\displaystyle\int_{X}\widetilde{f}d\mu$. $\square$

Corollary 4.2**.**

Let $\rho_{m}$, $m\in\mathbb{N}$, $\nu_{m}$, $m\in\mathbb{N}$, $h_{m}$, $m\in\mathbb{N}$ sequences as in Lemma 4.2 and $K\subset X$ Jordan measurable, i.e. $\mu(\partial K)=0$ ($\partial K$ the boundary of $K$) with $\mu(K)>a$, for some $0<a<1$. Then there exists an $m_{\ell_{0}}\in\mathbb{N}$ such that

[TABLE]

The proof of the corollary is similar to that of [5, Corollary 4.3], so we omit it.

5 Some technical lemmata

In the sequel, we assume the curriculum of notations and definitions of [4, Section 5]. For $A\subseteq\mathbb{Z}$, $pr_{A}:\mathbb{N}^{\mathbb{Z}}\rightarrow\mathbb{N}^{A}$ denotes the natural projection and for $k\in\mathbb{N}$, $Z_{k}:=\{-k,\ldots,0,\ldots,k\}$.

We recall from [4] and [5] the following lemmata.

Lemma 5.1**.**

Let $B\subseteq\mathbb{N}^{\mathbb{Z}}$ compact with $\lambda(B)>0$ and $\beta$ with $0<\beta<1$. Then there exists an $a=(a_{-k},\ldots,a_{-1},a_{0},a_{1},\ldots,a_{k})\in\mathbb{N}^{\mathbb{Z}_{k}}$, for $k\in\mathbb{N}$ such that

[TABLE]

Proof. See [4, Lemma 5.1]. $\square$

Lemma 5.2**.**

Let $F\subseteq Seq$ finite. Then there exists a $\beta$, $0<\beta<1$, such that, if $B\subseteq\mathbb{N}^{\mathbb{Z}}$ measurable, with $\lambda(B)>0$ and $a\in\mathbb{N}^{\mathbb{Z}_{k}}$ for some $k\in\mathbb{N}$ satisfying

[TABLE]

then for sufficiently large $n$ $(n\geq n_{1})$, there exists a $t_{n}\in\mathbb{N}^{n-2k-1}$ such that

[TABLE]

for all $z\in F$, (where $|z|$ denotes the length of $z$).

Proof. See [5, Lemma 6.1]. $\square$

The following lemma is highly technical and its meaning will be clear in the proof of Theorem 6.2.

Lemma 5.3**.**

Let $\nu$ be a Borel probability measure on $X\times Y$ singular with respect to $\mu\times\lambda$, such that the projection of $\nu$ on $Y$ coincides with $\lambda$. Then given $0<\omega<1$, $0<\theta<1$ and $h:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$ a non-decreasing function, there exist $Q_{k}$, $k=1,2,\ldots,s$, $s\in\mathbb{N}$, disjoint compact subsets of $X$, $K\subseteq X\backslash\bigcup\limits^{s}_{k=1}Q_{k}$ compact, and $B\subseteq Y$ compact, with $\lambda(B)>0$, such that

1. (i)

$\mu(K)>1-\omega$, $\mu(\partial K)=0$ $(\partial K$ the boundary) 2. (ii)

setting $e:=$ distance \Big{(}K,\bigcup\limits^{s}_{k=1}Q_{k}\Big{)}>0, we have

[TABLE] 3. (iii)

\nu_{y}\Big{(}\bigcup\limits^{s}_{k=1}Q_{k}\Big{)}>1-\theta,  for   $y\in B$ 4. (iv)

$|\nu_{y}(Q_{k})-\nu_{y^{\prime}}(Q_{k})|<\dfrac{\theta}{s}$  for every   $y,y^{\prime}\in B$,  $k=1,2,\ldots,s$

(where $\nu_{y}$ denotes the conditional measure induced by $\nu$ on the fiber $X\times\{y\}$).

Proof. See [4, Lemma 6.1]. $\square$

Note. Although the $\varPhi$’s in [4] are commutative, this is not used in the proof of [4, Lemma 6.1].

Under the assumptions of Lemma 5.3, we have the following

Corollary 5.1**.**

Let $y_{0}\in B$, $B^{\prime}\subset B$ measurable, with $\lambda(B^{\prime})>0$ and $\mathcal{P}\subset\{1,2,\ldots,s\}$, such that

[TABLE]

Then

[TABLE]

Proof. See [5, Corollary 5.1]. $\square$

6 The proof of Theorem 2.1

The proof of Theorem 2.1 will be given in two major steps. First, we shall prove that if $\tau$ is absolutely continuous with respect to $\mu\times\lambda$ then $\tau$ coincides with $\mu\times\lambda$. Second, we shall prove that $\tau$ has a trivial singular part with respect to $\mu\times\lambda$. These two steps are described in Theorems 6.1 and 6.2, respectively.

We have

Theorem 6.1**.**

The measure $\mu\times\lambda$ is the unique Borel probability measure on $X\times Y$, invariant under ${\varPsi}$ and absolutely continuous with respect to $\mu\times\lambda$.

Proof. This follows from the ergodicity of the skew product ${\varPsi}$, see the random ergodic theorem in [10]. $\square$

Remark. Note that the use of the random ergodic theorem of Ryll-Nardzewski (see [10]) gives immediately Theorem 6.1, so we can omit the lengthy proof of the “first step” that appears in [4, Proposition 5.1] and [5, Theorem 6.1].

The proof of the following theorem is an amalgamation of the proofs of [4, Theorem 7.1] and [5, Theorem 7.1].

Theorem 6.2**.**

Let $\nu$ be a Borel probability measure on $X\times Y$ singular with respect to $\mu\times\lambda$, such that the projection of $\nu$ on $Y$ coincides with $\lambda$. Then $\nu$ is not invariant under ${\varPsi}$.

Proof. Suppose that the conclusion of the theorem does not hold i.e. $\nu$ is invariant for ${\varPsi}$.

Since the semigroup $\mathcal{H}$ generated by $\varPhi_{1},\ldots,\varPhi_{n},\ldots$ acts equicontinuously on $X$ (by Corollary 4.1), if $\rho$ denotes the metric on $X$, then clearly there exists a non-decreasing $h:\mathbb{R}^{+}\rightarrow\mathbb{R}^{+}$, such that for every $f\in\mathcal{H}$ and $x,y\in X$ with $\rho(x,y)<h(\delta)$ $(\delta>0)$, then $\rho(f(x),f(y))<\delta$. Now given $0<\omega<\dfrac{1}{100}$, $0<\theta<\dfrac{1}{100}$ and $h$ as above, by Lemma 5.3 there exist $Q_{k}$, $k=1,\ldots,s$, disjoint compact subsets of $X$, $K\subseteq X\backslash\bigcup\limits^{s}_{k=1}Q_{k}$ compact and $B_{1}\subseteq Y:=\widetilde{\mathbb{N}}^{\mathbb{Z}}$ compact with $\lambda(B_{1})>0$ satisfying conditions (i), (ii), (iii), (iv) of the lemma, (with $B_{1}$ in place of $B$).

Let $B^{\prime}_{1}:=B_{1}\cap\mathbb{N}^{\mathbb{Z}}$. Then $\lambda(B^{\prime}_{1})=\lambda(B_{1})>0$ and by the regularity of $\lambda$, there exists some compact $B\subseteq B^{\prime}_{1}$, such that $\lambda(B)>0$. The set $B$ satisfies the conditions of Lemma 5.3

We consider $\rho_{m}$, $m\in\mathbb{N}$ a sequence in $conv({\mit\Theta})$ as in Proposition 4.1. Since $\rho_{m}\in conv({\mit\Theta})$, there exist a finite $F_{m}\subset Seq$ and $\theta_{z}(m)>0$ for $z\in F_{m}$, such that $\sum\limits_{z\in F_{m}}\theta_{z}(m)=1$ and $\rho_{m}=\sum\limits_{z\in F_{m}}\theta_{z}(m)\varphi_{z}$.

By Lemma 5.2 for each $F_{m}$, $(m\in\mathbb{N})$ there exists a $\beta_{m}$, $0<\beta_{m}<1$, satisfying the conclusion of that lemma.

Applying Lemma 5.1 repeatedly, we find for each couple

[TABLE]

a $k_{m}\in\mathbb{N}$ and an $a^{(m)}=(a^{(m)}_{-k_{m}},\ldots,a^{(m)}_{0},\ldots,a^{(m)}_{k_{m}})\in\mathbb{N}^{\mathbb{Z}_{k_{m}}}$ satisfying

[TABLE]

for $m=1,2,\ldots\;.$

Next, applying Lemma 5.2 repeatedly, taking in view of (9), we find for each quadruple

[TABLE]

an $n_{m}\in\mathbb{N}$ and a $t_{n_{m}}\in\mathbb{N}^{n_{m}-2k_{m}-1}$ such that, setting $t_{n_{m}}=t_{m}$ for brevity in the notation,

[TABLE]

for all $z\in F_{m}$.

In the sequel we fix some $y_{0}\in B$ and set

[TABLE]

We fix $x_{k}\in Q_{k}$, $k=1,2,\ldots,s$ and consider the probability measure

[TABLE]

At the present situation, we can apply Corollary 4.2 for the sequences $\rho_{m}$, $m\in\mathbb{N}$ (previously considered),

[TABLE]

and $K$, (where $a^{(m)}_{+}=(a_{1}^{(m)},\ldots,a^{(m)}_{k_{m}})$ and $a^{(m)}_{-}=(a^{(m)}_{-k_{m}},\ldots,a^{(m)}_{0})(=h_{m})$) and find an $m_{\ell_{0}}$ such that, setting $m_{\ell_{0}}=m_{0}$ for brevity in the notation

[TABLE]

Since $\rho_{m_{0}}$ is a convex combination, there exists a $z^{\ast}_{m_{0}}\in F_{m_{0}}$ such that

[TABLE]

i.e. by the form of $\nu_{m_{0}}$

[TABLE]

We set

[TABLE]

and since

[TABLE]

setting $\mathcal{P}:=\{k\in\{1,2,\ldots,s\}|\xi_{k}\in K\}$, by (11) we have

[TABLE]

So, by the definition of the $\gamma_{k}$’s

[TABLE]

and since by (iii) of Lemma 5.3 \nu_{y_{0}}\Big{(}\displaystyle\bigcup^{s}_{i=1}Q_{i}\Big{)}>1-\theta we have

[TABLE]

Claim. \Big{(}\varPhi_{(a_{+}^{(m_{0})},t_{m_{0}},z^{\ast}_{m_{0}},a_{-}^{(m_{0})})}\Big{(}\displaystyle\bigcup_{k\in\mathcal{P}}\overline{Q}_{k}\Big{)}\Big{)}\cap\Big{(}\displaystyle\bigcup^{s}_{k=1}Q_{k}\Big{)}=\emptyset.

Indeed, by (ii) of Lemma 5.3, diameter${(\overline{Q_{k}}})=$ diameter $(Q_{k})<h(e)$, for $k=1,2,\ldots,s$, where $e:=\,$distance\Big{(}K,\bigcup\limits^{s}_{k=1}Q_{k}\Big{)}, so we have

[TABLE]

On the other hand by the definition of $\mathcal{P}$, we have $\xi_{k}:=\varPhi_{(a_{+}^{(m_{0})},t_{m_{0}},z^{\ast}_{m_{0}},a_{-}^{(m_{0})})}(x_{k})\in K$, for $k\in\mathcal{P}$, where $x_{k}\in Q_{k}$. So for $k\in\mathcal{P}$

[TABLE]

i.e. the claim.

Next, we set

[TABLE]

(where $|z^{\ast}_{m_{0}}|$ denotes the length of $z^{\ast}_{m_{0}}$)

By (10) we have $\lambda(W^{\ast})>0$. Clearly, $T^{-(n_{m_{0}}+|z^{\ast}_{m_{0}}|)}(W^{\ast})\subseteq B$, so by (12) and Corollary 5.1 we have

[TABLE]

Clearly, by the form of $W^{\ast}$ we have

[TABLE]

which is measurable, since $\overline{Q}_{k}$ are compact sets.

By the invariance of $\nu$ under ${\varPsi}$ and (13) we have

[TABLE]

By (14) and (15) we have

[TABLE]

On the other hand, since clearly $W^{\ast}\subseteq B$, by (iii) of Lemma 5.3 we have \nu_{y}\Big{(}\bigcup\limits^{s}_{k=1}Q_{k}\Big{)}>1-\theta, for every $y\in W^{\ast}$ and intergrating the above inequality over $W^{\ast}$, we have

[TABLE]

Finally, (16), (17) and the claim give

[TABLE]

which obviously contradicts the fact that the projection of $\nu$ on $Y$ coincides with $\lambda$. $\square$

Finally, combining Theorems 6.1 and 6.2, we can conclude the proof of Theorem 2.1. For more details, see [5, Section 8].

Acknowledgements. We would like to express our gratitude to Professor Constantinos Gryllakis for his guidance during the preparation of this manuscript.