TITLE: Topological amenability vs amenability of an action
QUESTION [12 upvotes]: Let $G$ be a discrete group and let $X$ be a compact, Hausdorff space.
Assume that $G$ acts on $X$ by homeomorphisms.
Consider the following two definitions:

[$C^*$-algebras and finite dimensional approximations- Brown and Ozawa- Definition 4.3.5.]


The action is (topologically) amenable  if there exists a net of
  continuous maps $m_i: X\to Prob(G)$ s.t. for each $s\in G$: $$
 \lim_{i\to \infty} (\sup_{x\in X} \|s.m_i^x-m_i^{s.x}\|_1)=0$$ where
  $s.m_i^x(g)=m_i^x(s^{-1}g)$.

Now, regard $X$ as a $G$-set. Then one can define:
2.[Amenability of Groups and G-Sets, see definition in the introduction, for example]:

The action is amenable if there exists an invariant mean on the power
  set of $X$.

I would expect that (1) will imply (2). Namely, that if the action of $G$ on $X$ is topologically amenable then it is amenable (set-theoretically).
However, I know very little about amenable actions and I would appreciate any help. 
Thanks.

REPLY [9 votes]: The terminology is unfortunately confusing as these properties are in a sense "orthogonal". The definition in terms of existence of an invariant mean on the action space (your definition 2) goes back to von Neumann, whereas your definition 1 goes back to Zimmer who introduced it (in the measure category and in a somewhat different form though) in 1977. Amenability of the group $G$ is equivalent to the amenability in the sense of Definition 1 of its trivial action on the one-point space and to the amenability in the sense of Definition 2 of its action on itself.
For an explicit counterexample to your claim take the boundary action of a free group. It is topologically amenable in the sense of Definition 1, but not amenable in the sense of Definition 2. 
A counterexample in the opposite direction is provided just by the trivial action of any non-amenable group on a singleton.