Paradoxical decomposition modulo finite sets 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.
 A: 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.
