A. Appendix: Some Measurability Results A.I Sup-Measurable Operator Functions Let S be some measure space, and Y,Z be Banach spaces. Let A(s) : Y --+ Z be an (not necessarily linear) operator function for almost all s E S. One often has the problem to integrate the function s f-t A(s)x(s), where x : S --+ Y. Boundedness is usually trivial, thus it suffices to check measurability: Definition A.I.I. The operator function A is calledsup-measurable, if s f-t A( s)x( s) is measurable for any measurable x : S --+ Y. Theorem A.I.I. Let for almost all s the mapping A( s) : Y --+ Z be contin• uous. Then A is sup­measurable, if and only if s f-t A(s)y is measurable for any y E Y. Proof. We prove sufficiency. First, assume x(s) = LXkXEk(S) is simple, where Ek are pairwise disjoint with UEk = S. Since each Yk(S) = A(S)Xk is measurable, also LXEk(S)Yk(S) = A(s)x(s) is measurable. In the general case, let ES have finite measure. Choose a sequence of simple functions X n : S --+ Y, which converges to x a.e. on E. Then A(s)xn(s) --+ A(s)x(s) for almost all sEE. Thus, s f-t A(s)x(s) is measur- able on E, whence measurable. 0 In different terminology the main part of Theorem A.l.l may be reformu- lated as: The superposition operator generated by the Caratheodory function f(s, y) = A(s)y maps measurable functions into measurable functions. This is of course well-known for scalar Y (see e.g. [4]). Recall that an operator function A is said to be strongly continuous, if s f-t A(s)y is continuous for any fixed y. Corollary A.I.I. Let I be some compact interval, A : I --+ £(Y, Z) be a strongly continuous operator function. Then for any integrable x : I --+ Y also s f-t A(s)x(s) is integrable. Proof. A is sup-measurable, since s f-t A( s)y is continuous for all y E Y. By the uniform boundedness principle, IIA(s)11 ::; C for all s E I. Thus 128 A. Appendix: Some Measurability Results IIA(s)x(s)11 :S c Ilx(s)ll. 0 We remark that in Corollary A.1.1 it may happen that A is not measurable, even if Y = Z is separable: Example A.l.l. Let I = [0,1]' and Y = Z = LdI). Define A : I -+ £(Y) by A(s)x(t) = J; x(eJ)deJ. A is strongly continuous but not measurable, since it is not essentially separable valued by IA(s) - A(eJ)I = 1 for s =I- o and the following Lemma A.1.1. We remark that the proof of the following lemma is straightforward, if the axiom of choice may be used. But since we just want to use the principle of dependent choices, we have to argue in a different way: Lemma A.l.l. Let X be a metric space, M X. Assume, there exist <5 > 0 and an uncountable UM with d(x,y) 2: <5 (x,y E U, x =I- y). Then M is not separable in X. Proof. If M is separable in X, there exists an at most countable set E X, whose closure contains M, hence also U. Let D consist of all nEE with dist(n, U) < <5/2. Then also D is at most countable, and its closure contains U. For any nED choose some Xn E U with d(n, x n) < <5/2. Then the set C of all X n is at most countable. Hence there exists some y E U \ C. For any nED we have <5 :S d(xn, y) :s d(xn, n) + d(n, y) :S <5/2 + d(n, y), hence y does not belong to the closure of D, a contradiction. 0 Thus even in the special case of Example A.1.1 sup-measurability does not imply measurability in operator norm (although this is true for A : 8 -+ £(Y, Z) with finite-dimensional Y: Apply Theorem 5.1.5 for X = L oo(8, Y) and c = A). The converse usually is true: Theorem A.l.2. Let Y, Z be Banach spaces, and W be a Banach space of continuous mappings Y -+ Z. Assume, An -+ A in W implies that for all y E Y we have AnY -+ Ay. Then any measurable A : 8 -+ W is sup- measurable. Proof. Let A : 8 -+ W be measurable. Let y E Y. Let E 8 have finite mea- sure. Choose a sequence of simple functions An : 8 -+ W, converging to A a.e. on E. Then An(s)y -+ A(s)y for almost all sEE. Since each S I--t An(s)y is also simple, s I-t A(s)y is measurable on E, whence measurable. Now use Theorem A.I.1. 0 A.2 Majorising Principles for Measurable Operator Functions 129 A.2 Majorising Principles for Measurable Operator Functions We are now going to prove for a measurable operator function A the exis• tence of a measurable function x, such that s I-t A(s)x(s) has some majorising properties. The problem consists in finding a measurable such x. The main tool is the following lemma, which is quite an immediate conse• quence of Egorov's theorem: Lemma A.2.1. Let Y be a Banach space. Then for any measurable x: S -+ Y with a-finite support and any measurable function y with suppx suppy there exists a sequence of measurable sets Sn S with mes(S \ USn) = 0 and a sequence X n of simple functions with Ix(s) ­ xn(s)l :S ly(s)1 (A.I) Moreover, we may additionally satisfy Sl S2 ... and xklsk = xnlsk for k :S n. Proof. 1. We first drop the additional assumptions. Since we may put xn(s) = 0 for s suppx, we may assume without loss of generality that S = suppx. Let S = U En, where E 1 E 2 '" have finite measure. Choose any sequence Zk of simple functions, which converges a.e. to x. Now fix n. Since En suppy, there exists some N such that the measure of 1 1 M = {s E En : ly(s)1 :S N- } is less that n- . By Egorov's theorem there exists a measurable subset s; En \ M with mes( En \ Sn) < 2n ­1, such that 1 Zk -+ x uniformly on Sn' Hence some X n = Zk n satisfies Ix(s) ­ xn(s)1 :S N- for s E s.; Thus SnnM = 0implies (A.I). Moreover, each Qn = En \Uk Sk is a null set, since for each k 2 n we have a; e;\s; Hence U Qn = S\ U s, is a null set. 2. By 1. there exists a sequence of measurable sets Tn with mes(S\U Tn) = 0, and a sequence of simple functions Yn with Ix(s) ­ Yn(s)1 :S ly(s)1 for s E Tn· Now put En = t; \ UTk' s; = Ue, = Ur; k<n k:Sn k:Sn Since the sets En are pairwise disjoint, the simple functions n xn(s) =I:XEk (S)Yk(S) k=l satisfy (A.1). The additional assumptions are also satisfied. 0 With this lemma we may prove the first result: 130 A. Appendix: Some Measurability Results Theorem A.2.1. Let Y be a normed linear space, Z be a Banach space, and 5 be a measure space. Assume, A : 5 -+ £(Y, Z) is measurable with a-finite support. Then to any measurable y with suppA suppy and to any nonempty bounded set W Y there exists a measurable and essentially countable-valued function x : 5 -+ W with IA(s)x(s)1 2: IA(s)lw -ly(s)1 (s E 5), where IAlw = sUPwEW IAwl· Proof. 1. First assume additionally that 5 = suppA has finite measure and that ly(s)1 2: 3c: > 0 on 5. By Lemma A.2.1 there exists a sequence of measurable sets 5n 5 with mes(5 \ U5n) = 0 and a sequence of simple functions An : 5 -+ £(Y, Z) with IA(s) - An(s)lw ::; e Since we may additionally assume that 51 52 ... and Aklsk = Anlsk for k ::; n, we may assume without loss of generality that An(s) = LXDk(S)Ak, k=l where the pairwise disjoint measurable sets D k and A k E £(Y, Z) are inde- pendent of n with UD k = U5n and m1 < m2 < .... For any k there exists some Wk E W, with IAklw ::; IAkWkl + c. Now define 00 x(s) = L XDk (S)Wk' k=l Then x has the properties x(s) E W a.e. and IAn(s)lw ::; IAn(s)x(s)1 + e for s E 5 n . Particularly, for s E 5n we have IA(s)x(s)l2: IAn(s)x(s)I-1 [An(s) - A(s)]x(s)l2: IAn(s)lw-2c 2: IA(s)lw-3c. Since mes(5 \ U 5n ) = 0, and ly(s)1 2: 3c, this implies the statement (by modifying x on a proper null set). 2. Let suppA = US«, where 5n have finite measure, and put 1 Tnk = {s E 5n : ly(s)1 2: k- }. By 1. there exist measurable and essentially countable-valued functions Xnk : Tnk -+ W with IA(s)lw ::; IA(s)x(s)1 + ly(s)1 for s E Tnk.