Paradoxical decomposition

Remarks

Example: let $G = X = F_2$, the free group on $a, b$, acting by left multiplication. Let $W(x)$ be the set of reduced words that begin with the letter $x$. The four sets $W(a), W(a^{-1}), W(b), W(b^{-1})$ are pairwise disjoint. A reduced word $w$ not in $W(a)$ gives the reduced word $a^{-1} w \in W(a^{-1})$, so $F_2 = W(a) \cup a W(a^{-1})$, and likewise $F_2 = W(b) \cup b W(b^{-1})$. So $F_2$ is paradoxical, with $n = m = 2$.

Non-example: a finite nonempty set $E$ is never paradoxical, for any action. The pieces satisfy $\abs{E} \le \sum_i \abs{A_i}$ and $\abs{E} \le \sum_j \abs{B_j}$, so $2\abs{E} \le \sum_i \abs{A_i} + \sum_j \abs{B_j} \le \abs{E}$, which forces $E = \emptyset$.

In the definition one can ask the pieces $g_i A_i$ to be pairwise disjoint too, and likewise the $h_j B_j$; this gives the same notion.