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Appendix Nonatomic Measure Spaces Appendix Nonatomic Measure Spaces In this appendix we collect some facts about finite measure spaces without (and with) atoms, which are applied in Chapter 7. Let (Ω, , µ) be a σ-finite measure space. Then an atom of µ is a set A with Sµ(A) > 0 such that for all C with C A, either µ(C) = 0 or µ(C∈) S = µ(A). By σ-finiteness, we have µ∈(A S) < + .(⊂Ω, , µ) or µ is called nonatomic if it has no atoms. ∞ S Proposition A.1. Let (Ω, , µ) be a nonatomic finite measure space with µ(Ω) > 0. Then for any cSwith 0 < c < µ(Ω), there is an A with µ(A) = c. ∈ S Proof. We can assume that µ(Ω) = 1. It will first be shown that for some C , 1/3 µ(C) 2/3. Suppose not. Let p := sup µ(B): µ(B) < 1/3 1/3.∈ SThen p≤ > 0 since≤Ω is not an atomand if 2/3 < µ({D) < 1, let B := Ω }D ≤. Take B with µ(B ) p. Let E := n B . It will be shown by induction\ n ∈ S n ↑ n j=1 j that µ(En) < 1/3 for all n. This is true for n = 1. Assuming it holds for a given n, we have µ(E ) = µ(E BS ) < (1/3)+(1/3) = 2/3. Thus by the n+1 n ∪ n+1 assumption that µ takes no values in [1/3, 2/3], µ(En+1) < 1/3, completing ∞ the induction. Let E := n=1 Bn. Then µ(E) = p, so p < 1/3. Now, Ω E is not an atom, so take F Ω E with 0 < µ(F ) < 1 p. If µ(F ) (2/3)\ p, then pS < µ(E F ) 2/⊂3, contradicting\ the assumption− or the definition≤ − of p. But if µ(F ) >∪(2/3)≤ p, replacing F by Ω (E F ) leads to the same contradiction. So, it is shown− that for some C ,\ 1/3∪ µ(C) 2/3. ∈ S ≤ ≤ n Next, an inclusion-chain will be a finite sequence Cj j=0 with = n { } ⊂ S ∅ C0 C1 Cn = Ω. One inclusion-chain Cj j=0 will be a refinement ⊂ ⊂ · · · ⊂k { } of another Di i=0 iff for each i, Di = Cj for some j. Let 0 := ,Ω . Given { } n 2n C {∅ } an inclusion-chain n = Cj j=0, define a refinement n+1 of it by adjoining, C n { } C n n for each j = 1,..., 2 , a set Cj−1 Aj where Aj , Aj Cj Cj−1, and n n ∪ ∈ S ⊂ \ µ(Aj )/µ(Cj Cj−1) [1/3, 2/3]. Thus n are defined recursively for all n. \ ∈ C n Clearly, for each n and j = 1,..., 2n, if = Cn 2 , then µ(Cn Cn ) Cn { j }j=0 j \ j−1 ≤ R.M. Dudley and R. Norvaiša, Concrete Functional Calculus, Springer Monographs 645 in Mathematics, DOI 10.1007/978-1-4419-6950-7, © Springer Science+Business Media, LLC 2011 646 Appendix Nonatomic Measure Spaces n n 2n (2/3) 0 as n . Thus the values µ(Cj ) j=0 become dense in [0, 1] as → → ∞ { n } n . For any t [0, 1], let B := 2 Cn : µ(Cn) t . Then B → ∞ ∈ t n j=0{ j j ≤ } t ∈ S and µ(Bt) = t, completing the proof. 2 S S Proposition A.1 and induction on n give directly the following: Corollary A.2. Let (Ω, , µ) be a finite nonatomic measure space. Let ri S n for i = 1, . , n be numbers with ri > 0 and i=1 ri = µ(Ω). Then Ω can be decomposed as a union of disjoint sets Ai with µ(Ai) = ri for i = 1, . , n. ∈ SP Here are sufficient conditions under which atoms equal singletons up to sets of measure 0, as holds in familiar cases. A measurable space (Ω, ) is called separated iff for every x = y in Ω, there is some C containingS just one of x and y. 6 ∈ S Proposition A.3. Let (Ω, , µ) be a σ-finite measure space such that (Ω, ) is separated and is countablyS generated, i.e. for some countably many A S S j ∈ , is the smallest σ-algebra containing all Aj. Then for any atom A of µ thereS S is an x A such that µ(A x ) = 0. So, the singleton x is also an atom. This∈ holds in particular\{ if there} is a metric on Ω for{ which} it is separable and is its Borel σ-algebra. S Proof. Let A be an atom of µ. By σ-finiteness we have 0 < µ(A) < + . For each j, let B := A if µ(A A ) = µ(A) or B := Ac if µ(A A ) =∞ 0. Let j j ∩ j j j ∩ j C := A ∞ B . Then µ(C) = µ(A) and C is also an atom. If there exist ∩ j=1 j x = y in C then by the separated assumption there must exist Aj such that 6 T Aj contains one of x and y but not both, contradicting the definition of C. Thus C is a singleton x and µ(A x ) = 0 as stated. The Borel σ-algebra{ of} a separable\{ metric} space is generated by the count- ably many open balls with centers in a countable dense set and rational radii, and it clearly separates points. This completes the proof. 2 0 Let λ = λ(Ω, , µ) be the set of all functions G (Ω, , µ) such that µ G−1 Dhas aD boundedS density ξ with respect to Lebesgue∈ L measureS λ on R. ◦ G Such functions were characterized in Theorem 7.24. If the set λ(Ω, , µ) is nonempty then µ must be nonatomic as the following shows: D S Proposition A.4. Let (Ω, , µ) be a finite measure space and G 0(Ω, , µ). If µ has an atom then µ G−S1 has an atom y for some y. ∈L S ◦ { } Proof. Suppose that A is an atom of µ with c := µ(A) > 0. Restricting µ to A, we can assume∈ that S Ω = A. Let F (x) := µ(G−1(( , x])) [0, c] for each x R. Then F (x) 0 as x and F (x) −∞c as x ∈ + . If F (x) (0∈, c) for some x ↓ R, then→ the −∞ set G−1(( ↑, x]) →gives∞ a contradiction∈ to the fact that∈A is an atom. Thus F has−∞ only∈ one S point of Appendix Nonatomic Measure Spaces 647 increase, say y R, with F (x) = 0 for all x < y and F (x) = c for all x y, and so y is an∈ atom of µ G−1. This completes the proof. ≥2 { } ◦ On the other hand if (Ω, , µ) is nonatomic, then the set λ(Ω, , µ) is rich enough in the sense statedS by the following two propositions.D S Proposition A.5. For any nonatomic finite measure space (Ω, , µ) with S µ(Ω) > 0 and < a < b < , there is a G λ such that the den- sity ξ equals c1−∞, where c = µ(∞Ω)/(b a). ∈ D G [a,b] − Proof. If G exists such that the density ξG equals µ(Ω)1[0,1], then a+(b a)G satisfies the conclusion. So we can assume a = 0 and b = 1. Also, multiplying− µ by a constant, we can assume µ(Ω) = 1. By Proposition A.1, take A with µ(A ) = 1/2. Let A := Ω A , 11 ∈ S 11 12 \ 11 and G1 := (1/2)1A12 . At the jth stage we will have a decomposition of Ω j j j into 2 disjoint sets Ajk , k = 1,..., 2 , with µ(Ajk) = 1/2 and Aj,k = ∈ S j j Aj+1,2k−1 Aj+1,2k for each j 1 and k = 1,..., 2 . Let Gj := (k 1)/2 on ∪ j ≥ −1 − Ajk for k = 1,..., 2 . Then µ Gj is uniformly distributed over the points j j ◦ (k 1)/2 for k = 1,..., 2 . As j , Gj converges uniformly to a µ- − → ∞ −1 −1 measurable function G. So the probability laws µ Gj converge to µ G ◦ ◦ 2 with the density ξG = 1[0,1], completing the proof of the proposition. Proposition A.6. For any nonatomic finite measure space (Ω, , µ) and 1 s < , bounded elements of are dense in Ls(Ω, , µ). S ≤ ∞ Dλ S k Proof. We know that µ-simple functions h := i=1 ci1Ai are dense. By Propo- sition A.5, for any δ > 0 and i = 1, . , k, there is a µ-measurable function P −1 fi on Ai such that for the restriction µi of µ to Ai, µi fi has density ◦ k µ(Ai)1[ci,ci+δ]/δ with respect to Lebesgue measure. Let G := i=1 fi1Ai . Then G h δµ(Ω)1/s and µ G−1 has density ξ µ(Ω)/δ < . The s G P proof ofk the− propositionk ≤ is complete.◦ ≤ ∞ 2 The following is used to prove the necessity of a suitable measurability assumption on Nemytskii operators acting on Lp spaces (Theorem 7.13(c)). Proposition A.7. Let (Ω, , µ) be any nonatomic probability space and let Q be any probability measureS on the Borel sets of R. Then there is a G 0(Ω, , µ) such that µ G−1 = Q. ∈ L S ◦ Proof. By Proposition A.5, take H 0(Ω, , µ) such that µ H−1 = U[0, 1] (Lebesgue measure restricted to [0, 1]).∈ L Let FS(x) := Q(( , x◦]) for all x R and F ←(y) := inf x: F (x) y for 0 < y < 1.
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