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Let $G$ be a finitely generated group. Does the following condition imply the amenability of $G$: there is a function $\mu:\mathcal{P}(G) \to [0,1]$ such that:

  • (subadditive) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$, and $A \subset B$ implies $\mu(A) \leq \mu(B)$,

  • (invariant) $\mu(A \cdot g) = \mu(A)$ for all $g$ in $G$,

  • (exhaustive) if $\{A_n : n <\infty\}$ is pairwise disjoint, then $\inf_n \mu(A_n) = 0$.

    A group distinguishing this condition from amenability cannot contain $F_2$. I am aware of the relationship to the (now solved) Maharam problem.

    I am not expecting an answer so much as asking whether (and where) this question has been studied.

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    Hi Justin, Welcome! –  Andres Caicedo Nov 12 '10 at 2:52
    I second Andres' comment. Welcome. –  Stefan Geschke Nov 12 '10 at 8:28
    Do you assume $\mu(A) \leq \mu(B)$ if $A \subset B$? –  Andreas Thom Nov 12 '10 at 9:25
    That follows from subadditivity. –  Justin Moore Nov 12 '10 at 14:59
    Sorry, you cannot assign 2. OK. Then assign 1 to G, 2/3 to every 1-element subset, and 3/4 to the empty set. –  Mark Sapir Nov 12 '10 at 18:34
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    2 Answers

    Modulo a (possibly severe) measurability issue, the only groups which admit an invariant submeasure are the amenable groups.

    Let $G$ be a non-amenable group and $\mu$ be a $G$-invariant submeasure on $G$. Damien Gaboriau and Russell Lyons showed that every non-amenable group contains a random free subgroup. One way of putting this is to say that there exists a free and measure preserving Borel action of $G$ and $F_2$ on some standard probability space $(X,\alpha)$ and a cocycle $$c \colon F_2 \times X \to G$$ such that $c_x$ is injective for each $x \in X$ and $$c(g,x) \cdot x = g\cdot x, \quad \forall g \in F_2,x \in X.$$

    Considering now the measure space $Z=G \times X$, we see that the action of $F_2$ on $Z$ given by $g(t,x) = (c(g,x)t,g \cdot x)$ admits a fundamental domain $Y \subset Z$. (The measure of $Y$ is typically not finite. To get a picture, $Y$ plays the role of a set of representatives of cosets in case $F_2$ is an honest subgroup of $G$.)

    Let $W$ be a measurable subset of $Z$ and set $$\mu'(W) = \int_X \mu(W_x) d \alpha(x).$$ Moreover, for every element $z$ in the full group of the equivalence relation generated by $G$, we see that $\mu'(zA) = \mu'(A)$. Indeed, the action of each such element on $Z$ is given by a measure preserving shuffle of the $x$-coordinate and an $x$-dependent shift in the $G$-coordinate. Both operations preserve the integral.

    Let now $A \subset F_2$ and define $\mu''(A) := \mu'(A \cdot Y).$ Since $F_2$ lies in the full group of the equivalence relation generated by the action of $G$, we can conclude that $\mu''$ is $F_2$-invariant.

    It remains to analyze the implications of the assumption that $\mu$ is exhaustive. Let us assume that there exists an infinite disjoint family $(A_n)$ of subsets of $F_2$ such that $\mu''(A_n) \geq \delta>0$. Then, the functions $x \mapsto \mu(A_{n,x})$ are greater $\delta/2$ on a set of measure at least $\delta/2$. This, with an additional argument, (I did not check every detail) seems to contradict that $\mu$ is exhaustive. Hence, $\mu''$ is exhaustive as well.

    This is a contradiction. Hence, every group which admits a $G$-invariant submeasure is amenable.

    (The problem with this argument is that $x \mapsto \mu(W_x)$ need not be Borel, so that an additional argument is needed to make sense of the integral in the definition of $\mu'$. Maybe this can be cured, but I do not have time to think about it right now.)

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    This is probably a trivial observation, but maybe useful to track the difficult part of the problem.

    If instead of condition three you require:

    (strongly exhaustive) There exists some constant $C>0$, such that $\lbrace A_n \mid n< \infty\rbrace$ pairwise disjoint implies that $\sum_n \mu(A_n) < C.$

    then $G$ is amenable. Indeed,

    $$\sigma(A) := \sup\left\lbrace \sum \mu(A_i) \mid A = \cup A_i \mbox{ is a finite disjoint union} \right\rbrace$$

    defines an invariant mean on $G$. First of all, since $\mu$ is strongly exhaustive, $\sigma(A)$ is a well-defined real number for every $A \subset G$. Secondly, $G$-invariance is obvious from the definition and the $G$-invariance of $\mu$.

    Now, if $C = A \cup B$ is a disjoint union, then $\sigma(C) \leq \sigma(A) + \sigma(B)$ follows as before, since each partition of $C$ gives rise to a partition of $A$ and $B$. However, $\sigma(A) + \sigma(B) \leq \sigma(C)$ also holds since a partition of $A$ and a partition of $B$ clearly merge to give a partition of $C$.

    All in all, $\sigma(A) + \sigma(B) = \sigma(C)$ whenever $C$ is the disjoint union of $A$ and $B$. This implies that $\sigma$ is a finite $G$-invariant finitely additive measure on $G$, and hence $G$ is amenable.

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