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Extra info for Global aspects of ergodic group actions
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Next we will see that N [E] is a Polishable subgroup of (Aut(X, µ), w), if E is aperiodic. 6. AUTOMORPHISM GROUPS OF EQUIVALENCE RELATIONS 41 Note that each T ∈ N [E] induces by conjugation an isometry iT of ([E], δu ) : iT (S) = T ST −1 . Consider the Polish group Iso([E], δu ), with the pointwise convergence topology and the map i(T ) = iT . We first check that it is an algebraic isomorphism of N [E] with a subgroup of Iso([E], δu ), provided that E is aperiodic. It is clearly a homomorphism. 11.
Then if Fn = E ∪En , Fn is hyperfinite, measure preserving and has 2 ergodic invariant measures, so by the n = 2 case, we can find F ⊇ Fn , F hyperfinite, ergodic and measure preserving. Case 2. E is countably infinite. Say E = {e1 , e2 , . . ). By Case 1, there is hyperfinite, measure preserving, ergodic En on Xe1 ∪· · ·∪Xen with E|Xe1 ∪· · ·∪E|Xen ⊆ En and En ⊆ En+1 . Let F = n En . This clearly works. Case 3. E is uncountable and ν is non-atomic. Let then T ∈ Aut(E, ν) be ergodic. For each e ∈ E, let ϕe : Xe → XT (e) be a Borel bijection such that ϕ(e, x) = ϕe (x) is Borel and ϕe is an isomorphism of (E|Xe , e) with (E|XT (e) , T (e)) modulo null sets.
Case 1. E is finite. We will prove then the result by induction on card(E). If card(E) = 1, there is nothing to prove. Assume now card(E) = 2, say E = {e1 , e2 }, with µ(Xe1 ) ≥ µ(Xe2 ). Note that ei = (µ|Xei )/µ(Xei ). By Dye’s Theorem, we can find X1 ⊆ Xe1 , with µ(X1 ) = µ(Xe2 ) and a Borel isomorphism ϕ from E|X1 to E|Xe2 (modulo null sets) which preserves µ. Then define F to 34 I. MEASURE PRESERVING AUTOMORPHISMS be the equivalence relation generated by E and ϕ. Clearly F is measure preserving and ergodic.