= Solution
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 $A=\bigsqcup_iA_i$, $B=\bigsqcup_iB_i$ and elements $g_i\in G$ such that $g_iA_i=B_i$. 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 $F_2=\langle a,b\rangle$ 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 decompositions
$$
F_2=W(a)\sqcup aW(a^{-1}),
\qquad
F_2=W(b)\sqcup bW(b^{-1}).
$$
If an invariant measure $m$ existed, these would imply
$$
1=m(W(a))+m(W(a^{-1})),
\qquad
1=m(W(b))+m(W(b^{-1})).
$$
Every 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
$$
\boxed{F_2\text{ is not amenable}.}
$$
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