A stronger version of supramenability?

Question

A group $G$ is supramenable iff for all $\varnothing\ne A\subseteq G$ there is a finitely-additive left-$G$-invariant measure $\mu_A$ on $G$ with $\mu_A(A)=1$. I'm interested in a seemingly stronger condition that naturally connects up $\mu_A$ for different $A$s.

The condition I want is that one can additionally choose the $\mu_A$ so that $\mu_B(A)\mu_C(B)=\mu_C(A)$ whenever the lhs is defined (i.e., is not zero times infinity or infinity times zero). Say such a group is neatly supramenable (but if there is a term in the literature, I'll be happy to use that).

Questions:

1. Has anybody studied this property?

2. Is every supramenable group neatly supramenable?

3. Is the direct product of neatly supramenable groups neatly supramenable?

4. Is every neatly supramenable group exponentially bounded?

(3 and 4 are analogues of open problems for supramenable groups. But they might be easier for neatly supramenable ones.)

Here's what I have so far (a lot of this uses AC via ultrafilters):

A. Exponentially bounded groups are neatly supramenable.

B. The following are equivalent:

1. $G$ is neatly supramenable

2. There is a hyperreal-valued finitely-additive measure $\mu$ on $2^G$ that vanishes only on the empty set and is approximately left-$G$-invariant: $\operatorname{st}(\mu(gA)/\mu(A)) = 1$

3. There is a hyperreal-valued exactly left-$G$-invariant function $\mu$ on $2^G$ that vanishes on and only on the empty set and is approximately finitely-additive: $\operatorname{st}(\mu(A\cup B)/(\mu(A)+\mu(B)))=1$ if one of the sets is non-empty

4. There is a finitely-additive conditional probability $P:2^G\times (2^G-\{\varnothing\})$ that is strongly left-$G$-invariant: i.e., $P(gA|B)=P(A|B)$ whenever $A,gA\subseteq B$.

5. There is a finitely-additive conditional probability $P:2^X\times (2^X-\{\varnothing\})$ that is strongly left-$G$-invariant on every space $X$ on which $G$ acts.

(The equivalence of 2 and 4 is due to Krauss's result on strictly positive measures with values in extensions of the reals and conditional probabilities. The equivalence of 1 and 4 is by letting $P(A|B)=\mu_{B}(A\cap B)$ in one direction and $\mu_B(A)=P(A|A\cup B)/P(B|A\cup B)$ in the other.)

Answer 1

It looks like neat supramenability is equivalent to supramenability, at least given AC. This follows from Proposition 1.7 in the 1989 paper by Armstrong in this volume (page 7). The proof uses the existence of a maximal set $M$ of non-trivial Renyi-ordered finitely additive measures (two measures are Renyi-ordered iff one vanishes on all sets on which the other is finite), and then shows that for every non-empty subset $A$ of $G$, some measure in the set gives finite non-zero value to $A$, or else one could use supramenability to create a new measure that is Renyi-ordered to each measure in $M$. We then define $\mu_B(A) = \mu(A)/\mu(B)$ where $\mu$ is the unique measure in the maximal set for which $0<\mu(B)<\infty$.