1 Introduction
The ergodic theorems of Birkhoff, Hopf, von Neumann, Wiener, and Yoshida were generalized to the measure-free setting of vector lattices (Riesz spaces) in 2007, see [Reference Kuo, Labuschagne and Watson19]. The Poincaré recurrence theorem and Kac’s formula in vector lattices were published in 2023, see [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1]. In none of the above were the concept of Rokhlin towers/Kakutani–Rokhlin decomposition used and, up to the present, the concept of Rokhlin towers/Kakutani–Rokhlin decomposition had not been generalized to the vector lattice setting. In this paper, we present a Kakutani–Rokhlin decomposition for dynamical systems defined by iterates of a Riesz homomorphism acting on a vector lattice (Riesz space). This takes the work of [Reference Eisner, Farkas, Haase and Nagel5, Reference Petersen23, Reference Rokhlin24], and others, out of the realm of metric, topological, and measure spaces. We note here the contrast between the development in vector lattices and that of [Reference Bochi4], where the ergodic theorems were derived as a consequence of the Rokhlin towers/Kakutani–Rokhlin decomposition. Our generalization to vector lattices is with respect to a discrete-time process, but generalizes the underlying space of the process. Other generalizations, see for example [Reference Lindenstrauss22], have kept the underlying space as a measure space, but have extended the time-index space to amenable groups. We refer the reader to [Reference Kra14, Reference Weiss27] for some applications and other generalizations of the Rokhlin towers/Kakutani–Rokhlin decomposition.
The extension given here, when applied back to probabilistic systems gives the existence of a Kakutani–Rokhlin decomposition (also known in this context as Rokhlin towers) for conditionally ergodic processes. A consequence of this is that every conditional expectation preserving system that is aperiodic admits a Kakutani–Rokhlin decomposition. Examples are given in each stage of our development to show that the given result cannot be improved without additional assumptions.
 In the probability setting, the Kakutani–Rokhlin lemma gives that, for each 
 $n\in {\mathbb N}$
 and
$n\in {\mathbb N}$
 and 
 $\epsilon>0$
, each almost everywhere (a.e.) bijective ergodic aperiodic measure preserving transformation,
$\epsilon>0$
, each almost everywhere (a.e.) bijective ergodic aperiodic measure preserving transformation, 
 $\tau $
, on the probability space
$\tau $
, on the probability space 
 $(\Omega ,{\cal A},\mu )$
, there is a set
$(\Omega ,{\cal A},\mu )$
, there is a set 
 $B\in {\cal A}$
 so that
$B\in {\cal A}$
 so that 
 $\tau ^{-j}(B), j=0,\ldots ,n-1,$
 are disjoint and
$\tau ^{-j}(B), j=0,\ldots ,n-1,$
 are disjoint and 
 ${\mu (\Omega \setminus \bigcup _{j=1}^n \tau ^{1-j}(B))<\epsilon }$
. We refer the reader to [Reference Rokhlin24, §3.3] for the specific result and to [Reference Kornfeld13] for a survey of research around such decompositions. The generalization of the Kakutani–Rokhlin lemma to the topology-free, metric-free, measure-free setting of Riesz spaces (vector lattices) is given in Theorem 4.7.
${\mu (\Omega \setminus \bigcup _{j=1}^n \tau ^{1-j}(B))<\epsilon }$
. We refer the reader to [Reference Rokhlin24, §3.3] for the specific result and to [Reference Kornfeld13] for a survey of research around such decompositions. The generalization of the Kakutani–Rokhlin lemma to the topology-free, metric-free, measure-free setting of Riesz spaces (vector lattices) is given in Theorem 4.7.
 We begin by giving an 
 $\epsilon $
-free version of the Kakutani–Rokhlin lemma in the Riesz space setting, Theorem 3.2, see [Reference Weiss27] for the measure space version. This version applies to conditionally ergodic systems on Riesz spaces (in fact, each conditional expectation preserving system of a Riesz space gives rise to a conditionally ergodic system) and does not require aperiodicity. This then forms the foundation of Theorem 4.7, where aperiodicity is essential.
$\epsilon $
-free version of the Kakutani–Rokhlin lemma in the Riesz space setting, Theorem 3.2, see [Reference Weiss27] for the measure space version. This version applies to conditionally ergodic systems on Riesz spaces (in fact, each conditional expectation preserving system of a Riesz space gives rise to a conditionally ergodic system) and does not require aperiodicity. This then forms the foundation of Theorem 4.7, where aperiodicity is essential.
The remaining foundational aspects of ergodic theory in Riesz spaces needed for the current work can be found in [Reference Ben Amor, Homann, Kuo and Watson3, Reference Homann, Kuo and Watson9, Reference Kuo, Labuschagne and Watson16] for the general theory of conditional expectation operators in Riesz spaces. It should be noted that many other stochastic processes have been studied in the Riesz space (vector lattice) framework, for example, discrete [Reference Kuo, Labuschagne and Watson15] and continuous [Reference Grobler6, Reference Grobler7, Reference Stoica26] time martingale processes as well as mixing processes [Reference Ben Amor, Homann, Kuo and Watson3].
 In §2, we recall the basics of conditional expectation preserving systems, ergodic processes, Poincaré’s recurrence theorem, and Kac’s formula in Riesz spaces. In §3, we give a Riesz space version of the 
 $\epsilon $
-free Kakutani–Rokhlin type decompositions. In §4, we introduce aperiodicity in Riesz spaces, and use this concept together with the Kac formula and the
$\epsilon $
-free Kakutani–Rokhlin type decompositions. In §4, we introduce aperiodicity in Riesz spaces, and use this concept together with the Kac formula and the 
 $\epsilon $
-free Kakutani–Rokhlin type decomposition to give an
$\epsilon $
-free Kakutani–Rokhlin type decomposition to give an 
 $\epsilon $
-bound version of the Kakutani–Rokhlin decomposition in Riesz spaces. We conclude in §5 with an application of the Kakutani–Rokhlin theorem for aperiodic processes to show that every conditionally ergodic process which can be decomposed into aperiodic processes can be approximated by a periodic processes.
$\epsilon $
-bound version of the Kakutani–Rokhlin decomposition in Riesz spaces. We conclude in §5 with an application of the Kakutani–Rokhlin theorem for aperiodic processes to show that every conditionally ergodic process which can be decomposed into aperiodic processes can be approximated by a periodic processes.
2 Preliminaries
For Riesz space theory and associated terminology, we refer readers to [Reference Zaanen28]. The background material on ergodic theory can be found in [Reference Eisner, Farkas, Haase and Nagel5, Reference Petersen23]. Our current work builds on [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1], in which many of the foundational results can be found. The concept of a conditional expectation operator on a Riesz space is fundamental to the material presented here and, hence, we quote its definition from [Reference Kuo, Labuschagne and Watson16].
Definition 2.1. Let E be an Archimedean Riesz space with weak order unit. A positive order continuous projection 
 $T \colon E \rightarrow E$
 that maps weak order units to weak order units is called a conditional expectation if the range of T,
$T \colon E \rightarrow E$
 that maps weak order units to weak order units is called a conditional expectation if the range of T, 
 $R(T)$
, is a Dedekind complete Riesz subspace of E.
$R(T)$
, is a Dedekind complete Riesz subspace of E.
 Throughout this work, we will assume that the conditional expectation operator T is strictly positive, that is, if 
 $f\in E_+$
 and
$f\in E_+$
 and 
 $Tf=0$
, then
$Tf=0$
, then 
 $f=0$
.
$f=0$
.
The Riesz space analogue of a measure-preserving system was introduced in [Reference Homann, Kuo and Watson10] as a conditional expectation preserving system, see below. The concept was first used and studied in [Reference Kuo, Labuschagne and Watson19], but not given a name there.
Definition 2.2. The 4-tuple, 
 $(E,T,S,e)$
, is called a conditional expectation preserving system (CEPS) if E is a Dedekind complete Riesz space, e is a weak order unit of E, T is a conditional expectation operator on E with
$(E,T,S,e)$
, is called a conditional expectation preserving system (CEPS) if E is a Dedekind complete Riesz space, e is a weak order unit of E, T is a conditional expectation operator on E with 
 $Te=e$
, and S is an order continuous Riesz homomorphism on E with
$Te=e$
, and S is an order continuous Riesz homomorphism on E with 
 $Se=e$
 and
$Se=e$
 and 
 $TSf=Tf$
 for all
$TSf=Tf$
 for all 
 $f \in E$
.
$f \in E$
.
Remark 2.3. If 
 $(E,T,S,e)$
 is a conditional expectation preserving system, then
$(E,T,S,e)$
 is a conditional expectation preserving system, then 
 $$ \begin{align*}TS^jf=Tf\end{align*} $$
$$ \begin{align*}TS^jf=Tf\end{align*} $$
for all 
 $j \in {\mathbb N}_0$
 and
$j \in {\mathbb N}_0$
 and 
 $f \in E$
.
$f \in E$
.
 We also note that if S is a Riesz homomorphism with 
 $TS=T$
, where T is a strictly positive conditional expectation operator on a Dedekind complete Riesz space E, then S is order continuous, see [Reference Kalauch, Stennder and van Gaans12] for a more general study of order continuity Riesz homomorphism. To see this, we let
$TS=T$
, where T is a strictly positive conditional expectation operator on a Dedekind complete Riesz space E, then S is order continuous, see [Reference Kalauch, Stennder and van Gaans12] for a more general study of order continuity Riesz homomorphism. To see this, we let 
 $f_\alpha $
 be a downwards directed net in
$f_\alpha $
 be a downwards directed net in 
 $E^+$
 with
$E^+$
 with 
 $f_\alpha \downarrow 0$
, then
$f_\alpha \downarrow 0$
, then 
 $Sf_\alpha $
 is downwards directed with
$Sf_\alpha $
 is downwards directed with 
 $Sf_\alpha \downarrow h$
 for some
$Sf_\alpha \downarrow h$
 for some 
 $h\in E^+$
. However,
$h\in E^+$
. However, 
 $TS=T$
 so, as T is order continuous,
$TS=T$
 so, as T is order continuous, 
 $0\leftarrow Tf_\alpha =T(Sf_\alpha )\to Th$
. The strict positivity of T now gives
$0\leftarrow Tf_\alpha =T(Sf_\alpha )\to Th$
. The strict positivity of T now gives 
 $h=0$
. Hence,
$h=0$
. Hence, 
 $Sf_\alpha \to 0$
, making it order continuous.
$Sf_\alpha \to 0$
, making it order continuous.
 We recall, from [Reference Kuo, Labuschagne and Watson16], the concept of T-universal completeness, the T-universal completion, and, from [Reference Kuo, Rogans and Watson20], the 
 $R(T)$
-module structure of
$R(T)$
-module structure of 
 $L^1(T)$
, see also [Reference Azouzi and Trabelsi2].
$L^1(T)$
, see also [Reference Azouzi and Trabelsi2].
Definition 2.4. If T is a strictly positive conditional expectation operator on a Dedekind complete Riesz space, E with weak order unit 
 $e=Te$
, then the natural domain of T is
$e=Te$
, then the natural domain of T is 
 $$ \begin{align*}\mbox{dom}(T):=\{f\in E^u_+|\text{ there exists } \mbox{ net } f_\alpha\uparrow f \mbox{ in } E^u, (f_\alpha)\subset E_+, Tf_\alpha \mbox{ bounded in } E^u\},\end{align*} $$
$$ \begin{align*}\mbox{dom}(T):=\{f\in E^u_+|\text{ there exists } \mbox{ net } f_\alpha\uparrow f \mbox{ in } E^u, (f_\alpha)\subset E_+, Tf_\alpha \mbox{ bounded in } E^u\},\end{align*} $$
where 
 $E^u$
 denotes the universal completion of E. We define
$E^u$
 denotes the universal completion of E. We define 
 $$ \begin{align*}L^1(T):=\mbox{dom}(T)-\mbox{dom}(T)=\{f-g|f,g\in\mbox{dom}(T)\}\end{align*} $$
$$ \begin{align*}L^1(T):=\mbox{dom}(T)-\mbox{dom}(T)=\{f-g|f,g\in\mbox{dom}(T)\}\end{align*} $$
and say that E is T-universally complete if 
 $E=L^1(T)$
.
$E=L^1(T)$
.
 From the above definition, E is T-universally complete if, and only if, for each upwards directed net 
 $(f_{\alpha })_{\alpha \in \Lambda }$
 in
$(f_{\alpha })_{\alpha \in \Lambda }$
 in 
 $E_+$
 such that
$E_+$
 such that 
 $(Tf_{\alpha })_{\alpha \in \Lambda }$
 is order bounded in
$(Tf_{\alpha })_{\alpha \in \Lambda }$
 is order bounded in 
 $E^{u}$
, we have that
$E^{u}$
, we have that 
 $(f_{\alpha })_{\alpha \in \Lambda }$
 is order convergent in E.
$(f_{\alpha })_{\alpha \in \Lambda }$
 is order convergent in E.
 Here, 
 $E^u$
 has an f-algebra structure which can be chosen so that e is the multiplicative identity. For T acting on
$E^u$
 has an f-algebra structure which can be chosen so that e is the multiplicative identity. For T acting on 
 $E=L^1(T)$
,
$E=L^1(T)$
, 
 $R(T)$
 is a universally complete and thus an f-algebra, and, further,
$R(T)$
 is a universally complete and thus an f-algebra, and, further, 
 $L^1(T)$
 is an
$L^1(T)$
 is an 
 $R(T)$
-module. From [Reference Kuo, Labuschagne and Watson16, Theorem 5.3], T is an averaging operator, which means that if
$R(T)$
-module. From [Reference Kuo, Labuschagne and Watson16, Theorem 5.3], T is an averaging operator, which means that if 
 $f\in R(T)$
 and
$f\in R(T)$
 and 
 $g\in E$
, then
$g\in E$
, then 
 $T(fg)=fT(g)$
.
$T(fg)=fT(g)$
.
 From [Reference Kuo, Labuschagne and Watson19], for each 
 $f\in L^1(T)$
, the Cesàro mean
$f\in L^1(T)$
, the Cesàro mean 
 $$ \begin{align} L_Sf:=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}S^kf, \end{align} $$
$$ \begin{align} L_Sf:=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}S^kf, \end{align} $$
converges in order, in 
 $L^1(T)$
, for each Riesz homomorphism S on
$L^1(T)$
, for each Riesz homomorphism S on 
 $E=L^1(T)$
 with
$E=L^1(T)$
 with 
 $TS=T$
 and
$TS=T$
 and 
 $Se=e$
. We denote the invariant set of the Riesz homomorphism, S, by
$Se=e$
. We denote the invariant set of the Riesz homomorphism, S, by 
 $$ \begin{align*}{\cal I}_S:=\{f\in L^1(T) : Sf=f\}.\end{align*} $$
$$ \begin{align*}{\cal I}_S:=\{f\in L^1(T) : Sf=f\}.\end{align*} $$
 We say that 
 $p\in E_+$
 is a component of
$p\in E_+$
 is a component of 
 $q\in E_+$
 if
$q\in E_+$
 if 
 $p\le q$
 and
$p\le q$
 and 
 $(q-p)\wedge p=0$
. We denote the set of components of q by
$(q-p)\wedge p=0$
. We denote the set of components of q by 
 $C_q$
.
$C_q$
.
 The conditional expectation preserving system 
 $(E=L^1(T),T,S,e)$
 is said to be conditionally ergodic if
$(E=L^1(T),T,S,e)$
 is said to be conditionally ergodic if 
 $L_S=T$
, which is equivalent to
$L_S=T$
, which is equivalent to 
 ${\cal I}_S=R(T)$
, see [Reference Homann, Kuo and Watson9], in which case,
${\cal I}_S=R(T)$
, see [Reference Homann, Kuo and Watson9], in which case, 
 $ST=T$
 and hence
$ST=T$
 and hence 
 $S^jTf=Tf$
 for all
$S^jTf=Tf$
 for all 
 $j \in {\mathbb N}_0$
 and
$j \in {\mathbb N}_0$
 and 
 $f \in E$
.
$f \in E$
.
Lemma 2.5. If 
 $(E,T,S,e)$
 is a conditional expectation preserving system and T is strictly positive, then
$(E,T,S,e)$
 is a conditional expectation preserving system and T is strictly positive, then 
 $Sg=g$
 for all
$Sg=g$
 for all 
 $g\in R(T)$
. In the case of E being an
$g\in R(T)$
. In the case of E being an 
 $R(T)$
 module, this invariance gives that
$R(T)$
 module, this invariance gives that 
 $S(gf)=gSf$
 for all
$S(gf)=gSf$
 for all 
 $g\in R(T)$
 and
$g\in R(T)$
 and 
 $f\in E$
.
$f\in E$
.
Proof. Due to the order continuity of S and the order density of the linear combinations of components of e in E, it suffices to prove the result for 
 $g\in C_e\cap R(T)$
 and
$g\in C_e\cap R(T)$
 and 
 $f\in C_e$
.
$f\in C_e$
.
 For 
 $g\in C_e\cap R(T)$
, we have that
$g\in C_e\cap R(T)$
, we have that 
 $$ \begin{align*}g=Tg=TSg.\end{align*} $$
$$ \begin{align*}g=Tg=TSg.\end{align*} $$
The averaging property of conditional expectations operators in terms of band projections gives that 
 $P_{TSg}\ge P_{Sg}$
, where these are respectively the band projections generated by
$P_{TSg}\ge P_{Sg}$
, where these are respectively the band projections generated by 
 $TSg$
 and
$TSg$
 and 
 $Sg$
, see [Reference Kuo, Labuschagne, Watson, Liu, Chen and Ying17, Corollary 2.3] and [Reference Kuo, Labuschagne and Watson18, Lemma 2.3]. Here,
$Sg$
, see [Reference Kuo, Labuschagne, Watson, Liu, Chen and Ying17, Corollary 2.3] and [Reference Kuo, Labuschagne and Watson18, Lemma 2.3]. Here, 
 $Sg$
 is a component of e so
$Sg$
 is a component of e so 
 $P_{Sg}e=Sg$
. Further, as
$P_{Sg}e=Sg$
. Further, as 
 $g=TSg$
, which is a component of e, we have
$g=TSg$
, which is a component of e, we have 
 $P_{TSg}e=g$
. Thus,
$P_{TSg}e=g$
. Thus, 
 $g\ge Sg$
. As T is strictly positive and S is a Riesz homomorphism by [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1, Note 2.3], we have
$g\ge Sg$
. As T is strictly positive and S is a Riesz homomorphism by [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1, Note 2.3], we have 
 $Sg=g$
.
$Sg=g$
.
 For the second result, if 
 $f\in C_e$
, then
$f\in C_e$
, then 
 $fg=f\wedge g$
, so
$fg=f\wedge g$
, so 
 $$ \begin{align*}S(gf)=S(g)\wedge S(f)=g\wedge S(f) =gSf,\end{align*} $$
$$ \begin{align*}S(gf)=S(g)\wedge S(f)=g\wedge S(f) =gSf,\end{align*} $$
since 
 $Sf$
 is also a component of e.
$Sf$
 is also a component of e.
In [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1, Lemma 3.1], an equivalent formulation for the definition of recurrence in [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1, Definition 1.4] was proved. For convenience, here, we will take this equivalent statement as a our definition of recurrence below.
Definition 2.6. (Recurrence)
 Let 
 $(E,T,S,e)$
 be a conditional expectation preserving system with S bijective, then
$(E,T,S,e)$
 be a conditional expectation preserving system with S bijective, then 
 $p\in C_q$
 is recurrent with respect to
$p\in C_q$
 is recurrent with respect to 
 $q\in C_ e$
 if
$q\in C_ e$
 if 
 $$ \begin{align*}p\le\bigvee_{n\in{\mathbb N}} S^{-n} q.\end{align*} $$
$$ \begin{align*}p\le\bigvee_{n\in{\mathbb N}} S^{-n} q.\end{align*} $$
The following Riesz space generalization of the Poincaré recurrence theorem was proved in [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1, Theorem 3.2].
For brevity of notation, we define the supremum over an empty family of components of e to be zero, that is,
 $$ \begin{align*}\bigvee\limits_{j=1}^{0}p_j:=0\end{align*} $$
$$ \begin{align*}\bigvee\limits_{j=1}^{0}p_j:=0\end{align*} $$
for 
 $(p_j)\subset C_e$
.
$(p_j)\subset C_e$
.
Theorem 2.7. (Poincaré)
 Let 
 $(E,T,S,e)$
 be a conditional expectation preserving system with T strictly positive and S surjective, then each
$(E,T,S,e)$
 be a conditional expectation preserving system with T strictly positive and S surjective, then each 
 $p\in C_q$
 is recurrent with respect to q for each
$p\in C_q$
 is recurrent with respect to q for each 
 $q\in C_e$
.
$q\in C_e$
.
 For 
 $k \in {\mathbb N}$
, let
$k \in {\mathbb N}$
, let 
 $$ \begin{align*}q(p,k):=p \wedge (S^{-k}p) \wedge \bigg(e-\bigvee\limits_{j=1}^{k-1}S^{-j}p\bigg).\end{align*} $$
$$ \begin{align*}q(p,k):=p \wedge (S^{-k}p) \wedge \bigg(e-\bigvee\limits_{j=1}^{k-1}S^{-j}p\bigg).\end{align*} $$
 Here, 
 $q(p,k)$
 is the maximal component of p recurrent at exactly k iterates of S and
$q(p,k)$
 is the maximal component of p recurrent at exactly k iterates of S and 
 $$ \begin{align*}q(p,k)\wedge q(p,m)=0\quad \mbox{for } k\ne m,\, k,m\in{\mathbb N}.\end{align*} $$
$$ \begin{align*}q(p,k)\wedge q(p,m)=0\quad \mbox{for } k\ne m,\, k,m\in{\mathbb N}.\end{align*} $$
 Writing Theorem 2.7 in terms of 
 $q(p,k), k\in {\mathbb N}$
, we obtain the next corollary.
$q(p,k), k\in {\mathbb N}$
, we obtain the next corollary.
Corollary 2.8. Let 
 $(E,T,S,e)$
 be a conditional expectation preserving system with T strictly positive and S surjective, then for each component p of e, we have
$(E,T,S,e)$
 be a conditional expectation preserving system with T strictly positive and S surjective, then for each component p of e, we have 
 $$ \begin{align*}p=\sum\limits_{k=1}^\infty q(p,k).\end{align*} $$
$$ \begin{align*}p=\sum\limits_{k=1}^\infty q(p,k).\end{align*} $$
Here, this summation is order convergent in E.
 From the definition of 
 $q(p,k)$
, we have that
$q(p,k)$
, we have that 
 $$ \begin{align*}S^k q(p,k) \le p\end{align*} $$
$$ \begin{align*}S^k q(p,k) \le p\end{align*} $$
for all 
 $k \in {\mathbb N}$
.
$k \in {\mathbb N}$
.
Lemma 2.9. Let 
 $(E,T,S,e)$
 be a conditional expectation preserving system with T strictly positive and S surjective, then for all
$(E,T,S,e)$
 be a conditional expectation preserving system with T strictly positive and S surjective, then for all 
 $m,n \in {\mathbb N}$
 with
$m,n \in {\mathbb N}$
 with 
 $0 \le i \le m-1, \; 0\le j \le n-1$
, and
$0 \le i \le m-1, \; 0\le j \le n-1$
, and 
 $(i,m) \ne (j,n)$
, we have
$(i,m) \ne (j,n)$
, we have 
 $$ \begin{align} S^i q(p,m) \wedge S^j q(p,n)=0. \end{align} $$
$$ \begin{align} S^i q(p,m) \wedge S^j q(p,n)=0. \end{align} $$
Proof. Let 
 $i,j,m,n$
 be as above.
$i,j,m,n$
 be as above.
 
Case I: If 
 $m\le n$
 and
$m\le n$
 and 
 $n-1\ge j> i\ge 0$
, then as
$n-1\ge j> i\ge 0$
, then as 
 $S^j$
 is a Riesz homomorphism,
$S^j$
 is a Riesz homomorphism, 
 $$ \begin{align*} S^i q(p,m) \wedge S^j q(p,n)=S^j(S^{i-j}q(p,m) \wedge q(p,n)). \end{align*} $$
$$ \begin{align*} S^i q(p,m) \wedge S^j q(p,n)=S^j(S^{i-j}q(p,m) \wedge q(p,n)). \end{align*} $$
Here,
 $$ \begin{align*} S^{i-j}q(p,m) \wedge q(p,n)\le S^{i-j}p \wedge \bigg(e-\bigvee\limits_{k=1}^{n-1}S^{-k}p\bigg)=0 \end{align*} $$
$$ \begin{align*} S^{i-j}q(p,m) \wedge q(p,n)\le S^{i-j}p \wedge \bigg(e-\bigvee\limits_{k=1}^{n-1}S^{-k}p\bigg)=0 \end{align*} $$
since 
 $i-j\in \{-k|k=1,\ldots ,n-1\}$
.
$i-j\in \{-k|k=1,\ldots ,n-1\}$
.
 
Case II: If 
 $m<n$
 and
$m<n$
 and 
 $i=j$
, then
$i=j$
, then 
 $$ \begin{align*} S^i q(p,m) \wedge S^j q(p,n)=S^i(q(p,m) \wedge q(p,n))=0 \end{align*} $$
$$ \begin{align*} S^i q(p,m) \wedge S^j q(p,n)=S^i(q(p,m) \wedge q(p,n))=0 \end{align*} $$
as 
 $m\ne n$
 and
$m\ne n$
 and 
 $S^i$
 is a Riesz homomorphism.
$S^i$
 is a Riesz homomorphism.
 
Case III: If 
 $m<n$
 and
$m<n$
 and 
 $m-1\ge i> j\ge 0$
, then
$m-1\ge i> j\ge 0$
, then 
 $$ \begin{align*} S^i q(p,m) \wedge S^j q(p,n)=S^i(q(p,m) \wedge S^{j-i}q(p,n)). \end{align*} $$
$$ \begin{align*} S^i q(p,m) \wedge S^j q(p,n)=S^i(q(p,m) \wedge S^{j-i}q(p,n)). \end{align*} $$
Here,
 $$ \begin{align*} q(p,m) \wedge S^{j-i}q(p,n)\le \bigg(e-\bigvee\limits_{k=1}^{m-1}S^{-k}p\bigg)\wedge S^{j-i}p=0, \end{align*} $$
$$ \begin{align*} q(p,m) \wedge S^{j-i}q(p,n)\le \bigg(e-\bigvee\limits_{k=1}^{m-1}S^{-k}p\bigg)\wedge S^{j-i}p=0, \end{align*} $$
since 
 $j-i\in \{-k|k=1,\ldots ,m-1\}$
.
$j-i\in \{-k|k=1,\ldots ,m-1\}$
.
 Rewriting the expression for the first recurrence time for p a component of e, 
 $n(p)$
, from [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1], in terms of
$n(p)$
, from [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1], in terms of 
 $q(p,k)$
, we get
$q(p,k)$
, we get 
 $$ \begin{align*}n(p)=\sum\limits_{k=1}^\infty k q(p,k).\end{align*} $$
$$ \begin{align*}n(p)=\sum\limits_{k=1}^\infty k q(p,k).\end{align*} $$
 The conditional Kac formula of [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1] gives the conditional expectation of 
 $n(p)$
, as follows.
$n(p)$
, as follows.
Theorem 2.10. (Kac)
 Let 
 $(E,T,S,e)$
 be a conditionally ergodic conditional expectation preserving system, where T is strictly positive, E is T-universally complete, and S is surjective. For each p a component of e, we have that
$(E,T,S,e)$
 be a conditionally ergodic conditional expectation preserving system, where T is strictly positive, E is T-universally complete, and S is surjective. For each p a component of e, we have that 
 $$ \begin{align*}Tn(p)=P_{Tp}e,\end{align*} $$
$$ \begin{align*}Tn(p)=P_{Tp}e,\end{align*} $$
where 
 $P_{Tp}$
 is the band projection onto the band in E generated by
$P_{Tp}$
 is the band projection onto the band in E generated by 
 $Tp$
.
$Tp$
.
3 Kakutani–Rokhlin lemma—
 $\epsilon $
-free
$\epsilon $
-free
 We recall an 
 $\epsilon $
-free version of the Kakutani–Rokhlin decomposition for ergodic measure-preserving systems from [Reference Lehrer and Weiss21, Theorem 2] and [Reference Eisner, Farkas, Haase and Nagel5, Theorem 6.24]. We note here that these references state the bound
$\epsilon $
-free version of the Kakutani–Rokhlin decomposition for ergodic measure-preserving systems from [Reference Lehrer and Weiss21, Theorem 2] and [Reference Eisner, Farkas, Haase and Nagel5, Theorem 6.24]. We note here that these references state the bound 
 $1-n\mu (A)$
; however, their proofs yield the better bound given below.
$1-n\mu (A)$
; however, their proofs yield the better bound given below.
Theorem 3.1. Let 
 $(\Omega ,{\cal B},\mu ,\tau )$
 be an ergodic measure preserving system, let
$(\Omega ,{\cal B},\mu ,\tau )$
 be an ergodic measure preserving system, let 
 $A \in {\cal B}$
 with
$A \in {\cal B}$
 with 
 $\mu (A)>0$
 and
$\mu (A)>0$
 and 
 $n \in {\mathbb N}$
. Then, there is a set
$n \in {\mathbb N}$
. Then, there is a set 
 $B \in {\cal B}$
 such that
$B \in {\cal B}$
 such that 
 $B, \tau ^{-1} B, \ldots , \tau ^{1-n}B$
 are pairwise disjoint and
$B, \tau ^{-1} B, \ldots , \tau ^{1-n}B$
 are pairwise disjoint and 
 $$ \begin{align*}\mu\bigg(\bigcup\limits_{i=0}^{n-1}\tau^{-i} B\bigg) \ge 1-(n-1)\mu(A).\end{align*} $$
$$ \begin{align*}\mu\bigg(\bigcup\limits_{i=0}^{n-1}\tau^{-i} B\bigg) \ge 1-(n-1)\mu(A).\end{align*} $$
 We now give a conditional Riesz space version of the previous result. If, in the following result, p is taken as the characteristic function, 
 $\chi _A$
, with A of the above result, and T is the expectation with respect to a probability measure
$\chi _A$
, with A of the above result, and T is the expectation with respect to a probability measure 
 $\mu $
, then the below result yields immediately the above result. However, if T is a conditional expectation, then the below result yields the above, but with
$\mu $
, then the below result yields immediately the above result. However, if T is a conditional expectation, then the below result yields the above, but with 
 $\mu $
 being the conditional probability induced by T.
$\mu $
 being the conditional probability induced by T.
Theorem 3.2. (Kakutani–Rokhlin lemma)
 Let 
 $(E,T,S,e)$
 be a conditionally ergodic conditional expectation preserving system with S surjective. Let
$(E,T,S,e)$
 be a conditionally ergodic conditional expectation preserving system with S surjective. Let 
 $n \in {\mathbb N}$
 and p be a component of e, then there exists a component q of
$n \in {\mathbb N}$
 and p be a component of e, then there exists a component q of 
 $P_{Tp}e$
 such that
$P_{Tp}e$
 such that 
 $q, Sq, \ldots , S^{n-1}q$
 are pairwise disjoint and
$q, Sq, \ldots , S^{n-1}q$
 are pairwise disjoint and 
 $$ \begin{align} T\bigg(\bigvee\limits_{j=0}^{n-1}S^j q\bigg) \ge (P_{Tp}e - (n-1) Tp)^+. \end{align} $$
$$ \begin{align} T\bigg(\bigvee\limits_{j=0}^{n-1}S^j q\bigg) \ge (P_{Tp}e - (n-1) Tp)^+. \end{align} $$
Proof. By Corollary 2.8, p can be decomposed into a sum of disjoint components as follows:
 $$ \begin{align*}p=\sum\limits_{i=1}^\infty q(p,i).\end{align*} $$
$$ \begin{align*}p=\sum\limits_{i=1}^\infty q(p,i).\end{align*} $$
Let
 $$ \begin{align*}R_k=\sum\limits_{i=k}^\infty q(p,i)=\bigvee\limits_{i=k}^\infty q(p,i),\end{align*} $$
$$ \begin{align*}R_k=\sum\limits_{i=k}^\infty q(p,i)=\bigvee\limits_{i=k}^\infty q(p,i),\end{align*} $$
then 
 $R_k$
 is the maximal component of p with no component recurrent in under k steps.
$R_k$
 is the maximal component of p with no component recurrent in under k steps.
 For fixed 
 $n\in {\mathbb N}$
 and
$n\in {\mathbb N}$
 and 
 $k,j\in {\mathbb N}_0$
 with
$k,j\in {\mathbb N}_0$
 with 
 $k\ne j$
, from equation (2.2), we have
$k\ne j$
, from equation (2.2), we have 
 $$ \begin{align} S^{nj}R_{n(j+1)}\wedge S^{nk}R_{n(k+1)} =\bigvee\limits_{i\ge n(j+1)} \bigvee\limits_{r\ge n(k+1)} (S^{nj}q(p,i)\wedge S^{nk}q(p,r))=0, \end{align} $$
$$ \begin{align} S^{nj}R_{n(j+1)}\wedge S^{nk}R_{n(k+1)} =\bigvee\limits_{i\ge n(j+1)} \bigvee\limits_{r\ge n(k+1)} (S^{nj}q(p,i)\wedge S^{nk}q(p,r))=0, \end{align} $$
since 
 $nj<i$
,
$nj<i$
, 
 $nk<r$
 and
$nk<r$
 and 
 $nj\ne nk$
. Let
$nj\ne nk$
. Let 
 $$ \begin{align} q:=\sum\limits_{j=0}^{\infty} S^{nj}R_{n(j+1)}=\bigvee\limits_{j=0}^{\infty} S^{nj}R_{n(j+1)}= \sum\limits_{j=0}^{\infty}\sum\limits_{i \ge n(j+1)}S^{nj} q(p,i). \end{align} $$
$$ \begin{align} q:=\sum\limits_{j=0}^{\infty} S^{nj}R_{n(j+1)}=\bigvee\limits_{j=0}^{\infty} S^{nj}R_{n(j+1)}= \sum\limits_{j=0}^{\infty}\sum\limits_{i \ge n(j+1)}S^{nj} q(p,i). \end{align} $$
Here, q is a component of e.
 We now show that 
 $S^i q \wedge S^j q=0$
 for all
$S^i q \wedge S^j q=0$
 for all 
 $0\le i<j \le n-1$
. For this, it suffices to prove that
$0\le i<j \le n-1$
. For this, it suffices to prove that 
 $q \wedge S^k q=0$
 for all
$q \wedge S^k q=0$
 for all 
 $1 \le k \le n-1$
. If
$1 \le k \le n-1$
. If 
 $j,m\in {\mathbb N}_0$
, with
$j,m\in {\mathbb N}_0$
, with 
 $i \ge n(j+1)$
 and
$i \ge n(j+1)$
 and 
 $r \ge n(m+1)$
, then
$r \ge n(m+1)$
, then 
 $nj\ne nm+k$
,
$nj\ne nm+k$
, 
 $i>nj$
, and
$i>nj$
, and 
 $r>nm+k$
, so, by equation (2.2),
$r>nm+k$
, so, by equation (2.2), 
 $$ \begin{align} q\wedge S^k q= \sum\limits_{j,m=0}^{\infty}\sum\limits_{i \ge n(j+1)} \sum\limits_{r \ge n(m+1)}S^{nj} q(p,i)\wedge S^{nm+k} q(p,r)=0. \end{align} $$
$$ \begin{align} q\wedge S^k q= \sum\limits_{j,m=0}^{\infty}\sum\limits_{i \ge n(j+1)} \sum\limits_{r \ge n(m+1)}S^{nj} q(p,i)\wedge S^{nm+k} q(p,r)=0. \end{align} $$
Thus, 
 $q, Sq,\ldots ,S^{n-1}q$
 are disjoint.
$q, Sq,\ldots ,S^{n-1}q$
 are disjoint.
We now proceed to the proof of equation (3.1). From the definition of q in equation (3.3), we have
 $$ \begin{align*}\bigvee\limits_{i=0}^{n-1}S^i q=\sum\limits_{i=0}^{n-1}S^i q =\sum\limits_{i=0}^{n-1}\sum\limits_{j=0}^{\infty}\sum\limits_{k \ge n(j+1)}S^{nj+i} q(p,k).\end{align*} $$
$$ \begin{align*}\bigvee\limits_{i=0}^{n-1}S^i q=\sum\limits_{i=0}^{n-1}S^i q =\sum\limits_{i=0}^{n-1}\sum\limits_{j=0}^{\infty}\sum\limits_{k \ge n(j+1)}S^{nj+i} q(p,k).\end{align*} $$
Applying T to the above equation and using 
 $TS^i=T, i\in {\mathbb N}_0,$
 along with the order continuity of T, we have
$TS^i=T, i\in {\mathbb N}_0,$
 along with the order continuity of T, we have 
 $$ \begin{align*} T\bigg(\bigvee\limits_{k=0}^{n-1}S^k q\bigg) &=\sum\limits_{k=0}^{n-1}\sum\limits_{j=0}^{\infty}\sum\limits_{i= n(j+1)}^\infty T q(p,i)\\ &=\sum\limits_{j=0}^{\infty}\sum\limits_{i=n(j+1)}^\infty nTq(p,i)\\ &=\sum\limits_{i=0}^\infty n \bigg[\frac{i}{n}\bigg]Tq(p,i). \end{align*} $$
$$ \begin{align*} T\bigg(\bigvee\limits_{k=0}^{n-1}S^k q\bigg) &=\sum\limits_{k=0}^{n-1}\sum\limits_{j=0}^{\infty}\sum\limits_{i= n(j+1)}^\infty T q(p,i)\\ &=\sum\limits_{j=0}^{\infty}\sum\limits_{i=n(j+1)}^\infty nTq(p,i)\\ &=\sum\limits_{i=0}^\infty n \bigg[\frac{i}{n}\bigg]Tq(p,i). \end{align*} $$
However, by the Riesz space version of the Kac theorem, that is, Theorem 2.10, we have
 $$ \begin{align*}P_{Tp}e=Tn(p)=\sum\limits_{i=1}^\infty i Tq(p,i).\end{align*} $$
$$ \begin{align*}P_{Tp}e=Tn(p)=\sum\limits_{i=1}^\infty i Tq(p,i).\end{align*} $$
Therefore,
 $$ \begin{align*}P_{Tp}e-T\bigg(\bigvee\limits_{i=0}^{n-1}S^i q\!\bigg) =\sum\limits_{i=1}^\infty n\bigg(\frac{i}{n}-\bigg[\frac{i}{n}\bigg]\bigg) Tq(p,i) {\kern-1pt}\le{\kern-1pt} \sum\limits_{i=1}^\infty (n{\kern-1pt}-{\kern-1pt}1) Tq(p,i) =(n{\kern-1pt}-{\kern-1pt}1)Tp,\end{align*} $$
$$ \begin{align*}P_{Tp}e-T\bigg(\bigvee\limits_{i=0}^{n-1}S^i q\!\bigg) =\sum\limits_{i=1}^\infty n\bigg(\frac{i}{n}-\bigg[\frac{i}{n}\bigg]\bigg) Tq(p,i) {\kern-1pt}\le{\kern-1pt} \sum\limits_{i=1}^\infty (n{\kern-1pt}-{\kern-1pt}1) Tq(p,i) =(n{\kern-1pt}-{\kern-1pt}1)Tp,\end{align*} $$
concluding the proof of equation (3.1).
Example 3.3. We now give an example of where Theorem 3.2 cannot be improved to the 
 $\epsilon $
 approximation of Theorem 4.7. Consider the Riesz space
$\epsilon $
 approximation of Theorem 4.7. Consider the Riesz space 
 $E={\mathbb R}\times {\mathbb R}$
 with componentwise ordering and weak order unit
$E={\mathbb R}\times {\mathbb R}$
 with componentwise ordering and weak order unit 
 $e=(1,1)$
 and order continuous Riesz homomorphism
$e=(1,1)$
 and order continuous Riesz homomorphism 
 $S(x,y)=(y,x)$
. We take as our conditional expectation
$S(x,y)=(y,x)$
. We take as our conditional expectation 
 $T(x,y)=({x+y})/{2}(1,1)$
. It is easily verified that
$T(x,y)=({x+y})/{2}(1,1)$
. It is easily verified that 
 $(E,T,S,e)$
 is a conditionally ergodic conditional expectation preserving system. Taking
$(E,T,S,e)$
 is a conditionally ergodic conditional expectation preserving system. Taking 
 $p=(1,0)$
 in Theorem 3.2, we have that
$p=(1,0)$
 in Theorem 3.2, we have that 
 $Tp=\tfrac 12(1,1)$
 giving
$Tp=\tfrac 12(1,1)$
 giving 
 $$ \begin{align*}(P_{Tp}e-(n-1)Tp)^+=\bigg((1,1)-\frac{n-1}{2}(1,1)\bigg)^+ =\begin{cases}(1,1),&n=1,\\ \dfrac{1}{2}(1,1),&n=2,\\ (0,0),& n\ge 3.\end{cases}\end{align*} $$
$$ \begin{align*}(P_{Tp}e-(n-1)Tp)^+=\bigg((1,1)-\frac{n-1}{2}(1,1)\bigg)^+ =\begin{cases}(1,1),&n=1,\\ \dfrac{1}{2}(1,1),&n=2,\\ (0,0),& n\ge 3.\end{cases}\end{align*} $$
The required components of 
 $P_{Tp}e=e$
 for the respective values of n are
$P_{Tp}e=e$
 for the respective values of n are 
 $q_1=(1,1)$
,
$q_1=(1,1)$
, 
 $q_2=(0,1)$
, and
$q_2=(0,1)$
, and 
 $q=(0,0)$
 for
$q=(0,0)$
 for 
 $n\ge 3$
. Then,
$n\ge 3$
. Then, 
 $q_n,\ldots S^{n-1}q_n$
 are disjoint and
$q_n,\ldots S^{n-1}q_n$
 are disjoint and 
 $$ \begin{align*}T\bigg(\bigvee\limits_{j=0}^{n-1}S^j q_n\bigg) =\begin{cases}(1,1),&n=1,\\ (1,1),&n=2,\\ (0,0),&n\ge 3.\end{cases}\end{align*} $$
$$ \begin{align*}T\bigg(\bigvee\limits_{j=0}^{n-1}S^j q_n\bigg) =\begin{cases}(1,1),&n=1,\\ (1,1),&n=2,\\ (0,0),&n\ge 3.\end{cases}\end{align*} $$
For this example, the 
 $\epsilon $
 bound of Theorem 4.7 fails for
$\epsilon $
 bound of Theorem 4.7 fails for 
 $0<\epsilon <\tfrac 12$
 and
$0<\epsilon <\tfrac 12$
 and 
 $n=2$
.
$n=2$
.
Example 3.4. As in [Reference Azouzi, Ben Amor, Homann, Masmoudi and Watson1, §5], let 
 $(\Omega ,{\cal A},\mu )$
 be a probability space, where
$(\Omega ,{\cal A},\mu )$
 be a probability space, where 
 $\mu $
 is a complete measure. Let
$\mu $
 is a complete measure. Let 
 $\Sigma $
 be a sub-
$\Sigma $
 be a sub-
 $\sigma $
-algebra of
$\sigma $
-algebra of 
 ${\cal A}$
. As the Riesz space E, we take the space of a.e. equivalence classes of measurable functions
${\cal A}$
. As the Riesz space E, we take the space of a.e. equivalence classes of measurable functions 
 $f:\Omega \to {\mathbb R}$
 for which the sequence
$f:\Omega \to {\mathbb R}$
 for which the sequence 
 $({\mathbb E}[\min (|f(x)|, \mathbf { n})|\Sigma ])_{n\in {\mathbb N}}$
 is bounded above by an a.e. finite valued measurable function. Here,
$({\mathbb E}[\min (|f(x)|, \mathbf { n})|\Sigma ])_{n\in {\mathbb N}}$
 is bounded above by an a.e. finite valued measurable function. Here, 
 $\mathbf {n}$
 is the (equivalence class of the) function with value n a.e. For
$\mathbf {n}$
 is the (equivalence class of the) function with value n a.e. For 
 $f\in E$
 with
$f\in E$
 with 
 $f\ge 0$
, we define
$f\ge 0$
, we define 
 $$ \begin{align*}Tf=\lim_{n\to\infty} {\mathbb E}[\min(f(x), \mathbf{n})|\Sigma]\end{align*} $$
$$ \begin{align*}Tf=\lim_{n\to\infty} {\mathbb E}[\min(f(x), \mathbf{n})|\Sigma]\end{align*} $$
in the sense of a.e. pointwise limits. We now extend T to E by setting 
 $Tf=Tf^+-Tf^-$
. This T is the maximal extension of
$Tf=Tf^+-Tf^-$
. This T is the maximal extension of 
 ${\mathbb E}[\cdot |\Sigma ]$
 as a conditional expectation operator, and we will denote it again by
${\mathbb E}[\cdot |\Sigma ]$
 as a conditional expectation operator, and we will denote it again by 
 ${\mathbb E}[\cdot |\Sigma ]$
. The space E has the a.e. equivalence class of the constant
${\mathbb E}[\cdot |\Sigma ]$
. The space E has the a.e. equivalence class of the constant 
 $1$
 function as a weak order unit. The space E is a T-universally complete Riesz space with weak order unit
$1$
 function as a weak order unit. The space E is a T-universally complete Riesz space with weak order unit 
 $\mathbf {1}$
 and T is a strictly positive Riesz space conditional expectation operator on E having
$\mathbf {1}$
 and T is a strictly positive Riesz space conditional expectation operator on E having 
 $T\mathbf {1}=\mathbf { 1}$
. If we take
$T\mathbf {1}=\mathbf { 1}$
. If we take 
 $\tau :\Omega \to \Omega $
 to be a map with
$\tau :\Omega \to \Omega $
 to be a map with 
 $\tau ^{-1}(A)\in {\cal A}$
 and
$\tau ^{-1}(A)\in {\cal A}$
 and 
 ${\mathbb E}[\chi _{\tau ^{-1}(A)}|\Sigma ]={\mathbb E}[\chi _{A}|\Sigma ]$
 for all
${\mathbb E}[\chi _{\tau ^{-1}(A)}|\Sigma ]={\mathbb E}[\chi _{A}|\Sigma ]$
 for all 
 $A\in {\cal A}$
 and set
$A\in {\cal A}$
 and set 
 $Sf:=f\circ \tau $
, the Koopman map, then S is a Riesz homomorphism on E with
$Sf:=f\circ \tau $
, the Koopman map, then S is a Riesz homomorphism on E with 
 $S\mathbf {1}=\mathbf {1}$
 and
$S\mathbf {1}=\mathbf {1}$
 and 
 $TS=T$
. Further, if for each
$TS=T$
. Further, if for each 
 $A\in {\cal A}$
 there is
$A\in {\cal A}$
 there is 
 $B_A\in {\cal A}$
 so that
$B_A\in {\cal A}$
 so that 
 $\mu (A\Delta \tau ^{-1}(B_A))=0$
, then S is a surjective.
$\mu (A\Delta \tau ^{-1}(B_A))=0$
, then S is a surjective.
 The system 
 $(E,T,S,e)$
 is a conditional expectation preserving system, with S surjective and
$(E,T,S,e)$
 is a conditional expectation preserving system, with S surjective and 
 $$ \begin{align*}L_Sf=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ \tau^k\end{align*} $$
$$ \begin{align*}L_Sf=\lim_{n\to\infty}\frac{1}{n}\sum_{k=0}^{n-1}f\circ \tau^k\end{align*} $$
converges a.e. pointwise to a conditional expectation operator on E (which when restricted to 
 $L^1(\Omega ,{\cal A},\mu )$
, is a classical conditional expectation operator). The system
$L^1(\Omega ,{\cal A},\mu )$
, is a classical conditional expectation operator). The system 
 $(E,T,S,e)$
 is conditionally ergodic if and only if
$(E,T,S,e)$
 is conditionally ergodic if and only if 
 $L_S=T$
.
$L_S=T$
.
 Then, Theorem 3.2 gives that if 
 $n \in {\mathbb N}$
 and
$n \in {\mathbb N}$
 and 
 $A\in {\cal A}$
, then there exists
$A\in {\cal A}$
, then there exists 
 $B\in {\cal A}$
 with
$B\in {\cal A}$
 with 
 ${\mathbb E}[A|\Sigma ]>0$
 a.e. on B such that
${\mathbb E}[A|\Sigma ]>0$
 a.e. on B such that 
 $B, \tau ^{-1}(B), \ldots , \tau ^{1-n}(B)$
 are a.e. pairwise disjoint and
$B, \tau ^{-1}(B), \ldots , \tau ^{1-n}(B)$
 are a.e. pairwise disjoint and 
 $$ \begin{align} {\mathbb E}\bigg[\bigcup_{j=0}^{n-1} \tau^{-j}B\bigg|\Sigma\bigg] \ge (\chi_{\{\omega|{\mathbb E}[\chi_A|\Sigma](\omega)>0\}} - (n-1) {\mathbb E}[A|\Sigma])^+. \end{align} $$
$$ \begin{align} {\mathbb E}\bigg[\bigcup_{j=0}^{n-1} \tau^{-j}B\bigg|\Sigma\bigg] \ge (\chi_{\{\omega|{\mathbb E}[\chi_A|\Sigma](\omega)>0\}} - (n-1) {\mathbb E}[A|\Sigma])^+. \end{align} $$
4 Aperiodicity and an 
 $\epsilon $
-bounded decomposition
$\epsilon $
-bounded decomposition
 A probability space 
 $(\Omega ,{\cal B}, \mu )$
,
$(\Omega ,{\cal B}, \mu )$
, 
 $\mu $
 is non-atomic if for any
$\mu $
 is non-atomic if for any 
 $A \in {\cal B}$
 with
$A \in {\cal B}$
 with 
 $\mu (A)>0$
, there exists
$\mu (A)>0$
, there exists 
 $B \in {\cal B}$
 with
$B \in {\cal B}$
 with 
 $B \subset A$
 and
$B \subset A$
 and 
 $0< \mu (B) < \mu (A)$
. On non-atomic measure spaces, an
$0< \mu (B) < \mu (A)$
. On non-atomic measure spaces, an 
 $\epsilon $
-bounded version of the Kakutani–Rokhlin decomposition can be obtained, see [Reference Eisner, Farkas, Haase and Nagel5, Corollary 6.25] and [Reference Petersen23, Lemma 4.7].
$\epsilon $
-bounded version of the Kakutani–Rokhlin decomposition can be obtained, see [Reference Eisner, Farkas, Haase and Nagel5, Corollary 6.25] and [Reference Petersen23, Lemma 4.7].
Theorem 4.1. Let 
 $\tau :\Omega \rightarrow \Omega $
 be an ergodic measure preserving transformation on a non-atomic measure space
$\tau :\Omega \rightarrow \Omega $
 be an ergodic measure preserving transformation on a non-atomic measure space 
 $(\Omega , {\cal B}, \mu )$
,
$(\Omega , {\cal B}, \mu )$
, 
 $n\in {\mathbb Z}$
, and
$n\in {\mathbb Z}$
, and 
 $\epsilon>0$
, then there is a measurable set
$\epsilon>0$
, then there is a measurable set 
 $B \subset \Omega $
 such that
$B \subset \Omega $
 such that 
 $B, \; \tau ^{-1} B, \;\ldots \; \tau ^{1-n}B$
 are pairwise disjoint and cover
$B, \; \tau ^{-1} B, \;\ldots \; \tau ^{1-n}B$
 are pairwise disjoint and cover 
 $\Omega $
 up to a set of measure less than
$\Omega $
 up to a set of measure less than 
 $\epsilon $
.
$\epsilon $
.
 The original 
 $\epsilon $
-bounded version of the decomposition, as developed by Rokhlin, see [Reference Rokhlin24, p. 10], was posed for ergodic measure preserving systems which are aperiodic. See also [Reference Kakutani11].
$\epsilon $
-bounded version of the decomposition, as developed by Rokhlin, see [Reference Rokhlin24, p. 10], was posed for ergodic measure preserving systems which are aperiodic. See also [Reference Kakutani11].
 On a probability space 
 $(\Omega ,{\cal B}, \mu )$
, an aperiodic transformation is a transformation whose periodic points form a set of measure
$(\Omega ,{\cal B}, \mu )$
, an aperiodic transformation is a transformation whose periodic points form a set of measure 
 $0$
 (see [Reference Rokhlin24]), that is,
$0$
 (see [Reference Rokhlin24]), that is, 
 $\mu (\{x \in \Omega \; | \; \tau ^p x=x \mbox { for some } p \in {\mathbb N} \})=0$
. We recall Rokhlin’s 1943 version of the Kakutani–Rokhlin lemma requiring aperiodicity, quoted from [Reference Weiss27].
$\mu (\{x \in \Omega \; | \; \tau ^p x=x \mbox { for some } p \in {\mathbb N} \})=0$
. We recall Rokhlin’s 1943 version of the Kakutani–Rokhlin lemma requiring aperiodicity, quoted from [Reference Weiss27].
Theorem 4.2. If 
 $\tau $
 is an aperiodic automorphism, then for any natural number n and any positive
$\tau $
 is an aperiodic automorphism, then for any natural number n and any positive 
 $\epsilon $
, there exists a measurable set
$\epsilon $
, there exists a measurable set 
 $A \subset \Omega $
 such that the sets
$A \subset \Omega $
 such that the sets 
 $A, \tau ^{-1} A,\ldots ,\tau ^{1-n}A$
 are pairwise disjoint, and the complement of their union has measure less than
$A, \tau ^{-1} A,\ldots ,\tau ^{1-n}A$
 are pairwise disjoint, and the complement of their union has measure less than 
 $\epsilon $
.
$\epsilon $
.
 Let 
 $(\Omega ,{\cal B}, \mu )$
 be a probability space. The measure
$(\Omega ,{\cal B}, \mu )$
 be a probability space. The measure 
 $\mu $
 is said to be continuous if for any
$\mu $
 is said to be continuous if for any 
 $A \in {\cal B}$
 with
$A \in {\cal B}$
 with 
 $\mu (A)>0$
 and any
$\mu (A)>0$
 and any 
 $\alpha \in {\mathbb R}$
 with
$\alpha \in {\mathbb R}$
 with 
 $0< \alpha < \mu (A)$
, there exists
$0< \alpha < \mu (A)$
, there exists 
 $B \in {\cal B}$
 with
$B \in {\cal B}$
 with 
 $B \subset A$
 and
$B \subset A$
 and 
 $ \mu (B)=\alpha $
. Note that every continuous measure is non-atomic. If
$ \mu (B)=\alpha $
. Note that every continuous measure is non-atomic. If 
 $\mu $
 is a continuous measure and
$\mu $
 is a continuous measure and 
 $\tau $
 is an ergodic measure preserving transformation, then
$\tau $
 is an ergodic measure preserving transformation, then 
 $\tau $
 is aperiodic. Indeed, as
$\tau $
 is aperiodic. Indeed, as 
 $\tau $
 is ergodic, there exists
$\tau $
 is ergodic, there exists 
 $p \in {\mathbb N}$
 such that
$p \in {\mathbb N}$
 such that 
 $\mu (A_p)>0$
, where
$\mu (A_p)>0$
, where 
 $A_p=\{x \in \Omega \; | \;\tau ^p x=x \}$
. Choose
$A_p=\{x \in \Omega \; | \;\tau ^p x=x \}$
. Choose 
 $B \subset A_p$
 with
$B \subset A_p$
 with 
 $0 < \mu (B) < {1}/{p}$
, which is possible as
$0 < \mu (B) < {1}/{p}$
, which is possible as 
 $\mu $
 is a continuous measure. The set
$\mu $
 is a continuous measure. The set 
 $C=B \cup \tau ^{-1}B \cup \cdots \cup \tau ^{1-p}B$
 is
$C=B \cup \tau ^{-1}B \cup \cdots \cup \tau ^{1-p}B$
 is 
 $\tau $
-invariant and satisfies
$\tau $
-invariant and satisfies 
 $0< \mu (C) <1$
, contradicting the assumption of ergodicity.
$0< \mu (C) <1$
, contradicting the assumption of ergodicity.
 An aperiodic transformation on a continuous measure space need not be ergodic. For example, consider the unit square 
 $[0,1]\times [0,1]$
 with Lebesgue measure. The transformation
$[0,1]\times [0,1]$
 with Lebesgue measure. The transformation 
 $$ \begin{align*}\tau(x,y)=((x+\alpha)\mbox{ mod }1,y)\quad \text{for all } (x,y) \in [0,1]\times [0,1],\end{align*} $$
$$ \begin{align*}\tau(x,y)=((x+\alpha)\mbox{ mod }1,y)\quad \text{for all } (x,y) \in [0,1]\times [0,1],\end{align*} $$
with 
 $\alpha \in [0,1]$
 irrational, is aperiodic but not ergodic.
$\alpha \in [0,1]$
 irrational, is aperiodic but not ergodic.
We extend these decompositions to the measure-free context of Riesz spaces and begin by giving (a non-pointwise) definition of periodicity in the setting of Riesz spaces.
Definition 4.3. Let 
 $(E,T,S,e)$
 be a conditional expectation preserving system and v be a component of e. We say that
$(E,T,S,e)$
 be a conditional expectation preserving system and v be a component of e. We say that 
 $(S,v)$
 is periodic if there is
$(S,v)$
 is periodic if there is 
 $N \in {\mathbb N}$
 so that for all components c of v with
$N \in {\mathbb N}$
 so that for all components c of v with 
 $0 \ne c \ne v$
, we have that
$0 \ne c \ne v$
, we have that 
 $q(c,k)=0$
 for all
$q(c,k)=0$
 for all 
 $k \ge N$
.
$k \ge N$
.
The logic of this definition is that for all such c, we have
 $$ \begin{align*}c=\bigvee\limits_{k=1}^{N-1} q(c,k)\end{align*} $$
$$ \begin{align*}c=\bigvee\limits_{k=1}^{N-1} q(c,k)\end{align*} $$
and 
 $S^kq(c,k)\le c$
 for
$S^kq(c,k)\le c$
 for 
 $k=1,\ldots , N-1.$
$k=1,\ldots , N-1.$
We note that, as in the measure theoretic setting, aperiodicity is defined as a stronger constraint than the negation of periodicity.
Definition 4.4. Let 
 $(E,T,S,e)$
 be a conditional expectation preserving system and
$(E,T,S,e)$
 be a conditional expectation preserving system and 
 $v\ne 0$
 be a component of e. We say that
$v\ne 0$
 be a component of e. We say that 
 $(S,v)$
 is aperiodic, if for each
$(S,v)$
 is aperiodic, if for each 
 $N\in {\mathbb N}$
 and each component
$N\in {\mathbb N}$
 and each component 
 $c\ne 0$
 of v, there exists
$c\ne 0$
 of v, there exists 
 $k \ge N$
 and a component u of c with
$k \ge N$
 and a component u of c with 
 $q(u,k) \ne 0$
.
$q(u,k) \ne 0$
.
Theorem 4.5. Consider 
 $E=L^1(\Omega ,{\cal B},\mu )$
 a probability space with
$E=L^1(\Omega ,{\cal B},\mu )$
 a probability space with 
 $Tf:=\mathbb {E}[f]\mathbf {1}$
, where
$Tf:=\mathbb {E}[f]\mathbf {1}$
, where 
 $e:=\mathbf {1}$
 is the constant function with value
$e:=\mathbf {1}$
 is the constant function with value 
 $1$
 a.e., and
$1$
 a.e., and 
 $Sf:=f\circ \tau $
 is the von Neumann map generated by
$Sf:=f\circ \tau $
 is the von Neumann map generated by 
 $\tau $
, a measure-preserving transformation with
$\tau $
, a measure-preserving transformation with 
 $\tau $
 a.e. surjective. Further assume that there is
$\tau $
 a.e. surjective. Further assume that there is 
 $G\in {\cal B}$
 with
$G\in {\cal B}$
 with 
 $0<\mu (G)<1$
. In this case, the measure theoretic definition of aperiodicity of
$0<\mu (G)<1$
. In this case, the measure theoretic definition of aperiodicity of 
 $\tau $
 is equivalent to the Riesz spaces definition of
$\tau $
 is equivalent to the Riesz spaces definition of 
 $(S,e)$
 being aperiodic.
$(S,e)$
 being aperiodic.
Proof. Suppose that 
 $(S,e)$
 is aperiodic, that is, for each
$(S,e)$
 is aperiodic, that is, for each 
 $N \in {\mathbb N}$
 and each component c of e, there exists
$N \in {\mathbb N}$
 and each component c of e, there exists 
 $k \ge N$
 and a component u of c with
$k \ge N$
 and a component u of c with 
 $q(u,k) \ne 0$
. Let A denote the set of periodic points of
$q(u,k) \ne 0$
. Let A denote the set of periodic points of 
 $\tau $
. By the way of contradiction, suppose that
$\tau $
. By the way of contradiction, suppose that 
 $\mu (A)>0$
. Hence, there exists
$\mu (A)>0$
. Hence, there exists 
 $N \in {\mathbb N}$
 such that
$N \in {\mathbb N}$
 such that 
 $\mu (A_N)>0$
, where
$\mu (A_N)>0$
, where 
 $A_N$
 is the set of points of period N. Let
$A_N$
 is the set of points of period N. Let 
 $c:=\chi _{A_N}$
, then from the aperiodicity of
$c:=\chi _{A_N}$
, then from the aperiodicity of 
 $(S,v)$
, there is a component u of c and
$(S,v)$
, there is a component u of c and 
 $k>N$
 so that
$k>N$
 so that 
 $q(u,k)\ne 0$
. Here, there is a measurable subset B of
$q(u,k)\ne 0$
. Here, there is a measurable subset B of 
 $A_N$
 so that
$A_N$
 so that 
 $u=\chi _B$
. Here, all points of B are of period N, giving
$u=\chi _B$
. Here, all points of B are of period N, giving 
 $q(u,j)=0$
 for all
$q(u,j)=0$
 for all 
 $j\ge N$
, which contradicts
$j\ge N$
, which contradicts 
 $q(u,k)\ne 0$
.
$q(u,k)\ne 0$
.
 Conversely, if the set of periodic points of 
 $\tau $
 has measure zero, we show that
$\tau $
 has measure zero, we show that 
 $(S,e)$
 is aperiodic.
$(S,e)$
 is aperiodic.
 Developing on [Reference Rudolph25, Lemma 3.12], we give a meaning to the set of periodic points of 
 $\tau $
 having measure zero in a point-less setting. Let
$\tau $
 having measure zero in a point-less setting. Let 
 $p_k:=\chi _{A_k}$
, where
$p_k:=\chi _{A_k}$
, where 
 $A_k$
 is the a.e. maximal measurable set which has every measurable subset invariant under
$A_k$
 is the a.e. maximal measurable set which has every measurable subset invariant under 
 $\tau ^{-k}$
. In the Riesz space terminology,
$\tau ^{-k}$
. In the Riesz space terminology, 
 $p_k$
 is the maximal component of e with
$p_k$
 is the maximal component of e with 
 $S^kv=v$
 for each v a component of
$S^kv=v$
 for each v a component of 
 $p_k$
. Now,
$p_k$
. Now, 
 $\tau $
 being aperiodic gives that
$\tau $
 being aperiodic gives that 
 $p_k=0$
 for all
$p_k=0$
 for all 
 $k\in {\mathbb N}$
.
$k\in {\mathbb N}$
.
 Suppose that 
 $(S,e)$
 is not aperiodic, then there exist
$(S,e)$
 is not aperiodic, then there exist 
 $N\in {\mathbb N}$
 and a component
$N\in {\mathbb N}$
 and a component 
 $c\ne 0$
 of e, so that, for all
$c\ne 0$
 of e, so that, for all 
 $k \ge N$
 and components u of c, we have
$k \ge N$
 and components u of c, we have 
 $q(u,k) = 0$
. Hence,
$q(u,k) = 0$
. Hence, 
 $$ \begin{align*}u=\sum_{k=1}^{N-1}q(u,k)\end{align*} $$
$$ \begin{align*}u=\sum_{k=1}^{N-1}q(u,k)\end{align*} $$
for all 
 $k\ge N$
 and u a component of c. Thus,
$k\ge N$
 and u a component of c. Thus, 
 $$ \begin{align*}S^{N!}u=u\end{align*} $$
$$ \begin{align*}S^{N!}u=u\end{align*} $$
for all components u of c, making c a component of 
 $p_{N!}=0$
. Thus,
$p_{N!}=0$
. Thus, 
 $0< c\le p_{N!}=0$
, which is a contradiction. Thus,
$0< c\le p_{N!}=0$
, which is a contradiction. Thus, 
 $(S,e)$
 is aperiodic.
$(S,e)$
 is aperiodic.
Lemma 4.6. Let 
 $(E, T, S, e)$
 be a conditionally ergodic conditional expectation preserving system with
$(E, T, S, e)$
 be a conditionally ergodic conditional expectation preserving system with 
 $E T$
-universally complete and
$E T$
-universally complete and 
 $(S,e)$
 aperiodic, then, for each
$(S,e)$
 aperiodic, then, for each 
 $N\in {\mathbb N}$
, there is a component
$N\in {\mathbb N}$
, there is a component 
 $c_N$
 of e with
$c_N$
 of e with 
 $P_{Tc_N}e=e$
 and
$P_{Tc_N}e=e$
 and 
 $S^ic_N\wedge S^jc_N=0$
 for all
$S^ic_N\wedge S^jc_N=0$
 for all 
 $i,j=0,\ldots ,N$
 with
$i,j=0,\ldots ,N$
 with 
 $i\ne j$
.
$i\ne j$
.
Proof. For each component 
 $p\ne 0$
 of e, let
$p\ne 0$
 of e, let 
 $$ \begin{align*}K_N(p)=\sum_{k=N+1}^\infty q(p,k). \end{align*} $$
$$ \begin{align*}K_N(p)=\sum_{k=N+1}^\infty q(p,k). \end{align*} $$
Here, 
 $K_N(p)$
 is a component of p and, by Lemma 2.9,
$K_N(p)$
 is a component of p and, by Lemma 2.9, 
 $$ \begin{align*}0=S^iK_N(p)\wedge S^jK_N(p)\end{align*} $$
$$ \begin{align*}0=S^iK_N(p)\wedge S^jK_N(p)\end{align*} $$
for all 
 $0\le i< j\le N$
.
$0\le i< j\le N$
.
Let
 $$ \begin{align*}{\frak G}:=\{(p,TK_N(p))\,|\,p \mbox{ a component of } e\}.\end{align*} $$
$$ \begin{align*}{\frak G}:=\{(p,TK_N(p))\,|\,p \mbox{ a component of } e\}.\end{align*} $$
Here, 
 $(e,0)\in {\frak G}$
 so
$(e,0)\in {\frak G}$
 so 
 ${\frak G}$
 is non-empty. We partially order
${\frak G}$
 is non-empty. We partially order 
 ${\frak G}$
 by
${\frak G}$
 by 
 $(p,TK_N(p))\le (\tilde {p},TK_N(\tilde {p}))$
 if and only if
$(p,TK_N(p))\le (\tilde {p},TK_N(\tilde {p}))$
 if and only if 
 $p\le \tilde {p}$
 and
$p\le \tilde {p}$
 and 
 $TK_N(p)\le TK_N(\tilde {p})$
.
$TK_N(p)\le TK_N(\tilde {p})$
.
 If 
 $(p, Tk_N(p))_{p\in \Lambda }$
 is a chain (totally ordered subset) in
$(p, Tk_N(p))_{p\in \Lambda }$
 is a chain (totally ordered subset) in 
 ${\frak G}$
, let
${\frak G}$
, let 
 $$ \begin{align*}\hat{p}=\bigvee_{p\in \Lambda} p.\end{align*} $$
$$ \begin{align*}\hat{p}=\bigvee_{p\in \Lambda} p.\end{align*} $$
Here,
 $$ \begin{align*}\hat{p}=\lim_{p\in \Lambda} p,\end{align*} $$
$$ \begin{align*}\hat{p}=\lim_{p\in \Lambda} p,\end{align*} $$
where 
 $(p)_{p\in \Lambda }$
 is an upwards directed net, directed by the partial ordering in the Riesz space. By Lemma 2.8,
$(p)_{p\in \Lambda }$
 is an upwards directed net, directed by the partial ordering in the Riesz space. By Lemma 2.8, 
 $$ \begin{align*}K_N(p)=p-\sum_{k=1}^N q(p,k),\end{align*} $$
$$ \begin{align*}K_N(p)=p-\sum_{k=1}^N q(p,k),\end{align*} $$
making 
 $K_N(p)$
 order continuous in p, see the definition of
$K_N(p)$
 order continuous in p, see the definition of 
 $q(p,k)$
. Thus,
$q(p,k)$
. Thus, 
 $$ \begin{align*}TK_N(\hat{p})=\lim_{p\in\Lambda} TK_N(p).\end{align*} $$
$$ \begin{align*}TK_N(\hat{p})=\lim_{p\in\Lambda} TK_N(p).\end{align*} $$
Further, by the ordering on 
 ${\frak G}$
, the net
${\frak G}$
, the net 
 $(TK_N(p))_{p\in \Lambda }$
 is increasing and bounded, thus having
$(TK_N(p))_{p\in \Lambda }$
 is increasing and bounded, thus having 
 $$ \begin{align*}\lim_{p\in\Lambda} TK_N(p)=\bigvee_{p\in\Lambda} TK_N(p).\end{align*} $$
$$ \begin{align*}\lim_{p\in\Lambda} TK_N(p)=\bigvee_{p\in\Lambda} TK_N(p).\end{align*} $$
Hence, we have
 $$ \begin{align*}TK_N(\hat{p})=\bigvee_{p\in\Lambda} TK_N(p),\end{align*} $$
$$ \begin{align*}TK_N(\hat{p})=\bigvee_{p\in\Lambda} TK_N(p),\end{align*} $$
making 
 $(\hat {p}, Tk_N(\hat {p}))$
 an upper bound (in fact, the supremum) for
$(\hat {p}, Tk_N(\hat {p}))$
 an upper bound (in fact, the supremum) for 
 $(p, Tk_N(p))_{p\in \Lambda }$
.
$(p, Tk_N(p))_{p\in \Lambda }$
.
 Thus, Zorn’s lemma can be applied to 
 ${\frak G}$
 to give that it has a maximal element, say
${\frak G}$
 to give that it has a maximal element, say 
 $(p, Tk_N(p))$
.
$(p, Tk_N(p))$
.
 If 
 $P_{TK_N(p)}e\ne e$
, let
$P_{TK_N(p)}e\ne e$
, let 
 $u=e-P_{TK_N(p)}e$
, then u is a non-zero component of e, so by the aperiodicity of
$u=e-P_{TK_N(p)}e$
, then u is a non-zero component of e, so by the aperiodicity of 
 $(S,e)$
, there exists
$(S,e)$
, there exists 
 $k>N$
 and
$k>N$
 and 
 $c_k(u)$
 a component of u with
$c_k(u)$
 a component of u with 
 $q(c_k(u),k)\ne 0$
. As
$q(c_k(u),k)\ne 0$
. As 
 $u\in R(T)$
, we have
$u\in R(T)$
, we have 
 $S^ju=u$
 for all
$S^ju=u$
 for all 
 $j\in {\mathbb Z}$
 and
$j\in {\mathbb Z}$
 and 
 $p\le e-u\in R(T)$
, which give
$p\le e-u\in R(T)$
, which give 
 $$ \begin{align*}K_N(c_k(u)+p)=K_N(c_k(u))+K_N(p)>K_N(p).\end{align*} $$
$$ \begin{align*}K_N(c_k(u)+p)=K_N(c_k(u))+K_N(p)>K_N(p).\end{align*} $$
Thus,
 $$ \begin{align*}(p,TK_N(p))<(p+c_k(u),TK_N(p+c_k(u))\in {\frak G},\end{align*} $$
$$ \begin{align*}(p,TK_N(p))<(p+c_k(u),TK_N(p+c_k(u))\in {\frak G},\end{align*} $$
which contradicts the maximality of 
 $(p,TK_N(p))$
. Hence,
$(p,TK_N(p))$
. Hence, 
 $P_{TK_N(p)}=e$
 and setting
$P_{TK_N(p)}=e$
 and setting 
 $c_N=K_N(p)$
 concludes the proof.
$c_N=K_N(p)$
 concludes the proof.
 The Kakutani–Rokhlin lemma with 
 $\epsilon $
-bound can be formulated in a Riesz space as follows.
$\epsilon $
-bound can be formulated in a Riesz space as follows.
Theorem 4.7. (Riesz space Kakutani–Rokhlin)
 Let 
 $(E, T, S, e)$
 be a conditionally ergodic conditional expectation preserving system with
$(E, T, S, e)$
 be a conditionally ergodic conditional expectation preserving system with 
 $E T$
-universally complete and S surjective. If
$E T$
-universally complete and S surjective. If 
 $(S,e)$
 is aperiodic,
$(S,e)$
 is aperiodic, 
 $n \in {\mathbb N}$
 and
$n \in {\mathbb N}$
 and 
 $\epsilon>0$
, then there exists a component q of e in E with
$\epsilon>0$
, then there exists a component q of e in E with 
 $(S^i q)_{i=0,\ldots ,n-1}$
 disjoint and
$(S^i q)_{i=0,\ldots ,n-1}$
 disjoint and 
 $$ \begin{align} T\bigg(e-\bigvee\limits_{i=0}^{n-1}S^i q\bigg) \le \epsilon e. \end{align} $$
$$ \begin{align} T\bigg(e-\bigvee\limits_{i=0}^{n-1}S^i q\bigg) \le \epsilon e. \end{align} $$
Proof. Let 
 $n \in {\mathbb N}$
 and
$n \in {\mathbb N}$
 and 
 $\epsilon>0$
. Take
$\epsilon>0$
. Take 
 $N>({n-1})/{\epsilon }$
. By Lemma 4.6, there exists a component
$N>({n-1})/{\epsilon }$
. By Lemma 4.6, there exists a component 
 $c_N$
 of e with
$c_N$
 of e with 
 $P_{Tc_N}e=e$
 and
$P_{Tc_N}e=e$
 and 
 $S^ic_N\wedge S^jc_N=0$
 for all
$S^ic_N\wedge S^jc_N=0$
 for all 
 $i,j=0,\ldots ,N$
 with
$i,j=0,\ldots ,N$
 with 
 $i\ne j$
. Let
$i\ne j$
. Let 
 $p:=c_N$
. Then,
$p:=c_N$
. Then, 
 $q(p,k)=0$
 for all
$q(p,k)=0$
 for all 
 $k=1,\ldots ,N$
 giving that
$k=1,\ldots ,N$
 giving that 
 $$ \begin{align} Np \le n(p). \end{align} $$
$$ \begin{align} Np \le n(p). \end{align} $$
By Theorem 2.10, we have
 $$ \begin{align} e=P_{Tp}e=Tn(p). \end{align} $$
$$ \begin{align} e=P_{Tp}e=Tn(p). \end{align} $$
Combining equations (4.2) and (4.3), we get
 $$ \begin{align} NTp \le Tn(p)=e. \end{align} $$
$$ \begin{align} NTp \le Tn(p)=e. \end{align} $$
Since 
 $N>({n-1})/{\epsilon }$
, equation (4.4) yields
$N>({n-1})/{\epsilon }$
, equation (4.4) yields 
 $$ \begin{align} Tp \le \frac{\epsilon}{n-1}e. \end{align} $$
$$ \begin{align} Tp \le \frac{\epsilon}{n-1}e. \end{align} $$
 By Theorem 3.2, there exists a component q of 
 $P_{Tp}e=e$
 such that
$P_{Tp}e=e$
 such that 
 $q, Sq , \ldots ,S^{n-1}q$
 are pairwise disjoint and
$q, Sq , \ldots ,S^{n-1}q$
 are pairwise disjoint and 
 $$ \begin{align} T\bigg(\bigvee\limits_{j=0}^{n-1}S^j q\bigg) \ge P_{Tp}e - (n-1) Tp \ge e- \epsilon e, \end{align} $$
$$ \begin{align} T\bigg(\bigvee\limits_{j=0}^{n-1}S^j q\bigg) \ge P_{Tp}e - (n-1) Tp \ge e- \epsilon e, \end{align} $$
which gives the inequality of the theorem.
 On reading the works of Rokhlin, it appears that the requirement of conditional ergodicity is redundant and only aperiodicity is needed. As we know, every CEPS is conditionally ergodic with respect to 
 $L_S$
. So, in our case, conditional ergodicity can be dispensed with, but we need to be careful to use the conditional expectation operator
$L_S$
. So, in our case, conditional ergodicity can be dispensed with, but we need to be careful to use the conditional expectation operator 
 $L_S$
 and work in the
$L_S$
 and work in the 
 $L_S$
-universal completion of E, which we will denote by
$L_S$
-universal completion of E, which we will denote by 
 $\hat {E}$
.
$\hat {E}$
.
Corollary 4.8. Let 
 $(E, T, S, e)$
 be a conditional expectation preserving system with
$(E, T, S, e)$
 be a conditional expectation preserving system with 
 $E T$
-universally complete and S surjective, then
$E T$
-universally complete and S surjective, then 
 $(\hat {E}, L_S, S, e)$
 is a conditionally ergodic conditional expectation preserving system. If v is a component of e in
$(\hat {E}, L_S, S, e)$
 is a conditionally ergodic conditional expectation preserving system. If v is a component of e in 
 $R(L_S)$
 with
$R(L_S)$
 with 
 $(S,v)$
 aperiodic,
$(S,v)$
 aperiodic, 
 $n \in {\mathbb N}$
 and
$n \in {\mathbb N}$
 and 
 $\epsilon>0$
, then there exists a component q of v in E with
$\epsilon>0$
, then there exists a component q of v in E with 
 $(S^i q)_{i=0,\ldots ,n-1}$
 disjoint and
$(S^i q)_{i=0,\ldots ,n-1}$
 disjoint and 
 $$ \begin{align*}L_S\bigg(v-\bigvee\limits_{i=0}^{n-1}S^i q\bigg) \le \epsilon v.\end{align*} $$
$$ \begin{align*}L_S\bigg(v-\bigvee\limits_{i=0}^{n-1}S^i q\bigg) \le \epsilon v.\end{align*} $$
 Theorem 4.7 is the specific case of Corollary 4.8, where 
 $(E, T, S, e)$
 is conditionally ergodic and E is T-universally complete, as then
$(E, T, S, e)$
 is conditionally ergodic and E is T-universally complete, as then 
 $L_S=T$
.
$L_S=T$
.
Example 4.9. Continuing on Example 3.4, let 
 $n \in {\mathbb N}$
 and
$n \in {\mathbb N}$
 and 
 $A\in {\cal A}$
 be so that
$A\in {\cal A}$
 be so that 
 $L_S\chi _A = \chi _A$
. If
$L_S\chi _A = \chi _A$
. If 
 $\epsilon>0$
 and, in addition,
$\epsilon>0$
 and, in addition, 
 $\tau $
 is an a.e. aperiodic map on A, then Corollary 4.8 gives that there exists
$\tau $
 is an a.e. aperiodic map on A, then Corollary 4.8 gives that there exists 
 $B\in {\cal A}$
 with
$B\in {\cal A}$
 with 
 $B\subset A$
 such that
$B\subset A$
 such that 
 $B, \tau ^{-1}(B), \ldots , \tau ^{1-n}(B)$
 are a.e. pairwise disjoint and
$B, \tau ^{-1}(B), \ldots , \tau ^{1-n}(B)$
 are a.e. pairwise disjoint and 
 $$ \begin{align} 0\le L_S(\chi_A-\chi_C) \le \epsilon\chi_A, \end{align} $$
$$ \begin{align} 0\le L_S(\chi_A-\chi_C) \le \epsilon\chi_A, \end{align} $$
where 
 $C=\bigcup _{j=0}^{n-1} \tau ^{-j}B$
.
$C=\bigcup _{j=0}^{n-1} \tau ^{-j}B$
.
 To highlight the need for aperiodicity, we now give an example of a conditionally ergodic preserving system 
 $(E,T,S,e)$
, which is T-universally complete and is neither periodic nor aperiodic, and for which the
$(E,T,S,e)$
, which is T-universally complete and is neither periodic nor aperiodic, and for which the 
 $\epsilon $
 approximation of Theorem 4.7 fails.
$\epsilon $
 approximation of Theorem 4.7 fails.
Example 4.10. Let 
 $E_n=\ell (n)$
, the space of real finite sequences of length n with componentwise ordering. On
$E_n=\ell (n)$
, the space of real finite sequences of length n with componentwise ordering. On 
 $E_n$
, we introduce the conditional expectation
$E_n$
, we introduce the conditional expectation 
 $$ \begin{align*}T_n(f_n)(i)=\frac{1}{n}\sum_{j=1}^n f_n(j)\mathbf{1}_n, \, f_n\in E_n,\end{align*} $$
$$ \begin{align*}T_n(f_n)(i)=\frac{1}{n}\sum_{j=1}^n f_n(j)\mathbf{1}_n, \, f_n\in E_n,\end{align*} $$
where 
 $\mathbf {1}_n(j)=1$
 for all
$\mathbf {1}_n(j)=1$
 for all 
 $j=1,\ldots , n$
. Further,
$j=1,\ldots , n$
. Further, 
 $\mathbf {1}_n$
 is a weak order unit for
$\mathbf {1}_n$
 is a weak order unit for 
 $E_n$
. On each
$E_n$
. On each 
 $E_n$
, we take
$E_n$
, we take 
 $S_n$
 as the Riesz homomorphism given by
$S_n$
 as the Riesz homomorphism given by 
 $S_1(f_1)=f_1$
, and for
$S_1(f_1)=f_1$
, and for 
 $n\ge 2$
,
$n\ge 2$
, 
 $$ \begin{align*}S_nf_n(j)=\begin{cases}f_n((j-1)), &j=2,\ldots,n,\\ f_n(n),&j=1.\end{cases}\end{align*} $$
$$ \begin{align*}S_nf_n(j)=\begin{cases}f_n((j-1)), &j=2,\ldots,n,\\ f_n(n),&j=1.\end{cases}\end{align*} $$
Clearly, 
 $(E_n,T_n,S_n,\mathbf {1}_n)$
 is a
$(E_n,T_n,S_n,\mathbf {1}_n)$
 is a 
 $T_n$
-universally complete ergodic conditional expectation preserving system. We take
$T_n$
-universally complete ergodic conditional expectation preserving system. We take 
 $(E,T,S,\mathbf {1})$
 as the direct product of the spaces
$(E,T,S,\mathbf {1})$
 as the direct product of the spaces 
 $(E_n,T_n,S_n,\mathbf {1}_n), n\in {\mathbb N}$
. Now,
$(E_n,T_n,S_n,\mathbf {1}_n), n\in {\mathbb N}$
. Now, 
 $$ \begin{align*}E=\prod_{n=1}^\infty E_n, \quad T=\prod_{n=1}^\infty T_n,\quad S=\prod_{n=1}^\infty S_n, \quad \mathbf{1}=(\mathbf{1}_1,\mathbf{1}_2,\ldots). \end{align*} $$
$$ \begin{align*}E=\prod_{n=1}^\infty E_n, \quad T=\prod_{n=1}^\infty T_n,\quad S=\prod_{n=1}^\infty S_n, \quad \mathbf{1}=(\mathbf{1}_1,\mathbf{1}_2,\ldots). \end{align*} $$
The resulting space 
 $(E, T, S, e)$
 is a conditionally ergodic conditional expectation preserving system with
$(E, T, S, e)$
 is a conditionally ergodic conditional expectation preserving system with 
 $E T$
-universally complete. Further,
$E T$
-universally complete. Further, 
 $(S,e)$
 is neither periodic nor aperiodic. If we take
$(S,e)$
 is neither periodic nor aperiodic. If we take 
 $n\in {\mathbb N}\setminus \{1\}$
 and
$n\in {\mathbb N}\setminus \{1\}$
 and 
 $0<\epsilon <1/n$
 in Theorem 4.7 and assume that there exists component p of
$0<\epsilon <1/n$
 in Theorem 4.7 and assume that there exists component p of 
 $\mathbf {1}$
 exhibiting the required properties of the theorem, then the disjointness of
$\mathbf {1}$
 exhibiting the required properties of the theorem, then the disjointness of 
 $S^jp$
 for
$S^jp$
 for 
 $j=0,\ldots ,n$
 gives that the components
$j=0,\ldots ,n$
 gives that the components 
 $p_j=0$
 for
$p_j=0$
 for 
 $j\le n$
, but then
$j\le n$
, but then 
 $S_j^kp_j=0$
 for all k, and so equation (4.1) fails.
$S_j^kp_j=0$
 for all k, and so equation (4.1) fails.
5 Approximation of aperiodic maps
 Rokhlin proved an interesting consequence of his lemma stating that any aperiodic transformation 
 $\tau $
 can be approximated by periodic ones, see [Reference Halmos8, pp. 75] and [Reference Weiss27]. That is, for any positive integer n and any
$\tau $
 can be approximated by periodic ones, see [Reference Halmos8, pp. 75] and [Reference Weiss27]. That is, for any positive integer n and any 
 $\epsilon>0$
, there exists a periodic transformation
$\epsilon>0$
, there exists a periodic transformation 
 $\tau '$
 of period n such that
$\tau '$
 of period n such that 
 $d(\tau , \tau ' ) \le {1}/{n}+\epsilon $
, where
$d(\tau , \tau ' ) \le {1}/{n}+\epsilon $
, where 
 $d(\tau , \tau ' )=\mu \{x : \tau x \ne \tau ' x\}$
.
$d(\tau , \tau ' )=\mu \{x : \tau x \ne \tau ' x\}$
.
In this section, we apply Theorem 4.7 to obtain an approximation of aperiodic conditional expectation preserving transformations by periodic ones in the conditional Riesz space setting.
Theorem 5.1. Let 
 $(E,T, S, e)$
 be a T-universally complete conditionally ergodic preserving system, where S is a surjective Riesz homomorphism and
$(E,T, S, e)$
 be a T-universally complete conditionally ergodic preserving system, where S is a surjective Riesz homomorphism and 
 $(S,e)$
 aperiodic. For each
$(S,e)$
 aperiodic. For each 
 $1>\epsilon >0$
, there exists a surjective Riesz homomorphism
$1>\epsilon >0$
, there exists a surjective Riesz homomorphism 
 $S'$
 such that
$S'$
 such that 
 $(E,T, S', e)$
 is a conditional expectation preserving system with
$(E,T, S', e)$
 is a conditional expectation preserving system with 
 $(S',e)$
 periodic and
$(S',e)$
 periodic and 
 $$ \begin{align} \sup_{u\in C_e} T |(S-S')u| \le \epsilon e. \end{align} $$
$$ \begin{align} \sup_{u\in C_e} T |(S-S')u| \le \epsilon e. \end{align} $$
Proof. Let 
 $\epsilon>0$
 and
$\epsilon>0$
 and 
 $n>4/\epsilon $
. By Theorem 4.7, there is a component p of e such that
$n>4/\epsilon $
. By Theorem 4.7, there is a component p of e such that 
 $(S^i p)_{i=0,\ldots ,n-1}$
 are disjoint and
$(S^i p)_{i=0,\ldots ,n-1}$
 are disjoint and 
 $$ \begin{align} T(e-h) \le \frac{\epsilon}{4} e, \end{align} $$
$$ \begin{align} T(e-h) \le \frac{\epsilon}{4} e, \end{align} $$
where
 $$ \begin{align} h=\sum_{i=0}^{n-1} S^ip=\bigvee_{i=0}^{n-1} S^ip\in C_e. \end{align} $$
$$ \begin{align} h=\sum_{i=0}^{n-1} S^ip=\bigvee_{i=0}^{n-1} S^ip\in C_e. \end{align} $$
Hence,
 $$ \begin{align} \bigg(1-\frac{\epsilon}{4}\bigg)e \le Th. \end{align} $$
$$ \begin{align} \bigg(1-\frac{\epsilon}{4}\bigg)e \le Th. \end{align} $$
Further, applying T to equation (5.3) gives
 $$ \begin{align} nTp=Th\le Te=e, \end{align} $$
$$ \begin{align} nTp=Th\le Te=e, \end{align} $$
so
 $$ \begin{align} Tp\le \frac{1}{n}e\le\frac{\epsilon}{4}e. \end{align} $$
$$ \begin{align} Tp\le \frac{1}{n}e\le\frac{\epsilon}{4}e. \end{align} $$
We now give a decomposition of e which will be the basis for the decomposition of E into bands. Let
 $$ \begin{align} q=\sum_{i=0}^{n-2} S^ip=\bigvee_{i=0}^{n-2} S^ip\in C_e. \end{align} $$
$$ \begin{align} q=\sum_{i=0}^{n-2} S^ip=\bigvee_{i=0}^{n-2} S^ip\in C_e. \end{align} $$
Here,
 $$ \begin{align} h=S^{n-1}p+q=(S^{n-1}p)\vee q. \end{align} $$
$$ \begin{align} h=S^{n-1}p+q=(S^{n-1}p)\vee q. \end{align} $$
Hence, we have the following disjoint decomposition of e by its components 
 $q, S^{n-1}p, e-h$
,
$q, S^{n-1}p, e-h$
, 
 $$ \begin{align} q+({S^{n-1}p})+(e-h)=e. \end{align} $$
$$ \begin{align} q+({S^{n-1}p})+(e-h)=e. \end{align} $$
We define the approximation Riesz to S as
 $$ \begin{align} S'=SP_{q}+S^{1-n}P_{S^{n-1}p}+P_{e-h}. \end{align} $$
$$ \begin{align} S'=SP_{q}+S^{1-n}P_{S^{n-1}p}+P_{e-h}. \end{align} $$
Here, 
 $S'$
 is a finite sum of order continuous maps and is thus order continuous.
$S'$
 is a finite sum of order continuous maps and is thus order continuous.
 We begin by verifying that 
 $(E,T,S',e)$
 is a conditional expectation preserving system with
$(E,T,S',e)$
 is a conditional expectation preserving system with 
 $S'$
 surjective. Additionally,
$S'$
 surjective. Additionally, 
 $S'$
 is a sum of compositions of Riesz homomorphisms and is thus a Riesz homomorphism. From equations (5.9) and (5.10), we get
$S'$
 is a sum of compositions of Riesz homomorphisms and is thus a Riesz homomorphism. From equations (5.9) and (5.10), we get 
 $$ \begin{align*}S'e=Sq+p+(e-h)=e.\end{align*} $$
$$ \begin{align*}S'e=Sq+p+(e-h)=e.\end{align*} $$
 From equations (5.10) and (5.9), as 
 $TS=T$
,
$TS=T$
, 
 $$ \begin{align*}TS'= TSP_{q}+TS^{1-n}P_{S^{n-1}p}+TP_{e-h}=T(P_{e-h+q+S^{n-1}p})=T.\end{align*} $$
$$ \begin{align*}TS'= TSP_{q}+TS^{1-n}P_{S^{n-1}p}+TP_{e-h}=T(P_{e-h+q+S^{n-1}p})=T.\end{align*} $$
As T is strictly positive, the condition 
 $TS'=T$
 ensures that
$TS'=T$
 ensures that 
 $S'$
 is injective.
$S'$
 is injective.
 We now prove that 
 $S'$
 is surjective. In particular, for
$S'$
 is surjective. In particular, for 
 $f\in E$
, set
$f\in E$
, set 
 $$ \begin{align*}\hat{f}=P_{e-h}f+P_{q}S^{-1}f+P_{S^{n-1}p}S^{n-1}f.\end{align*} $$
$$ \begin{align*}\hat{f}=P_{e-h}f+P_{q}S^{-1}f+P_{S^{n-1}p}S^{n-1}f.\end{align*} $$
We show that 
 $S'\hat {f}=f$
. To see this
$S'\hat {f}=f$
. To see this 
 $$ \begin{align*} S'\hat{f}&=( SP_{q}+S^{1-n}P_{S^{n-1}p}+P_{e-h}) (P_{e-h}f+P_{q}S^{-1}f+P_{S^{n-1}p}S^{n-1}f)\\ &=P_{e-h}f+SP_{q}S^{-1}f+S^{1-n}P_{S^{n-1}p}S^{n-1}f. \end{align*} $$
$$ \begin{align*} S'\hat{f}&=( SP_{q}+S^{1-n}P_{S^{n-1}p}+P_{e-h}) (P_{e-h}f+P_{q}S^{-1}f+P_{S^{n-1}p}S^{n-1}f)\\ &=P_{e-h}f+SP_{q}S^{-1}f+S^{1-n}P_{S^{n-1}p}S^{n-1}f. \end{align*} $$
Here,
 $$ \begin{align*}SP_qS^{-1}f=SP_{S^{-1}Sq}S^{-1}f=SS^{-1}P_{Sq}f=P_{Sq}f\end{align*} $$
$$ \begin{align*}SP_qS^{-1}f=SP_{S^{-1}Sq}S^{-1}f=SS^{-1}P_{Sq}f=P_{Sq}f\end{align*} $$
and
 $$ \begin{align*}S^{1-n}P_{S^{n-1}p}S^{n-1}f=S^{1-n}S^{n-1}P_{p}f=P_pf.\end{align*} $$
$$ \begin{align*}S^{1-n}P_{S^{n-1}p}S^{n-1}f=S^{1-n}S^{n-1}P_{p}f=P_pf.\end{align*} $$
Thus, by equation (5.10),
 $$ \begin{align*} S'\hat{f}= P_{e-h}f+P_{Sq}f+P_{p}f=P_{(e-h)+Sq+p}f \end{align*} $$
$$ \begin{align*} S'\hat{f}= P_{e-h}f+P_{Sq}f+P_{p}f=P_{(e-h)+Sq+p}f \end{align*} $$
giving 
 $S'\hat {f}=f$
, showing that
$S'\hat {f}=f$
, showing that 
 $S'$
 is surjective.
$S'$
 is surjective.
 We now show that 
 $(S',e)$
 is periodic. It suffices to show that, for each
$(S',e)$
 is periodic. It suffices to show that, for each 
 $u\in C_e$
 with
$u\in C_e$
 with 
 $u\ne 0$
, we have that
$u\ne 0$
, we have that 
 $$ \begin{align*}\bigvee_{k=1}^{n} (S')^k u \ge u.\end{align*} $$
$$ \begin{align*}\bigvee_{k=1}^{n} (S')^k u \ge u.\end{align*} $$
Let 
 $u\in C_e$
 with
$u\in C_e$
 with 
 $u\ne 0$
. Here,
$u\ne 0$
. Here, 
 $$ \begin{align} S'u\ge S'(u\wedge (e-h))=u\wedge (e-h). \end{align} $$
$$ \begin{align} S'u\ge S'(u\wedge (e-h))=u\wedge (e-h). \end{align} $$
 Let 
 $0\le j\le n-1$
 and
$0\le j\le n-1$
 and 
 $0\le i\le n-1-j.$
 We show inductively with respect to i that
$0\le i\le n-1-j.$
 We show inductively with respect to i that 
 $$ \begin{align} (S')^i(u\wedge S^jp)=S^i(u\wedge S^jp)\le S^{i+j}p\le q. \end{align} $$
$$ \begin{align} (S')^i(u\wedge S^jp)=S^i(u\wedge S^jp)\le S^{i+j}p\le q. \end{align} $$
For 
 $i=0$
,
$i=0$
, 
 $$ \begin{align} (S')^0(u\wedge S^jp)=u\wedge S^jp=S^0(u\wedge S^jp)\le S^jp\le q. \end{align} $$
$$ \begin{align} (S')^0(u\wedge S^jp)=u\wedge S^jp=S^0(u\wedge S^jp)\le S^jp\le q. \end{align} $$
Now, suppose 
 $0\le i\le n-1-j-1$
 and that
$0\le i\le n-1-j-1$
 and that 
 $$ \begin{align} (S')^i(u\wedge S^jp)=S^i(u\wedge S^jp)\le S^{i+j}p\le q, \end{align} $$
$$ \begin{align} (S')^i(u\wedge S^jp)=S^i(u\wedge S^jp)\le S^{i+j}p\le q, \end{align} $$
then applying 
 $S'$
 to equation (5.14), from the definition of
$S'$
 to equation (5.14), from the definition of 
 $S'$
, we get
$S'$
, we get 
 $$ \begin{align*} S'(S')^{i}(u\wedge S^jp)=S(S')^{i}(u\wedge S^jp)=S^{i+1}(u\wedge S^jp)\le S^{i+j+1}p\le q, \end{align*} $$
$$ \begin{align*} S'(S')^{i}(u\wedge S^jp)=S(S')^{i}(u\wedge S^jp)=S^{i+1}(u\wedge S^jp)\le S^{i+j+1}p\le q, \end{align*} $$
from which equation (5.12) holds by induction.
 In particular, for 
 $i=n-j-1$
,
$i=n-j-1$
, 
 $$ \begin{align} (S')^{n-1-j}(u\wedge S^jp)=S^{n-1-j}(u\wedge S^jp)\le S^{n-1}p. \end{align} $$
$$ \begin{align} (S')^{n-1-j}(u\wedge S^jp)=S^{n-1-j}(u\wedge S^jp)\le S^{n-1}p. \end{align} $$
Now, applying 
 $S'$
 to the above gives
$S'$
 to the above gives 
 $$ \begin{align} (S')^{n-j}(u\wedge S^jp)=S^{1-n}S^{n-1-j}(u\wedge S^jp)=S^{-j}(u\wedge S^jp)\le p=S^0p. \end{align} $$
$$ \begin{align} (S')^{n-j}(u\wedge S^jp)=S^{1-n}S^{n-1-j}(u\wedge S^jp)=S^{-j}(u\wedge S^jp)\le p=S^0p. \end{align} $$
Hence, equation (5.16) can be written as
 $$ \begin{align} (S')^{n-j}(u\wedge S^jp)=(S^{-j}u)\wedge S^0p. \end{align} $$
$$ \begin{align} (S')^{n-j}(u\wedge S^jp)=(S^{-j}u)\wedge S^0p. \end{align} $$
So by equation (5.17), we have
 $$ \begin{align*}(S')^n(u\wedge S^jp)=(S')^j(S')^{n-j}(u\wedge S^jp)=(S')^j((S^{-j}u)\wedge S^0p),\end{align*} $$
$$ \begin{align*}(S')^n(u\wedge S^jp)=(S')^j(S')^{n-j}(u\wedge S^jp)=(S')^j((S^{-j}u)\wedge S^0p),\end{align*} $$
and now applying equation (5.12) (with replacing j by 
 $0$
, i by j, and u by
$0$
, i by j, and u by 
 $S^{-j}u$
), we have
$S^{-j}u$
), we have 
 $$ \begin{align} (S')^n(u\wedge S^jp)=(S')^j((S^{-j}u)\wedge S^0p)=S^j((S^{-j}u)\wedge S^0p)=u\wedge S^jp. \end{align} $$
$$ \begin{align} (S')^n(u\wedge S^jp)=(S')^j((S^{-j}u)\wedge S^0p)=S^j((S^{-j}u)\wedge S^0p)=u\wedge S^jp. \end{align} $$
Taking the supremum of equation (5.18) over 
 $j=0,\ldots ,n-1$
 gives
$j=0,\ldots ,n-1$
 gives 
 $$ \begin{align} (S')^nu\ge (S')^n(u\wedge h)=\bigvee_{j=0}^{n-1}(S')^n(u\wedge S^jp)=u\wedge \bigg(\bigvee_{j=0}^{n-1}S^jp\bigg)=u\wedge h. \end{align} $$
$$ \begin{align} (S')^nu\ge (S')^n(u\wedge h)=\bigvee_{j=0}^{n-1}(S')^n(u\wedge S^jp)=u\wedge \bigg(\bigvee_{j=0}^{n-1}S^jp\bigg)=u\wedge h. \end{align} $$
Combining equations (5.11) and (5.19) gives
 $$ \begin{align*}(S')^nu \vee S'u\ge (u\wedge h)\vee (u\wedge (e-h))=u\wedge e=u,\end{align*} $$
$$ \begin{align*}(S')^nu \vee S'u\ge (u\wedge h)\vee (u\wedge (e-h))=u\wedge e=u,\end{align*} $$
showing that 
 $(S',e)$
 is periodic.
$(S',e)$
 is periodic.
 Finally, we show that 
 $S'$
 obeys the bound in equation (5.1). For
$S'$
 obeys the bound in equation (5.1). For 
 $u\in {\cal C}_e$
, we have
$u\in {\cal C}_e$
, we have 
 $$ \begin{align*}(S-S')u=(S-S^{1-n})P_{S^{n-1}p}u+(S-I)P_{e-h}u.\end{align*} $$
$$ \begin{align*}(S-S')u=(S-S^{1-n})P_{S^{n-1}p}u+(S-I)P_{e-h}u.\end{align*} $$
Here, as 
 $TS^j=T, j\ge 0,$
 by equation (5.6),
$TS^j=T, j\ge 0,$
 by equation (5.6), 
 $$ \begin{align*}T|(S-S^{1-n})P_{S^{n-1}p}u|\le TS^np+Tp= 2Tp\le \frac{\epsilon}{2} e\end{align*} $$
$$ \begin{align*}T|(S-S^{1-n})P_{S^{n-1}p}u|\le TS^np+Tp= 2Tp\le \frac{\epsilon}{2} e\end{align*} $$
and, by equation (5.2),
 $$ \begin{align*}T|(S-I)P_{e-h}u|\le TS(e-h)+T(e-h)=2T(e-h)\le \frac{\epsilon}{2}e,\end{align*} $$
$$ \begin{align*}T|(S-I)P_{e-h}u|\le TS(e-h)+T(e-h)=2T(e-h)\le \frac{\epsilon}{2}e,\end{align*} $$
giving
 $$ \begin{align*}T |(S-S')u| \le T|(S-S^{1-n})P_{S^{n-1}p}u|+T|(S-I)P_{e-h}u|\le \epsilon e.\end{align*} $$
$$ \begin{align*}T |(S-S')u| \le T|(S-S^{1-n})P_{S^{n-1}p}u|+T|(S-I)P_{e-h}u|\le \epsilon e.\end{align*} $$
Thus, equation (5.1) holds.
Acknowledgements
This research was funded in part by the joint South Africa–Tunisia Grant (South African National Research Foundation Grant Number SATN180717350298, grant number 120112) and CoE-MaSS.
 
 

 
 
 
 
