Disintegration theory for von Neumann algebras

Measure theoretic results
Before the proof of the theorem, it will be convenient to prove a few lemmas. The following notation
will be usefull. If (X, Ω, µ) is a measure space, define µ ∗ on all subsets A of X by
µ ∗ (A) = inf{µ(S ) | A ⊆ S, S ∈ Ω}.
It is obvious that µ ∗ is monotone in the sense that if A ⊆ B ⊆ X then µ ∗ (A) ≤ µ ∗ (B).
Lemma A.20. For any A ⊆ X there is S ∈ Ω containing A such that µ ∗ (A) = µ(S ).
Proof. Let S n be a measurable set containing A such that µ(S n ) ≤ µ ∗ (A) + 1 n , and let S = ⋂ n S n .
Then S ∈ Ω and µ(S ) ≤ µ ∗ (A). Since A ⊆ S , it is clear that µ ∗ (A) ≤ µ(S ).
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Lemma A.21. If A 1 ⊆ A 2 ⊆ · · · is an ascending chain with union A, then µ ∗ (A) = lim µ ∗ (A n ).
Proof. Obviously lim µ ∗ (A n ) ≤ µ ∗ (A), since each A n ⊆ A. Choose measurable sets S n such that A n ⊆
S n and µ ∗ (A n ) = µ(S n ) as garuanteed by the previous lemma. Let T n = ⋂ k≥n S k . Then A n ⊆ T n ⊆ S n ,
and hence µ(T n ) = µ(S n ) = µ ∗ (A n ). Notice that (T n ) is an ascending chain of measurable sets with
union T containing A. Thus µ ∗ (A) ≤ µ(T). And also µ(T) = lim µ(T n ) = lim µ ∗ (A n ). This proves the
lemma.
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Lemma A.22. Suppose µ is complete and finite. If A ⊆ X, and for each m ∈ N there is a measurable
subset C m of A such that µ ∗ (A) − 1 m ≤ µ(C m), then A is measurable.
Proof. First choose S ∈ Ω such that A ⊆ S and µ ∗ (A) = µ(S ). Let C = ⋃ m C m . Then C is a
measurable subset of A and µ(C) ≥ µ ∗ (A) by the assumptions on C m . Also µ(C) ≤ µ(S ) = µ ∗ (A) and
C ⊆ A ⊆ S , and hence µ(S \ C) = 0, since µ is finite. So A is the union of the measurable set C and
the null set A \ C.
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We are almost ready to prove Theorem A.19, but first we will observe that the proof can be reduced
to a simpler case. Let A be an analytic subset of X. Since K n is analytic (being compact) the set
A ∩ K n is analytic. Since A = ⋃ n(A ∩ K n ), it will suffice to prove the theorem in the case where X is
compact and of finite measure. These assumptions are in force during the following two proofs.
Lemma A.23. Let A be an analytic subset of X. There is a compact Polish space Y, a descending
sequence (Y n ) of σ-compact subsets of Y with intersection B and a continuous map f : Y → X such
that f (B) = A.
Proof. Let R n denote a copy of the one-point compactification of R for each n ∈ N. Let Z = ∏ ∞
n=1
R n
be the countable product of the one-point compactifications. We may view N N as a subset of Z. Let
Y = Z × X. Then Y is a compact Polish space (X is assumed to be compact).
As mentioned after Definition A.16 there is a continuous map g : N N → X with image A. Let B
denote the graph of g in Z × X, and let f : Z × X → X be the projection. Then f is continuous and
f (B) = A.
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