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Consider a measure-preserving action Γ (X, µ) of a countable group Γ and a measurable cocycle α : X × Γ → Aut(Y) with countable image , where (X, µ) is a standard Lebesgue space and (Y, ν) is any probability space. We prove that if the Koopman representation associated to the action Γ X is non-amenable, then there does not exist a countable-to-one Borel(More)
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