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.