Let $G$ be a locally compact, second countable, non-amenable group, let $X$ be a Haudorff space that is not necessarily compact, and let $G \curvearrowright X$ be a topological action that is free (i.e., all stabilizers are trivial) and minimal (i.e., every orbit is dense). I would like to show (or disprove) that if there exists a $G$-invariant mean on $X$ then there exists an $G$-invariant Borel probability measure on $X$.

This obviously holds if the action $G \curvearrowright X$ is transitive, since then it is isomorphic to the left translation action of $G$ on itself. It should also be easy to show that this is true when $G$ has property (T).

If needed, we can also assume that there exists an invariant $\sigma$-finite Borel measure on $X$, and that $X$ is "nice": Polish, locally compact, or whatever.


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    $\begingroup$ Isn't there a version of the Riesz Representation theorem that equates bounded, finitely additive measures with the dual of C(X)? $\endgroup$ – Ben Willson May 28 '15 at 5:06
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    $\begingroup$ rather of $\mathcal{L}^\infty(X)$, the space of bounded measurable functions. $\endgroup$ – YCor May 28 '15 at 5:43

Let $G$ be a non-compact locally compact amenable group which contains a dense countable group $\Gamma$ which is non-amenable as a discrete group (e.g., $G = E(3)$ by Banach-Tarski). Then the action of $\Gamma$ on $G$ is free and minimal since $\Gamma$ is dense, has an invariant mean since $G$ is amenable, and does not have an invariant Borel probability measure since $\Gamma$ is dense and hence such a measure would be $G$-invariant, but $G$ is non-compact.

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  • $\begingroup$ Thanks! This works and is very simple (unfortunately for me). $\endgroup$ – Vladimir May 28 '15 at 20:29

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