I have the following statement: if all groups in the directed system $\{G_{i}\}_{i\in I}$ are amenable, then so is their directed union $G:= \bigcup_{i\in I}G_{i}$ (Remember that a group is amenable if it admits a finitely aditive and left invariant measure. In the proof below, each $G_{i}$ has measure $\mu_{i}$.)
Let $M_{i} := \{\mu:\mathit{P}(G)\longrightarrow [0,1] \mbox{ such that } \mu \mbox{ is finitely aditive measure and } \mu(gA)=\mu(A) \forall g\in G_{i}\}$
Note that each $M_{i}$ is nonempty since we can define $\mu(A):= \mu_{i}(A\cap G_{i})$. Also, we have that $[0,1]^{\mathit{P}(G)}$ is compact by Tychonoff's theorem and that each $M_{i}$ is closed in $[0,1]^{\mathit{P}(G)}$.
I do not see why the $M_{i}$ are closed in $[0,1]^{\mathit{P}(G)}$.