The free group on two generators is not amenable

Suppose a left-invariant mean $m$ exists. For each subset $E\subseteq F_2$ put $\nu (E):=m({\mathbf{1}}_E)$, which is defined by [F1]. Positivity gives $\nu (E)\ge 0$ and monotonicity under inclusion; complex linearity gives finite additivity on disjoint sets. Since $L_g{\mathbf{1}}_E={\mathbf{1}}_{gE}$, left invariance gives $\nu (gE)=\nu (E)$ for every $g\in F_2$, and $\nu (F_2)=m(1)=1$.

Let $A$ be the set of reduced words whose initial maximal block is $a^k$ for some nonzero integer $k$; the empty word is not in $A$. Every word outside $A$ is either empty or begins with a nonzero power of $b$. In the first case $a^{-1}\in A$; in the second case the reduced word $a^{-1}w$ begins with $a^{-1}$ and is in $A$. Thus $F_2=A\cup aA$. By [F2], $\nu (aA)=\nu (A)$; monotonicity and finite subadditivity from step 1.1 give $1=\nu (F_2)\le \nu (A)+\nu (aA)=2\nu (A)$, hence $\nu (A)\ge \frac{1}{2}$.

The sets $A$, $bA$, and $b^{2}A$ are pairwise disjoint: their reduced words have initial maximal blocks respectively a nonzero power of $a$, exactly one $b$ followed by a nonzero power of $a$, and exactly two $b$'s followed by a nonzero power of $a$; no cancellation occurs at these joins. Therefore monotonicity, finite additivity, and left invariance from step 1.1 yield $1=\nu (F_2)\ge \nu (A)+\nu (bA)+\nu (b^{2}A)=3\nu (A)\ge \frac{3}{2}$, a contradiction. Hence no invariant mean exists, and the amenability definition shows $F_2$ is not amenable. This proves the claim and its stated discrete counterpart.