Let $X$ be a subset of a group $G$. We say that $X$ is *left amenable* with respect to $G$ if there is a function $\mu:\mathcal P(G)\to [0,\infty]$ with the following three properties.


1. $\mu(A\cup B)=\mu(A)+\mu(B)$ for every pair of disjoint subsets $A, B$ of $G$.
2. $\mu(gA)=\mu(A)$ for all $g\in G$ and $A\in \mathcal{P}(G)$.
3. $\mu(X)=1$.


The subset $X$ is *right amenable* with respect to $G$ if there is a function $\mu:\mathcal P(G)\to [0,\infty]$ that has properties (1) and (3) above as well as property (2') below.


2') $\mu(Ag)=\mu(A)$ for all $g\in G$ and $A\in \mathcal{P}(G)$.


It is easy to see that $X$ is right amenable with respect to $G$ if and only if $X^{-1}$ is left amenable with respect to $G$. Moreover, $G$ being either left or right amenable with respect to $G$ is equivalent to $G$ being an amenable group in the usual sense.


**Question.** What is an example of a subset $X$ of an amenable group $G$ such that $X$ is left amenable with respect to $G$ but not right amenable with respect to $G$?


Tarski proved that $X$ is left (right) amenable with respect to $G$ if and only if $X$ fails to admit a left (right) $G$-paradoxical decomposition. One current reference for this result is the book "The Banach-Tarski Paradox" by G. Tomkowicz and S. Wagon (Corollary 11.2, page 197 in the second edition).


Hence the question posed is asking for an example of a subset $X$ of an amenable group $G$ that admits a right $G$-paradoxical decomposition but no left $G$-paradoxical decomposition.



Amenable subsets of groups
Here is a slightly modified version of an example that was communicated to the author of the question by Nicolas Monod.


Let $k>1$ be an integer and $G$ the Baumslag-Solitar group $\mathrm{BS}(1,k)$; that is,
$$G:=\langle a, b\ :\ bab^{-1}=a^k\rangle.$$
Let $A$ and $B$ be the infinite cyclic subgroups generated by $a$ and $b$, respectively. Notice that $G=A^G\rtimes B$, where $A^G\cong \mathbb Z[1/k]$ is the normal closure of $A$ in $G$.
Since $G$ is metabelian, it is an amenable group.


Put $X:=AB$. We claim that, relative to $G$, the subset $X$ is right amenable but not left amenable.
To prove the latter assertion, write $X\_0:=\langle a^k\rangle B$. Observe that $X\_0$ and $aX\_0$ are disjoint subsets of $X$, and that $bX=X\_0$. Hence the existence of a measure $\mu:\mathcal{P}(G)\to [0,\infty]$ satisfying properties (1), (2), and (3) above would imply
$$2=\mu(X)+\mu(X)=\mu(X\_0)+\mu(X\_0)=\mu(X\_0)+\mu(aX\_0)\leq \mu(X)=1.$$
Therefore $X$ is not left amenable with respect to $G$.


To establish the right amenability of $X$, we invoke the following proposition, due to J.M. Rosenblatt and appearing as Corollary 1.5 in his paper "A generalization of Følner's condition (*Math. Scand*. **33** (1973), 153-170)." Rosenblatt states the proposition in terms of left amenability; we have repurposed it for right amenability. Following Rosenblatt, we employ the following notation: for any $n$-tuple $(u\_1,\dots,u\_n)$ and any set $X$,
$$\|X\cap(u\_1,\dots,u\_n)\|:=|\{ i=1,\dots, n:u\_i\in X\}|,$$
where the '$|\ \ |$' notation on the right is for cardinality.


**Proposition** (Rosenblatt)**.** Let $G$ be an amenable group. A subset $X$ of $G$ is right amenable with respect to $G$ if and only if, for every $m$-tuple $(u\_1,\dots, u\_m)$ and $n$-tuple $(v\_1,\dots,v\_n)$ from $G$ such that $m<n$, there exists $g\in G$ such that
$$\|gX\cap (u\_1,\dots,u\_m)\|<\|gX\cap (v\_1,\dots, v\_n)\|.$$


Let $\{t\_{\alpha}:\alpha\in I\}$ be a complete set of coset representatives of $A$ in $A^G$. Then
$\{t\_{\alpha}X:\alpha \in I\}$ is a partition of $G$. In order to apply Rosenblatt's proposition, we let $(u\_1,\dots,u\_m)$ and $(v\_1,\dots, v\_n)$ be finite sequences from $G$ with $m<n$. Then
$$m=\sum\_{\alpha\in I} \|t\_{\alpha}X\cap (u\_1,\dots u\_m)\|\ \ \mbox{and}\ \ n=\sum\_{\alpha\in I} \|t\_{\alpha}X\cap (v\_1,\dots, v\_n)\|.$$
Thus it must be the case that, for some $\alpha\in I$,
$$\|t\_{\alpha}X\cap (u\_1,\dots,u\_m)\|<\|t\_{\alpha}X\cap (v\_1,\dots, v\_n)\|.$$
Hence $X$ is right amenable with respect to $G$.


The set $X^{-1}$ is, then, left amenable but not right amenable, answering the question as it was formulated.


