Disintegration theory for von Neumann algebras

A.2 Measure theoretic technicalities
the sets in Ω. Of course, Borel sets are measurable, but the converse is not always true.
If (X, Ω) and (Y, Σ) are measurable spaces, a map f : X → Y is called measurable, if the preimage
of any set from Σ is in Ω, i.e f −1 (B) ∈ Ω for any B ∈ Σ. In the case where (X, Ω, µ) is a Borel
measure space, the map f is called a Borel map if f −1 (B) ∈ Ω for any B ∈ Σ, and a measurable map
if f −1 (B) ∈ Ω for any B ∈ Σ.
To check measurability of a map f : (X, Ω) → (Y, Σ) it suffices to check that f −1 (B) ∈ Ω for any B in
a generating family for the σ-algebra Σ.
A.2 Measure theoretic technicalities
Lemma A.1. Let (X, µ) be a complete Borel measure space (so that µ is the completion of a Borel
measure on X), and let (Y, Σ) be a measurable space, where Σ is a countably generated σ-algebra on
Y. If f : X → Y is a measurable map, there is a Borel null set N of X such that f |(X \ N) is a Borel
map.
Proof. Let {Y n } be a countable family of measurable sets of Y generating Σ. Then f −1 (Y n ) is measurable
in X and is therefore the union of a Borel set X n and a null set M n . Let N n be a Borel null set
containing the null set. Let N = ⋃ n N n . Of course N is a Borel null set, and also
f −1 (Y n ) ∩ (X \ N) = X n ∩ (X \ N).
The set X n ∩ (X \ N) is a Borel subset of X \ N, and since {Y n } generates Σ it follows that f |(X \ N) is
a Borel map.
□
Lemma A.2. Let (X, µ) be a complete measure space, f : X → C and g : X → C complex-valued
functions on X. Suppose f is measurable, and f and g agree almost everywhere. Then g is also
measurable.
Proof. Let N denote the null set {x ∈ X | f (x) g(x)}. Without loss of generality, it may be assumed
that f is constantly zero. Let U ⊂ C be a measurable set. If 0 U, then g −1 (U) ⊆ N is measurable,
because µ is complete. If 0 ∈ U, let V = U \ {0}. Then g −1 (V) is measurable and
g −1 (U) = g −1 (V) ∪ g −1 ({0}) = g −1 (V) ∪ (X \ N).
Hence g −1 (U) is measurable in this case as well.
□
Lemma A.3. Let (X, µ) be a measure space, and let f : X → C be a measurable, integrable function.
If
∫
f dµ = 0
for each measurable subset Y of X, then f (x) = 0 almost everywhere.
Y
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