I am currently trying to learn Patterson-Sullivan theory, but I am getting stuck on basic questions about ergodic theory. Here is one of them, given as an exercise in one of the texts I am trying to read. If you need wider context, the text (in French) is [here](http://www.numdam.org/issue/MSMF_2003_2_95__1_0.pdf), with the relevant question towards the bottom of page 17 (next-to-last sentence of the long paragraph).


Let ($\Omega$, $\Gamma$, $\mu$) be a dynamical system satisfying the following assumptions:


* $\Omega$ is a Hausdorff, locally compact and $\sigma$-compact topological space;
* $\Gamma$ is a countable group acting on $\Omega$;
* $\mu$ is a Radon measure on $\Omega$ that is quasi-invariant by $\Gamma$ (for every $g \in \Gamma$, $g\_\* \mu$ is absolutely continuous with respect to $\mu$).


Let us recall some definitions:


* a subset $W \subset \Omega$ is called *wandering* if, for $\mu$-almost all $w \in W$, the intersection of the orbit $\Gamma w$ with $W$ is finite;
* we say that the system is *completely conservative* if $\Omega$ has no wandering subsets of positive measure;
* we say that the system is *completely dissipative* if $\Omega$ has some wandering subset $W$ such that $\Omega = \bigcup\_{g \in \Gamma} g W$.


**Statement to prove: If the system is ergodic and the measure $\mu$ does not have any atoms, then the system is completely conservative.**


Even though it is supposed to be an easy exercise ("on vérifie aisément que..."), I am at a loss. I have found a partial proof, that works only if you can assume that the set $W$ introduced in the proof is closed. Unfortunately, I have no idea whether it is true, in general, that every completely dissipative dynamical system is generated by a closed wandering set. **Can someone give me a proof of this last statement, or find some other way to fix my proof presented below?**


In fact, I have remarkably little intuition about what "completely dissipative" actually means. I am tempted to think that a completely dissipative action should in particular be properly discontinuous, but apparently this is false (though I have never seen an actual counterexample). I would be very happy if I could find, somewhere, an arsenal of examples and counterexamples of completely dissipative and completely conservative systems, that is sufficiently well-stocked to forge an intuition about these properties and to be able to test whether some reasonable-sounding statement is actually true.


**The partial proof**: Let $\Omega = \Omega\_C \sqcup \Omega\_D$ be the Hopf decomposition of $\Omega$, i.e. a partition of $\Omega$ into two $\Gamma$-invariant subsets such that $\Omega\_C$ is completely conservative and $\Omega\_D$ is completely dissipative. Since the system is ergodic, they cannot both have positive measure: this means that $\Omega$ is either completely conservative or completely dissipative. By contradiction, assume the latter: let $W \subset \Omega$ be a wandering subset such that
$$\Omega = \bigcup\_{g \in \Gamma} g W.$$
Note that the set
$$\{x \in W \mid \Gamma x \cap W \text{ is finite}\} \cap \operatorname{Supp}(\mu)$$
has the same (positive) measure as $W$, hence is nonempty. Let $x\_0$ be any point in this set. Let $F := \{ g \in \Gamma \mid g x\_0 \in W \}$; this is a finite set by construction.


Since $x\_0 \in \operatorname{Supp}(\mu)$, every neighborhood of $x\_0$ has positive measure; on the other hand, since the measure is Radon and without atoms, any point has neighborhoods of arbitrarily small measure. Let $U$ be some neighborhood of $x\_0$ such that for every $g \in F$, we have $\mu(g U) < \frac{\mu(W)}{\# F}$. We can also assume (**provided that $W$ is closed!**) that for all $g \not\in F$, the image $g U$ is disjoint from $W$. Then consider the set $X := \bigcup\_{g \in \Gamma} g U$:


* this set is $\Gamma$-invariant by construction;
* it has positive measure by construction;
* its complement $\Omega \setminus X$ contains in particular $W \setminus X$, the complement of $W \cap X$ in $W$. But by construction, the intersection $W \cap X$ is contained in $\bigcup\_{g \in F} g U$, so that we have
$$\mu(W \cap X) \leq \sum\_{g \in F} \mu(g U) < \mu(W).$$
Hence the complement $W \setminus X$, and hence also the complement $\Omega \setminus X$, has positive measure. But these three properties of the set $X$ contradict ergodicity, QED.



Ergodic without atoms implies completely conservative?
The desired statement is entirely measure-theoretic, so it's not really necessary to think about the topology of $X$. By standard descriptive set theory, there exists a Borel linear ordering $\preceq$ on $X$. (This means that the graph of $\preceq$ is a Borel subset of $X \times X$.) 


Suppose toward a contradiction that $\mu$ is nonatomic and the $\Gamma$-action is ergodic but that there exists a wandering set $W$ of positive measure. Without loss of generality we may assume that the intersection of every $\Gamma$-orbit with $W$ is finite. The ordering $\preceq$ restricts to a linear ordering on each $\Gamma$-orbit. If a $\Gamma$-orbit $\mathcal{O}$ intersects $W$ then $\mathcal{O} \cap W$ is nonempty and finite, so that the $\preceq$-least element of $\mathcal{O} \cap W$ is well-defined. Since $\preceq$ is Borel, the set $V$ consisting of the $\preceq$-least element of each such set $\mathcal{O} \cap W$ is Borel. The set $V$ contains at most one point from each $\Gamma$-orbit. Since $W \subseteq \bigcup\_{\gamma \in \Gamma} V$ and $\mu$ is $\Gamma$-quasi-invariant it follows that $\mu(V) > 0$.


Since $\mu$ is nonatomic, there exists a Borel set $V\_1 \subseteq V$ such that if we write $V\_2 = V \setminus V\_1$ then $\min(\mu(V\_1),\mu(V\_2)) > 0$. Let $\widetilde{V}\_j = \bigcup\_{\gamma \in \Gamma} \gamma V\_j$. For each $j \in \{1,2\}$ the set $\widetilde{V}\_j$ is $\Gamma$-invariant and since $\widetilde{V}\_j$ contains $V\_j$ we have $\mu(\widetilde{V}\_j) > 0$. By ergodicity it follows that $\mu(X \setminus \widetilde{V}\_j) = 0$ for each $j \in \{1,2\}$.


We claim that $\widetilde{V}\_1$ and $\widetilde{V}\_2$ are pairwise disjoint, which will lead to the contradiction $$0 < \mu(\widetilde{V}\_2) \leq \mu(X \setminus \widetilde{V}\_1) =0$$


We will establish the claim by showing that that existence of $x \in \widetilde{V}\_1 \cap \widetilde{V}\_2$ leads to a further contradiction. Suppose that there exists such a point $x$. Then there exist $\gamma\_1,\gamma\_2 \in \Gamma$ with $\gamma\_1x \in V\_1$ and $\gamma\_2x \in V\_2$. Since $V\_1 \cap V\_2 = \emptyset$ we have $\gamma\_1x \neq \gamma\_2x$. Since $\gamma\_1x = \gamma\_1\gamma\_2^{-1}\gamma\_2x$ we see that $\gamma\_1x$ and $\gamma\_2x$ are in the same $\Gamma$-orbit. This cannot be the case since $\gamma\_1x$ and $\gamma\_2x$ are both elements of $V$ and $V$ contains at most one point from each $\Gamma$-orbit.


$\mathbf{Response\,\, to \,\,comment}$


The way to see that the set $V$ is Borel is to regard Borel sets as being defined by logical formulas. We say that a formula 'is Borel' if it defines a Borel subset of $X$. Countable existential quantifiers correspond to countable unions and not corresponds to complements. Thus these logical operations preserve the property of being a Borel formula.


It is implicit in the hypotheses that $\Gamma$ acts by Borel automorphisms. Since $\preceq$ is Borel, it follows that the predicate $\gamma x \prec x$ is Borel for any $\gamma \in \Gamma$. The set $V$ can be defined as $$\{x \in W: (\not \exists)(\gamma \in \Gamma)((\gamma x \in W) \wedge (\gamma x \prec x))\}$$.


