3 shortened the title

2 made a shorter version of the question

Quick version of the question. Let $(X, \mu)$ be a probability measure space and let $Z$, the group of integers, act on $X$ in a measure preserving way. How can I decompose $X$ into ergodic componenets? More precisely, can $X$ be equivariantly decomposed into a countable union of subspaces $U_i$, each of which is isomorphic to a product $A_i\times B_i$, such that action on $U_i$ is a product of an ergodic action on $A_i$ and the trivial action on $B_i$?

One can ask the same question also for groups other than integers.

Mymotivation

I'm currently learning basics of ergodic theory. More precisely, I'm interested in the notion of cost. Let me recall it for group actions: Let a countable discrete group $G$ act on a probability measure space $(X,\mu)$ in a free and probability measure preserving (pmp) manner. Call the action $\rho$. Let $\mathcal R(\rho)$ be the equivalence realtion on $X$ given by $\rho$ (i.e. two points of $X$ are equivalent iff there's a group element which sends one point to the other). Let $F=(U_i,g_i)_{i=1}^\infty$ be a countable family of pairs, where each $U_i$ is a measurable set, and each $g_i$ is an element of $G$. Let $\mathcal R(F)$ be the equivalence relation on $X$ generated by the relation $x \sim y$ iff for some $i$ we have $x\in U_i$ and $\rho(g_i)(x)=y$. Define $$cost(F) = \sum \mu(U_i),$$ and let cost of the action $\rho$ be the infimum of numbers $cost(F)$ over all families $F$ such that $\mathcal R(F) = \mathcal R(\rho)$, perhaps after restricting both relations to subsets of measure $1$.

Theorem. Let $\rho$ be a free pmp action of $\mathbb Z \times H$, where $H$ is any countable group. Then $cost(\rho)=1$

Suppose first that restriction of the action $\rho$ to $\mathbb Z$ is ergodic. Fix $\varepsilon$. Then for the family $F$ choose pairs $(X, t), (A_1, h_1), (A_2,h_2) \ldots$, where $t$ is a generator of $\mathbb Z$, $h_i$ is an enumeration of elements of $H$, and $A_i$ is any set such that $\mu(A_i)= \frac{\varepsilon}{2^i}$.

Clearly $cost(F) \le 1 + \varepsilon$, so it's enough to see that $\mathcal R(F) = \mathcal R(\rho)$. Take a point $x$ of $X$ and fix $h_i\in H$. We're gonna show that, with probability $1$, $x$ is in relation with $\rho(h_i)(x)$. By the ergodic theorem, since we assume action of $\mathbb Z$ is ergodic, with probability $1$ for some $j$ we have $\rho(t^j)(x)\in A_i$, so we have $x \sim \rho(t^j)(x) \sim \rho(h_it^j)(x) \sim \rho(h_i)(x)$.

When I heard the argument it wasn't even mentioned that we assume that restricion to $\mathbb Z$ is ergodic. Intuitively it's clear what to do - choose $A_i$ more cleverly, "perpendicular to ergodic components of $\mathbb Z$".

Question. Which theorem from ergodic theory allows to make this choice of $A_i$ "perpendicular to ergodic components" precise?

1

Which theorem about decomposition of a dynamical system into ergodic componenents makes this proof of a theorem about cost correct?

I'm currently learning basics of ergodic theory. More precisely, I'm interested in the notion of cost. Let me recall it for group actions: Let a countable discrete group $G$ act on a probability measure space $(X,\mu)$ in a free and probability measure preserving (pmp) manner. Call the action $\rho$. Let $\mathcal R(\rho)$ be the equivalence realtion on $X$ given by $\rho$ (i.e. two points of $X$ are equivalent iff there's a group element which sends one point to the other). Let $F=(U_i,g_i)_{i=1}^\infty$ be a countable family of pairs, where each $U_i$ is a measurable set, and each $g_i$ is an element of $G$. Let $\mathcal R(F)$ be the equivalence relation on $X$ generated by the relation $x \sim y$ iff for some $i$ we have $x\in U_i$ and $\rho(g_i)(x)=y$. Define $$cost(F) = \sum \mu(U_i),$$ and let cost of the action $\rho$ be the infimum of numbers $cost(F)$ over all families $F$ such that $\mathcal R(F) = \mathcal R(\rho)$, perhaps after restricting both relations to subsets of measure $1$.

Theorem. Let $\rho$ be a free pmp action of $\mathbb Z \times H$, where $H$ is any countable group. Then $cost(\rho)=1$

Suppose first that restriction of the action $\rho$ to $\mathbb Z$ is ergodic. Fix $\varepsilon$. Then for the family $F$ choose pairs $(X, t), (A_1, h_1), (A_2,h_2) \ldots$, where $t$ is a generator of $\mathbb Z$, $h_i$ is an enumeration of elements of $H$, and $A_i$ is any set such that $\mu(A_i)= \frac{\varepsilon}{2^i}$.

Clearly $cost(F) \le 1 + \varepsilon$, so it's enough to see that $\mathcal R(F) = \mathcal R(\rho)$. Take a point $x$ of $X$ and fix $h_i\in H$. We're gonna show that, with probability $1$, $x$ is in relation with $\rho(h_i)(x)$. By the ergodic theorem, since we assume action of $\mathbb Z$ is ergodic, with probability $1$ for some $j$ we have $\rho(t^j)(x)\in A_i$, so we have $x \sim \rho(t^j)(x) \sim \rho(h_it^j)(x) \sim \rho(h_i)(x)$.

When I heard the argument it wasn't even mentioned that we assume that restricion to $\mathbb Z$ is ergodic. Intuitively it's clear what to do - choose $A_i$ more cleverly, "perpendicular to ergodic components of $\mathbb Z$".

Question. Which theorem from ergodic theory allows to make this choice of $A_i$ "perpendicular to ergodic components" precise?