Disintegration theory for von Neumann algebras

A.5 Measurable selections
form a basis. If
s = (n 1 , n 2 , . . . , n k , 1, 1, . . .) and t = (n 1 , n 2 , . . . , n k + 1, 1, 1, . . .)
then B = {x ∈ N N | s ≤ x < t}, and we notice that p ∈ g −1 (B) if and only if f −1 (p) has its smallest
element in B. This happens if and only if there is an x < t such that f (x) = p but there are no y < s
such that f (y) = p. If B s denotes the set {x ∈ N N | x < s}, then we observe that g −1 (B) = f (B t )\ f (B s ).
If we write s = (s 1 , s 2 , . . .) then for any x = (x 1 , x 2 , . . .) ∈ B s it holds for some k that x j = s j for all
j < k and x k < s k . Then the set U x = {x 1 } × · · · × {x k } × N × N × · · · is open in N N , contains x and is
contained in B s . This proves that B s is open. Hence B s is analytic, and so is f (B s ). Similarly, f (B t )
is analytic, and it follows from Theorem A.19 that f (B t ) \ f (B s ) is measurable. This proves that g is
measurable.
□
Theorem A.27 (Measurable selection theorem). Let X and Y be a Polish spaces, and let µ be the
completion of a Borel measure on X. Let π X : X × Y → X denote the projection. Suppose there is a
sequence (K n ) of compact subsets of X with finite µ-measure and union X. If B is an analytic subset
of X × Y and A = π X (B), there is a measurable map η : A → Y such that (p, η(p)) ∈ B for every
p ∈ A.
Proof. Let Y p = {q ∈ Y | (p, q) ∈ B}. Then since A is the projection of B, it follows that A is the
set of points p such that Y p is non-empty. The aim is to find a measurable map η : A → Y such that
η(p) ∈ Y p for every p.
Since B is analytic, there is a continuous map h : N N → X × Y with image B. Composing with π X we
get a continuous map π X ◦h of N N onto A. From Lemma A.26 there is a measurable map g : A → N N
such that π X ◦ h ◦ g = id A . Let η = π Y ◦ h ◦ g, where π Y : X × Y → Y is the projection onto Y. The
map η is measurable A → Y.
It remains to check that η(p) ∈ Y p for each p ∈ A. Let h(g(p)) be denoted (p ′ , q) ∈ B. Then η(p) = q,
and since p ′ = π X (h(g(p))) = id A (p) = p, it follows that q ∈ Y p . This completes the proof. □
42