MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 ?

share|cite|improve this question
up vote 5 down vote accepted

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 $p\neq q$.

share|cite|improve this answer
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
Hello Vaughn. I think there's a mistake in your exercise 2.1 p10. If $\wedge \xi_n$ denotes the set-theoretic intersection, then it is not measurable in general: Vershik's adic representation theorem show that the orbital partition of any ergodic automorphism is such an intersection. But in the Rokhlin school, this intersection is rather denoted by $\cap \xi_n$, whereas $\wedge \xi_n$ denotes its measurable hull. – Stéphane Laurent Aug 18 '15 at 13:38
@StéphaneLaurent, thank you for pointing this out -- it seems this is a mistake in a version of our notes that is still posted on A. Katok's website (the link in Jon's answer), but was corrected when we produced a version for the arXiv, which is at -- in the corresponding exercise there we removed the false claim about the intersection. It seems there were other very substantial changes too, since the arxiv version is about 12 pages longer. – Vaughn Climenhaga Aug 23 '15 at 23:32
Ok, thanks ! :) – Stéphane Laurent Aug 24 '15 at 5:49

It seems that the nice set of notes "Measure theory through dynamical eyes" by Katok and Climenhaga here (see, for example, Section 1.4) will be useful for the first question (1). It is also mentioned that if the ergodic system has more than one orbit, then ergodicity is sufficient for nonmeasurability of the partition (2). I hope this is helpful. Perhaps Vaughan Climenhaga will see this and write a nice answer.

share|cite|improve this answer
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
Great! I'm glad I could help. – Jon Bannon Feb 12 '12 at 15:42
I'm delighted to see those notes being useful! @Stéphane: I've taken the liberty of fleshing out the consequences of your comments in another answer, as much to remind myself of the details of this as anything else. – Vaughn Climenhaga Feb 13 '12 at 4:49

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.