I was asking if it is possible to extend the definition of topological Friedland entropy for $\mathbb{Z}^d$ continuos actions to measure preserving actions.

The topologica Friedland entropy is constructed as follows. Let $T$ be a $\mathbb{Z}^d$ action on a compact topological space $X$. Consider then teh sequence space $$ \mathcal{X}=\mathcal{X}_T=\left\{(x_n)_{n\in \mathbb{N}} \in \prod_{n \in \mathbb{N}}X: T_i(x_n)=x_{n+1} \text{ for some } i =1,\dots,d\right\}. $$ This is a compact space. There is a natural shift on $\mathcal{X}$, given by $\sigma((x_n)_{n\in \mathbb{N}})=(x_{n+1})_{n \in \mathbb{N}}$. We can see that $\sigma$ is continuos and then define the entropy of the action $T$, $ent_{top}(T)=h_{top}(\sigma)$.

This definition is motivated from the fact that when $d=1$, $h_{top}(\sigma)=h_{top}(T)$. This comes from the fact that the map $\phi:X \ni x \mapsto orb(x)=(T^{i}(x))_{i \in \mathbb{N}}$ is a topological conjugacy map.

This definition can be given also for metric non compact spaces, considering Bowen topological entropy.

Is it possible to define an extension of the metric entropy of Kolmogorov using this method? Consider the case of measure preserving actions on probability spaces, the problem I am facing are the following

- How can define a probability measure on $\mathcal{X}$? I have the product probability on $\prod_{n \in \mathbb{N}}X$, but i cannot see a way to restric it $\mathcal{X}$.
- How can I see that $\sigma$ is measure preserving?
- How can I see that those definition are the same in the case $d=1$, using a conjugacy argument?

Thank you for you suggestions!