Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 143 / 4 / a

An amenable group is a group $G$ admitting a left-invariant finitely additive probability measure $m:\mathcal{P}(G)\to [0,1]$. For a $G$-set $X$, subsets $A,B\subseteq X$ are equidecomposable subsets under a group action if there are finite partitions $A={⨆}_i{A}_i$, $B={⨆}_i{B}_i$ and elements $g_i\in G$ such that $g_i{A}_i=B_i$. The action is a paradoxical group action if $X$ contains two disjoint subsets, each $G$-equidecomposable with $X$.

To prove the nonamenability of a nonabelian free group, write $F_2=⟨a,b⟩$ and let $W(s)$ be the set of nonempty reduced words beginning with $s\in \{a,a^{-1},b,b^{-1}\}$. Cancellation of the first letter gives the disjoint decompositionsIf an invariant measure $m$ existed, these would implyEvery singleton has measure zero: invariance gives all singletons the same measure, and finite additivity over arbitrarily many distinct points forces that measure to vanish. The four sets $W(s)$ partition $F_2\setminus \{1\}$, so their measures sum to one. The two displayed equations say that the same sum is two, a contradiction. Hence