Proof. We compute that

$$ \begin{align*} \int_{X}\bigg|\frac{d\mu\circ T_{g}^{-1}}{d\mu}-1\bigg|\,d\mu&= \sum_{n\in\Bbb N}\int_{Y}|f_{n}\circ S_{g}^{-1}- f_{n}|\frac {a_{n}}{\kappa(F_{n})}\,d\kappa\\ &= \sum_{n\in\Bbb N}\int_{Y}|\unicode{x3bb}_{n}-1|\cdot |1_{F_{n}}\circ S_{g}^{-1}-1_{F_{n}}|\frac{a_{n}}{\kappa(F_{n})}\,d\kappa\\ &= \sum_{n\in\Bbb N}a_{n}|\unicode{x3bb}_{n}-1|\frac{\kappa (S_{g}F_{n}\triangle F_{n})}{\kappa(F_{n})}. \end{align*} $$

Hence, (3.1) follows. In a similar way, we obtain that

$$ \begin{align*} \chi(T_{g})=\int_{X}\bigg(\frac{d\mu\circ T_{g}^{-1}}{d\mu}-1\bigg)\,d\mu= \sum_{n\in\Bbb N}\frac{a_{n}}{\nu(F_{n})}\int_{Y}(f_{n}\circ S_{g}^{-1}- f_{n})\,d\nu=0.\\[-3.8pc] \end{align*} $$

Proof. We observe that

$$ \begin{align*} \int_{X}\bigg|\bigg(\frac{d\mu}{d\mu\circ T_{g}^{-1}}\bigg)^{2}-1\bigg|\,d\mu &=\sum_{n=1}^{\infty} \frac{a_{n}}{\kappa(F_{n})}\int_{Y}\bigg|\frac{f^{3}}{(f\circ T_{g}^{-1})^{2}}-f\bigg|\,d\kappa\\ &=\sum_{n=1}^{\infty} a_{n} \frac{|1-\unicode{x3bb}_{n}^{2}|(1+\unicode{x3bb}_{n}^{3})}{\unicode{x3bb}_{n}^{2}}\frac{\kappa (S_{g}F_{n}\triangle F_{n}) }{\kappa(F_{n})} \end{align*} $$

and the first claim of Lemma 3.2 follows. In a similar way,

$$ \begin{align*} \int_{X}\bigg(\bigg(\frac {d\mu}{d\mu\circ T_{g}^{-1}}\bigg)^{2}-1\bigg)\,d\mu &=\sum_{n=1}^{\infty} \frac{a_{n}}{\kappa(F_{n})}( (\unicode{x3bb}_{n}^{3}-\unicode{x3bb}_{n})\kappa(F_{n}\setminus S_{g}F_{n})\\ &\quad+(\unicode{x3bb}_{n}^{-2}-1)\kappa(S_{g} F_{n}\setminus F_{n}))\\ &=\sum_{k=1}^{\infty} a_{n}c(\unicode{x3bb}_{n})\frac{\kappa (S_{g}F_{n}\triangle F_{n}) }{2\kappa(F_{n})}, \end{align*} $$

as desired.

Proof. Since $(X^{*},\mu ^{*},T^{*})$ is isomorphic to $\bigotimes _{n\in \Bbb N}(Y^{*},\mu _{n}^{*},S^{*})$ and $\mu _{n}^{*}\sim \nu _{n}^{*}$ for each $n\in \Bbb N$ , it follows that $(Y^{*},\nu _{1}^{*},S^{*})$ is a factor of $(X^{*},\mu ^{*},T^{*})$ . Hence, it suffices to show that $(Y^{*},\nu _{1}^{*},S^{*})$ is not strongly ergodic. We note that $\nu _{1}^{*}([B_{n}]_{0})=e^{-\nu _{1}(B_{n})}$ and

$$ \begin{align*} \nu_{1}^{*}([B_{n}]_{0}\cap S_{g}^{*}[B_{n}]_{0})=\nu_{1}^{*}([B_{n}\cup S_{g}B_{n}]_{0})=e^{-\nu_{1}(B_{n}\cup S_{g}B_{n})} \end{align*} $$

for each $n\in \Bbb N$ . Since $(B_{n})_{n=1}^{\infty }$ is S-Følner, $\sup _{g\in K}|{\nu _{1}(B_{n}\cup S_{g}B_{n})}/{\nu _{1}(B_{n})}-1|\to 0$ as $n\to \infty $ for each compact subset $K\subset G$ . Hence, $\sup _{g\in K}|\nu _{1}(B_{n}\cup S_{g}B_{n})-\nu _{1}(B_{n})|\to 0$ as $n\to \infty $ . It follows that $([B_{n}]_{0})_{n=1}^{\infty }$ is an asymptotically invariant sequence in $(Y^{*},\mathfrak {Y}^{*},\nu _{1}^{*})$ . Since $\nu _{1}(B_{n})={a_{1}c}/{\kappa (F_{1})}$ , the sequence $([B_{n}]_{0})_{n=1}^{\infty }$ is non-trivial. Hence, $T^{*}$ is not strongly ergodic.

Proof. Since $(X^{*},\mu ^{*},T^{*})=\bigotimes _{n=1}^{\infty } (Y^{*},\mu _{n}^{*},S^{*})$ and $\mu _{1}^{*}\sim \nu _{1}^{*}$ , by Fact 2.18(ii), it suffices to show that the action $S^{*}$ on $(Y^{*},\nu _{1}^{*})$ is free. Since Q preserves $\nu _{1}$ , it follows that the Poisson suspension $Q^{*}$ of Q preserves $\nu _{1}^{*}$ . We deduce from (ii) that $Q^{*}$ is ergodic. Moreover, (i) yields that $Q^{*}$ commutes with $S_{g}^{*}$ for each $g\in G$ .

Denote by $\mathcal {G}$ the space of all closed subgroups of G and endow $\mathcal {G}$ with the Fell topology [Reference FellFe]. Then, $\mathcal {G}$ is a compact metric space [Reference FellFe]. Given $\omega \in Y^{*}$ , let $G_{\omega }:=\{g\in G\mid S^{*}_{g}\omega =\omega \}$ stand for the stability group of $S^{*}$ at $\omega $ . Then, $G_{\omega }\in \mathcal {G}$ at each $\omega $ [Reference Auslander and MooreAuMo, I, Proposition 3.7]. The mapping $ \eta :Y^{*}\ni \omega \mapsto G_{\omega }\in \mathcal {G} $ is Borel [Reference Auslander and MooreAuMo, II, Proposition 2.3]. Since $Q^{*}$ commutes with $S^{*}$ , it is straightforward to verify that $\eta $ is invariant under $Q^{*}$ . As $Q^{*}$ is ergodic, we obtain that $\eta $ is constant. Thus, there is a subgroup $H\in \mathcal {G}$ such that $G_{\omega }=H$ at a.e. $\omega \in Y^{*}$ . Therefore, $S^{*}_{g}=\text {Id}$ for each $g\in H$ . This implies that $S_{g}=\text {Id}$ for each $g\in H$ . Since S is effective, H is trivial. Thus, we obtain that $S^{*}$ (and hence $T^{*}$ ) is free, as desired.

Proof of Theorem 4.2

As we noted in §3, the dynamical system $(X^{*},\mathfrak {B}^{*},\mu ^{*},T^{*})$ is canonically isomorphic to the infinite direct product $\bigotimes _{n\in \Bbb N}(Y^{*},\mathfrak {Y}^{*},\mu _{n}^{*},S^{*})$ . Let $B_{n}:=[F_{n}]_{0}$ and $\boldsymbol {B}:=( B_{n})_{n=1}^{\infty }$ .

We first recall that $\nu _{n}:=({a_{n}}/{\kappa (F_{n})})\,\kappa =\kappa $ in view of $(\alpha _{1})$ and $(\alpha _{4})$ . Therefore, $\nu _{n}(F_{n})=1$ and $\nu _{n}^{*}(B_{n})=e^{-\nu _{n}(F_{n})}=e^{-1}$ . It follows now from Fact 2.18(iii) that if $\omega \in B_{n}$ then

$$ \begin{align*} \frac{d\mu_{n}^{*}}{d\nu_{n}^{*}}(\omega)=e^{-(\unicode{x3bb}_{n}-1)\nu_{n}(F_{n})}\sum_{k=0}^{\infty}\unicode{x3bb}_{n}^{k}1_{[F_{n}]_{k}}(\omega)=e^{1-\unicode{x3bb}_{n}}. \end{align*} $$

Hence,

$$ \begin{align*} H^{2}\bigg(\frac1{\nu_{n}^{*}(B_{n})}\nu_{n}^{*}\restriction B_{n},\mu_{n}^{*}\bigg)& =1-\int_{B_{n}} \frac{1}{\sqrt{\nu_{n}^{*}(B_{n})}}\sqrt{\frac{d\mu^{*}_{n}}{d\nu_{n}^{*}}} \,d\nu_{n}^{*} \\ &=1- e^{-1/2}e^{({1-\unicode{x3bb}_{n}})/2} \\ &=1-e^{-{\unicode{x3bb}_{n}}/{2}}. \end{align*} $$

Since $\sum _{n=1}^{\infty }\unicode{x3bb} _{n}<\infty $ by $(\alpha _{4})$ , Claim A follows.

Fact 2.14(i) and Claim A yield that $\mu ^{*}$ is equivalent to the restricted infinite product of $(\nu _{n}^{*})_{n=1}^{\infty }$ with respect to $\boldsymbol {B}$ . We will denote this restricted product by $(\nu ^{*})^{\boldsymbol {B}}$ . Since $\prod _{n=1}^{\infty }\nu _{n}^{*}(B_{n})=0$ , it follows that $(\nu ^{*})^{\boldsymbol {B}}(X^{*})=\infty $ . Since

$$ \begin{align*} \sum_{n=1}^{\infty}\frac{\nu_{n}^{*}(B_{n}\triangle S_{g}^{*}B_{n})}{\nu_{n}^{*}(B_{n})}= 2e\sum_{n=1}^{\infty}(\nu_{n}^{*}(B_{n})-\nu_{n}^{*}(B_{n}\cap S_{g}^{*}B_{n}))= 2\sum_{n=1}^{\infty}(1-e^{1-\nu_{n}(F_{n}\cap S_{g}F_{n})}) \end{align*} $$

and

$$ \begin{align*} \sum_{n=1}^{\infty} (1-\nu_{n}(F_{n}\cap S_{g}F_{n}))=\frac12\sum_{n=1}^{\infty} \frac{\kappa(F_{n}\triangle S_{g}F_{n})}{\kappa(F_{n})}, \end{align*} $$

it follows from this and $(\alpha _{3})$ that

(4.1) $$ \begin{align} \sup_{g\in K}\sum_{n=1}^{\infty}\frac{\nu_{n}^{*}(B_{n}\triangle S_{g}^{*}B_{n})}{\nu_{n}^{*}(B_{n})}<\infty\quad\text{ for each compact }K\subset G\text{.} \end{align} $$

Therefore, by Fact 2.11, $T^{*}$ preserves $(\nu ^{*})^{\boldsymbol {B}}$ .

Thus, $(X^{*},\mathfrak {B}^{*},\mu ^{*},T^{*})$ admits an equivalent invariant infinite $\sigma $ -finite measure $(\nu ^{*})^{\boldsymbol {B}}$ . Since S is free, S is effective. Therefore, Proposition 3.4 and $(\alpha _{5})$ yield that $(X^{*},\mathfrak {B}^{*},\mu ^{*},T^{*})$ is free.

Unfortunately, we do not know whether $(X^{*},\mathfrak {B}^{*},\mu ^{*},T^{*})$ is ergodic or not. However, we observe that, for each increasing sequence of positive integers $(k_{n})_{n=1}^{\infty }$ , the conditions $(\alpha _{1})$ – $(\alpha _{5})$ hold if we replace $\boldsymbol {F}$ , $(a_{n})_{n=1}^{\infty }$ and $(\unicode{x3bb} _{n})_{n=1}^{\infty }$ with the subsequences $\boldsymbol {F}^{\prime }:=(F_{k_{n}})_{n=1}^{\infty }$ , $(a_{k_{n}})_{n=1}^{\infty }$ and $(\unicode{x3bb} _{k_{n}})_{n=1}^{\infty }$ , respectively. Hence, given $(Y,\mathfrak {Y},\kappa , S)$ and these three subsequences, the general construction from §3 yields a new dynamical system which we denote by $(\widetilde X,\widetilde {\mathfrak {B}},\widetilde \mu ,\widetilde T)$ . As we have shown above, $(\alpha _{1})$ – $(\alpha _{5})$ imply that the non-singular Poisson suspension $(\widetilde X^{*},{\widetilde {\mathfrak {B}}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ of $(\widetilde X,\widetilde {\mathfrak {B}},\widetilde \mu ,\widetilde T)$ is well defined. Moreover, the dynamical system $(\widetilde X^{*},{\widetilde {\mathfrak {B}}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ is free and IDPFT. Of course, $(\widetilde X^{*},{\widetilde {\mathfrak {B}}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ is canonically isomorphic to $\bigotimes _{n\in \Bbb N}(Y^{*},\mathfrak {Y}^{*},\mu _{k_{n}}^{*},S^{*})$ . If we let $\boldsymbol B^{\prime }:=(B_{k_{n}})_{n=1}^{\infty }$ , then the measure $(\nu ^{*})^{\boldsymbol B^{\prime }}$ is infinite, $\sigma $ -finite, $\widetilde \mu ^{*}$ -equivalent and $\widetilde T^{*}$ -invariant.

To simplify the argument, we only show how to choose $(k_{n})_{n=1}^{\infty }$ so that the Cartesian square of $\bigotimes _{n\in \Bbb N}(Y^{*},\mathfrak {Y}^{*},\mu _{k_{n}}^{*},S^{*})$ is ergodic. The sequence $(k_{n})_{n=1}^{\infty }$ will be defined inductively. First, we set $k_{1}:=1$ . Suppose that we have already specified $(k_{j})_{j=1}^{n}$ . Our purpose is to define $k_{n+1}$ . For an arbitrary $j\in \Bbb N$ , let $U_{j,S^{*}}$ denote the Koopman unitary representation of G associated with the probability preserving system $(Y^{*},\nu _{j}^{*}, S^{*})$ . We deduce from $(\alpha _{2})$ and Facts 2.2 and 2.21 that the unitary representation $U_{j,S^{*}}\restriction L^{2}_{0}(Y^{*},\nu _{j}^{*})$ of G is weakly mixing. Hence, the dynamical system $(Y^{*},\nu _{j}^{*}, S^{*})$ is weakly mixing by Fact 2.4(ii). It follows that

1. (⊳) the dynamical system $\bigotimes _{l=1}^{n}(Y^{*}\times Y^{*},\nu _{k_{l}}^{*}\otimes \nu _{k_{l}}^{*}, (S^{*}_{g}\times S^{*}_{g})_{g\in G})$ is ergodic.

Let $(\mathcal {P}_{j,l})_{l=1}^{\infty }$ be a sequence of finite partitions of $Y^{*}$ into Borel subsets of equal measure $\nu _{j}^{*}$ such that

• (∘) $(\mathcal {P}_{j,l})_{l=1}^{\infty }$ approximates $(\mathfrak {Y}^{*},\nu _{j}^{*})$ .

It follows from $(\triangleright )$ that there is a finite subset $H_{n}\subset G$ such that, for every two atoms $P,Q$ of the partition $\bigotimes _{j=1}^{n}(\mathcal {P}_{k_{j},n})^{\otimes 2}$ of $(Y^{*}\times Y^{*})^{n}$ , there is a family of measured subsets $(P_{f})_{f\in H_{n}}$ of P such that:

1. (i) $P_{f}\cap P_{h}=\emptyset $ and $(S_{f}^{*})^{\times 2n}P_{f}\cap (S_{f}^{*})^{\times 2n}P_{h}=\emptyset $ for all $f,h\in H$ with $f\ne h$ ;

2. (ii) $\bigsqcup _{f\in F}(S_{f}^{*})^{\times 2n}P_{f}\subset Q$ ; and

3. (iii) $(\bigotimes _{j=1}^{n}(\nu _{k_{j}}^{*})^{\otimes 2})(\bigsqcup _{f\in H_{n}}P_{f})>0.5(\bigotimes _{j=1}^{n}(\nu _{k_{j}}^{*})^{\otimes 2})(P)$ .

Utilizing (4.1), we can select $k_{n+1}\in \Bbb N$ large so that $k_{n+1}>k_{n}$ and

(4.2) $$ \begin{align} \min_{f\in H_{n}}\prod_{j\ge k_{n+1}}\frac{\nu_{j}^{*}(S_{f}^{*}B_{j}\cap B_{j})}{\nu_{j}^{*}(B_{j})}>0.5. \end{align} $$

Repeating this process infinitely many times, we construct the entire sequence $(k_{n})_{n=1}^{\infty }$ . Since $\widetilde \mu ^{*}\sim (\nu ^{*})^{\boldsymbol {B}^{\prime }}$ , it suffices to verify that the Cartesian square of $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},(\nu ^{*})^{{\boldsymbol {B}}^{\prime }}, \widetilde T^{*})$ is ergodic. As $\widetilde T^{*}$ preserves $(\nu ^{*})^{{\boldsymbol {B}}^{\prime }}$ , we will use Lemma 4.3 for this verification. Let

(4.3) $$ \begin{align} \widetilde {\boldsymbol{B}}^{n}:=(Y^{*})^{n}\times B_{k_{n+1}}\times B_{k_{n+2}}\times\cdots\subset \widetilde X^{*}\quad\text{for each }n\in\Bbb N\text{.} \end{align} $$

For each pair of atoms $P,Q\in \bigotimes _{j=1}^{n}(\mathcal {P}_{k_{j},n})^{\otimes 2}$ and $f\in H_{n}$ , we set

$$ \begin{align*} P^{\prime} &:=P\times (B_{k_{n+1}}\times B_{k_{n+2}}\times\cdots)^{\times 2}\in (\mathcal{P}_{n}^{\prime})^{\otimes 2},\\ Q^{\prime} &:=Q\times(B_{k_{n+1}}\times B_{k_{n+2}}\times\cdots)^{\times 2}\in (\mathcal{P}_{n}^{\prime})^{\otimes 2} \quad \text{and}\\ P_{f}^{\prime} &:=P_{f}\times((B_{k_{n+1}}\cap (S^{*}_{f})^{-1}B_{k_{n+1}})\times (B_{k_{n+2}}\cap (S^{*}_{f})^{-1}B_{k_{n+2}})\times\cdots)^{\times 2}, \end{align*} $$

where $(P_{f})_{f\in H_{n}}$ are mutually disjoint subsets of P satisfying $(i)$ – $(iii)$ . Then, $\mathcal {P}_{n}^{\prime }:=(P^{\prime })_{P\in \bigotimes _{j=1}^{n}(\mathcal {P}_{k_{j},n})^{\otimes 2}}$ is a finite partition of $\widetilde {\boldsymbol {B}}^{n}\times \widetilde {\boldsymbol {B}}^{n}$ into subsets of equal measure $(\nu ^{*})^{{\boldsymbol {B}}^{\prime }}\otimes (\nu ^{*})^{{\boldsymbol {B}}^{\prime }}$ . Moreover, $(\circ )$ yields that

• (•) $(\mathcal {P}_{n}^{\prime })_{n=1}^{\infty }$ approximates $((\widetilde {\mathfrak {B}}^{*}\otimes \widetilde {\mathfrak {B}}^{*})_{0},(\nu ^{*})^{{\boldsymbol {B}}^{\prime }}\otimes (\nu ^{*})^{{\boldsymbol {B}}^{\prime }})$ ,

where $(\widetilde {\mathfrak {B}}^{*}\otimes \widetilde {\mathfrak {B}}^{*})_{0}$ denotes the ring of subsets of finite measure in $\widetilde {\mathfrak {B}}^{*}\otimes \widetilde {\mathfrak {B}}^{*}$ . We also note that $(P^{\prime }_{f})_{f\in H_{n}}$ are mutually disjoint subsets of $P^{\prime }$ and $(( \widetilde T_{f}^{*}\times \widetilde T_{f}^{*})P^{\prime }_{f})_{f\in H_{n}}$ are mutually disjoint subsets of $Q^{\prime }$ . It follows from (4.2) that, for each $f\in H_{n}$ ,

$$ \begin{align*} \prod_{j\ge n+1}\frac{\nu_{k_{j}}^{*}(S_{f}^{*}B_{k_{j}}\cap B_{k_{j}})}{\nu_{k_{j}}^{*}(B_{k_{j}})}>0.5. \end{align*} $$

Using this inequality and $(iii)$ , we obtain that

$$ \begin{align*} ((\nu^{*})^{{\boldsymbol{B}}^{\prime}}\otimes (\nu^{*})^{{\boldsymbol{B}}^{\prime}})\bigg(\bigsqcup_{f\in H_{n}}P^{\prime}_{f}\bigg)&> \frac{1/2(\bigotimes_{j=1}^{n}(\nu_{k_{j}}^{*})^{\otimes 2})(P)}{(\prod_{j=1}^{n}\nu^{*}_{k_{j}}(B_{k_{j}}))^{2}}\cdot \frac14 \\ &=\frac 18 ((\nu^{*})^{{\boldsymbol{B}}^{\prime}}\otimes (\nu^{*})^{{\boldsymbol{B}}^{\prime}})(P^{\prime}) \end{align*} $$

for every pair of $\mathcal {P}_{n}^{\prime }$ -atoms $P^{\prime },Q^{\prime }$ . Therefore, it follows from Lemma 4.3 that the G-action $(\widetilde T_{g}^{*}\times \widetilde T_{g}^{*})_{g\in G}$ on $(\widetilde {X}^{*}\otimes \widetilde {X}^{*},(\nu ^{*})^{{\boldsymbol {B}}^{\prime }}\otimes (\nu ^{*})^{{\boldsymbol {B}}^{\prime }})$ is ergodic. Thus, Claim B is proved.

It follows that $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ is of type $II_{\infty }$ . We deduce from Proposition 3.3 and $(\alpha _{1})$ that $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ is not strongly ergodic. Since $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ is IDPFT, Proposition 2.25 yields that $T^{*}$ is amenable in the Greenleaf sense. Moreover, $\widetilde T^{*}$ is amenable if and only if G is amenable.

Thus, Theorem 4.2 is proved completely.

Proof. In view of [Reference Delabie, Jolissaint and ZumbrunnenDeJoZu] (see also [Reference DanilenkoDa]), there is an infinite $\sigma $ -finite measure-preserving system $(Y,\mathfrak {Y},\kappa , S)$ of 0-type satisfying $(\alpha _{1})$ – $(\alpha _{5})$ . Then, we repeat the proof of Theorem 4.2 verbatim to obtain a non-strongly ergodic free IDPFT system $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ of infinite ergodic index and of type $II_{\infty }$ . The $\widetilde \mu ^{*}$ -equivalent infinite $\sigma $ -finite $\widetilde T^{*}$ -invariant measure is $(\nu ^{*})^{{\boldsymbol {B}}^{\prime }}$ . Since the dynamical system $(Y,\nu _{n},S)$ is of 0-type, the Poisson suspension $(Y^{*},\nu _{n}^{*},S^{*})$ is mixing in view of Fact 2.21 for each $n\in \Bbb N$ . Then, by Corollary 2.13, $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},(\nu ^{*})^{{\boldsymbol {B}}^{\prime }},\widetilde T^{*})$ is of 0-type. Hence, $(\widetilde X^{*},\widetilde {\mathfrak {B}}^{*},\widetilde \mu ^{*},\widetilde T^{*})$ is of 0-type too.

Proof. It was shown in §3 that $(X^{*},\mathfrak {B}^{*},\mu ^{*},T^{*})$ is IDPFT. Hence, $T^{*}$ is amenable in the Greenleaf sense by Proposition 2.25. Proposition 3.4 and $(\alpha _{4})$ imply that $T^{*}$ is free. Proposition 3.3 and $(\alpha _{1})$ imply that $T^{*}$ is not strongly ergodic.

We now prove that $T^{*}$ is weakly mixing. We note that (5.3), Lemma 3.2 and $(\alpha _{6})$ yield that

$$ \begin{gather*} \bigg(\frac {d\mu}{d\mu\circ T_{g}^{-1}}\bigg)^{2}-1\in L^{1}(X,\mu)\quad\text{and}\\ \int_{X}\bigg(\bigg(\frac{d\mu}{d\mu\circ T_{g}^{-1}}\bigg)^{2}-1\bigg)\,d\mu=\sum_{n=1}^{\infty} a_{n}c(\unicode{x3bb}_{n})\frac{\kappa (S_{g}F_{n}\triangle F_{n}) }{2\kappa(F_{n})}=\sum_{n=1}^{\infty}\frac1{2n}{\kappa (S_{g}F_{n}\triangle F_{n}) } \end{gather*} $$

for every $g\in G$ . Hence, for each $k\in \Bbb N$ ,

$$ \begin{align*} \int_{X}\bigg(\bigg(\frac{d\mu}{d\mu\circ T_{g_{k}}^{-1}}\bigg)^{2}-1\bigg)\,d\mu \le\frac 12\sum_{n=1}^{k}\frac 1n+ \sum_{n=k+1}^{\infty}{\kappa ( S_{g_{k}} F_{n}\triangle F_{n} )}\le \frac 12\log k+3 \end{align*} $$

in view of (5.1). This implies that

$$ \begin{align*} \sum_{k=1}^{\infty} \frac 1{k^{2}}\,e^{\int_{X}(({d\mu}/({d\mu\circ T_{g_{k}}^{-1}}))^{2}-1)\,d\mu}<+\infty. \end{align*} $$

Therefore, by Proposition 2.20,

$$ \begin{align*} \sum_{k=1}^{\infty}\frac{d\mu^{*}\circ (T_{g_{k}}^{*})^{-1}}{d\mu^{*}}(\omega)=+\infty\quad\text{at a.e. }\omega\text{.} \end{align*} $$

On the other hand, in view of Fact 2.21, $(Y^{*},\mathfrak {Y}^{*},\nu _{k}^{*}, S^{*})$ is mixing along H for each $k\in \Bbb N$ because $(\alpha _{5})$ holds. (We refer to §3 for the definitions of $(\nu _{k})_{k=1}^{\infty }$ and $(\mu _{k})_{k=1}^{\infty }$ .) Hence, by Proposition 2.16, $T^{*}$ is weakly mixing.

Suppose, by contraposition, that $T^{*}$ is of type $II$ . Let $\eta $ stand for a $T^{*}$ -invariant $\sigma $ -finite measure equivalent to $\mu ^{*}$ . Fix $n\in \Bbb N$ . Since $\bigotimes _{k=1}^{n}\mu _{k}^{*}\sim \bigotimes _{k=1}^{n}\nu _{k}^{*}$ , it follows that the projection of $\eta $ to $(Y^{*})^{n}$ along the mapping

$$ \begin{align*} \pi_{n}:X^{*}\ni(x^{*}_{k})_{k=1}^{\infty}\mapsto (x^{*}_{k})_{k=1}^{n}\in( X^{*})^{n} \end{align*} $$

has the same collection of 0-measure subsets as the measure $\bigotimes _{k=1}^{n}\nu _{k}^{*}$ . Hence, we can disintegrate $\eta $ over $\bigotimes _{k=1}^{n}\nu _{k}^{*}$ along $\pi _{n}$ . Thus, there is measurable field $(Y^{*})^{n}\ni y\mapsto \eta _{y}$ of $\sigma $ -finite measures on $(Y^{*})^{\Bbb N}$ such that

(5.4) $$ \begin{align} \eta=\int_{(Y^{*})^{n}}\delta_{y}\otimes\eta_{y}\,d\bigg(\bigotimes_{k=1}^{n}\nu_{k}^{*}\bigg)(y). \end{align} $$

Since Q preserves $\nu _{k}$ , we obtain that the Poisson suspension $Q^{*}$ of Q preserves $\nu _{k}^{*}$ . It follows from $(\alpha _{4})$ that $Q^{*}$ is ergodic with respect to $\nu _{k}^{*}$ for each $k\in \Bbb N$ . Moreover, $Q^{*}$ commutes with $S^{*}$ . Let

$$ \begin{align*} \Gamma_{n}:=\{(Q^{*})^{l_{1}}\otimes\cdots\otimes (Q^{*})^{l_{n}}\mid l_{1},\ldots,l_{n}\in\Bbb Z\}. \end{align*} $$

Then, $\Gamma _{n}$ is a countable group (isomorphic to $\Bbb Z^{n}$ ) of measure-preserving transformations of the product space $((Y^{*})^{n},(\mathfrak {Y}^{*})^{\otimes n},\bigotimes _{k=1}^{n}\nu _{k}^{*})$ . Of course, $\Gamma _{n}\otimes 1$ is a group of $\mu ^{*}$ -non-singular transformations commuting with $T^{*}$ . Since $\eta \sim \mu ^{*}$ , $\Gamma _{n}\otimes 1$ is also $\eta $ -non-singular. Since $T^{*}$ is ergodic and $\eta $ -preserving, it follows that $\eta \circ \gamma =a_{\gamma }\eta $ for some $a_{\gamma }>0$ , for each transformation $\gamma \in \Gamma _{n}\otimes 1$ . Since $\gamma $ preserves a finite measure $(\bigotimes _{k=1}^{n}\nu _{k}^{*})\otimes \bigotimes _{k>n}\mu _{n}^{*}$ which is equivalent to $\eta $ , we obtain that $\gamma $ is conservative with respect to $\eta $ . Hence, $a_{\gamma }=1$ , that is, $\gamma $ preserves $\eta $ for each $\gamma \in \Gamma _{n}\otimes 1$ . We deduce from this fact and (5.4) that the mapping

$$ \begin{align*} \Psi:(Y^{*})^{n}\ni y\mapsto \eta_{y} \end{align*} $$

is invariant under $\Gamma _{n}$ . Since $\Gamma _{n}$ is ergodic with respect to $\bigotimes _{k=1}^{n}\nu _{k}^{*}$ , it follows that $\Psi $ is constant almost everywhere, that is, there exists a $\sigma $ -finite measure $\eta ^{n}$ on $({({Y}^{*})}^{\Bbb N},(\mathfrak {Y}^{*})^{\otimes \Bbb N}),$ such that $\eta _{y}=\eta ^{n}$ at a.e. $y\in (Y^{*})^{n}$ . We thus obtain that $\eta =(\bigotimes _{k=1}^{n}\nu _{k}^{*})\otimes \eta ^{n}$ . Since n is arbitrary, $\eta $ is an Moore–Hill product of $(\nu _{k}^{*})_{k=1}^{\infty }$ (see Definition 2.10(ii)). Hence, by Fact 2.14(ii), $\eta $ is proportional to the restricted product of $(\nu ^{*}_{k})_{k=1}^{\infty }$ with respect to some sequence $\boldsymbol {C}=(C_{n})_{n=1}^{\infty }$ of subsets $C_{n}\subset Y^{*}$ . In particular,

(5.5) $$ \begin{align} \eta(C_{1}\times C_{2}\times\cdots)>0. \end{align} $$

If $\prod _{n=1}^{\infty }\nu _{n}^{*}(C_{n})>0$ , then $\eta \sim \bigotimes _{n=1}^{\infty }\nu _{n}^{*}$ and hence $\mu ^{*}\sim \nu ^{*}$ . The latter happens if and only if $ \sqrt {{d\mu }/{d\nu }}-1\in L^{2}(X,\nu ) $ by Fact 2.18(ii). However,

$$ \begin{align*} \int_{X}\bigg(\sqrt{\frac{d\mu}{d\nu}}-1\bigg)^{2}\,d\nu=\sum_{n=1}^{\infty}(\sqrt{\unicode{x3bb}_{n}}-1)^{2}a_{n}\asymp \sum_{n=1}^{\infty}\frac1{n\unicode{x3bb}_{n}^{2}}=\sum_{n=1}^{\infty}\frac 1{n4^{l_{n}}}=+\infty \end{align*} $$

by choice of $(\unicode{x3bb} _{n})_{n=1}^{\infty }$ , which is a contradiction. Hence, $\prod _{n=1}^{\infty }\nu _{n}^{*}(C_{n})=0$ or, equivalently, $\sum _{n=1}^{\infty }\nu _{n}^{*}(Y_{n}^{*}\setminus C_{n})=+\infty $ . Since $\eta \sim \mu ^{*}$ , we deduce from (5.5) that

$$ \begin{align*} 0<\mu^{*}(C_{1}\times C_{2}\times\cdots)=\prod_{n=1}^{\infty}\mu_{n}^{*}(C_{n}) \end{align*} $$

or, equivalently, $\sum _{n=1}^{\infty }\mu _{n}^{*}(Y_{n}^{*}\setminus C_{n})<+\infty $ . On the other hand, it follows from Fact 2.18(iii) that

$$ \begin{align*} \frac{d\nu_{n}^{*}}{d\mu_{n}^{*}}(\omega)=e^{-((1/{\unicode{x3bb}_{n}})-1)\mu_{n}(F_{n})}\unicode{x3bb}_{n}^{-\omega(F_{n})} \end{align*} $$

at $\mu _{n}^{*}$ -a.e. $\omega \in Y_{n}^{*}$ . Since $\unicode{x3bb} _{n}\ge 1$ and $\mu _{n}(F_{n})=a_{n}\unicode{x3bb} _{n}={\unicode{x3bb} _{n}}/{nc(\unicode{x3bb} _{n})}\le 1$ , we obtain that ${d\nu _{n}^{*}}/{d\mu _{n}^{*}}(\omega )\le e$ for a.e. $\omega \in X_{n}^{*}$ . Hence,

$$ \begin{align*} +\infty=\sum_{n=1}^{\infty}\nu_{n}^{*}(Y_{n}^{*}\setminus C_{n})\le e^{}\sum_{n=1}^{\infty}\mu_{n}^{*}(Y_{n}^{*}\setminus C_{n})<+\infty, \end{align*} $$

which is a contradiction. Thus, Claim A is proved.

Fix $n\in \Bbb N$ . Let $\mu ^{*,n}$ denote the infinite product measure $(\bigotimes _{k=1}^{n}\nu _{n}^{*})\otimes \bigotimes _{k>n}\mu _{k}^{*}$ . Of course, $\mu ^{*,n}\sim \mu ^{*}$ . Hence, $r(T^{*},\mu ^{*})=r(T^{*},\mu ^{*,n})$ . We recall that

$$ \begin{align*} f_{n}:=\frac{d\mu_{n}}{d\nu_{n}}=1+(\unicode{x3bb}_{n}-1)1_{F_{n}}. \end{align*} $$

From Fact 2.18(iv), we deduce that

$$ \begin{align*} \frac{d\mu_{k}^{*}\circ S_{g}^{*}}{d\mu_{k}^{*}}(\omega)=\prod_{\{y\in Y\mid \omega(\{y\})=1\}}\frac{d\mu_{k}\circ S_{g}}{d\mu_{k}}(y)= \prod_{\{y\in Y\mid \omega(\{y\})=1\}}\frac{f_{k}(S_{g}y)}{f_{k}(y)} \in\{\unicode{x3bb}_{k}^{m}\mid m\in\Bbb Z\} \end{align*} $$

for each $g\in G$ at $\mu _{k}^{*}$ -a.e. $\omega \in Y^{*}$ . We deduce from this, $(\alpha _{6})$ and the fact that $(X^{*},\mathfrak {B}^{*},\mu ^{*,n}, T^{*})$ is IDPFT that

$$ \begin{align*} \frac{d\mu^{*,n}\circ T_{g}^{*}}{d\mu^{*,n}}(\omega)=\prod_{k>n}^{\infty}\frac{d\mu_{k}^{*}\circ S_{g}^{*}}{d\mu_{k}^{*}}(\omega_{k})\in\{2^{kl_{n+1}}\mid k\in\Bbb Z\} \end{align*} $$

for each $g\in G$ at a.e. $\omega \in X^{*}$ . Hence, $r(T^{*},\mu ^{*,n})\subset \{2^{kl_{n+1}}\mid k\in \Bbb Z\}$ . This yields that $r(T^{*},\mu ^{*})\subset \bigcap _{n=1}^{\infty }\{2^{kl_{n+1}}\mid k\in \Bbb Z\}=\{1\}$ . Since $T^{*}$ is not of type $II$ , it follows that $T^{*}$ is of type $III_{0}$ . Thus, Claim B is proved.

Proof. Since G is Haagerup, there is a measure-preserving G-action $S=(S_{g})_{g\in G}$ on an infinite $\sigma $ -finite standard measure space $(Y,\mathfrak {Y},\kappa )$ such that $(\alpha _{1})$ – $(\alpha _{6})$ hold and, moreover, S is of 0-type (see [Reference Delabie, Jolissaint and ZumbrunnenDeJoZu] or [Reference DanilenkoDa]). Then, following the proof of Theorem 5.1 verbatim, we construct a dynamical system $( X,\mathfrak {B},\mu , T)$ whose Poisson suspension $( X^{*},\mathfrak {B}^{*},\mu ^{*}, T^{*})$ possesses all the properties listed in the statement of Theorem 5.1. It remains only to prove that $( X^{*},\mathfrak {B}^{*},\mu ^{*}, T^{*})$ is of 0-type. By [Reference Danilenko, Kosloff and RoyDaKoRo1, Proposition 6.9], $T^{*}$ is of 0-type if and only if $\int _{X}(\sqrt {({d\mu \circ T_{g}})/{d\mu }}-1)^{2}\,d\mu \to 0$ as $g\to \infty $ . Since