By Ollav Kallenberg

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The statements in (ii) and (iii) are easy consequences. 27 (µ ⊗ ν)s f = µ(s, dt) ν(s, t, du)f (t, u), s ∈ S, or simply (µ ⊗ ν)f = µ(νf ). By iteration we may combine any kernels µk from S0 × · · · × Sk−1 to Sk , k = 1, . . , n, into a kernel µ1 ⊗ · · · ⊗ µn from S0 to S1 × · · · × Sn , given by (µ1 ⊗ · · · ⊗ µn )f = µ1 (µ2 (· · · (µn f ) · · ·)) for any measurable function f ≥ 0 on S1 × · · · × Sn . In applications we may often encounter kernels µk from Sk−1 to Sk , k = 1, . . , n, in which case the composition µ1 · · · µn is deﬁned as a kernel from S0 to Sn given for measurable B ⊂ Sn by (µ1 · · · µn )s B = (µ1 ⊗ · · · ⊗ µn )s (S1 × · · · × Sn−1 × B) = µ1 (s, ds1 ) ··· µ2 (s1 , ds2 ) · · · µn−1 (sn−2 , dsn−1 )µn (sn−1 , B).

22 (kernels and randomization) Let µ be a probability kernel from a measurable space S to a Borel space T . Then there exists some measurable function f : S × [0, 1] → T such that if ϑ is U (0, 1), then f (s, ϑ) has distribution µ(s, ·) for every s ∈ S. Proof: We may assume that T is a Borel subset of [0, 1], in which case we may easily reduce to the case when T = [0, 1]. 12, and the supremum in (9) can be restricted to rational x. 3. 22 there exist some measurable functions fn : [0, 1] → Sn such that λ ◦ fn−1 = µn .

D. U (0, 1). 34 Foundations of Modern Probability Proof: Introduce for every x ∈ [0, 1] the associated binary expansion g1 (x), g2 (x), . . , and note that the gk are measurable. Rearrange the gk into a two-dimensional array hnj , n, j ∈ N, and deﬁne fn (x) = j 2−j hnj (x), x ∈ [0, 1], n ∈ N. 20 the random variables gk (ϑ) form a Bernoulli sequence with rate 12 , and the same result shows that the variables ϑn = fn (ϑ) are U (0, 1). 7. ✷ Finally, we need to construct a random element with arbitrary distribution from a given randomization variable.

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