Halmos recurrence theorem for a locally compact group

Question

The recurrence theorem of Halmos is well known in the case of a non-singular endomorphism $T$ of a measured space $(X,\mathcal B,\mu)$. A measurable subset $A$ is contained in the conservative part (mod $\mu$) if and only if $$ \sum_{n \geq 0} 1_B \circ T^n = \infty $$ holds a.e. in $B$ (where $1_B$ stands for the characteristic function) for any measurable $B \subset A$.

See e.g. Aaronson's book. Now I have never seen a proof of this theorem in the case of a non-singular group operation (where the group is locally compact and second countable). In fact I have never seen it stated in this form but I believe it must be true.

Does someone know of a reference?

A related theorem is Hopf's theorem which states that the conservative part is (mod $\mu$) the set of all $x$ such that $$ \int f(hx) d\eta(h) = \infty $$ where $f \in L^1(\mu)$ is $>0$ and $\eta$ is some Haar measure on $H$. I would be equally happy with a proof of Hopf's theorem instead.

Thank you.

Answer 1

Assume that $A$ is contained in the conservative part and $\mu(A)>0$.

We argue by contradiction so let us pretend there is some measurable subset $A' \subset A$ such that for any $x \in A'$, $\eta(H(x))$ is finite, where $H(x)=\{h \in H ; hx \in A \}$. We may also assume that there is some constant $C>0$ such that $\eta(H(x) < C$ for any $x \in A'$. Then there must be some relatively compact neighbourhood $V$ of the identity element of $G$ such that $\mu(H(x) \cap V) > \frac{1}{2} \mu(H(x))$ ($x \in A'$).

Now let $x\in A',h \in H$ be such that $hx \in A'$. We have $$ \frac{\mu(H(hx) \cap V)}{\mu(H(hx))} > 1/2$$ and $H(hx)=H(x)h^{-1}$ so we get $$ \frac{\mu(H(x) \cap Vh)}{\mu(H(x))} > 1/2 $$ (thanks to relative right invariance of left Haar measure). Since we also have the same inequality with $Vh$ replaced by $V$, we deduce that $Vh \cap V \neq \varnothing$, that is, $h \in V^{-1} V$: $h$ belongs to some compact set. This contradicts the fact that $A$ is contained in the conservative part.

Answer 2

Regarding what you wrote I would be equally happy with a proof of Hopf's theorem instead, it seems that the very fact that there is "conservative part" for a nonsingular action is unavailable for general lcsc groups. However, there is a "version" of the Hopf Decomposition for the purpose of checking the divergence $$(*)\quad \intop_{G}\frac{d\mu\circ g^{-1}}{d\mu}f\left(gx\right)dm\left(g\right)=\infty.$$

This can be done as follows. Pick a lattice in $G$ (if there exists any!) and take the Hopf Decomposition $D\cup C$ w.r.t. this lattice (regarding this lattice as an acting group on its own right, and using the Hopf Decomposition for the countable case as in [1, Proposition 1.6.2]). It is then true that $(*)$ holds for $x\in D$ and, moreover, the set $D$ is independent on the lattice up to null set. See [1, Theorem 1.6.4] and [2, Corollary 2.2].

References

[1] Aaronson, Jon, An introduction to infinite ergodic theory, Mathematical Surveys and Monographs. 50. Providence, RI: American Mathematical Society (AMS). xii, 284 p. (1997). ZBL0882.28013.

[2] Roy, Parthanil, Nonsingular group actions and stationary S(\alpha )S random fields, Proc. Am. Math. Soc. 138, No. 6, 2195-2202 (2010). ZBL1196.60093.