# partition into the orbits of a dynamical system

Let $T$ be a measure-preserving invertible transformation of a Lebesgue space, and let $P$ be the partition of the Lebesgue space into the orbits of $T$. 1) Is it true that $P$ is nonmeasurable (in the sense of Rokhlin) when $T$ is ergodic ? Why ? 2) If true, is it a necessary and sufficient condition for ergodicity ? 3) Is it true that the (complete) $\sigma$-field generated by $P$ is the $\sigma$-field of $T$-invariant sets ?

-

Although it appears you've already settled matters with the information in Jon's answer, I'll offer a quick summary and elaboration.

Let $(X,\mathcal{B},\mu)$ be a Lebesgue space (set + $\sigma$-algebra + probability measure) and $P$ the partition into orbits $\mathcal{O}(x) = \{T^n x \mid n\in \mathbb{Z}\}$ for an invertible measure-preserving transformation $T$.

For (3), it's exactly as you say: $\mathcal{B}$ contains each orbit $\mathcal{O}(x)$, and since a set $A$ is $T$-invariant if and only if it is a union of complete $T$-orbits, we get that the $\sigma$-algebra generated by $P$ is precisely the collection of $T$-invariant sets. The completed $\sigma$-algebra generated by $P$ is the collection of sets that are $T$-invariant mod $0$.

Consequently the answer to (1) is yes unless a single orbit carries full measure: ergodicity implies that every element of the $\sigma$-algebra generated by $P$ has measure $0$ or $1$, and consequently this $\sigma$-algebra is equivalent mod $0$ to the trivial $\sigma$-algebra. Thus for an ergodic transformation, $P$ is measurable if and only if a single partition element has full measure, which happens exactly when $\mu$ is supported on a single periodic orbit.

Finally, for (2), you observe correctly that non-measurability of $P$ follows as soon as $\mu$ has an ergodic component that is not a periodic orbit. For a concrete example, one may consider the map $T\colon [0,1]\to[0,1]$ that takes $x$ to $2x \pmod 1$, and the measure $\mu = \frac 12(\delta_0 + \lambda)$, where $\delta_0$ is the point mass on the fixed point at $0$, and $\lambda$ is Lebesgue measure. In this case $\mu$ is non-ergodic but $P$ is non-measurable. Or, if you prefer a completely non-atomic example, you can let $\nu_p$ denote the $T$-invariant measure on $[0,1]$ that comes from the $(p,1-p)$-Bernoulli measure on the full two-shift (where $p\in (0,1)$) and let $\mu$ be any convex combination of $\nu_p$ and $\nu_q$ for $p\neq q$.

-
Thank you very much :) –  Stéphane Laurent Feb 13 '12 at 8:41
Thanks for the notes, and the answer, Vaughn! –  Jon Bannon Feb 13 '12 at 15:28
Thanks, very nice notes ! I believe these notes provide all the answers to my questions. If I well understand, the argument which shows that the orbit partition of an ergodic $T$ is nonmeasurable can be applied to see that ergodicity is not a necessary condition for nonmeasurability: whenever a nonergodic $T$ has more than one orbit in an ergodic component, its orbit partition is nonmeasurable. –  Stéphane Laurent Feb 12 '12 at 15:30