A. Appendix: Some Measurability Results

A.I Sup-Measurable Operator Functions

Let S be some , 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 (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 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 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. Let M: N -+ N x N be onto, and define En = TM(n) \ UTM(k)' k

For sEEn put x(s) = XM(n)(S), and for s E 5 \ UEn put x(s) = w, where w E W is fixed. Since the En are pairwise disjoint, x is well defined, A.2 Majorising Principles for Measurable Operator Functions 131 measurable and essentially countable-valued. We have x(s) E W, and for sEE = UEn = Unk Tnk we have IA(s)l w ::; IA(s)x(s) 1+ ly(s)l. Thus, if we modify x on the null set (suppA) \ E, we have the statement. 0

We emphasize that the following consequence holds also for complex Banach spaces:

Corollary A.2.1. Let Y be a Banach space, Y' be its , and S be a measure space. Let y : S -+ Y' be measurable. Then for any measurable z , w with a-finite M = suppy nsuppw and M suppz, there exists a measurable x: S -+ Y with Ix(s)1 == Iw(s)l, y(s)x(s) 2: 0, and

y(s)x(s) 2: ly(s)llw(s)I-lz(s)1 (s E S). Proof. Since outside M you may define x(s) == w(s)sgn[y(s)w(s)] (for com- plex z :j:. °put sgnz = z] Izl), we may assume without loss of generality that M = S. Theorem A.2.1 implies for A = y that there exists a measurable u : S -+ Y with lui = 1 and

Iy(s)u(s)12: Iy(s)1- min{Iw(s) r 1 Iz(s) I,ly(s)J}. Now just put x(s) = u(s) Iw(s)1 sgn[y(s)u(s)]. 0

For w == 1 and z == E. an analogous result may be found in [51, Lemma 2] (for finite measure spaces).

The 'dual' version of Corollary A.2.1 uses the fact that the canonical embed- ding of Y into Y" is norm-preserving (see Definition 2.1.8).

Corollary A.2.2. Let Y be a Banach space with the bidual property, Y' be its dual space, and S be a measure space. Let x : S -+ Y be measurable. Then to any measurable z,w with a-finite M = suppx n suppw and M suppz, there exists a measurable y : S -+ Y' with ly(s)1 = Iw(s)1, y(s)x(s) 2: 0, and

y(s)x(s) 2: Iw(s)llx(s)1 -lz(s)1 (s E S).

Proof. Let i : Y -+ Y" be the canonical embedding, and let u == i 0 x, i.e. u(s)l = l(x(s)). u is measurable, since i is continuous. Corollary A.2.1 applied for Y' instead of Y and for u instead of y yields that there exists a measurable y : S -+ Y' with ly(s)1 = lu(s)l, u(s)y(s) 2: 0, and u(s)y(s) 2: lu(s)llw(s)1 - Iz(s)1 (s E S). This implies the statement, since u(s)y(s) == y(s)x(s), and since lu(s)1 Ix(s)l, because Y has the bidual property. 0 132 A. Appendix: Some Measurability Results

Remark A.2.1. If we assume the axiom of choice and are just looking for a measurable (i.e. not necessarily essentially countable-valued) function x, the condition of M = suppA being o-finite in Theorem A.2.1 (and thus of course also in Corollaries A.2.1 and A.2.2) can be weakened to the assumption that M has the direct sum property: Definition A.2.1. A measurable set M has the direct sum property, if there exists a family of pairwise disjoint sets Ma of finite measure with M = UMa, such that a function x is measurable on M, if and only if it is measurable on each Mer.. The extension of Theorem A.2.l is straightforward: Apply the o-finite version of the theorem on each Mer. to find a corresponding measurable function Xer., and then define x = Xer. on Mer. (this step needs the axiom of choice).

The direct sum property coincides with the direct sum property given in literature (like e.g. in [14]):

Proposition A.2.1. A measurable set M = UM a has the direct sum prop• erty with respect to the family (Mer.)aEA of pairwise disjoint sets of finite measure, if and only if for any set E M of finite measure there exists an at most countable number of indices an, such that E \ UMer. n is a null set. Proof. First, let M have the direct sum property with respect to (Mer.)aEA. We have to show that the above property holds true. For this purpose observe that there exists a Banach space Y, and for any a E A some Yer. E Y, such that IYa - Y.BI 2: 1 for a f. (3. For example you might choose Y = L2(A) being the of square integrable functions A -+ lR with the counting measure on the index set A and let Ya = X{er.} be the canonical orthonormal base of Y. Now, define a mapping x : M -+ Y by x(s) = Yer. for s E Mer.. Since M has the direct sum property, x must be measurable. In particular, x must be essentially separable valued on each set E of finite measure, thus in view of Lemma A.I.l even essentially countable valued on E. But by definition of x this means that E is contained in the union of countable many Mer. (up to a null set). Conversely, let M have the property of the statement with a family of sets Mer.. Assume that x : M -+ Y is measurable on each Mer.. We have to prove that x is the limit of simple functions on each set EM of finite measure, i.e, we have to prove that x is measurable on E. But by assumption there exist a countable number of indices an such that E \ U Mer. n is a null set.

Thus it suffices to observe that each xn(s) = XMa n (s)x(s) is measurable with x = L X n a.e. on E. 0

The direct sum property is a very weak property. It does not even imply the finite subset property: A.2 Majorising Principles for Measurable Operator Functions 133

Example A.2.1. Let S be the following measure space on [0, 1J x [0, 1J: Call a set ES measurable, if each Et = {s : (t, s) E E} is Lebesgue measurable. In this case let mesE = 00 if there is an uncountable number of nonempty Et, otherwise let mesE = Lt /-l(Et) , /-l being the Lebesgue measure. This measure space is discussed in [48, §10 Example 6], and it does not have the finite subset property and is not localizable (see [48, §34,§35]). However, M = S has the direct sum property, since you may choose M a = {a} x [O,lJ for a E [0,1]. This example shows also that, even if M has the direct sum property, it is not enough just to choose M a of positive and finite measure such that M is 'exhausted': If you choose the family M a = ({a} x [0,1]) U {(l,a)} for a E (0,1), M a = ({O} x [0,1]) U {(I, 0), (1, I)} for a = 0, the measurability of x on each M a does not imply the measurability of x on M, since x(l,') need not be Lebesgue measurable.

The following measure space may or may not have the direct sum property, depending on the system of axioms we use: Example A.2.2. We consider the measure space discussed in [48, §10 Example 7J over M = S = [O,lJ x [O,lJ. There a set ES is measurable if each E, = {s : (t, s) E E} and each ES = {t : (t, s) E E} is Lebesgue measurable. In this case define mesE in the following way: Ifthe number of nonempty Et is at most countable, let mesE = Lt /-l(Et) (/-l being the Lebesgue measure), and if the number of nonempty E S is at most countable, let mesE = Ls /-leeS). Otherwise let mesE = 00. Assume now that IvI = UM a with M a as in Definition A.2.1. Consider the sets Et = {(t, s) : s E [0, I]} of finite measure. By Proposition A.2.1 for any t there is a countable number of set Man such that E, \ UMan is a null set. In particular, if U denotes the union of all sets M a such that {s : (t, s) E M a} is of positive measure for at least one t, then U; = {s : (t, s) E U} has measure 1 for any t. Similarly, if V is the union of all sets M a such that {t : (t, s) E M a } is of positive measure for at least one s we have that V S = {t : (t, s) E V} has measure 1 for any s. Since all M a have finite measure and are pairwise disjoint, also U and V are disjoint. But if U or V is Lebesgue measurable on [O,lJ x [0,1] this is not possible by the Cavalieri principle. Thus the assumption that M has the direct sum property allows us to con• struct a non Lebesgue measurable set on [O,lJ x [O,lJ without referring to the axiom of choice or the continuum hypothesis. But this is not possible in Solovay's model [44] (see also [23]). Hence (provided Solovay's model exists) we have shown that at least within Zermelo's set theory with the principle of dependent choices it is not possible to prove that M has the direct sum property. But on the other hand, assuming the continuum hypothesis, Sierpinski showed [42] that there exists a set D [0,1] x [0,1]' such that each s Dt = {s: (t,s) E D}isanullsetandsuchthateachD = {t: (t,s) E D}con- tains almost all points of [0, 1]' see [40, 8.9(c)]. Now let M a = ({a} x [0, 1])\D 134 A. Appendix: Some Measurability Results for a E [0,1], and M a = ([0,1] x {a - 2}) n D for a E [2,3]. Then the M a are pairwise disjoint and UM a = M. Moreover, if E has finite measure then either just a countable number of sets E, = {s : (t, s) E E} or a count- able number of sets ES = {t : (t, s) E E} is nonempty. In the first case put 1= {t : E; i:- 0}, in the second case put I = {s + 2 : ES i:- 0}. In both cases I is at most countable, and E \ UaEI M a is a null set. Thus M has the direct sum property by Proposition A.2.1.

We now want to give an 'integral version' of Theorem A.2.1 and its corol- laries. The a-finite case is a straightforward application of the theorem. But surprisingly for integrals we may drop this assumption always. The reason for this is the following tricky Lemma A.2.2. Let S be some measure space, and x, y : S -+ [0,00] be measurable. Then there exists a sequence of measurable functions 0 :::; X n :::; x with xn(s) < xes) for xes) i:- 0, such that l xn(s)y(s)ds -+ l x(s)y(s)ds.

Moreover, if y : S -+ [0, (0) and F = suppx n suppy has the finite subset property, you can choose X n in such a way that each xnY is integrable. 1 Proof. For the first statement choose X n = min{n, max{x - n- , On and use the monotone convergence theorem. Now, assume y is finite, and F has the finite subset property. Let f(s) = x(s)y(s). It suffices to consider the case IF f(s)ds = 00. 1. First, assume additionally that also x is finite everywhere. Let T be the set of all measurable E with IE f (s)ds < 00. Choose a sequence En E r with

r f(s)ds -+ sup r f(s)ds = a. i; EalE

Without loss of generality we may assume that E 1 E2 .. , (otherwise replace En by E k ). Let U = UEn. By the monotone convergence the- orem, we have i;r f(s)ds -+ i;r f(s)ds. If a < 00, this implies U E r. Thus F \ U must have positive measure. But then for some k the set {s E F \ U : f (s) :::; k} has positive measure, hence contains a subset D of positive but finite measure. But then also U U D E r, which is by a = lurf(s)ds < luuDr f(s)ds a contradiction to the definition of a. Thus a = 00. Now let xn(s) = 1 XE n (s) max{x(s) - n- , O}. Then we have by the monotone convergence the- orem A.2 Majorising Principles for Measurable Operator FUnctions 135

00 > hxn(s)y(s)ds --+ i f(s)ds = a = 00. 2. Now we drop the assumption that x is everywhere finite. But we may apply what we have shown for W n = min{x,n} instead of x. Observing that Jwn(s)y(s)ds --+ 00 by the monotone convergence theorem, we have that for any n there is some 0 X n x with xn(s) < x(s) for x(s) ¥- 0 with

n

Now the sequence Xk is the sequence, we are looking for. 0

Now we can prove the integral version of Theorem A.2.1 on arbitrary measure spaces:

Theorem A.2.2. Let Y be a normed linear space, Z be a Banach space, and S be a measure space. Assume, A : S --+ 'c(Y,Z) is measurable. Let y : S --+ [0,00] be measurable, and My consist of all measurable functions x: S --+ Y with Ix(s)1 = y(s) for y(s) < 00. Then

sup {IA(s)x(s)j ds = (jA(s)j y(s)ds, xEMy is is where the case that a side is infinite is not excluded.

Proof. We prove the nontrivial inequality. Let f(s) = IA(s)1 y(s). 1. First, assume T = suppf has the finite subset property. Then by Lemma A.2.2 there exists a sequence ° Yn ::; Y with Yn < Y on suppy and

00 > hIA(s)! Yn(s)ds --+ hf(s)ds. Thus it remains to prove for any n that for g(s) = IA(s)1 Yn(S)

sup {IA(s)x(s)1 ds 2: { g(s)ds. (A.2) xEMy is is Observe that suppg is a-finite, because it is the union of the sets Sk = {s : 1 g(s) 2: k- }, which have by 00 > Jg(s)ds 2: k-lmesSk finite measure. Without loss of generality we may assume that suppA suppg. Fix n, and let w(s) = y(s) for y(s) < 00, w(s) = Yn(s) + 1 for y(s) = 00. For y(s) ¥- 0 let z(s) = IA(s)1 (1 - w(S)-lYn(s)), otherwise z(s) = IA(s)l. Then suppz = suppA. If we apply Theorem A.2.1 for W = {y EY: lyl = I}, we find that there exists a measurable u : S --+ W with

IA(s)u(s)1 2: IA(s)l- z(s) (s E S). Putting x(s) = u(s)w(s), we have x E My and 136 A. Appendix: Some Measurability Results

IA(s)x(s)1 2: IA(s)1 w(s) - z(s)w(s) = o9(s) (s E S). Integrating this inequality yields (A.2). 2. Now, assume suppj has not the finite subset property. This means, it contains a measurable set E of infinite measure, such that each measur• able subset of E has either infinite measure or is a null set. But this implies that any function defined on E is measurable. Hence there ex• ists some x E My with A(s)x(s) f 0 on E. Since for some k the set W = {s E E: JA(s)x(s)j 2: k-1} has positive (whence infinite) measure, we thus have Is IA(s)x(s)1 ds 2: Iw k-1ds = 00. 0

Again, the following corollaries continue to hold for complex Banach spaces: Corollary A.2.3. Let Y be some Banach space, Y* its dual space, and S be some measure space. Let M be a set oj measurable junctions S -+ Y with the property that the conditions x EM, Iz I = Ix I jar some measurable z imply z EM. Then jar any measurable y : S -+ Y*we have

sup r y(s)x(s)ds = sup r ly(s)llx(s)1 ds, xEM, yx;::o:o J5 xEM J5 where the case that a side is infinite is not excluded. Proof. Again, we just show the nontrivial inequality. Thus fix some x EM. Applying Theorem A.2.2 for A = y and y = [z], we find a sequence of functions Zn E M with

Is ly(s)zn(s)1 ds -+ Is ly(s)llx(s)1 ds.

Putting xn(s) = zn(s)sgn[y(s)zn(s)J (for complex z f 0 let sgnz = z] Izl), we have X n E M, y(s)xn(s) = ly(s)zn(s)l, hence I y(s)xn(s)ds -+ I ly(s)llx(s)1 ds. o

For finite measure spaces, an analogous result may be found in [51, Lemma 3J.

The 'dual' version of Corollary A.2.3 again uses the fact that the canonical embedding of Y into Y** is norm­preserving (see Definition 2.1.8): Corollary A.2.4. Let Y be some Banach space with the bidual property, Y* its dual space, and S be some measure space. Let M be a set oj y* -valued measurable junctions with the property that the conditions y E M, Izi = Iyl jar some measurable z imply z EM. Then jar any measurable x : S -+ Y we have sup r y(s)x(s)ds = sup r ly(s)llx(s)1 ds, yEM, yx;::o:o J5 yEM J5 where the case that a side is infinite is not excluded. A.2 Majorising Principles for Measurable Operator Functions 137

Proof. Let Z* = Y** be the dual space of Z = Y*. Let i : Y -+ Z* be the canonical embedding. If we define z : S -+ Z* by z = iox, i.e. z(s)l = l(x(s)), z is measurable, because i is continuous. If we apply Corollary A.2.3 for Z instead of Y, and for z instead of y, we find

sup r z(s)y(s)ds = sup j' Iz(s)lly(s)1 ds. yEM, } 5 yEM 5 This is the stated equality, since z(s)y(s) = y(s)x(s) and Iz(s)1 = Ix(s)l, because Y has the bidual property. 0 138 A. Appendix: Some Measurability Results A.3 A Generalization of a Theorem of Luxemburg-Gribanov

At first, we recall a theorem of Luxemburg [31] and Gribanov [21] (see [49, Theorem 99.2 and Corollary 99.3]). In the following, let T and S be rr-finite measure spaces.

Theorem A.3.1. Let B be some set of measurable nonnegative functions over S, and x be measurable and nonnegative on T x S. Then there exists a countable set A S;; B, such that for almost all t

sup rx(t, s)u(s)ds = sup rx(t, s)u(s)ds. uEBJS uEAJS

In particular, t 1---7 SUPuEB Is x(t, s )u(s)ds is measurable. Proof. 1. We first show that we additionally may assume that there exists some set of finite measure E S;; S, such that suppx S;; T x S, and that all u E B and x are uniformly bounded by some C > O. Indeed, assume the statement has been shown for this case. Let S = USn, where S1 S;; S2 S;; ... have finite measure, and put X n = PT x S n min(x, n), and B n = {min(u, n) : u E B}. Since for any u E B by the monotone convergence theorem (for almost all t)

Is xn(t, s) min(u(s), n)ds -t Is x(t, s)u(s)ds, we have sup sup rxn(t, s)u(s)ds = sup rx(t, s)u(s)ds. n uEBn J 5 uEB J 5 By assumption, each B n contains a countable An satisfying sup rxn(t, s)u(s)ds = sup rxn(t, s)u(s)ds. uEBn J s uEAn J s Putting A* = UAn, we thus have by taking the supremum over all n:

sup rx(t, s)u(s)ds = sup rx(t, s)u(s)ds. UEA·JS uEBJS By construction, for each u* E A * there exists a function u E B with u 2: u*. Thus the statement is true for A being the set of all that u, 2. Now, assume that x is elementary, i.e. (Lemma 4.3.2)

m

x(t, s) = L XTk (t)xds), k=1 with simple functions Xk 2: 0 and measurable pairwise disjoint T k C T, UTi; = T. For any k and any n there exists some Ukn E B satisfying A.3 A Generalization of a Theorem of Luxemburg-Gribanov 139

1 sup r xk(s)u(s)ds::; r Xk(S)Ukn(s)ds + n- . uEB is is Then for almost all t E Tk, we have by x(t,s) = Xk(S) that

sup r x(t,s)u(s)ds = sup r x(t,s)ukn(s)ds, UEBis n is i.e. the statement is true for A being the set of all Ukn' 3. We now put the additional assumptions of 1. on Band x. Let Yn be elemen- tary, converging to x a.e. (Lemma 4.3.3). Then X n = PTxEmin(C, IYnl) 2: 0 converges to x, but additionally satisfies 0 ::; x., ::; C and sUPPX n T x E. Thus Lebesgue's dominated convergence theorem implies (for almost all t)

lim sup I r x(t, s)u(s)ds - r xn(t, S)U(S)dS! n-+oouEB is is ::; lim r1x(t,s)-xn(t,s)ICds=0. (A.3) n-+oo is By 2., for any n, there exists some countable An B satisfying

sup r xn(t, s)u(s)ds::; sup r xn(t, s)u(s)ds. (A.4) uEB is uEA n is

Now, given some E: > 0, by (A.3) there exists some n with

Is xn(t, s)u(s)ds - E: ::; Is x(t, s)u(s)ds

sup r x(t, s)u(s)ds ::; sup r xn(t, s)u(s)ds + E: UEBis UEBis

::; sup r xn(t, s)u(s)ds + E: ::; sup r x(t, s)u(s)ds + 2 . UEAis UEAis This shows the statement, since e was arbitrary. 0

We need the generalization, that for any error-function Y, one can choose a measurable majorizing function: Theorem A.3.2. Let B be some set of measurable nonnegative functions over T, and x be measurable and nonnegative on T x S. Assume, Y is non- negative and measurable with yet) > 0 for x(t,·) -I- O. Then there exists a nonnegative measurable function u on T x S, such that for almost all t we have u(t,·) E Band

r x(t, s)u(t, s)ds 2: sup r x(t, s)u(s)ds - yet). is uEB is Moreover, the function t f-t u(t,') is just countable-valued. 140 A. Appendix: Some Measurability Results

Proof. We write Fu(t) = Is x(t, s)u(s)ds and F(t) = sUPuEB Fu(t) for short. By Theorem A.3.1, F(t) is measurable, and there exists a sequence Un E B with F(t) = sup Fun(t). (A.5) n Now, let En = {t E T : Fun(t) 2: F(t) - y(t)}. Then En belongs to the Lebesgue extension of the measure, and T \ UEn is a null set by (A.5). Now put o; = En \ Ue; k

The previous theorems also have obvious countable generalizations: Theorem A.3.3. Let B(-) be a family of sets of nonnegative measurable functions over S, which is countable measurable on T. Let x be measurable and nonnegative on T x S. Then there exists a countable measurable family A(t) B(t) of countable sets, such that for almost all t:

sup rx(t, s)u(s)ds = sup rx(t, s)u(s)ds. uEB(t) is uEA(t) is

In particular, t r-+ sUPuEB(t) Is x(t, s )u(s )ds is measurable. Theorem A.3.4. Let B(·) be a family of sets of nonnegative measurable functions over S, which is countable measurable on T. Let x be measurable and nonnegative on T x S. Assume, y is nonnegative and measurable with y(t) > 0 for x(t,') =j:. O. Then there exists a nonnegative measurable function u on T x S, such that for almost all t we have u(t,') E B(t) and

r x(t, s)u(t, s)ds 2: sup r x(t, s)u(s)ds - y(t). is uEB(t) is Moreover, the function t r-+ u(t,') is just countable-valued.

For the proofs, observe that there exist a countable number of sets B n and measurable pairwise disjoint 'E; T with UTn = T, such that B(t) = B n for almost all t E Tn- Thus, it suffices to apply Theorem A.3.1 on each Tn instead of T to find countable An B n, and to put A(t) = An for t E Tn. This shows Theorem A.3.3. To prove Theorem A.3.4, similarly apply Theorem A.3.2 for each Tn instead of T to find a function Un, and then put u(t, s) = un(t, s) for t E Tn. References

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K,9 - strong, 11 £(Y),9 closed on rays, 46 PD,48 completely regular space, 48 X(w),82 contain a complete ray, 46 XL,44 continuum hypothesis, 7, 91, 98, 133 XIS',82 convergence X' 59 - extended, 20 X,}, 65 - extended', 20 Y[k], 119 - in the extended sense, 17 [U -+ V], 80 a.e., 17 [U +- V], 80 -- in measure, 19 [U(-) -+ V], 85 -- uniformly, 19 Co, 48 countable measurable, 85 d(x,y),17 Ip,9 defined L p , 9 - integral operator, 105 Lp(S, Y), 9 - partial integral operator, 110 Lp,9 dependent choices, principle of, 7, 69, 1I·II 128 L,44 direct sum property, 132 11·ll x , , 59 11·ll x " , 62 duality map, 11 - E-, 14 absolute continuous norm, 48 - fully defined, 11 - uniformly, 50 - nontrivial, 11 absolutely defined - integral operator, 105 e-simple set, 86 - partial integral operator, 110 elementary function, 86 accumulation point, essential, 78 elementary set, 86 almost a-perfect, 39 s-duality map, 14 almost perfect, 39 equivalent kernels, 105, 110 a-perfect, 39 essential accumulation point, 78 associate space, 59 esslim, 78 axiom of choice, 7, 14, 15, 31, 69, 70, ess lim inf, 78 75, 95, 105, 106, 108, 128, 132 ess lim sup, 78 axiom of dependent choice, 7, 69, 128 extended convergence, 20 extended' convergence, 20 bidual property, 14 F-space, 29, ]06 Caratheodory condition, 112 Fatou property, 40 - strict, 115 - weak, 40 characterization property, 10 finite atomic free, 54 Index 145 finite subset property, 19 pre-ideal space, 8 FTC, 101 pre-ideal' space, 9 fully defined map, 11 principle of dependent choices, 7, 69, function 128 - elementary, 86 proper space, 106 - FTC satisfying, 101 property - real, 8 - bidual, 14 - strongly continuous operator, 127 - characterization, 10 - sup-measurable, 12 -- strong, 11 - sup-measurable operator, 127 - direct sum, 132 - support of a, 19 - Fatou,40 - vanishing at 00 in norm, 48 -- weak, 40 - with absolute continuous norm, 48 - finite subset, 19 - Young, 9 - Levi, 41 functional - Riesz-Fischer, 35 - integral, 60 - W-, 29 - positive, 70 - singular, 71 ray - closed on, 46 Hammerstein operator, 120 - contain a complete, 46 real form of an ideal space, 8 ideal space, 8 real function, 8 - real form of, 8 regular part, 48 ideal, t.;», 9 regular space, 48 ideal' space, 9 Riesz-Fischer property, 35 inner-regular part, 48 inner-regular space, 48 S-measurable space, 75 integral functional, 60 semi-perfect, 38 integral operator simple-regular support, 76 - absolutely defined, 105 singular functional, 71 - absolutely defined partial, 110 space - defined, 105 - almost a-perfect, 39 - defined partial, 110 - almost perfect, 39 - a-perfect, 39 Levi property, 41 - associate, 59 Loo-ideal, 9 -- natural, 94 Lorentz seminorm, 44 - completely regular, 48 Lorentz space, 44 - F-, 29, 106 Luxemburg norm, 9 - ideal, 8 - ideal' , 9 measure space, 7 - inner-regular, 48 mixed family norm, 85 - outer-regular, 48 mixed norm, 80 - perfect, 39 - pre-ideal, 8 natural associate, 94 - pre-ideal' , 9 nontrivial map, 11 - proper, 106 normal, 46 - real-valued, 9 - regular, 48 , 9 - S-measurable, 75 outer-regular part, 48 - semi-perfect, 38 outer-regular space, 48 T-elementary measurable family of, 86 perfect, 39 - T-measurable, 75 positive functional, 70 146 Index

-T -rneasurable family of, 85 - T-simple measurable, 76 -T -simple measurable family of, 86 - weighted, 82 - weighted projection, 82 - with mixed family-norm, 85 - with simple-regular support, 76 - with the bidual property, 14 - with the Riesz-Fischer property, 35 - with the W-property, 29 - with uniformly-simple regular support, 89 -Y-valued, 9 strict Caratheodory condition, 115 strong characterization property, 11 strongly continuous operator function, 127 sup-measurable - function, 12 - operator function, 127 superposition operator, 112 supp, 19, 24 supp, 25 supp, 25 support - of a function, 19 - of a set of functions, 24 - simple-regular, 76 - uniformly-simple regular, 89

T -elementary measurable family, 86 T-measurable family, 85 T-measurable space, 75 T-simple measurable family, 86 T-simple measurable space, 76 uniformly-simple regular support, 89 vanish at 00 in norm, 48 - uniformly, 50

W-property, 29 weak Fatou property, 40 weighted function space, 82 weighted projection, 82

Young function, 9