Extensions of amenable groups are amenable

Let $m_N$ be a left-invariant mean on $N$. For bounded $f:G\to \mathbb{R}$ and $g\in G$, define ${\mathrm{\Phi }}_f(gN):=m_N(n\mapsto f(gn))$. If $g^{\mathrm{\prime }}=gh$ with $h\in N$, then the integrand for $g^{\mathrm{\prime }}$ is $n\mapsto f(ghn)$, which is the left translate of $n\mapsto f(gn)$ by $h^{-1}$ in the $N$-variable. Thus the value of ${\mathrm{\Phi }}_f(gN)$ does not depend on the chosen representative of the right coset $gN$. The same formula also shows $\mathrm{\mid }{\mathrm{\Phi }}_f(gN)\mathrm{\mid }\le \mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$, so ${\mathrm{\Phi }}_f$ is bounded on $G\mathrm{/}N$.

Let $m_Q$ be a left-invariant mean on $G\mathrm{/}N$, and set $m_G(f)=m_Q({\mathrm{\Phi }}_f)$. Positivity and $m_G({\mathbf{1}}_G)=1$ are immediate. For $x\in G$, one has ${\mathrm{\Phi }}_{x\cdot f}(gN)=m_N(n\mapsto f(x^{-1}gn))={\mathrm{\Phi }}_f(x^{-1}gN)$, so ${\mathrm{\Phi }}_{x\cdot f}=xN\cdot {\mathrm{\Phi }}_f$. Therefore $m_G(x\cdot f)=m_Q(xN\cdot {\mathrm{\Phi }}_f)=m_Q({\mathrm{\Phi }}_f)=m_G(f)$.

Thus $m_G$ is a left-invariant mean on $G$, so [L1] makes $G$ amenable.