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Justin Moore
Justin Moore
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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.

    • 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) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$,  
  • (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.

    • 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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    Justin Moore
    Justin Moore
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    Justin Moore
    Justin Moore
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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) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$,
  • (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.

    • 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) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$,
  • (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 expecting an answer so much as asking whether (and where) this question has been studied.

    • 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) $\mu(G) = 1$, $\mu(A \cup B) \leq \mu(A) + \mu(B)$,
  • (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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    Justin Moore
    Justin Moore
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    Justin Moore
    Justin Moore
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