. ?w-y-!-prob, Gr d .E// is T 0 . We claim that there exists 2 Prob.Gr d .E// such that the set ? 1 .G/ has full measure in Y . The standard argument is similar to the prof of Proposition . : for a countable basis B i for the topology of Gn Prob.Gr d .E//, the set \ ¹B i j ? 1 .B i / is full in Y º \ \ ¹B c i j ? 1 .B i / is null in Y º is clearly a singleton, whose preimage is of full measure in Y . Let be a preimage of this singleton in Prob, Gr d .E//. By Theorem . the action of G on Prob.Gr d .E// is almost algebraic (as the action of G on Gr d .E/ is almost algebraic by Proposition . ), and the quotient topology on Gn Prob

;. By and H. Oe-g,-as-h-normalizes-h-0-;-e/-d-i.e/, Using the ?-measure embedding I.E/ ,! Prob.I.E// and obtain a new f -map Y ! Prob.I.E//. We are then reduced to the case d D 0, to be discussed below. We consider now the case d D 0, that is we assume having an f -map Y ! Prob.I.E//. We set V D Prob.I.E//. By Lemma . , G acts isometrically and with bounded stabilizers on I.E/. By Lemma . , G acts isometrically on V . Let us check that stabilizers are bounded. Fix 2 Prob.I.E//, and let L be its stabilizer in G. Since I.E/ is Polish there is a ball B of I.E/ such that .B/ > 1=2. It follows that for any g 2 L, gB intersects B. Thus the set LB is bounded in I.E/, and by Lemma . its stabilizer is bounded in G, Theorem AG

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