# Halmos recurrence theorem for a locally compact group

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.

## 1 Answer

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.