Suppose a group $G$ acts on an infinite set $X$ and $X$ has no non-empty $G$-paradoxical subsets. Is it possible for $X$ to have non-trivial $G$-paradoxical subsets modulo finite sets? I.e., can there be infinite subsets $E\_1,...,E\_n$ such that $E\_i \cap E\_j$ is finite when $i\ne j$, and $\tau\_1,\dots,\tau\_n \in G$ such that $E\_1\cup \dots \cup E\_n$, $\tau\_1 E\_1 \cup \dots \cup \tau\_m E\_m$ and $\tau\_{m+1} E\_{m+1} \cup \dots \cup \tau\_n E\_n$ differ only by finite sets for some $m$?


**Note:** When $G$ acts transitively (for any $x$ and $y$ there is a $\tau\in G$ such that $y=\tau x)$, the answer is negative. For let $E=\bigcup E\_i$. Because $G$ has no paradoxical sets, there is a $G$-invariant finitely additive measure $\mu:2^X \to [0,\infty]$ with $0<\mu(E)<\infty$. Because $G$ acts transitively, all singletons have equal measure, and since $E$ is infinite and has finite measure, all finite sets must have measure zero. Paradoxicality of $E$ modulo finite sets would then imply that $\mu(E)=\mu(E)+\mu(E)$, a contradiction.



Paradoxical decomposition modulo finite sets
No it's not possible.


Recall the notation. For $G$ acting on a set $X$, subsets $Y,Y'$ of $X$ are called $G$-congruent if $\exists g\in G$ such that $Y'=gY$, and piecewise $G$-congruent if there are partitions $(Y\_i)\_{1\le i\le n}$ of $Y$, $(Y'\_i)\_{1\le i\le n}$ of $Y$ such that $Y\_i$ and $Y'\_i$ are $G$-congruent for all $i$. The subset $Y$ of $X$ is $G$-paradoxical if there is a partition $Y=Y\_1\sqcup Y\_2$ such that $Y$ is piecewise $G$-congruent to both $Y\_1$ and $Y\_2$.


Then Tarski's theorem says that $Y$ is not $G$-paradoxical iff there exists a finitely additive $G$-invariant measure $\mathcal{P}(X)\to [0,\infty]$ mapping $Y$ to $1$.


Next one can define near (piecewise) $G$-congruent, near $G$-paradoxical by allowing "modulo finite subsets".


**Proposition** $Y\subset X$ is near $G$-paradoxical iff it can be written as union of a $G$-paradoxical subset and a finite set.


Proof: $\Leftarrow$ is trivial. Let us prove $\Rightarrow$ by contraposition. That is, assume that $Y\smallsetminus F$ is not $G$-paradoxical for every finite subset $F$.


Let $Y\_0$ be the set of $y\in Y$ such that $Gy\cap Y$ is finite. Say that a finite subset $F$ of $Y$ is $G$-saturated if $GF\cap Y\subset F$ (this forces $F\subset Y\_0$).


For each $G$-saturated $F$, by Tarski's theorem, let $m\_F:\mathcal{P}(X\smallsetminus GF)\to [0,\infty]$ be a finitely additive $G$-invariant measure mapping $Y\smallsetminus F$ to 1; extend it to $\mathcal{P}(X)$ as being zero on $GF$.


Let $m$ be a limit point of $m\_F$ when $F$ tends to $Y\_0$. So $m$ is a finitely additive $G$-invariant measure mapping $Y\smallsetminus F$ to 1; moreover $m$ vanishes on every singleton of $Y\_0$. Then $m$ vanishes on every singleton (if $y\in Y\smallsetminus Y\_0$ then $Gy\cap Y$ is infinite).
So $Y$ is not near $G$-paradoxical. The proof is complete.


**Corollary** If every nonempty $Y\subset X$ is non-paradoxical, then every infinite $Y\subset X$ is non-near-paradoxical.


This answers the question.


