Posted by: yglima | March 20, 2012

CF2: essential values

We continue the discussion of CF1. The goal of the present post is to identify when a cylinder flow is ergodic. Such will be attained via the notion of essential values, introduced by Klaus Schmidt.

We work in greater generality than in CF1. Let ${(X,\mathcal B,\nu,T)}$ be an ergodic probability measure-preserving system. Let ${G}$ be an abelian locally compact topological group endowed with a metric ${\|\cdot\|}$, ${\mathcal C}$ its Borel sigma-algebra and ${\lambda}$ its Haar measure. For every ${\phi: X\rightarrow G}$, we define the cylinder flow ${(Z,\mathcal A,\mu,F)}$ by:

• ${Z=X\times G}$,
• ${\mathcal A=\mathcal B\times\mathcal C}$,
• ${\mu=\nu\times\lambda}$, and
• ${F:Z\rightarrow Z}$ defined as ${F(x,y)=(Tx,y+\phi(x))}$.

It is clear that ${\mu}$ is preserved by ${F}$. For each ${g\in G}$, let

$\displaystyle \begin{array}{rcccl} V_g&:&Z &\longrightarrow &Z\\ & &(x,y)&\longmapsto &(x,y+g) \end{array}$

be the vertical translation. Because ${G}$ is abelian, ${F}$ commutes with each ${V_g}$:

$\displaystyle \begin{array}{rcl} FV_g(x,y)&=&F(x,y+g)\\ &=&(Tx,y+g+\phi(x))\\ &=&V_g(Tx,y+\phi(x))\\ &=&V_gF(x,y). \end{array}$

In particular, if ${A\in\mathcal A}$ is ${F}$-invariant, then so is ${V_gA}$.

1. Periods

Considering

$\displaystyle \mathcal I(F)=\{A\in\mathcal A\,;\,F^{-1}A=A\},$

the above discussion motivates us to introduce the following subgroup of ${G}$.

Definition 1 Define the collection of periods for ${F}$ as

$\displaystyle {\rm Per}(F)=\{g\in G\,;\,V_gA=A\text{ for all }A\in\mathcal I(F)\}.$

Saying that ${g\in{\rm Per}(F)}$ is exactly the same as saying that the behavior of ${F}$ is periodic with period ${g}$. Of course, ${{\rm Per}(F)}$ is a subgroup of ${G}$, because ${V_gV_h=V_{g+h}=V_hV_g}$ for any ${g,h\in G}$. It is also closed, as we’ll prove below. In particular, if ${G=\mathbb R}$, then ${{\rm Per}(F)}$ is either ${\mathbb R}$ or ${\alpha\mathbb Z}$ for some ${\alpha\in\mathbb R}$. In the second situation, we can see ${F}$ as defined in the torus ${X\times [0,\alpha]/\sim}$, where ${(x,0)\sim(x,\alpha)}$ for ${x\in X}$. In the first situation, when ${{\rm Per}(F)=\mathbb R}$, ${F}$ is ergodic. This is indeed the general case, as stated below.

Lemma 2 In the above notations,

• ${{\rm Per}(F)}$ is a closed subgroup of ${G}$, and
• ${F}$ is ergodic if and only if ${{\rm Per}(F)=G}$.

Proof:
For the first item, it remains to prove ${{\rm Per}(F)}$ is closed. This follows from the continuous dependence of ${V_g}$ in ${g}$: if ${g_n\rightarrow g}$ in ${G}$, then for any ${A\in\mathcal A}$

$\displaystyle V_gA=\lim_{n\rightarrow\infty}V_{g_n}A.$

If in particular each ${g_n}$ belongs to ${{\rm Per}(F)}$, then for ${A\in\mathcal I(F)}$ we have

$\displaystyle V_gA=\lim_{n\rightarrow\infty}V_{g_n}A=A,$

thus proving that ${g\in{\rm Per}(F)}$.

We postpone the proof of the second item to Section 2. $\Box$

Even though the above lemma gives a good criterium for ergodicity, it is not clear when an element ${g\in G}$ belongs to ${{\rm Per}(F)}$, mainly because the condition has to be checked along elements of ${\mathcal I(F)}$ only, and this set will eventually contain only the trivial elements of ${\mathcal A}$. What we want now is to get rid of this soft description and to define a more quantitative criterium for ergodicity that does not only depend on the elements of ${\mathcal I(F)}$. This new notion, which in principle seems to make things more difficult – but does not, is that of

2. Essential values

From now and on, ${B_\varepsilon(g)}$ is the ball of ${G}$ of radius ${\varepsilon>0}$ centered in ${g}$, ${S_n\phi}$ is the ${n}$-th Birkhoff sum of ${\phi}$ with respect to ${T}$ and set-theoretical equalities mean set-theoretical equalities modulo zero.

Definition 3 An element ${g\in G}$ is an essential value for ${F}$ if, for any ${B\in\mathcal B}$ with ${\nu(B)>0}$ and any ${\varepsilon>0}$, there is ${n\in\mathbb Z}$ such that

$\displaystyle \nu(B\cap T^{-n}B\cap\{x\in X\,;\,\|S_n\phi(x)-g\|<\varepsilon\})>0.$

The set of all essential values of ${F}$ is denoted by ${E(F)}$.

This, on the contrary of the geometrical notion of period, is more quantitative and translates the fact that, in order to ${F}$ be ergodic, the ${F}$-iterates of every set of positive ${\mu}$-measure has to saturate the whole cylinder ${Z}$. As claimed above, this is just an alternative definition for periods, according to the

Theorem 4 ${{\rm Per}(F)=E(F)}$. In particular, ${F}$ is ergodic if and only if ${E(F)=G.}$

Proof: Let ${g\not\in E(F)}$. By definition, there are ${B\in\mathcal B}$ with ${\nu(B)>0}$ and ${\varepsilon>0}$ such that

$\displaystyle \nu(B\cap T^{-n}B\cap\{x\in X\,;\,\|S_n\phi(x)-g\|<\varepsilon\})=0\,,\ \forall\,n\in\mathbb Z.$

This implies that if

$\displaystyle B_1=B\times B_{\varepsilon/2}(0)\ \text{ and }\ B_2=B\times B_{\varepsilon/2}(g)$

then ${F^n B_1\cap B_2=\emptyset}$ for all ${n\in\mathbb Z}$. This in turn guarantees that if we define ${A_1,A_2\in\mathcal I(F)}$ by

$\displaystyle A_1=\bigcup_{n\in\mathbb Z}F^nB_1\ \ \ \text{ and }\ \ \ A_2=\bigcup_{n\in\mathbb Z}F^nB_2\,,$

then ${A_1\cap A_2=\emptyset}$. But ${A_2=V_gA_1}$ and so

$\displaystyle V_g A_1\cap A_1=\emptyset\ \ \Longrightarrow\ \ g\not\in{\rm Per}(F).$

This proves that ${{\rm Per}(F)\subset E(F)}$.

For the reverse inequality, let ${g\not\in{\rm Per}(F)}$. We claim this implies the existence of disjoint ${A_1,A_2\in\mathcal I(F)}$ with ${\mu(A_1)=\mu(A_2)>0}$ such that

$\displaystyle V_gA_1=A_2.$

Indeed, take ${A\in\mathcal I(F)}$ such that ${V_gA\not=A}$ and define

$\displaystyle A_1=A\backslash V_gA\ \ \text{ and }\ \ A_2=V_gA\backslash A.$

For each ${x\in X}$, define ${A_1^x\subset\{x\}\times G}$ by the equality

$\displaystyle A_1=\bigcup_{x\in X}A_1^x.$

Then ${A_2=\bigcup_{x\in X}(A_1^x+g)}$. Furthermore, the ${F}$-invariance of ${A_1}$ gives

$\displaystyle FA_1^x=A_1^{Tx}\ \ \Longrightarrow\ \ A_1^{Tx}=A_1^x+\phi(x)$

and so ${\lambda(A_1^{Tx})=\lambda(A_1^x+\phi(x))=\lambda(A_1^x)}$, thus proving that the map ${x\mapsto\lambda(A_1^x)}$ is ${T}$-invariant. Because ${T}$ is ergodic,

$\displaystyle \lambda(A_1^x)=c>0$

for ${\nu}$-almost every ${x\in X}$.

The idea now is to define a measurable function ${\omega:X\times X\times G\rightarrow\mathbb R}$ by

$\displaystyle \omega(x,y,g)=\lambda(A_1^x\cap(A_1^y+g)).$

As defined above, ${\omega}$ may not be measurable, but we can apply a trick – to be explained below – in order to guarantee measurability. Assume we have done this and that ${\omega}$ as above is measurable. We have ${\omega(x,x,0)=c>0}$ and thus, by Lusin’s theorem, there are sets ${X_0\subset X}$ with ${\nu(X_0)>0}$ and ${B_\varepsilon(0)\subset G}$ such that

$\displaystyle \omega(x,y,g)>0,\ \ \forall\,(x,y,g)\in X_0\times X_0\times B_\varepsilon(0).$

We claim the pair ${X_0\in\mathcal B,\varepsilon>0}$ contradicts the assumption that ${g\in E(F)}$. In other words, we claim

$\displaystyle \nu(X_0\cap T^{-n}X_0\cap\{x\in X\,;\,\|S_n\phi(x)-g\|<\varepsilon\})=0,\ \ \forall\,n\in\mathbb Z,$

which is equivalent to the implication

$\displaystyle x\in X_0\,,\, T^nx\in X_0\ \ \Longrightarrow\ \ \|S_n\phi(x)-g\|\ge\varepsilon$

and this in turn follows if we prove that

$\displaystyle x\in X_0\,,\,T^nx\in X_0\ \ \Longrightarrow\ \ \omega(x,T^nx,g-S_n\phi(x))=0. \ \ \ \ \ (1)$

And indeed, as ${A_1\cap A_2=\emptyset}$, we have

$\displaystyle \begin{array}{rrcl} &A_1^{T^nx}\cap A_2^{T^nx}&=&\emptyset\\ &&&\\ \Longrightarrow &(A_1^x+S_n\phi(x))\cap(A_1^{T^nx}+g)&=&\emptyset\\ &&&\\ \Longrightarrow &A_1^x\cap(A_1^{T^nx}+g-S_n\phi(x))&=&\emptyset\\ &&&\\ \Longrightarrow &\omega(x,T^nx,g-S_n\phi(x))&=&0, \end{array}$

thus establishing (1).

The above proof has two flaws. Firstly, as defined above, ${\omega}$ may not be measurable. Secondly, we used Lusin’s theorem, which assumes that ${\omega}$ is defined in a Polish space. We bypass these problems in three steps:

1. In general, ${A_1^x,A_1^y}$ may have infinite ${\lambda}$-measure and then the intersection ${A_1^x\cap(A_1^y+g)}$ does not behave measurably. We fix this by considering a measurable subset ${\theta\subset A}$ such that ${\theta^x=\theta\cap A_1^x}$ satisfies

$\displaystyle 0<\lambda(\theta^x)<\infty.$

2. ${\lambda}$ is not a probability measure, but there exists a probability measure (not necessarily ${G}$-invariant) equivalent to ${\lambda}$. Indeed, by the ${\sigma}$-finiteness of ${\lambda}$ there are disjoint sets ${(A_n)_{n\ge 1}}$ with ${0<\lambda(A_n)<\infty}$ and ${G=\bigcup_{n\ge 1}A_n}$. The probability measure ${\tilde\lambda}$ in ${G}$ defined by

$\displaystyle \tilde\lambda(A)=\sum_{n\ge 1}\dfrac{1}{2^n\lambda(A_n)}\cdot\lambda(A\cap A_n)$

is clearly equivalent to ${\lambda}$.

3. ${(X,\mathcal B,\nu)}$ can be assumed to be Polish. Indeed, if ${(\mathcal{\hat B},\hat\nu)}$ is the measure algebra of ${(X,\mathcal B,\nu)}$ (see ERT11),

$\displaystyle d(\hat{B_1},\hat{B_2})=\nu(B_1\bigtriangleup B_2).$

define a distance in ${\mathcal{\hat B}}$.

We can now define ${\omega}$ in the probability Polish space ${X\times X\times G}$ by

$\displaystyle \omega(x,y,g)=\lambda(\theta^x\cap(\theta^y+g)).$

By Fubini’s theorem, the map ${x\mapsto \hat{\theta^x}}$ is measurable and, by the ${\lambda}$-finiteness of each ${\theta^x}$ (see Exercise 1 below), this new map is measurable. Furthermore, the space of definition of ${\omega}$ being a Polish space, Lusin’s theorem can be applied to guarantee the existence of the neighborhood ${X_0\times X_0\times B_\varepsilon(0)\subset X\times X\times G}$ in which ${\omega}$ is bounded away from zero. We leave the details to the reader. $\Box$

Exercise 1 Let ${(X,\mathcal B,\nu)}$ be a measure space. Prove that if

$\displaystyle \mathcal B_0=\{B\in\mathcal B\,;\,\nu(B)<\infty\},$

then the maps from ${\mathcal B_0\times \mathcal B_0}$ to ${\mathcal B_0}$ defined by

$\displaystyle (B_1,B_2)\mapsto B_1\cap B_2\ \text{ and }\ (B_1,B_2)\mapsto B_1\cup B_2$

are measurable.

The end of this section is devoted to the proof of the second part of Lemma 2. We first need Steinhaus’ theorem.

Lemma 5 (Steinhaus) Let ${A_1,A_2\subset G}$ be Borel sets with ${\lambda(A_1),\lambda(A_2)>0}$. Then there is an open set ${G_0\subset G}$ such that

$\displaystyle \lambda(A_1\cap(A_2+g))>0,\ \ \forall\,g\in G_0.$

Proof: Obviously, we can assume ${\lambda(A_1),\lambda(A_2)<\infty}$. Then ${\psi:G\rightarrow\mathbb R}$ defined by

$\displaystyle \psi(g)=\lambda(A_1\cap(A_2+g))$

is measurable. Note that

$\displaystyle \begin{array}{rcl}\displaystyle \int_G\psi(g)d\lambda(g)&=&\displaystyle\int_G\displaystyle\int_G\chi_{A_1}(h)\chi_{A_2+g}(h)d\lambda(h)d\lambda(g)\\&&\\ &=&\displaystyle\int_G\chi_{A_1}(h)\left(\displaystyle\int_G\chi_{A_2+g}(h)d\lambda(g)\right)d\lambda(h)\\&&\\ &=&\displaystyle\int_G\chi_{A_1}(h)\left(\displaystyle\int_G\chi_{h-A_2}(g)d\lambda(g)\right)d\lambda(h)\\&&\\ &=&\displaystyle\int_G\chi_{A_1}(h)\lambda(h-A_2)d\lambda(h)\\&&\\ &=&\lambda(A_1)\lambda(A_2)\\&&\\ &>&0. \end{array}$

By Lusin’s theorem, there is an open set ${G_0\subset G}$ such that ${\psi(g)>0}$ for all ${g\in G_0}$. $\Box$

We finally arrive to the proof of the second part of Lemma 2.

${(\Longrightarrow)}$ Assume ${F}$ is ergodic. For ${B\subset X}$ with ${\nu(B)>0}$ and ${\varepsilon>0}$, let ${A=B\times B_\varepsilon(0)}$. For ${g\in G}$, the ergodicity of ${F}$ guarantees the existence of ${n}$ such that ${A}$ intersects ${B\times B_\varepsilon(g)}$ in a set of positive ${\mu}$-measure. This implies that

$\displaystyle \nu(B\cap T^{-n}B\cap\{x\in X\,;\,\|S_n\phi(x)-g\|<\varepsilon\})>0.$

${(\Longleftarrow)}$ Assume ${{\rm Per}(F)=G}$. Let ${A\in\mathcal I(F)}$ and define ${A^x\subset\{x\}\times G}$ by the equality

$\displaystyle A=\bigcup_{x\in X}A^x.$

Using the same argument as in the proof of Theorem 4, ${\nu(A^x)=c>0}$ for ${\nu}$-almost every ${x\in X}$. For a fixed ${g\in G}$, we have by assumption that ${V_gA=A}$ and so ${A^x=A^x+g}$ for ${\nu}$-almost every ${x\in X}$. Let ${\{g_1,g_2,\ldots\}}$ be a dense subset of ${G}$. For each ${n\ge 1}$, let

$\displaystyle N_n=\{x\in X\,;\,A^x\not=A^x+g_n\}$

and ${N=\bigcup_{n\ge 1}N_n}$. ${N}$ has zero ${\nu}$-measure. We will prove that ${A^x=G}$ for every ${x\in X\backslash N}$. Assume this is not the case. By Steinhaus’ theorem, there is an open set ${G_0}$ such that

$\displaystyle \lambda(A^x\cap(G\backslash A^x+g))>0\,,\ \ \forall\,g\in G_0.$

This is a contradiction: for ${g_n\in G_0}$,

$\displaystyle A^x=A^x+g_n\ \Longrightarrow\ A^x\cap (G\backslash A^x+g_n)=\emptyset.$

This completes the proof.

3. An example

In this section we study a particular class of cylinder flow. Let ${\phi:\mathbb T\rightarrow\mathbb Z}$ be the Haar function, defined as

$\displaystyle \phi(x)=\left\{ \begin{array}{rl} 1\ ,&\text{ if }x\in\left[0,\dfrac{1}{2}\right)\\ &\\ -1\ ,&\text{ if }x\in\left[\dfrac{1}{2},1\right)\,, \end{array} \right.$

and

$\displaystyle \begin{array}{rcrcl} F&:&\mathbb T\times\mathbb Z&\longrightarrow &\mathbb T\times\mathbb Z\\ & & (x,y)&\longmapsto &(x+\alpha,y+\phi(x)) \end{array}$

the associate cylinder flow. These examples are classical in the literature and were studied by many authors, from which we mention J. Aaronson, J.-P. Conze, M. Keane and K. Schmidt.

Up to my knowledge, Schmidt was the first one to be interested in these cylinder flows, because they encode the properties of deterministic random walks in the following sense: if, instead of the irrational rotation, the basis transformation of ${F}$ is the doubling map, i.e. ${F(x,y)=(2x,y+\phi(x))}$, then

$\displaystyle F^n(x,y)=(2^nx,y+S_n\phi(x)),$

where ${(S_n\phi)}$ is statistically the same as the classical random walk on the integers (starting at zero, at each step we toss a coin and the particle moves one step to the right in case of ‘heads’ and one step to the left in case of `tails’). In this case, the basis transformation has positive topological entropy. In ours, the random walk ${(S_n\phi)}$ is driven by a zero topological entropy transformation.

The goal of this section is to prove, via the notion of essential values, that ${F}$ is ergodic. Because ${E(F)}$ is a subgroup of ${\mathbb Z}$, it is enough to show that ${1\in E(F)}$. Still with the notation of the previous sections, let ${\nu}$, ${\lambda}$ and ${\mu=\nu\times\lambda}$ be the Lebesgue measures in ${\mathbb T}$, ${\mathbb Z}$ and ${\mathbb T\times\mathbb Z}$, respectively. We first, following the notation of Section 2 of CF1, prove a lemma on continued fractions.

Lemma 6 For every irrational number ${\alpha}$, there are infinitely many odd positive integers ${q}$ such that

$\displaystyle q\|q\alpha\|<\dfrac{1}{2}\,\cdot$

Proof: Let ${\alpha=[a_0;a_1,a_2,\ldots]}$ be the continued fraction expansion of ${\alpha}$ and

$\displaystyle \dfrac{p_n}{q_n}=[a_0;a_1,\ldots,a_n].$

We use the fact (and leave the proof to the reader) that ${q_n\|q_n\alpha\|<\alpha_{n+1}}$, where ${\alpha_{n+1}=[0;a_{n+1},a_{n+2},\ldots]}$, and consequently

$\displaystyle q_n\|q_n\alpha\|<\dfrac{1}{a_{n+1}}\,\cdot$

In particular, every odd ${q_n}$ with ${a_{n+1}>1}$ satisfies the lemma. By contradiction suppose that, for large ${n}$, ${a_{n+1}=1}$ whenever ${q_n}$ is odd. Fix one such ${n}$. If ${q_{n-1}}$ is even, ${q_{n+1}=a_{n+1}q_n+q_{n-1}=q_n+q_{n-1}}$ is odd and so ${a_{n+2}=1}$. Then ${\alpha_{n+1}=[0;1,1,a_{n+3},\ldots]<1/2}$. If ${q_{n-1}}$ is odd, then both ${a_n=a_{n+1}=1}$ and again ${\alpha_n=[0;1,1,a_{n+2},\ldots]<1/2}$. $\Box$

Lemma 7 If ${q}$ is an odd positive integer such that ${q\|q\alpha\|<1/2}$, then ${|S_q\phi(x)|=1}$. Furthermore, there is a partition of ${\mathbb T}$ into consecutive intervals ${I_1,\ldots,I_{2q}}$ each of length ${1/2q}$ such that ${S_q\phi}$ is ${1}$ in ${I_1,I_3,\ldots,I_{2q-1}}$ and ${-1}$ in ${I_2,I_4,\ldots,I_{2q}}$.

Proof: We proceed as in the proof of Denjoy-Koksma’s inequality. Consider the partition of ${\mathbb T}$ in the intervals ${[i/q,(i+1)/q)}$, ${i=0,1,\ldots,q-1}$. ${\phi}$ is equal to ${1}$ in ${(q-1)/2}$ of them and to ${-1}$ in ${(q-1)/2}$ of them, so the first claim follows if we prove that for each ${k=0,1,\ldots,q-1}$ there is one and exactly one integer ${i=0,1,\ldots,q-1}$ for which

$\displaystyle \{x+k\alpha\}\in\left[\dfrac{i}{q}\,,\,\dfrac{i+1}{q}\right)\,\cdot$

Let ${x\in [j/q,(j+1)/q)}$. Because ${S_q\phi(x+1/2)=-S_q\phi(x)}$, we can take ${x\in [j/q,j/q+1/2q)}$. Without loss of generality, assume that

$\displaystyle 0<\alpha-\dfrac{p}{q}<\dfrac{1}{2q^2}\,\cdot$

Then, for ${k=0,1,\ldots,q-1}$,

$\displaystyle 0

and so

$\displaystyle \{x+k\alpha\}\in \left[\dfrac{i+j}{q},\dfrac{i+j+1}{q}\right)\,,$

where ${i}$ is the residue of ${kp}$ modulo ${q}$. As ${kp}$ runs overs all residues modulo ${q}$, the first claim is proved. The case ${0 is similar, in which case

$\displaystyle \{x+k\alpha\}\in\left[\dfrac{i+j-1}{q},\dfrac{i+j}{q}\right)\,\cdot$

Now decompose the interval ${I=[(q-1)/2q,(q+1)/2q)}$ in which ${\phi}$ changes sign into two subintervals

$\displaystyle I_+=\left[\dfrac{q-1}{2q}\,,\,\dfrac{1}{2}\right)\ \ \text{ and }I_-=\left[\dfrac{1}{2}\,,\,\dfrac{q+1}{2q}\right).$

By the first part, to each ${x\in\mathbb T}$ there is a unique ${k=k(x)\in\{0,1,\ldots,q-1\}}$ such that ${\{x+k\alpha\}\in I}$, and ${S_q\phi(x)=1}$ if and only if ${\{x+k\alpha\}\in I_+}$, which happens for an interval of length ${1/2q}$ containing ${x}$. $\Box$

Theorem 8 For any irrational number ${\alpha}$, the cylinder flow ${F}$ is ergodic.

Proof: Let ${(q_n)}$ be a sequence of odd positive integers such that ${q_n\|q_n\alpha\|<1/2}$. We claim Lemma 7 implies that, for any ${B\subset\mathbb T}$,

$\displaystyle \lim_{n\rightarrow\infty}\mu(B\cap\{x\in\mathbb T\,;\,S_{q_n}\phi(x)=1\})=\dfrac{\mu(B)}{2}\,\cdot \ \ \ \ \ (2)$

To prove this, let ${\varepsilon>0}$ and consider a cover of ${B}$ by disjoint intervals ${J_1,\ldots,J_j}$ such that

$\displaystyle \mu\left(B\Delta\bigcup_{i=1}^j J_i\right)<\varepsilon.$

To each ${i=1,\ldots,j}$, at least ${(2q_n-2)/2\cdot |J_i|}$ of the intervals ${I_1,I_3,\ldots,I_{2q_n-1}}$ are contained in ${J_i}$ and so

$\displaystyle \mu(J_i\cap\{x\in\mathbb T\,;\,S_{q_n}\phi(x)=1\})\ge \dfrac{2q_n-2}{4q_n}\cdot|J_i|.$

Summing up this in ${i}$,

$\displaystyle \mu\left(\bigcup_{i=1}^j J_i\cap\{x\in\mathbb T\,;\,S_{q_n}\phi(x)=1\}\right)\ge \dfrac{2q_n-2}{4q_n}\cdot\mu\left(\bigcup_{i=1}^j J_i\right).$

Taking limits in both sides, (2) follows.

Now it is to see that ${1}$ is an essential value. Let ${B\subset\mathbb T}$ with ${\nu(B)>0}$. Because ${\lim_{n\rightarrow\infty}\nu(B\cap R_\alpha^{q_n}B)=\nu(B)}$, (2) implies that

$\displaystyle \lim_{n\rightarrow\infty}\nu(B\cap R_\alpha^{-q_n}B\cap \{x\in\mathbb T\,;\,S_{q_n}\phi(x)=1\})=\dfrac{1}{2}>0\,.$

$\Box$

Previous posts: CF0CF1.