A (discrete) group is amenable if it admits a finitely additive probability measure (on all its subsets), invariant under left translation. It is a basic fact that every abelian group is amenable. But the proof I know is surprisingly convoluted. I'd like to know if there's a more direct proof.

The proof I know runs as follows.

1. Every finite group is amenable (in a unique way). This is trivial.

2. $\mathbb{Z}$ is amenable. This is not trivial as far as I know; the proof I know involves choosing a non-principal ultrafilter on $\mathbb{N}$. This means that $\mathbb{Z}$ is amenable in many different ways, i.e. there are many measures on it, but apparently you can't write down any measure 'explicitly' (without using the Axiom of Choice).

3. The direct product of two amenable groups is amenable. This isn't exactly trivial, but the measure on the product is at least constructed canonically from the two given measures.

4. Every finitely generated abelian group is amenable. This follows from 1--3 and the classification theorem.

5. The class of amenable groups is closed under direct limits (=colimits over a directed poset). This is like step 2: it seems that there's no canonical way of constructing a measure on the direct limit, given measures on each of the groups that you start with; and the proof involves choosing a non-principal ultrafilter on the poset.

6. Every abelian group is amenable. This follows from 4 and 5, since every abelian group is the direct limit of its finitely generated subgroups.

Is there a more direct proof? Is there even a one-step proof?

Apologies if this is well-known to those in the field.