Generic properties of extensions

TL;DR

This paper demonstrates that in a broad class of measure-preserving systems, weakly mixing extensions are typical, and such generic extensions lack intermediate nilfactors, extending classical results on mixing properties.

Contribution

It establishes the genericity of weakly but not strongly mixing extensions on fixed product spaces with non-atomic measures, generalizing classical results.

Findings

Weakly but not strongly mixing extensions are generic.

Generic extensions do not possess intermediate nilfactors.

Extends classical results by Halmos, Rokhlin, and Furstenberg.

Abstract

Motivated by the classical results by Halmos and Rokhlin on the genericity of weakly but not strongly mixing transformations and the Furstenberg tower construction, we show that weakly but not strongly mixing extensions on a fixed product space with both measures non-atomic are generic. In particular, a generic extension does not have an intermediate nilfactor.

Figures

Equations

$$N_{\epsilon }(T;E)=\{S\in \mathcal{G}(Z):\mu (TE△SE)<\epsilon \},$$

$$d(S,T)$$

$$d^{\prime }(S,T)$$

$$d^{\prime }(RS,RT)=d^{\prime }(S,T)=d^{\prime }(SR,TR).$$

$$f\in L^2(Z|X)\text{ if and only if }\mathbb{E}({\left|f\right|}^2|X{)}^{1/2}\in L^{\infty }(X),$$

$${\Vert f\Vert }_{L^2(Z|X)}:=\mathbb{E}({\left|f\right|}^2|X{)}^{1/2}$$

$$⟨f,g{⟩}_{L^2(Z|X)}:=\mathbb{E}(f\overline{g}|X).$$

$$\left|⟨f,g{⟩}_{L^2(Z|X)}\right|\le {\Vert f\Vert }_{L^2(Z|X)}{\Vert g\Vert }_{L^2(Z|X)}\text{ a.e.}$$

$$\underset{N\to \infty }{\lim }\frac{1}{N}\sum _{n=0}^{N-1}{\Vert \mathbb{E}(T^{n}f\overline{g}|X)-(T^{\prime }{)}^n\mathbb{E}(f|X)\mathbb{E}(g|X)\Vert }_{L^2(X)}=0.$$

$$\sum _{n=1}^{\infty }\mu (TE\backslash C_n)<\infty .$$

$$\hat{C}:=\bigcup _{n=1}^{\infty }\bigcap _{m\ge n}{C}_m.$$

$$\hat{C}\backslash TE=\left(\bigcup _{n=1}^{\infty }\bigcap _{m\ge n}{C}_m\right)\backslash TE=\bigcup _{n=1}^{\infty }\bigcap _{m\ge n}(C_m\backslash TE).$$

$$\mu (\hat{C}\backslash TE)\le \sum _{n=1}^{\infty }\mu \left(\bigcap _{m\ge n}(C_m\backslash TE)\right)=0.$$

$$TE\backslash \hat{C}=TE\backslash \left(\bigcup _{n=1}^{\infty }\bigcap _{m\ge n}{C}_m\right)=\bigcap _{n=1}^{\infty }\bigcup _{m\ge n}(TE\backslash C_m).$$

$$N_{\epsilon }(T)=\{S\in \mathcal{G}(Z):\mu (T{E}_i△S{E}_i)<\epsilon ,i=1,\dots ,n\},$$

$$\underset{E}{\sup }\mu (TE△SE)=d(T,S)\le d^{\prime }(T,S)=\mu (A)<\epsilon .$$

$$\underset{N\to \infty }{\lim }\frac{1}{N}\sum _{n=0}^{N-1}{\left({\int }_X{\left|\mathbb{E}(T^{n}f\overline{f}|X)\right|}^{2}dm\right)}^{1/2}\ne 0.$$

$$f(x,1)=\frac{-{\sum }_{i=2}^L{w}_i}{w_1}$$

$$\mathbb{E}(T^{n}f\overline{f}|X)(x)=\sum _{j=1}^L{w}_{j}f(x,j)f((T^{\prime }{)}^{n}x,{\sigma }_{n,x}(j))=\sum _{j=1}^L{w}_{j}f(x,j)f(x,{\sigma }_{n,x}(j)),$$

$$\sum _{j=1}^L{w}_{j}f(x,j)f(x,{\sigma }_{n,x}(j))=f(x,1)(w_1+w_i)+\left(\sum _{j=2}^L{w}_j\right)-w_i.$$

$$f(x,1)(w_1+w_i)=\frac{-{\sum }_{j=2}^L{w}_j}{w_1}(w_1+w_i)=\left(-\sum _{j=2}^L{w}_j\right)\left(1+\frac{w_i}{w_1}\right).$$

$$f(x,1)(w_1+w_i)+\left(\sum _{j=2}^L{w}_j\right)-w_i\le \left(-\sum _{j=2}^L{w}_j\right)+\left(\sum _{j=2}^L{w}_j\right)-w_i=-w_i<0.$$

$$N_{\epsilon }(T)=\{S\in {\mathcal{G}}_X:\mu (T{D}_l△S{D}_l)<\epsilon ,l=1,\dots ,L(2^n)\},$$

$$\mu ({\hat{A}}_{\lambda ,k}△A_{\lambda ,k})<\frac{\epsilon }{2\left|\Lambda \right|},\text{ and }\mu (T^{-1}{\hat{A}}_{\lambda ,k_2}△B_{\lambda ,k_1})<\frac{\epsilon }{2\left|\Lambda \right|}.$$

$$\mu \left(D_j△\bigcup B_{\lambda ,k}\right)=\mu \left(\bigcup T^{-1}{\hat{A}}_{\lambda ,i_{\lambda ,j}}△\bigcup B_{\lambda ,k}\right)\le \sum \mu (T^{-1}{\hat{A}}_{\lambda ,i_{\lambda ,j}}△B_{\lambda ,k}).$$

$$\mu \left(Q{D}_j△\bigcup A_{\lambda ,i_{\lambda ,j}}\right)=\mu \left(D_j△\bigcup B_{\lambda ,k}\right)\le \frac{\epsilon }{2}.$$

$$\mu \left(\bigcup A_{\lambda ,i_{\lambda ,j}}△T{D}_j\right)=\mu \left(\bigcup A_{\lambda ,i_{\lambda ,j}}△\bigcup {\hat{A}}_{\lambda ,i_{\lambda ,j}}\right)\le \sum (A_{\lambda ,i_{\lambda ,j}}△{\hat{A}}_{\lambda ,i_{\lambda ,j}}).$$

$$\mu (T{D}_j△Q{D}_j)\le \mu \left(T{D}_j△\bigcup A_{\lambda ,i_{\lambda ,j}}\right)+\left(\bigcup A_{\lambda ,i_{\lambda ,j}}△Q{D}_j\right)\le \frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon .$$

$$\sum _{j=1}^{2^K}m(A_{ij})<2\epsilon$$

$$\sum _{j}m(A_{ij})=\sum _j\left|m({\tilde{F}}_{ij})-r_{ij}\right|\le \sum _j\left|m(E_{ij})-r_{ij}\right|+\sum _j\left|m(E_{ij})-m({\tilde{F}}_{ij})\right|.$$

$$\sum _j\left|m(E_{ij})-m({\tilde{F}}_{ij})\right|\le \sum _{j}m(E_{ij}△{\tilde{F}}_{ij})=m\left(\bigcup _j(E_{ij}△{\tilde{F}}_{ij})\right).$$

Full text

1. Introduction

The classical results by Halmos [22] and Rokhlin [32] state that a “typical” measure-preserving transformation on a probability space $\displaystyle(X,\mu)$ is weakly but not strongly mixing. More precisely, the set of weakly mixing transformations is a dense, $\displaystyle G_{\delta}$ (hence residual) set for the weak topology and the set of strongly mixing transformations are of first category. The combination of these two results proved the existence of a weakly but not strongly mixing transformation, without providing a concrete example (for such an example, see [9]). Since then, much research has been done in finding “typical” properties of measure-preserving dynamical systems. See, e.g., [28], [26], [1], [27], [2], [5], [3], [12], [4], [33], [21].

More than thirty years later, in completely unrelated efforts, Furstenberg [17] presented his celebrated ergodic theoretic proof of Szemeredi’s Theorem on the existence of arbitrarily long arithmetic progressions in large subsets of $\displaystyle\mathbb{N}.$ While the original proof by Furstenberg used diagonal measures, the alternative proof by him, Katznelson, and Ornstein [18] building up a tower of so-called compact and weakly mixing extensions had much greater impact on further development of ergodic theory. This method of finding the characteristic factor has been extended to various ergodic theorems and is an active area of research, see, e.g., [10], [11], [7], [15], [8], [16], [14], [35] [6], [30]. For example, the correct characteristic factor for norm convergence of multiple ergodic averages was identified by Host and Kra [24] and has the structure of an inverse limit of nilsystems, see also Ziegler [37].

The purpose of this paper is to prove analogues of the Halmos and Rokhlin category theorems for extensions (see Theorems 5 and 6), extending a result of Robertson on compact group extensions [31]. Inspired by Rokhlin’s skew product representation theorem (see, for example, [19, p.69]), we consider extensions defined on product spaces with the natural projection as the factor map. We show that for a fixed product space, where both measures are non-atomic, a “typical” extension is weakly but not strongly mixing. Here, by an extension we mean an invertible extension of some (non-fixed), invertible, measure-preserving transformation on the factor. Note that the set of extensions is a closed, nowhere dense set in all invertible transformations on the product space (see Proposition 2), so the classical Halmos and Rokhlin results cannot be applied. The proof for weakly mixing extensions is a non-trivial adaptation of the original construction by Halmos. In particular, a “typical” extension does not have an intermediate nilfactor. For examples of systems lacking non-trivial nilfactors, see [25].

The paper is organized as follows. After discussing some preliminaries in Section 2, we consider the case of discrete extensions in Section 3, in particular showing that there are no weak mixing extensions on these spaces (Proposition 3), but that permutations are dense (Theorem 1). We then prove the Weak Approximation Theorem for Extensions on the unit square (Theorem 3) in Section 4 using the density result for discrete extensions mentioned above. In Section 5 we generalize a few results, including Halmos’ Uniform Approximation Theorem, which are necessary to prove our Conjugacy Lemma for Extensions (Lemma 5) in Section 6. Section 7 is devoted to the proof that weakly mixing extensions on the unit square are residual and Section 8 addresses the case of general vertical measure. In Section 9 we define strongly mixing extensions, and show that such extensions are of first category (Theorem 6). Finally in Section 10 we formulate some open questions. After this paper was made public, the first part of Question 2 was answered by Eli Glasner and Benjamin Weiss, see [20].

Acknowledgments. The question of “typical” behavior of extensions was asked by Terence Tao for a fixed factor, motivated by [25], cf. Question 2 and note in particular the following discussion on the difficulties with fixed factors. The author is very grateful to him for the inspiration. The author also thanks Tanja Eisner for introducing him to the problem and for many helpful discussions. The author is further thankful to Ben Stanley for being available to exchange ideas, the referee of this paper for careful reading and valuable comments which improved the paper, and to Bryna Kra, Michael Lin, Philipp Kunde, and Yonatan Gutman for helpful remarks. Lastly, the support of the Max Planck Institute is greatly acknowledged.

2. Preliminaries

As explained in the introduction, in this paper we will be working with extensions on product spaces through the natural projection. To be more precise, we let $\displaystyle(X,m)$ be a non-atomic standard probability space, $\displaystyle(Y,\eta)$ be a probability space, $\displaystyle(Z,\mu)=(X\times Y,m\times\eta),$ and $\displaystyle T,T^{\prime}$ be measure-preserving transformations on $\displaystyle(Z,\mu),(X,m)$ respectively, such that $\displaystyle(Z,\mu,T)$ is an extension of $\displaystyle(X,m,T^{\prime})$ through the natural projection map $\displaystyle\pi:Z\to X$ onto the first coordinate. We will assume throughout that $\displaystyle T,T^{\prime}$ are invertible, and will identify two transformations if they differ on a set of measure zero. We will say “$\displaystyle T$ is an extension of $\displaystyle T^{\prime}$” or “$\displaystyle T$ extends $\displaystyle T^{\prime}$” if $\displaystyle T$ and $\displaystyle T^{\prime}$ satisfy all conditions stated above. Throughout this paper, we will assume without loss of generality that $\displaystyle X$ is the unit interval and $\displaystyle m$ is the Lebesgue measure. We can assume this because all non-atomic standard probability spaces are isomorphic (see [22, p. 61]).

Let $\displaystyle\mathcal{G}(Z)$ denote the set of all invertible, measure-preserving transformations on $\displaystyle(Z,\mu)$ and let $\displaystyle\mathcal{G}_{X}=\{T\in\mathcal{G}(Z):\exists\ T^{\prime}\in\mathcal{G}(X)\text{ s.t. }T\text{ extends }T^{\prime}\}$. Note that if we say $\displaystyle T\in\mathcal{G}_{X},$ we assume that the transformation on the factor will be notated by $\displaystyle T^{\prime}$. Further note that we will also write $\displaystyle\mathcal{G}_{X}$ to denote the corresponding set of Koopman operators.

The weak topology on $\displaystyle\mathcal{G}(Z)$ is the topology defined by the subbasic neighborhoods

[TABLE]

where $\displaystyle\varepsilon>0$ and $\displaystyle E$ is some measurable subset of $\displaystyle Z$. Note that if $\displaystyle Z$ is, say, the unit square with the Lebesgue measure, then it is sufficient for a subbasis to consider only dyadic sets (i.e., a finite union of dyadic squares). See [23] for discussions of this topology. It is helpful to note that the weak topology happens to coincide with the weak (and strong) operator topology for the corresponding Koopman operators. Further, in this paper we will be interested in the weak topology on $\displaystyle\mathcal{G}_{X},$ by which we mean the subspace topology inherited by the weak topology.

We will need the following two metrics on $\displaystyle\mathcal{G}(Z)$ defined by

[TABLE]

where the $\displaystyle\sup$ in the first definition is taken over all measurable sets $\displaystyle E$. These metrics were used by Halmos in his proof of the category theorem, see [23]. We note that both metrics induce the same topology on $\displaystyle\mathcal{G}(Z)$, but that topology is not the weak topology. Moreover, they satisfy $\displaystyle d(S,T)\leq d^{\prime}(S,T)$ for all $\displaystyle S,T\in\mathcal{G}(Z).$ The last important note is that $\displaystyle d^{\prime}$ is invariant under multiplication by transformations. That is to say, for all $\displaystyle R,S,T\in\mathcal{G}(Z),$

[TABLE]

Let $\displaystyle L^{2}(Z|X)$ denote the Hilbert module over $\displaystyle L^{\infty}(X).$ More precisely, for $\displaystyle f\in L^{2}(Z),$

[TABLE]

where $\displaystyle\mathbb{E}(f|X)$ is the conditional expectation of $\displaystyle f$ with respect to $\displaystyle X$. More specifically, it is the conditional expectation with respect to $\displaystyle\mathcal{A}:=\{\pi^{-1}(A):A\in\mathcal{L}\},$ where $\displaystyle\mathcal{L}$ is the Lebesgue sigma algebra on $\displaystyle X$. Let

[TABLE]

and

[TABLE]

For more on $\displaystyle L^{2}(Z|X),$ see [34]. One important result of $\displaystyle L^{2}(Z|X)$ that we do wish to emphasize for later is the Cauchy-Schwarz Inequality.

Proposition 1**.**

Let $\displaystyle f,g\in L^{2}(Z|X).$ Then

[TABLE]

Next we give a definition for weakly mixing extensions, cf. [34].

Definition 1**.**

An extension $\displaystyle T$ of $\displaystyle T^{\prime}$ is said to be weakly mixing if for all $\displaystyle f,g\in L^{2}(Z|X),$

[TABLE]

For other possible (equivalent) definitions of weakly mixing extensions, see [19, p.192].

We denote by $\displaystyle\mathcal{W}_{X}\subset\mathcal{G}_{X}$ the set of weakly mixing extensions on $\displaystyle Z$.

We finally prove that the Baire Category Theorem is applicable to $\displaystyle\mathcal{G}_{X},$ and further that $\displaystyle\mathcal{G}_{X}$ is topologically a small subset of $\displaystyle\mathcal{G}(Z).$

Proposition 2**.**

Suppose $\displaystyle Y$ has more than one point. Then $\displaystyle\mathcal{G}_{X}$ is closed and nowhere dense in $\displaystyle\mathcal{G}(Z).$

Proof.

We first prove that $\displaystyle\mathcal{G}_{X}$ is closed. Let $\displaystyle T\in\mathcal{G}(Z)\backslash\mathcal{G}_{X}.$ We wish to find a neighborhod of $\displaystyle T$ that is disjoint from $\displaystyle\mathcal{G}_{X}.$ To this end, let $\displaystyle E\subset Z$ be a cylinder set (to be precise, $\displaystyle E$ is of the form $\displaystyle D\times Y$ for some measurable $\displaystyle D\subset X$), such that $\displaystyle TE$ is not a cylinder set, even up to measure zero. Define $\displaystyle M:=\mu(E).$ Then for all cylinder sets $\displaystyle C$ with $\displaystyle\mu(C)=M,\mu(TE\triangle C)>0.$

We claim that indeed $\displaystyle\inf_{C}\mu(TE\triangle C)>0,$ where the $\displaystyle\inf$ is taken over all cylinder sets with measure exactly $\displaystyle M$. Suppose to the contrary that $\displaystyle\inf_{C}\mu(TE\triangle C)=0$. Let $\displaystyle C_{n}$ be a sequence of such cylinder sets such that not only $\displaystyle\mu(TE\triangle C_{n})\to 0,$ but further such that

[TABLE]

Define

[TABLE]

We claim that $\displaystyle\mu(TE\triangle\hat{C})=0.$ As $\displaystyle\hat{C}$ is clearly a cylinder set, we will arrive at a contradiction.

First consider

[TABLE]

Now because $\displaystyle\mu(C_{m}\triangle TE)\to 0,\mu\left(\bigcap_{m\geq n}(C_{m}\backslash TE)\right)=0$ for all $\displaystyle n.$ But then

[TABLE]

On the other hand,

[TABLE]

But by assumption, $\displaystyle\sum_{n=1}^{\infty}\mu(TE\backslash C_{n})<\infty,$ so by the Borel-Cantelli lemma, $\displaystyle\mu(TE\backslash\hat{C})=0.$

Now, let $\displaystyle\varepsilon:=\inf_{C}\mu(TE\triangle C).$ We claim that for any $\displaystyle S\in\mathcal{G}_{X},S\notin N_{\varepsilon}(T;E).$ Indeed, $\displaystyle SE$ is (up to a null set) a cylinder set, and $\displaystyle\mu(SE)=M,$ so $\displaystyle\mu(TE\triangle SE)\geq\varepsilon$ by definition of $\displaystyle\varepsilon$.

Now because $\displaystyle\mathcal{G}_{X}$ is closed, in order to prove that it is nowhere dense, it is sufficient to show that $\displaystyle\mathcal{G}(Z)\backslash\mathcal{G}_{X}$ is dense. Fix $\displaystyle T\in\mathcal{G}_{X}$, let $\displaystyle\varepsilon>0$ and let

[TABLE]

where $\displaystyle E_{i}$ are measurable sets. Now let $\displaystyle A\subset Z$ be a measurable set such that $\displaystyle 0\mu(A)<\varepsilon,$ and $\displaystyle A$ is not a cylinder set. Further let $\displaystyle B\subset Z$ be a cylinder set such that $\displaystyle\mu(A\cap B)=\mu(A\cap B^{c}),$ and define $\displaystyle A_{1}:=A\cap B,A_{2}:=A\cap B^{c}.$

We now take $\displaystyle S\in\mathcal{G}(Z)$ with the following properties: if $\displaystyle z\in Z\backslash A,Sz:=Tz,SA_{1}=TA_{2},$ and $\displaystyle SA_{2}=TA_{1}.$ Note that because $\displaystyle T$ is an extension and $\displaystyle A$ is not a cylinder set, $\displaystyle S\notin\mathcal{G}_{X}.$ Further note that $\displaystyle\{z\in Z:Sz\neq Tz\}=A.$ Therefore,

[TABLE]

So $\displaystyle S\in N_{\varepsilon}(T).$ ∎

By Proposition 2, $\displaystyle\mathcal{G}_{X}$ is a closed subset of a Baire space, so $\displaystyle\mathcal{G}_{X}$ is itself a Baire space. Further, because $\displaystyle\mathcal{G}_{X}$ is nowhere dense, the classical Halmos and Rokhlin results can provide no information about $\displaystyle\mathcal{G}_{X}.$

3. Discrete Extensions

As stated in Section 2, throughout this paper we will let $\displaystyle(X,m)$ be the unit interval with Lebesgue measure. For this section, let $\displaystyle Z=X\times\{1,\ldots,L\},$ with $\displaystyle L\geq 2$, $\displaystyle w$ be a probability measure on $\displaystyle\{1,\ldots,L\}$, and $\displaystyle w_{i}:=w(i)$ (without loss of generality, $\displaystyle w_{i}\neq 0$ for all $\displaystyle i$). Let $\displaystyle\mu$ be the product measure of $\displaystyle m$ and $\displaystyle w$ on $\displaystyle Z$.

In this section we will be exploring some results regarding these discrete extension measure spaces. We begin by showing that such systems can never be weakly mixing extensions.

Proposition 3**.**

Let $\displaystyle(Z,\mu),(X,m)$ be as above. Then $\displaystyle\mathcal{W}_{X}=\emptyset.$

Proof.

Fix $\displaystyle T\in\mathcal{G}_{X}.$ It suffices to show that there exists an $\displaystyle f\in L^{2}(Z|X)$ with relative mean zero (that is, $\displaystyle\mathbb{E}(f|X)=0$ $\displaystyle m-$almost everywhere) such that

[TABLE]

In particular, we will construct $\displaystyle f$ such that $\displaystyle\mathbb{E}(T^{n}f\overline{f}|X)(x)$ can take only a finite number of possible values, none of which are 0. Thus $\displaystyle\left|\mathbb{E}(T^{n}f\overline{f}|X)\right|^{2}(x)$ is always positive, $\displaystyle\frac{1}{N}\sum_{n=0}^{N-1}{\left|\mathbb{E}(T^{n}f\overline{f}|X)\right|^{2}(x)}$ is bounded away from 0, and $\displaystyle\frac{1}{N}\sum_{n=0}^{N-1}{\left(\int_{X}\left|\mathbb{E}(T^{n}f\overline{f}|X)\right|^{2}dm\right)^{1/2}}$ cannot converge to the zero function on $\displaystyle X$.

Consider $\displaystyle f(x,y)$, where $\displaystyle f(x,i)=1$ for all $\displaystyle x\in X,i=2,\ldots,L$, and

[TABLE]

for all $\displaystyle x\in X$. It is easy to see that $\displaystyle f$ has relative mean zero when $\displaystyle L\geq 2$,which is why we made this assumption at the beginning of the section.

Let $\displaystyle\sigma_{n,x}(i)$ be such that $\displaystyle T^{n}(x,i)=((T^{\prime})^{n}x,\sigma_{n,x}(i))$ for all $\displaystyle(x,i)\in Z$. Now,

[TABLE]

with the last equality because $\displaystyle f$ is constant on any given level. Thus we see that because $\displaystyle T$ is invertible, $\displaystyle\sigma_{n,x}$ is a permutation on an $\displaystyle L$ element set, and the value of $\displaystyle\mathbb{E}(T^{n}f\overline{f}|X)(x)$ is completely determined by the specific permutation $\displaystyle\sigma_{n,x}$. As there are $\displaystyle L!$ permutations of the $\displaystyle L$ levels, there are finitely many possible values of $\displaystyle\mathbb{E}(T^{n}f\overline{f}|X)(x)$.

To see $\displaystyle\mathbb{E}(T^{n}f\overline{f}|X)(x)\neq 0$, consider 2 cases. In the first case, we have $\displaystyle\sigma_{n,x}(1)=1$. In this case it is easy to see that every summand of $\displaystyle\sum_{j=1}^{L}{w_{j}f(x,j)f(x,\sigma_{n,x}(j))}$ is positive, and thus the sum is positive (in particular, nonzero). So now suppose $\displaystyle\sigma_{n,x}(i)=1,i\neq 1$. In this case we have

[TABLE]

Consider

[TABLE]

Note that $\displaystyle\left(-\sum_{j=2}^{L}w_{j}\right)\left(1+\frac{w_{i}}{w_{1}}\right)\leq-\sum_{j=2}^{L}w_{j}$. Thus,

[TABLE]

So $\displaystyle\mathbb{E}(T^{n}f\overline{f}|X)(x)$ is always nonzero, and $\displaystyle\left|\mathbb{E}(T^{n}f\overline{f}|X)\right|^{2}(x)$ is always positive, as desired. ∎

We make two notes here. First, the proof of Proposition 3 never used any assumptions on the factor, $\displaystyle(X,m,T^{\prime}),$ and thus it will hold when the factor is any probability space, with any measure preserving transformation on that space. Second, the proof is still valid in the case that $\displaystyle Z$ has countably many levels instead of finitely many. The key observation is that for almost all $\displaystyle z$, if $\displaystyle z,Tz$ are on levels $\displaystyle k_{1},k_{2}$ respectively, then $\displaystyle w(k_{1})=w(k_{2}).$ As for any fixed $\displaystyle\alpha\in(0,1)$, there can be only finitely many levels $\displaystyle k$ with $\displaystyle w(k)=\alpha,T$ decomposes into invariant subsystems, to each of which we can apply Proposition 3.

Though $\displaystyle\mathcal{W}_{X}$ is empty on these discrete extension spaces, they are still worth exploring. But before we can proceed, we will henceforth suppose that the probability measure $\displaystyle w$ is the normalized counting measure. That is, $\displaystyle w_{i}=\frac{1}{L}$ for all $\displaystyle i$. With this assumption, we extend the notion of dyadic sets and permutations on $\displaystyle X$ to dyadic sets and permutations on $\displaystyle Z.$

Definition 2**.**

If $\displaystyle D$ is a dyadic interval of rank $\displaystyle k$ in $\displaystyle X$, then a dyadic square of rank $\displaystyle k$ in $\displaystyle Z$ is a set of the form $\displaystyle D\times\{i\}$. A dyadic set in $\displaystyle Z$ is a union of dyadic squares. A dyadic permutation of rank $\displaystyle k$ on $\displaystyle Z$ is a permutation of the dyadic squares of rank $\displaystyle k$. A column-preserving (dyadic) permutation (of rank $\displaystyle k$) on $\displaystyle Z$ is a dyadic permutation on $\displaystyle Z$ which is an extension of a dyadic permutation on $\displaystyle X$.

We wish to generalize the fact that dyadic permutations are dense in $\displaystyle\mathcal{G}(X)$ to density of column-preserving permutations in $\displaystyle\mathcal{G}_{X}$. To this end, we make a couple notes. First we introduce the following notation: we write $\displaystyle A\subset i$ if there exists $\displaystyle A^{\prime}\subset X$ such that $\displaystyle A=A^{\prime}\times\{i\}$.

Second, we will require the use of the following lemma by Halmos (for proof, see [23, p.67]).

Lemma 1**.**

Let $\displaystyle\{E_{i}:i=1,\ldots n\}$ partition the unit interval, and $\displaystyle r_{i}$ be dyadic rationals such that $\displaystyle\sum_{i=1}^{n}r_{i}=1$ and $\displaystyle\left|m(E_{i})-r_{i}\right|<\delta$ for some $\displaystyle\delta>0$ and for all $\displaystyle i$. Then there exists $\displaystyle\{F_{i}:i=1,\ldots n\}$, dyadic sets that partition the unit interval such that $\displaystyle m(F_{i})=r_{i}$ and $\displaystyle m(E_{i}\triangle F_{i})<2\delta$ for all $\displaystyle i$.

We now move to the main result of this section.

Theorem 1** (Density of column-preserving permutations).**

Column-preserving permutations are dense in $\displaystyle\mathcal{G}_{X}.$ More precisely, let $\displaystyle T\in\mathcal{G}_{X}.$ Given $\displaystyle N_{\varepsilon}(T)$, a dyadic neighborhood of $\displaystyle T$, there exists $\displaystyle Q\in N_{\varepsilon}(T)$, a column-preserving permutation.

Proof.

Without loss of generality, assume

[TABLE]

where $\displaystyle D_{l}$ are every dyadic square of some fixed rank $\displaystyle n$ (note that $\displaystyle D_{l_{1}},D_{l_{2}}$ are disjoint up to boundary points).

Let $\displaystyle k\in\{1,\ldots,L\}$, and let $\displaystyle P_{k}:=\{D_{i}\cap TD_{j}|D_{i}\subset k,j=1,\ldots,L(2^{n})\}$. Note that $\displaystyle P_{k}$ partitions level $\displaystyle k$. If $\displaystyle\pi$ is the natural projection onto $\displaystyle X$, then let $\displaystyle P^{\prime}_{k}:=\pi P_{k}=\{\pi E|E\in P_{k}\}$. $\displaystyle P^{\prime}_{k}$ is a partition of $\displaystyle X$. Let $\displaystyle P^{\prime}=\{\hat{A}_{\lambda}|\lambda\in\Lambda\}$ be a common refinement of $\displaystyle P^{\prime}_{k}$ for $\displaystyle k=1,\ldots,L$, and let $\displaystyle P=\{\hat{A}_{\lambda,k}|\lambda\in\Lambda,k=1,\ldots,L\}$ be a partition of $\displaystyle Z$ obtained by lifting every element of $\displaystyle P^{\prime}$ to every level ($\displaystyle\hat{A}_{\lambda,k}\subset k$).

Applying a weaker version of Lemma 1 (one where we do not care about the value of $\displaystyle\left|m(E_{i})-r_{i}\right|$ in the formulation of the lemma) to the partition $\displaystyle P^{\prime}$, we obtain a partition $\displaystyle\{A_{\lambda}\}$ of $\displaystyle X$ into dyadic sets so that $\displaystyle m(\hat{A}_{\lambda}\triangle A_{\lambda})<\frac{L\varepsilon}{2\left|\Lambda\right|}$. Applying Lemma 1 again, we get a partition of $\displaystyle X$ into dyadic sets $\displaystyle B_{\lambda}$ so that $\displaystyle m((T^{\prime})^{-1}\hat{A}_{\lambda}\triangle B_{\lambda})<\frac{L\varepsilon}{2\left|\Lambda\right|}$. Note the full strength of Lemma 1 guarantees we can select this partition so that $\displaystyle m(A_{\lambda})=m(B_{\lambda})$ (as $\displaystyle m(\hat{A}_{\lambda})=m((T^{\prime})^{-1}\hat{A}_{\lambda})$). We can now lift $\displaystyle A_{\lambda},B_{\lambda}$ to sets $\displaystyle A_{\lambda,k},B_{\lambda,k}$ so that $\displaystyle A_{\lambda,k},B_{\lambda,k}\subset k$. Note that

[TABLE]

where $\displaystyle k_{1},k_{2}$ are such that if $\displaystyle i,j$ are such that $\displaystyle\hat{A}_{\lambda,k_{2}}\subset D_{i}\cap TD_{j},$ then $\displaystyle D_{i}\subset k_{2},D_{j}\subset k_{1}$.

We will now define $\displaystyle Q$ of some rank $\displaystyle r\in\mathbb{N}$ where $\displaystyle r$ is at least as large as the ranks of $\displaystyle D_{i}$ for every $\displaystyle i$, and $\displaystyle A_{\lambda},B_{\lambda}$ for every $\displaystyle\lambda$. We first define $\displaystyle Q^{\prime}$ a dyadic permutation on $\displaystyle X$ as any dyadic permutation which maps $\displaystyle B_{\lambda}$ to $\displaystyle A_{\lambda}$ for every $\displaystyle\lambda$. Next we define $\displaystyle Q$, a column preserving permutation of rank $\displaystyle r$. First let $\displaystyle k_{1},k_{2}$ be as before: if $\displaystyle i,j$ are such that $\displaystyle\hat{A}_{\lambda,k_{2}}\subset D_{i}\cap TD_{j}$, then $\displaystyle D_{i}\subset k_{2},D_{j}\subset k_{1}$. Then $\displaystyle Q$ will be the extension of $\displaystyle Q^{\prime}$ such that $\displaystyle B_{\lambda,k_{1}}\mapsto A_{\lambda,k_{2}}$. Note that for all $\displaystyle\lambda,k,$ we have that $\displaystyle\text{level }Q^{-1}A_{\lambda,k}=\text{level }T^{-1}\hat{A}_{\lambda,k},$ where $\displaystyle\text{level }A:=k$ if and only if $\displaystyle A\subset k.$

We now show that $\displaystyle\mu(TD_{j}\triangle QD_{j})<\varepsilon$ for all $\displaystyle j$. Fix $\displaystyle j\in 1,\ldots,L(2^{n}),$ and define $\displaystyle k$ so that $\displaystyle D_{j}\subset k$. Let $\displaystyle\Lambda_{j}:=\{\lambda\in\Lambda|(T^{\prime})^{-1}\hat{A}_{\lambda}\subset\pi D_{j}\}$. For $\displaystyle\lambda\in\Lambda_{j},$ let $\displaystyle i_{\lambda,j}$ be such that $\displaystyle T^{-1}\hat{A}_{\lambda,i_{\lambda,j}}\subset D_{j}$. Then $\displaystyle D_{j}=\bigcup_{\lambda\in\Lambda_{j}}{T^{-1}\hat{A}_{\lambda,i_{\lambda,j}}}.$

Further, by the definitions of $\displaystyle Q$ and $\displaystyle\Lambda_{j},$ as well as the previous note, we have that $\displaystyle Q^{-1}A_{\lambda,i_{\lambda,j}}=B_{\lambda,k}.$

Note all unions and sums will be taken over $\displaystyle\lambda\in\Lambda_{j}.$ We have

[TABLE]

But by (1), $\displaystyle\mu(T^{-1}\hat{A}_{\lambda,i_{\lambda,j}}\triangle B_{\lambda,k})<\frac{\varepsilon}{2\left|\Lambda\right|}$ so $\displaystyle(\ref{Eq:DiscreteFinal1})<\sum\frac{\varepsilon}{2\left|\Lambda\right|}=\frac{\varepsilon}{2}.$ Therefore

[TABLE]

On the other hand,

[TABLE]

Again by (1), $\displaystyle\,u(A_{\lambda,i_{\lambda,j}}\triangle\hat{A}_{\lambda,i_{\lambda,j}})<\frac{\varepsilon}{2\left|\Lambda\right|}$ so $\displaystyle(\ref{Eq:DiscreteFinal2})<\sum\frac{\varepsilon}{2\left|\Lambda\right|}=\frac{\varepsilon}{2}$. Finally,

[TABLE]

And because this holds for all $\displaystyle j,$ we have that $\displaystyle Q\in N_{\varepsilon}(T).$ ∎

4. Weak Approximation Theorem for Extensions on the Unit Square

Now we let $\displaystyle(Z,m_{2})$ be $\displaystyle X\times X$ with the Lebesgue measure. If we need further clarity, we will write the Lebesgue measure on $\displaystyle X$ as $\displaystyle m_{1}$, but in general we will denote both Lebesgue measures by $\displaystyle m$.

We begin by drawing some connections to Section 3. First, however, we need some more notation. For $\displaystyle L\in\mathbb{N},$ define $\displaystyle Z_{L}:=\bigcup_{j=0}^{L-1}{\left(X\times\left\{\frac{j}{L}\right\}\right)}\subset Z,\mu_{L},$ a measure on $\displaystyle Z_{L},$ to be the product of the Lebesgue measure with a normalized counting measure on $\displaystyle L$ points. Further, $\displaystyle\pi_{L}:Z\to Z_{L}$ be the natural projection onto $\displaystyle Z_{L}.$ That is, if $\displaystyle z=\left(x,\frac{j}{L}+\gamma\right)$ for $\displaystyle\gamma\in\left[0,\frac{1}{L}\right),$ then $\displaystyle\pi_{L}(z)=\left(x,\frac{j}{L}\right).$

Definition 3**.**

Let $\displaystyle T\in\mathcal{G}(Z)$. We say that $\displaystyle T$ is discrete equivalent if there exists $\displaystyle L$ and $\displaystyle T_{L}\in\mathcal{G}(Z_{L})$, such that $\displaystyle(Z,m,T)$ is an extension of $\displaystyle(Z_{L},\mu_{L},T_{L})$ through the factor map $\displaystyle\pi_{L}.$ Further, we say that $\displaystyle T$ is simply discrete equivalent if $\displaystyle T$ is an identity extension. That is, if we write $\displaystyle Z$ as $\displaystyle Z_{L}\times\left[0,\frac{1}{L}\right),$ then $\displaystyle T=T_{L}\times I.$ If we wish to emphasize the number of levels, $\displaystyle L$, we will say $\displaystyle T$ is $\displaystyle L$-(simply) discrete equivalent.

Definition 3 is fairly easy to visualize. We take the square and divide it into $\displaystyle L$ equal measure horizontal pieces. Then $\displaystyle T$ is discrete equivalent if $\displaystyle T$ moves fibers on each small piece to other such fibers, and is simply discrete equivalent if it does not move any points within the fiber. Note that in general, a discrete equivalent $\displaystyle T$ need not be in $\displaystyle\mathcal{G}_{X}.$ However, if $\displaystyle T\in\mathcal{G}_{X},$ then $\displaystyle T_{L}$ is also an extension of $\displaystyle T^{\prime}.$

Our goal for this section is to provide a version of Halmos’ Weak Approximation Theorem (see [23, p.65]) when restricted to $\displaystyle\mathcal{G}_{X}$. Mostly this will mean proving a result equivalent to Theorem 1. However we first need to lay some ground work. Definitions of dyadic squares, sets, and permutations are all standard in this case, so we do not redefine them. Column-preserving permutations are defined just as they are in Definition 2.

Before moving on, we make a few remarks.

Remark 1*.*

Lemma 1 holds on $\displaystyle(Z,m),$ because $\displaystyle(Z,m),$ like $\displaystyle(X,m),$ is a non-atomic standard probability space, and Lemma 1 holds for all such spaces (replacing “dyadic sets” in the statement of Lemma 1 with a class $\displaystyle\mathcal{B}$ which is isomorphic to the class of dyadic sets). Alternatively, one can simply prove Lemma 1 again in the context of the square. No part of the proof relies on the the fact that we were on the unit interval, so nothing changes in the proof.

Remark 2*.*

We will make the following notational convenience. If $\displaystyle T\in\mathcal{G}(Z)$ and $\displaystyle S^{\prime}\in\mathcal{G}(X),$ then we will write $\displaystyle S^{\prime}T$ in place of $\displaystyle(S^{\prime}\times I)T.$

Remark 3*.*

If $\displaystyle Q\in\mathcal{G}(Z)$ is a dyadic permutation of rank $\displaystyle K$, then $\displaystyle Q$ is $\displaystyle L$-simply discrete equivalent with $\displaystyle L=2^{K}.$ Further if $\displaystyle S^{\prime}\in\mathcal{G}(X),$ then $\displaystyle S^{\prime}Q$ is also $\displaystyle L$-simply discrete equivalent. If $\displaystyle Q$ is further an extension of $\displaystyle Q^{\prime}\in\mathcal{G}(X),$ then $\displaystyle S^{\prime}Q$ is an extension of $\displaystyle S^{\prime}Q^{\prime}.$

The key to our goal is the following strengthening of Lemma 1.

Lemma 2**.**

Let $\displaystyle\{E_{1},\ldots E_{N}\}$ be a finite partition of $\displaystyle Z$, $\displaystyle\varepsilon>0,$ and suppose $\displaystyle\{\tilde{F}_{1},\ldots,\tilde{F}_{N}\}$ is another partition of $\displaystyle Z,$ where $\displaystyle\tilde{F}_{i}$ are all dyadic sets, and $\displaystyle m(E_{i}\triangle\tilde{F}_{i})<\varepsilon.$ Let $\displaystyle K:=\text{\emph{max rank} }\tilde{F}_{i}$ and let $\displaystyle E_{ij}:=E_{i}\cap\pi^{-1}C_{j},j\in\{1,\ldots,2^{K}\},$ where $\displaystyle C_{j}$ is a dyadic interval of rank $\displaystyle K$. Let $\displaystyle r_{ij}$ be dyadic rationals (possibly zero) such that $\displaystyle\sum_{i=1}^{N}r_{ij}=\frac{1}{2^{K}}$ for all $\displaystyle j$ and $\displaystyle\left|m(E_{ij})-r_{ij}\right|<\frac{\varepsilon}{2^{K}}$ for all $\displaystyle i,j.$ Then there exists $\displaystyle\{F_{1},\ldots,F_{N}\}$ a partition of $\displaystyle Z$ such that $\displaystyle F_{i}$ is dyadic set for all $\displaystyle i,m(F_{ij})=r_{ij}$ (with $\displaystyle F_{ij}$ similarly defined as $\displaystyle E_{ij}$) for all $\displaystyle i,j$, and $\displaystyle m(E_{i}\triangle{F_{i}})<3\varepsilon$ for all $\displaystyle i.$

Proof.

Similar to the definition of $\displaystyle E_{ij},$ define $\displaystyle\tilde{F}_{ij}:=\tilde{F}_{i}\cap\pi^{-1}C_{j}.$ Note that by choice of $\displaystyle K,\tilde{F}_{ij}$ is of the product of $\displaystyle C_{j}$ and a dyadic set for all $\displaystyle i,j$. Now, for all $\displaystyle\tilde{F}_{ij}$ with $\displaystyle m(\tilde{F}_{ij})>r_{ij},$ let $\displaystyle A_{ij}\subset\tilde{F}_{ij}$ of the form $\displaystyle A_{ij}=C_{j}\times B_{ij}$ with $\displaystyle B_{ij}$ a dyadic set, and $\displaystyle m(A_{ij})=m(\tilde{F}_{ij})-r_{ij}.$ Define $\displaystyle F_{ij}:=\tilde{F}_{ij}\backslash A_{ij}.$ Let $\displaystyle A$ be the union of all $\displaystyle A_{ij}$ chosen up to this point. Now for $\displaystyle\tilde{F}_{ij}$ with $\displaystyle m(\tilde{F}_{ij})<r_{ij},$ let $\displaystyle A_{ij}\subset A$ of the same form as above, this time with $\displaystyle m(A_{ij})=r_{ij}-m(\tilde{F}_{ij}).$ In this case, define $\displaystyle F_{ij}:=\tilde{F}_{ij}\cup A_{ij}.$ Now let $\displaystyle F_{i}:=\bigcup_{j=1}^{2^{K}}F_{ij}.$ Note that some $\displaystyle F_{ij}$ may be empty. In particular, $\displaystyle F_{ij}=\emptyset$ if and only if $\displaystyle r_{ij}=0$.

Note that by definition, $\displaystyle m(F_{ij})=r_{ij}$ and note further that $\displaystyle\tilde{F}_{ij}\triangle F_{ij}=A_{ij}.$ We claim

[TABLE]

for all $\displaystyle i$. Let $\displaystyle i$ be fixed, and consider

[TABLE]

We have $\displaystyle\left|m(E_{ij})-r_{ij}\right|<\frac{\varepsilon}{2^{K}}$ so $\displaystyle\sum_{j}\left|m(E_{ij})-r_{ij}\right|<\varepsilon.$ On the other hand,

[TABLE]

Now, because $\displaystyle m(E_{ij_{1}}\cap\tilde{F}_{ij_{2}})=0$ if $\displaystyle j_{1}\neq j_{2},$ we have that

[TABLE]

Therefore, $\displaystyle\sum_{j}m(A_{ij})<2\varepsilon.$

We will now show $\displaystyle m(E_{i}\triangle F_{i})<3\varepsilon.$ Firstly, we have $\displaystyle m(E_{i}\triangle F_{i})\leq m(E_{i}\triangle\tilde{F}_{i})+m(\tilde{F}_{i}\triangle F_{i}).$ But $\displaystyle m(E_{i}\triangle\tilde{F}_{i})<\varepsilon.$ Further,

[TABLE]

But as previously noted, $\displaystyle\tilde{F}_{ij}\triangle F_{ij}=A_{ij},$ and we already showed $\displaystyle\sum_{j}m(A_{ij})<2\varepsilon$. Thus, $\displaystyle m(E_{i}\triangle F_{i})<3\varepsilon$ as desired. ∎

With Lemma 2, we can now prove the equivalent version of Theorem 1 for the unit square, which will be the core result for proving our version of the Weak Approximation Theorem.

Theorem 2** (Density of column-preserving permutations).**

Column-preserving permutations are dense in $\displaystyle\mathcal{G}_{X}.$ More precisely, let $\displaystyle T\in\mathcal{G}_{X}.$ Given $\displaystyle N_{\varepsilon}(T)$, a dyadic neighborhood of $\displaystyle T$, there exists $\displaystyle Q\in N_{\varepsilon}(T)$, a column-preserving permutation.

Proof.

We may assume without loss of generality that

[TABLE]

where $\displaystyle D_{i}$ are dyadic squares of some fixed rank $\displaystyle N$. We start with the case where $\displaystyle T^{\prime}=I_{X}.$

Let $\displaystyle D_{ij}:=D_{i}\cap TD_{j}.$ Note that $\displaystyle\{D_{ij}\}$ partitions $\displaystyle Z$. By Lemma 1, there exists a partition of $\displaystyle Z$ into dyadic sets, $\displaystyle\{\tilde{E}_{ij}\},$ such that $\displaystyle m(D_{ij}\triangle\tilde{E}_{ij})<\frac{\varepsilon}{6M},$ where $\displaystyle M:=2^{2N}.$ Further by Lemma 1, we can find a dyadic partition of $\displaystyle Z$ into sets $\displaystyle\{\tilde{F}_{ij}\}$ where $\displaystyle m(T^{-1}D_{ij}\triangle\tilde{F}_{ij})<\frac{\varepsilon}{6M}.$ Note that because $\displaystyle m(D_{ij})=m(T^{-1}D_{ij}),$ we can assume that $\displaystyle m(\tilde{E}_{ij})=m(\tilde{F}_{ij})$ Let $\displaystyle K=\text{max rank}\{\tilde{E}_{ij},\tilde{F}_{ij}\}.$ We can now apply Lemma 2 to both $\displaystyle\tilde{E}_{ij}$ and $\displaystyle\tilde{F}_{ij}$ to get dyadic partitions $\displaystyle\{E_{ij}\}$ and $\displaystyle\{F_{ij}\}$ such that

[TABLE]

Recall that if $\displaystyle C_{k}$ is a dyadic interval of rank $\displaystyle K$, then in the notation of Lemma 2, $\displaystyle E_{ijk}:=E_{ij}\cap\pi^{-1}C_{k}$ and $\displaystyle F_{ijk}:=F_{ij}\cap\pi^{-1}C_{k}.$ Note that not only do we have $\displaystyle m(D_{ij})=m(T^{-1}D_{ij}),$ but because $\displaystyle T$ is an extension of identity, $\displaystyle m(T^{-1}D_{ij}\cap\pi^{-1}C_{k})=m(T^{-1}(D_{ij}\cap\pi^{-1}C_{k}))=m(D_{ij}\cap\pi^{-1}C_{k}).$ Thus we are able to choose the same dyadic rationals in both applications of Lemma 2, and subsequently have that $\displaystyle m(E_{ijk})=m(F_{ijk})$ for $\displaystyle i,j=1,\ldots 2^{2N},k=1,\ldots,2^{K}.$

We now define $\displaystyle Q$ as the permutation which maps $\displaystyle F_{ijk}$ to $\displaystyle E_{ijk}.$ Note that in particular, $\displaystyle Q$ will map $\displaystyle F_{ij}$ to $\displaystyle E_{ij}.$ Further note that $\displaystyle Q$ will be an extension of the identity.

Let $\displaystyle j$ be fixed. We will now show $\displaystyle m(QD_{j}\triangle TD_{j})<\varepsilon$. Recall $\displaystyle D_{ij}=D_{i}\cap TD_{j},$ so $\displaystyle T^{-1}D_{ij}=T^{-1}D_{i}\cap D_{j}$ and $\displaystyle D_{j}=\bigcup_{i}T^{-1}D_{ij}.$ We have

[TABLE]

But per (4), $\displaystyle m(T^{-1}D_{ij}\triangle F_{ij})<\frac{\varepsilon}{2M},$ so $\displaystyle(\ref{Eq:SquareFinal1})<\sum_{i}\frac{\varepsilon}{2M}\leq\frac{\varepsilon}{2}$. Therefore

[TABLE]

On the other hand,

[TABLE]

Again, per (4), $\displaystyle m(D_{ij}\triangle E_{ij})<\frac{\varepsilon}{2M},$ so $\displaystyle(\ref{Eq:SquareFinal2})<\sum_{i}\frac{\varepsilon}{2M}=\frac{\varepsilon}{2}.$ Therefore,

[TABLE]

As this holds for all $\displaystyle j,$ we have that $\displaystyle Q\in N_{\varepsilon}(T).$

Now suppose $\displaystyle T$ is an extension of some invertible $\displaystyle T^{\prime}.$ Define $\displaystyle\tilde{T}:=(T^{\prime})^{-1}T.$ Then $\displaystyle\tilde{T}$ is an extension of the identity, so there exists a column-preserving permutation $\displaystyle\tilde{Q}\in N_{\varepsilon/2}(\tilde{T}).$ But then $\displaystyle T^{\prime}\tilde{Q}\in N_{\varepsilon/2}(T)$ as $\displaystyle m(T^{\prime}\tilde{Q}D_{i}\triangle TD_{i})=m(\tilde{Q}D_{i}\triangle\tilde{T}D_{i})<\frac{\varepsilon}{2}.$ By Remark 3, $\displaystyle T^{\prime}\tilde{Q}$ is $\displaystyle L$-simply discrete equivalent, with $\displaystyle L=2^{\text{rank }\tilde{Q}}$. If we let $\displaystyle G_{i}:=\pi_{L}D_{i}$ and let $\displaystyle\tilde{N}_{\varepsilon/2}(\pi_{L}(T^{\prime}\tilde{Q})):=\{S_{L}:\mu_{L}(\pi_{L}(T^{\prime}\tilde{Q})G_{i}\triangle S_{L}G_{i})<\frac{\varepsilon}{2}\forall i\},$ then by Theorem 1 there exists a column-preserving dyadic permutation $\displaystyle\hat{Q}\in\tilde{N}_{\varepsilon/2}(\pi_{L}(T^{\prime}\tilde{Q})).$ Now we define $\displaystyle Q$ to be the simply discrete equivalent extension of $\displaystyle\hat{Q}.$ Note that because $\displaystyle L$ was dyadic, $\displaystyle Q$ is a (column-preserving) dyadic permutation. Further, $\displaystyle Q\in N_{\varepsilon/2}(T^{\prime}\tilde{Q})$ as $\displaystyle m(QD_{i}\triangle T^{\prime}\tilde{Q}D_{i})=\mu_{L}(\hat{Q}G_{i}\triangle\pi_{L}(T^{\prime}\tilde{Q})G_{i})<\frac{\varepsilon}{2}.$ So

[TABLE]

for all $\displaystyle i$, and thus $\displaystyle Q\in N_{\varepsilon}(T).$ ∎

We close this section with the promised version of Halmos’ Weak Approximation Theorem for extensions.

Theorem 3** (Weak Approximation Theorem for Extensions).**

Let $\displaystyle T\in\mathcal{G}_{X}$, and let $\displaystyle N_{\varepsilon}(T)$ be a dyadic neighborhood of $\displaystyle T.$ Then for any $\displaystyle k_{0}\in\mathbb{N},$ there exists $\displaystyle k\geq k_{0}$ and $\displaystyle Q\in\mathcal{G}_{X}$such that the following hold:

• •

$\displaystyle Q,Q^{\prime}$* are dyadic permutations of rank $\displaystyle k$ on $\displaystyle Z,X$ respectively,*

• •

$\displaystyle Q^{\prime}$* is cyclic,*

• •

$\displaystyle Q$* is periodic with period $\displaystyle 2^{k}$ everywhere,*

• •

$\displaystyle Q\in N_{\varepsilon}(T).$**

Proof.

Because Theorem 2 tells us that $\displaystyle N_{\varepsilon/2}(T)$ will contain $\displaystyle P\in\mathcal{G}_{X},$ ($\displaystyle P$ a column-preserving permutation) we need only prove the case where $\displaystyle T$ is a permutation itself (we will therefore proceed using $\displaystyle P,P^{\prime}$ in place of $\displaystyle T,T^{\prime}$).

Fix $\displaystyle k_{0}\in\mathbb{N}.$ Because $\displaystyle P$ is a permutation and $\displaystyle D_{i}$ is a dyadic set, $\displaystyle PD_{i}$ is also a dyadic set. Let $\displaystyle M$ be the maximum rank of $\displaystyle PD_{i}$ (so that $\displaystyle P$ is a permutation of rank $\displaystyle M$), $\displaystyle K$ be the number of disjoint cycles in $\displaystyle P^{\prime},$ and $\displaystyle k$ be chosen to be greater than both $\displaystyle M$ and $\displaystyle k_{0},$ and such that $\displaystyle\frac{K}{2^{k-1}}<\varepsilon$.

We will now construct $\displaystyle Q$ of rank $\displaystyle k.$ Note that following the proof there will be an example of this construction. To start, let $\displaystyle E_{1}$ be any dyadic square of rank $\displaystyle M$. If $\displaystyle\pi E_{1}$ is not a fixed point of $\displaystyle P^{\prime},$ we have $\displaystyle Q$ map the “first” rank $\displaystyle k$ dyadic square (which we will henceforth refer to as a $\displaystyle k$-square) of $\displaystyle E_{1}$ to the “first” $\displaystyle k$-square in $\displaystyle PE_{1}.$ By “first” $\displaystyle k$-square, we mean the top left $\displaystyle k$-square. Now, if $\displaystyle(P^{\prime})^{2}\pi E_{1}\neq\pi E_{1},$ we continue to map to the first $\displaystyle k$-square in $\displaystyle(P^{\prime})^{2}\pi E_{1}.$ Eventually, however, we reach a point where $\displaystyle(P^{\prime})^{l}\pi E_{1}=\pi E_{1}.$ From where we are in $\displaystyle(P)^{l-1}E_{1},$ we continue to map to the “second” $\displaystyle k$-square in $\displaystyle P^{l}E_{1}$ (by “second” we mean the one to the right of the first). Note that $\displaystyle P^{l}E_{1}\neq E_{1}$ in general.

We now repeat the entire process, replacing “first” for “second,” eventually “third” and so on, as well as replacing $\displaystyle E_{1}$ with $\displaystyle P^{l}E_{1}$. Eventually we will arrive at a $\displaystyle k$-square whose projection is at the far right of $\displaystyle(P^{\prime})^{l-1}\pi E_{1}.$ At this point, we choose an $\displaystyle M$-square $\displaystyle E_{2}$ such that $\displaystyle\pi E_{2}$ is not in the $\displaystyle P^{\prime}$ cycle of $\displaystyle\pi E_{1}$ (assuming such an $\displaystyle E_{2}$ exists). Then from our current position, we map to the first $\displaystyle k$-square of $\displaystyle E_{2},$ and repeat the process.

We continue on like this until we have exhausted every $\displaystyle P^{\prime}$ cycle (including fixed points), at which point we return to the the first $\displaystyle k$-square of $\displaystyle E_{1}.$ Note that we have visited every $\displaystyle k$-column exactly once. We are not quite done yet, though. We now choose a $\displaystyle k$-square on the same column as the first $\displaystyle k$-square in $\displaystyle E_{1},$ and we repeat the entire process. Now shifting to rows within the $\displaystyle M$-squares that correspond to our new choice of starting point. That is, in the original process, we were in the top row of every $\displaystyle M$-square, because our original $\displaystyle k$-square was in the top row. If our new $\displaystyle k$-square is in the 3rd row within its $\displaystyle M$-square, say, all our choices will be in the 3rd row of the respective $\displaystyle M$-squares. Repeating this process, we eventually define $\displaystyle Q$ for all $\displaystyle k$-squares.

We now find a bound for $\displaystyle m(PD_{i}\triangle QD_{i}).$ Note that by our construction the only points that can be in $\displaystyle PD_{i}\triangle QD_{i}$ come from $\displaystyle k$-squares in $\displaystyle D_{i}$ whose projections are in the last $\displaystyle k$-interval in each $\displaystyle P^{\prime}$ cycle. Let $\displaystyle E_{j}$ be such a $\displaystyle k$-square. Then $\displaystyle m(PE_{j}\triangle QE_{j})\leq\frac{2}{2^{2k}}=\frac{1}{2^{2k-1}}.$ There are $\displaystyle 2^{k}$ such $\displaystyle E_{j}$ per $\displaystyle k$-column, and there are $\displaystyle K$ such $\displaystyle k$-columns. Thus, $\displaystyle m(PD_{i}\triangle QD_{i})\leq\bigcup_{j}m(PE_{j}\triangle QE_{j})\leq\frac{K2^{k}}{2^{2k-1}}=\frac{K}{2^{k-1}}<\varepsilon$. ∎

The construction in the proof of Theorem 3 can be difficult to follow closely, so we provide an example of the construction. We first provide a $\displaystyle P$ which, in this case, will be of rank $\displaystyle 2$. See Figure 1 for reference on how we label the $\displaystyle 2$-squares. Note that we will define $\displaystyle P$ using cycle decomposition notation. That is, if we write $\displaystyle R=(1\ 2\ 3),$ then we mean that the image under $\displaystyle R$ of the square labeled $\displaystyle 1$ is the square labeled $\displaystyle 2$. Similarly the image of “$\displaystyle 2$” is “$\displaystyle 3$” and the image of “$\displaystyle 3$” is “$\displaystyle 1$”. Any squares not written explicitly in the decomposition are fixed points. Now, we let $\displaystyle P:=(1\ 11\ 5\ 3)(13\ 15)(9\ 7)(2\ 6\ 14)(4\ 16\ 12\ 8).$ Note that $\displaystyle P$ extends $\displaystyle P^{\prime}:=(1\ 3)$ on $\displaystyle X$.

Suppose we were to construct $\displaystyle Q$ to be a rank $\displaystyle 3$ permutation. Rather than write the entire cycle decomposition of $\displaystyle Q$ (as it would involve writing all $\displaystyle 64$ $\displaystyle 3$-squares), we label Figure 2 to define $\displaystyle Q$. Here we have labeled the $\displaystyle 3$-squares such that for a square labeled $\displaystyle(n,k),$ we have that $\displaystyle Q(n,k)=(n,k+1),(k\mod 8)$ (for consistency, here we have $\displaystyle 8\mod 8:=8$ instead of $\displaystyle 0$ as it typically would be). Further, if $\displaystyle n_{1}\neq n_{2},$ then $\displaystyle(n_{1},k_{1}),(n_{2},k_{2})$ are in independent cycles. It is easy to see with this notation that $\displaystyle Q$ is an extension of a cyclic permutation $\displaystyle Q^{\prime}$ on $\displaystyle X.$ We also note that the $\displaystyle Q$ we constructed is not the only possible $\displaystyle Q$ we could have constructed, as we have many free choices in the construction.

To close this section, we note that a very simple modification of the proof of Theorem 3 would yield a column preserving permutation $\displaystyle Q$ such that not only $\displaystyle Q^{\prime}$ is cyclic, but $\displaystyle Q$ is cyclic as well. In our example seen in Figure 2, this modification would be accomplished by changing the definition of $\displaystyle Q$ slightly so that $\displaystyle Q(n,8)=(n+1,1),(n\mod 8)$. This formulation is more akin to the classical theorem. However, we choose the formulation given in Theorem 3 as it is this formulation we need for further results.

5. Uniform Approximation

Our goal in this section is to prove results that are generalizations of those needed for Halmos’ classical Conjugacy Lemma (the key lemma for proving that weakly mixing transformations on $\displaystyle X$ are dense in $\displaystyle\mathcal{G}(X)$), and whose proofs quickly follow from the classical results and their proofs.

Lemma 3**.**

Let $\displaystyle T\in\mathcal{G}_{X}$ where $\displaystyle T^{\prime}$ is periodic of period $\displaystyle n$ (almost) everywhere. Then there exists a set $\displaystyle E$ such that $\displaystyle E=\pi^{-1}E^{\prime}$ for some $\displaystyle E^{\prime}\subset X,$ and $\displaystyle\{E,TE,\ldots,T^{n-1}E\}$ partition $\displaystyle Z.$

Proof.

Because $\displaystyle T^{\prime}$ is has period $\displaystyle n$ everywhere, there exists $\displaystyle E^{\prime}$ such that $\displaystyle\{E^{\prime},T^{\prime}E^{\prime},\ldots,(T^{\prime})^{n-1}E^{\prime}\}$ partitions $\displaystyle X$. Setting $\displaystyle E:=\pi^{-1}E^{\prime}$ we have

[TABLE]

are pairwise disjoint because $\displaystyle T$ extends $\displaystyle T^{\prime}$. Further, because $\displaystyle m(E)=m(E^{\prime})=\frac{1}{n},$ we have $\displaystyle m\left(\bigcup_{i=0}^{n-1}T^{i}E\right)=\sum_{i=0}^{n-1}m(T^{i}E)=1,$ or $\displaystyle\bigcup_{i=0}^{n-1}T^{i}E=X$. ∎

Next we move to a version of Rokhlin’s lemma (see, for example, [23, p.71]).

Lemma 4**.**

Let $\displaystyle T\in\mathcal{G}_{X}$ where $\displaystyle T^{\prime}$ is antiperiodic. Then for every $\displaystyle n\in\mathbb{N}$ and $\displaystyle\varepsilon>0$ there exists $\displaystyle E$ such that $\displaystyle E=\pi^{-1}E^{\prime}$ for some $\displaystyle E^{\prime},\{E,TE,\ldots,T^{n-1}E\}$ are pairwise disjoint, and $\displaystyle m\left(\bigcup_{i=0}^{n-1}T^{i}E\right)>1-\varepsilon.$

Proof.

Let $\displaystyle n\in\mathbb{N}$ and $\displaystyle\varepsilon>0$. Because $\displaystyle T^{\prime}$ is antiperiodic, there exists $\displaystyle E^{\prime}\subset X$ such that $\displaystyle\{E^{\prime},T^{\prime}E^{\prime},\ldots,(T^{\prime})^{n-1}E^{\prime}\}$ are pairwise disjoint and $\displaystyle m\left(\bigcup_{i=0}^{n-1}(T^{\prime})^{i}E^{\prime}\right)>1-\varepsilon$. Let $\displaystyle E:=\pi^{-1}E^{\prime}$. Because $\displaystyle T$ extends $\displaystyle T^{\prime},\{E,TE,\ldots,T^{n-1}E\}$ are pairwise disjoint. Further $\displaystyle m\left(\bigcup_{i=0}^{n-1}T^{i}E\right)=\sum_{i=0}^{n-1}m(T^{i}E)=\sum_{i=0}^{n-1}m((T^{\prime})^{i}E^{\prime})=m\left(\bigcup_{i=0}^{n-1}(T^{\prime})^{i}E^{\prime}\right)>1-\varepsilon.$ ∎

We conclude this section with a version of Halmos’ Uniform Approximation Theorem (see [23, p.75]).

Theorem 4** (Uniform Approximation Theorem for Extensions).**

Let $\displaystyle T\in\mathcal{G}_{X}$ where $\displaystyle T^{\prime}$ is antiperiodic. Then for every $\displaystyle n\in\mathbb{N}$ and $\displaystyle\varepsilon>0$ there exists $\displaystyle R\in\mathcal{G}_{X},$ such that both $\displaystyle R$ and $\displaystyle R^{\prime}$ are periodic with period $\displaystyle n$ almost everywhere, and $\displaystyle d^{\prime}(R,T)\leq\frac{1}{n}+\varepsilon.$

Proof.

By Lemma 4, there exists $\displaystyle E$ a cylinder set, such that $\displaystyle\{E,TE,\ldots,T^{n-1}E\}$ are pairwise disjoint, and $\displaystyle m\left(\bigcup_{i=0}^{n-1}T^{i}E\right)>1-\varepsilon$. If $\displaystyle z\in\bigcup_{i=0}^{n-2}T^{i}E,$ define $\displaystyle Rz:=Tz,$ and if $\displaystyle z\in T^{n-1}E,$ define $\displaystyle Rz:=T^{-(n-1)}z,$ thus making $\displaystyle R$ have period $\displaystyle n$ for all points on which we have thus far defined it. Further, because $\displaystyle T$ extends $\displaystyle T^{\prime},R$ is also an extension. And for any definition of $\displaystyle R$ on the remainder of $\displaystyle Z$, we have $\displaystyle d^{\prime}(R,T)\leq m(T^{n-1}E)+\varepsilon\leq\frac{1}{n}+\varepsilon.$

All that remains is to define $\displaystyle R$ on the remainder of $\displaystyle Z$ so that $\displaystyle R$ is an extension, and $\displaystyle R,R^{\prime}$ have period $\displaystyle n.$ Since the remainder is a cylinder set, this can be done by defining $\displaystyle R^{\prime}$ on the projection of the remainder, as you would in the classical case, and then letting $\displaystyle R=R^{\prime}\times I$ on this set of measure $\displaystyle\varepsilon$. ∎

6. Conjugacy Lemma

We now prove a generalization of Halmos’ Conjugacy Lemma (see [23, p.77]), using the same techniques as Halmos’ original proof.

Lemma 5** (Conjugacy Lemma for Extensions).**

Let $\displaystyle T\in\mathcal{G}_{X},T_{0}\in\mathcal{G}_{X}$ such that $\displaystyle T^{\prime}_{0}$ is antiperiodic, and let $\displaystyle N_{\varepsilon}(T)=\{V\in\mathcal{G}_{X}:m(VD_{i}\triangle TD_{i})<\varepsilon,i=1,\ldots,N\}$ be a dyadic neighborhood of $\displaystyle T$. Then there exists $\displaystyle S\in\mathcal{G}_{X}$ such that $\displaystyle S^{-1}T_{0}S\in N_{\varepsilon}(T).$

Proof.

Let $\displaystyle k_{0}\in\mathbb{N}$ be greater than the ranks of all $\displaystyle D_{i}$ and $\displaystyle\frac{1}{2^{k_{0}-2}}<\varepsilon.$ Further, let $\displaystyle Q\in N_{\varepsilon/2}(T)$, a dyadic permutations of rank $\displaystyle k\geq k_{0}$ with all properties guaranteed by the Weak Approximation Theorem for Extensions, Theorem 3 ($\displaystyle Q^{\prime}$ is cyclic, $\displaystyle Q$ is $\displaystyle 2^{k}$ periodic). Applying the Uniform Approximation Theorem for Extensions, Theorem 4, with $\displaystyle 2^{k}$ in place of $\displaystyle n$ and $\displaystyle\frac{1}{2^{k}}$ in place of $\displaystyle\varepsilon$, there exists $\displaystyle R\in\mathcal{G}_{X}$ such that $\displaystyle R,R^{\prime}$ are have period $\displaystyle 2^{k}$ almost everywhere, and $\displaystyle d^{\prime}(R,T_{0})\leq\frac{1}{2^{k}}+\frac{1}{2^{k}}<\frac{\varepsilon}{2}$.

We will show $\displaystyle Q$ and $\displaystyle R$ are conjugate by some $\displaystyle S\in\mathcal{G}_{X}.$ Let $\displaystyle q=2^{k}$ and $\displaystyle E_{0},\ldots,E_{q-1}$ be cylinder sets of dyadic intervals of rank $\displaystyle k$ in $\displaystyle X,$ arranged so that $\displaystyle QE_{i}=E_{i+1}~{}(i\mod q).$ Note that $\displaystyle m(E_{i})=\frac{1}{q}$ By Lemma 3, there exists $\displaystyle F_{0}$, a cylinder set, such that $\displaystyle m(F_{0})=\frac{1}{q}$ and $\displaystyle F_{0},RF_{0},\ldots,R^{q-1}F_{0}$ partition $\displaystyle Z$. Let $\displaystyle F_{i}:=R^{i}F_{0}$. Let $\displaystyle S$ be any measure preserving transformation which maps $\displaystyle E_{0}$ to $\displaystyle F_{0}$ as an extension of some $\displaystyle S^{\prime}.$ Then for $\displaystyle z\in E_{i},$ let $\displaystyle Sz:=R^{i}SQ^{-i}z.$ This can be seen in the following diagram:

[TABLE]

Commutation of the diagram shows that $\displaystyle Q=S^{-1}RS.$ Further, because $\displaystyle Q,R,S\restriction_{E_{0}}$ are extensions of $\displaystyle Q^{\prime},R^{\prime},S^{\prime}\restriction_{\pi E_{0}},S$ is an extension of $\displaystyle S^{\prime}$.

Now, because $\displaystyle d^{\prime}$ is invariant under group operations, we have

[TABLE]

Thus, for any $\displaystyle D_{i}$ we have:

[TABLE]

But $\displaystyle d\leq d^{\prime},$ so $\displaystyle m(TD_{i}\triangle S^{-1}T_{0}SD_{i})\leq\frac{\varepsilon}{2}+d^{\prime}(Q,S^{-1}T_{0}S)<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.$ ∎

This lemma is so important because as we will see in our main result in Theorem $\displaystyle\ref{Thm:CategoryTheorem},$ the conjugacy class of $\displaystyle\mathcal{W}_{X}$ is $\displaystyle\mathcal{W}_{X}$ itself. Thus, in proving Lemma 5, we have indeed proven half of Theorem 5.

7. Category Theorem

We are fast approaching our main goal: that $\displaystyle\mathcal{W}_{X}$ is a dense, $\displaystyle G_{\delta}$ subset of $\displaystyle\mathcal{G}_{X}$. Before we can prove it, we need to prove a few technical results. First we have a quick consequence of the Cauchy-Schwarz Inequality, Proposition 1, but it will be important enough to make a special note of it.

Proposition 4**.**

Let $\displaystyle f,g\in L^{2}(Z|X).$ Then

[TABLE]

Proof.

By the Cauchy-Schwarz Inequality for $\displaystyle L^{2}(Z|X)$, we have

[TABLE]

pointwise. Now, by definition of $\displaystyle L^{2}(Z|X),\mathbb{E}(\left|f\right|^{2}|X)^{1/2}\in L^{\infty}(X).$ Letting $\displaystyle M:=\left\lVert\left\lVert f\right\rVert_{L^{2}(Z|X)}\right\rVert_{L^{\infty}(X)},$ we have

[TABLE]

Now note that by Fubini,

[TABLE]

where $\displaystyle Y=X,\mu_{X}=\mu_{Y}=m_{1}$ (the notation here was changed to clarify what integrals were intended). And so taking $\displaystyle L^{2}(X)$ norm on both sides of (7), we arrive at the desired inequality. ∎

Our next goal is to prove that $\displaystyle T\in\mathcal{W}_{X}$ is equivalent to the existence of a subsequence $\displaystyle n_{k}$ such that for all $\displaystyle f,g\in L^{2}(Z|X),$

[TABLE]

To this end, we first show that if (8) holds for an $\displaystyle L^{2}(Z)$-dense subset of $\displaystyle L^{2}(Z|X),$ it holds for all of $\displaystyle L^{2}(Z|X).$

Lemma 6**.**

Let $\displaystyle T\in\mathcal{G}_{X},$ and let $\displaystyle D\subset L^{2}(Z|X),$ with $\displaystyle\left\lVert\left\lVert\overline{f}\right\rVert_{L^{2}(Z|X)}\right\rVert_{L^{\infty}(X)}\leq 1$ for all $\displaystyle f\in D,$ such that $\displaystyle D$ is dense in the unit ball of $\displaystyle L^{2}(Z|X),$ but with respect to the $\displaystyle L^{2}(Z)$ norm topology. Further suppose that there exists a subsequence $\displaystyle n_{k}$ such that for all $\displaystyle f_{i},f_{j}\in D,$

[TABLE]

Then for all $\displaystyle h,g\in L^{2}(Z|X)$

[TABLE]

Proof.

Fix $\displaystyle h,g\in L^{2}(Z|X)$ and let $\displaystyle\{h_{j}\},\{g_{j}\}\subset D$ such that

[TABLE]

We first claim that

[TABLE]

uniformly with respect to $\displaystyle n$. Indeed, we have

[TABLE]

Now by Proposition 4

[TABLE]

In turn, as $\displaystyle\left\lVert\left\lVert T^{n}h_{j}\right\rVert_{L^{2}(Z|X)}\right\rVert_{L^{\infty}(X)}=\left\lVert\left\lVert h_{j}\right\rVert_{L^{2}(Z|X)}\right\rVert_{L^{\infty}(X)}$ and $\displaystyle T$ is an isometry, we get

[TABLE]

Because $\displaystyle h_{j}\to h,g_{j}\to g$ in $\displaystyle L^{2}(Z),$ we have the desired result. Note that a similar argument using Proposition 4 will show $\displaystyle\mathbb{E}(\overline{g_{j}}|X)\to\mathbb{E}(\overline{g}|X),(T^{\prime})^{n}\mathbb{E}(h_{j}|X)\to(T^{\prime})^{n}\mathbb{E}(h|X))$ in $\displaystyle L^{2}(X).$

Now,

[TABLE]

By hypothesis, there is a subsequence $\displaystyle n_{k}$ (independent of $\displaystyle j$) such that the middle term converges to 0. Further, the first and third terms converge to 0 as $\displaystyle j\to\infty$ uniformly in $\displaystyle n$, so

[TABLE]

as desired. ∎

Next, recall that a function $\displaystyle f\in L^{2}(Z|X)$ is called generalized eigenfunction for a given $\displaystyle T\in\mathcal{G}_{X}$ (see [19, p.179]) if the $\displaystyle L^{\infty}(X)$-module spanned by $\displaystyle\{T^{n}f:n\in\mathbb{N}\}$ has finite rank. In other words, there exists $\displaystyle g_{1},\ldots,g_{l}\in L^{2}(Z|X)$ such that for all $\displaystyle n$, there exists $\displaystyle c^{n}_{j}\in L^{\infty}(X),1\leq j\leq l$ such that

[TABLE]

Lemma 7**.**

Let $\displaystyle T\in\mathcal{G}_{X}$. Then $\displaystyle T\in\mathcal{W}_{X}$ if and only if there exists a subsequence $\displaystyle n_{k}$ such that for all $\displaystyle f,g\in L^{2}(Z|X)$

[TABLE]

Proof.

Let $\displaystyle T\in\mathcal{W}_{X}.$ Lemma 6 tells us that we need only show that there exists a subsequence such that (9) holds for an $\displaystyle L^{2}(Z)$-dense subset of $\displaystyle L^{2}(Z|X)$. Let $\displaystyle(f_{i})_{i=1}^{\infty}$ be an enumeration of that dense subset, and let $\displaystyle l\mapsto(f_{i},f_{j}),1\leq l<\infty$ order the set $\displaystyle\{(f_{i},f_{j})\}$ arbitrarily. Now, by the definition of a weakly mixing extension and the Koopman-von Neumann Lemma (see, e.g., [13, p.54]), for each $\displaystyle l$ there exists a subsequence $\displaystyle n^{\prime l}_{m}$ of upper asymptotic density 1 such that

[TABLE]

Stated differently, for the pair $\displaystyle f_{i},f_{j}$ corresponding to $\displaystyle l$, we have the desired convergence along the dense subsequence $\displaystyle n^{\prime l}_{m}$. We now define a new density 1 subsequence for each $\displaystyle l$ inductively. Let $\displaystyle n^{1}_{m}:=n^{\prime 1}_{m},$ and for $\displaystyle l>1,$ define $\displaystyle n^{l}_{m}$ to be a common density 1 subsequence of $\displaystyle n^{l-1}_{m}$ and $\displaystyle n^{\prime l}_{m}.$ Now define $\displaystyle n_{k}$ be a diagonal sequence obtained from $\displaystyle n^{l}_{m},$ i.e., $\displaystyle n_{k}:=n^{k}_{k}.$ It is easy to see that for all $\displaystyle f_{i},f_{j},$ (9) holds.

To prove the converse, suppose $\displaystyle T\in\mathcal{G}_{X}\backslash\mathcal{W}_{X}$. Then there exists $\displaystyle f\in L^{2}(Z|X)\backslash L^{\infty}(X)$ that is a generalized eigenfunction for $\displaystyle T$, see [19, p.192]. Without loss of generality, $\displaystyle\left\lVert f\right\rVert_{L^{1}(Z)}=1.$ Let $\displaystyle g_{1},\ldots,g_{l}\in L^{2}(Z|X)$ be a basis for the module spanned by $\displaystyle T^{n}f.$ We want $\displaystyle g_{i}$ to be “relatively orthonormal”. That is, we want $\displaystyle\mathbb{E}(g_{i}\overline{g}_{j}|X)=0\ a.e.$ when $\displaystyle i\neq j$ and $\displaystyle\mathbb{E}(\left|g_{j}\right|^{2}|X)=1\ a.e.$ This can be accomplished with a relative Gram-Schmidt process. We start by defining $\displaystyle h^{\prime}_{1}:=g_{1}.$ For any $\displaystyle x$ such that $\displaystyle\mathbb{E}(\left|h^{\prime}_{1}\right|^{2}|X)(x)=0,$ we have that $\displaystyle h^{\prime}_{1}(x,y):=h^{\prime}_{1,x}(y)\equiv 0.$ Thus by setting the corresponding $\displaystyle c^{n}_{1}(x)=0$ for all $\displaystyle n$ we can define $\displaystyle h_{1,x}(y)$ arbitrarily, so long as it’s not identically 0. For all other $\displaystyle x$, define $\displaystyle h_{1,x}(y):=h^{\prime}_{1,x}(y)$. Now, having defined $\displaystyle h_{j},1\leq j\leq i-1,$ we define $\displaystyle h^{\prime}_{i}$ by

[TABLE]

Similar to the above, if there are any $\displaystyle x$ such that $\displaystyle\mathbb{E}(\left|h^{\prime}_{i}\right|^{2}|X)(x)=0,$ we define $\displaystyle h_{i,x}(y)\not\equiv 0$ (again changing $\displaystyle c^{n}_{i}(x)$ to 0 for all $\displaystyle n$) and for all other $\displaystyle x,h_{i,x}(y):=h^{\prime}_{i,x}(y)$. Finally we normalize and redefine $\displaystyle g_{j}$ so that

[TABLE]

Now define a function $\displaystyle j:\mathbb{N}\to\{1,\ldots,l\}$ such that $\displaystyle\left\lVert c^{n}_{j(n)}\right\rVert_{L^{2}(Z)}\geq\left\lVert c^{n}_{i}\right\rVert_{L^{2}(Z)},1\leq i\leq l.$ Note that for each $\displaystyle n,\left\lVert c^{n}_{j(n)}\right\rVert_{L^{2}(Z)}\geq 1/l$ as else

[TABLE]

Fix $\displaystyle n$ and suppose for now that $\displaystyle\mathbb{E}(f|X)\equiv 0$ almost everywhere. Note that this is guaranteed to be possible because if $\displaystyle f=f_{0}+\mathbb{E}(f|X)$ is a generalized eigenfunction with basis $\displaystyle\{g_{1},\ldots g_{l}\},$ then $\displaystyle f_{0}$ is a generalized eigenfunction with spanning set $\displaystyle\{g_{1},\ldots,g_{l},\mathbb{E}(f|X)\},$ and $\displaystyle\mathbb{E}(f_{0}|X)\equiv 0$ by design. Now, by relative orthonormality we have

[TABLE]

Define $\displaystyle B_{i}:=j^{-1}(i)$. By the above work, if $\displaystyle n\in B_{i},$

[TABLE]

Now given any subsequence $\displaystyle(n_{k})$ there will be at least one $\displaystyle i\in\{frm[o]{--},\ldots,l\}$ such that $\displaystyle(n_{k})$ intersects $\displaystyle B_{i}$ infinitely often. Thus, $\displaystyle\left\lVert\mathbb{E}(T^{n_{k}}f\overline{g_{i}}|X)\right\rVert$ does not converge to 0 as $\displaystyle k\to\infty$.

If $\displaystyle\mathbb{E}(f|X)\neq 0,$ we write $\displaystyle f=f_{0}+h$ where $\displaystyle\mathbb{E}(f_{0}|X)=0$ and $\displaystyle h:=\mathbb{E}(f|X)$. Then

[TABLE]

By linearity of the conditional expectation, this is the same as

[TABLE]

The second term is 0 as $\displaystyle\mathbb{E}(f_{0}|X)\equiv 0.$ Further, because $\displaystyle h\in L^{\infty}(X),\mathbb{E}(T^{n}h\overline{g}|X)=(T^{\prime})^{n}h\mathbb{E}(\overline{g}|X)=(T^{\prime})^{n}\mathbb{E}(h|X)\mathbb{E}(\overline{g}|X),$ which cancels with the fourth term. We are left with $\displaystyle\mathbb{E}(T^{n}f_{0}\overline{g}|X)$ and have reduced this to the previous case. ∎

Finally, we arrive at our goal.

Theorem 5** (Weakly Mixing Extensions are Residual).**

$\displaystyle\mathcal{W}_{X}$* is a dense, $\displaystyle G_{\delta}$ subset of $\displaystyle\mathcal{G}_{X}$.*

Proof.

We begin by proving that if $\displaystyle T\in\mathcal{W}_{X}$ and $\displaystyle S\in\mathcal{G}_{X},$ then $\displaystyle S^{-1}TS\in\mathcal{W}_{X}.$ With the note that there are weakly mixing extensions of antiperiodic factors, we then use Lemma 5 to conclude $\displaystyle\mathcal{W}_{X}$ is dense.

We need to prove that there exists a subsequence of

[TABLE]

which converges to 0. Indeed, the above equals

[TABLE]

But $\displaystyle T$ is a weakly mixing extension of $\displaystyle T^{\prime}$ so there exists a subsequence for which the above converges to 0.

To prove that $\displaystyle\mathcal{W}_{X}$ is $\displaystyle G_{\delta},$ let $\displaystyle\{f_{i}\}\subset L^{2}(Z|X)$ be dense with respect to $\displaystyle L^{2}(Z)$ and for $\displaystyle i,j,k,n\in\mathbb{N},$ consider the sets

[TABLE]

Due to Lemmas 6, 7, we see that $\displaystyle\bigcap_{i,j,k}\bigcup_{n\geq k}A_{i,j,k,n}=\mathcal{W}_{X}.$

Thus it is sufficient to prove that each $\displaystyle A_{i,j,k,n}$ is open. For this, we show that for fixed $\displaystyle n\in\mathbb{N},f,g\in L^{2}(Z|X)$ and $\displaystyle\varepsilon>0,$ the set

[TABLE]

is open in the weak topology. To this end, we show that the complement

[TABLE]

is closed. Let $\displaystyle(S_{m})\subset V(n,f,g,\varepsilon)$ be a sequence of Koopman operators with $\displaystyle(S_{m})$ converging weakly to a Koopman operator $\displaystyle S$. Note that this implies that $\displaystyle S_{m}\to S$ strongly (as Koopman operators are all isometries).

First note that in general, if we have functions $\displaystyle g,h,h_{1},h_{2},\ldots\in L^{2}(Z|X),$ and $\displaystyle h_{m}\to h$ in $\displaystyle L^{2}(Z),$ then $\displaystyle\mathbb{E}(h_{m}g|X)\to\mathbb{E}(hg|X)$ in $\displaystyle L^{2}(X).$ Indeed, by Proposition 4,

[TABLE]

Second, note that $\displaystyle S_{m}\to S$ strongly implies $\displaystyle(S_{m})^{\prime}\to S^{\prime}$ strongly. To see this, let $\displaystyle h\in L^{2}(X)$ and let $\displaystyle\hat{h}\in L^{2}(Z)$ be defined so that $\displaystyle\hat{h}(x,y):=h(x)$ (that is, $\displaystyle\hat{h}$ is constant on fibers). Then by Fubini, $\displaystyle\left\lVert(S_{m})^{\prime}h-S^{\prime}h\right\rVert_{L^{2}(X)}=\left\lVert S_{m}\hat{h}-S\hat{h}\right\rVert_{L^{2}(Z)}.$

Lastly note that if $\displaystyle S_{m}\to S$ strongly, then $\displaystyle S_{m}^{n}\to S^{n}$ strongly.

With these facts, we see that $\displaystyle\mathbb{E}(S_{m}^{n}f\overline{g}|X)-(S_{m}^{\prime})^{n}\mathbb{E}(f|X)\mathbb{E}(\overline{g}|X)$ converges to $\displaystyle\mathbb{E}(S^{n}f\overline{g}|X)-(S^{\prime})^{n}\mathbb{E}(f|X)\mathbb{E}(\overline{g}|X)$ strongly. Thus, $\displaystyle S\in V(n,f,g,\varepsilon),$ and so $\displaystyle V(n,f,g,\varepsilon)$ is closed. ∎

Remark 4*.*

Our assumption that $\displaystyle(X,m)$ was the Lebesgue measure on the unit interval was only important for the proof of density, where we need a non-atomic probability space to have an antiperiodic factor. Proving $\displaystyle\mathcal{W}_{X}$ is $\displaystyle G_{\delta}$ never required anything of $\displaystyle X$, and so the proof will hold for $\displaystyle(X,m)$ being replaced by any probability space.

8. General Case

There is still a case left to consider. Namely, the case where the “vertical” measure is neither purely non-atomic nor purely discrete. Let $\displaystyle(X,m,T^{\prime})$ be as before with $\displaystyle T^{\prime}$ invertible, and let $\displaystyle(Y,\eta)$ be a probability space where $\displaystyle Y=A\dot{\cup}B$, where $\displaystyle B$ is an at most countable set, each of which is an atom of $\displaystyle\eta$, and $\displaystyle\eta\restriction_{A}$ is non-atomic (in particular, $\displaystyle 0<\eta(A)<1$). Let $\displaystyle(Z,\mu,T)=(X\times Y,m\times\eta,T),$ where $\displaystyle T\in\mathcal{G}_{X}$ is an extension of $\displaystyle T^{\prime}.$ Let $\displaystyle C:=X\times A$ and $\displaystyle D:=X\times B$

We first show that $\displaystyle T$ cannot mix points on the discrete and non-discrete parts of $\displaystyle Y$.

Proposition 5**.**

Let $\displaystyle C,D\subset Z$ be as defined above. Then up to a set of measure zero, $\displaystyle TC\subset C$ and $\displaystyle TD\subset D.$

Proof.

First, suppose there exists $\displaystyle D^{\prime}\subset D$ such that $\displaystyle\mu(D^{\prime})>0$ and $\displaystyle TD^{\prime}\subset C$. We can assume without loss of generality that there exists $\displaystyle k$, a level of $\displaystyle D,$ such that $\displaystyle D^{\prime}\subset k$ (see Section 3 for an explanation of this notation). We claim that $\displaystyle m(TD^{\prime})=0,$ so that $\displaystyle T$ is not measure preserving. Note that because $\displaystyle T$ is an extension of invertible $\displaystyle T^{\prime}$ and $\displaystyle D^{\prime}$ is contained to a single level of $\displaystyle D$, then for each fixed $\displaystyle x\in X,TD^{\prime}\cap\pi^{-1}(x)$ contains at most one point. Therefore, by Fubini

[TABLE]

But by our previous note and the fact that $\displaystyle(A,\eta\restriction_{A})$ is non-atomic, $\displaystyle\int_{Y}\chi_{TD^{\prime}}(x,y)d\eta=0$ for all $\displaystyle x$, and thus $\displaystyle\mu(TD^{\prime})=0.$

Now suppose there exists $\displaystyle C^{\prime}\subset C$ such that $\displaystyle\mu(C^{\prime})>0$ and $\displaystyle TC^{\prime}\subset D.$ Note that there exists $\displaystyle x_{0}\in X$ such that $\displaystyle\pi^{-1}(x_{0})\cap C^{\prime}$ is uncountable (else $\displaystyle\mu(C^{\prime})=0$ with a similar argument as above). But $\displaystyle T(\pi^{-1}(x_{0})\cap C^{\prime})\subset T^{\prime}x_{0}\times B$ is a countable set. This contradicts the invertibility of $\displaystyle T$. ∎

A quick consequence of Proposition 5 is that there are no weakly mixing extensions on $\displaystyle Z$.

Corollary 1**.**

Let $\displaystyle(X,m,T^{\prime}),(Y,\eta),(Z,\mu,T),C,$ and $\displaystyle D$ be as defined above, with $\displaystyle A$ and $\displaystyle B$, continuous and discrete parts of $\displaystyle Y$, respectively, nonempty. Then $\displaystyle T\notin\mathcal{W}_{X}$.

Proof.

Note that we can assume without loss of generality that $\displaystyle TC\subset C$ and $\displaystyle TD\subset D$ (that is, not up to a set of measure 0, but rather everywhere). Define $\displaystyle f(z)$ as

\displaystyle f(z):=\left\{\begin{array}[]{cc}\frac{1}{\eta(A)}&\text{if }z\in C\\ \frac{-1}{\eta(B)}&\text{if }z\in D\end{array}\right..

Clearly $\displaystyle f\in L^{2}(Z|X)$ and simple calculation shows that $\displaystyle f$ has relative mean zero. Further, by Proposition 5, $\displaystyle f$ is $\displaystyle T$-invariant, and so

[TABLE]

Therefore, $\displaystyle T$ is not a weakly mixing extension of $\displaystyle T^{\prime}$. ∎

Remark 5*.*

As with the $\displaystyle G_{\delta}$ part of Theorem 5, Proposition 5 and Corollary 1 hold when $\displaystyle(X,m)$ is replaced with any standard probability space.

9. Strongly Mixing Extensions

In this section we first extend the notion of strongly mixing transformations to extensions, just as the notions of ergodic and weakly mixing transformations were extended to extensions. Afterwards we will show that the set of strongly mixing extensions form a set of first category in $\displaystyle\mathcal{G}_{X}$.

Definition 4**.**

Let $\displaystyle(X,\nu),(Z,\mu)$ be probability spaces. We say that $\displaystyle T\in\mathcal{G}_{X}$ is a (strongly) mixing extension of $\displaystyle T^{\prime}$ or $\displaystyle T$ is (strongly) mixing relative to $\displaystyle T^{\prime}$ if for all $\displaystyle f,g\in L^{2}(Z|X),$

[TABLE]

Let $\displaystyle\mathcal{S}_{X}\subset\mathcal{G}_{X}$ denote the set of strongly mixing extensions.

Definition 4 yields some of the properties one would hope to have from the extension of the notion of strongly mixing transformations. For example, $\displaystyle\mathcal{S}_{X}$ is in general not empty. Indeed, any direct product transformation where the second component is strongly mixing will be a strongly mixing extension. We also have that if $\displaystyle X$ is a single point, then the definition coincides with classical strongly mixing transformations. Further, it is clear that $\displaystyle\mathcal{S}_{X}\subset\mathcal{W}_{X}$.

We once again return to the case where $\displaystyle(X,m)$ is the unit interval with the Lebesgue measure, and $\displaystyle(Z,m)$ is the unit square with the Lebesgue measure. Analogous to Rokhlin’s result and its proof, we now show that $\displaystyle\mathcal{S}_{X}$ is a first category subset of $\displaystyle\mathcal{G}_{X}$.

Theorem 6** (Strongly Mixing Extensions are of First Category).**

$\displaystyle\mathcal{S}_{X}\subset\mathcal{G}_{X}$* is of first category.*

Proof.

For $\displaystyle k\in\mathbb{N},$ let $\displaystyle P_{k}:=\{T\in\mathcal{G}_{X}|T^{k}=I_{Z}\}.$ Note in particular that if $\displaystyle T\in P_{k}$ then $\displaystyle(T^{\prime})^{k}=I_{X}.$ For $\displaystyle n\in\mathbb{N},$ let $\displaystyle\hat{P}_{n}:=\bigcup_{k>n}P_{k}.$ Note that the Weak Approximation Theorem for Extensions (Theorem 3) implies that $\displaystyle\hat{P}_{n}$ is dense in $\displaystyle\mathcal{G}_{X}$.

Let $\displaystyle A:=[0,1]\times[0,1/2]$ (the bottom half of $\displaystyle Z$). Note that $\displaystyle\mathbb{E}(\chi_{A}|X)=1/2$ for all $\displaystyle x\in X$. We now define new sets,

[TABLE]

Using the same arguments as used in the proof of Theorem 5 for the sets $\displaystyle V(n,f,g,\varepsilon),$ we see that $\displaystyle M_{k}$ is closed for all $\displaystyle k$. Now let

[TABLE]

It is easy to see that $\displaystyle\mathcal{S}_{X}\subset M$. Thus it is sufficient to show that $\displaystyle M$ is of first category. It is in turn sufficient to show that $\displaystyle\bigcap_{k>n}M_{k}$ is nowhere dense for all $\displaystyle n$, and further given that $\displaystyle\bigcap_{k>n}M_{k}$ is closed for all $\displaystyle n$, it is thus sufficient to show that $\displaystyle\mathcal{G}_{X}\backslash\bigcap_{k>n}M_{k}$ is dense. Lastly, as $\displaystyle\mathcal{G}_{X}\backslash\bigcap_{k>n}M_{k}=\bigcup_{k>n}(\mathcal{G}_{X}\backslash M_{k}),$ it will suffice to show that $\displaystyle P_{k}\subset(\mathcal{G}_{X}\backslash M_{k})$ for all $\displaystyle k$ as then $\displaystyle\hat{P}_{n}=\bigcup_{k>n}P_{k}\subset\bigcup_{k>n}(\mathcal{G}_{X}\backslash M_{k})$ and $\displaystyle\hat{P}_{n}$ is dense.

Now, if $\displaystyle T\in P_{k},$ then $\displaystyle T^{k}=I_{Z},(T^{\prime})^{k}=I_{X},$ so

[TABLE]

Thus, $\displaystyle T\notin M_{k}$. ∎

Corollary 2**.**

$\displaystyle\mathcal{S}_{X}$* is a proper subset of $\displaystyle\mathcal{W}_{X}.$*

10. Further Questions

To conclude this paper, we formulate some open questions.

Question 1*.*

Let $\displaystyle(X,\nu)$ be any probability space, $\displaystyle(Y,\eta)$ be a non-atomic probability space and $\displaystyle(Z,\mu)=(X\times Y,\nu\times\eta).$ Is $\displaystyle\mathcal{W}_{X}$ a dense, $\displaystyle G_{\delta}$ subset of $\displaystyle\mathcal{G}_{X}$?

It would be sufficient for Question 1 to consider the case where $\displaystyle(X,\nu)$ is purely atomic. We cannot use the Conjugacy Lemma in this case because there are no antiperiodic transformations on a discrete set.

Question 2*.*

Let $\displaystyle(X,m)$ be a (potentially non-atomic) probability space, $\displaystyle(Y,\eta)$ be a non-atomic probability space and $\displaystyle(Z,\mu)=(X\times Y,\nu\times\eta).$ Let $\displaystyle R\in\mathcal{G}(X)$ be fixed. Define

[TABLE]

and

[TABLE]

Is $\displaystyle\mathcal{W}_{R}$ a dense, $\displaystyle G_{\delta}$ subset of $\displaystyle\mathcal{G}_{R}$?

Similarly let

[TABLE]

Is $\displaystyle\mathcal{S}_{R}$ a first category subset of $\displaystyle\mathcal{G}_{R}?$

The main difficulty as of now in answering Question 2 for weakly mixing extensions is proving the density. The freedom of having a non-fixed factor allowed a generalization of Halmos’ Conjugacy Lemma. One cannot conjugate in $\displaystyle\mathcal{G}_{R}$ unless $\displaystyle R$ is the identity on $\displaystyle X$, as $\displaystyle\mathcal{G}_{R}$ is not closed under inverses for any other invertible $\displaystyle R$. That is not even to mention lack of dyadic permutations or other tools which we had at our disposal throughout this paper. Even strongly mixing extensions at first present some trouble, as the proof relied on period $\displaystyle T$, and if $\displaystyle R$ is not periodic, then neither is $\displaystyle T$.

We note that in the case of compact group extensions of a fixed weakly mixing $\displaystyle R$, Robinson gave a positive answer to the question of genericity of weakly mixing extensions [31].

Further, we see from the following proposition that these Question 2 cannot be derived from Theorems 5 and 6. Recall that $\displaystyle(X,m)$ is the unit interval with the Lebesgue measure.

Proposition 6**.**

For all $\displaystyle T^{\prime}\in\mathcal{G}(X),\mathcal{G}_{T^{\prime}}$ is a closed, nowhere dense subset of $\displaystyle\mathcal{G}_{X}.$

Proof.

Fix $\displaystyle T^{\prime}\in\mathcal{G}(X).$ We first show that $\displaystyle\mathcal{G}_{T^{\prime}}$ is closed. Let $\displaystyle S\in\mathcal{G}_{X}\backslash\mathcal{G}_{T^{\prime}}.$ As $\displaystyle S^{\prime}\neq T^{\prime},$ there exists $\displaystyle E\subset X$ such that $\displaystyle m(S^{\prime}E\triangle T^{\prime}E)>0.$ Let $\displaystyle\varepsilon:=m(S^{\prime}E\triangle T^{\prime}E)/2.$

Consider

[TABLE]

For any $\displaystyle T\in\mathcal{G}_{T^{\prime}},\mu(T\pi^{-1}E\triangle S\pi^{-1}E)=m(T^{\prime}E\triangle S^{\prime}E)=2\varepsilon>\varepsilon,$ so $\displaystyle T\notin N_{\varepsilon}(S),$ and thus $\displaystyle\mathcal{G}_{T^{\prime}}$ is closed.

Now, to show $\displaystyle\mathcal{G}_{T^{\prime}}$ is nowhere dense, we show $\displaystyle\mathcal{G}_{X}\backslash\mathcal{G}_{T^{\prime}}$ is dense. Fix $\displaystyle T\in\mathcal{G}_{T^{\prime}},\varepsilon>0$ and let

[TABLE]

where $\displaystyle D_{i}$ are all dyadic squares of some rank $\displaystyle N$. Fix $\displaystyle i$ and let $\displaystyle E\subset\pi D_{i}$ with $\displaystyle m(E)<\varepsilon$. Let $\displaystyle E_{1},E_{2}$ be disjoint sets whose union is $\displaystyle E$ and $\displaystyle m(E_{1})=m(E_{2}).$

Select $\displaystyle R\in\mathcal{G}(X)$ with the following properties: for $\displaystyle x\in X\backslash E,Rx=x,RE_{1}=E_{2},RE_{2}=E_{1}.$ Define $\displaystyle\tilde{T}:=T(R\times I)$. Note that for $\displaystyle j$ such that $\displaystyle\pi D_{j}\neq\pi D_{i},(R\times I)D_{j}=D_{j},$ so for such $\displaystyle j,\mu(\tilde{T}D_{j}\triangle TD_{j})=0.$ On the other hand, if $\displaystyle\pi D_{j}=\pi D_{i},$ then $\displaystyle\mu(\tilde{T}D_{j}\triangle TD_{j})<m(E)=\varepsilon.$ Thus $\displaystyle\tilde{T}\in N_{\varepsilon}(T),$ but $\displaystyle\tilde{T}\notin\mathcal{G}_{T^{\prime}}.$ ∎

Question 3*.*

Can one find an extension analogue of Katok’s result (see, for example, [28]) that rigid transformations form a residual set?

Note that is not completely clear how one should define a rigid extension. The author currently has no notion of how it should be defined.