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.
My motivation
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?
Best Answer
Answer to the quick version. Yes it is true as soon as $(X,\mu)$ is a Lebesgue space. Beware that the transformation on the product $A_i\times B_i$ is not necessarily a true product, but instead it is a skew-product of the form $(x,y)\mapsto (T_x(y),y)$. This follows from the ergodic decomposition theorem, together with the classification of measurable partitions.
Recall that if T is an invertible measurable transformation acting on a Lebesgue space X, then there is a measurable partition $C_i$ (which may have uncountably many elements) and probability measures $\mu_i$ on $C_i$ such that all $C_i$ are invariant by T, T is ergodic w.r.t $\mu_i$ and $\mu$ is obtained by integrating the $\mu_i$. $$\mu(A) = \int_X \mu_i(A) d\mu$$ There are two kinds of ergodic components $C_i$. The one of positive measure, there are at most countably many such components. Let us remove these components from $X$, together with the periodic points, which are easily dealt with. Rohlin structure theorem on measurable partitions (1947) now says that there is a isomorphism between $([0,1]\times [0,1], \lambda)$ and $(X,\mu)$ such that the pullback of the measurable partition $(C_i)$ is the decomposition into horizontal lines $([0,1]\times \{i\})_{i\in [0,1]}$. A reference is the book of Parry, "entropy generators in ergodic theory".
Here is how the ergodic decomposition is often used. If it happens that a result is true for an ergodic transform, then it is true for an arbitrary transform in restriction to its ergodic components, and you (may) recover the result on the whole space $X$ just by using the integral formula given above. A reference for the ergodic decomposition for countable groups action is Glasner, "ergodic theory via joinings" th. 3.22.
Finally the result you are alluding in your last question is a section theorem. Given a measure preserving transform between two Lebesgue spaces X and Y, there is a section from Y to X, up to null sets, and some warning is in order here because this is not true in the Borel category. I think this is again due to Rohlin, and it can be deduced from its structure theorem for measurable partitions. Have a look at the book of Parry, but really this is overkill.
EDIT: following the comments of R.W., here is a counterexample to having a true product, instead of just a skew-product. On $[0,1]\times [0,1]$ take $(x,y)\mapsto (x+y\ \ mod\ \ 1,y)$, together with Lebesgue measure. The restrictions to the fibers $[0,1]\times \{y\}$ are ergodic for a.e. y, and give uncountably many different isomorphic systems, as can be shown by looking at their spectra.