Amenable group $G$ and a (left) invariant mean on it.

Question

While reading "The Banach-Tarski Paradox" by S. Wagon, I faced the definition of a amenable group: a group that possesses a finitely additive, $G$-invariant measure on all subsets, with total measure one. (From now on, I will use the word 'measure' with this meaning only.) After some googling, I noticed this is actually what people call a 'discrete' amenable group, since no topology is considered on $G$.

After the definition, the book proceeds to explain that $G$ is amenable if and only if it possesses a left invariant mean. A left invariant mean is, by definition, a linear functional (denoted by $\int \,d\mu$) on $B(G)$, the collection of bounded real-valued functions on $G$, with the properties;

1. $\int af+bg \,\,d\mu=a\int f \,d\mu+b\int g \,d\mu$ if $a,b\in\mathbb{R}$;
2. $\int f \,d\mu\geq0$ if, for all $x\in G$, $f(x)\geq 0$;
3. $\int 1 \,d\mu=1$;
4. For each $g\in G, f\in B(G)$, $\int {}_{g}f \,d\mu=\int f \,d\mu$, where $({}_{g}f)(x)=f(g^{-1}x)$; that is, the integral is left-invariant.

If a left invariant mean exists, then using characteristic functions we immediately get a measure for $G$. BUT the problem I'm having is the opposite direction, constructing a left invariant mean from a measure, say $\mu$.

Obviously, we can try following the usual construction of Lebesgue integration, and define integral for simple functions, non-negative functions, and so on. But to my knowledge, the additivity of the integral requires Lebesgue's monotone convergence theorem, which relies on the countable additivity of the measure. Since the measure we have is only finitely additive, convergence theorems fail to hold and it seems like we have a serious problem with this approach!

To summarize, my questions are ..

1. Why does the existence of measure implies that of a left invariant mean?
2. Is there a way to show that integrals are linear functionals, avoiding the well known convergence theorems?

Please enlighten me.

Answer 1

Given a left invariant finite additive probability measure $\mu$, we define $\Lambda$ on simple functions by $$ \Lambda \left( \sum_{k=1}^n a_k \mathbb{I}_{E_k} \right) = \sum_{k=1}^n a_k \mu \left( E_k \right) $$ if $E_k$ is disjoint. Noticed that simple functions is dense in $B(G)$ and $\Lambda$ is a non-negetive, left invariant and bounded linear functional, so $\Lambda$ can be extended on $B(G)$ by continuity.