By Sergey V. Ludkovsky

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**Extra resources for Quasi-invariant and pseudo-differentiable measures in Banach spaces **

**Example text**

1) Then (Mµ , r) is the complete pseudo-metric space. Proof. Suppose that an ∈ Mµ and limn→∞,m→∞ r(an , am ) = 0. We show that there exists a ∈ Mµ for which limn→∞ r(an , a) = 0. Mention that the sequence {an : n} is bounded in (X, ∗ ). If ank → ∞ with k → ∞, choose nk such that r(ank , ank+1 ) < 3−k . For B -measurable function with f L1 (µ) = 1 and with a bounded support we obtain N−1 X f (x + anN )µ(dx) ≥ X f (x + an )µ(dx) − ∑ r(an , an k k+1 ). k=1 Choosing f so that X f (x + an1 )µ(dx) > 1/2 and taking the limit with N → ∞ we infer that limN→∞ X f (x + anN )µ(dx) > 0, that is impossible for f with the bounded support, when anN → ∞.

Remark. From the proof of the preceding lemma we get that X φn+1 (y)µ(dy, U n |x) ≥ φn (x) (1) for each n. In the martingale case we have here the equality instead of the inequality. If X |φn (x)|µ(dx) < ∞ for each n, then {φn : n} is the sub-martingale. 39, then X gn (x)µ(dx) = X [gn (x) − φn (x)]µ(dx) + X φn (x)µ(dx). 38(2) ψ(x) = 1 we get that X fn (x)µ(dx) = X f1 (x)µ(dx) for the martingale { fn (x) = gn (x) − φn (x) : n}. Therefore, X gn (x)µ(dx) = X [g1 (x) − φ1 (x)]µ(dx) + X φn (x)µ(dx) = X φn (x)µ(dx) − X φ1 (x)µ(dx).

35. Theorem. Non-Archimedean analog of the Minlos-Sazonov theorem. 6(3, 4, 5) and for each c > 0 there exists a compact operator Sc : X → X such that |Re(θ(y) − θ(x))| < c for |˜z(Sc z)| < 1; (II) θ is a characteristic functional of a probability Radon measure µ on E, where z˜ is an element z ∈ X ֒→ X ∗ considered as an element of X ∗ under the natural embedding associated with the standard base of c0 (ω0 , K), z = x − y, x and y are arbitrary elements of X. Proof. (II → I). For a positive definite function θ generated by a probability measure µ in view of the inequality |θ(y) − θ(x)|2 ≤ 2θ(0)(θ(0) − Re(θ(y − x)) (see also Propositions 20 Sergey V.