If $\mu$ is the normalized counting measure on a finite group $G$, then $\mu(G)=1$ and $\mu(C)=1/n$ for every coset $C$ of a subgroup of index $n$. Let's ask for the same for infinite groups:

Question: When $G$ is any group, is there a finitely additive measure $\mu$ on $G$ such that for every positive integer $n$ one gets $\mu(G)=1$ and $\mu(C)=1/n$ for every coset $C$ of every subgroup of index $n$?

(Question edited for clarity.)

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    $\begingroup$ Is the subgroup meant to be fixed, or is this condition means to hold for every coset of every subgroup of index $n$? For that matter, is $n$ fixed, or is this meant to hold for every finite-index subgroup? $\endgroup$ – LSpice Jun 17 '19 at 22:19
  • $\begingroup$ @LSpice Yes, for every coset of of every subgroup of finite index. $\endgroup$ – user95282 Jun 18 '19 at 1:30
  • $\begingroup$ You might find the keyword "invariant random subgroup" useful. $\endgroup$ – Jalex Stark Jun 18 '19 at 5:50
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    $\begingroup$ Use Hahn-Banach on $\ell^\infty(G)$, for the obvious functional defined on the subspace consisting of functions which are invariant under some finite index normal subgroup. $\endgroup$ – Uri Bader Jun 18 '19 at 7:19
  • $\begingroup$ @JalexStark How does randomness help here? Or is this a general comment, not specifically for this question? $\endgroup$ – user95282 Jun 18 '19 at 11:00

This should be possible for any group, assuming I have facts about extending measures correct. The idea is to first construct such a measure on a smaller Boolean algebra of subsets of $G$, and then use general measure theory facts to extend this measure to $\mathcal{P}(G)$.

So first, let $\mathcal{B}\subseteq\mathcal{P}(G)$ be the Boolean algebra generated by cosets of finite index subgroups of $G$. Define a measure $\mu$ on $\mathcal{B}$ as follows. Suppose $A\in\mathcal{B}$. Then there is some finite index subgroup $H$ of $G$ such that $A$ is a union of cosets of $H$. If $n$ is the index of $H$, and $m$ is the number of cosets in the union, then define $\mu(A)=m/n$. It can be directly checked that $\mu$ is well-defined. Moreover, if $C$ is a coset of subgroup of index $n<\infty$, then $\mu(C)=1/n$ by construction. Note that $\mu$ is also translation-invariant.

Now one can (non-uniquely) extend $\mu$ to some finitely additive measure on $\mathcal{P}(G)$ (which won't be translation-invariant necessarily). I believe this follows from Section 457 of Fremlin's "Measure Theory". Specifically, any finitely additive probability measure on a Boolean algebra (of subsets of some fixed set $X$) can be extended to such a measure on any larger Boolean algebra.

Remark 1: The initial measure $\mu$ is in fact the unique $G$-invariant finitely additive probability measure on $\mathcal{B}$, and can be constructed from the Haar measure on the profinite completion of $G$. In particular, if $\mathcal{N}$ denotes the collection of finite index normal subgroups of $G$, then the profinite completion is $\hat{G}=\varprojlim_{\mathcal{N}}G/N$. We can write elements of $\hat{G}$ as $(C_N)_{N\in\mathcal{N}}$, where $C_N$ is a coset of $N$. Given a set $A$ in $\mathcal{B}$, define $X_A$ to be the set of $(C_N)_{N\in\mathcal{N}}\in \hat{G}$ such that $C_N\cap A\neq\emptyset$ for all $N\in\mathcal{N}$. Then $X_A$ is closed, and it can be shown that $\mu(A)$ is the Haar measure of $X_A$.

Remark 2: Perhaps it's also worth mentioning that if $G$ is amenable (e.g., abelian) then there is a translation-invariant finitely additive probability measure on $\mathcal{P}(G)$, which must satisfy the desired conditions outright by finite additivity and invariance.

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    $\begingroup$ I'd be curious if we can impose in addition the condition that the measure of any left coset of any infinite index subgroup is zero (which is automatic in the invariant case, but working only for amenable groups), i.e., assume the condition for $n=\infty$ as well. This would retrieve B.H. Neumann's result that no group is covered by finitely many cosets of infinite index subgroups. $\endgroup$ – YCor Jun 18 '19 at 9:21
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    $\begingroup$ Yves, one can certainly impose such a condition using B.H. Neumann's result: denote by $I<\ell^\infty(G)$ the closed ideal consisting of functions $f$ such that for every $\epsilon>0$ there exists a finite set of cosets of infinite index subgroups outside of its union $|f|<\epsilon$. Then $I$ is a proper ideal and you can use Hahn-Banach for $\ell^\infty(G)/I$. $\endgroup$ – Uri Bader Jun 18 '19 at 10:48
  • $\begingroup$ @UriBader OK fine (but I was hoping to get a measure-like proof of B.H. Neumann"s result). $\endgroup$ – YCor Jun 18 '19 at 11:57
  • $\begingroup$ @YCor I believe one could instead use the algebra $\mathcal{B}^*$ generated by the cosets of all subgroups of $G$. Then define $\mu$ directly (giving any infinite index coset measure $0$); or obtain $\mu$ as the pullback of the Haar measure on $\hat{G}$ in the same way. In any case, I am certain such a measure exists because I know how to construct it on yet a larger Boolean algebra. So the question is how construct it on $\mathcal{B}^*$ in the simplest way. $\endgroup$ – Gabe Conant Jun 18 '19 at 13:30
  • $\begingroup$ @GabeConant Yes, such a measure exists, but is there a proof that does not appeal to B.H. Neumann's theorem? The profinite completion doesn't lead to subgroups of infinite index, or does it? $\endgroup$ – user95282 Jun 18 '19 at 14:36

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