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Let $G$ be an amenable group acting on a space $X$.

Amenability means there is a $G$-invariant mean on $L^\infty(G,{\mathbf R})$. Given a bounded function $f\colon X\to {\mathbf R}$ one can use the mean to define a $G$-invariant function $\overline{f}\colon X\to {\mathbf R}$: one just uses the mean to average over the $G$-orbits.

Question: If $f$ was Borel-measurable, then so is $\overline{f}$ Borel-measurable? (At least if $G$ is countable?) This is true for finite groups because sums of measurable functions are measurable.

This was asked on math.stackexchange but got no attention.

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  • $\begingroup$ To be more precise: group actions are meant to be continuous and "measurability" means just Borel-measurability. $\endgroup$
    – ThiKu
    Nov 9 '15 at 4:26
  • $\begingroup$ Could you please clarify if you merely wanted the existence of an averaging map which preserved measurability, or if you wanted to know whether all invariant means on $\ell^\infty(G)$ produce "measurability-preserving averaging maps" for functions on $X$? It seems to me that the former question is answered below, but the second one is not. $\endgroup$
    – Yemon Choi
    Nov 9 '15 at 15:12
  • $\begingroup$ For My Applications, I am happy with the existence of such an averaging map indeed. $\endgroup$
    – ThiKu
    Nov 9 '15 at 16:22
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I imagine that in your setting the group $G$ is assumed to preserve some measure $\mu$ on $X$, or at least the measure class of $\mu$ (you are not being very precise about what "space" means, or in what sense are things measurable).

The answer is "yes" (at least for discrete groups), and here is one way to see this. Let's denote by $F_n$ a sequence of Folner sets in $G$, so that for any $g\in G$ the symmetric differences $F_n \Delta g F_n$ satisfy $| F_n \Delta g F_n | / | F_n | \to 0$. Regard $M=L^\infty(X,\mathbb{R})$ as the dual of $M_* = L^1(X,\mu)$ and equip it with the weak-* topology given by point wise convergence on $L^1$. Then the unit ball of $M$ is weak-* compact. Given a function $\zeta\in M$, consider the functions $\zeta_n = (1/F_n) ( \sum_{g\in F_n} g\zeta )$ and let $\bar\zeta$ be a weak-* limit point of that sequence. It is not hard to verify that $\bar\zeta$ is $G$-invariant. Indeed, for any $\phi \in L^1(X,\mu)$ you get that $$\left|\int \phi(x) (g\cdot \zeta_n(x) -\zeta_n) d\mu(x)\right|= \left|\int \frac{1}{|F_n|} \left(\sum_{h\in gF\Delta F} h\cdot \zeta(x)\right) \phi(x) d\mu(x)\right|$$ $$\leq \Vert \zeta\Vert_\infty \Vert\phi\Vert_1 \frac{|F\Delta gF|}{|F|}\to 0$$ so that $\langle g\bar\zeta -\bar\zeta,\phi\rangle=0$ for any $\phi$.

The function $\bar\zeta$ is measurable, since it remains in $M$.

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  • $\begingroup$ Thanks, that was actually easier than I thought. It's perhaps worth mentioning that this doesn't reprove the claim for finite groups because measurability of finite sums is already used in the argument. $\endgroup$
    – ThiKu
    Nov 9 '15 at 5:03
  • $\begingroup$ Strictly speaking, the original question seemed to ask if applying a given invariant mean to a Borel function resulted in something Borel; what is shown here is that if there is an invariant mean on a countable group, then using Folner sequences one can build an averaging map on Borel functions that preserves measurability. $\endgroup$
    – Yemon Choi
    Nov 9 '15 at 15:09
  • $\begingroup$ If you have acces to MathSciNet, what do you make of the reviewer's remarks in ams.org/mathscinet-getitem?mr=1772597 ? $\endgroup$
    – Yemon Choi
    Nov 9 '15 at 15:10
  • $\begingroup$ I must admit that I don't understand the details in the reviewers argument. Even if the set of means has cardinality $2^{\mathfrak c}$, how do we know that this is also true for the set of invariant means? And also, how do we know that different means necessarily give different functions? $\endgroup$
    – ThiKu
    Nov 18 '15 at 8:15
  • $\begingroup$ In any case I agree that we do not know whether every averaged function is Borel-measurable. But for my purposes the existence of one invariant Borel-measurable function (provided by the above argument) is enough. $\endgroup$
    – ThiKu
    Nov 18 '15 at 8:17

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