Integrals for functions with values in a partially ordered vector space
IIntegrals for functions with values in a partially orderedvector space
A.C.M. van Rooij I W.B. van Zuijlen II October 21, 2018
Abstract
We consider integration of functions with values in a partially ordered vectorspace, and two notions of extension of the space of integrable functions. Applyingboth extensions to the space of real valued simple functions on a measure spaceleads to the classical space of integrable functions.
Key words and phrases.
Partially ordered vector space, Riesz space, Bochner inte-gral, Pettis integral, integral, vertical extension, lateral extension.
For functions with values in a Banach space there exist several notions of integration.The best known are the Bochner and Pettis integrals (see [2] and [14]). These havebeen thoroughly studied, yielding a substantial theory (see Chapter III in the book byE. Hille and R.S. Phillips, [10]).As far as we know, there is no notion of integration for functions with values in apartially ordered vector space; not necessarily a σ -Dedekind complete Riesz space. Inthis paper we present such a notion. The basic idea is the following. (Here, E is apartially ordered vector space in which our integrals take their values.)In the style of Daniell [5] and Bourbaki [4, Chapter 3,4], we do not start from ameasure space but from a set X , a collection Γ of functions X → E , and a functional ϕ : Γ → E , our “elementary integral”. We describe two procedures for extending ϕ to a larger class of functions X → E . The first (see § § § E to be σ -Dedekind complete or even Archimedean. I Radboud University Nijmegen, Department of Mathematics, P.O. Box 9010, 6500 GL Nijmegen,the Netherlands. II Leiden University, Mathematical Institute, P.O. Box 9512, 2300 RA, Leiden, the Netherlands. a r X i v : . [ m a t h . F A ] M a y owever, under some mild conditions on E one can embed E into a σ -Dedekind com-plete space. In § § § E -valued functions on a measurespace. (In § E = R .) In § E is a Banach lattice. In § N is { , , , . . . } .Let X be a set. We write P ( X ) for the set of subsets of X . For a subset A of X : A ( x ) = (cid:40) x ∈ A, x / ∈ A. As a shorthand notation we write = X .Let E be a vector space. We write x = ( x , x , . . . ) for functions x : N → E (i.e.,elements of E N ) and we define c [ E ] = { x ∈ E N : ∃ N ∀ n ≥ N [ x n = 0] } , c = c [ R ]We write c for the set of sequences in R that converge to 0, c for the set of convergentsequences in R , (cid:96) ∞ ( X ) for the set of bounded functions X → R , (cid:96) ∞ for (cid:96) ∞ ( N ), and (cid:96) for the set of absolutely summable sequences in R . We write e n for the element { n } of R N .For a complete σ -finite measure space ( X, A , µ ) we write L ( µ ) for the space of integrablefunctions, L ( µ ) = L ( µ ) / N where N denotes the space of functions that are zero µ -a.e.Moreover we write L ∞ ( µ ) for the space of equivalence classes of measurable functionsthat are almost everywhere bounded.For a subset Γ of a partially ordered vector space Ω, we write Γ + = { f ∈ Γ : f ≥ } .If Λ , Υ ⊂ Ω and f ≤ g for all f ∈ Λ and g ∈ Υ we write Λ ≤ Υ; if Λ = { f } we write f ≤ Υ instead of { f } ≤ Υ etc. For a sequence ( h n ) n ∈ N in a partially ordered vectorspace we write h n ↓ h ≥ h ≥ h ≥ · · · and inf n ∈ N h n = 0. Throughout this section, E and Ω are partially ordered vector spaces, Γ ⊂ Ω is a linear subspace and ϕ : Γ → E is order preserving and linear. Additionalassumptions are given in 3.14. Definition 3.1.
DefineΓ v = (cid:110) f ∈ Ω : sup σ ∈ Γ: σ ≤ f ϕ ( σ ) = inf τ ∈ Γ: τ ≥ f ϕ ( τ ) (cid:111) , (1)2nd ϕ v : Γ v → E by ϕ v ( f ) = sup σ ∈ Γ: σ ≤ f ϕ ( σ ) ( f ∈ Γ v ) . (2)Note: If f ∈ Ω and there exist subsets Λ , Υ ⊂ Γ with Λ ≤ f ≤ Υ such thatsup ϕ (Λ) = inf ϕ (Υ), then f ∈ Γ v and ϕ v ( f ) = inf ϕ (Υ). The following observations are elementary.(a) Γ ⊂ Γ v and ϕ v ( τ ) = ϕ ( τ ) for all τ ∈ Γ.(b) Γ v is a partially ordered vector space and ϕ v is a linear order preserving map .(c) (Γ v ) v = Γ v and ( ϕ v ) v = ϕ v .(d) If Π is a subset of Γ, then Π v ⊂ Γ v .Of more importance to us then Γ v and ϕ v is the following variation in which weconsider only countable subsets of Γ. Definition 3.3.
Let Γ V be the set consisting of those f for which there exist countablesets Λ , Υ ⊂ Γ with Λ ≤ f ≤ Υ such thatsup ϕ (Λ) = inf ϕ (Υ) . (3)From the remark following Definition 3.1 it follows that Γ V is a subset of Γ v and that(for f and Λ as above) ϕ v ( f ) is equal to sup ϕ (Λ). We will write ϕ V = ϕ v | Γ V . We callΓ V the vertical extension under ϕ of Γ and ϕ V the vertical extension of ϕ .In what follows we will only consider ϕ V and not ϕ v . However, most of the theorypresented can be developed similarly for ϕ v . (For comments see 11.2.) Example 3.4. Γ V is the set of Riemann integrable functions on [0 ,
1] and ϕ V is theRiemann integral in case E = R and Γ is the linear span of { I : I is an interval in [0 , } and ϕ is the Riemann integral on Γ. In analogy with 3.2 we have the following.(a) Γ ⊂ Γ V and ϕ V ( τ ) = ϕ ( τ ) for all τ ∈ Γ.(b) Γ V is a partially ordered vector space and ϕ V is a linear order preserving map.(c) (Γ V ) V = Γ V and ( ϕ V ) V = ϕ V .(d) If Π ⊂ Γ, then Π V ⊂ Γ V . Definition 3.6.
Let D be a linear subspace of E . D is called mediated in E if thefollowing is true:If A and B are countable subsets of D such that inf A − B = 0 in E , then A has an infimum (and consequently B has a supremum and inf A = sup B ) . (4) This follows from the following fact: Let
A, B ⊂ E . If A and B have suprema (infima) in E , thenso does A + B and sup( A + B ) = sup A + sup B (inf( A + B ) = inf A + inf B ). One could also define the vertical extension in case E , Ω, Γ ⊂ Ω are partially ordered sets (notnecessarily vector spaces) and ϕ : Γ → E is an order preserving map. is mediated in E if and only if the following requirement (equivalent with ordercompleteness in the sense of [6], for D = E ) is satisfiedIf A and B are countable subsets of D such that inf A − B = 0 in E , thenthere exists an h ∈ E with B ≤ h ≤ A . (5)We say that E is mediated if E is mediated in itself.Note: if D is mediated in E , then so is every linear subspace of D . Every σ -Dedekindcomplete E is mediated, but so is R , ordered lexicographically. Also, c and c aremediated in c , but c is not mediated.With this the following lemma is a tautology. Lemma 3.7.
Suppose ϕ (Γ) is mediated in E . Let f ∈ Ω . Then f ∈ Γ V if and only ifthere exist countable sets Λ , Υ ⊂ Γ with Λ ≤ f ≤ Υ such that inf τ ∈ Υ ,σ ∈ Λ ϕ ( τ − σ ) = 0 . (6)The next example shows that Γ V is not necessarily a Riesz space even if E and Γare. However, see Corollary 3.10. Example 3.8.
Consider E = c , Γ = c × c , Ω = (cid:96) ∞ × (cid:96) ∞ . Let ϕ : Γ → c be given by ϕ ( f, g ) = f + g . For all f ∈ (cid:96) ∞ there are h , h , · · · ∈ c with h n ↓ f . It follows that,Γ V = { ( f, g ) ∈ (cid:96) ∞ × (cid:96) ∞ : f + g ∈ c } . Note that Γ V is not a Riesz space since for every f ∈ (cid:96) ∞ with f ≥ f / ∈ c we have ( f, − f ) ∈ Γ V but ( f, − f ) + = ( f, / ∈ Γ V . Lemma 3.9.
Suppose ϕ (Γ) is mediated in E . Let Θ : Ω → Ω be an order preservingmap with the properties: • if σ, τ ∈ Γ and σ ≤ τ , then ≤ Θ( τ ) − Θ( σ ) ≤ τ − σ ; • Θ(Γ) ⊂ Γ V .Then Θ(Γ V ) ⊂ Γ V .Proof. Let f ∈ Γ V and let Λ , Υ ⊂ Γ be countable sets with Λ ≤ f ≤ Υ satisfying (6).Then Θ(Λ) ≤ Θ( f ) ≤ Θ(Υ) andinf τ ∈ Θ(Υ) ,σ ∈ Θ(Λ) ϕ ( τ − σ ) = inf τ ∈ Υ ,σ ∈ Λ ϕ (cid:0) Θ( τ ) − Θ( σ ) (cid:1) ≤ inf τ ∈ Υ ,σ ∈ Λ ϕ ( τ − σ ) = 0 . (7) Corollary 3.10.
Suppose that ϕ (Γ) is mediated in E . Suppose Ω is a Riesz space and Γ is a Riesz subspace of Ω . Then so is Γ V .Proof. Apply Theorem 3.9 with Θ( ω ) = ω + . If Γ is a directed set, i.e., Γ = Γ + − Γ + , then so is Γ V . Indeed, if f ∈ Γ V , thenthere exist σ, τ ∈ Γ + such that f ≥ τ − σ and thus f = ( f + σ ) − σ ∈ Γ + V − Γ + V .4 .12. In the last part of this section we will consider a situation in which Ω has someextra structure. But first we briefly consider the case where E is a Banach lattice with σ -order continuous norm. As it turns out, such an E is mediated (see Theorem 4.24),but is not necessarily σ -Dedekind complete (consider the Banach lattice C ( X ) where X is the one-point compactification of an uncountable discrete space). For such E wedescribe Γ V in terms of the norm. Theorem 3.13.
Let E be a Banach lattice with a σ -order continuous norm. Let Ω bea Riesz space and Γ be a Riesz subspace of Ω . For f ∈ Ω we have: f ∈ Γ V if and onlyif for every ε > there exist σ, τ ∈ Γ with σ ≤ f ≤ τ and (cid:107) ϕ ( τ ) − ϕ ( σ ) (cid:107) < ε .Proof. First, assume f ∈ Γ V . As Γ is a Riesz subspace of Ω there exist sequences( σ n ) n ∈ N and ( τ n ) n ∈ N in Γ such that σ n ↑ , τ n ↓ , σ n ≤ f ≤ τ n ( n ∈ N ) , sup n ∈ N ϕ ( σ n ) = inf n ∈ N ϕ ( τ n ) . (8)Then ϕ ( τ n − σ n ) ↓ E , so (cid:107) ϕ ( τ n ) − ϕ ( σ n ) (cid:107) ↓ n ∈ N , choose σ n , τ n ∈ Γ for which σ n ≤ f ≤ τ n , (cid:107) ϕ ( τ n ) − ϕ ( σ n ) (cid:107) ≤ n − . (9)Setting σ (cid:48) n = σ ∨ · · · ∨ σ n and τ (cid:48) n = τ ∧ · · · ∧ τ n we have, for each n ∈ N σ (cid:48) n , τ (cid:48) n ∈ Γ , σ (cid:48) n ≤ f ≤ τ (cid:48) n . (10)If n ≥ N , then 0 ≤ σ (cid:48) n − σ (cid:48) N ≤ f − σ N ≤ τ N − σ N , whence (cid:107) ϕ ( σ (cid:48) n ) − ϕ ( σ (cid:48) N ) (cid:107) ≤(cid:107) ϕ ( τ N ) − ϕ ( σ N ) (cid:107) ≤ N − . Thus, the sequence ( ϕ ( σ (cid:48) n )) n ∈ N converges in the sense ofthe norm. So does ( ϕ ( τ (cid:48) n )) n ∈ N . Their limits are the same element a of E , and, since σ (cid:48) n ↑ , τ (cid:48) m ↓ , we see that a = sup n ∈ N ϕ ( σ (cid:48) n ) = inf m ∈ N ϕ ( τ (cid:48) m ). In the rest of this section Ω is the collection F X of all maps of a set X into a partially ordered vector space F . A function g : X → R determines a multiplication operator f (cid:55)→ gf in Ω. Weinvestigate the collection of all functions g for which f ∈ Γ V = ⇒ gf ∈ Γ V , (11)and, for given f , the behaviour of the map g (cid:55)→ ϕ V ( gf ). For an algebra of subsets of X , A ⊂ P ( X ) we write [ A ] for the Riesz space of all A -step functions, i.e., functions of the form (cid:80) ni =1 λ i A i for n ∈ N , λ i ∈ R , A i ∈ A for i ∈ { , . . . , n } . Define the collection of functions [ A ] o by[ A ] o = { f ∈ R X : there are ( s n ) n ∈ N in [ A ] and ( j n ) n ∈ N in [ A ] + (12)for which | f − s n | ≤ j n and j n ↓ } . A ] o is the vertical extension of [ A ] obtained by, in Definition 3.3, choosing E = R X , Ω = R X , Γ = [ A ] , ϕ ( f ) = f ( f ∈ Γ).) Note that [ A ] and [ A ] o are Riesz spaces,and uniform limits of elements of [ A ] are in [ A ] o . (Actually, [ A ] o is uniformly complete.)Furthermore, [ A ] o contains every bounded function f with { x ∈ X : f ( x ) ≤ s } ∈ A for all s ∈ R . In case A is a σ -algebra, [ A ] o is precisely the collection of all bounded A -measurable functions. Lemma 3.17.
Let
A ⊂ P ( X ) be an algebra of subsets of a set X . Suppose that ( g n ) n ∈ N is a sequence in [ A ] o for which g n ↓ pointwise. Then there exists a sequence ( j n ) n ∈ N in [ A ] with j n ≥ g n and j n ↓ pointwise.Proof. For all n ∈ N there exists a sequence ( s nk ) k ∈ N in [ A ] with s nk ≥ g n for all k ∈ N and s nk ↓ k g n pointwise. Since ( g n ) n ∈ N is a decreasing sequence, we have s mk ≥ g n forall m ≤ n and all k ∈ N . Hence j n := inf m,k ≤ n s mk is an element in [ A ] with j n ≥ g n .Clearly j n ↓ and inf n ∈ N j n = inf n ∈ N inf m,k ≤ n s mk = inf n ∈ N inf k ∈ N s nk = inf n ∈ N g n =0. The following lemma is a consequence of Theorem 3.9. Lemma 3.18.
Define the algebra A = { A ⊂ X : f A ∈ Γ for f ∈ Γ } . (13) If ϕ (Γ) is mediated in E , then f A ∈ Γ V ( f ∈ Γ V , A ∈ A ) . (14) Definition 3.19. E is called integrally closed (see Birkhoff [1]) if for all a, b ∈ E thefollowing holds: if na ≤ b for all n ∈ N , then a ≤ Definition 3.20.
A sequence ( a n ) n ∈ N in E is called order convergent to an element a ∈ E if there exists a sequence ( h n ) n ∈ N in E + with h n ↓ − h n ≤ a − a n ≤ h n .Notation: a n o −→ a . Theorem 3.21.
Let A be as in (13) . Suppose that E is integrally closed, Γ is directedand ϕ (Γ) is mediated in E . Furthermore assume ϕ has the following continuity property.If A , A , . . . in A are such that A ⊃ A ⊃ · · · and (cid:84) n ∈ N A n = ∅ , (15) then ϕ ( f A n ) ↓ for all f ∈ Γ + . (a) gf ∈ Γ V for all g ∈ [ A ] o and all f ∈ Γ V . (b) Let g ∈ [ A ] o and let ( g n ) n ∈ N be a sequence in [ A ] o for which there is a sequence ( j n ) n ∈ N in [ A ] o + with − j n ≤ g n − g ≤ j n and j n ↓ pointwise. Then ϕ V ( g n f ) o −→ ϕ V ( gf ) ( f ∈ Γ V ) . (16) (Order convergence in the sense of E .) roof. We first prove the following:( (cid:63) ) Let f ∈ Γ + V . Let ( g n ) n ∈ N be a sequence in [ A ] o for which g n f ∈ Γ V for all n ∈ N and g n ↓ ϕ V ( g n f ) ↓ . (17)Let σ ∈ Γ + , σ ≥ f . It follows from Lemma 3.17 that we may assume g n ∈ [ A ] for all n ∈ N . For all n ∈ N we have 0 ≤ ϕ V ( g n f ) ≤ ϕ V ( g n σ ), so we are done if ϕ V ( g n σ ) ↓ h ∈ E , h ≤ ϕ V ( g n σ ) for all n ∈ N ; we prove h ≤ ε >
0. For each n ∈ N , set A n = { x ∈ X : g n ( x ) ≥ ε } . Then A n ∈ A for n ∈ N and A ⊃ A ⊃ · · · and (cid:84) n ∈ N A n = ∅ . Putting M = (cid:107) g (cid:107) ∞ we see that g n ≤ ε X + M A n ( n ∈ N ) , (18)whence h ≤ ϕ V ( g n σ ) ≤ εϕ ( σ ) + M ϕ ( A n σ ) ( n ∈ N ) . (19)By the continuity property of ϕ , h ≤ εϕ ( σ ). As this is true for each ε > E isintegrally closed, we obtain h ≤ V is directed (see 3.11) it is sufficient to consider f ∈ Γ + V . Let g ∈ [ A ] o .There are sequences of step functions ( h n ) n ∈ N and ( j n ) n ∈ N for which h n ↑ g , j n ↓ g andthus j n − h n ↓
0. By Lemma 3.18 h n f, j n f ∈ Γ V for all n ∈ N . Then h n f ≤ gf ≤ j n f for n ∈ N and inf n ∈ N ϕ V (( j n − h n ) f ) = 0 by ( (cid:63) ). By Lemma 3.7 and 3.5(c) we obtainthat gf ∈ Γ V .(b) It is sufficient to consider f ∈ Γ + V . By (a) we may also assume g = 0. But then(b) follows from ( (cid:63) ). Remark 3.22.
Consider the situation in Theorem 3.21. Suppose
B ⊂ A is a σ -algebra.Then all bounded B -measurable functions lie in [ A ] o . If ( g n ) n ∈ N is a bounded sequenceof bounded B -measurable functions that converges pointwise to a function g , then thecondition of Theorem 3.21(b) is satisfied. Remark 3.23.
In the next section we will consider a situation similar to the one ofTheorem 3.21, in which A is replaced by a subset I that is closed under taking finiteintersections. We will also adapt the continuity property on ϕ (see 4.3). The construction described in Definition 3.3 is reminiscent of the Riemann integral and,indeed, the Riemann integral is a special case (see Example 3.4).In the present section we consider a type of extension, analogous to the improper
Riemann integral. One usually defines the improper integral of a function f on [0 , ∞ )to be lim s →∞ (cid:90) s f ( x ) d x, (20)7pproximating the domain, not the values of f .For our purposes a more convenient description of the same integral would be ∞ (cid:88) n =1 (cid:90) a n +1 a n f ( x ) d x, (21)where 0 = a < a < · · · and a n → ∞ . Here the domain is split up into manageablepieces.Splitting up the domain is the basic idea we develop in this section. (This mayexplain our use of the terms “vertical” and “lateral”.) Throughout this section, E and F are partially ordered vector spaces, Γ is a directed linear subspace of F X , and ϕ is a linear order preserving map Γ → E . (With Ω = F X , all considerations of § Γ ⊂ F X E ϕ Furthermore, I is a collection of subsets of X , closed under taking finiteintersections. See Definition 4.1 and Definition 4.2 for two more assump-tions. As a shorthand notation, if ( a n ) n ∈ N is a sequence in E + and { (cid:80) Nn =1 a n : N ∈ N } has a supremum, we denote this supremum by (cid:88) n a n . (22) Definition 4.1.
A disjoint sequence ( A n ) n ∈ N of elements in I whose union is X is calleda partition . If ( A n ) n ∈ N and ( B n ) n ∈ N are partitions and for all n ∈ N there exists an m ∈ N for which B n ⊂ A m , then ( B n ) n ∈ N is called a refinement of ( A n ) n ∈ N . Note that if( A n ) n ∈ N and ( B n ) n ∈ N are partitions then there exists a refinement of both ( A n ) n ∈ N and( B n ) n ∈ N (e.g., a partition that consists of all sets of the form A n ∩ B m with n, m ∈ N ). We assume that there exists at least one partition.
Definition 4.2.
We call a linear subspace ∆ of F X stable (under I ) if f A ∈ ∆ ( f ∈ ∆ , A ∈ I ) . (23)If ∆ is a stable space, then a linear and order preserving map ω : ∆ → E is said to be laterally extendable if for all partitions ( A n ) n ∈ N ω ( f ) = (cid:88) n ω ( f A n ) (see(22)) ( f ∈ ∆ + ) . (24) We assume Γ is stable and ϕ is laterally extendable. For the construction of the lateral extension, one does not need to assume that Γ is directed.However, as one can see later on in the construction, the only part of Γ that matters for the extensionis Γ + − Γ + . .3. In the situation of Theorem 3.21 we can choose I = A ; then (15) is precisely thelateral extendability of ϕ . Example 4.4.
For any partially ordered vector space F and a linear subspace E ⊂ F ,the following choices lead to a system fulfilling all of our assumptions: X = N , I = P ( N ),Γ = c [ E ] (see § ϕ ( f ) = (cid:80) n ∈ N f ( n ) for f ∈ Γ. Definition 4.5.
Let ∆ be a stable subspace of F X and let ω : ∆ → E be a laterallyextendable linear order preserving map. Let ( A n ) n ∈ N be a partition, and f : X → F .We call ( A n ) n ∈ N a partition for f (occasionally ∆ -partition for f ) if f A n ∈ ∆ ( n ∈ N ) . (25)A function f : X → F is said to be a partially in ∆ if there exists a partition for f . For f : X → F + , ( A n ) n ∈ N is called a ω -partition for f if it is a partition for f and if (cid:88) n ω ( f A n ) exists . (26)A function f : X → F + that is partially in ∆ is called laterally ω -integrable if thereexists a ω -partition for f . Example 4.6.
Consider the situation of Example 4.4. A function x : N → F is partiallyin Γ if and only if x n ∈ E for every n ∈ N . If x ≥
0, then x is laterally integrable if x n ∈ E for every n ∈ N and (cid:80) n x n exists in E . Naturally, we wish to use (26) to define an integral for f . For that we have to showthe supremum to be independent of the choice of the partition ( A n ) n ∈ N . Lemma 4.8. (a)
Let f : X → F and let ( A n ) n ∈ N be a partition for f . If ( B n ) n ∈ N is a partitionthat is a refinement of ( A n ) n ∈ N , then ( B n ) n ∈ N is a partition for f . (b) Let f : X → F + and let ( A n ) n ∈ N and ( B m ) m ∈ N be partitions for f . Then the sets (cid:40) N (cid:88) n =1 ϕ ( f A n ) : N ∈ N (cid:41) and (cid:40) M (cid:88) m =1 ϕ ( f B m ) : M ∈ N (cid:41) (27) have the same upper bounds in E .Proof. We leave the proof of (a) to the reader. Let u be an upper bound for theset { (cid:80) Nn =1 ϕ ( f A n ) : N ∈ N } ; it suffices to prove that u is an upper bound for { (cid:80) Mm =1 ϕ ( f B m ) : M ∈ N } . Take M ∈ N ; we are done if u ≥ (cid:80) Mm =1 ϕ ( f B m ), i.e.,if u ≥ ϕ ( f B ) where B = B ∪ · · · ∪ B M . But f B ∈ Γ so ϕ ( f B ) = (cid:80) n ϕ ( f B A n ) =sup N ∈ N (cid:80) Nn =1 ϕ ( f B A n ), whereas, for each N ∈ N N (cid:88) n =1 ϕ ( f B A n ) ≤ N (cid:88) n =1 ϕ ( f A n ) ≤ u. (28)9 heorem 4.9. Let f : X → F + be laterally ϕ -integrable. Then every partition for f isa ϕ -partition for f . There exists an a ∈ E + such that for every partition ( A n ) n ∈ N for f , a = (cid:88) n ϕ ( f A n ) . (29) If f ∈ Γ + , then a = ϕ ( f ) .Proof. This is a consequence of Lemma 4.8(b).
Definition 4.10.
For a laterally ϕ -integrable f : X → F + we call the element a ∈ E + for which (29) holds its ϕ L -integral and denote it by ϕ L ( f ). For the moment, denote by(Γ + ) L the set of all laterally ϕ -integrable functions f : X → F + . We proceed to extend ϕ L to a linear function defined on the linear hull of (Γ + ) L , see Definition 4.14. The assumptions that Γ is stable and ϕ is laterally extendable are crucial for thefact that the ϕ L -integral of a laterally ϕ -integrable function is independent of the choiceof a ϕ -partition (see Lemma 4.8(b)). We will use the following rules for a partially ordered vector space E : a n ↑ a, b n ↑ b = ⇒ a n + b n ↑ a + b ( a n , b n , a, b ∈ E ) , (30) a n ↑ , b n ↑ b, a n + b n ↑ a + b = ⇒ a n ↑ a ( a n , b n , a, b ∈ E ) . (31) ϕ L ) Define Γ L = { f − f : f , f ∈ (Γ + ) L } . Step 1.
Let f, g ∈ (Γ + ) L . There exists an ( A n ) n ∈ N that is a ϕ -partition for f and for g . By defining a N = (cid:80) Nn =1 ϕ ( f A n ) and b N = (cid:80) Nn =1 ϕ ( g A n ) for N ∈ N , by (30) weobtain f + g ∈ (Γ + ) L with ϕ L ( f + g ) = ϕ L ( f ) + ϕ L ( g ).Consequently, Γ L is a vector space, containing (Γ + ) L . Step 2. If g , g , h , h ∈ (Γ + ) L and g − g = h − h , then g + h = g + h so that,by the above, ϕ L ( g ) − ϕ L ( h ) = ϕ L ( g ) − ϕ L ( h ).Hence, ϕ L extends to a linear function Γ L → E (also denoted by ϕ L ). Step 3.
Let f, g ∈ (Γ + ) L and f ≤ g . By defining a N and b N as in step 1 and c N = b N − a N , by (31) we infer that g − f ∈ (Γ + ) L .Thus, if f ∈ Γ L and f ≥
0, then f ∈ (Γ + ) L . Briefly: (Γ + ) L is Γ + L , the positive partof Γ L . Definition 4.14.
A function f : X → F is called laterally ϕ -integrable if f ∈ Γ L (see4.13), i.e., if there exist f , f ∈ (Γ + ) L for which f = f − f . The ϕ L -integral of sucha function is defined by ϕ L ( f ) = ϕ L ( f ) − ϕ L ( f ). ϕ L is a function Γ L → E and is called the lateral extension of ϕ . The set of laterally ϕ -integrable functions, Γ L , is called the lateral extension of Γ under ϕ .Note that, thanks to Step 3 of 4.13, this definition of “laterally ϕ -integrable” does notconflict with the one given in Definition 4.10. Like for the vertical extension, we have the following elementary observations:10a) Γ ⊂ Γ L and ϕ L ( τ ) = ϕ ( τ ) for all τ ∈ Γ.(b) Γ L is a directed partially ordered vector space and ϕ L is a linear order preservingfunction on Γ L .(c) If Π is a directed linear subspace of F X and Π ⊂ Γ, then Π L ⊂ Γ L .((Γ L ) L is not so easy. See Theorem 4.18 and Example 4.19.)In case E is a Banach lattice with σ -order continuous norm, for Γ + L we have ananalogue of Theorem 3.13. Lemma 4.16.
Suppose E is a Banach lattice with σ -order continuous norm. Let f : X → F + . Then f lies in Γ + L if and only if there exists a Γ -partition ( A n ) n ∈ N for f such that the sequence ( ϕ ( f A n )) n ∈ N has a sum in the sense of the norm, in which case ϕ L ( f ) is this sum.Proof. The “only if” part follows by definition of Γ L and the σ -order continuity of thenorm. For the “if” part; this follows from the fact that if a n ↑ and (cid:107) a n − a (cid:107) → a, a , a , · · · ∈ E , then a n ↑ a .We will now investigate conditions under which ϕ L and ϕ V themselves are laterallyextendable. (For that, their domains have to be able to play the role of Γ, so they haveto be stable.) First a useful lemma: Lemma 4.17.
Let f ∈ Γ L . Then there exists a partition ( A n ) n ∈ N for f such that everyrefinement ( B m ) m ∈ N of it (is a partition for f and) has this property: h ∈ E, h ≥ M (cid:88) m =1 ϕ ( f B m ) for all M ∈ N = ⇒ h ≥ ϕ L ( f ) . (32) Proof.
Write f = f − f with f , f ∈ Γ + L . Let ( A n ) n ∈ N be a partition for f and f ,and let ( B m ) m ∈ N be a refinement of ( A n ) n ∈ N . Note that ( B m ) m ∈ N is a partition for f and f . Let h be an upper bound for { (cid:80) Mm =1 ϕ ( f B m ) : M ∈ N } in E . For all M ∈ N , h + M (cid:88) m =1 ϕ ( f B m ) ≥ M (cid:88) m =1 ϕ ( f B m ) + M (cid:88) m =1 ϕ ( f B m ) = M (cid:88) m =1 ϕ ( f B m ) . (33)Taking the supremum over M yields h + ϕ L ( f ) ≥ ϕ L ( f ), i.e., h ≥ ϕ L ( f ). Theorem 4.18. (a)
Suppose Γ L is stable. Then ϕ L is laterally extendable, i.e., ϕ L ( f ) = (cid:88) n ϕ L ( f A n ) (34) for every f ∈ Γ + L and every ϕ L -partition ( A n ) n ∈ N for f . Therefore (Γ L ) L = Γ L and ( ϕ L ) L = ϕ L . Note that for this inclusion it is necessary that Γ be directed.
Suppose Γ V is stable. Then ϕ V is laterally extendable. (For (Γ V ) L see § (a) Let f ∈ Γ + L and let ( B n ) n ∈ N be a ϕ L -partition for f . Let ( A n ) n ∈ N be thepartition for f as in Lemma 4.17. Then form a common refinement of ( B n ) n ∈ N and( A n ) n ∈ N and apply Lemma 4.17.(b) Let f ∈ Γ + V and let ( A n ) n ∈ N be a partition. Let h ∈ E, h ≥ (cid:80) Nn =1 ϕ V ( f A n ) forevery N ∈ N . We wish to prove h ≥ ϕ V ( f ), which will be the case if h ≥ ϕ ( σ ) for every σ ∈ Γ with σ ≤ f . For that apply Lemma 4.17 to σ .The following shows that Γ L may not be stable, in which case there is no (Γ L ) L .(However, see Theorem 4.25(a).) Example 4.19.
Consider the situation in Example 4.4 and assume there is an a : N → E + such that (cid:80) n a n exists in F and (cid:80) n a n does not (e.g. E = F = c and a n = e n = { n } ). By Example 4.6 a lies in Γ L but b = (0 , a , , a , . . . ) does not; but b = a { , , ,... } and { , , , . . . } ∈ I . (Actually, the existence of such an a : N → E + isequivalent to E not being “splitting” in F ; see Definition 4.21 and (36).) Remark 4.20. Γ V may not be stable either. With E = c , F = (cid:96) ∞ , X = { , } , Γ = c × c and ϕ ( f, g ) = f + g (as in Example 3.8), the space Γ V is not stable for I = P ( X ). Definition 4.21.
Let D be a linear subspace of E . D is called splitting in E if thefollowing is true:If ( a n ) n ∈ N and ( b n ) n ∈ N are sequences in D with 0 ≤ a n ≤ b n for n ∈ N and (cid:80) n b n exists in E , then so does (cid:80) n a n . (35)It is not difficult to see that D is splitting in E if and only ifIf ( a n ) n ∈ N is a sequence in D + and (cid:80) n a n exists in E ,then so does (cid:80) n A ( n ) a n for all A ⊂ N . (36)If D is splitting in E , then so is every linear subspace of D . If E is σ -Dedekind complete,then E is also splitting. More generally, D is splitting in E if every bounded increasingsequence in D has a supremum in E . Also, R with the lexicographical ordering issplitting.In Theorem 4.25 we will see what is the use of this concept. First, we have a lookat the connection between “splitting” and “mediated”. Lemma 4.22.
Suppose D is a linear subspace of E . Consider the condition:For all sequences ( a n ) n ∈ N , ( b n ) n ∈ N in D : a n ↓ , b n ↑ , inf n ∈ N a n − b n = 0 = ⇒ inf n ∈ N a n = sup n ∈ N b n . (37) (The infima and suprema in (37) are to be taken in E .) If D is either splitting ormediated in E , then (37) holds. Conversely, (37) implies that D is splitting if D = E ,whereas (37) implies that D is mediated in E if E is a Riesz space and D is a Rieszsubspace of E . roof. It will be clear that mediatedness implies (37) and vice versa if E is a Riesz spaceand D a Riesz subspace of E .If D is splitting in E and a n ↓ , b n ↑ and inf a n − b n = 0, then (cid:80) n b n +1 − b n + a n − a n +1 = a − b . Hence (37) holds.Suppose D = E and (37) holds. Let ( a n ) n ∈ N and ( b n ) n ∈ N be sequences in D with0 ≤ a n ≤ b n for n ∈ N such that (cid:80) n b n exists. Let z = (cid:80) n b n , A n = (cid:80) ni =1 a i , C n = (cid:80) ni =1 b i − a i for n ∈ N . Then A n ↑ , C n ↑ and z − C n − A n ↓ z − C n ∈ D ). Hence sup n ∈ N A n = (cid:80) n a n exists. (a) If E is a Riesz space, then every splitting Riesz subspace is mediated in E .(b) If E is mediated, then it is splitting. The converse is also true if E is a Rieszspace.(c) c is mediated in c , not splitting in c (with D = E = c also (37) is not satisfied)).(d) If D is the space of all polynomial functions on [0 ,
1] with degree at most 2 and E = C [0 , D is splitting in E , but not mediated in E . (Actually, D issplitting, but not mediated.) D is splitting (and satisfies (37) with E = D ): If u n ∈ E + , u n ↑ and u n ≤ , then | u n ( x ) − u n ( y ) | ≤ | x − y | as can be concluded from the postscript in Example 5.15.Therefore the pointwise supremum is continuous. It is even in D since u n ( x ) = a n x + b n x + c n , where a n , b n , c n are linear combinations of u n (0) , u n ( ) , u n (1)(see also the postscript in Example 5.15). D is not mediated: For example one can find countable A, B ⊂ E for which [ , is pointwise the infimum of A and ( , is pointwise the supremum of B , theninf A − B = 0, but there is no h ∈ E with B ≤ h ≤ A .) Theorem 4.24.
Let E be a Banach lattice with σ -order continuous norm. Then E isboth mediated and splitting.Proof. Suppose a n , b n ∈ E with 0 ≤ a n ≤ b n for n ∈ N . Suppose that { (cid:80) Nn =1 b n : N ∈ N } has a supremum s in E . We prove that { (cid:80) Nn =1 a n : N ∈ N } has a supremum in E .Since the norm is σ -order continuous, we have (cid:107) s − (cid:80) Nn =1 b n (cid:107) →
0. In particular weget that for all ε > N ∈ N such that for all n, m ≥ N with m > n wehave (cid:107) (cid:80) mi = n b i (cid:107) < ε and thus (cid:107) (cid:80) mi = n a i (cid:107) < ε . From this we infer that ( (cid:80) Nn =1 a n ) N ∈ N converges in norm. Therefore it has a supremum in E . Thus E is splitting. By Lemma4.22 E is mediated. Theorem 4.25. (a) ϕ (Γ) splitting in E = ⇒ Γ L is stable and ϕ L is laterally extendable. (b) ϕ (Γ) mediated in E = ⇒ Γ V is stable and ϕ V is laterally extendable. (c) ϕ (Γ) splitting in E and ϕ L (Γ L ) mediated in E = ⇒ (Γ L ) V is stable and ( ϕ L ) V islaterally extendable.Proof. (a) Let f ∈ Γ L , B ∈ I ; we prove f B ∈ Γ L . (This is sufficient by Theorem4.18(a).) Without loss of generality, assume f ≥
0. Choose a ϕ -partition ( A n ) n ∈ N for13 . Now apply (35) to a n := ϕ ( f A n ∩ B ) , b n := ϕ ( f A n ) ( n ∈ N ) . (38)(b) follows from Lemma 3.18 and Theorem 4.18(b).(c) By (a) Γ L is stable and ϕ L is laterally extendable. Hence we can apply (b) to Γ L and ϕ L (instead of Γ and ϕ ) and obtain (c). To some extent, the assumption of Theorem 4.25(a) is minimal.Indeed, in the situation of Example 4.4, we see that Γ L is stable if and only if E (whichis ϕ (Γ)) is splitting in F (see (36)).In Theorem 4.25(c) we assumed that ϕ L (Γ L ) (and thus also ϕ (Γ)) was mediated in E . It may happen that ϕ (Γ) is mediated in E , but ϕ L (Γ L ) is not, as Example 4.27illustrates. However, splitting is preserved under the lateral extension and mediation ispreserved under the vertical extension, see Theorem 4.28. Example 4.27.
Let X = N , I = P ( N ), E = F = c . Let Γ = c [ c ] (see §
2) and ϕ : Γ → E be given by ϕ ( f ) = (cid:80) n ∈ N f ( n ). Then ϕ (Γ) = c , which is mediated in c . Afunction f : N → c is partially in Γ if and only if f ( N ) ⊂ c . For x ∈ c + the functiongiven by f ( n ) = x ( n ) { n } for n ∈ N lies in Γ L , and ϕ L ( f ) = x . It follows that ϕ L (Γ L )is c , which is not mediated in c . Theorem 4.28. (a) If ϕ (Γ) is splitting in E , then so is ϕ L (Γ L ) . (b) If ϕ (Γ) is mediated in E , then so is ϕ V (Γ V ) .Proof. (a) Suppose a n ∈ ϕ L (Γ L ) + for n ∈ N and (cid:80) n a n exists. Let A ⊂ N . For all n ∈ N there exist b n , b n , · · · ∈ ϕ (Γ) + with a n = (cid:80) m b nm . Hence (cid:80) n a n = (cid:80) n,m b nm and so (cid:80) n,m A × N ( n, m ) b nm = (cid:80) n A ( n ) a n exists in E .(b) Suppose A, B ⊂ ϕ V (Γ V ) are countable sets with inf A − B = 0. For all a ∈ A and b ∈ B there exist countable sets Υ a , Λ b ⊂ Γ with a = inf ϕ (Υ a ) , b = sup ϕ (Λ b ). Theninf ϕ ( (cid:83) a ∈ A Υ a − (cid:83) b ∈ B Λ b ) = 0 and thus inf A = inf ϕ ( (cid:83) a ∈ A Υ a ) = sup ϕ ( (cid:83) b ∈ B Λ b ) =sup B . For a Riesz space F we will now investigate under which conditions the space Γ L is a Riesz subspace of F X . The next example shows that even if E is a Riesz space andΓ is a Riesz subspace of F X , Γ L may not be one. However, see Theorem 4.32. Example 4.30.
Let a, b be as in Example 4.19; this time put d = (0 , a + a , , a + a , . . . ). Then a, d ∈ Γ L but a ∧ d = b / ∈ Γ L .Hence, in Example 4.4, if F is a Riesz space and E is not splitting in F , then Γ L isnot a Riesz subspace of F X . As we will see in Theorem 4.32, considering the situationof Example 4.4: Γ L is a Riesz subspace of F X if and only if E is splitting in F . Lemma 4.31.
Let f : X → F be partially in Γ . If f is in Γ LV , then f ∈ Γ L . (b) Suppose ϕ (Γ) is splitting in E . If g ≤ f ≤ h for certain g, h ∈ Γ L , then f ∈ Γ L .Proof. (a) By the definition of Γ LV there exists a ρ ∈ Γ L with ρ ≤ f . Then f − ρ ispartially in Γ, f − ρ ∈ Γ LV , and we are done if f − ρ ∈ Γ L . Hence we may assume f ≥ A n ) n ∈ N be a partition for f ; we prove (cid:80) n ϕ ( f A n ) = ϕ LV ( f ). It will be clear that (cid:80) Nn =1 ϕ ( f A n ) ≤ ϕ LV ( f ) for N ∈ N . For the reverse inequality let h ∈ E be an upperbound for { (cid:80) Nn =1 ϕ ( f A n ) : N ∈ N } . It suffices to show that h must be an upper boundfor { ϕ L ( σ ) : σ ∈ Γ L , σ ≤ f } .Take a σ ∈ Γ L with σ ≤ f . If ( B n ) n ∈ N is any refinement of ( A n ) n ∈ N that is a ϕ -partitionfor σ , then for all M ∈ N there exists an N ∈ N with B ∪· · ·∪ B M ⊂ A ∪· · ·∪ A N , so that h ≥ (cid:80) Nn =1 ϕ ( f A n ) ≥ (cid:80) Mm =1 ϕ ( f B m ) ≥ (cid:80) Mm =1 ϕ ( σ B m ). It follows from Lemma 4.17,applied to σ , that the partition ( B m ) m ∈ N can be chosen so that this implies h ≥ ϕ L ( σ ).(b) As h − g ∈ Γ L and 0 ≤ f − g ≤ h − g , we may (and do) assume g = 0. Let( A n ) n ∈ N be a partition for f that is also a ϕ -partition for h . Now just apply (35) to a n := ϕ ( f A n ) , b n := ϕ ( h A n ) ( n ∈ N ) . (39)As a consequence of Lemma 4.31: Theorem 4.32.
Let F be a Riesz space and Γ be a Riesz subspace of F X . The functions X → F that are partially in Γ form a Riesz space, Ξ . If ϕ (Γ) is splitting in E , then Γ L is a Riesz ideal in Ξ , in particular, Γ L is a Riesz space. In the classical integration theory and the Bochner integration theory one startswith considering a measure space ( X, A , µ ) and simple functions on X with values in R or in a Banach space. One defines an integral on these simple functions using themeasure and extends this integral to a larger class of integrable functions. In 4.33 wewill follow a similar procedure, replacing R or the Banach space with E and applyingthe lateral extension. In Section 8 we will treat such extensions in more detail. Suppose ( X, A , µ ) is a σ -finite complete measure space and suppose E is directed.Let F = E . For I we choose { A ∈ A : µ ( A ) < ∞} . The σ -finiteness of µ guaranteesthe existence of a partition (and vice versa).We say that a function f : X → E is simple if there exist N ∈ N , a , . . . , a N ∈ E , A , . . . , A N ∈ I for which f = N (cid:88) n =1 a n A n . (40)The simple functions form a stable directed linear subspace S of E X , which is a Rieszsubspace of E X in case E is a Riesz space.For a given f in S one can choose a representation (40) in which the sets A , . . . , A N are pairwise disjoint; thanks to the σ -finiteness of µ one can choose them in such a waythat they occur in a partition ( A n ) n ∈ N . 15his S is going to be our Γ. We define ϕ : S → E by ϕ ( f ) = N (cid:88) n =1 µ ( A n ) a n , (41)where f, N, A n , a n are as in (40). The σ -additivity of µ is (necessary and) sufficient toshow that S is laterally extendable.A function f : X → E is partially in S if and only if there exist a partition ( A n ) n ∈ N and a sequence ( a n ) n ∈ N in E for which f = (cid:88) n ∈ N a n A n . (42)An f as in (42) with f ≥ S is an element of S L if and only if (cid:80) n µ ( A n ) a n exists in E . (See Theorem 4.9.) In this section
E, F, X, I , Γ , ϕ are as in Section 4. As we have seen, the lateral extension differs from the vertical extension in the sensethat the vertical extensions of Γ and ϕ can always be made, but for lateral extensionwe had to assume the space Γ to be stable and ϕ to be laterally extendable (see 4.11).In this section we investigate when one can make a lateral extension of another (sayvertical) extension. Furthermore we will compare different extensions and combinationsof extensions.Instead of (Γ L ) V and ((Γ L ) V ) L we write Γ LV and Γ LV L ; similarly ϕ LV = ( ϕ L ) V etc. By Theorem 4.18 the following holds for a stable directed linear subspace ∆ of F X and a laterally extendable order preserving linear map ω : ∆ → E : If ∆ L is stable,then ω L is laterally extendable (and so ∆ LL exists). If ∆ V is stable, then ω V is laterallyextendable (and so ∆ V L exists). We will use these facts without explicit mention.
The following statements follow from the definitions and theorems we have:(a) Γ V ⊂ Γ LV and ϕ LV = ϕ V on Γ V .(b) Γ L ⊂ Γ LV and ϕ LV = ϕ L on Γ L .(c) ϕ V = ϕ L on Γ L ∩ Γ V .For (d), (e) and (f) let Γ V be stable.(d) Γ LV ⊂ Γ V LV and ϕ V LV = ϕ LV on Γ LV .(e) Γ V L ⊂ Γ V LV and ϕ V LV = ϕ V L on Γ
V L .(f) ϕ LV = ϕ V L on Γ LV ∩ Γ V L .Observe that as a consequence of (a) and (b): If f ∈ Γ L and g ∈ Γ V and f ≤ g (or f ≥ g ), then ϕ L ( f ) ≤ ϕ V ( g ) (or ϕ L ( f ) ≥ ϕ V ( g )). Moreover, as a consequence of(c) and (d); if Γ V is stable: If f ∈ Γ LV and g ∈ Γ V LV and f ≤ g (or f ≥ g ), then ϕ LV ( f ) ≤ ϕ V LV ( g ) (or ϕ LV ( f ) ≥ ϕ V LV ( g )).16 .3. Note that if Γ is stable and ϕ is laterally extendable, then we can extend Γ toΓ V , Γ L and Γ LV . If, moreover, Γ V is stable, then we can also extend Γ to Γ V L andΓ
V LV . However, “more stability” will not give us larger extensions than Γ
V LV . Indeed,if Γ LV is stable then Γ LV ⊂ Γ LV L = Γ
V L (see Theorem 5.8). If moreover Γ
V LV is stable,then even Γ
V LV L = Γ
V LV = Γ
V L . Lemma 5.4. (a) If f ∈ Γ + LV , then there exists a countable Λ ⊂ Γ with Λ ≤ f and ϕ LV ( f ) =sup ϕ (Λ) . (b) If Γ V is stable and f ∈ Γ + V L , then there exists a countable Λ ⊂ Γ with Λ ≤ f and ϕ V L ( f ) = sup ϕ (Λ) .Proof. (a) There exist σ , σ , . . . in Γ L with σ n ≤ f for all n ∈ N and sup n ∈ N ϕ L ( σ n ) = ϕ LV ( f ). Hence, we are done if for every σ in Γ L with σ ≤ f there is a countable setΛ σ ⊂ { ρ ∈ Γ : ρ ≤ f } such that every upper bound for ϕ (Λ σ ) majorizes ϕ L ( σ ). But thatis not hard to prove. For such a σ , by Lemma 4.17 there exists a partition ( B m ) m ∈ N forwhich (32) holds. Now let Λ σ be { (cid:80) Mm =1 σ B m : M ∈ N } .(b) Suppose Γ V is stable. Let ( A n ) n ∈ N be a ϕ V -partition for f . Then the set Λ f = { (cid:80) Nn =1 f A n : N ∈ N } is a countable subset of Γ V and sup ϕ V (Λ f ) = ϕ V L ( f ). Moreover,for every N ∈ N there is a countable set Λ N ⊂ { σ ∈ Γ : σ ≤ (cid:80) Nn =1 f A n } for whichsup ϕ (Λ N ) = ϕ V ( (cid:80) Nn =1 f A n ). Take Λ = (cid:83) N ∈ N Λ N . Theorem 5.5.
For (b),(c),(d) and (e) let Γ V be stable and f be partially in Γ V . (a) If f ∈ Γ LV , then f ∈ Γ V ⇐⇒ there exist π, ρ ∈ Γ with π ≤ f ≤ ρ . (b) If f ∈ Γ V L , then f ∈ Γ V ⇐⇒ there exist π, ρ ∈ Γ with π ≤ f ≤ ρ . (c) f ∈ Γ LV ⇐⇒ f ∈ Γ V L and there exist π, ρ ∈ Γ L with π ≤ f ≤ ρ. (d) If ϕ V (Γ V ) is splitting in E , then f ∈ Γ V L ⇐⇒ there exist π, ρ ∈ Γ V L with π ≤ f ≤ ρ. (e) If ϕ V (Γ V ) is splitting in E , then f ∈ Γ V L ∩ Γ LV ⇐⇒ there exist π, ρ ∈ Γ L with π ≤ f ≤ ρ. By the definition of ideal in [3] or [7] (note that Γ V is directed) this means that Γ V is the smallestideal in Γ LV (and for (b); in Γ V L ) that contains Γ. roof. The proofs of (a) and (b) are similar to the proof of (c) and therefore omitted.(c) ⇐ : By Lemma 5.4 (b) there exist countable sets Λ , Υ ⊂ Γ with Λ ≤ f − π andΥ ≤ ρ − f for which sup ϕ (Λ) = ϕ V L ( f − π ) and sup ϕ (Υ) = ϕ V L ( ρ − f ). Then Λ + π and ρ − Υ are countable subsets of Γ L with Λ + π ≤ f ≤ ρ − Υ and sup ϕ L (Λ + π ) = ϕ V L ( f ) = inf ϕ L ( ρ − Υ). Hence f ∈ Γ LV . ⇒ : Let f ∈ Γ LV and be partially in Γ V . There exists a π ∈ Γ L for which f − π ∈ Γ + LV ,hence we may assume f ≥
0. Let ( A n ) n ∈ N be a Γ V -partition for f , i.e., f A n ∈ Γ V and thus ϕ LV ( f A n ) = ϕ V ( f A n ) for all n ∈ N (see 5.2(a)). Then ϕ LV ( f ) ≥ (cid:80) Nn =1 ϕ V ( f A n ) for all N ∈ N . Let h ∈ E be such that h ≥ (cid:80) Nn =1 ϕ V ( f A n ) for all N ∈ N . From Lemma 4.17 we infer that h ≥ ϕ L ( σ ) for every σ ∈ Γ L with σ ≤ f . Weconclude that (cid:80) n ϕ V ( f A n ) = ϕ LV ( f ), i.e., f ∈ Γ V L .(d) ⇐ : We may assume π = 0. Let ( A n ) n ∈ N be a ϕ V -partition for ρ with f A n ∈ Γ V for all n ∈ N . Then 0 ≤ ϕ V ( f A n ) ≤ ϕ V ( ρ A n ) for all n ∈ N and (cid:80) n ϕ V ( ρ A n ) existsin E . Hence, so does (cid:80) n ϕ V ( f A n ), i.e., f ∈ Γ V L .(e) is a consequence of (c) and (d).In the following example all functions in Γ LV are partially in Γ V . Example 5.6.
Consider X = N , I = P ( N ), E = F ; let D be a linear subspace of E andlet D V be the vertical extension of D with respect to the inclusion map D → E . LetΓ = c [ D ] and ϕ : Γ → E be ϕ ( f ) = (cid:80) n ∈ N f ( n ). Then Γ V = c [ D V ]. Let f ∈ Γ LV .We will show that f ( k ) ∈ D V and thus that f is partially in Γ V . Let σ n , τ n ∈ Γ L besuch that σ n ≤ f ≤ τ n and inf n ∈ N ϕ ( τ n ) = sup n ∈ N ϕ ( σ n ). Then inf n ∈ N ( τ n ( k ) − σ n ( k )) ≤ inf n ∈ N ϕ ( τ n − σ n ) = 0. Since σ n ( k ) , τ n ( k ) ∈ D for all n ∈ N , we have f ( k ) ∈ D V .Thus every f ∈ Γ LV is partially in Γ V . Since Γ V is stable, by Theorem 5.5(c) weconclude that Γ LV ⊂ Γ V L . Lemma 5.7.
Suppose that Γ LV is stable. Then every f ∈ Γ LV is partially in Γ V .Proof. Let f ∈ Γ LV and let π, ρ ∈ Γ L be such that π ≤ f ≤ ρ . Let ( A n ) n ∈ N be a ϕ -partition for both π and ρ . Then f A n ∈ Γ LV and π A n ≤ f A n ≤ ρ A n for all n ∈ N . By Theorem 5.5(a) we conclude that f A n ∈ Γ V . Theorem 5.8.
Suppose that Γ V and Γ LV are stable. Then Γ LV ⊂ Γ V L = Γ
LV L . Write
Γ = Γ
V L and ϕ = ϕ V L . If Γ is stable, then Γ L = Γ and ϕ L = ϕ . If Γ V is stable, then Γ V = Γ and ϕ V = ϕ .In particular, if ϕ L (Γ L ) is mediated in E and ϕ V (Γ V ) is splitting in E , then Γ V , Γ LV and Γ V L are stable (see Theorem 4.25) and thus Γ LV ⊂ Γ , Γ = Γ V = Γ L , ϕ = ϕ V = ϕ L ,so Γ = Γ (and ϕ = ϕ ).Proof. The inclusion Γ LV ⊂ Γ V L follows by Theorem 5.5(c) and Lemma 5.7. We proveΓ
LV L ⊂ Γ V L . For f ∈ Γ + LV L there is a ϕ LV -partition for f and since Γ LV ⊂ Γ V L this isalso a ϕ V L -partition for f , hence there exists a ϕ V -partition for f , i.e., f ∈ Γ V L .Suppose Γ is stable. Then Γ L = (Γ V L ) L = Γ V L = Γ and ϕ L = ϕ by Theorem4.18(a). 18uppose Γ V to be stable. As Γ V is stable we can apply the first part of the theoremto Γ V instead of Γ. Indeed, (Γ V ) V and (Γ V ) LV are stable, since (Γ V ) V = Γ V and(Γ V ) LV = Γ V . Hence, (Γ V ) LV ⊂ (Γ V ) V L = Γ
V L , i.e., Γ V ⊂ Γ (and ϕ V = ϕ ).Suppose ϕ L (Γ L ) is mediated in E and ϕ V (Γ V ) is splitting in E . Then Γ L , Γ V and Γ LV are stable by Theorem 4.25(a),(b) and (c). Consequently, again by Theorem4.25(b) Γ V L is stable.
Corollary 5.9.
Suppose E is mediated (and thus splitting), Γ = Γ
V L . Then
Γ = Γ V =Γ L , so Γ = Γ (and ϕ = ϕ ). At the end of § V L (cid:40) Γ LV (Example 5.14) andsometimes Γ LV (cid:40) Γ V L (Example 5.15). Note that this implies that Γ
V LV can bestrictly larger then either Γ
V L or Γ LV .Theorem 5.8 raises the question whether stability of Γ V entails Γ V L ⊂ Γ LV . Ingeneral the answer is negative; see Example 5.15. In Theorem 5.10 we give conditionssufficient for the inclusion. Theorem 5.10.
Suppose Γ V is stable. Consider these two statements. (a) For every f ∈ Γ + V L there is a ρ in Γ + L with f ≤ ρ . (b) E satisfies:If Y , Y , · · · ⊂ E are nonempty countable with inf Y n = 0 for all n ∈ N , then there exist y ∈ Y , y ∈ Y , . . . such that (cid:88) n y n exists in E. (43) If (a) is satisfied, then Γ V L ⊂ Γ LV . (b) implies (a).Proof. If (a) is satisfied, then by Theorem 5.5(c) follows that Γ
V L ⊂ Γ LV .Suppose (b). Let f ∈ Γ + V L . Let ( A n ) n ∈ N be a ϕ V -partition for f . For n ∈ N , let Υ n ⊂ Γbe a countable set with f A n ≤ Υ n and ϕ V ( f A n ) = inf ϕ (Υ n ) . (44)We may assume σ A n = σ for all σ ∈ Υ n . Choose σ n ∈ Υ n for n ∈ N such that (cid:80) n ( ϕ ( σ n ) − ϕ V ( f A n )) and thus (cid:80) n ϕ ( σ n ) exist in E . Then ρ := (cid:80) n ∈ N σ n is in Γ + L with f ≤ ρ . We will discuss examples of spaces E for which (43) holds.(I) If E is a Banach lattice with σ -order continuous norm, then E satisfies (43) (onecan find y n ∈ Y n with (cid:107) y n (cid:107) ≤ − n ).(II) Let ( X, A , µ ) be a complete σ -finite measure space and assume there exists a g ∈ L ( µ ) with g > µ -a.e.. Then the space E of equivalence classes of measurable functions X → R satisfies (43): It is sufficient to prove that if Z , Z , · · · ⊂ E are nonemptycountable with inf Z n = 0 for all n ∈ N , then there exists z ∈ Z , z ∈ Z , . . . and a z ∈ E such that z n ≤ z for all n ∈ N (for Z n take 2 n Y n ). One can prove that such a z exists by mapping the equivalence classes of measurable functions into L ( µ ) by theorder isomorphism f (cid:55)→ (arctan ◦ f ) g .(III) R N is a special case of (II), therefore satisfies (43).19 heorem 5.12. Let E be mediated and splitting and satisfy (43) (e.g. E be a Banachlattice with σ -order continuous norm (Theorem 4.24), or E is the space mentioned in5.11(II)). Then Γ V is stable and Γ V L = Γ LV , ϕ V L = ϕ LV .Proof. This is a consequence of Theorem 5.8 and Theorem 5.10.For a Riesz space F and a Riesz subspace Γ of F X we will now investigate underwhich conditions on ϕ (Γ), ϕ L (Γ L ) and ϕ V (Γ V ) the spaces Γ LV and Γ V L are Rieszsubspaces of F X . Theorem 5.13.
Suppose F is a Riesz space and Γ is a Riesz subspace of F X . If ϕ (Γ) is splitting in E and ϕ L (Γ L ) is mediated in E , then Γ LV is a Riesz subspace of F X . If ϕ (Γ) is mediated in E and ϕ V (Γ V ) is splitting in E , then Γ V L is a Riesz subspace of F X .In particular, if E is mediated (and thus splitting), then both Γ LV and Γ V L are Rieszsubspaces of F X .Proof. Note first that if ϕ (Γ) is mediated in E , then Γ V is stable by Theorem 4.25(b).For a proof, combine Theorem 4.32 and Corollary 3.10.The next example illustrates that Γ LV is not always included in Γ V L (given that Γ V is stable) even if E and F are Riesz spaces and Γ , Γ LV , Γ V L
Riesz subspaces of F X . Example 5.14. [ Γ V L (cid:40) Γ LV = Γ V LV ] For an element b = ( β , β , . . . ) of R N we write b = (cid:80) n ∈ N β n e n .Consider X = { , , , . . . } and I = P ( X ). Let E = c , F = R N , Ω = F X . We view theelements of Ω as sequences ( a, b , b , . . . ) with a, b , b , · · · ∈ R N .Define sets Γ ⊂ Θ ⊂ Ω and a map Φ : Θ → R N byΘ = { ( a, β e , β e , . . . ) : a ∈ c, β , β , · · · ∈ R } , (45)Φ( a, β e , β e , . . . ) = a + (cid:88) n ∈ N β n e n ( a ∈ c, β , β , · · · ∈ R ) , (46)Γ = { ( a, β e , β e , . . . ) : a ∈ c, ( β , β , . . . ) ∈ c } . (47)Then Φ(Γ) = c = E ; let ϕ = Φ | Γ . From the definition it is easy to see that Γ is stableand ϕ is laterally extendable. We leave it to the reader to verify that Γ V = Γ,Γ L = { ( a, β e , β e , . . . ) : a ∈ c, ( β , β , . . . ) ∈ c } (48)and ϕ L = Φ on Γ L .It follows that Γ V is stable and Γ V L = Γ L ⊂ Γ LV = Γ V LV . We prove Γ
V L (cid:54) = Γ LV .To this end, define h ∈ Ω by (cid:40) h ( n ) = ( − n e n ( n = 1 , , . . . ) ,h (0) = − (cid:80) n ∈ N h ( n ) = − (cid:80) n ∈ N ( − n e n . (49)20s h (0) / ∈ c we have h { } / ∈ Γ; in particular, h is not partially in Γ, so h / ∈ Γ L = Γ V L .It remains to prove h ∈ Γ LV .For k ∈ N , define τ k , σ k : X → R N : τ k (0) = − (cid:80) kn =1 ( − n e n + (cid:80) ∞ n = k +1 e n ,τ k ( n ) = h ( n ) = ( − n e n ( n = 1 , . . . , k ) ,τ k ( n ) = e n ( n = k + 1 , k + 2 , . . . ) , (50) σ k (0) = − (cid:80) kn =1 ( − n e n − (cid:80) ∞ n = k +1 e n ,σ k ( n ) = h ( n ) = ( − n e n ( n = 1 , . . . , k ) ,σ k ( n ) = − e n ( n = k + 1 , k + 2 , . . . ) . (51)Then τ k , σ k ∈ Γ L , τ k ≥ h ≥ σ k , ϕ L ( τ k ) = Φ( τ k ) = 2 (cid:80) n>k e n , ϕ L ( σ k ) = − (cid:80) n>k e n , soinf k ∈ N ϕ L ( τ k ) = sup k ∈ N ϕ L ( σ k ) = 0, and h ∈ Γ LV .The next example illustrates that Γ V L is not always included in Γ LV ; it providesan example of an f ∈ Γ + V L for which there exist no ρ ∈ Γ + L with f ≤ ρ (see Theorem5.5(c)). Example 5.15. [ Γ LV (cid:40) Γ V L ] Let E = C [0 ,
1] and let D ⊂ C [0 ,
1] be the set of polynomials of degree ≤
2. The set D is order dense in C [0 ,
1] (see [9, Example 4.4]). Hence, for all f ∈ E there exist( g n ) n ∈ N , ( h n ) n ∈ N in D with f = inf n ∈ N g n = sup n ∈ N h n . Therefore E is the verticalextension of D with respect to the inclusion map D → E .Take X = N , I = P ( N ) , F = E = C [0 , , Γ = c [ D ] ⊂ F N = E N and let ϕ : Γ → E be given by ϕ ( f ) = (cid:80) n ∈ N f ( n ). Since this situation is the same as in Example 5.6 with D V = E , we have Γ V = c [ E ] and Γ LV ⊂ Γ V L .Furthermore (see 4.6)Γ + L = { f ∈ ( D + ) N : (cid:88) n f ( n ) exists in E } , (52)Γ + V L = { f ∈ ( E + ) N : (cid:88) n f ( n ) exists in E } . (53)We construct an f ∈ Γ + V L that is not in Γ LV . For n ∈ N let f n be the ‘tent’ functiondefined by f n (0) = 0; f n ( n ) = 1; f n ( i ) = 0 if i ∈ N , i (cid:54) = n ; f n is affine on the interval [ i , i ] for all i ∈ N . (54)Then (cid:80) ∞ n =1 f n = (0 , pointwise, so (cid:80) n f n = in C [0 , f = ( f , f , f , . . . ) ∈ Γ + V L .We will prove that f / ∈ Γ LV ; by showing there exists no ρ ∈ Γ L for which f ≤ ρ . A subspace D of a partially ordered vector space E is called order dense in E if x = sup { d ∈ D : d ≤ x } (and thus x = inf { d ∈ D : d ≥ x } ) for all x ∈ E . f f n ( n > n +1 1 n n − Figure 1: Graph of f n .Suppose ρ ∈ Γ L and f ≤ ρ . Then ρ = ( ρ , ρ , . . . ) where ρ , ρ , . . . are elements of D + and j = (cid:80) n ρ n exists in E = C [0 , M be the largest value of j . Every ρ n isa quadratic function that maps [0 ,
1] into [0 , M ]. Consequently (see the postscript) | ρ n ( x ) − ρ n ( y ) | ≤ M | x − y | ( x, y ∈ [0 , , n ∈ N ) . (55)In particular, ρ n (0) ≥ ρ n ( n ) − M n ≥ f n ( n ) − M n = 1 − M n ≥ for n ≥ M . As j (0) ≥ (cid:80) n ≥ N ρ n (0) for all N ∈ N , this is a contradiction.Postscript. Let h : x (cid:55)→ ax + bx + c be a quadratic function on [0 ,
1] and 0 ≤ h ( x ) ≤ M for all x ; we prove | h (cid:48) ( x ) | ≤ M for all x ∈ [0 , | h (cid:48) ( x ) | ≤ max {| h (cid:48) (0) | , | h (cid:48) (1) |} . Now h (cid:48) (0) = b = 4 h ( ) − h (1) − h (0) and h (cid:48) (1) = 2 a + b = 3 h (1) + h (0) − h ( ). Since | h ( x ) − h ( y ) | ≤ M for all x, y ∈ [0 , | h (cid:48) (0) | ≤ M and | h (cid:48) (1) | ≤ M as desired. Observe that Γ
V L in Example 5.15 is not stable since ( f , , f , , . . . ) / ∈ Γ V L . E in a (slightly) larger space In this section
E, F, X, I , Γ , ϕ are as in Section 4. Suppose E • is another partially ordered vector space and E ⊂ E • . Consider ϕ • :Γ → E • , where ϕ • ( f ) = ϕ ( f ) for f ∈ Γ.Write Γ • V for the vertical extension of Γ with respect to ϕ • . If ϕ • is laterally ex-tendable, write Γ • L for the lateral extension of Γ with respect to ϕ • , Γ • LV for the verticalextension of Γ • L with respect to ϕ • L . Similarly, if Γ • V is stable, we introduce the notationsΓ • V L and Γ • V LV .It is not generally the case that Γ V ⊂ Γ • V or Γ L ⊂ Γ • L , but a natural restriction on E • helps; see Theorem 6.2.For E • we can choose to be a Dedekind complete Riesz space in which countablesuprema of E are preserved, in case E is integrally closed and directed (see 6.3). In thissituation, in some sense, Γ • V L is the largest extension one can obtain.
Definition 6.1.
Let D be a subspace of a partially ordered vector space P . Then wesay that countable suprema in D are preserved in P if the following implication holdsfor all a ∈ D and all countable A ⊂ DA has supremum a in D = ⇒ A has supremum a in P. (56)Note that the reverse implication holds always.22he following theorem is a natural consequence. Theorem 6.2.
Suppose that countable suprema in E are preserved in E • . Then ϕ • islaterally extendable and f ∈ Γ V ⇐⇒ f ∈ Γ • V and ϕ • V ( f ) ∈ E, (57) f ∈ Γ L ⇐⇒ f ∈ Γ • L and ϕ • L ( f ) ∈ E, (58) ϕ • V ( f ) = ϕ V ( f ) for f ∈ Γ V , ϕ • L ( f ) = ϕ L ( f ) for f ∈ Γ L , (59)Γ LV ⊂ Γ • LV , ϕ • LV ( f ) = ϕ LV ( f ) for f ∈ Γ LV . (60) Suppose Γ V and Γ • V are stable. Then Γ V L ⊂ Γ • V L , ϕ • V L ( f ) = ϕ V L ( f ) for f ∈ Γ V L , (61)Γ V LV ⊂ Γ • V LV , ϕ • V LV ( f ) = ϕ V LV ( f ) for f ∈ Γ V LV . (62) Under the assumptions made in § L , Γ V (see 3.11) andΓ LV (etc.). Hence ϕ V (Γ V ), ϕ L (Γ L ), ϕ LV (Γ LV ) (etc.) are all subsets of E + − E + . Forthis reason we may assume that E itself is directed.Then under the (rather general) assumption that E is also integrally closed (seeDefinition 3.19), E can be embedded in a Dedekind complete Riesz space such thatsuprema and infima in E are preserved, as we state in Theorem 6.4.Consequently, choosing such a Dedekind complete Riesz space for E • one has thefollowing: Γ • V , Γ • LV , Γ • V L , Γ • V LV are stable and Γ • LV ⊂ Γ • V L =: Γ • , Γ • L = Γ • V = Γ • and ϕ • L = ϕ • V = ϕ • , where ϕ • := ϕ • V L (see 5.8). Moreover, one has (60) and if Γ V is stable;(61) and (62). For this reason one may consider Γ • and ϕ • instead of Γ LV and ϕ LV ,instead of Γ • LV and ϕ • LV or instead of Γ V LV and ϕ V LV , indeed Γ • contains all of theother extensions and ϕ • agrees with all integrals. Theorem 6.4. [13, Chapter 4, Theorem 1.19]Let E be an integrally closed directed partially ordered vector space. Then E can beembedded in a Dedekind complete Riesz space ˆ E :There exists an injective linear γ : E → ˆ E for which (a) a ≥ ⇐⇒ γ ( a ) ≥ , (b) γ ( E ) is order dense in ˆ E (for the definition of order dense see the sixth footnote).Consequently, suprema in γ ( E ) are preserved in ˆ E . R In this section ( X, A , µ ) is a complete σ -finite measure space and E = F = R . We write S for the vector space of simple functions from X to R (see 4.33). Since R is a Banach lattice with σ -order continuous norm, S V is stable and S LV = S V L , ϕ LV = ϕ V L (by Theorem 5.12). We write S = S V L and ϕ = ϕ V L . Theorem 7.1. S = L ( µ ) and ϕ ( f ) = (cid:82) f d µ for all f ∈ S . roof. We prove that S + V L ⊂ L ( µ ) + ⊂ S + LV and that ϕ LV ( f ) = (cid:82) f d µ for all f ∈L + ( µ ). S V consists of the bounded integrable functions f for which { x ∈ X : f ( x ) (cid:54) = 0 } hasfinite measure. By monotone convergence, we have f ∈ L ( µ ) for every f ∈ S + V L .Conversely, let f ∈ L ( µ ) + ; we prove f ∈ S + LV and ϕ LV ( f ) = (cid:82) f d µ . Let t ∈ (1 , ∞ ).For n ∈ Z , put A n = { x ∈ X : t n ≤ f ( x ) < t n +1 } . Then ( A n ) n ∈ Z forms a partition.Define g := (cid:80) n ∈ Z t n A n and h := tg ; then g ≤ f ≤ h . Since (cid:88) n ∈ Z t n µ ( A n ) ≤ (cid:88) n ∈ Z (cid:90) f A n d µ = (cid:90) f d µ, (63)we have g ∈ S L and ϕ L ( g ) ≤ (cid:82) f d µ . Also, h = tg ∈ S L , and ϕ L ( h ) − ϕ L ( g ) =( t − ϕ L ( g ) ≤ ( t − (cid:82) f d µ . By this and Lemma 3.7 it follows that f ∈ S LV and ϕ LV ( f ) = (cid:82) f d µ . In this section E is a directed partially ordered vector space, ( X, A , µ ) is acomplete σ -finite measure space and I , S, ϕ are as in 4.33 ( F = E ). In 8.1–8.8 for f in S LV or S V L we discuss the relation between f being almosteverywhere equal to zero and f having integral zero (i.e., either ϕ LV ( f ) = 0 or ϕ V L ( f ) =0). In 8.9 we show that under some conditions a function in S V multiplied with anintegrable function with values in R is a function in S LV .In 8.11–8.13 we investigate the relation between the “ LV ”-extension on simple func-tions with respect to µ and ν , where ν = hµ for some measurable h : X → [0 , ∞ ).In 8.14 we discuss the relation between the “ LV ”-extension simple functions withvalues in E or in another partially ordered vector space F , when one makes the com-position of a function in the extension with a σ -order continuous linear map E → F .In 8.15–8.17 we will prove that under certain conditions on X the function x (cid:55)→ F ( x, · ) is in S V for all F ∈ C ( X × T ) and we relate that to convolution of certain finitemeasures with continuous functions on a topological group. Theorem 8.1.
Let f : X → E and f = 0 a.e.. If f ∈ S LV , then ϕ LV ( f ) = 0 . If S V isstable and f ∈ S V LV , then ϕ V LV ( f ) = 0 .Proof. Let B = { x ∈ X : f ( x ) (cid:54) = 0 } . Then B ∈ A and µ ( B ) = 0.(I) Assume f ∈ S V . Choose σ, τ ∈ S with σ ≤ f ≤ τ . Then σ B , τ B ∈ S , σ B ≤ f ≤ τ B , and ϕ ( σ B ) = ϕ ( τ B ) = 0. Hence ϕ V ( f ) = 0.(II) Suppose σ ∈ S + L and ( A n ) n ∈ N is a ϕ -partition for σ . Then σ A n ∩ B ∈ S + for all n ∈ N and (cid:80) n ϕ ( σ A n ∩ B ) = 0, i.e., σ B ∈ S + L with ϕ L ( σ B ) = 0. In particular, if f ∈ S L then ϕ L ( f ) = 0.(III) Assume f ∈ S LV . With (II) one can repeat the argument of (I) with S replacedby S L and conclude ϕ LV ( f ) = 0. 24IV) Suppose S V is stable and f ∈ S V LV . One can repeat the argument in (III) with S replaced by S V and conclude ϕ V LV ( f ) = 0. Definition 8.2.
A subset D ⊂ E is called order bounded if there are a, b ∈ E for which a ≤ D ≤ b . Theorem 8.3.
Let f ∈ S LV or (assuming S V is stable) f ∈ S V LV . Then there exists apartition ( A n ) n ∈ N such that each set f ( A n ) is order bounded.Proof. There exists a partition ( A n ) n ∈ N such that for all n ∈ N there exist h n , g n ∈ S for which h n ≤ f A n ≤ g n . Choose a n , b n ∈ E for which a n ≤ h n ( x ) and g n ( x ) ≤ b n forall x ∈ X . Then a n ≤ f ( x ) ≤ b n for n ∈ N , x ∈ A n . Theorem 8.4.
Let f : X → E and f = 0 a.e.. Suppose there exists a partition ( A n ) n ∈ N such that for every n ∈ N the subset f ( A n ) of E is order bounded. Then f ∈ S LV andif S V is stable then also f ∈ Γ V L .Proof.
Choose a , a , . . . and b , b , . . . in E such that a n ≤ f ( x ) ≤ b n ( n ∈ N , x ∈ A n ) . (64)Let B = { x ∈ X : f ( x ) (cid:54) = 0 } . Then B ∈ A and µ ( B ) = 0. Hence g := (cid:80) n ∈ N a n A n ∩ B and h := (cid:80) n ∈ N b n A n ∩ B are elements of S L with ϕ ( g ) = 0 and ϕ L ( h ) = 0. As g ≤ f ≤ h ,we get f ∈ S LV and if S V is stable also f ∈ S V L .For a real valued function f : X → R with f ≥ (cid:82) f d µ = 0 we have f = 0a.e.. We will give an example of a f ∈ S + V with ϕ V ( f ) = 0 but which is nowhere zero(Example 8.8). On the positive side, in Theorem 8.7 we show that f = 0 a.e. if f ∈ S + LV and ϕ LV ( f ) = 0 provided that E satisfies a certain separability condition. Definition 8.5.
We call a subset D of E + \ { } pervasive in E if for all a ∈ E with a > d ∈ D such that 0 < d ≤ a . We say that E possesses a pervasivesubset if there exists a pervasive D ⊂ E + \ { } . Example 8.6.
The Riesz spaces R N , (cid:96) ∞ , c, c , (cid:96) and c possess countable pervasivesubsets. Indeed, in each of them the set { λe n : λ ∈ Q + , λ > , n ∈ N } is pervasive.If X is a completely regular topological space, then C ( X ) has a countable pervasive sub-set if and only if X has a countable base. (If D ⊂ E + \ { } is countable and pervasive,then U = { f − (0 , ∞ ) : f ∈ D } is a countable base; vise versa if U is a countable basethen with choosing an f U in C ( X ) + for each U ∈ U with f U = 0 on U c and f U ( x ) = 1for some x ∈ U , the set D = { εf U : ε ∈ Q , ε > , U ∈ U } is pervasive.) L ( λ ) and L ∞ ( λ ) do not possess countable pervasive subsets, considering the Lebesguemeasure space ( R , M , λ ). (Suppose one of them does. Then one can prove the existenceof non-negligible measurable sets A , A , · · · ∈ M such that every non-negligible measur-able set contains an A n , whereas λ ( A n ) < − n for all n ∈ N . Putting C = R \ (cid:83) n ∈ N A n we have a non-negligible measurable set that contains no A n : a contradiction.) Our use of the term is similar to the one of O. van Gaans and A. Kalauch do in [8, Definition 2.3]. heorem 8.7. Let E possess a countable pervasive subset D . Let f ∈ S LV . Let Λ , Υ ⊂ S L be countable sets such that Λ ≤ f ≤ Υ and sup ϕ L (Λ) = inf ϕ L (Υ) . Then foralmost all x ∈ X sup g ∈ Λ g ( x ) = f ( x ) = inf h ∈ Υ h ( x ) . (65) Consequently, if f ∈ S + LV and ϕ LV ( f ) = 0 , then f = 0 a.e.. (However, see Example8.8.)Proof. (I) First, as a special case (namely f = 0), let ( τ n ) n ∈ N be a sequence in S L with τ n ≥ n ∈ N and inf n ∈ N ϕ L ( τ n ) = 0. We prove that inf n ∈ N τ n ( x ) = 0 foralmost all x ∈ X , by proving that µ ( A ) = 0, where A is the complement of the set { x ∈ X : inf n ∈ N τ n ( x ) = 0 } . Indeed, for this A we have A = (cid:91) d ∈ D A d , with A d = (cid:92) n ∈ N { x ∈ X : d ≤ τ n ( x ) } . (66)Note that for all n ∈ N and d ∈ D the set { x ∈ X : d ≤ τ n ( x ) } is measurable.Furthermore, for all d ∈ D we have: dµ ( A d ) = ϕ ( d A d ) ≤ ϕ L ( τ n ) ( n ∈ N ) . (67)Hence µ ( A d ) = 0 for all d ∈ D and thus µ ( A ) = 0.(II) Suppose that Λ , Υ ⊂ Γ L are countable sets such that Λ ≤ f ≤ Υ, sup ϕ L (Λ) =inf ϕ L (Υ). Then inf ϕ L (Υ − Λ) = 0, so by (I) inf g ∈ Υ ,h ∈ Λ ( g ( x ) − h ( x )) = 0 for almost all x ∈ X . Example 8.8.
We give an example of a f ∈ S + V with ϕ V ( f ) = 0, where f (cid:54) = 0 every-where. Let ([0 , , M , λ ) be the Lebesgue measure space with underlying set [0 , E = (cid:96) ∞ ([0 , § f : R → E + be defined by f ( t ) = { t } for t ∈ [0 , f is not partially in S . We will show f ∈ S V . For n ∈ N make τ n ∈ S : τ n ( t ) = [ i − n , in ) if i ∈ { , . . . , n } , t ∈ [ i − n , in ) . (68)Then ϕ ( τ n ) = n [0 , and 0 ≤ f ≤ τ n for n ∈ N , so f ∈ S V and ϕ V ( f ) = 0. But f ( t ) (cid:54) = 0for all t . Theorem 8.9.
Let E be integrally closed and mediated. Let f : X → E and g : X → R .We write gf for the function x (cid:55)→ g ( x ) f ( x ) . Then (a) f ∈ S V and g is bounded and measurable = ⇒ gf ∈ S V . (b) f is partially in S V and g is measurable = ⇒ gf is partially in S V . (c) f ∈ S V and g ∈ L ( µ ) = ⇒ gf ∈ S LV . (d) f ∈ S V L and g is bounded and measurable = ⇒ gf ∈ S V L . (e) f ∈ S V L , f ( X ) is order bounded and g ∈ L ( µ ) = ⇒ gf ∈ S V L . roof. E is splitting (see 4.23(b)).(a) is a consequence of Theorem 3.21(a) (see also Remark 3.22).(b) Let ( A n ) n ∈ N be a partition such that f A n ∈ S V and g A n is bounded for all n ∈ N .By (a) every gf A n lies in S V . Then gf is partially in S V .(c) Assume f ≥ g ≥
0. Choose (see the proof of Theorem 7.1) a partition ( A n ) n ∈ N and numbers λ , λ , . . . in [0 , ∞ ) with τ := (cid:88) n ∈ N λ n A n ≥ g, (cid:88) n ∈ N λ n µ ( A n ) < ∞ . (69)Then τ s ∈ S L for all s ∈ S . Choose s ∈ S with s ≥ f . Then 0 ≤ gf ≤ τ s . FromTheorem 5.5(e) and (b) it follows that gf ∈ S LV .(d) Assume f ≥ ≤ g ≤ . Using (b), choose a partition ( A n ) n ∈ N with f A n ∈ S V and gf A n ∈ S V for all n ∈ N . Then0 ≤ ϕ V ( gf A n ) ≤ ϕ V ( f A n ) ( n ∈ N ) . (70)Since (cid:80) n ϕ V ( f A n ) exists and E is splitting, (cid:80) n ϕ V ( gf A n ) exists.(e) Assume f ≥ g ≥
0. Choose a ∈ E + with f ( x ) ≤ a for all x ∈ X . Choose apartition ( A n ) n ∈ N and λ , λ , · · · ∈ [0 , ∞ ) with gf A n ∈ S V ( n ∈ N ) , (71) g ≤ (cid:88) n ∈ N λ A n , (cid:88) n ∈ N λ n µ ( A n ) < ∞ (see the proof of Theorem 7.1) . (72)Then gf A n ≤ λ n a A n ( n ∈ N ) , (73) ϕ V ( λ n a A n ) = ϕ ( λ n a A n ) = λ n µ ( A n ) a ( n ∈ N ) , (74)so (cid:80) n ϕ V ( λ n a A n ) exists and so does (cid:80) n ϕ V ( gf A n ). In Lemma 8.11, Theorem 8.12 and Theorem 8.13 we investigate the relationbetween the extensions S LV generated by two different measures, namely µ and hµ fora measurable function h : X → [0 , ∞ ).Note that for such a function h and all s ∈ (1 , ∞ ) there exists a j : X → [0 , ∞ ) thatis partially in the space of simple functions X → [0 , ∞ ), i.e., j = (cid:80) n ∈ N α n A n for apartition ( A n ) n ∈ N and ( α n ) n ∈ N in [0 , ∞ ) (or in the language of 3.16 j is partially in [ A ])for which j ≤ h ≤ sj . In the following (8.11, 8.12 and 8.13) we will write I µ , S µ and ϕ µ instead of I , S and ϕ and, similarly for another measure ν on ( X, A ), we write I ν , S ν and ϕ ν according to 4.33 with ν instead of µ . Lemma 8.11.
Suppose E is splitting. Let h : X → [0 , ∞ ) be measurable, ν := hµ . Let s ∈ (1 , ∞ ) and let j : X → [0 , ∞ ) be partially in [ A ] and such that j ≤ h ≤ sj . Let f ∈ S ν + L . Then jf ∈ S µL and ϕ µL ( jf ) ≤ ϕ νL ( f ) ≤ sϕ µL ( jf ) . roof. Assume ( A n ) n ∈ N is a partition for j and a ϕ µ -partition for f (so ( A n ) n ∈ N is in I ν ∩ I µ , i.e., µ ( A n ) , ν ( A n ) < ∞ for all n ∈ N ). Choose ( α n ) n ∈ N in [0 , ∞ ) and ( b n ) n ∈ N in E + such that j = (cid:88) n ∈ N α n A n , f = (cid:88) n ∈ N b n A n . (75)Then jf = (cid:80) n ∈ N α n b n A n and thus is in S µL if (cid:80) n µ ( A n ) α n β n exists in E . For each n ∈ N ≤ µ ( A n ) α n = (cid:90) j A n d µ ≤ (cid:90) h A n d µ = ν ( A n ) , (76)whence 0 ≤ µ ( A n ) α n b n ≤ ν ( A n ) b n . Because f ∈ S ν + L , (cid:80) n ν ( A n ) b n exists in E . Since E is splitting also (cid:80) n µ ( A n ) α n b n exists in E , i.e., jf ∈ S µL .Furthermore, ϕ µL ( jf ) = (cid:80) n µ ( A n ) α n b n ≤ (cid:80) n ν ( A n ) b n = ϕ νL ( f ). On the other hand,we get µ ( A n ) α n = (cid:82) j A n d µ ≥ s (cid:82) h A n d µ = s ν ( A n ) for each n ∈ N : it follows that ϕ µL ( jf ) ≥ s ϕ νL ( f ). Theorem 8.12.
Let E be integrally closed and splitting. Let h : X → [0 , ∞ ) be mea-surable, ν := hµ . (a) f ∈ S νLV = ⇒ hf ∈ S µLV , ϕ µLV ( hf ) = ϕ νLV ( f ) , (b) f ∈ S νV L = ⇒ hf ∈ S µV L , ϕ µV L ( hf ) = ϕ νV L ( f ) .Proof. Since both S νLV and S νV L are directed, we assume f ≥ f ∈ S ν + LV . For n ∈ N let j n be partially in [ A ] and such that j n ≤ h ≤ (1 + n ) j n . Let Λ , Υ ⊂ S νL be countable sets with Λ ≤ f ≤ Υ be such that sup ϕ νL (Λ) = ϕ νLV ( f ) = inf ϕ νL (Υ). Then for all σ ∈ Λ (note that σ ∈ S ν + L − S ν + L ), τ ∈ Υ and n ∈ N we have j n σ ≤ hf ≤ (1 + n ) j n τ and by Lemma 8.11 j n σ and (1 + n ) j n τ arein S µL . Therefore we are done if both inf n ∈ N ,σ ∈ Λ ,τ ∈ Υ ϕ µL ((1 + n ) j n τ − j n σ ) = 0 and ϕ µL ( j n σ ) ≤ ϕ νLV ( f ) ≤ ϕ µL ((1 + n ) j n τ ) for all n ∈ N and all σ ∈ Λ , τ ∈ Υ. By Lemma8.11 applied repeatedly we have0 ≤ ϕ µL ((1 + n ) j n τ − j n σ ) = ϕ µL ( j n τ − j n σ ) + n ϕ µL ( j n τ ) ≤ ϕ νL ( τ − σ ) + n ϕ νL ( τ ) , (77)which has infimum 0 since E is integrally closed and inf τ ∈ Υ ,σ ∈ Λ ϕ νL ( τ − σ ) = 0. On theother hand, by Lemma 8.11, ϕ µ ( j n σ ) ≤ ϕ νL ( σ ) ≤ ϕ νLV ( f ) ≤ ϕ νL ( τ ) ≤ (1 + n ) ϕ µL ( j n τ ) ( n ∈ N , σ ∈ Λ , τ ∈ Υ) . (78)(b) Let f ∈ S ν + V L . Choose a partition ( A n ) n ∈ N with f A n ∈ S νV for n ∈ N . By (a), hf A n ∈ S µLV for n ∈ N ; by Lemma 5.7 hf A n is partially in S µV .Therefore we can choose a partition ( B n ) n ∈ N with f B n ∈ S νV , hf B n ∈ S µV ( n ∈ N ) . (79)28y (a), ϕ νV ( f B n ) = ϕ µV ( hf B n ) for all n ∈ N . But f ∈ S ν + V L , so ϕ νV L ( f ) = (cid:88) n ϕ νV ( f B n ) = (cid:88) n ϕ µV ( hf B n ) . (80)Then hf ∈ S µV L and ϕ µV L ( hf ) = ϕ νV L ( f ). Theorem 8.13.
Let E be integrally closed and splitting. Let h : X → [0 , ∞ ) be mea-surable, ν := hµ , A = { x ∈ X : h ( x ) > } . Let f : X → E be such that hf ∈ S µLV .Then f A ∈ S νLV .Proof. Define h ∗ : X → [0 , ∞ ) by h ∗ ( x ) = (cid:40) h ( x ) if x ∈ A, x / ∈ A. (81)Then h ∗ is measurable and hh ∗ = A and A = ν -a.e.. hf is in S µL and thus in S A µL , and since A µ = h ∗ ν , also hf ∈ S h ∗ νL . By Theorem8.12, applied to h ∗ , h ∗ ν, ν, hf instead of h, ν, µ, f , the function h ∗ hf is an element of S νLV . But h ∗ hf = A f .In Theorem 8.14 we show that extensions of simple functions with values in E com-posed with a σ -order continuous linear map E → F are extensions of simple functionswith values in F (where E and F are Riesz spaces). Theorem 8.14.
Let E and F be Riesz spaces. Let S E and ϕ E be as in 4.33, and let S F and ϕ F be defined analogously. Let L c ( E, F ) denote the set of σ -order continuouslinear functions E → F and E ∼ c = L c ( E, R ) (definition and notation as in Zaanen [16,Chapter 12, § f ∈ S ELV . Then α ◦ f ∈ S FLV for all α ∈ L c ( E, F ) and α (cid:0) ϕ ELV ( f ) (cid:1) = ϕ FLV ( α ◦ f ) . (82) In particular, α ◦ f is integrable for all α ∈ E ∼ c , and α ( ϕ ELV ( f )) = (cid:82) α ◦ f d µ .Proof. Suppose α ∈ L c ( E, F ) + . Let τ ∈ S E + L . Suppose τ = (cid:80) n ∈ N a n A n for somepartition ( A n ) n ∈ N and a sequence ( a n ) n ∈ N in E + . Then α ( ϕ EL ( τ )) = α ( (cid:80) n µ ( A n ) a n ) = (cid:80) n µ ( A n ) α ( a n ). Thus α ◦ τ is in S FLV with α ( ϕ EL ( τ )) = ϕ FL ( α ◦ τ ). Let ( σ n ) n ∈ N , ( τ n ) n ∈ N be sequences in S EL with σ n ≤ f ≤ τ n , σ n ↑ , τ n ↓ and ϕ ELV ( f ) = sup n ∈ N ϕ EL ( σ n ) =inf n ∈ N ϕ EL ( τ n ). Then we have α ( ϕ ELV ( f )) = sup n ∈ N α ( ϕ EL ( σ n )) = sup n ∈ N ϕ FL ( α ◦ σ n ) and α ( ϕ ELV ( f )) = inf n ∈ N α ( ϕ EL ( τ n )) = inf n ∈ N ϕ FL ( α ◦ τ n ). Since α ◦ σ n ≤ α ◦ f ≤ α ◦ τ n for all n ∈ N , we conclude that α ◦ f ∈ ( S F ) LV (see Theorem 7.1) with α ( ϕ ELV ( f )) = ϕ FLV ( f ).Theorem 8.14 will be used in § ϕ LV and ϕ V L with thePettis integral.Before proving Theorem 8.16 we state (in Theorem 8.15) that there is an equivalentformulation for a function F to be in C ( X × T ) whenever X, T are topological spacesand X is compact. 29 heorem 8.15. [15, Theorem 7.7.5] Let X be a compact and let T be a topological space.Let F : X × T → R be such that F ( · , t ) ∈ C ( X ) for all t ∈ T . Then F ∈ C ( X × T ) ifand only if t (cid:55)→ F ( · , t ) is continuous, where C ( X ) is equipped with the supremum norm.Consequently, if A ⊂ X is a compact set, then t (cid:55)→ sup F ( A, t ) and t (cid:55)→ inf F ( A, t ) arecontinuous. Theorem 8.16.
Let ( X, d, µ ) be a compact metric probability space. Let T be a topo-logical space and F ∈ C ( X × T ) . The function H : X → C ( T ) given by H ( x ) = F ( x, · ) is an element of S V . Furthermore, for t ∈ T , x (cid:55)→ F ( x, t ) is integrable and [ ϕ V ( H )] ( t ) = (cid:90) F ( x, t ) d µ ( x ) ( t ∈ T ) . (83) Proof.
For k ∈ N let A k , . . . , A kn k be a partition of X with diam A ki ≤ k − . Define∆ k ( t ) = sup x,y ∈ X,d ( x,y ) Consider a metrisable locally compact group G . Let X ⊂ G be acompact set and µ be a finite (positive) measure on B ( X ), the Borel- σ -algebra of X .Let g ∈ C ( G ). Define the convolution of g and µ to be the function g ∗ µ : G → R givenby g ∗ µ ( t ) = (cid:82) g ( tx − ) d µ ( x ) for t ∈ G . For x ∈ X , let L x g ∈ C ( G ) be the function t (cid:55)→ g ( tx − ). Then by Theorem 8.16, the function f : X → C ( G ) given by f ( x ) = L x g is in S V and g ∗ µ = ϕ V ( f ) ∈ C ( G ). 30 Comparison with Bochner- and Pettis integral We consider the situation of § 8, with an E that has the structure of a Banachlattice. We write (cid:107) · (cid:107) for the norm on E and E (cid:48) for the dual of E . Then, next to our ϕ LV (and other extensions) there are the Bochner and the Pettis integrals. (We referthe reader to Hille and Phillips [10, Section 3.7] for background on both integrals.) Wedenote the set of Bochner (Pettis) integrable functions from the measure space ( X, A , µ )into the Banach lattice E by B ( P ) and the Bochner (Pettis) integral of an integrablefunction f by b ( f ) ( p ( f )). By definition of the Bochner integral, where one also starts with defining theintegral on simple functions: S ⊂ B and ϕ = b on S . Since B ⊂ P and b = p on B wealso have S ⊂ P with ϕ = p on S . The following is used in this section. The Banach dual of E is equal to the orderdual, i.e., E (cid:48) = E ∼ . Moreover, for x, y ∈ E (see de Jonge and van Rooij [12, Theorem10.2]) x ≤ y ⇐⇒ α ( x ) ≤ α ( y ) for all α ∈ E ∼ + . (87)This implies that for a sequence ( y n ) n ∈ N and x, y in E :inf n ∈ N α ( y n ) = 0 for all α ∈ E ∼ + = ⇒ inf n ∈ N y n = 0 . (88) Theorem 9.3. Let f ∈ P + and f be partially in S . Then f ∈ S + L and p ( f ) = ϕ L ( f ) .Proof. Let ( A n ) n ∈ N be a partition for which f n := f A n ∈ S . Then for every α ∈ E ∼ + α ( p ( f )) = (cid:90) α ◦ f d µ = (cid:88) n ∈ N (cid:90) α ◦ f n d µ = (cid:88) n ∈ N α ( ϕ ( f n )) . (89)Hence inf N ∈ N α ( p ( f ) − (cid:80) Nn =1 ϕ ( f n )) = 0 and thus p ( f ) = (cid:80) n ϕ ( f n ) (see (88)). Theorem 9.4. Let f ∈ P . Then the following holds. (a) If g ∈ S LV and f ≤ g , then p ( f ) ≤ ϕ LV ( g ) . (b) If S V is stable, g ∈ S V LV and f ≤ g , then p ( f ) ≤ ϕ V LV ( g ) .Consequently, p = ϕ LV on P ∩ S LV , and p = ϕ V LV on P ∩ S V LV if S V is stable.The statements in (a) and (b) remain valid by replacing all “ ≤ ” by “ ≥ ”.Proof. It will be clear that if g ∈ S and f ≤ g , then g ∈ P and hence p ( f ) ≤ p ( g ) = ϕ ( g ).If g ∈ S V and f ≤ g , then there exists an Υ ⊂ S with g ≤ Υ and ϕ V ( g ) = inf ϕ (Υ) =inf p (Υ) ≥ p ( f ).Let g ∈ S L and assume f ≤ g . Let g , g ∈ S + L be such that g = g − g . Let ( B i ) i ∈ N be a ϕ -partition for both g and g . Write A n = (cid:83) ni =1 B i for n ∈ N . Let α ∈ E ∼ + .31 ◦ ( f A ) = ( α ◦ f ) A for every A ∈ A , so that α ◦ ( f A ) is integrable. Thus, for n ∈ N we have (cid:90) ( α ◦ f ) A n d µ = (cid:90) α ◦ ( f A n ) d µ ≤ (cid:90) α ◦ ( g A n ) d µ = (cid:90) α ◦ g A n d µ − (cid:90) α ◦ g A n d µ = α ( ϕ ( g A n )) − α ( ϕ ( g A n )) ≤ α ( ϕ ( g A m )) − α ( ϕ ( g A k )) ( k, m ∈ N , k < n < m ) . (90)Which implies that (cid:82) ( α ◦ f ) A n d µ + α ( ϕ ( g A k )) ≤ α ( ϕ L ( g )) as soon as k < n . Byletting n tend to ∞ (as (cid:82) ( α ◦ f ) A n d µ → (cid:82) α ◦ f d µ = α ( p ( f ))), for each k ∈ N weobtain α ( p ( f )) ≤ α ( ϕ L ( g ) − ϕ ( g A k )) . (91)This holds for all α ∈ E ∼ + , so p ( f ) ≤ ϕ L ( g ) − ϕ ( g A k ) . (92)This, in tern is true for every k , so p ( f )) ≤ ϕ L ( g ).We leave it to check that the preceding lines can be repeated with S V , S L or S V L insteadof S . Theorem 9.5. Suppose (cid:107) · (cid:107) is σ -order continuous. Write S = S LV = S V L and ϕ = ϕ LV = ϕ V L (see Theorem 5.12). (a) Then S ⊂ P . Consequently, if f is essentially separably valued and in S , then f ∈ B . In particular, S L ⊂ B . (b) Suppose there exists an α ∈ E ∼ + c with the property that if b ∈ E and b > , then α ( b ) > . Then B V ⊂ B . Consequently, S ⊂ B .Proof. (a) Because (cid:107)·(cid:107) is σ -order continuous, E (cid:48) = E ∼ c . Therefore Theorem 8.14 impliesthat S ⊂ P .Note that S L ⊂ B . Since B is a Riesz ideal in the space of strongly measurablefunctions X → E , an f ∈ S is an element of B if it is essentially separably valued, sincethere are elements σ, τ ∈ S L with σ ≤ f ≤ τ and f is weakly measurable since f ∈ P .(b) Suppose f ∈ B V and σ n , τ n ∈ B are such that σ n ≤ f ≤ τ n for n ∈ N , σ n ↑ , τ n ↓ and sup n ∈ N b ( σ n ) = b V ( f ) = inf n ∈ N b ( τ n ). Then inf n ∈ N (cid:82) α ◦ ( τ n − σ n ) d µ = α (inf n ∈ N b ( τ n − σ n )) = 0 and therefore α (inf n ∈ N ( τ n − σ n )) = inf n ∈ N α ◦ ( τ n − σ n ) isintegrable with integral equal to zero. Therefore inf n ∈ N ( τ n − σ n ) = 0 a.e., hence τ n → f a.e.. Therefore f is strongly measurable and thus f ∈ B by (a). By (a) S L ⊂ B , hence S = S LV ⊂ B . Lemma 9.6. Let E be a Banach lattice with an abstract L-norm (i.e., (cid:107) a + b (cid:107) = (cid:107) a (cid:107) + (cid:107) b (cid:107) for a, b ∈ E + ). Then (cid:107) b ( f ) (cid:107) = (cid:90) (cid:107) f (cid:107) d µ ( f ∈ B + ) . (93)(b) B L = B . (c) There exist an α ∈ E ∼ + c as in Theorem 9.5(b). Consequently B V = B .Proof. (a) It is clear that (cid:107) b ( f ) (cid:107) = (cid:82) (cid:107) f (cid:107) d µ for f ∈ S + , hence by limits for all f ∈ B + .(b) Suppose f ∈ B + L . Let ( A n ) n ∈ N be a b -partition for f , write f n = f A n . Then (cid:107) (cid:80) Nn =1 f n − f (cid:107) → 0, hence f is strongly measurable. Moreover, since (cid:107) · (cid:107) is σ -ordercontinuous (cid:107) (cid:80) Nn =1 b ( f n ) − b L ( f ) (cid:107) → 0, hence (cid:80) Nn =1 (cid:107) b ( f n ) (cid:107) → b L ( f ). Using (a) weobtain (cid:82) (cid:107) f (cid:107) d µ = (cid:80) n ∈ N (cid:82) (cid:107) f n (cid:107) d µ = (cid:80) n ∈ N (cid:107) b ( f n ) (cid:107) < ∞ , i.e., f ∈ B .(c) Extend α : E + → R given by α ( b ) = (cid:107) b (cid:107) to a linear map on E . Examples 9.7. (I) Take X = N , A = P ( N ), and let µ be the counting measure. Wehave S = c [ E ]; S V = c [ E ]; all functions N → E are partially in S ; S := S LV = S V L = S L (see Theorem 5.5(c)) and S + consists precisely of the functions f : N → E + for which (cid:80) n f ( n ) exists in the sense of the ordering. On the other hand, f : N → E isBochner integrable if and only if (cid:80) ∞ n =1 (cid:107) f ( n ) (cid:107) < ∞ . • If (cid:107) · (cid:107) is a σ -order continuous norm, then B ⊂ S . • Moreover (cid:107) · (cid:107) is equivalent to an abstract L-norm if and only if B = S (since, if B = S , the following holds: if x , x , · · · ∈ E + and (cid:80) n x n exists, then (cid:80) n ∈ N (cid:107) x n (cid:107) < ∞ ,see Theorem A.1). • For E = c there exists an f ∈ P that is not in S . For example f : N → c givenby f = ( e , − e , e , − e , e , − e , . . . ) (94)is Pettis integrable since c (cid:48) ∼ = (cid:96) has basis { δ n : n ∈ N } where δ n ( x ) = x ( n ) and (cid:80) m ∈ N δ n ( f ( m )) = 0 for all m ∈ N . c is σ -Dedekind complete and thus by Theorem4.32 the set S is a Riesz space. However, | f | is not in S and therefore neither f is. • For E = c there exists an f ∈ S that is not in B and not in P : Considerfor example f : n (cid:55)→ e n . It is an element of S but not of B . It is not even Pettisintegrable. (Suppose it is, and its integral is a . Then for all u ∈ c (cid:48) we have u ( a ) = (cid:82) u ◦ f d µ = (cid:80) ∞ n =1 u ( f ( n )) = (cid:80) ∞ n =1 u ( e n ). Letting u be the coordinate functions, wesee that a ( n ) = 1 for all n ∈ N ; letting u be x (cid:55)→ lim n →∞ x ( n ) we have a contradiction.)(II) B (cid:54)⊂ S V LV . Let ( R , M , λ ) be the Lebesgue measure space. Let E be the σ -Dedekind complete Riesz space L ( λ ). Let g ∈ L ( λ ) be the equivalence class of thefunction that equals t − for 0 < t ≤ t . Let L x g ( t ) = g ( t − x )for x ∈ R . Then the function f : R → L ( λ ) for which f ( x ) = [0 , ( x ) L x g is Bochnerintegrable ( f is continuous in the (cid:107) · (cid:107) norm (because (cid:107) L ε g − g (cid:107) = 2 √ ε for ε > 0) and (cid:82) (cid:107) f ( x ) (cid:107) d λ ( x ) = (cid:82) (cid:82) | g ( t − x ) | d λ ( t ) d λ ( x ) = (cid:107) g (cid:107) < ∞ ) but no element of S V LV (byTheorem 8.3). 33 Consider the situation of § As we have seen in Examples 9.7, e.g., (94), the set of Pettis integrable functionsneed not be stable. We show that B is stable and b is laterally extendable. Furthermorewe give an example of an f ∈ B LV that is neither in S V LV , nor in B L or B V . Theorem 10.1. B is stable and b is laterally extendable.Proof. Note that f B ∈ B for all f ∈ B and B ∈ A (since f B is strongly measurableand (cid:107) f B (cid:107) is integrable), i.e., B is stable. Let ( A n ) n ∈ N be a partition in A of X .Let f : X → E + be a Bochner integrable function. Then (cid:82) (cid:107) f (cid:107) d µ < ∞ and with B n = A ∪ · · · ∪ A n and Lebesgue’s Dominated Convergence Theorem we obtain (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) b ( f − N (cid:88) n =1 f A n ) (cid:13)(cid:13)(cid:13)(cid:13)(cid:13) ≤ (cid:90) (cid:107) f ( x ) − B N ( x ) f ( x ) (cid:107) d µ ( x ) → . (95)Thus b ( f ) = lim N →∞ N (cid:88) n =1 b ( f A n ) = (cid:88) n b ( f A n ) . (96)We conclude that b is laterally extendable. Consider the situation of Example 8.8. Since S ⊂ B and ϕ ( h ) = b ( h ) for h ∈ S : f ∈ B V . The function f is not essentially separably-valued (i.e., f ( X \ A ) is notseparable for all null sets A ∈ A ), hence f (and thus g ) is not strongly measurable (see[10, Theorem 3.5.2]). Hence f is not Bochner integrable, i.e., f ∈ B V but f / ∈ B .In a similar way as has been shown in Example 8.8, one can show that g : R → E + defined by g ( t ) = { t } for t ∈ R is in S LV . Then g ∈ B LV but g / ∈ B V . All f ∈ B L are strongly measurable. Therefore for f ∈ B L we have f / ∈ B if andonly if (cid:82) (cid:107) f (cid:107) d µ = ∞ .The following example illustrates that by extending the Bochner integrable functionsone can obtain more than by extending the simple functions. Example 10.4. [ ψ ∈ B V , ψ / ∈ B ] Let X = [2 , A be the set of Lebesgue measurable subsets of X and µ be theLebesgue measure on X . Let M denote the set of equivalence classes of measurablefunctions R → R . Let E = (cid:110) f ∈ M : sup x ∈ R (cid:90) x +1 x | f | < ∞ (cid:111) , (cid:107) · (cid:107) : E → [0 , ∞ ) , (cid:107) f (cid:107) = sup x ∈ R (cid:90) x +1 x | f | . (97)Then E equipped with the norm (cid:107) · (cid:107) is a Banach lattice. E is an ideal in M andtherefore σ -Dedekind complete (hence S V is stable; 4.25). The norm (cid:107) · (cid:107) is not σ -ordercontinuous. 34or a ∈ R , c > S a,c : X → E + by S a,c ( x ) = ( a + cx, ∞ ) . If x, y ∈ X with y > x then (cid:107) S a,c ( x ) − S a,c ( y ) (cid:107) ≤ (cid:107) ( a + cx,a + cy ] (cid:107) ≤ c | x − y | , so S a,c is continuous and thereforestrongly measurable. Furthermore (cid:107) S a,c ( x ) (cid:107) = 1 for all x ∈ X , i.e., x (cid:55)→ (cid:107) S a,c ( x ) (cid:107) isintegrable. Thus S a,c is Bochner integrable. For d, e ∈ R with e > d the map E → R , f (cid:55)→ (cid:82) ed f is a continuous linear functional. Therefore (cid:90) ed b ( S a,c ) = (cid:90) X (cid:90) ed ( S a,c ( x ))( t ) d t d x = (cid:90) ed (cid:90) X ( S a,c ( x ))( t ) d x d t. (98)Since this holds for all d, e ∈ R with e > d , for t ∈ R we have( b ( S a,c )) ( t ) = (cid:90) X ( S a,c ( x ))( t ) d x = (cid:90) ( a + cx, ∞ ) ( t ) d x = (cid:0) t − ac ∧ − (cid:1) ∨ . (99)For k ∈ N define r k , R k : X → E by R k := S ,k , r k := S ,k − S ,k . (100)For x ∈ X and k ∈ N , r k ( x ) = ( kx,kx +1] and kx + 1 < ( k + 1) x . Define ψ ( x ) := (cid:83) k ∈ N ( kx,kx +1] = (cid:88) k ∈ N r k ( x ) , σ n := n (cid:88) k =1 r k , τ n := n (cid:88) k =1 r k + R n +1 . (101)Note that σ n ≤ ψ ≤ τ n and σ n , τ n ∈ B all for n ∈ N . Since E is σ -Dedekind completeand therefore mediated, from the fact thatinf n ∈ N b ( τ n − σ n ) = inf n ∈ N b ( R n +1 ) = 0 , (102)it follows that ψ ∈ B V . However, ψ / ∈ B since ψ is not essentially separably valued:Let x, y ∈ X , x < y . We prove (cid:107) ψ ( x ) − ψ ( y ) (cid:107) ≥ 1. For k ∈ N : k − ≤ y − x < k = ⇒ (cid:40) k − y ≤ kx, kx < ky, = ⇒ ( kx, kx + 1] ∩ (cid:91) i ∈ N ( iy, iy + 1] = ∅ . (103)Hence (cid:107) ψ ( x ) − ψ ( y ) (cid:107) ≥ x, y ∈ X with x (cid:54) = y .So ψ is an element of B V but not of B (and neither of B L ). Example 10.5. [ f ∈ B LV , f / ∈ B L , f / ∈ B V , f / ∈ S V LV ] Let ( X, A , µ ) be the Lebesgue measure space ( R , M , λ ). Let E and ψ be as in Example10.4. Define u : R → E by u ( x ) = (cid:40) ψ ( x ) x ∈ [2 , , . (104)35hen u is an element of B V and not of B L . As we have seen in Examples 9.7(II) thereexists a g in L ( λ ) and thus in E such that v : x (cid:55)→ [0 , ( x ) L x g is an element of B that is not an element of S V LV . Furthermore w : R → E given by w ( x ) = ( n,n +1] for x ∈ ( n, n + 1] is an element of B L and not of B V . Therefore f = u + v + w is an elementof B LV (and thus of B V L ; see Theorem 5.8) but is neither an element of S V LV nor of B V or B L . 11 Discussion Of course, to some extent our approach is arbitrary. We mention some alternatives,with comments. The reader may have wondered why in our definition of the lateral extension thesets A n are required not only to be disjoint but also to cover X (i.e., to form a partition).Without the covering of X the definition remains perfectly meaningful, but the sum oftwo positive laterally integrable functions need not be laterally integrable, even in quitenatural situations. (E.g., take E = F = R and X = [0 , I be the ring generatedby the open intervals, Γ the space of all Riemann integrable functions on [0 , ϕ the Riemann integral. If f is the indicator of the Cantor set, then − f is laterallyintegrable but 2 − f is not.) For the vertical extension we have, somewhat artificially, introduced a countabilityrestriction leading us from ϕ v to ϕ V ; see Definition 3.3. In some sense, ϕ v would haveserved as well as ϕ V . In order to get a non-void theory, however, we would need amuch stronger (but analogous) condition than “mediatedness”, restricting our worlddrastically. A different approach to both the vertical and the lateral extension, closer toDaniell and Bourbaki, could run as follows. Starting from the situation of 3.14, call afunction X → F + “integrable” if there exist f , f , · · · ∈ Γ + such that (cid:40) f n ↑ f in F X , sup n ∈ N ϕ ( f n ) exists in E, (105)then define the “integral” ϕ ( f ) of f by ϕ ( f ) := sup n ∈ N ϕ ( f n ) . (106)This definition is meaningful only if, in the above situation g ∈ Γ + , g ≤ f, = ⇒ ϕ ( g ) ≤ sup n ∈ N ϕ ( f n ) (107)which in a natural way leads to the requirement that Γ be a lattice and that ϕ becontinuous in the following sense: h , h , · · · ∈ Γ + , h n ↓ ⇒ ϕ ( h n ) ↓ . (108)36hese conditions lead to a sensible theory, but again we consider them as too restrictive.(See Example II.2.4 in the thesis of G. Jeurnink [11] for an example of a Γ that consistsof simple functions on a measure space with values in a C ( X ) for which (108) does nothold for the standard integral on simple functions (see 4.33).) Acknowledgements The authors are grateful to O. van Gaans for valuable discussions. W.B. van Zuijlen issupported by ERC Advanced Grant VARIS-267356. References [1] G. Birkhoff. Lattice Theory , volume XXV. American Mathematical Society, thirdedition, 1967.[2] S. Bochner. Integration von Funktionen, deren Werte die Elemente eines Vektor-raumes sind. Fund. Math. , 20:262–276, 1933.[3] F.F. Bonsall. Sublinear functionals and ideals in partially ordered vector spaces. Proc. Lond. Math. Soc. , 4(3):402–418, 1954.[4] N. Bourbaki. Livre VI Int´egration . Hermann, 1952.[5] P. J. Daniell. A general form of integral. Annals of Mathematics, Second series ,19:279–294, 1918.[6] B. de Pagter. On z-ideals and d-ideals in Riesz spaces III. Indag. Math. , 43:409–422,1981.[7] L. Fuchs. Riesz Vector Spaces and Riesz Algebras. S´eminaire Dubreil. Alg`ebre etth´eorie des nombres , 19(23-24):1–9, 1965-1966.[8] O. van Gaans and A. Kalauch. Bands in Pervasive Pre-Riesz Spaces. Operatorsand Matrices , 2(2):177–191, 2008.[9] O. van Gaans and A. Kalauch. Ideals and bands in pre-Riesz spaces. Positivity ,12:591–611, 2008.[10] E. Hille and R.S. Phillips. Functional Analysis and Semi-Groups , volume 31. Amer-ican Mathematical Society, 1957.[11] G.A. Jeurnink. Integration of functions with values in a Banach lattice . PhD thesis,University of Nijmegen, 1982.[12] E. de Jonge and A.C.M. van Rooij. Introduction to Riesz spaces . MathematischCentrum, 1977. 3713] Anthony L. Peressini. Ordered topological vector spaces . Harper & Row, Publishers,1967.[14] B.J. Pettis. On integration in vector spaces. Trans. Amer. Math. Soc. , 44:277–304,1938.[15] Zbigniew Semadeni. Banach Spaces of Continuous Functions: Vol. I . PWN PolishScientific Publishers, Warszawa, 1971.[16] A.C. Zaanen. Riesz Spaces II . North-Holland Publishing Company, 1983. A Appendix Theorem A.1. Let E be a Banach lattice with the propertyIf x , x , · · · ∈ E + and (cid:88) n x n exists, then (cid:88) n ∈ N (cid:107) x n (cid:107) < ∞ . (109) Then the norm (cid:107) · (cid:107) is equivalent to an L-norm. The proof uses the following lemma. Lemma A.2. Let E be a Banach lattice that satisfies (109) . Then there exists a C > such that x , x , · · · ∈ E + , (cid:88) n x n exists = ⇒ (cid:88) n ∈ N (cid:107) x n (cid:107) ≤ C (cid:13)(cid:13)(cid:13) (cid:88) n x n (cid:13)(cid:13)(cid:13) . (110) Proof. Suppose not. For i ∈ N let x i , x i , · · · ∈ E + , (cid:80) n x in = b i and (cid:80) n ∈ N (cid:107) x in (cid:107) > i (cid:107) b i (cid:107) and (cid:107) b i (cid:107) = 2 − i . Then (cid:80) i ∈ N (cid:107) b i (cid:107) < ∞ , so (cid:80) i b i exists. As (cid:80) i b i = (cid:80) i (cid:80) n x in ,by (109) we get ∞ > (cid:80) i ∈ N (cid:80) n ∈ N (cid:107) x in (cid:107) > (cid:80) i ∈ N i (cid:107) b i (cid:107) = ∞ . Proof of Theorem A.1. By Lemma A.2 we can define p : E → [0 , ∞ ), p ( x ) = sup (cid:40)(cid:88) n ∈ N (cid:107) x n (cid:107) : x , x , · · · ∈ E + , (cid:88) n x n ≤ | x | (cid:41) , (111)obtaining p ( x ) = p ( | x | ) , p ( tx ) = | t | p ( x ) , (cid:107) x (cid:107) ≤ p ( x ) ≤ C (cid:107) x (cid:107) for all x ∈ E , t ∈ R (with C as in Lemma A.2) and p ( x ) ≤ p ( y ) for x, y ∈ E + with x ≤ y .Let x, y ∈ E + ; we prove p ( x + y ) = p ( x ) + p ( y ). • For ε > x , x , . . . , y , y , · · · ∈ E + , (cid:80) n x n ≤ x, (cid:80) n y n ≤ y , (cid:80) n ∈ N (cid:107) x n (cid:107) ≥ p ( x ) − ε , (cid:80) n ∈ N (cid:107) y n (cid:107) ≥ p ( y ) − ε . Considering the sequence x , y , x , y , . . . we find (cid:80) n ∈ N ( (cid:107) x n (cid:107) + (cid:107) y n (cid:107) ) ≤ p ( x + y ). Hence p ( x + y ) ≥ p ( x ) + p ( y ). • On the other hand: Let z , z , · · · ∈ E + , (cid:80) n z n ≤ x + y ; we prove (cid:80) n ∈ N (cid:107) z n (cid:107) ≤ p ( x ) + p ( y ). Define u n , v n by u + · · · + u n = ( z + · · · + z n ) ∧ x, v n = z n − u n ( n ∈ N ) . (112)38hen ( z + · · · + z n ) ∧ x − z n = ( z + · · · + z n − z n ) ∧ ( x − z n ) ≤ ( z + · · · + z n − ) ∧ x ,implying u n − z n ≤ 0; and ( z + · · · + z n ) ∧ x ≥ ( z + · · · + z n − ) ∧ x , implying u n ≥ u n ≥ , v n ≥ n ∈ N ) , (113) (cid:80) n ∈ N (cid:107) u n (cid:107) ≤ (cid:80) n ∈ N (cid:107) z n (cid:107) < ∞ , so (cid:80) n u n exists; (cid:80) n u n ≤ x , and (cid:80) n ∈ N (cid:107) u n (cid:107) ≤ p ( x ). (cid:80) n ∈ N (cid:107) v n (cid:107) ≤ (cid:80) n ∈ N (cid:107) z n (cid:107) < ∞ , so (cid:80) n v n exists. For every n ∈ N , z + · · · + z n ≤ ( z + · · · + z n + y ) ∧ ( x + y ) = ( z + · · · + z n ) ∧ x + y = u + · · · + u n + y , so v + · · · + v n ≤ y ;then (cid:80) n v n ≤ y and (cid:80) n ∈ N (cid:107) v n (cid:107) ≤ p ( y ).Thus (cid:80) n ∈ N (cid:107) z n (cid:107) ≤ (cid:80) n ∈ N (cid:107) u n (cid:107) + (cid:80) n ∈ N (cid:107) v n (cid:107) ≤ p ( x ) + p ( yy