Proof that Theorem 3.1 implies Theorem 1.6.

Let $A\subset G$ have positive left upper Banach density, so that there exists some Følner sequence $\Phi $ such that $d_\Phi (A)=\lim _{N\to \infty }\frac {|A\cap \Phi _N|}{|\Phi _N|}>0$ , (where we have passed to a subsequence). Then consider $(X,\mu ,T)$ , E, $\Psi $ and a, as ensured by Theorem 2.15, satisfying $\mu (E)\geq d_\Phi (A)>0$ and $\{h\in G: T_ha\in E\}=A$ . It follows then by Theorem 3.1 that there exist an infinite sequence $B=(b_n)_{n\in \mathbb {N}}\subset A$ and some $t\in G$ such that .

Proof of Lemma 3.4.

By assumption, there exists an infinite sequence $(g_n)_{n\in \mathbb {N}}$ in G such that $(T_{g_n}\times T_{g_n}) (a,x_1)\to (x_1,x_2)$ . Since $T_{g_n}a\to x_1\in E$ and E is open, we get that $T_{g_n}a\in E$ for n sufficiently large, so we may assume without loss of generality that $(g_n)_{n\in \mathbb {N}}\subset \{g\in G: T_ga\in E\}$ . Therefore, we will construct the sequence B to be a subset of $(g_n)_{n\in \mathbb {N}}$ .

We construct the sequence $B=(b_n)_{n\in \mathbb {N}}$ inductively.

• • Since $T_{g_n}x_1\to x_2\in F$ and F is open, we can pick $b_1\in (g_n)_{n\in \mathbb {N}}$ such that $T_{b_1}x_1\in F$ . Then $(x_1,x_2)\in (T^{-1}_{b_1}F)\times F$ which is open.

• • Since $(T_{g_n}\times T_{g_n})(a,x_1)\to (x_1,x_2)$ , we pick $b_2\in (g_n)_{n\in \mathbb {N}}$ such that $(T_{b_2}\times T_{b_2})(a,x_1)\in (T^{-1}_{b_1}F)\times F$ and $b_2\neq b_1$ (this is possible since there are infinitely many choices). Then

  $$ \begin{align*}a\in T^{-1}_{{b_1}{b_2}}F \quad \text{and} \quad x_1\in T^{-1}_{b_2}F.\end{align*} $$

• • Induction step: Assume we have found $b_1, b_2, \dots , b_n\in (g_n)_{n\in \mathbb {N}}$ all distinct to each other such that

  (3.2) $$ \begin{align} a\in\bigcap_{1\leq i<j\leq n}T^{-1}_{{b_i}{b_j}}F \quad \text{and} \quad x_1\in\bigcap_{1\leq m \leq n}T^{-1}_{b_m}F. \end{align} $$
  Since $(T_{g_n}\times T_{g_n})(a,x_1)\to (x_1,x_2)\in \Big (\bigcap _{1\leq m \leq n}T^{-1}_{b_m}F\Big )\times F$ and this set is open, we can pick $b_{n+1}\in (g_n)_{n\in \mathbb {N}}$ such that
  $$ \begin{align*}(T_{b_{n+1}}\times T_{b_{n+1}})(a,x_1)\in \bigg(\bigcap_{1\leq m \leq n}T^{-1}_{b_m}F\bigg)\times F\end{align*} $$
  and $b_{n+1}\not \in \{b_m: 1\leq m\leq n\}$ (since there are infinitely many choices). Combining this with the inductive hypothesis, we obtain that
  $$ \begin{align*}a\in\bigcap_{1\leq i<j\leq n+1}T^{-1}_{{b_i}{b_j}}F \quad \text{and} \quad x_1\in\bigcap_{1\leq m \leq n+1}T^{-1}_{b_m}F.\end{align*} $$

Taking $B=(b_n)_{n\in \mathbb {N}}$ , we clearly have an infinite subset of $(g_n)_{n\in \mathbb {N}}$ , and since (3.2) holds for any $n\in \mathbb {N}$ by construction, we get that , as desired.

Proof. The proof of this lemma is given in the Appendix A.

Proof. The proofs of (i) and (ii) are identical to those in the case G is the semigroup $(\mathbb {N},+)$ , and they can be found in [Reference Kra, Moreira, Richter and Robertson14, Proposition 3.20]. Therefore, here we only provide a sketch of the proof.

Let $(Z,m,R)$ be the Kronecker factor of $(X,\mu ,T)$ , and let $\pi :X\to Z$ be a factor map. Define $\widetilde {X}=X\times Z$ and $\widetilde {T}=T\times R$ , consider the map $\rho :X\to \widetilde {X}$ , given by $\rho (x)=(x,\pi (x))$ , and then define $\widetilde {\mu }=\rho \mu $ . Then the map $\rho : X \to \widetilde {X}$ is an isomorphism of the G-systems $(\widetilde {X},\widetilde {\mu },\widetilde {T})$ and $(X,\mu ,T)$ , and therefore, since $(X,\mu ,T)$ is ergodic, we get that $(\widetilde {X},\widetilde {\mu },\widetilde {T})$ is also ergodic. In addition, the projection on the first coordinate $\widetilde {\pi }:\widetilde {X} \to X$ is a continuous factor map of the systems.

By Remark 3.6, we may assume that the point a is transitive. Then we can use Lemma 3.5 to find a point $z\in Z$ and Følner sequence $\widetilde {\Phi }$ such that (3.3) holds for all $f_1\in C(X)$ and $f_2 \in C(Z)$ . Using the definition of the measure $\widetilde {\mu }$ , it is not too difficult to see that for any continuous function $F\in C(\widetilde {X})$ ,

$$ \begin{align*}\lim_{N\to \infty} \frac{1}{|\widetilde{\Phi}_N|} \sum_{g\in \widetilde{\Phi}_N} F(T_g \times R_g)(a,z) = \int_{\widetilde{X}} F {~\mathrm{d}} \widetilde{\mu},\end{align*} $$

which means that the point $\widetilde {a}= (a,z)$ is in $\texttt {{gen}}(\widetilde {\mu },\widetilde {\Phi })$ . In addition, $\widetilde {\pi }(\widetilde {a})=a$ .

Now, as the systems $(X,\mu ,T)$ and $(\widetilde {X},\widetilde {\mu },\widetilde {T})$ are isomorphic, their Kronecker factors are also isomorphic, so we may assume that $(Z,m,R)$ is the Kronecker factor of $(\widetilde {X},\widetilde {\mu },\widetilde {T})$ . Then the projection on the second coordinate $p: \widetilde {X} \to Z$ is a continuous factor map from $(\widetilde {X},\widetilde {\mu },\widetilde {T})$ to its Kronecker factor.

Note that (iii) is immediate from the definition of factors and factor maps. Hence, it remains to prove (iv). By assumption, $(\widetilde {T}_{g_n}\times \widetilde {T}_{g_n}) (\widetilde {a},\widetilde {x}_1) \to (\widetilde {x}_1,\widetilde {x}_2)$ for some $(g_n)_{n\in \mathbb {N}}$ in G. We now notice that

$$ \begin{align*}T_{g_n}(a)=T_{g_n}(\widetilde{\pi}(\widetilde{a})) = \widetilde{\pi}(\widetilde{T}_{g_n}) \to\widetilde{\pi}(\widetilde{x}_1)=x_1,\end{align*} $$

since $\widetilde {\pi }$ is a continuous factor map. Similarly, we get $T_{g_n}(x_1)\to x_2$ , and the result follows.

Proof. The space of Borel probability measures on K is weak $^{\ast }$ compact and metrizable, so in order to prove the result, it suffices to prove that if $(\nu _{N_j})_{j\in \mathbb {N}}$ is a convergent subsequence of $(\nu _N)_{N\in \mathbb {N}}$ , then it converges to the Haar measure $m_K$ .

Let $(\nu _{N_j})_{j\in \mathbb {N}}$ be a subsequence of $(\nu _N)_{N\in \mathbb {N}}$ which converges in the weak $^{\ast }$ topology to a measure $\nu $ on K. To prove that $\nu $ is the Haar measure on K, it suffices to prove that for all continuous functions h on K and all $k\in K$ , we have that $\int _K h(k y) {~\mathrm {d}} \nu (y) = \int _K h(y) {~\mathrm {d}} \nu (y)$ .

Let $d_K$ be a translation invariant metric on K. Let also $h: K \to \mathbb {C}$ be continuous, $k \in K$ and $\epsilon>0$ . Since K is compact, we have that h is uniformly continuous, so there is $\delta>0$ such that if $d_K(y_1, y_2)< \delta $ , then $| h(y_1) - h(y_2)| < \epsilon $ . Recall that $(\alpha (g))_{g\in G}$ is dense in K, so there is $g_0$ such that $d_K(k, \alpha (g_0)) < \delta $ . We then have that for all $y \in K$ , $d_K(ky, \alpha (g_0) y) < \delta $ , so $|h(ky)-h(\alpha (g_0)y)| < \epsilon $ , and we obtain that

$$ \begin{align*} \bigg| \int_K h(y) {~\mathrm{d}} \nu(y) - \int_K h(ky) {~\mathrm{d}} \nu(y) \bigg| & \leq \bigg| \int_K h(y) {~\mathrm{d}} \nu(y) - \int_K h(\alpha (g_0)y) {~\mathrm{d}} \nu(y) \bigg| \\ &\hspace{.09cm} + \bigg| \int_K h(\alpha (g_0)y) {~\mathrm{d}} \nu(y) - \int_K h(ky) {~\mathrm{d}} \nu(y) \bigg| \\ & \leq \bigg| \int_K h(y) {~\mathrm{d}} \nu(y) - \int_K h(\alpha (g_0)y) {~\mathrm{d}} \nu(y) \bigg| + \epsilon. \end{align*} $$

From the continuity of h and the definition of $ \nu $ , we have that

$$ \begin{align*} \int_K h(\alpha(g_0)y) {~\mathrm{d}} \nu (y) &= \lim_{j\to \infty} \frac{1}{|\Psi_{N_j}|} \sum_{g\in \Psi_{N_j}} h(\alpha(g_0) \alpha(g)) = \lim_{j\to \infty} \frac{1}{|\Psi_{N_j}|} \sum_{g\in \Psi_{N_j}} h(\alpha(g_0 g)) \\ &= \lim_{j \to \infty} \frac{1}{|\Psi_{N_j}|} \sum_{g\in g_0 \Psi_{N_j}} h(\alpha(g)) = \lim_{j \to \infty} \frac{1}{|\Psi_{N_j}|} \sum_{g\in \Psi_{N_j}} h(\alpha(g)) \\ &= \int_K h(y) {~\mathrm{d}} \nu (y) , \end{align*} $$

so combining with the previous, we get that $\big | \int _K h(y) {~\mathrm {d}} \nu (y) - \int _K h(ky) {~\mathrm {d}} \nu (y) \big | \leq \epsilon $ , and since $\epsilon $ was arbitrary, we obtain that $ \int _K h(y) {~\mathrm {d}} \nu (y) = \int _K h(ky) {~\mathrm {d}} \nu (y) $ , which proves that $ \nu = m_K$ and concludes the proof.

Proof of Lemma 4.2.

As K is compact and metrizable, the measures $m_K, m_{K^2}$ are regular. In particular, for each Borel A, we have

$$ \begin{align*} m_K(A) = \sup_{\substack{C\subset A \\ C \: \text{compact}}} m_K(C) = \inf_{\substack{O \supset A \\ O \: \text{open}}} m_K(O) \:\: \text{ and } \:\: m_{K^2}(A) = \sup_{\substack{C\subset A \\ C \: \text{compact}}} m_{K^2}(C) = \inf_{\substack{O \supset A \\ O \: \text{open}}} m_{K^2}(O). \end{align*} $$

Therefore, to prove that $m_{K^2}$ is absolutely continuous with respect to $m_K$ , it suffices to prove that for each compact set $C \subset K$ , if $m_K (C) =0$ , then $m_{K^2}(C)=0$ .

Let $C \subset K$ be a nonempty (for otherwise the result is trivial) compact with $m_K(C) =0$ and let $\epsilon>0$ . As G is square absolutely continuous, we know that there are two Følner sequences $\Phi $ and $\Psi $ in G and a $\delta>0$ such that for any $u:G\to [0,1]$ satisfying $\limsup _{N\to \infty }\frac {1}{|\Phi _N|}\sum _{g\in \Phi _N} u(g) < \delta ,$ we have that $\limsup _{N\to \infty }\frac {1}{|\Psi _N|}\sum _{g\in \Psi _N} u(g^2) < \epsilon .$ Since

$$ \begin{align*}0=m_K(C)= \inf_{\substack{O \supset C \\ O \: \text{open}}} m_K(O),\end{align*} $$

we can pick an open set $O \supset C$ with $m_K(O) < \delta $ . By Urysohn’s lemma, we know that there is a continuous function $f : K \to [0,1]$ such that $f=1$ on C and $f=0$ outside O. By Lemma 4.3, we then have that

$$ \begin{align*} \delta> m_K(O) \geq \int_K f(k) {~\mathrm{d}} m_K(k) = \lim_{N\to \infty} \frac{1}{|\Phi_N|} \sum_{g\in \Phi_N} f(\alpha(g)), \end{align*} $$

and by the choice of $\delta $ , we get that

$$ \begin{align*} \limsup_{N\to \infty} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} f(\alpha(g^2)) <\epsilon. \end{align*} $$

From the definition of $m_{K^2}$ and the continuity of $k \mapsto f(k^2)$ , we then obtain

$$ \begin{align*} \int_K f(k) {~\mathrm{d}} m_{K^2}(k) = \int_K f(k^2) {~\mathrm{d}} m_K (k) & = \lim_{N\to \infty} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} f(\alpha(g)^2) \\ & = \lim_{N\to \infty} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} f(\alpha(g^2)) < \epsilon, \end{align*} $$

where for the third equality, we use that $\alpha $ is a group homomorphism. So after all, we have $m_{K^2}(C) \leq \int _K f(k) {~\mathrm {d}} m_{K^2}(k) < \epsilon $ , and as $\epsilon $ was arbitrary, we have that $m_{K^2}(C) =0$ . This concludes the proof.

Proof of Proposition 4.5.

(i) Using (4.2), we have that

$$ \begin{align*}\pi_1\sigma = \int_K \eta_{kH} {~\mathrm{d}} m_K(k) = \int_Z \eta_z {~\mathrm{d}} m(z) = \mu.\end{align*} $$

(ii) From the definition of $\sigma $ , we have that $\pi _2 \sigma = \int _{K} \eta _{k^2\pi (a)} {~\mathrm {d}} m_K(k).$ Fix $k_0$ such that $\pi (a)=k_0 H$ .

Let $A\subset X$ with $\mu (A)=0$ . Then $\mu (A)=\int _Z \eta _z(A) {~\mathrm {d}} m(z)$ , so we get that there is a set $Z'\subset Z$ with $m(Z')=1$ such that for all $z\in Z'$ , $\eta _z(A)=0$ . As $pm_K =m$ , we have $m_K(p^{-1}(Z'))=1$ and then also $m_K ((p^{-1}Z') k_0 ^{-1} )=1 $ . Finally, using Lemma 4.2, we get that $m_{K^2}((p^{-1}Z') k_0 ^{-1} )= m_K (s_K ^{-1}((p^{-1}Z') k_0 ^{-1} ))=1 $ . For each $k \in s_K ^{-1}((p^{-1}Z') k_0 ^{-1} )$ , we have that $k^2 \pi (a) \in Z'$ , so $\eta _{k^2 \pi (a)}(A)=0$ , and therefore, $\pi _2 \sigma (A)=0$ .

Proof. Let $E\subset X$ with $\mu (E)>0$ and recall that we want to show that the set $E\times \bigcup _{t\in G}T^{-1}_tE$ has positive measure with respect to $\sigma $ . We begin by expressing this set as

$$ \begin{align*}E\times \bigcup_{t\in G}T^{-1}_tE = (E\times X)\cap\bigg(X\times\bigcup_{t\in G}T^{-1}_tE\bigg).\end{align*} $$

By Proposition 4.5 (i), we have that $\sigma (E\times X)=\mu (E)>0$ . Therefore, it is enough to show that

(4.3) $$ \begin{align} \sigma\bigg(X\times\bigcup_{t\in G}T^{-1}_tE\bigg)=1. \end{align} $$

Notice that the set $\bigcup _{t\in G} T_t^{-1}E$ is clearly T-invariant, and since $\mu $ is ergodic and $\mu (E)>0$ , it follows that $\mu \big (\bigcup _{t\in G} T_t^{-1}E\big )=1$ . By Proposition 4.5 (ii), $\pi _2 \sigma $ is absolutely continuous with respect to $\mu $ , so

$$ \begin{align*} 1 = \pi_2 \sigma\bigg(\bigcup_{t\in G} T_t^{-1}E\bigg) = \sigma\bigg(X \times \bigcup_{t\in G} T_t^{-1}E\bigg), \end{align*} $$

which concludes the proof.

Proof. For the proof of (i) and (ii), we refer to [Reference Kra, Moreira, Richter and Robertson14, Proposition 3.11], as the proof there can be directly adapted to our case. We will now prove (iii). It is not too difficult to see that for all $(x_1, x_2) \in X\times X$ and for all $g \in G$ , $(T_g \times T_g) \lambda _{(x_1, x_2)} = \lambda _{(x_1, x_2)}$ (i.e., $\lambda _{(x_1, x_2)}$ is $T\times T$ -invariant). Therefore, to prove (iii), it suffices to prove that there is some Følner sequence $\Psi $ in G such that for $(\mu \times \mu )$ -almost every $(x_1,x_2)\in X\times X$ and all bounded and measurable functions F on $X\times X$ ,

$$ \begin{align*}\lim_{N\to\infty}\frac{1}{|\Psi_N|}\sum_{g\in\Psi_N}(T_g\times T_g)F = \int_{X\times X}F{~\mathrm{d}}\lambda_{(x_1,x_2)},\end{align*} $$

in $L^2(X\times X,\lambda _{(x_1,x_2)})$ .

Now, since X is a compact metric space, there is a countable family of continuous functions $(f_k)_{k\in \mathbb {N}}$ which is dense in $L^p(\nu )$ for all $p \in [1, + \infty )$ and all Borel probability measures $\nu $ on X. Then, it is not too difficult to see that the set consisting of finite linear combinations of functions the form $(f_{j_1} \otimes f_{j_2})_{j_1, j_2 \in \mathbb {N}}$ is dense in $L^2(\rho )$ for all Borel probability measures $\rho $ on $X\times X$ . Hence, using an approximation argument, it suffices to prove that there is a Følner $\Psi $ in G and a set $W\subset X \times X$ with $(\mu \times \mu )(W) =1 $ such that for all $(x_1, x_2) \in W$ and for all $j_1, j_2 \in \mathbb {N}$ ,

$$ \begin{align*} \lim_{N\to\infty}\frac{1}{|\Psi_N|}\sum_{g\in\Psi_N}(T_g\times T_g) (f_{j_1} \otimes f_{j_2}) = \int_{X\times X} f_{j_1} \otimes f_{j_2} {~\mathrm{d}}\lambda_{(x_1,x_2)}, \end{align*} $$

in $L^2(X\times X,\lambda _{(x_1,x_2)})$ .

Step 1. Let $\Phi $ be any Følner sequence in G. Then, using Theorem 2.13, we get that for each $j_1, j_2 \in \mathbb {N}$ ,

$$ \begin{align*} \lim_{N\to\infty}\frac{1}{|\Phi_N|}\sum_{g\in\Phi_N} (T_{g} \times T_{g}) (f_{j_1}\otimes f_{j_2}) = \lim_{N\to\infty}\frac{1}{|\Phi_N|}\sum_{g\in\Phi_N} (T_g \times T_g) (\mathbb{E}_{\mu}(f_{j_1}\: | \: Z)\otimes \mathbb{E}_{\mu}(f_{j_2}\: | \: Z)) \end{align*} $$

in $L^2(\mu \times \mu )$ . Combining this with (ii) yields

$$ \begin{align*} & \lim_{N\to \infty} \int_{X\times X} \int_{X\times X} \bigg|\frac{1}{|\Phi_N|} \sum_{g\in \Phi_N} (T_g \times T_g)(f_{j_1}\otimes f_{j_2})(y_1, y_2) \\ &\:\:\:\:\: - \frac{1}{|\Phi_N|} \sum_{g\in \Phi_N} (T_g \times T_g)(\mathbb{E}_{\mu}(f_{j_1}\: | \: Z)\otimes \mathbb{E}_{\mu}(f_{j_2}\: | \: Z))(y_1, y_2)\bigg| ^2 {~\mathrm{d}} \lambda_{(x_1, x_2)}(y_1, y_2) {~\mathrm{d}} (\mu \times \mu)(x_1, x_2) = 0. \end{align*} $$

Then for each $j_1,j_2\in \mathbb {N}$ , we can find a sub-Følner sequence $\widetilde {\Phi }$ of $\Phi $ , depending on $j_1,j_2$ , such that for $(\mu \times \mu )$ -almost every $(x_1, x_2)\in X \times X$ , we have

$$ \begin{align*} &\lim_{N\to\infty}\int_{X\times X} \bigg|\frac{1}{|\widetilde{\Phi}_N|} \sum_{g\in\widetilde{\Phi}_N} (T_g \times T_g)(f_{j_1}\otimes f_{j_2}) \\ &- \frac{1}{|\widetilde{\Phi}_N|} \sum_{g\in\widetilde{\Phi}_N} (T_g \times T_g)(\mathbb{E}_{\mu}(f_{j_1}\: | \: Z)\otimes \mathbb{E}_{\mu}(f_{j_2}\: | \: Z))\bigg|^2 {~\mathrm{d}}\lambda_{(x_1,x_2)} = 0, \end{align*} $$

and since the limits of both averages above exist by Theorem 2.1, we have that, for $(\mu \times \mu )$ -almost every $(x_1,x_2)\in X\times X$ ,

$$ \begin{align*} \lim_{N\to\infty}\frac{1}{|\widetilde{\Phi}_N|} \sum_{g\in\widetilde{\Phi}_N} (T_g \times T_g)(f_{j_1}\otimes f_{j_2}) = \lim_{N\to\infty}\frac{1}{|\widetilde{\Phi}_N|} \sum_{g\in\widetilde{\Phi}_N} (T_g \times T_g)(\mathbb{E}_{\mu}(f_{j_1}\: | \: Z)\otimes \mathbb{E}_{\mu}(f_{j_2}\: | \: Z)) \end{align*} $$

in $L^2(X\times X,\lambda _{(x_1,x_2)})$ . Since the family $(f_{j_1} \otimes f_{j_2})_{j_1, j_2 \in \mathbb {N}}$ is countable, then using a diagonal argument, one can find a Følner sequence $\Psi $ and a set $W_1 \subset X \times X$ with $(\mu \times \mu )(W_1) =1$ such that for all $(x_1, x_2)\in W_1$ and $j_1, j_2 \in \mathbb {N}$ , we have that

(4.7) $$ \begin{align} \lim_{N\to\infty}\frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (T_g \times T_g)(f_{j_1}\otimes f_{j_2}) = \lim_{N\to\infty}\frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (T_g \times T_g)(\mathbb{E}_{\mu}(f_{j_1}\: | \: Z)\otimes \mathbb{E}_{\mu}(f_{j_2}\: | \: Z)) \end{align} $$

in $L^2(X\times X,\lambda _{(x_1,x_2)})$ .

Step 2. Consider the sequence of probability measures on K defined by $\nu _N := \frac {1}{|\Psi _N|} \sum _{g\in \Psi _N} \delta _{\alpha (g)}$ . From Lemma 4.3, we know that $\nu _N \to m_K$ as $N \to \infty $ in the weak $^{\ast }$ topology.

Let $\phi _1, \phi _2 : Z \to \mathbb {C}$ be continuous. For each $z_1, z_2 \in Z$ , consider the function $\phi _{z_1, z_2} : K \to \mathbb {C}$ defined by $\phi _{z_1, z_2}(k)= \phi _1(k z_1) \phi _2(k z_2)$ . Then, $\phi _{z_1, z_2}$ is continuous, so we have that

$$ \begin{align*} \int_K \phi_1(k z_1) \phi_2(k z_2) {~\mathrm{d}} m_K(k) &= \int_K \phi_{z_1, z_2}(k) {~\mathrm{d}} m_K(k) = \lim_{N\to \infty} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} \phi_{z_1, z_2}(\alpha(g))\\ &= \lim_{N\to \infty} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} \phi_1(\alpha(g) z_1) \phi_2(\alpha(g) z_2)\\ &= \lim_{N\to \infty} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (R_g \times R_g)(\phi_1\otimes \phi_2)( z_1, z_2). \end{align*} $$

Since the previous holds for all $z_1, z_2 \in Z$ , using the dominated convergence theorem, we get that

$$ \begin{align*} \lim_{N \to \infty} \int_{Z \times Z} \bigg| \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (R_g \times R_g)(\phi_1\otimes \phi_2)(z_1, z_2) - \int_K \phi_1(k z_1) \phi_2(k z_2) {~\mathrm{d}} m_K(k) \bigg|^2 {~\mathrm{d}} (m\times m) (z_1, z_2)=0 \end{align*} $$

(i.e., the sequence $\frac {1}{|\Psi _N|} \sum _{g\in \Psi _N} (R_g \times R_g)(\phi _1\otimes \phi _2)$ converges in $L^2(Z \times Z, m\times m )$ as $N\to \infty $ to the function $(z_1, z_2) \mapsto \int _K \phi _1(k z_1) \phi _2(k z_2) {~\mathrm {d}} m_K(k)$ ).

Step 3. Given two bounded measurable maps $h_1, h_2 : Z \to \mathbb {C}$ , approximating them in $L^2(Z,m)$ by two continuous functions $\phi _1, \phi _2$ and using Step 2, one can prove that $\frac {1}{|\Psi _N|}\sum _{g\in \Psi _N} (R_g \times R_g)(h_1\otimes h_2)$ converges in $L^2(Z \times Z, m\times m )$ as $N\to \infty $ to the function $(z_1, z_2) \mapsto \int _K h_1(k z_1) h_2(k z_2) {~\mathrm {d}} m_K(k)$ . Since $\pi : (X,\mu ,T) \to (Z, m, R)$ is a factor map, it is not too difficult then to see that

(4.8) $$ \begin{align} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (T_g \times T_g) (h_1 \circ \pi \otimes h_2 \circ \pi) \to \Big[(x_1, x_2) \mapsto \int_K h_1(k \pi(x_1)) h_2(k \pi(x_2)) {~\mathrm{d}} m_K(k)\Big] \end{align} $$

as $N\to \infty $ in $L^2(X \times X, \mu \times \mu )$ .

For each $j \in \mathbb {N}$ , $\mathbb {E}_{\mu }(f_j\: | \: Z)$ can be viewed either as a function on Z or as a function on X measurable with respect to $\pi ^{-1}(Z)$ . In this proof, we always view $\mathbb {E}_{\mu }(f_j \: |\: Z)$ as a function on X measurable with respect to $\pi ^{-1}(Z)$ . For each $j \in \mathbb {N}$ , let $\psi _j $ be $\mathbb {E}_{\mu }(f_j \: | \: Z)$ when viewed as a function on Z, so we have that for $\mu $ -almost every $x \in X$ , $\psi _j \circ \pi (x) = \mathbb {E}_{\mu }(f_j \: | \: Z)(x)$ . Then for each $j_1, j_2 \in \mathbb {N}$ and $(x_1, x_2) \in X\times X$ , we have

(4.9) $$ \begin{align} \int_{X \times X} f_{j_1} (y_1) f_{j_2}(y_2) {~\mathrm{d}} \lambda_{(x_1, x_2)} (y_1, y_2) &= \int_K \int_{X \times X} f_{j_1} (y_1) f_{j_2}(y_2) {~\mathrm{d}} (\eta_{k\pi(x_1)} \times\eta_{k\pi(x_2)})(y_1,y_2) {~\mathrm{d}} m_K(k) \nonumber \\ &= \int_K \int_{X} f_{j_1} (y_1) {~\mathrm{d}} \eta_{k\pi(x_1)} (y_1) \int_{X} f_{j_2}(y_2) {~\mathrm{d}} \eta_{k\pi(x_2)} (y_2) {~\mathrm{d}} m_K(k) \nonumber \\ & = \int_K \psi_{j_1}(k\pi(x_1)) \psi_{j_1}(k\pi(x_1)) {~\mathrm{d}} m_K(k) \end{align} $$

where the last equality follows using (2.1). After all, combining (4.8) and (4.9), we get that for each $j_1, j_2 \in \mathbb {N}$ ,

$$ \begin{align*} \frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (T_g \times T_g) (\mathbb{E}_{\mu}(f_{j_1} \: | \: Z) \otimes \mathbb{E}_{\mu}(f_{j_2} \: | \: Z)) \to \Big[(x_1, x_2) \mapsto \int_{X \times X} f_{j_1} \otimes f_{j_2}{~\mathrm{d}} \lambda_{(x_1, x_2)} \Big] \end{align*} $$

as $N\to \infty $ in $L^2(X \times X, \mu \times \mu )$ . Now, since the family $f_{j_1}\otimes f_{j_2}$ is countable, using (ii) and a diagonal argument as in Step 1, one can find a sub-Følner sequence of $\Psi $ , which by abuse of notation we again denote by $\Psi $ , and a set $W_2 \subset X \times X$ with $(\mu \times \mu )(W_2) =1$ such that for all $(x_1, x_2)\in W_2$ and $j_1, j_2 \in \mathbb {N}$ ,

(4.10) $$ \begin{align} \lim_{N\to\infty}\frac{1}{|\Psi_N|} \sum_{g\in \Psi_N} (T_g \times T_g)(\mathbb{E}_{\mu}(f_{j_1}\: | \: Z)\otimes \mathbb{E}_{\mu}(f_{j_2}\: | \: Z)) = \int_{X \times X} f_{j_1} \otimes f_{j_2}{~\mathrm{d}} \lambda_{(x_1, x_2)} \end{align} $$

in $L^2(X\times X, \lambda _{(x_1, x_2)})$ .

Let $W = W_1 \cap W_2$ . Then $(\mu \times \mu )(W) =1$ , and combining (4.7) and (4.10), we get that for all $(x_1, x_2) \in W$ and all $j_1, j_2 \in \mathbb {N}$ ,

$$ \begin{align*}\lim_{N\to\infty}\frac{1}{|\Psi_N|}\sum_{g\in\Psi_N} (T_g\times T_g)(f_{j_1} \otimes f_{j_2}) = \int_{X\times X} f_{j_1} \otimes f_{j_2} {~\mathrm{d}}\lambda_{(x_1,x_2)}\end{align*} $$

in $L^2(X\times X,\lambda _{(x_1,x_2)})$ , which was to be proved.

To conclude the proof, we are left with showing (4.6). To this end, using the invariance of $m_K$ , for any $g\in G$ , we have that

$$ \begin{align*} \lambda_{(T_gx_1,T_gx_2)} = \int_Z \eta_{k\alpha(g)\pi(x_1)}\times \eta_{k\alpha(g)\pi(x_2)}{~\mathrm{d}} m_K(k) & = \int_Z \eta_{k\pi(x_1)}\times \eta_{k\pi(x_2)}{~\mathrm{d}} m_K(k\alpha(g)^{-1}) \\ & = \int_Z \eta_{k\pi(x_1)}\times \eta_{k\pi(x_2)}{~\mathrm{d}} m_K(k) = \lambda_{(x_1,x_2)}. \end{align*} $$

The proof of the theorem is now complete.

Proof. Let $F: K\to \mathcal {F}(X)$ given by $F(k) = \mathtt {{supp}}(\eta _{kH})$ . The natural projection $p:K\to Z$ is continuous, hence Borel measurable, the map $z\mapsto \eta _z$ is Borel measurable, as $(\eta _z)_{z\in Z}$ is a disintegration, and by Lemma 4.11, the map $\nu \mapsto \mathtt {{supp}}(\nu )$ is also Borel measurable; thus, we obtain that their composition F is also Borel measurable. By Lusin’s theorem [Reference Aliprantis and Border2, Theorem 12.8], for any $j\in \mathbb {N}$ , there exists a closed $K_j\subset K$ with

(4.12) $$ \begin{align} m_K(K_j)>1-2^{-j} \end{align} $$

such that $F|_{K_j}$ is continuous. For $j\in \mathbb {N}$ , using the fact that K, and so $K_j$ , is compact, we obtain that $F|_{K_j}$ is uniformly continuous. Therefore, for any $j\in \mathbb {N}$ , there exists some $\delta _j>0$ such that for any $k_1,k_2\in K_j$ we have

$$ \begin{align*}d_K(k_1,k_2)\leq \delta_j \Longrightarrow \mathtt{{D}}(F(k_1),F(k_2)) <\frac{1}{j}.\end{align*} $$

Fix $j \in \mathbb {N}$ . Then by the invariance of $m_K$ , there is $c_j>0$ such that for all $k \in K$ , $m_K(\mathtt {{B}}_K(k, \delta _j)) = c_j$ . The regularity of the measure $m_K$ implies that there is a compact set $C_j \subset \mathtt {{B}}_K(e_K, \delta _j)$ such that $m_K(C_j) \geq c_j - \frac {c_j}{2^j}$ . Now, by Urysohn’s lemma, there is a continuous function $f_j: K \to [0, 1]$ such that $f_j = 1$ on $C_j$ and $f_j =0 $ outside $\mathtt {{B}}_K(e_K, \delta _j)$ .

Consider the set

$$ \begin{align*}W_j = \bigg\{k \in K_j : \int_{K_j} f_j (w k^{-1}) {~\mathrm{d}} m_K(w) \geq \bigg(1-\frac{1}{j} \bigg) \int_K f_j (w ) {~\mathrm{d}} m_K(w)\bigg\}.\end{align*} $$

Note that

$$ \begin{align*}\int_K f_j (w) {~\mathrm{d}} m_K(w) \geq m_K(C_j) \geq c_j - \frac{c_j}{2^j}>0.\end{align*} $$

Then we can consider the function

$$ \begin{align*} \chi_j : K \to [0,1], \: \: \chi_j (k) = \frac{\int_{K_j} f_j (w k^{-1}) {~\mathrm{d}} m_K(w)}{\int_K f_j (w) {~\mathrm{d}} m_K(w)}, \end{align*} $$

and moreover, we let

$$ \begin{align*}A_j =\bigg \{ k \in K : \chi_j (k) \geq 1 - \frac{1}{j} \bigg\}.\end{align*} $$

Then we see that

(4.13) $$ \begin{align} W_j = K_j\cap A_j. \end{align} $$

We will show that the set $W_j$ is closed. Using the dominated convergence theorem and the fact that $f_j$ is continuous, one can show that if $(k_{\ell })_{\ell \in \mathbb {N}} $ is a sequence in K and $k_{\ell } \to k$ , then $\int _{K_j} f_j (w k_{\ell }^{-1}) {~\mathrm {d}} m_K(w) \to \int _{K_j} f_j (w k^{-1}) {~\mathrm {d}} m_K(w)$ , and this proves the continuity of $\chi _j$ . As a result, the set $A_j$ is a closed subset of K, and since $K_j$ is also closed, it follows that $W_j$ is closed.

Now, using Fubini’s theorem and the invariance of $m_K$ , we deduce that

and then we have that

$$ \begin{align*}1-\frac{1}{2^j}< \int_{A_j} \chi_j(k) {~\mathrm{d}} m_K(k) + \int_{A_j^{\text{c}}} \chi_j(k) {~\mathrm{d}} m_K(k) \leq m_K(A_j) + \bigg(1-\frac{1}{j}\bigg)m_K(A_j^{\text{c}}) = 1 - \frac{1}{j} + \frac{m_K(A_j)}{j},\end{align*} $$

which gives that

(4.14) $$ \begin{align} m_K(A_j)>1-\frac{j}{2^j}. \end{align} $$

Combining (4.12), (4.13) and (4.14), we obtain that

$$ \begin{align*}\sum_{j\in\mathbb{N}} m_K(K\setminus W_j) = \sum_{j\in\mathbb{N}} m_K(K\setminus(K_j\cap A_j)) < \sum_{j\in\mathbb{N}} \frac{1+j}{2^j} < \infty.\end{align*} $$

Let $W = \bigcup _{J\in \mathbb {N}}\bigcap _{j\geq J} W_j$ . It follows by the Borel-Cantelli lemma, using the last equation above, that $m_K(W) = 1$ . Now let $L:=\{x\in X\colon x\in \mathtt {{supp}}(\eta _{\pi (x)})\}\cap \pi ^{-1}(p(W))$ . Observe that $p(W)= \bigcup _{J\in \mathbb {N}} p\big (\bigcap _{j\geq J} W_j\big )$ . For each $J \in \mathbb {N}$ , $\bigcap _{j\geq J} W_j$ is a closed subset of K and thus is compact, and since p is continuous, we get that $p\big (\bigcap _{j\geq J} W_j\big )$ is also compact; thus it is Borel measurable. As a result, we get that $p(W)$ is indeed a Borel subset of Z. In addition, $p^{-1}(p(W))\supset W$ , and since $m_K(W)=1$ , we have that $m_K(p^{-1}(p(W))=1$ . Therefore, $m(p(W))=1$ and hence, $\mu (\pi ^{-1}(p(W)))=1$ . Then, in view of Lemma 4.12, it follows that L is a Borel subset of X and $\mu (L)=1$ .

We now show that elements of L satisfy (4.11), and this will conclude the proof. Let $x\in L= \{x\in X\colon x\in \mathtt {{supp}}(\eta _{\pi (x)})\} \cap \pi ^{-1}(p(W))$ . Then $x\in \pi ^{-1}(p(W))$ so there is $w\in W$ such that $\pi (x)=p(w)=wH$ . Let U be an open neighborhood of x. Then we have that there exists $J\in \mathbb {N}$ such that for any $j\geq J$ , $w\in W_j$ and $\mathtt {{B}}(x,\tfrac {1}{j})\subset U$ .

We now claim that

$$ \begin{align*}\mathtt{{B}}_K(w,\delta_j)\cap K_j\subset \{k\in K\colon \eta_{kH}(U)>0\}.\end{align*} $$

To prove this, we let $w'\in \mathtt {{B}}_K(w,\delta _j)\cap K_j$ . Then $d_K(w',w)<\delta _j$ , and so, $\mathtt {{D}}(F(w'),F(w)) < \tfrac {1}{j}$ . Notice now that $F(w) = \mathtt {{supp}}(\eta _{wH}) = \mathtt {{supp}}(\eta _{\pi (x)})$ , and so, $x\in F(w)$ , as $x\in L$ . Then, by the definition of the Hausdorff metric, there exists $x'\in F(w')$ with $d_X(x,x')<\tfrac {1}{j}$ , and so, $x'\in U$ , which, combined with the fact that $x'\in F (w')$ , yields $U\cap F(w')\neq \emptyset $ . It follows that $\eta _{w'H}(U)>0$ .

It follows from the above claim that

(4.15)

The denominator in the right-most term in (4.15) is smaller or equal to $\frac {2^j}{2^j -1} m_K(C_j)$ , which then is smaller or equal to $\frac {2^j}{2^j -1} \int _K f_j(u) {~\mathrm {d}} m_K(u)$ , and therefore, the expression in (4.15) is greater or equal to

(4.16)

Observe that for all $u \in K$ ,

, so we have that

Combining the last equation with (4.15) and (4.16), we get that

$$ \begin{align*} \frac{m_K(\{k\in K\colon \eta_{kH}(U)>0\}\cap\mathtt{{B}}_K(w,\delta_j))} {m_K(\mathtt{{B}}_K(w,\delta_j))} &\geq \frac{\int_{K_j} f_j(uw^{-1}) {~\mathrm{d}} m_K(u)}{\int_K f_j(u) {~\mathrm{d}} m_K(u)} \bigg(1-\frac{1}{2^j}\bigg) \\ &\geq \bigg(1- \frac{1}{j}\bigg) \bigg(1-\frac{1}{2^j}\bigg), \end{align*} $$

where the least inequality is due to the fact that $w \in W_j$ . Then, taking the limit as $j \to \infty $ , we obtain that

$$ \begin{align*}\lim_{j\to\infty}\frac{m_K(\{k\in K\colon \eta_{kH}(U)>0\}\cap \mathtt{{B}}_K(w,\delta_j))} {m_K(\mathtt{{B}}_K(w,\delta_j))} = 1,\end{align*} $$

and this concludes the proof.

Proof of Theorem 4.9.

Let $S=\{(x_1,x_2)\in X\times X\colon (x_1,x_2)\in \mathtt {{supp}}(\lambda _{(x_1,x_2)})\}$ . By Lemma 4.11, S is a Borel subset of $X\times X$ . Consider a sequence $\delta _j\to 0$ such that Proposition 4.13 is satisfied, and let $L\subset X$ be the set of $x\in X$ that satisfy (4.11). By Proposition 4.13, $\mu (L)=1$ . Following the argument in [Reference Kra, Moreira, Richter and Robertson15, Proposition 3.11], we will show that $\sigma (L\times L)=1$ and $L\times L\subset S$ . Consequently, we will have that $\sigma (S)=1$ , concluding the proof.

We start by showing that $\sigma (L\times L)=1$ . We write $L\times L = (L\times X)\cap (X\times L)$ , and so it is enough to show that both sets in this intersection have full measure $\sigma $ . By Proposition 4.5 (i), we have that $\sigma (L\times X)=\pi _1\sigma (L)=\mu (L)=1$ . By Proposition 4.5 (ii), the measure $\pi _2\sigma $ is absolutely continuous with respect to $\mu $ , and since $\mu (L)=1$ , it follows that $\sigma (X\times L)=\pi _2\sigma (L)=1$ .

To conclude the proof, we show that $L\times L\subset S$ . Let $(x_1,x_2)\in L\times L$ . To show that $(x_1,x_2)\in S$ , it is enough to show that for all open neighborhoods $U_1, U_2$ of $x_1, x_2$ , we have $\lambda _{(x_1,x_2)}(U_1\times U_2)>0$ .

Let $U_1,U_2$ be open neighborhoods of $x_1,x_2$ , respectively. By writing $\lambda _{(x_1,x_2)}= \int _K \eta _{kH}\times \eta _{kk_1^{-1} k_2 H} {~\mathrm {d}} m_K(k)$ , where $k_1,k_2\in K$ are such that $\pi (x_1)=k_1H$ , $\pi (x_2)=k_2H$ , we see that it suffices to show that the set $W=W(k_1,k_2):=\{k\in K: \eta _{kH}(U_1)>0 \text { and } \eta _{kk_1^{-1}k_2H }(U_2)>0\}$ has positive measure $m_K$ , for some choice of the $k_1,k_2$ as above. By Proposition 4.13, we can choose the elements $k_1, k_2 \in K$ such that

(4.17) $$ \begin{align} \frac{m_K(\{k\in K:\eta_{kH}(U_1)>0\}\cap\mathtt{{B}}_K(k_1,\delta))} {m_K(\mathtt{{B}}_K(k_1,\delta))} \geq \frac{3}{4} \end{align} $$

and

(4.18) $$ \begin{align} \frac{m_K(\{k\in K:\eta_{kH}(U_2)>0\}\cap\mathtt{{B}}_K(k_2,\delta))} {m_K(\mathtt{{B}}_K(k_2,\delta))} \geq \frac{3}{4}, \end{align} $$

for some $\delta>0$ . Now using (4.18) along with the bi-invariance of both $d_K$ and $m_K$ , we have that

(4.19) $$ \begin{align} & \frac{m_K(\{k \in K:\eta_{kk_1^{-1}k_2 H} (U_2)>0\}\cap\mathtt{{B}}_K(k_1,\delta))}{m_K(\mathtt{{B}}_K(k_1,\delta))} \notag \\ & \quad = \frac{m_K(\{k \in K:\eta_{kH}(U_2)>0\}\cdot k_2^{-1}k_1\cap\mathtt{{B}}_K(k_2,\delta)\cdot k_2^{-1}k_1)} {m_K(\mathtt{{B}}_K(k_2,\delta)\cdot k_2^{-1}k_1)} \notag \\ & \quad = \frac{m_K(\{k \in K:\eta_{kH}(U_2)>0\}\cap\mathtt{{B}}_K(k_2,\delta))} {m_K(\mathtt{{B}}_K(k_2,\delta))} \geq \frac{3}{4}. \end{align} $$

Combining (4.17) and (4.19) yields $\frac {m_K(W)}{m_K(\mathtt {{B}}_K (k_1, \delta ))} \geq \tfrac {1}{2}$ . This implies that $m_K(W)>0$ and concludes the proof.

Proof. By the definition of $\sigma $ and the fact that $z\mapsto \eta _z$ is a disintegration, it follows that for $\sigma $ -almost every $(x_1,x_2)$ , we have $\pi (x_1)=w\pi (a)$ and $\pi (x_2)=w\pi (x_1)$ , for some $w\in K$ . For any such $(x_1,x_2)$ , using the right invariance of $m_K$ , we have

$$ \begin{align*} \lambda_{(x_1,x_2)} & = \int_{X\times X} \eta_{k\pi(x_1)}\times\eta_{k\pi(x_2)}{~\mathrm{d}} m_K(k) = \int_{X\times X} \eta_{kw\pi(a)}\times\eta_{kw\pi(x_1)}{~\mathrm{d}} m_K(k) \\ & = \int_{X\times X} \eta_{k\pi(a)}\times\eta_{k\pi(x_1)}{~\mathrm{d}} m_K(kw^{-1}) = \int_{X\times X} \eta_{k\pi(a)}\times\eta_{k\pi(x_1)}{~\mathrm{d}} m_K(k) = \lambda_{(a,x_1)}. \end{align*} $$

This concludes the proof.

Proof of Theorem 4.10.

Consider the measure $\nu _a := \delta _a \times \mu $ , where $\delta _a$ denotes the Dirac mass at a.

Claim. There exists some Følner sequence $\Psi $ such that

$$ \begin{align*}\nu_a(\{(x_0,x_1)\in X\times X: (x_0,x_1)\in\mathtt{{gen}}(\lambda_{(x_0,x_1)},\Psi)\})=1.\end{align*} $$

Proof of Claim.

In this proof, we follow the argument used in the proof of [Reference Kra, Moreira, Richter and Robertson14, Theorem 7.6]. Apply Lemma 2.8 for the ergodic decomposition $(x_0,x_1)\mapsto \lambda _{(x_0,x_1)}$ to obtain a Følner sequence $\Phi $ such that

(4.20) $$ \begin{align} (\mu\times\mu)(\{(x_0,x_1)\in X\times X: (x_0,x_1)\in\mathtt{{gen}}(\lambda_{(x_0,x_1)},\Phi)\})=1. \end{align} $$

Consider the map $X \to M(X)$ , $s \mapsto \nu _s =\delta _s \times \mu $ , where $\delta _s $ denotes the Dirac mass at s. It is quite straightforward to see that $s\mapsto \nu _s$ is a continuous disintegration of $\mu \times \mu $ and, moreover, satisfies

(4.21) $$ \begin{align} (T_g\times T_g)\nu_s = \nu_{T_gs} \end{align} $$

for any $s\in X$ . By (4.20) and the fact that $s\mapsto \nu _s$ is a disintegration of $\mu \times \mu $ , it follows that for $\mu $ -almost every $s\in X$ , $\nu _s$ -almost every $(x_0,x_1)\in X\times X$ is in $\mathtt {{gen}}(\lambda _{(x_0,x_1)},\Phi )$ . Fix $b\in \mathtt {{supp}}(\mu )$ . Then

(4.22) $$ \begin{align} \nu_b\text{-almost every } (x_0,x_1) \text{ is in } \mathtt{{gen}}(\lambda_{(x_0,x_1)},\Phi). \end{align} $$

By Lemma 2.5, there exists some sequence $(g_n)_{n\in \mathbb {N}}$ in G such that $T_{g_n}a\to b$ , and now by continuity of the disintegration $s\mapsto \nu _s$ , combined with (4.21), it follows that

(4.23) $$ \begin{align} (T_{g_n}\times T_{g_n})\nu_a\to\nu_b. \end{align} $$

Now, let $\mathcal {F}=(F_k)_{k\in \mathbb {N}}$ be a dense subset of $C(X\times X)$ and for $k,N\in \mathbb {N}$ , consider the sets

$$ \begin{align*}A_{k,N}=\bigg\{(x_0,x_1)\in X\times X: \max_{1\leq j\leq k}\bigg|\frac{1}{|\Phi_N|}\sum_{g\in\Phi_N} F_j((T_g\times T_g)(x_0,x_1)) - \int_{X\times X}F_j{~\mathrm{d}}\lambda_{(x_0,x_1)}\bigg|\leq\frac{1}{k}\bigg\}.\end{align*} $$

Using (4.22) and the monotone convergence theorem, it follows that for any $k\in \mathbb {N}$ , there exists some $N(k)\in \mathbb {N}$ , such that

(4.24) $$ \begin{align} \nu_b(A_{k,N(k)})\geq 1-2^{-k}. \end{align} $$

For $k,N\in \mathbb {N}$ , we define

$$ \begin{align*}B_{k,N}=\bigg\{(x_0,x_1)\in X\times X: d_{X\times X}((x_0,x_1),A_{k,N})<\frac{1}{k}\bigg\},\end{align*} $$

where $d_{X\times X}$ is the metric on $X\times X$ . Then for k sufficiently large, we have

(4.25) $$ \begin{align} \max_{1\leq j\leq k} \bigg|\frac{1}{|\Phi_{N(k)}|}\sum_{g\in\Phi_{N(k)}}F_j((T_g\times T_g)(x_0,x_1))-\int_{X\times X}F_j{~\mathrm{d}}\lambda_{(x_0,x_1)}\bigg| \leq \frac{2}{k}, \end{align} $$

for any $(x_0,x_1)\in B_{k,N(k)}$ . The sets $A_{k,N}$ are open, while the sets $B_{k,N}$ are closed subsets of $X\times X$ , and also $A_{k,N(k)}\subset B_{k,N(k)}$ , so by Urysohn’s lemma, we can find, for all $k\in \mathbb {N}$ , continuous functions $f_k:X\times X\to [0,1]$ such that

$$ \begin{align*}f_k|_{A_{k,N(k)}}=1 \quad \text{and}\quad f_k|_{(X\times X)\setminus B_{k,N(k)}}=0.\end{align*} $$

By (4.22), for each $k\in \mathbb {N}$ , there exists $n(k)\in \mathbb {N}$ such that

$$ \begin{align*}\bigg|\int_{X\times X}(T_{g_{n(k)}}\times T_{g_{n(k)}})f_k{~\mathrm{d}}\nu_a - \int_{X\times X}f_k{~\mathrm{d}}\nu_b\bigg| \leq 2^{-k}. \end{align*} $$

Let $(h_k)_{k\in \mathbb {N}}$ be the subsequence of $(g_n)_{n\in \mathbb {N}}$ defined by $h_k=g_{n(k)}$ , for $k\in \mathbb {N}$ . Then, by the equation above, we have

$$ \begin{align*} \nu_a\big((T_{h_k}\times T_{h_k})^{-1}B_{k,N(k)}\big) &\geq \int_{X\times X}(T_{h_k}\times T_{h_k})f_k{~\mathrm{d}}\nu_a \geq \int_{X\times X}f_k{~\mathrm{d}}\nu_b - 2^{-k} \\ &\geq \nu_b(A_{k,N(k)}) - 2^{-k} \geq 1-2^{-k+1} \quad \text{(by (4.24))}, \end{align*} $$

for any $k\in \mathbb {N}$ . Therefore, it holds that

$$ \begin{align*}\sum_{k\geq1}\nu_a\big((T_{h_k}\times T_{h_k})^{-1}B_{k,N(k)}\big) =\infty,\end{align*} $$

and then, by the Borel-Cantelli lemma, it follows that $\nu _a$ -almost every $(x_0,x_1)\in \mathtt {{supp}}(\nu _a)$ belong to all, but finitely many, sets $(T_{h_k}\times T_{h_k})^{-1}B_{k,N(k)}$ . Then by (4.25), it follows that for $\nu _a$ -almost every $(x_0,x_1)\in X\times X$ and k sufficiently large, we have

$$ \begin{align*}\max_{1\leq j\leq k}\bigg|\frac{1}{|\Psi_k|}\sum_{g\in\Psi_k} F_j((T_g\times T_g)(x_0,x_1))- \int_{X\times X} F_j{~\mathrm{d}}\lambda_{(T_{h_k}\times T_{h_k})(x_0,x_1)}\bigg| \leq \frac{2}{k},\end{align*} $$

where $\Psi $ is the Følner sequence defined by $\Psi _k = \Phi _{N(k)}h_k$ , for $k\in \mathbb {N}$ . Using (4.6), the above equation becomes

$$ \begin{align*}\max_{1\leq j\leq k}\bigg|\frac{1}{|\Psi_k|}\sum_{g\in\Psi_k} F_j((T_g\times T_g)(x_0,x_1))- \int_{X\times X} F_j{~\mathrm{d}}\lambda_{(x_0,x_1)}\bigg| \leq \frac{2}{k}.\end{align*} $$

Sending $k\to \infty $ , we have shown that

$$ \begin{align*}\lim_{k\to\infty}\frac{1}{|\Psi_k|}\sum_{g\in\Psi_k} F((T_g\times T_g)(x_0,x_1))= \int_{X\times X} F{~\mathrm{d}}\lambda_{(x_0,x_1)}\end{align*} $$

holds for $\nu _a$ -almost every $(x_0,x_1)\in X\times X$ and for any $F\in \mathcal {F}$ . An approximation argument concludes the proof of the Claim.