# Is there always an G-invariant uniform measure preserving map for countable G?

Suppose $\circ:G \times X \to X$ with $X$ countable. There is a measure preserving dynamical system $([0,1]^X, \lambda^X, G)$ where $\lambda$ is the uniform measure on $[0,1]$ and for $(a_x)_{x \in X} \in [0,1]^X$, $g$ takes $(a_x)_{x \in X}$ to $(a_{\circ(g,x)})_{x \in X}$.

If $X$ has a finite orbit $Y$ then there is an invariant measure preserving map from $([0,1]^X, \lambda^X)$ to $([0,1], \lambda)$. For example

$$(a_x)_{x \in X} \mapsto \sum_{y \in Y} a_y \mod 1.$$

If $X$ has no finite orbits will there still always be an invariant measure preserving map from $([0,1]^X, \lambda^X)$ to $([0,1], \lambda)$?

As a concrete example, is there such an invariant measure preserving map when $X = (\mathbb{Z}, S)$ and $G =$Aut$(\mathbb{Z}, S)$, where $S$ is the successor relation (and there is no other structure on $\mathbb{Z}$)?

• What action do you mean when talking about "a finite orbit $H$"? – R W Sep 15 '17 at 0:32
• Sorry, I actually realized I didn't want to require the power to be G. This should be fixed now. – Nate Ackerman Sep 15 '17 at 0:43
• It's not clear what you mean by $1/|Y|\sum_{y\in Y} a_y \bmod1$. Maybe you just wanted to sum, not average? – Anthony Quas Sep 15 '17 at 5:10
• @AnthonyQuas: Yes, thanks. I fixed this. – Nate Ackerman Sep 15 '17 at 5:24
• What do you mean by $\mathsf{Aut}(\mathbb{Z})$? automorphism in which sense? as ring, it's the trivial group. As group, it's cyclic of order 2. As measurable space, or as measure space (with counting measure), it's the group of all permutations, more commonly denoted $\mathsf{Sym}(\mathbb{Z})$. – YCor Sep 15 '17 at 13:46

I presume that by an "invariant" map you mean one which is constant on the orbits of the action of $G$ on the configuration space. Therefore, it should factorize through the space of ergodic components of your action, and your question is equivalent to asking whether the space of ergodic components is purely non-atomic.
In the case when $X=\mathbb Z$ and $G=Aut(\mathbb Z)$ already the action of $\mathbb Z$ is ergodic (essentially by Kolmogorov's 0-1 law), so that the answer is "no".