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Suppose that $G$ is a discrete countable group and $\mu$ is an IRS (invariant random subgroup) of $G$: $\mu$ is conjugation invariant as a probability measure on the subgroups of $G$.

Since all the $G$-conjugation-invariant probability measures on the subgroups of $G$, IRS$(G)$,which is a convex, compact space, is a Choquet simplex, $\mu$ admits an ergodic decomposition: there is a unique measure $\upsilon$ on the extreme points of IRS$(G)$, or on the ergodic IRS's, such that for any continuous function $f: \text{Subgroups}(G) \to \mathbb{R}$ it holds $\int f d\mu = \int_{\text{ergodic IRS's}}[\int_{\text{Subgroups}(G)}fdm]d\upsilon(m)$.

Now suppose that there is a measurable $E \subset \text{Subgroups}(G)$ such that $\mu(E)=1$.

Define $E^* = \{\lambda \in \text{ergodic IRS's}: \lambda(E) = 1\}$

Can something be said about $\upsilon(E^*)$?

I'm attempting to show that $\upsilon(E^*) = 1$, in fact.

This may follow if we can show that, defining $\mu_o(A) = \mu(A\cap E)$, then the ergodic decomposition of $\mu_0$ is $\upsilon(A^* \cap E^*)$ for any measurable $A^* \subset$ Ergodic IRS's.

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What you are asking about has nothing to do with the space of subgroups and is true for any measure class preserving action. One just has to write the definition of the ergodic composition in the measure category: $$ \mu(E) = \int m(E) \,d v(m) \;. $$

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  • $\begingroup$ Can you please elaborate on why this equals $\upsilon(E^*)$? I guess I'm missing some background. $\endgroup$
    – Mariah
    Commented Apr 30, 2020 at 21:42
  • $\begingroup$ To reformulate: $\mu(E)=1$ iff $m(E)=1$ for $v$-a.e. $m$. $\endgroup$
    – R W
    Commented Apr 30, 2020 at 21:58

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