Conservative system

Remarks

Equivalently, $T$ has no wandering set of positive measure, where $W \in \mathcal{A}$ is wandering if the sets $T^{-n}(W)$, $n \ge 0$, are pairwise disjoint. For a conservative $T$, almost every point of each $A \in \mathcal{A}$ returns to $A$ infinitely often. The Hopf decomposition splits $X$, up to null sets, into an invariant part on which $T$ is conservative and a part that is a union of wandering sets.

Example: every measure-preserving map of a finite measure space is conservative, by the Poincaré recurrence theorem. For instance, an irrational rotation of the circle with Lebesgue measure is conservative.

Non-example: the translation $T(x) = x + 1$ on $\R$ with Lebesgue measure preserves the measure, but for $A = [0, 1)$ one has $T^{-n}(A) = [-n, 1 - n)$, which is disjoint from $A$ for every $n \ge 1$. So $\mu(A \cap T^{-n}(A)) = 0$ for all $n \ge 1$, and $T$ is not conservative.