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I should probably start with a warning that this is my first post in this board and that I am sorry, if it is not up to standards. It would be great, if you could let me know how to improve the post.

I am currently working towards my master thesis, which is mainly concerned with the following paper: Chatterji, Iozzi, Fernós - The median class and superrigidity of actions on CAT(0) cube complexes. More precisely, I try to understand the construction of the (measurable, $\Gamma$-equivariant) boundary map $$ \varphi\colon B \to \partial X, $$ where $B$ is a strong $\Gamma$-boundary for a group $\Gamma$ and $\partial X$ is the Roller boundary of a finite dimensional, locally countable CAT(0) cube complex $X$. The group action on $X$ is assumed to be essential and non-elementary. However, I only mention this for completeness sake. I think the precise setting should not enter in my actual question, because as far as I can see it is completely measure theoretic.

During the construction, there is a particular recurring argument I just cannot get my head around. I think, it should follow from general measure theoretic arguments, but my background in that field is not strong enough to pin it down. If I would have to formulate it as a Lemma, it would take the following form:

Let $(X,\mu)$ be a probability space on which a group $\Gamma$ acts ergodically. Let $B \subset X$ be measurable and $\mu(X \setminus B)=0$. Then $B$ contains a $\Gamma$ orbit.

I would be interested in any hint pointing towards a proof of this statement or towards any counterexample. I would also be interested in any restrictions necessary to make the above statement work. In particular, I was unable so far to collect all the assumptions I have on the group $\Gamma$. However, I am certain that a locally compact and a second countable topology are necessary and can be used, if necessary. Additionally, $(X,\mu)$ can be assumed to be a Lebesgue space.

Your help is very much appreciated. Thank you very much for your time and effort.

Best regards,

Tim

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  • $\begingroup$ I guess that the $\Gamma$-action on $X$ is probability-preserving? (and that $\Gamma$ is countable) $\endgroup$ – YCor Nov 20 '17 at 13:05
  • $\begingroup$ Probability preserving, yes, but I am fairly certain that I cannot assume $\Gamma$ to be countable. $\endgroup$ – Emrys-Merlin Nov 20 '17 at 13:22
  • $\begingroup$ Without a countability assumption on $\Gamma$, this is certainly not true : take $X=\Gamma=\mathbf R/\mathbf Z$. $\endgroup$ – Mikael de la Salle Nov 20 '17 at 13:42
  • $\begingroup$ On the other hand, if $\Gamma$ is countable, then you are looking for an $x \in X$ such that $\gamma x \in B$ for every $\gamma \in \Gamma$, \emph{i.e.} for an x in $\cap_{\gamma \in \Gamma} \gamma^{-1} B$, which is a countable intersection of subsets of full measure, and hence has full measure (by $\sigma$-additivity of the measure). $\endgroup$ – Mikael de la Salle Nov 20 '17 at 13:46
  • $\begingroup$ @MikaeldelaSalle: Thank you very much for your comments, especially for the counterexample in the uncountable case. It seems that I have to check the conditions again in order to rule it out... $\endgroup$ – Emrys-Merlin Nov 20 '17 at 14:04
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Formally, your question was answered in the comments, and I will not repeat the answer here, as it simply indicates that your initial question was "wrong". Possibly you just missed some "up to null set" comment (or hidden assumption) somewhere in the text. Nevertheless, if you are genuinely interested in delicate and fundamental questions of the type indicated in the OP, you may find some answers reading

Feldman, Jacob; Hahn, Peter; Moore, Calvin C. Orbit structure and countable sections for actions of continuous groups. Adv. in Math. 28 (1978), no. 3, 186–230.

Edit: In respond to a comment by YCor, I explain how applying the results of the above mentioned paper answers the OPed question.

A main result of the FHM paper is Theorem 2.8 which states that for every lcsc group $\Gamma$, for every analytic action of $\Gamma$ on a space $X$, and for every invariant measure class $\mu$ on $X$, there exists a measurable subset $N$ in $X$ and an identity neighborhood $U$ in $\Gamma$ such that $\Gamma N$ is conull in $X$ and for every $x\in N$, $Ux\cap N=\{x\}$. Replacing $X$ with $\Gamma N$ we may assume that $X=\Gamma N$. It is now an easy exercise to see that, unless the $\Gamma$ action on $X$ has discrete orbits, $N$ must be a null set. In this case, setting $B=X-N$ we get a contradiction to the boldfaced statement in the OP.

I didn't elaborate on this before, as Mikael de la Salle already gave an explicit counter-example in a comment: the action of $\mathbb{R}/\mathbb{Z}$ on itself (and $N=\{1\}$). Another, slightly less transparent, example to keep in mind is the following: Consider an irrational flow of $\mathbb{R}$ on the two torus $\mathbb{R}/\mathbb{Z}\times \mathbb{R}/\mathbb{Z}$ and let $N=\mathbb{R}/\mathbb{Z}\times \{0\}$.

But, let me emphasize: I do not think these type of questions (important us they are) are really relevant for the understanding of the Chatterji-Iozzi-Fernós paper.

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  • $\begingroup$ It may sound stupid but I'll reiterate my request: you need to show that every lcsc group $G$ there exists such a $(X,\mu)$ with $\mu$ non-atomic. This is obvious in almost any example I can think of (e.g., when $G$ has a non-discrete quotient with a cocompact lattice - this encompasses all virtually connected Lie groups). $\endgroup$ – YCor Nov 21 '17 at 10:06
  • $\begingroup$ @YCor, of course you are right on a formal level: in order to fully justify my original comment to the OP I should add an argument which explains how to construct an appropriate $X$ for any non-discrete $\Gamma$. I didn't elaborate on this, as I feel that this will not necessary lead the discussion in the right direction... $\endgroup$ – Uri Bader Nov 21 '17 at 10:20
  • $\begingroup$ But for completeness: for any $\Gamma$ one can construct the Gausiaan space $X$ associated with its left regular representation and it will do the job (a formal proof is needed here, of course, but we already spent enough time on this and I have no indication that any one else is interested in this discussion). $\endgroup$ – Uri Bader Nov 21 '17 at 10:20
  • $\begingroup$ @UriBader: Thank you very much for your elaboration. I agree with you that your argument probably goes too far afield for my particular problem. I will have to think more on the particular structure of the argument in the paper. I marked your post as my answer, but would also like to thank MikaeldelaSalle for his quick reply and very helpful comments. $\endgroup$ – Emrys-Merlin Nov 21 '17 at 10:24
  • $\begingroup$ Yes I agree that all this discussion (for actions of non-discrete lcsc groups) is useless for understanding the CFI paper, but it's of independent interest (of course I'm asking because I'm interested). Your last comment is fine, I'm not asking for details :) $\endgroup$ – YCor Nov 21 '17 at 13:40

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