Disintegration theory for von Neumann algebras

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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 34
Measure theoretic results Proof. Suppose first that f is real-valued. Define measurable sets A n = {x ∈ X | f (x) ≥ 1 n } where n ∈ N. Since ∫ ∫ 1 0 = f dµ ≥ A n A n n dµ = µ(A n) n it follows that each A n is a null set. Hence the union A = ⋃ n A n = {x ∈ X | f (x) > 0} is a null set. Similarly, B = {x ∈ X | f (x) < 0} is a null set, and then f (x) = 0 almost everywhere. If f is arbitrary, then write f = g + ih for some real-valued, measurable functions g, h. Then f (x) = 0 if and only if g(x) = h(x) = 0. It follows by the previous case, that g(x) = h(x) = 0 almost everywhere, and hence f (x) = 0 almost everywhere. □ A.3 Polish spaces Definition A.4 (Polish space). A topological space X is topologically complete, if it is homoemorphic to a complete metric space or, alternatively, if X admits a metric for the topology with which it is complete. A topological space is called a Polish space, if it is separable and topologically complete. Example A.5. The real line R and the closed unit interval [0, 1] are standard examples of Polish spaces. The open unit interval ]0, 1[ is also Polish, since it it homeomorphic to R. Any closed subspace of a Polish space is itself a Polish space, and any countable direct sum (disjoint union) of Polish spaces is Polish. Example A.6. If H is a separable Hilbert space, the closed unit ball in B(H) endoved with its weak operator, strong operator or strong ∗ operator topology is a Polish space. For example, the strong operator topology on the unit ball is induced by the complete metric d(S, T) = ∑ ∞ n=1 2 −n ||(S − T)e n ||, where {e n } ∞ n=1 is an orthonormal basis for H. The details of verifying this are left to the reader. Remark A.7. Whenever d is a metric for X, the standard bounded metric d(x, y) = min{d(x, y), 1} induces the same topology, and X is complete under d if and only if X is complete under d. So in our context we can always assume that our metrics are bounded with bound 1. Proposition A.8. Any countable product of Polish spaces is a Polish space. Proof. Let (X n , d n ) ∞ n=1 be Polish spaces. We assume each d n is a bounded metric, bounded by 1. Let d be the metric on the product space X = ∏ X n defined by d(x, y) = ∞∑ 2 −n d n (x n , y n ), where x = (x n ), y = (y n ). n=1 The metric d induces the product topology on X, and (X, d) is complete. To see that X is separable, choose a point p n in each X n , and let {p (n) 1 , p(n) 2 , . . .} be a countable dense subset of X n. Then D = {(p (1) k 1 , p (2) k 2 , . . . , p (n) k n , p n+1 , p n+2 , . . .) | n, k 1 , . . . , k n ∈ N} 35
Page 1: D E P A R T M E N T O F M A T H E M
Page 4 and 5: Abstract The main theorem is the ce
Page 6 and 7: Neumann algebras intensively in the
Page 8 and 9: 1.1 Maximal abelian algebras acting
Page 10 and 11: 1.1 Maximal abelian algebras acting
Page 12 and 13: 1.2 The commutant of an abelian von
Page 14 and 15: 2.1 Disintegration of Hilbert space
Page 16 and 17: 2.1 Disintegration of Hilbert space
Page 18 and 19: 2.2 Decomposable operators Now we r
Page 20 and 21: 2.2 Decomposable operators Defining
Page 22 and 23: 2.2 Decomposable operators From Lem
Page 24 and 25: 2.3 Decomposable representations Ex
Page 26 and 27: 2.4 Decomposable von Neumann algebr
Page 28 and 29: 2.5 Decompositions and the commutan
Page 36 and 37: 3.1 Decomposition relative to an ab
Page 38 and 39: 3.2 Type of the components Theorem
Page 42 and 43: A.3 Polish spaces is countable and
Page 44 and 45: A.4 Analytic sets Remark A.15. The
Page 46 and 47: A.4 Analytic sets For n ∈ N consi
Page 48 and 49: A.5 Measurable selections form a ba

Disintegration theory for von Neumann algebras

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