# Conservative Measures under a group action (reference request)

I was reading a paper and the author define the concept of conservative measure:

Let $(X,\mathcal{B})$ a measurable space and $G$ a group that acts on $X$ by $$G\times X:(g,x)\mapsto T_g(x)$$ where all the functions $T_g$ are measurable.

A measure $\mu$ on $(X,\mathcal{B})$ is called conservative under $G$ if for every $A\in \mathcal{B}$ with $\mu(A) > 0$ exists $g \in G\setminus\{1\}$ with $$\mu(A\cap T_g^{-1}A) > 0$$ I was wondering if this definition is known in the literature or if it is specific to this paper?.

One can notice that if the action of $G$ is given by an invertible measurable function $T$ and the measure $\mu$ is ergodic, then by the von Neumann ergodic theorem (on $L^2(\mu)$) we have that for every $A \in \mathcal{B}$ with $\mu(A) > 0$ $$\frac{1}{n} \sum_{k=0}^{n-1} \mu(A \cap T^{-k}A) = \left\langle 1_A , \frac{1}{n} \sum_{k=0}^{n-1} 1_A \circ T^k \right\rangle_{L^2(\mu)} \stackrel{n \to \infty}{\longrightarrow} \mu(A)^2 > 0$$ So it must exists $k \in \mathbb{N}$ such that $$\mu(A \cap T^{-k}A) > 0$$ Hence $\mu$ is conservative by the group action $(T^n)_n$.

Edit: More generally if the measure is only invariant by the action $(T^n)_n$ we have the same result thanks to the Poincare Recurrence Theorem .

There is a more general connection between conservative measures and ergodic measures?

Any comment or reference will be appreciated.

• If $G \times X \to X, (g,x) \mapsto T_g(x)$ really is a group action, we could always take $g = 1_G$ to achieve $\mu(A \cap T_g^{-1}A) > 0$, couldn't we? – PhoemueX Oct 15 '14 at 21:31
• I forgot to put the restriction $g \in G \setminus \{1\}$. Thanks for your comment – user90803 Oct 15 '14 at 21:54

Conservativity of the action is not necessarily immediate for infinite measure spaces, and this is where this definition is useful (see Aaronson's book), for example $x\mapsto x+1$ is not conservative on $\mathbb{R}$ with the Haar measure (but certianly measure-preserving).