2.1. Non-singular actions of rank one

Let G be a discrete infinite countable group. Let $T=(T_g)_{g\in G}$ be a free non-singular action of G on a standard $\sigma $ -finite non-atomic measure space . By a Rokhlin tower for T, we mean a pair $(B,F)$ , where with $0<\mu (B)<\infty $ and F is a finite subset of G with $1_G\in F$ such that:

• • the subsets $T_fB$ , $f\in F$ , are mutually disjoint;

• • the Radon–Nikodym derivative ${d\mu \circ T_f}/{d\mu }$ is constant on B for each $f\in F$ .

Given a Rokhlin tower $(B,F)$ , we let . Of course, ${\mu (X_{B,F})<\infty }$ . By $\xi _{B,F}$ , we mean the finite partition of $X_{B,F}$ into the subsets $T_fB$ , $f\in F$ . If $x\in T_fB$ , then we set $O_{B,F}(x):=\{T_gx\mid g\in Ff^{-1}\}$ .

It follows from property (i) that and ${\bigcup _{n=0}^\infty X_{B_n,F_n}=X}$ . The piecewise constant property of the Radon–Nikodym derivative on the Rokhlin towers yields that:

1. (iii) if $T_cB_{n+1}\subset B_{n}$ for some $c\in F_{n+1}$ , then

   $$ \begin{align*} \frac{\mu(T_{fc}B_{n+1})}{\mu(T_fB_n)}=\frac{\mu(T_cB_{n+1})}{\mu(B_n)}\quad\text{for each } f\in F_n \text{ and } n\ge 0. \end{align*} $$

Proof. Let two subsets be of positive measure. It follows from condition (i) that there are $n>0$ and $f_1,f_2\in F_n$ such that

$$ \begin{align*} \mu (A_1\cap T_{f_1}B_n)> 0.9 \mu (T_{f_1}B_n)\quad\text{and}\quad \mu (A_2\cap T_{f_2}B_n) > 0.9\mu(T_{f_2}B_n). \end{align*} $$

As $(B_n,F_n)$ is a Rokhlin tower, $T_{f_2f_1^{-1}}T_{f_1}B_n=T_{f_2}B_n$ and

$$ \begin{align*} \frac{d\mu\circ T_{f_2f_1^{-1}}}{d\mu}(x) = \frac{\mu(T_{f_2}B_n)}{\mu(T_{f_1}B_n)}\quad\text{at a.e. } x\in A_1. \end{align*} $$

It follows that

$$ \begin{align*} \begin{aligned} \mu(T_{f_2f_1^{-1}}A_1\cap T_{f_2}B_n)&= \mu (T_{f_2f_1^{-1}}(A_1\cap T_{f_1}B_n))\\ &= \mu(A_1\cap T_{f_1}B_n) \frac{\mu(T_{f_2}B_n)}{\mu(T_{f_1}B_n)}\\ &>0.9\mu(T_{f_2B_n}). \end{aligned} \end{align*} $$

Therefore, $\mu (T_{f_2f_1^{-1}}A_1\cap T_{f_2}B_n\cap A_2)>0.8\mu (T_{f_2}B_n)$ . Hence, $\mu (T_{f_2f_1^{-1}}A_1\cap A_2)>0$ , as desired.

2.2. $(C,F)$ -equivalence relations and non-singular $(C,F)$ -measures

Fix two sequences $(F_n)_{n\geq 0}$ and $(C_n)_{n\geq 1}$ of finite subsets in G such that $F_{0} = \{1_G\}$ and for each $n>0$ ,

(2.1) $$ \begin{align} \begin{aligned} & 1_G\in F_n\cap C_n, \ \#C_{n}> 1, \\ &F_{n} C_{n+1}\subset F_{n+1},\\ &F_{n} c\cap F_{n} c' = \emptyset\quad\text{if } c, c'\in C_{n+1}\text{ and } c \neq c'. \end{aligned} \end{align} $$

We let and endow this set with the infinite product topology. Then $X_n$ is a compact Cantor space. The mapping

$$ \begin{align*} X_n \ni (f_n,c_{n+1},c_{n+2},\ldots) \mapsto (f_n c_{n+1},c_{n+2},\ldots) \in X_{n+1} \end{align*} $$

is a continuous embedding of $X_n$ into $X_{n+1}$ . Therefore, the topological inductive limit X of the sequence $(X_n)_{n\geq 0}$ is well defined. Moreover, X is a locally compact Cantor space. Given a subset $A\subset F_n$ , we let

$$ \begin{align*} [A]_n := \{x=(f_n,c_{n+1},\ldots)\in X_n, f_n\in A\} \end{align*} $$

and call this set an n-cylinder in X. It is open and compact in X. For brevity, we will write $[f]_n$ for $[\{f\}]_n$ for an element $f\in F_n$ .

We remind that two points $x=(f_n,c_{n+1},\ldots )$ and $x'=(f_n',c_{n+1}',\ldots )$ of $X_n$ are tail equivalent if there is $N>n$ such that $c_l=c_{l}'$ for each $l>N$ . We thus obtain the tail equivalence relation on $X_n$ .

The following properties of $\mathcal R$ are easy to check:

• • each $\mathcal R$ -class is countable;

• • $\mathcal R$ is minimal, that is, the $\mathcal R$ -class of every point is dense in X;

• • $\mathcal R$ is hyperfinite, that is, there is a sequence $(\mathcal S_n)_{n=1}^\infty $ of subrelations of $\mathcal R$ such that , $\bigcup _{n=1}^\infty \mathcal S_n=\mathcal R$ and $\#\mathcal S_n(x)<\infty $ for each $x\in X$ and $n > 0$ . Indeed, we can define $\mathcal S_n$ by the following: $(x,y)\in \mathcal S_n$ if either $x,y\not \in X_n$ and $x=y$ or $x=(f_n,c_{n+1},\ldots )\in X_n,y=(f_n',c_{n+1}',\ldots )\in X_n$ and $c_m=c_m'$ for all $m>n$ .

We recall that the full group $[\mathcal R]$ of $\mathcal R$ is the group of all Borel bijections $\gamma :X\to X$ such that $(x,\gamma x)\in \mathcal R$ for each $x\in X$ . A Borel measure $\mu $ on X is called $\mathcal R$ -quasi-invariant if $\mu \circ \gamma \sim \mu $ for each $\gamma \in [\mathcal R]$ . Then there is a Borel mapping $\rho _\mu :\mathcal R\to \mathbb R^*_+$ such that

$$ \begin{align*} \rho_\mu(x,y)\rho_\mu(y,z)=\rho_\mu(x,z)\quad\text{for all } (x,y), (y,z)\in\mathcal R \end{align*} $$

and $\rho _\mu (\gamma x,x)= ({d\mu \circ \gamma }/{d\mu })(x)$ at a.e. $x\in X$ for each $\gamma \in [\mathcal R]$ . The mapping $\rho _\mu $ is called the Radon–Nikodym cocycle of $(\mathcal R,\mu )$ .

Suppose that for each $n\in \mathbb N$ , a non-degenerated probability measure $\kappa _n$ on $C_n$ is given. We now let , where $\nu _0$ is the Dirac measure supported at $1_G$ . Then, $\mu _0$ is an $(\mathcal R\restriction X_0)$ -quasi-invariant probability on $X_0$ . Of course, $\mu _0$ is non-atomic if and only if

(2.2) $$ \begin{align} \prod_{n>0}\max_{c\in C_n}\kappa_n(c)=0. \end{align} $$

By the Kolmogorov 0-1 law, $(\mathcal R\restriction X_0)$ is ergodic on the probability space $(X_0,\mu _0)$ . There are many ways to extend $\mu _0$ to an $\mathcal R$ -quasi-invariant measure on X. However, all such measures will be mutually equivalent. Select for each $n\in \mathbb N$ a non-degenerated finite measure $\nu _n$ on $F_n$ such that

(2.3) $$ \begin{align} \nu_{n+1}(fc)=\nu_n(f)\kappa_{n+1}(c)\quad\text{for each } f\in F_n\text{ and } c\in C_{n+1}. \end{align} $$

It is often convenient to consider $\nu _n$ and $\kappa _n$ as finite measures on G supported on $F_n$ and $C_{n}$ , respectively. Then equation (2.3) means that $\nu _{n+1}\restriction F_nC_{n+1}=\nu _n*\kappa _{n+1}$ , where the symbol $*$ means the convolution. We now define a Borel measure $\mu $ on X by setting

$$ \begin{align*} \mu([f]_n):=\nu_n(f)\quad\text{for each } g\in F_n \text{ and every } n\in\mathbb N. \end{align*} $$

It is straightforward to verify that $\mu $ is a well-defined $\sigma $ -finite Radon measure. Moreover, $\mu $ is $\mathcal R$ -quasi-invariant and

$$ \begin{align*} \rho_\mu(x,y)=\frac{\nu_n(f_n)}{\nu_n(f^{\prime}_n)}\prod_{m>n}\frac{\kappa_m(c_m)}{\kappa_m(c_m')}, \end{align*} $$

whenever $x=(f_n,c_{n+1},\ldots )$ and $y=(f_n',c_{n+1}',\ldots )$ are $\mathcal R$ -equivalent points that belong to for some $n>0$ .

The following definition extends [Reference DanilenkoDa1, Definition 4.2], where the case of Abelian G was considered.

Consider another sequence $(\nu _n')_{n=0}^\infty $ of non-degenerated measures on $(F_n)_{n=0}^\infty $ (in n) such that $\nu _0'$ is the Dirac measure supported at $1_G$ and $\nu _{n+1}'(fc)=\nu _n'(f)\kappa _{n+1}(c)$ for each $f\in F_n$ and $c\in C_{n+1}$ for each $n>0$ . Then, the $(C,F)$ -measure $\mu '$ determined by $(\kappa _n)_{n=1}^\infty $ , and $(\nu _n')_{n=0}^\infty $ is equivalent to $\mu $ and

$$ \begin{align*} \frac{d\mu'}{d\mu}(x)=\frac{\nu_n'(f_n)}{\nu_n(f_n)}\quad\text{if } x=(f_n,\ldots)\in X_n. \end{align*} $$

Another useful observation is that given $(\kappa _n)_{n=1}^\infty $ , we can always find $(\nu _n)_{n=0}^\infty $ satisfying equation (2.3). Thus, the equivalence class of a non-singular $(C,F)$ -measure is completely determined by $(\kappa _n)_{n=1}^\infty $ alone. In particular, we may always replace a $\sigma $ -finite non-singular $(C,F)$ -measure with an equivalent finite non-singular $(C,F)$ -measure.

Remark 2.5. We note that $\mathcal R$ is Radon uniquely ergodic, that is, there is a unique $\mathcal R$ -invariant Radon measure $\xi $ on X such that $\xi (X_0)=1$ . We call it the Haar measure for $\mathcal R$ . It is $\sigma $ -finite. Let $k_n$ be the equidistribution on $C_n$ and let $\nu _n(f)=\prod _{k=1}^n\kappa _k(1_G)$ for each $f\in F_n$ and $n\ge 0$ . Then, equations (2.2) and (2.3) hold for $(\kappa _n)_{n=1}^\infty $ and $(\nu _n)_{n=0}^\infty $ . Of course, the Haar measure for $\mathcal R$ is a $(C,F)$ -measure determined by $(\kappa _n)_{n=1}^\infty $ and $(\nu _n)_{n=0}^\infty $ . The Haar measure is finite if and only if

$$ \begin{align*} \prod_{n=1}^\infty\frac{\# F_{n+1}}{\# F_n\# C_{n+1}}<\infty. \end{align*} $$

It is easy to verify that $\mathcal R$ is conservative and ergodic on the $\sigma $ -finite measure space $(X,\mu )$ . This means that for each $\mathcal R$ -invariant subset $A\subset X$ , either $\mu (A)=0$ or $\mu (X\setminus A)=0$ .

Since the set of quasi-invariant probability measures with a fixed Radon–Nikodym derivative is a simplex [Reference Greschonig and SchmidtGrSc], it makes sense to introduce the following definition.

Proof. Let $\unicode{x3bb} $ be a Radon measure on X such that $\rho _\unicode{x3bb} =\rho _\mu $ and $\unicode{x3bb} (X_0)=1$ . We will prove that $\unicode{x3bb} =\mu $ . For that, it suffices to show that $\unicode{x3bb} ([f]_n)=\mu ([f]_n)$ for all $f\in F_n$ and $n\ge 0$ . As

$$ \begin{align*} \mu([f]_n)=\frac{\nu_n(f)}{\nu_n(1_G)}\mu([1_G]_n)\quad\text{and}\quad \unicode{x3bb}([f]_n)=\frac{\nu_n(f)}{\nu_n(1_G)}\unicode{x3bb}([1_G]_n), \end{align*} $$

it is enough to prove that $\mu ([1_G]_n)=\unicode{x3bb} ([1_G]_n)$ for each $n\ge 0$ . This will be done inductively. Of course, $\mu (X_0)=\mu ([1_G]_0)=\unicode{x3bb} ([1_G]_0)=\unicode{x3bb} (X_0)=1$ . Suppose that $\mu ([1_G]_n)=\unicode{x3bb} ([1_G]_n)$ for some n. Then, for each $c\in C_{n+1}$ ,

$$ \begin{align*} \unicode{x3bb}([c]_{n+1})=\frac{\nu_n(c)}{\nu_n(1_G)}\unicode{x3bb}([1_G]_{n+1}) =\frac{\nu_n(1_G)\kappa_{n+1}(c)}{\nu_n(1_G)\kappa_{n+1}(1_G)}\unicode{x3bb}([1_G]_{n+1}). \end{align*} $$

Since $[1_G]_n=\bigsqcup _{c\in C_{n+1}}[c]_{n+1}$ , we obtain that

$$ \begin{align*} \frac{\unicode{x3bb}([1_G]_{n})}{\unicode{x3bb}([1_G]_{n+1})} =\frac{\sum_{c\in C_{n+1}}\unicode{x3bb}([c]_{n+1})}{\unicode{x3bb}([1_G]_{n+1})} =\frac{\sum_{c\in C_{n+1}}\kappa_{n+1}(c)}{\kappa_{n+1}(1_G)} =\frac{1}{\kappa_{n+1}(1_G)} =\frac{\mu([1_G]_n)}{\mu([1_G]_{n+1})}. \end{align*} $$

Hence, $\unicode{x3bb} ([1_G]_{n+1})=\mu ([1_G]_{n+1})$ , as desired.

2.3. Non-singular $(C,F)$ -actions

Non-singular $(C,F)$ -actions were defined in [Reference DanilenkoDa1, Reference DanilenkoDa2] for Abelian groups only. We extend this definition to arbitrary countable groups. Given $g\in G$ , let

$$ \begin{align*} X_n^g:=\{(f_n, c_{n+1}, c_{n+2},\ldots)\in X_n\mid gf_n\in F_n\}. \end{align*} $$

Then, $X_n^g$ is a compact open subset of $X_n$ and $X_n^g\subset X^g_{n+1}$ . Hence, the union ${X^g:=\bigcup _{n\ge 0}X_n^g}$ is an open subset of X. Let $X^G:=\bigcap _{g\in G}X^g$ . Then, $X^G$ is a $G_\delta $ -subset of X. Hence, $X^G$ is Polish and totally disconnected in the induced topology. Given $g\in G$ and $x\in X_G$ , there is $n>0$ such that $x= (f_n, c_{n+1},\ldots )\in X_n$ and $gf_n\in F_n$ . We now let $T_gx:= (gf_n, c_{n+1}, \ldots )\in X_n\subset X$ . It is straightforward to verify that:

1. (i) $T_gx\in X^G$ ;

2. (ii) the mapping $T_g:X^G\ni x\mapsto T_gx\in X^G$ is a homeomorphism of $X^G$ ; and

3. (iii) $T_gT_{g'}=T_{gg'}$ for all $g, g'\in G$ .

Hence, $T:=(T_g)_{g\in G}$ is a continuous G-action on $X^G$ .

This action is free. The subset $X^G$ is $\mathcal R$ -invariant. The T-orbit equivalence relation coincides with the restriction of $\mathcal R$ to $X^G$ .

Proposition 2.9. [Reference DanilenkoDa3, Proposition 1.2]

$X^G=X$ if and only if for each $g\in G$ and $n>0$ , there is $m> n$ such that

(2.4)

Thus, if equation (2.4) holds, then T is a minimal continuous G-action on a locally compact Cantor space X. Moreover, T is Radon uniquely ergodic, that is, there exists a unique T-invariant Radon measure $\xi $ on X such that $\xi (X_0)=1$ .

From now on, T is a topological $(C,F)$ -action of G on $X^G$ and $\mu $ is the non-singular $(C,F)$ -measure on X determined by $(\kappa _n)_{n=1}^\infty $ and $(\nu _n)_{n=0}^\infty $ satisfying equations (2.2) and (2.3). Since $X^G$ is $\mathcal R$ -invariant, we obtain that either $\mu (X^G)=0$ or $\mu (X\setminus X^G)=0$ . In the latter case, T is $\mu $ -non-singular, conservative and ergodic.

Proof. (i) $\Leftrightarrow $ (ii) Since $\mu (X\setminus X^G)=0$ if and only if $ \mu (X_n\cap X^g_m)\to \mu (X_n)$ as $m\to \infty $ for each $g\in G$ and $n \ge 0$ , it suffices to note that

and $\mu (X_n)=\mu ([F_n]_n)=\nu _n(F_n)$ .

(ii) $\Rightarrow $ (iii) We set . Then,

Hence, according to item (ii). As $\kappa _{1,m}$ is supported on , it follows that

as desired.

(iii) $\Rightarrow $ (i) Fix $g\in G$ . Take arbitrary $n\ge 0$ and $f\in F_n$ . Then, it follows from property (iii) that for $\mu $ -a.e. $x= (1_G, c_{n+1}, c_{n+2},\ldots)\in [1_G]_n$ , there exists $m>0$ such that . This means that for $\mu $ -a.e. $y= (f_n, c_{n+1}, c_{n+2},\ldots )\in X_n$ ,

that is, $y\in X^g$ . Hence, $\mu (X\setminus X^g)=0$ and property (i) follows.

In the case where $\mu $ is the Haar measure for $\mathcal R$ , the equivalence (i) $\Leftrightarrow $ (ii) of Proposition 2.10 was proved in [Reference DanilenkoDa3].

Proof. (i) We note that $\nu _n(F_n)=\mu ([F_n]_n)=\mu (X_n)\to \mu (X)$ as $n\to \infty $ . Hence, it follows from Proposition 2.10(ii) that for each $\epsilon>0$ , there is $n>0$ such that if $m>n$ , then

Hence, $\nu _m(F_m\cap gF_m)>(1-2\epsilon )\nu _m(F_m)$ . It follows that $\lim _{m\to \infty }\nu _m(F_m\triangle gF_m)=0$ , as desired.

(ii) Since $\mu $ is the Haar measure for $\mathcal R$ , it follows that for each subset $A\subset F_n$ . Since $\mu $ is finite, there exists a limit

This fact and condition (i) yield that for each $g\in G$ ,

Hence, $(F_n)_{n=1}^\infty $ is a left Følner sequence in G. Therefore, G is amenable.

(iii) Suppose that H is of infinite index in G. We first prove an auxiliary claim.

Claim A. For each finite subset $S\subset G$ , there exists an element $g\in G$ such that $g\not \in \bigcup _{a,b\in S}aHb^{-1}$ .

By condition (ii), G is amenable. Hence, there exists a left-invariant finitely additive measure $\xi $ on the $\sigma $ -algebra of all subsets of G such that $\xi (G)=1$ . We first observe that since H is of infinite index, $\xi (H)=0$ . Indeed, for each $n>0$ , there are elements $g_1,\ldots ,g_n\in G$ such that the cosets $g_1H,\ldots ,g_nH$ are mutually disjoint. Hence,

$$ \begin{align*} 1\ge \xi\bigg(\bigsqcup_{j=1}^ng_jH\bigg)=\sum_{j=1}^n\xi(g_jH)=\sum_{j=1}^n\xi(H)=n\xi(H). \end{align*} $$

This yields that $\xi (H)=0$ , as desired. As $g^{-1}Hg$ is also a subgroup of infinite index in G, it follows that $\xi (g^{-1}Hg)=0$ for each $g\in G$ . Since $\xi $ is left-invariant, $\xi (kHg)=0$ for all $k,g\in G$ . This implies that $ \xi (\bigcup _{a,b\in S}aHb^{-1})=0. $ Therefore, $G\ne \bigcup _{a,b\in S}aHb^{-1}$ . Thus, Claim A is proved.

Since $\mu (X)<\infty $ , there exists n such that $\nu _n(F_n)>0.5\nu _m(F_m)$ and $C_m\subset H$ for each $m\ge n$ . Hence,

for each $m\ge n$ . By Claim A, there is $g\in G$ such that $gF_nH\cap F_nH=\emptyset $ . Since $\mu $ is the Haar measure, it follows that

This inequality and condition (i) yield that $\nu _m ((gF_nH)\cap F_m)>0.5\nu _m(F_m)$ eventually in m. Therefore, $\nu _m(F_nH\cap gF_nH)>0$ eventually in m, which is a contradiction.

From now on, we consider only the case where $\mu (X\setminus X^G)=0$ . As $X=X^G$ mod 0, we will assume that T is defined on the entire space X. Then for each n and every two elements $g,h\in F_n$ , we have that $T_{hg^{-1}}[g]_n=[h]_n$ and the Radon–Nikodym derivative of the transformation $T_{hg^{-1}}$ is constant on the subset $[g]_n$ . More precisely, this constant equals ${\nu _n(h)}/{\nu _n(g)}$ .

We now prove the main result of this section.

Proof. Let a sequence $(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ satisfy equations (2.1)–(2.3) and Proposition 2.10(ii). We claim that the $(C,F)$ -action $T=(T_g)_{g\in G}$ associated with this sequence is of rank one along $(F_n)_{n=0}^\infty $ . Let X be the space of this action and let $\mu $ be the $(C,F)$ -measure on X determined by $(\kappa _n)_{n=1}^\infty $ and $(\nu _n)_{n=0}^\infty $ . Then, $X=\bigcup _{n\ge 0}X_n$ and $\mathcal R=\bigcup _{n\ge 1}\mathcal S_n$ , where $X_n$ and $\mathcal S_n$ were introduced in §2.2. Of course, for each $n\in \mathbb N$ , the pair $([1_G]_n,F_n)$ is a Rokhlin tower for T. Moreover:

1. (a) $X_{[1_G]_n,F_n}=X_n$ ;

2. (b) $\xi _{[1_G]_n,F_n}$ is the partition of $X_n$ into cylinders $[f]_n$ , $f\in F_n$ ; and

3. (c) if $x=(f_n,c_{n+1},\ldots )\in X_n\cap X^G$ , then $O_{[1_G]_n,F_n}(x)=\{T_gx\mid g\in {F_nf_n^{-1}}\}=\mathcal S_n(x)$ .

We note that items (a) and (b) imply that Definition 2.1(i) holds. It follows from Proposition 2.10 that for a.e. $x\in X$ (or, more precisely, for each $x\in X^G$ ), the T-orbit of x equals $\mathcal R(x)$ . As $\mathcal R(x)=\bigcup _{n=1}^\infty \mathcal S_n(x)$ , it follows that item (c) implies condition (ii) from Definition 2.1. Hence, T is of rank one along $(F_n)_{n=1}^\infty $ .

Conversely, suppose that T is a non-singular G-action of rank one along an increasing sequence $(Q_n)_{n=0}^\infty $ of finite subsets in G with $Q_0=\{1_G\}$ . Let $(B_n,Q_n)_{n=0}^\infty $ be the corresponding generating sequence of Rokhlin towers such that conditions (i) and (ii) of Definition 2.1 hold. We have to define a sequence $(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ , satisfying equations (2.1)–(2.3) and Proposition 2.10(ii) such that the associated $(C,F)$ -action is isomorphic to T. We first set $F_n:=Q_n$ for each $n\ge 0$ . By Definition 2.1(i), for each $n\ge 0$ , there is a subset $R_{n+1}\subset Q_{n+1}$ such that $B_n=\bigsqcup _{f\in R_{n+1}}T_f B_{n+1}$ . Without loss of generality, we may assume that $1_G\in R_{n+1}$ . Indeed, if this is not the case, we replace $(B_{n+1},Q_{n+1})$ with another Rokhlin tower $(T_sB_{n+1}, Q_{n+1}s^{-1})$ for an element ${s\in R_{n+1}}$ . Then, $\xi _{B_{n+1},Q_{n+1}}=\xi _{T_sB_{n+1}, Q_{n+1}s^{-1}}$ and $O_{B_{n+1},Q_{n+1}}(x)= O_{T_sB_{n+1}, Q_{n+1}s^{-1}}(x)$ for each $x\in X$ . Hence, such replacements will not affect conditions (i) and (ii) of Definition 2.1. We now set $C_{n+1}:=R_{n+1}$ . Thus, we defined the entire sequence $(C_n,F_{n-1})_{n\ge 1}$ . It is straightforward to verify that equation (2.1) holds. Let $\nu _0$ be the Dirac measure supported at $1_G$ . Next, for each $n> 0$ and $f\in F_n$ , we let $\nu _n(f):=\mu (T_fB_n)$ . Thus, we obtain a non-degenerated measure $\nu _n$ on $F_n$ . Finally, we define a probability $\kappa _{n+1}$ on $C_{n+1}$ by setting

$$ \begin{align*} \kappa_{n+1}(c):=\frac{\nu_{n+1}(c)}{\nu_n(1_G)}\quad\text{for each } c\in C_{n+1}\text{ and } n\ge 0. \end{align*} $$

Thus, the entire sequence of measures $(\kappa _n,\nu _{n-1})_{n\ge 1}$ is defined. It follows from condition (iii) which is below Definition 2.1 that

$$ \begin{align*} \frac{\nu_{n+1}(fc)}{\nu_n(f)}=\frac{\nu_{n+1}(c)}{\nu_n(1_G)}=\kappa_{n+1}(c)\quad\text{for each } c\in C_{n+1} \text{ and } f\in F_n, \end{align*} $$

that is equation (2.3) holds. We note that for each $n>0$ , the restrictions of $\xi _{B_n,Q_n}$ to the subset $X_{B_0,Q_0}=B_0$ is the finite partition of $B_0$ into subsets , where $(c_1,\ldots ,c_n)$ runs the subset . As $\xi _{B_n,Q_n}\restriction B_0$ converges to the partition into singletons, we obtain that

Since equation (2.2) follows.

Fix $n>0$ , $g\in G$ and $\epsilon>0$ . It follows from Definition 2.1(ii) that there exists $M>n$ such that for each $m>M$ , there is a subset $A\subset X_{B_n,Q_n}$ such that $\mu ( X_{B_n,Q_n}\setminus A)<\epsilon $ and $T_gx\in O_{B_m,Q_m}(x)$ for each $x\in A$ . Hence, there exist $f_1,f_2\in Q_m$ such that ${T_gx=T_{f_1f_2^{-1}}x}$ and $T_{f_2^{-1}}x\in B_m$ . As T is free, $gf_2=f_1\in Q_m=F_m$ . It follows that ${T_{gf_2}B_m=T_{f_1}B_m\subset X_{B_m,Q_m}}$ . Since $T_{f_2}B_m\ni x$ and $x\in X_{B_n,Q_n}$ , we obtain that $T_{f_2}B_m\subset X_{B_n,Q_n}$ because $\xi _{B_m,Q_m}$ is finer than $\xi _{B_n,Q_n}$ . Thus, without loss of generality, we may assume that A is measurable with respect to the partition $\xi _{B_m,Q_m}$ . Since

it follows that $T_{f_2}B_m\subset X_{B_n,Q_n}$ if and only if . Hence,

This implies Proposition 2.10(ii).

Thus, $(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ , satisfies equations (2.1)–(2.3) and Proposition 2.10(ii). Denote by R the non-singular $(C,F)$ -action associated with $(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ . Let $(Y,\nu )$ be the space of this action. The correspondence

$$ \begin{align*} T_fB_n \longleftrightarrow R_f[1_G]_n\quad\text{where } f \text{ runs } F_n \text{ and } n \text{ runs } \mathbb N, \end{align*} $$

gives rise to a Boolean measure preserving isomorphism of the underlying algebras of measurable subsets on X and Y. The Boolean isomorphism is generated by a certain pointwise measure preserving isomorphism $\theta $ of $(X,\mu )$ onto $(Y,\nu )$ . We claim that $\theta $ intertwines T with R. Indeed, take $g\in G$ . As was shown above, for each $n>0$ and $\epsilon>0$ , there exists $M>n$ such that for each $m>M$ , there is a subset $Q'\subset Q_m$ such that $\bigsqcup _{f\in Q'}T_fB_m\subset X_{B_n,Q_n}$ , $\mu (X_{B_n,Q_n}\setminus \bigsqcup _{f\in Q'}T_fB_m)<\epsilon $ and $gQ'\subset Q_m$ . Hence,

$$ \begin{align*} \theta(T_gT_fB_m)=R_g\theta(T_fB_m)\quad\text{for all } f\in Q'. \end{align*} $$

Passing to the limit as $\epsilon \to 0$ and using the fact that $\nu \circ \theta =\mu $ , we obtain that ${\theta (T_gx)=R_g\theta x}$ for a.e. $x\in X_n$ . Since n is arbitrary, $\theta T_g=R_g\theta $ , as claimed.

2.4. Telescopings and reductions

Let a sequence $\mathcal T=(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ satisfy equations (2.1)–(2.3) and Proposition 2.10(ii). Denote by $T=(T_g)_{g\in G}$ the $(C,F)$ -action of G on X associated with $\mathcal T$ . Let $\mu $ stand for the non-singular $(C,F)$ -measure on X determined by $(\kappa _n)_{n=1}^\infty $ and $(\nu _n)_{n=0}^\infty $ .

Given a strictly increasing infinite sequence of integers $\boldsymbol l=(l_n)_{n=0}^\infty $ such that $l_0=0$ , we let

for each $n\ge 0$ .

It is easy to check that $\widetilde {\mathcal T}$ satisfies equations (2.1)–(2.3) and Proposition 2.10(ii). Hence, a non-singular $(C,F)$ -action $\widetilde T=(\widetilde T_g)_{g\in G}$ of G associated with $\widetilde {\mathcal T}$ is well defined. Let $\widetilde X$ denote the space of $\widetilde T$ and let $\widetilde \mu $ denote the non-singular $(C,F)$ -measure on $\widetilde X$ determined by $(\widetilde \kappa _n)_{n=1}^\infty $ and $(\widetilde \nu _n)_{n=0}^\infty $ . There is a canonical measure preserving isomorphism $\iota _{\boldsymbol l}$ of $(X,\mu )$ onto $(\widetilde X,\widetilde \mu )$ that intertwines T with $\widetilde T$ . Indeed, if $x\in X$ , then we select the smallest $n\ge 0$ such that $x=(f_{l_n},c_{l_n+1},c_{l_n+2},\ldots )\in X_{l_n}$ . Let

where . It is a routine to verify that $\iota _{\boldsymbol l}$ is a homeomorphism of X onto $\widetilde X$ such that $\iota _{\boldsymbol l} T_g=\widetilde T_g\iota _{\boldsymbol l}$ for each $g\in G$ , as desired.

Let $\boldsymbol l=(l_n)_{n=0}^\infty $ and $\boldsymbol m=(m_n)_{n=0}^\infty $ be two strictly increasing sequences of integers such that $l_0=m_0=0$ . If $\widetilde {\mathcal T}$ is the $\boldsymbol l$ -telescoping of ${\mathcal T}$ and $\mathcal S$ is the $\boldsymbol m$ -telescoping of $\widetilde {\mathcal T}$ , then $\mathcal S$ is the $\boldsymbol l\circ \boldsymbol m$ -telescoping of $\mathcal T$ , where $\boldsymbol l\circ \boldsymbol m:=(l_{m_n})_{n=1}^\infty $ and $\iota _{\boldsymbol m}\circ \iota _{\boldsymbol l}=\iota _{\boldsymbol l\circ \boldsymbol m}$ .

Given a sequence $\boldsymbol A=(A_n)_{n=1}^\infty $ of subsets $A_n\subset C_{n}$ such that $1_G\in A_n$ for each $n\in \mathbb N$ and $\sum _{n=1}^\infty (1- \kappa _n(A_n))<\infty $ , we let

$$ \begin{align*} \kappa^{\prime}_n(a):=\frac{\kappa_n(a)}{\kappa_n(A_n)},\quad a\in A_n. \end{align*} $$

Then, $ \kappa ^{\prime }_n$ is a non-degenerated probability on $A_n$ for each $n\in \mathbb N$ . We also define a measure $\nu _n'$ on $F_n$ by setting

if $n>0$ and $\nu _0':=\nu _0$ . Let $\mathcal T':=(A_n,F_{n-1},\kappa _n',\nu _{n-1}')_{n=1}^\infty $ .

It is easy to check that $\mathcal T'$ satisfies equations (2.1)–(2.3). We note that

for each $g\in G$ . Passing to the limit and using Proposition 2.10(iii) for $\mathcal T$ , we obtain that as $m\to \infty $ . In other words,

that is, Proposition 2.10(iii) holds for $\mathcal T'$ . Hence, a non-singular $(C,F)$ -action $ {T'=(T_g')_{g\in G}}$ of G associated with $\mathcal T'$ is well defined. Let $X'$ denote the space of $T'$ and let $\mu '$ denote the non-singular $(C,F)$ -measure on $X'$ determined by $(\kappa _n')_{n=1}^\infty $ and $(\nu _n')_{n=0}^\infty $ .

Proof. Indeed, fix $n\in \mathbb N$ . Since , it follows from the Borel–Cantelli lemma that for a.e. $x=(f_n,c_{n+1},c_{n+2},\ldots )\in X_n$ , there is $N=N_x>n$ such that $c_m\in A_m$ for each $m>N$ . We then let

It is routine to verify that $\iota _{\boldsymbol A,n}:X_n\ni x\mapsto \iota _{\boldsymbol A,n}(x)\in X'$ is a well-defined non-singular mapping and

Moreover, $\iota _{\boldsymbol A,n+1}\restriction X_n=\iota _{\boldsymbol A,n}$ for each $n\in \mathbb N$ . Hence, a measurable mapping $\iota _{\boldsymbol A}:X\to X'$ is well defined by the restrictions $\iota _{\boldsymbol A}\restriction X_n=\iota _{\boldsymbol A,n}$ for all $n\in \mathbb N$ . It is straightforward to verify that $\iota _{\boldsymbol A}$ is an isomorphism of $(X,\mu )$ onto $(X',\mu ')$ with and $\iota _{\boldsymbol A}T_g=T_g'\iota _{\boldsymbol A}$ for each $g\in G$ .

2.5. Locally compact models for rank-one non-singular systems

Let Z be a locally compact Polish G-space. We remind that a Borel mapping $\rho :G\times Z\to \mathbb R^*_+$ is called a G-cocycle if

$$ \begin{align*} \rho(g_2,g_1z)\rho(g_1,z)=\rho(g_2g_1,z)\quad\text{for all } g_1,g_2\in G\text{ and } z\in Z. \end{align*} $$

The following definition is a dynamical analogue of Definition 2.6.

Definition 2.18. Fix a G-cocycle $\rho $ . We say that the G-action on Z is Radon $\rho $ -uniquely ergodic if there exists a unique (up to scaling) Radon G-quasi-invariant measure $\gamma $ on Z such that

$$ \begin{align*} \frac{d\gamma\circ g}{d\gamma}(z)=\rho(g,z)\quad\text{for all } g\in G \text{ and } z\in Z. \end{align*} $$

We now show that each rank-one non-singular action has a uniquely ergodic continuous realization on a locally compact Cantor space.

Proof. By Theorem 2.13, there is a sequence $\mathcal T=(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ satisfying equations (2.1)–(2.3) and Proposition 2.10(ii) such that the $(C,F)$ -action T of G associated with $\mathcal T$ is isomorphic to R via a measure preserving isomorphism. Denote by $(X,\mu )$ the space of T. Let $G=\{g_j\mid j\in \mathbb N\}$ . It follows from Proposition 2.10(ii) that there are an increasing sequence $\boldsymbol l=(l_n)_{n=0}^\infty $ of integers and a sequence $(D_n)_{n=1}^\infty $ of subsets in G such that $l_0=0$ , and

for each $n\ge 0$ . Denote by $\widetilde { \mathcal T}=(\widetilde C_n,\widetilde F_{n-1},\widetilde \kappa _n,\widetilde \nu _{n-1})_{n=1}^\infty $ the $\boldsymbol l$ -telescoping of $\mathcal T$ . Let $(\widetilde X,\widetilde \mu , \widetilde T)$ stand for the $(C,F)$ -action of G associated with $\widetilde { \mathcal T}$ . Then, $D_n\subset \widetilde C_n$ and $\widetilde \kappa _n(D_n)>1-n^{-2}$ for each $n>0$ . Denote by $\iota _{\boldsymbol l}$ the canonical measure preserving isomorphism intertwining T with $\widetilde T$ . In general, $1_G\not \in D_n$ . Therefore, we need to modify the $(C,F)$ -parameters $\widetilde {\mathcal T}$ . First, we choose, for each $n>0$ , an element $c_n\in D_n$ . Then, we let $z_0:=1_G$ and for each $n>0$ . Finally, we define a new sequence $\widehat { \mathcal T}=(\widehat C_n,\widehat F_{n-1},\widehat \kappa _n,\widehat \nu _{n-1})_{n=1}^\infty $ by setting

$$ \begin{align*} \widehat C_n:=z_{n-1}\widetilde C_nz_n^{-1}, \quad \widehat F_{n-1}:=\widetilde F_{n-1}z_{n-1}^{-1}, \end{align*} $$

$\widehat \kappa _n$ is the image of $\kappa _n$ under the bijection $\widetilde C_n\ni c\mapsto z_{n-1}cz_n^{-1}\in \widehat C_n $ and $\widehat \nu _{n-1}$ is the image of $\nu _{n-1}$ under the bijection $\widetilde F_{n-1}\ni f\mapsto fz_{n-1}^{-1}\in \widehat F_{n-1}$ . It is straightforward to verify that $\widehat { \mathcal T}$ satisfies equations (2.1)–(2.3) and Proposition 2.10(ii). Denote by $(\widehat X,\widehat \mu ,\widehat T)$ the $(C,F)$ -action of G associated with $\widehat { \mathcal T}$ . Then there is a canonical continuous measure preserving isomorphism $\vartheta : (\widetilde X,\widetilde \mu )\to ( \widehat X,\widehat \mu )$ that intertwines $\widetilde T$ with $\widehat T$ :

$$ \begin{align*} \widetilde X\supset \widetilde X_n\ni(f_n,c_{n+1},\ldots)\mapsto(f_nz_{n}^{-1}, z_{n}c_{n+1}z_{n+1}^{-1}, z_{n+1}c_{n+2}z_{n+2}^{-1},\ldots)\in\widehat X_n\subset\widehat X. \end{align*} $$

Let $\widehat D_n$ be the image of $D_n$ under the bijection $\widetilde C_n\ni c\mapsto z_{n-1}cz_n^{-1}\in \widehat C_n$ . Then, $1\in \widehat D_n$ and $\widehat \kappa _n(\widehat D_n)>1-n^{-2}$ for each $n\in \mathbb N$ . Hence, $\sum _{n=1}^\infty (1-\widehat \kappa _n(\widehat D_n))<\infty $ . We now set $\boldsymbol D:=(D_n)_{n=1}^\infty $ . Denote by $\mathcal T'$ the $\boldsymbol D$ -reduction of $\widehat { \mathcal T}$ . Then, $\mathcal T'$ satisfies not only equations (2.1)–(2.3) and Proposition 2.10(ii), but also equation (2.4). Let $( X',\mu ', T')$ denote the $(C,F)$ -action of G associated with $\mathcal T'$ and let $\iota _{\boldsymbol D}$ stand for the canonical measure scaling isomorphism of $(\widehat X,\widehat \mu )$ onto $( X',\mu ')$ that intertwines $\widehat T$ with $T'$ . Then, $\iota _{\boldsymbol D}\circ \vartheta \circ \iota _{\boldsymbol l}$ is a measure-scaling isomorphism of $(X,\mu , T)$ onto $(X',\mu ',T')$ . Replacing $\mu '$ with for an appropriate $a>0$ , we obtain that $\iota _{\boldsymbol D}\circ \vartheta \circ \iota _{\boldsymbol l}$ is measure preserving. It follows from Proposition 2.9 that $T'$ is a Radon uniquely ergodic minimal continuous action of G on the locally compact Polish space $X'$ . Thus, we proved conditions (i), (iii), (iv) and (v). The condition (ii) follows easily from equation (2.4) and the definition of $\mu '$ .

The final claim of the theorem follows from condition (iv) and Proposition 2.7.

2.6. Non-singular ${\mathbb Z}$ -actions of rank one along intervals and $(C,F)$ -construction

Let $G={\mathbb Z}$ . Suppose that a sequence $\mathcal T=(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ satisfies equations (2.1)–(2.3) and Proposition 2.10(ii), and there is a sequence $(h_n)_{n=0}^\infty $ of positive integers such that $F_n=\{0,1,\ldots ,h_n-1\}$ . Denote by $(X,\mu ,T)$ the $(C,F)$ -dynamical system associated with $\mathcal T$ .

We now show how to obtain $(X,\mu ,T)$ via the classical inductive geometric cutting-and-stacking in the case of ${\mathbb Z}$ -actions. On the initial step of the construction, we define a column $Y_0$ consisting of a single interval $[0,1)$ equipped with Lebesgue measure. Assume that at the nth step, we have a column

$$ \begin{align*} Y_n = \{I(i,n) \mid i=0,\ldots,h_n-1 \} \end{align*} $$

consisting of disjoint intervals $I(i,n) \subset \mathbb R$ such that $\bigsqcup _{i=0}^{h_n-1} I(i,n) = [0,\nu _n(F_n))$ . Then we define a continuous nth column mapping

$$ \begin{align*} T^{(n)}: [0,\nu_n(F_n))\setminus I(h_n-1,n)\to [0,\nu_n(F_n))\setminus I(0,n) \end{align*} $$

such that $T^{(n)}\restriction I(i,n)$ is the orientation preserving affine mapping of $I(i,n)$ onto $I(i+1,n)$ for $i=0,\ldots ,h_n-2$ . It is convenient to think of $I(i,n)$ as a level of $Y_n$ . The levels may be of different length, but they are parallel to each other and the ith level is above the jth level if $i>j$ . The nth column mapping moves every level, except the highest one, one level up. By the move here, we mean the orientation preserving affine mapping. On the highest level, $T^{(n)}$ is not defined.

On the $(n+1)$ th step of the construction, we first cut each $I(i, n)$ into subintervals $I(i+c, n+1)$ , $c\in C_{n+1}$ , such that $I(i+c, n+1)$ is from the left of $I(i+c', n+1)$ whenever $c<c'$ and

$$ \begin{align*} \frac{\text{the length of}\ I(i+c,n+1)}{\text{the length of}\ I(i,n)} =\kappa_{n+1}(c)\quad\text{for each } c\in C_{n+1}. \end{align*} $$

Hence, we obtain that $\bigsqcup _{j\in F_n+C_{n+1}}I(j,n+1)=[0,\nu _{n}(F_n))$ . Next, we cut the interval $[\nu _{n}(F_n),\nu _{n+1}(F_{n+1}))$ into subintervals $I(j, n+1)$ , $j\in F_{n+1}\setminus (F_n+C_{n+1})$ , such that the length of $I(j, n+1)$ is $\nu _{n+1}(j)$ for each j. These new levels are called spacers. Thus, we obtain a new column $Y_{n+1}=\{I(j,n+1) \mid j \in F_{n+1}\}$ with $\bigsqcup _{i\in F_{n+1}} I(i,n+1) = [0,\nu _{n+1}(F_{n+1}))$ . If an element $c\in C_{n+1}$ is not maximal in $C_{n+1}$ , then we denote by $c^+$ the least element of $C_{n+1}$ that is greater than c. We define a spacer mapping ${s_{n+1}:C_{n+1}\to {\mathbb Z}_+}$ by setting

$$ \begin{align*} s_{n+1}(c):= \begin{cases} c^+-c-h_n &\text{if } c\ne \max C_{n+1},\\ h_{n+1}-c-h_n&\text{if } c= \max C_{n+1}. \end{cases} \end{align*} $$

The subcolumn $Y_{n,c}:=\{I(i+c,n+1)\mid i\in F_n\}\subset Y_{n+1}$ is called the c-copy of $Y_n$ , ${c\in C_{n+1}}$ . Thus, $Y_{n+1}$ consists of $\# C_{n+1}$ copies of $Y_{n}$ , and spacers between them and above the highest copy of $Y_n$ . More precisely, there are exactly $s_{n+1}(c)$ spacers above the c-copy of $Y_n$ in $Y_{n+1}$ . The $(n+1)$ th column mapping

$$ \begin{align*} T^{(n+1)}: [0,\nu_{n+1}(F_{n+1}))\setminus I(h_{n+1}-1,n+1)\to [0,\nu_{n+1}(F_{n+1}))\setminus I(0,n+1) \end{align*} $$

is defined in a similar way as $T^{(n)}$ . Of course,

$$ \begin{align*} T^{(n+1)}\restriction ( [0,\nu_n(F_n))\setminus I(h_n-1,n))=T^{(n)}. \end{align*} $$

Passing to the limit as $n\to \infty $ , we obtain a well-defined non-singular (piecewise affine) transformation Q of the interval $[0,\lim _{n\to \infty }\nu _n(F_n))\subset \mathbb R$ equipped with Lebesgue measure such that

$$ \begin{align*} Q\restriction ( [0,\nu_n(F_n))\setminus I(h_n-1,n))=T^{(n)}\quad\text{for each } n\in\mathbb N. \end{align*} $$

It is possible that $\lim _{n\to \infty }\nu _n(F_n)=\infty $ and then T is defined on $[0,+\infty )$ . Of course, this transformation (or, more precisely, the ${\mathbb Z}$ -action generated by Q) is isomorphic to $(X,\mu , T)$ . The according non-singular isomorphism is generated by the following correspondence:

$$ \begin{align*} X\supset [i]_n \longleftrightarrow I(i,n)\subset [0,\lim_{n\to\infty}\nu_n(F_n)),\quad i\in F_n,n\in\mathbb N. \end{align*} $$

Without loss of generality, we may assume $s_n(\max C_n)=0$ for each $n>0$ (see, for instance, [Reference DanilenkoDa4]). This means that there are no spacers over the highest copy of $Y_n$ in $Y_{n+1}$ .

2.7. Non-singular rank-one ${\mathbb Z}$ -actions as transformations built under function over non-singular odometer base

Let $(X,\mu , T)$ be as in the previous subsection. We will assume that $\max F_{n+1}=\max F_n+\max C_{n+1}$ for each $n\ge 0$ . In terms of the geometrical cutting-and-stacking (see §2.6), this means exactly that there are no spacers on the top of the $(n\kern1.5pt{+}\kern1.5pt1)$ th column. We remind that ,  and

$$ \begin{align*} c^+:=\min\{d\in C_n\mid d>c\} \end{align*} $$

for each $n>0$ and $c\in C_n$ such that $c\ne \max C_n$ . Denote by R the transformation induced by $T_1$ on $(X_0,\mu \restriction X_0)$ . Since $T_1$ is conservative, R is a well-defined non-singular transformation of $X_0$ . Take $x=(c_1,c_2,\ldots )\in X_0$ . Choose $n\ge 0$ such that $c_i=\max C_i$ for each $i=1,\ldots ,n$ and $c_{n+1}\ne \max C_{n+1}$ . It is straightforward to verify that

$$ \begin{align*} Rx:= (\underbrace{0,\ldots,0}_{n\text{ times}}, c_{n+1}^+, c_{n+2}, c_{n+3}\ldots) \end{align*} $$

Thus, R is a classical non-singular odometer of product type (see [Reference AaronsonAa, Reference Danilenko, Silva and MeyersDaSi] and references therein). Let $x^{\max }:=(\max C_1,\max C_2,\ldots )\in X_0$ . We now define a function $\vartheta :X_0\setminus \{x^{\max }\}\to {\mathbb Z}_+$ by setting

$$ \begin{align*} \vartheta(x):=s_n(c_n) \end{align*} $$

if $x=(c_1,c_2,\ldots )$ , $c_i=\max C_i$ for $i=1,\ldots ,n-1$ and $c_n\ne \max C_n$ , where $s_n$ is the spacer mapping (see §2.6). Of course, $\vartheta $ is continuous. Then $(X,\mu ,T_1)$ is isomorphic to the transformation $R^\theta $ built under $\vartheta $ over the base R. We do not provide a proof of this fact which is essentially folklore.

4.1. Non-singular odometers

From now on, G is residually finite. If $\Gamma $ is a cofinite subgroup in G, then the largest subgroup $\widetilde \Gamma $ of $\Gamma $ which is normal in G is also cofinite in G. Of course, $\widetilde \Gamma =\bigcap _{g\in G}g\Gamma g^{-1}$ . If $\Xi $ is cofinite subgroup in $\Gamma $ , then $\widetilde \Xi \subset \widetilde \Gamma $ . We now fix a decreasing sequence of cofinite subgroups $\Gamma _n$ in G such that

(4.1) $$ \begin{align} \bigcap_{n=1}^\infty\bigcap_{g\in G}g\Gamma_ng^{-1}=\{1_G\}. \end{align} $$

It exists because G is residually finite. We note that equation (4.1) means that the intersection of the maximal normal (in G) subgroups of $\Gamma _n$ , $n\in \mathbb N$ , is trivial. At the same time, the intersection of all $\Gamma _n$ can be non-trivial. Consider the natural inverse sequence of homogeneous G-spaces and G-equivariant mappings intertwining them:

(4.2)

Denote by Y the projective limit of this sequence. A point of Y is a sequence $(g_n\Gamma _n)_{n=1}^\infty $ such that $g_n\Gamma _n=g_{n+1}\Gamma _n$ , that is, $g_n^{-1}g_{n+1}\in \Gamma _n$ for each $n>0$ . Endow Y with the topology of projective limit. Then Y is a compact Cantor G-space. Of course, the G-action on Y is minimal and uniquely ergodic. Denote this action by $O=(O_g)_{g\in G}$ . It follows from equation (4.1) that O is faithful, that is, $O_g\ne I$ if $g\ne 1_G$ . We note that a faithful action is not necessarily free.

In the finite measure preserving case, one can find the above definition in [Reference Danilenko and LemańczykDaLe] (see also [Reference Lightwood, Şahin and UgarcoviciLiSaUg], where odometers are called ‘subodometers’.)

We note that equation (4.1) is in no way restrictive. Indeed, let $\bigcap _{n=1}^\infty \bigcap _{g\in G}g\Gamma _ng^{-1}=N\ne \{1_G\}$ . Define $(Y,O)$ as above. Then, N is a proper normal subgroup of G and $N=\{g\in G\mid O_g=I\}$ . We now let $\widetilde G:=G/N$ and $\widetilde \Gamma _n:=\Gamma _n/N$ . Then, $\widetilde \Gamma _n$ is a cofinite subgroup in $\widetilde G$ for each $n\in \mathbb N$ , and $\bigcap _{n=1}^\infty \bigcap _{g\in \widetilde G}g\widetilde \Gamma _ng^{-1}=\{1_{\widetilde G}\}$ . Let $(\widetilde Y,\widetilde O)$ denote the topological $\widetilde G$ -odometer associated with the sequence $(\widetilde \Gamma _n)_{n=1}^\infty $ . Then, of course, $Y=\widetilde Y$ and $O_g=\widetilde O_{gN}$ for each $g\in G$ .

We now isolate a class of non-singular odometers of rank one. For each $n>0$ , we choose a finite subset $D_n\subset \Gamma _{n-1}$ such that $1_G\in D_n$ and each $\Gamma _{n}$ -coset in $\Gamma _{n-1}$ intersects $D_n$ exactly once. (For consistency of the notation, we let $\Gamma _0:=G$ .) We then call $D_n$ a $\Gamma _n$ -cross-section in $\Gamma _{n-1}$ . Then, the product is a $\Gamma _n$ -cross-section in G. Hence, there is a unique bijection

such that $\omega _n(g\Gamma _n)\Gamma _n=g\Gamma _n$ for each $g\in G$ and $\omega _n(\Gamma _n)=1_G$ . It follows, in particular, that

(4.3) $$ \begin{align} \text{if}\ \omega_n(g\Gamma_n)=h\omega_n(g'\Gamma_n)\ \text{for some } g,g',h\in G,\ \text{then}\ g\Gamma_n=hg'\Gamma_n. \end{align} $$

It is straightforward to verify that the diagram

(4.4)

commutes. The horizontal arrows in the upper line denote the natural projections. The other mappings in the diagram are defined as follows:

for each and $n\ge 0$ . It follows from equation (4.3) that there exists a natural homeomorphism of Y onto the infinite product space . (The homeomorphism pushes down to a bijection between $G/\Gamma _n$ and for each n.)

Proof. We set $F_0:=\{1_G\}$ , , $C_n:=D_n$ and for each $n\in \mathbb N$ . Then equations (2.1)–(2.3) hold for the sequence $(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ . Moreover, Proposition 2.10(iii) is exactly property (iii) in the case under consideration. Hence, the $(C,F)$ -action T of G associated with $(C_n,F_{n-1},\kappa _n,\nu _{n-1})_{n=1}^\infty $ is well defined. Let $(X,\mu )$ stand for the space of this action. It follows from the $(C,F)$ -construction that

$\mu $ is non-atomic and T is free (mod $\mu $ ). In view of equation (4.3), we can identify X with Y. Hence, we consider $\mu $ as a probability on Y. Moreover, equation (4.2) yields that T is conjugate to O. Thus, $(Y,\mu ,O)$ is a non-singular odometer. It remains to apply Theorem 2.13.

We now show that the classical non-singular ${\mathbb Z}$ -odometers of product type are covered by Definition 4.1.

Example 4.3. Let $G={\mathbb Z}$ and let $(a_n)_{n=1}^\infty $ be a sequence of integers such that $a_n>1$ for each $n\in \mathbb N$ . We set . Then, and $\bigcap _{n=1}^\infty \Gamma _n=\{0\}$ . The set is a $\Gamma _n$ -cross-section in $\Gamma _{n-1}$ . Hence, in view of equation (4.4), the space Y of the ${\mathbb Z}$ -odometer $O=(O_n)_{n\in {\mathbb Z}}$ associated with $(\Gamma _n)_{n=1}^\infty $ is homeomorphic to the infinite product . We identify $D_n$ naturally with the set $\{0,1,\ldots ,a_n-1\}$ . Then,

To define O explicitly on this space, we take $y=(y_n)_{n=1}^\infty \in D$ . It is a routine to check that if there is $k> 0$ such that $y_j=a_{j}-1$ for each $j<k$ and $y_k\ne a_k-1$ , then

$$ \begin{align*} O_1y=(0,\ldots,0,y_k+1,y_{k+1}, y_{k+2},\ldots). \end{align*} $$

If such a k does not exist, that is, $y_j=a_j-1$ for each $j>0$ , then $O_1y=(0,0,\ldots )$ . Let $\kappa _n$ be a non-degenerated probability measure on $\{0,1,\ldots ,a_n-1\}$ and let $\prod _{n>0}\max _{0\le d<a_n}\kappa _n(d)=0$ . This means that properties (i) and (ii) of Proposition 4.2 hold. Of course, Proposition 4.2(iii) holds also. Hence, by Proposition 4.2, the non-singular odometer

$$ \begin{align*} \bigg(\bigotimes_{n=1}^\infty \{0,\ldots,a_n-1\}, \bigotimes_{n=1}^\infty\kappa_n, O\bigg) \end{align*} $$

is of rank one. Thus, in this case, our definition of non-singular odometer coincides with the classical definition of non-singular ${\mathbb Z}$ -odometers of product type (see [Reference AaronsonAa, Reference Danilenko, Silva and MeyersDaSi]). Moreover, the ${\mathbb Z}$ -odometers of product type are of rank one.

It is routine to verify that if $G={\mathbb Z}^d$ with $d\in \mathbb N$ , then each probability preserving G-odometer is of rank one. This follows from Proposition 4.2 if one chooses the $\Gamma _n$ -cross-sections $D_n$ in $\Gamma _{n-1}$ in such a way that the sum is a parallelepiped for some $a_{1,n},\ldots ,a_{d,n}\in \mathbb N$ with $\lim _{n\to \infty }a_{j,n}=\infty $ for each j. We leave details to the reader (see also [Reference Johnson and McClendonJoMc, Theorem 2.11]).

Note, however, that there exist probability preserving free G-odometers which are not of rank one.

We also provide two examples of non-rank-one odometer actions for amenable groups G. In the first example, G is locally finite, and in the second one, G is non-locally finite periodic.

Example 4.5. Let $Z=\{0,1\}^{\mathbb {N}}$ . Endow Z with the infinite product $\eta $ of the equidistributions on $\{0,1\}$ . Fix a sequence $\boldsymbol {s}=(s_n)_{n=0}^\infty $ of mappings

$$ \begin{align*} s_n:\{0,1\}^n\to \text{Homeo}(\{0,1\}),\quad n\ge 0. \end{align*} $$

Consider the following transformation $T_{\boldsymbol {s}}$ of Z:

$$ \begin{align*} T_{\boldsymbol{s}}(z_1,z_2,\ldots):=(s_0z_1, s_1(z_1)z_2,s_2(z_1,z_2)z_3,s_3(z_1,z_2,z_3)z_4,\ldots). \end{align*} $$

Of course, $T_{\boldsymbol {s}}$ preserves $\eta $ . Let

$$ \begin{align*} G:=\{T_{\boldsymbol{s}}\mid \boldsymbol{s}=(s_n)_{n=0}^\infty \text{ with}\ s_n\equiv I\ \text{eventually}\}. \end{align*} $$

Then, G is a locally finite (and hence amenable) countable group. We claim that the dynamical system $(Z,\eta , G)$ is a G-odometer. Indeed, for each $n\in \mathbb N$ , we denote by $\pi _n:Z\to \{0,1\}^{n}$ the projection to the first n coordinates. Of course, there is a natural transitive action of G on $\{0,1\}^{n}$ :

$$ \begin{align*} T_{\boldsymbol{s}}*(z_1,\ldots,z_n):=(s_0z_1, s_1(z_1)z_2,\ldots, s_{n-1}(z_1,\ldots, z_{n-1})z_n). \end{align*} $$

Then, $\pi _n$ is a G-equivariant mapping. Thus, $\{0,1\}^n$ is a finite factor of $(Z,\eta , G)$ . Let

$$ \begin{align*} \Gamma_n:=\{g\in G\mid g*(0,\ldots,0)=(0,\ldots,0)\}. \end{align*} $$

In other words, $\Gamma _n$ is the stabilizer of a point $(0,\ldots ,0)\in \{0,1\}^n$ . It is straightforward to verify that

Hence, $\Gamma _n$ is a cofinite subgroup in G for each n. Moreover, . Thus, we obtain, for each n, a G-equivariant bijection $\phi _n:\{0,1\}^n\to G/\Gamma _n$ such that the following diagram commutes:

(4.5)

where $\tau _n(z_1,\ldots ,z_{n+1}):=(z_1,\ldots ,z_{n})$ for each $(z_1,\ldots ,z_{n+1})\in \{0,1\}^{n+1}$ and all ${n\in \mathbb N}$ . It is routine to check that

$$ \begin{align*} \widetilde\Gamma_n:=\bigcap_{g\in G}g\Gamma_n g^{-1}= \{T_{\boldsymbol{s}}\mid \boldsymbol{s}=(s_k)_{k=0}^\infty\text{ with } s_0=I, s_1\equiv I,s_2\equiv I,\ldots, s_{n-1}\equiv I\} \end{align*} $$

and $\bigcap _{n=1}^\infty \widetilde \Gamma _n=\{I\}$ . Denote by $(Y,O)$ the topological G-odometer associated with the sequence $(\Gamma _n)_{n=1}^\infty $ . Furnish it with the Haar measure $\nu $ . Then, equation (4.5) yields a G-equivariant isomorphism $\phi :(Z,\eta )\to (Y,\nu )$ , as desired.

We now show that $(Z,\eta ,G)$ is not free. Take a point $z=(z_n)_{n=1}^\infty \in Z$ . Then, the G-stabilizer $G_z$ of z is the group

Let $r_1:\{0,1\}\to \text {Homeo}(\{0,1\})$ be the only mapping such that $r_1(z_1)=I$ but $r_1\not \equiv I$ . We define a transformation R of $(Z,\eta )$ by setting

$$ \begin{align*} R(z_1,z_2,z_3,\ldots):=(z_1,r_1(z_1)z_2, z_3,z_4,\ldots). \end{align*} $$

Then, $G_z\ni R\ne I$ . Hence, O is not free. Therefore, O is not of rank one.