# Topological Properties of the Complex Vector Lattice of Bounded Measurable Functions

Authors: Marian Nowak
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

Let Σ be aσ-algebra of subsets of a nonempty set Ω. Let BΣ be the complex vector lattice of bounded Σ-measurable complex-valued functions on Ω and let caΣ be the Banach space of all bounded countably additive complex-valued measures on Ω. We study locally solid topologies on BΣ. In particular, it is shown that the Mackey topology τBΣ,caΣ is the finest locally convex-solid σ-Lebesgue topology on BΣ.


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## Body

## 1. Introduction and Terminology

For terminology concerning complex vector lattices we refer the reader to [1, Chapter 12, Section  91], [2, Chapter 16], [3]. We denote by σ(L,K) and τ(L,K) the weak topology and the Mackey topology with respect to a dual pair 〈L,K〉. For a topological vector space (L,η) by (L,η)′ or Lη′, we denote its topological dual.We assume thatΣ is a σ-algebra of subsets of a non-empty set Ω. Let B(Σ) denote the Banach space of all bounded Σ-measurable functions f:Ω→ℂ, provided with the uniform norm ∥·∥. Then, B(Σ) with the positive cone B(Σ)+={f∈B(Σ):f(ω)≥0 for all ω∈Ω} and a valuation B(Σ)∋f↦|f|∈B(Σ)+ is a complex vector lattice (in the sense of [2, Section  16-2.2]). By Bℝ(Σ) we denote the Dedekind σ-complete Banach lattice of all bounded Σ-measurable functions u:Ω→ℝ, with the natural ordering u≤v if u(ω)≤v(ω) for all ω∈Ω (see [4, Chapter 13.1]). Then, B(Σ)=Bℝ(Σ)+iBℝ(Σ), that is, B(Σ) is a complexification of Bℝ(Σ) and B(Σ) has the dominated decomposition property, that is, if |f|≤u1+u2, where f∈B(Σ), u1,u2∈B(Σ)+, then there exist f1,f2∈B(Σ) such that f=f1+f2 and |f1|≤u1, |f2|≤u2 (see [1, Theorem 91.3], [3]). This means that B(Σ) is a breakable complex vector lattice (see [2, Section 16-3]).Denote byba(Σ) the Banach space of all bounded finitely additive measures μ:Σ→ℂ with the norm ∥μ∥=|μ|(Ω), where |μ|(A) denotes the variation of μ on A∈Σ. It is well known that the Banach dual B(Σ)* of B(Σ) can be identified with ba(Σ) through the integration mapping ba(Σ)∋μ↦Φμ∈B(Σ)*, where Φμ(f)=∫Ω‍fdμ for all f∈B(Σ) and ∥Φμ∥=|μ|(Ω) (see [5, Chapter 1, Theorem 13, p.6]. Let ca(Σ) stand for the Banach space of all countably additive measures μ:Σ→ℂ.It turns out that the basic notions and results from the theory of vector lattices (in particular, locally solid topologies) can be carried over to the complex vector latticesB(Σ) and ba(Σ).A subsetH of B(Σ) is said to be solid if it follows from |f|≤|g| and f∈B(Σ), g∈H that f∈H. Every subset H of B(Σ) is included in the smallest (with respect to the inclusion) solid set, called the solid hull of H and denoted by sol(H). Clearly,
(1)sol(H)={g∈B(Σ):|g|≤|f|forsomef∈H}.A linear Hausdorff topologyτ on B(Σ) is said to be locally solid if it has a local base at 0 consisting of solid sets in B(Σ).In Section2, we study locally solid topologies on B(Σ). In Section 3, we develop the duality theory of locally convex-solid topologies on B(Σ). In Section 4, it is shown that the natural Mackey topology τ(B(Σ),ca(Σ)) is the finest locally convex-solid σ-Lebesgue topology on B(Σ) (see Theorem 16 below).


## 2. Locally Solid Topologies onB(Σ)

Using the dominated decomposition property forB(Σ) we can easily obtain the following result (see [6, Theorem 1.11]).Proposition 1.
The convex hullconvH of a solid subset H of B(Σ) is solid.Proposition 2.
Letτ be a locally solid topology on B(Σ). Then, the τ-closure clτH of a solid subset Hof B(Σ) is solid.Proof.
Letℬτ be a local base at 0 for τ consisting of solid sets. Then clτH=⋂{H+V:V∈ℬτ}. Assume that |f|≤|g|, where f∈B(Σ), g∈clτH, and let V0∈ℬτ. Then, g=g1+g2, where g1∈H and g2∈V0. Since |f|≤|g|≤|g1|+|g2| (by the dominated decomposition property for B(Σ)), there exist f1,f2∈B(Σ) such that f=f1+f2 and |fi|≤|gi| for i=1,2. Hence, f1∈H and f2∈V0 because both sets H and V are solid. Thus, f∈H+V0 for every V0∈ℬτ, so f∈clτH. This means that clτH is solid, as desired.Following the notion of a modulus of an operator between vector lattices (see [7, Definition 1.8 and Theorem 1.10]), for a linear functional Φ:B(Σ)→ℂ let us put
(2)|Φ|(u):=sup{|Φ(h)|:h∈B(Σ),|h|≤u}
for all u∈B(Σ)+.Proposition 3.
For a linear functionalΦ on B(Σ) the following statements are equivalent:(i)
Φ∈B(Σ)*.(ii)
|Φ|(u)<∞for  allu∈B(Σ)+.Proof.
(i)⇒(ii) By way of contradiction, assume that there exists Φ0∈B(Σ)* such that |Φ0|(u0)=∞ for some u0∈B(Σ)+. Hence, there exists a sequence (hn) in B(Σ) such that |hn|≤u0 and |Φ0(hn)|≥n for n∈ℕ. Since ∥(1/n)hn∥→0, we get (1/n)|Φ0(hn)|→0, which is in contradiction with |Φ0(hn)|≥n.
(ii)⇒(i) Assume that there exists a linear functional Φ0 on B(Σ) such that |Φ0|(u)<∞ for all u∈B(Σ)+ and Φ0∉B(Σ)*. Then, there exists a sequence (fn) in B(Σ) such that ∥fn∥=1 and |Φ0(fn)|>n3 for all n∈ℕ. Since ∑n=1∞‍(1/n2)∥fn∥<∞ and the space (B(Σ),∥·∥) is complete, there exists u0∈B(Σ)+ such that ∑n=1∞‍(1/n2)|fn|=u0. Then, (1/n2)|fn|≤u0, and; hence, n<|Φ0((1/n2)fn)|<∞ for all n∈ℕ, which is impossible.
Thus, the proof is complete.ForΦ1,Φ2∈B(Σ)* we will write |Φ1|≤|Φ2| whenever |Φ1|(u)≤|Φ2|(u) for all u∈B(Σ)+. A subset 𝒦 of B(Σ)* is said to be solid if |Φ1|≤|Φ2|, where Φ1∈B(Σ)*, Φ2∈𝒦 implies that Φ1∈𝒦. A linear subspace I of B(Σ)* is said to be an ideal if I is a solid set in B(Σ)*.Note that, if|Φ1|≤|Φ2|, then ∥Φ1∥≤∥Φ2∥.Proposition 4.
Letτ be a locally solid topology on B(Σ). Then, (B(Σ),τ)′ is an ideal of B(Σ)*.Proof.
To show that(B(Σ),τ)′⊂B(Σ)*, by the way of contradiction, assume that for some Φ0∈(B(Σ),τ)′ we have Φ0∉B(Σ)*, so in view of Proposition 3, we obtain that |Φ0|(u0)=∞ for some u0∈B(Σ)+. Hence, there exists a sequence (hn) in B(Σ) such that |hn|≤u0 and |Φ0(hn)|≥n for all n∈ℕ. Since (1/n)u0→0 for τ and τ is locally solid, we get (1/n)hn→0 for τ. Hence, Φ0((1/n)hn)→0, which is in contradiction with |Φ0(hn)|≥n for all n∈ℕ.
To see that(B(Σ),τ)′ is an ideal of B(Σ)*, assume that |Φ1|≤|Φ2|, where Φ1∈B(Σ)* and Φ2∈(B(Σ),τ)′. Let fα→0 for τ and ε>0 be given. Then, there exists a net (hα) in B(Σ)such that |hα|≤|fα|for each αand |Φ2|(|fα|)≤|Φ2(hα)|+ε. Clearly, hα→0for τ because τ is locally solid, so Φ2(hα)→0. Since |Φ1(fα)|≤|Φ1|(|fα|)≤|Φ2|(|fα|)≤|Φ2(hα)|+ε, we get Φ1(fα)→0, so Φ1∈(B(Σ),τ)′, as desired.


## 3. Duality of the SpaceB(Σ)

A locally convex Hausdorff topologyτ on B(Σ) is said to be locally convex-solid if it has a local base at 0 consisting of convex and solid sets in B(Σ). In view of Propositions 1 and 2, we see that for a locally convex-solid topology τ on B(Σ) the collection of all τ-closed convex and solid τ-neighborhoods of 0 forms a local base at 0 for τ.In this section, we develop the duality theory of locally convex-solid topologies onB(Σ). We start with the following lemma.Lemma 5.
Letμ∈ba(Σ). Then, for each u∈B(Σ)+ we have
(3)|Φμ|(u)=∫Ω‍ud|μ|=sup{|∫Ω‍hdμ|:h∈B(Σ),|h|≤u}.
In particular, we have for A∈Σ(4)|μ|(A)=sup{|∫A‍hdμ|:h∈B(Σ),∥h∥≤1}.Proof.
Letu∈B(Σ)+ and h∈B(Σ) with |h|≤u. Then, we have
(5)|∫Ω‍hdμ|≤∫Ω‍|h|d|μ|≤∫Ω‍ud|μ|,
so |Φμ|(u)≤∫Ω‍ud|μ|. Now we shall show that ∫Ω‍ud|μ|≤|Φμ|(u). Indeed, let ε>0 be given. Then, there exists a Σ-simple function s=∑i=1k‍αi1Ai such that αi>0, s≤u and
(6)|∫Ω‍ud|μ|-∑i=1k‍αi|μ|(Ai)|≤ε2.
Let α=∑i=1k‍αi. Then for each i=1,2,…,k one can choose a finite Σ-partition {Aij:j=1,…,ki} of Ai and εij∈ℂ with |εij|=1, εijμ(Aij)=|μ(Aij)|for j=1,…,ki such that
(7)|μ|(Ai)≤∑i=1ki‍εijμ(Aij)+ε2α.
Let s0=∑i=1k‍∑j=1ki‍αiεij1Aij. Then |s0|=s≤u and we have
(8)|∫Ω‍ud|μ|-∫Ω‍s0dμ|≤|∫Ω‍ud|μ|-∑i=1k‍αi|μ|(Ai)|+|∑i=1k‍αi|μ|(Ai)-∑i=1k‍∑j=1ki‍αiεijμ(Aij)|≤ε2+∑i=1k‍αi||μ|(Ai)-∑j=1ki‍εijμ(Aij)|≤ε2+∑i=1k‍αiε2α=ε2+ε2=ε.
Hence, ∫Ω‍ud|μ|≤|∫Ω‍s0dμ|+ε, and it follows that ∫Ω‍ud|μ|≤|Φμ|(u).Since|Φμ|:B(Σ)+→ℝ+ is additive, |Φμ| can be uniquely extended to a positive linear functional (denoted by |Φμ| again) |Φμ|:B(Σ)→ℂ (see [2, Theorem 16-4.7]). Then, for f∈B(Σ) we have
(9)|Φμ|(f)=|Φμ|((Ref)+)-|Φμ|((Ref)-)|Φμ|(f)+i|Φμ|((Imf)+)-i|Φμ|((Imf)-)|Φμ|(f)=∫Ω‍(Ref)+d|μ|-∫Ω‍(Ref)-d|μ||Φμ|(f)+i∫Ω‍(Imf)+d|μ|-i∫Ω‍(Imf)-d|μ||Φμ|(f)=∫Ω‍fd|μ|=Φ|μ|(f).
Hence, |Φμ|=Φ|μ|.Proposition 6.
Forμ,ν∈ba(Σ) the following statements are equivalent: (i)
|μ|≤|ν|.(ii)
|Φμ|≤|Φν|.Proof.
(i)⇒(ii) Assume that |μ|≤|ν|. Then, by Lemma 5, for u∈B(Σ)+, |Φμ|(u)=∫Ω‍ud|μ|≤∫Ω‍ud|ν|=|Φν|(u), that is, |Φμ|≤|Φν|.
(ii)⇒(i) Assume that |Φμ|≤|Φν|. Then, for A∈Σ we have |μ|(A)=|Φμ|(1A)≤|Φν|(1A)=|ν|(A).Now we consider the concept of solidness in the spaceba(Σ). A subset ℳ of ba(Σ) is said to be solid if |ν|≤|μ|, where ν∈ba(Σ), μ∈ℳ implies that ν∈ℳ.For a subsetℳ of ba(Σ) let
(10)sol(ℳ)={ν∈ba(Σ):|ν|≤|μ|forsomeμ∈ℳ}.
Then, sol(ℳ) is the solid hull of ℳ, that is, sol(ℳ) is the smallest solid set in ℳ that contains ℳ.A linear subspaceI of ba(Σ) is called an ideal in ba(Σ) if I is solid. Note that ca(Σ) is an ideal of ba(Σ). In view of Proposition 6, we see that a subset ℳ of ba(Σ) is solid if and only if the set {Φμ:μ∈ℳ} is solid in B(Σ)*.Corollary 7.
Letμ∈ba(Σ). Then, for f∈B(Σ) we have
(*)|Φμ|(|f|)=∫Ω‍|f|d|μ||Φμ|(|f|)=sup{|∫Ω‍fdν|:ν∈ba(Σ),|ν|≤|μ|}.
Moreover, for a subset ℳ of ba(Σ) and each f∈B(Σ) we have
(**)sup{|Φμ|(|f|):μ∈ℳ}=sup{∫Ω‍|f|d|μ|:μ∈ℳ}=sup{|∫Ω‍fdν|:ν∈sol(ℳ)}=sup{|∫Ω‍fdν|:ν∈conv(sol(ℳ))}.Proof.
In view of Proposition6 and [2, Theorem 16-5.6] we get
(11)|Φμ|(|f|)=sup{|∫Ω‍fdν|:ν∈ba(Σ),|Φν|≤|Φμ|}|Φμ|(|f|)=sup{|∫Ω‍fdν|:ν∈ba(Σ),|ν|≤|μ|}.
Next, we shall show that for each f∈B(Σ),
(12)sup{|Φμ|(|f|):μ∈ℳ}≤sup{|∫Ω‍fdν|:ν∈sol(ℳ)}.
Indeed, let f∈B(Σ) be given. Then, for μ∈ℳ and ε>0 by (*) there exists ν∈ba(Σ) with |ν|≤|μ| and such that |Φμ|(|f|)≤|∫Ω‍fdν|+ε. Since ν∈sol(ℳ), we get |Φμ|(|f|)≤sup{|∫Ω‍fdν|:ν∈sol(ℳ)}+ε, and it follows that (12) holds.
It is enough to show that for eachf∈B(Σ)we have
(13)sup{|∫Ω‍fdν|:ν∈conv(sol(ℳ))}≤sup{|Φμ|(|f|):μ∈ℳ}.
Indeed, let ν∈conv(sol(ℳ)). Then, ν=∑i=1n‍αiνi with νi∈sol(ℳ), αi≥0,i=1,…,n and ∑i=1n‍αi=1. Hence, |νi|≤|μi| for some μi∈ℳ, i=1,2,…,n. Hence, for f∈B(Σ) we get
(14)|∫Ω‍fdν|≤∫Ω‍|f|d|ν||∫Ω‍fdν|≤∑i=1n‍αi∫Ω‍|f|d|νi|≤∑i=1n‍αi∫Ω‍|f|d|μi||∫Ω‍fdν|≤(∑i=1n‍αi)sup{∫Ω‍|f|d|μ|:μ∈ℳ}|∫Ω‍fdν|=sup{|Φμ|(|f|):μ∈ℳ}.
It follows that (13) holds, as desired.LetI be an ideal of ba(Σ) separating the points of B(Σ). Then, the pair 〈B(Σ),I〉, under its natural duality
(15)〈f,μ〉:=∫Ω‍fdμforf∈B(Σ),μ∈I
will be referred to as a solid dual system (see [7, Definition 11.7]). For subsets H of B(Σ) and ℳ of I, let us put
(16)H0={μ∈I:|∫Ω‍fdμ|≤1∀f∈H},ℳ0={f∈B(Σ):|∫Ω‍fdμ|≤1∀μ∈ℳ}.Proposition 8.
Let〈B(Σ),I〉 be a solid dual system.(i)
IfHis a solid subset ofB(Σ), thenH0is a solid subset ofI.(ii)
Ifℳis a solid subset ofI, thenℳ0is a solid subset ofB(Σ).Proof.
(i) Let|ν|≤|μ| with ν∈I and μ∈H0. Assume that f∈H and let h∈B(Σ) with |h|≤|f|. Then, h∈Hbecause His solid, so |∫Ω‍hdμ|≤1. Hence ∫Ω‍|f|d|μ|=|Φμ|(|f|)|≤1(see Lemma 5). Thus |∫Ω‍fdν|≤∫Ω‍|f|d|ν|≤∫Ω‍|f|d|μ|≤1, that is, ν∈H0. This means that H0 is a solid subset of I.
(ii) Let|f|≤|g|, where f∈B(Σ) and g∈ℳ0. Since ℳ is a solid set in ba(Σ), by Corollary 7 for μ∈ℳ, we have
(17)∫Ω‍|g|d|μ|=sup{|∫Ω‍gdν|:ν∈ℳ,|ν|≤|μ|}.
Then, for every ν∈ℳ with |ν|≤|μ| we have |∫Ω‍gdν|≤1, so ∫Ω‍|g|d|μ|≤1. It follows that
(18)|∫Ω‍fdμ|≤∫Ω‍|f|d|μ|≤∫Ω‍|g|d|μ|≤1,
that is, f∈ℳ0.Observe that for eachμ∈ba(Σ),  by Corollary 7 the functional ϱμ:B(Σ)→[0,∞) defined by
(19)ϱμ(f)=|Φμ|(|f|)=sup{|∫Ω‍hdμ|:h∈B(Σ),|h|≤|f|}ρμ(f)=sup{|∫Ω‍fdν|:ν∈ba(Σ),|ν|≤|μ|},
is a solid seminorm on B(Σ), that is, ϱμ(f1)≤ϱμ(f2) if |f1|≤|f2|.LetI be an ideal of ba(Σ) separating the points of B(Σ). The absolute weak topology |σ|(B(Σ),I) generated by I on B(Σ) is the locally convex-solid topology on B(Σ) generated by the family of solid seminorms {ϱμ:μ∈I}. Note that |σ|(B(Σ),I) is the topology of uniform convergence on all sets Iμ={ν∈I:|ν|≤|μ|}, where μ∈I.Proposition 9.
For eachμ∈I, the set Iμ is convex, solid, and σ(I,B(Σ))-compact in I.Proof.
Clearly,Iμ is a convex and solid subset of I. We shall show that Iμ is σ(ba(Σ),B(Σ))-closed subset of ba(Σ). To this end, assume that (να) is a net in Iμ such that ∫Ω‍fdνα→∫Ω‍fdν for all f∈B(Σ), where ν∈ba(Σ). Then
(20)|∫Ω‍fdνα|⟶|∫Ω‍fdν|,|∫Ω‍fdνα|≤∫Ω‍|f|d|να|≤∫Ω‍|f|d|μ|.
Hence, for each A∈Σ, we get |∫A‍fdν|≤∫A‍|f|d|μ|for all f∈B(Σ), and by Lemma 5,
(21)|ν|(A)=sup{|∫A‍fdν|:f∈B(Σ),∥f∥≤1}≤∫A‍1Ωd|μ|=|μ|(A),
that is, |ν|≤|μ|. Thus, ν∈Iμ, as desired.
In view of the Banach-Alaoglu theorem,Iμ is a relatively σ(ba(Σ),B(Σ))-compact subset of ba(Σ). Since Iμ is σ(ba(Σ),B(Σ))-closed in ba(Σ), Iμ is σ(ba(Σ),B(Σ))-compact, and it follows that Iμ is σ(I,B(Σ))-compact.We definethe absolute Mackey topology  |τ|(B(Σ),I) as the topology of uniform convergence on all convex, solid and σ(I,B(Σ))-compact subsets of ba(Σ). By Proposition 8  |τ|(B(Σ),I) is a locally convex-solid topology on B(Σ). Thus, in view of Proposition 9 we have
(22)σ(B(Σ),I)⊂|σ|(B(Σ),I)σ(B(Σ),I)⊂|τ|(B(Σ),I)⊂τ(B(Σ),I).As a consequence, we obtain the following Mackey-Arens type theorem for locally convex-solid topologies onB(Σ).Proposition 10.
Letτ be a locally convex-solid topology on B(Σ). Then, |σ|(B(Σ),B(Σ)τ′)⊂τ⊂|τ|(B(Σ),B(Σ)τ′).Proof.
Note that the familyℬτ of all convex, solid and τ-closed τ-neighborhoods of 0 forms a local base at 0 for τ. Then, τ is the topology of uniform convergence on the family {V0:V∈ℬτ}. By Proposition 8 and the Banach-Alaoglu theorem, for each V∈ℬτ, V0 is convex, solid, and σ(B(Σ)τ′,B(Σ))-compact in ba(Σ). It follows that τ⊂|τ|(B(Σ),B(Σ)τ′).
To show that|σ|(B(Σ),B(Σ)τ′)⊂τ, assume that fα→0 in B(Σ) for τ. Then, |fα|→0 for τ because τ is locally solid. Let μ∈ba(Σ) be such that Φμ∈B(Σ)τ′. Then, |Φμ|∈B(Σ)τ′ because B(Σ)τ′ is an ideal of B(Σ)* (see Proposition 4). Hence, ρμ(fα)=|Φμ|(|fα|)→0, and it follows that fα→0 for |σ|(B(Σ),B(Σ)τ′), as desired.


## 4. The Mackey Topologyτ(B(Σ),ca(Σ))

In this section we study the Mackey topologyτ(B(Σ),ca(Σ)) on B(Σ). It is known that (B(Σ),τ(B(Σ),ca(Σ))) is a generalized DF-space, that is, τ(B(Σ),ca(Σ)) is the finest locally convex topology agreeing with itself on norm-bounded sets in B(Σ) (see [8, Corollary 11.8], [9, Section  4], [10, Theorem 2]).Letcaℝ(Σ) stand for the Dedekind complete Banach lattice of all countably additive measures μ:Σ→ℝ. Then, caℝ(Σ) is a band of baℝ(Σ). Moreover, the σ-order continuous dual Bℝ(Σ)c* of Bℝ(Σ) can be identified with caℝ(Σ) through the integration mapping caℝ(Σ)∋μ↦φμ∈Bℝ(Σ)c*, where φμ(u)=∫Ω‍udμ for all u∈Bℝ(Σ) (see [4, Theorem 13.5]).Recall that a sequence(un) in Bℝ(Σ) is said to be order convergent to u∈Bℝ(Σ) (in symbols,  un→(o)u) if there exists a sequence (vn) in Bℝ(Σ)+ such that |un-u|≤vn↓0 (in the vector lattice sense; see [6, Definition  1.12]).Definition 11.
A linear functionalΦ  on B(Σ) is said to be σ-order continuous if for each sequence (fn) in B(Σ), |fn|⟶(o)0 in Bℝ(Σ) implies Φ(fn)→0.Proposition 12.
Forμ∈ba(Σ) the following statements are equivalent:(i)
Φμisσ-order continuous.(ii)
Φμ(fn)→0for each uniformly bounded sequence(fn)inB(Σ)such thatfn(ω)→0for allω∈Ω.(iii)
Φμ(un)→0for each sequence(un)inB(Σ)+such thatun(ω)↓0for allω∈Ω.(iv)
μ∈ca(Σ).Proof.
(i)⇒(ii) Assume that Φμ is σ-order continuous and let fn(ω)→0 for all ω∈Ω and supn∥fn∥<∞. Let un(ω)=supk≥n|fk(ω)| for ω∈Ω, n∈ℕ. Then, un∈Bℝ(Σ) and |fn(ω)|≤un(ω)↓0 for all ω∈Ω and n∈ℕ. It follows that |fn|→(o)0 in Bℝ(Σ), and hence, ∫Ω‍fndμ→0.
(ii)⇒(iii) It is obvious.
(iii)⇒(iv) Assume that (iii) holds, and let An↓∅, (An)⊂Σ. Then, 1An(ω)↓0 for all ω∈Ω. Hence, μ(An)=∫Ω‍1Andμ→0.
(iv)⇒(i) Assume that μ∈ca(Σ). Let (fn) be a sequence in B(Σ) such that |fn|→(o)0 in Bℝ(Σ), that is, |fn|≤un↓0 for some sequence (un) in B(Σ)+. Then
(23)|Φμ(fn)|=|∫Ω‍fndμ|≤∫Ω‍|fn|d|μ|≤∫Ω‍und|μ|.
Note that un(ω)↓0 for |μ|-almost all ω∈Ω. Indeed, let un(ω)↓u(ω) for each ω∈Ω. Then, u∈Bℝ(Σ), so A={ω∈Ω:u(ω)>0}∈Σ. If μ(A)>0, then there exists ε>0 such that Aε={ω∈Ω:u(ω)≥ε}∈Σ satisfies |μ|(Aε)>0. But then un≥ε1Aε for each n∈ℕ, contradicting un↓0 in Bℝ(Σ). Hence, by Lebesgue Dominated Convergence theorem, we get ∫Ω‍und|μ|→0, so ∫Ω‍fndμ→0.ByB(Σ)c* we will denote the space of all σ-order continuous functionals on B(Σ). For ω∈Ω let δω denote the evaluation functional on B(Σ), that is, δω(f)=f(ω) for all f∈B(Σ). Then δω∈B(Σ)c* and it follows that B(Σ)c* separates the points of B(Σ).From Proposition12 and Lemma 5 if follows that for μ∈ba(Σ), Φμ∈B(Σ)c* if and only if |Φμ|∈B(Σ)c*. Since ca(Σ) is an ideal of ba(Σ), we see that B(Σ)c* is an ideal of B(Σ)*. For a subset ℳ of ba(Σ) let
(24)|ℳ|={|μ|:μ∈ℳ}.Proposition 13.
For a subsetℳ of ca(Σ) the following statements are equivalent:(i)
ℳis relativelyσ(ca(Σ),B(Σ))-compact.(ii)
ℳis relativelyσ(ca(Σ),B(Σ))-countably compact.(iii)
supμ∈ℳ∥μ∥<∞andℳis uniformly countably additive.(iv)
supν∈conv(sol(ℳ))∥ν∥<∞andconv(sol(ℳ))is uniformly countably additive.(v)
conv(sol(ℳ))is relatively weakly compact.(vi)
ℳis relatively weakly compact.Proof.
(i)⇒(ii) It is obvious.
(ii)⇒(iii) Assume that ℳ is relatively σ(ca(Σ),B(Σ))-countably compact. It follows that ℳ is σ(ca(Σ),B(Σ))-bounded, and since {Φμ:μ∈ℳ}⊂B(Σ)*, by the uniform boundedness theorem, supμ∈ℳ∥Φμ∥<∞, that is, supμ∈ℳ∥μ∥<∞.
Assume, on the contrary, thatℳ is not uniformly countably additive. Then there exist a sequence (An) in Σ, a sequence (μn) in ℳ and δ>0 such that An↓∅ and |μn(An)|≥δ for all n∈ℕ. Let 𝒞={μn:n∈ℕ}. Putting λ=∑n=1∞‍(1/2n)(|μn|/(1+∥μn∥)), we see that λ∈ca(Σ)+ and μn≪λ for all n∈ℕ. Hence, 𝒞 is a relatively σ(caλ(Σ),B(Σ))-countably compact subset of caλ(Σ) (= the space of all λ-absolutely continuous measures in ca(Σ)). In view of the Radon-Nikodym theorem, for each μ∈caλ(Σ) there exists uniquely fμ∈L1(λ) such that μ(A)=∫A‍fμdλ for each A∈Σ. It follows that for every g∈L∞(λ) we have ∫Ω‍fμgdλ=∫Ω‍gdμ. Hence, the linear mapping Fλ:caλ(Σ)∋μ↦fμ∈L1(λ) is (σ(caλ(Σ),B(Σ)),σ(L1(λ),L∞(λ)))-continuous. Thus, the set Fλ(𝒞)={fμn:n∈ℕ} is relatively σ(L1(λ),L∞(λ))-countably compact in L1(λ). Hence, by the Eberlein-Smulian theorem {fμn:n∈ℕ} is relatively σ(L1(λ),L∞(λ))-compact in L1(λ). According to the Dunford-Pettis theorem (see [5, Chapter 3, Theorem  15]), the set {fμn:n∈ℕ} in L1(λ) is uniformly integrable. Since An↓∅, we have that λ(An)→0, and hence μ(An)=∫An‍fμndλ→0, contradicting |μn(An)|≥δ for n∈ℕ.
(iii)⇒(iv) Assume that supμ∈ℳ∥μ∥<∞ and ℳ is uniformly countably additive. Then, |ℳ| is uniformly countably additive (see [5, Chapter 1, Proposition  17]). Note that for A∈Σ, supν∈sol(ℳ)|ν|(A)=supμ∈ℳ|μ|(A). It follows that supν∈sol(ℳ)∥ν∥<∞ and sol(ℳ) is uniformly countably additive. Moreover, one can easily observe that for A∈Σ, supν∈conv(sol(ℳ))|ν(A)|=supν∈sol(ℳ)|ν(A)|and supν∈conv(sol(ℳ))∥ν∥=supν∈sol(ℳ)∥ν∥. It follows that conv(sol(ℳ)) is uniformly bounded and uniformly countably additive.
(iv)⇒(v) It is well known (see [11, Chapter 7, Theorem  13]).
(v)⇒(vi)⇒(i) It is obvious.Recall that a locally convex space(L,η) is said to be strongly Mackey if every relatively σ(Lη′,L)-countably compact set in Lη′ is η-equicontinuous.Corollary 14.
The space(B(Σ),τ(B(Σ),ca(Σ))) is strongly Mackey.Proof.
Letℳ be a relatively σ(ca(Σ),B(Σ))-countably compact set in ca(Σ). Then, by Proposition 13, the set clσ(absconvℳ) is absolutely convex and σ(ca(Σ),B(Σ))-compact. Hence, 0(clσ(absconvℳ)) is a τ(B(Σ),ca(Σ))-neighborhood of 0. It follows that ℳ is τ(B(Σ),ca(Σ))-equicontinuous.Now we distinguish three natural classes of linear topologies onB(Σ).Definition 15.
A locally solid topologyτ on B(Σ) is said to be(i)
σ-Lebesgue if un→0 for τwhenever (un) is a sequence in B(Σ)+ such that un↓0 in Bℝ(Σ) (in the vector lattice sense).(ii)
aσ-smooth topology if fn→0 for τ whenever (fn) is uniformly bounded sequence in B(Σ) such that fn(ω)→0 for all ω∈Ω.(iii)
aσ-Dini topology if un→0 for τ whenever (un) is a sequence in B(Σ)+ such that un(ω)↓0 for all ω∈Ω.Note that everyσ-Lebesgue topology on B(Σ) is σ-smooth, and every σ-smooth topology on B(Σ) is σ-Dini.Now, we characterize the Mackey topologyτ(B(Σ),ca(Σ)) on B(Σ).Theorem 16.
(i)τ(B(Σ),ca(Σ)) is a locally convex-solid topology.
(ii)τ(B(Σ),ca(Σ)) is the finest locally convex σ-Lebesgue topology on B(Σ).
(iii)τ(B(Σ),ca(Σ)) is the finest locally convex σ-smooth topology on B(Σ).
(iv)τ(B(Σ),ca(Σ)) is the finest locally convex σ-Dini topology on B(Σ).Proof.
(i) LetV be an absolutely convex and closed neighborhood of 0 for τ(B(Σ),ca(Σ)). Then, the polar V0 of V (with respect to the solid dual system 〈B(Σ),ca(Σ)〉) is σ(ca(Σ),B(Σ))-compact. In view of Proposition 13  conv(sol(V0)) is relatively σ(ca(Σ),B(Σ))-compact and hence clσ(conv(sol(V0))) (the closure is taken in σ(ca(Σ),B(Σ))) is absolutely convex and σ(ca(Σ),B(Σ))-compact. Hence,
(25)0(clσ(conv(sol(V0))))⊂0(sol(V0))⊂0(V0)=V.
It follows that 0(sol(V0)) is a convex and solid neighborhood of 0for τ(B(Σ),ca(Σ)) (see Proposition 8). This means that τ(B(Σ),ca(Σ)) is a locally convex-solid topology on B(Σ).
(ii) Now we shall show thatτ(B(Σ),ca(Σ)) is a σ-Lebesgue topology. Assume that (un) is a sequence in B(Σ)+ such that un↓0 in the vector lattice Bℝ(Σ). Let ℳ be a relatively σ(ca(Σ),B(Σ))-compact set in ca(Σ). Then ℳ is uniformly bounded and uniformly countably additive (see Proposition 13), and it follows that |ℳ| is uniformly countably additive. Hence, by [10, Theorem 3] |ℳ| is τ(Bℝ(Σ),caℝ(Σ))-equicontinuous. Hence, the polar ℝ0|ℳ| of |ℳ| (with respect to the dual pair 〈Bℝ(Σ),caℝ(Σ)〉) is a τ(Bℝ(Σ),caℝ(Σ)-neighborhood of 0 in Bℝ(Σ). Let ε>0 be given. Since τ(Bℝ(Σ),caℝ(Σ)) is a σ-Lebesgue topology on Bℝ(Σ) (see [6, Example 18, page 178], [6, Corollary 6.29 and Theorem 3.20]), there exists nε∈ℕ such that |un|∈ε(ℝ0|ℳ|) for n≥nε. Hence for n≥nε we get
(26)pℳ(un)=sup{|∫Ω‍undμ|:μ∈ℳ}pM(un)≤sup{∫Ω‍und|μ|:μ∈ℳ}≤ε.
It follows that fn→0 for τ(B(Σ),ca(Σ)), and this means that τ(B(Σ),ca(Σ)) is a σ-Lebesgue topology on B(Σ).
Now assume thatτ is a locally convex-solid σ-Lebesgue topology on B(Σ). Then B(Σ)τ′⊂B(Σ)c*, and hence, τ⊂τ(B(Σ),B(Σ)τ′)⊂τ(B(Σ),B(Σ)c*)=τ(B(Σ),ca(Σ)).
(iii) From (ii) it follows thatτ(B(Σ),ca(Σ)) is a σ-smooth topology.
Now letτ be a locally convex σ-smooth topology on B(Σ). Then by Proposition 12 we get B(Σ)τ′⊂B(Σ)c*. Hence, τ⊂τ(B(Σ),B(Σ)τ′)⊂τ(B(Σ),B(Σ)c*)=τ(B(Σ),ca(Σ)).
(iv) Clearlyτ(B(Σ),ca(Σ)) is a σ-Dini topology on B(Σ). Assume that τ is a locally convex σ-Dini topology on B(Σ). In view of Proposition 12, B(Σ)τ′⊂B(Σ)c* and it follows that τ⊂τ(B(Σ),ca(Σ)).A characterization of relative weak compactness in the Banach spaceca(Σ) in terms of uniform boundedness and uniform countable additivity of measures belongs to the classical results in measure theory (see [11, Chapter 7, Theorem 13], [8, Theorem 11.6], [12, Theorem 1.1]). Now, using the properties of the Mackey topology τ(B(Σ),ca(Σ)) we are ready to add Grothendieck type conditions to this characterization.Corollary 17.
For a subsetℳ of ca(Σ) the following statements are equivalent:(i)
ℳ is a relatively weakly compact.(ii)
supμ∈ℳ∥μ∥<∞ and ∫Ω‍fndμ→0 uniformly for μ∈ℳ whenever (fn) is a uniformly bounded sequence in B(Σ) such that fn(ω)→0 for all ω∈Ω.(iii)
supμ∈ℳ∥μ∥<∞ and ∫Ω‍undμ→0 uniformly for μ∈ℳ whenever (un) is a sequence in B(Σ)+ such that un(ω)↓0 for all ω∈Ω.Proof.
(i)⇒(ii) Assume that ℳ is relatively weakly compact. Then, in view of Corollary 14  ℳ is bounded and τ(B(Σ),ca(Σ))-equicontinuous. Let (fn) be a uniformly bounded sequence in B(Σ) such that fn(ω)→0 for all ω∈Ω. Since τ(B(Σ),ca(Σ)) is a σ-smooth topology on B(Σ) (see Theorem 16), we obtain that fn→0 for τ(B(Σ),ca(Σ)). Now let ε>0 be given. Then, there exists a τ(B(Σ),ca(Σ))-neighborhood V of 0 such that supμ∈ℳ|∫Ω‍fdμ|≤ε for all f∈V. Hence, there exists nε∈ℕ such that fn∈V for n≥nε, and it follows that supμ∈ℳ|∫Ω‍fndμ|≤ε for n≥nε, as desired.
(ii)⇒(iii) It is obvious.
(iii)⇒(i) Assume that (iv) holds and let An↓∅, (An)⊂Σ. Then, 1An(ω)↓0 for all ω∈Ω. Hence supμ∈ℳ|μ(An)|=supμ∈ℳ|∫Ω‍1Andμ|→0, that is, ℳ is uniformly countably additive. Hence, ℳ is relatively weakly compact (see [11, Chapter 7, Theorem 13]).Remark 18.
One can observe that Corollary17 is closely related to a Grothendieck’s characterization of relative weak compactness in the space of bounded complex Radon measures on a locally compact space (see [13, Theorem 2]).Now we define a class of linear operators onB(Σ).Definition 19.
Let(E,ξ) be a locally convex Hausdorff space (briefly, lcHs). A linear operator T:B(Σ)→E is said to be σ-order continuous if for each sequence (fn) in B(Σ),|fn|→(ο)0 in Bℝ(Σ) implies T(fn)→0 for ξ.For terminology and basic results concerning the integration with respect to vector measures, we refer the reader to [14–17].Let(E,ξ) be a quasicomplete lcHs. Let m:Σ→E be a ξ-bounded measure (i.e., the range of m is ξ-bounded in E). Given f∈B(Σ), let (sn) be a sequence of Σ-simple scalar functions that converges uniformly to f on Ω. Following [14, Definition 1] we say that f is m-integrable and define
(27)∫Ω‍fdm:=ξ-lim∫Ω‍sndm.
The ∫Ω‍fdm is well defined (see [14, Lemma 5]) and the map Tm:B(Σ)→E given by Tm(f)=∫Ω‍fdm is (∥·∥,ξ)-continuous and linear, and for each e′∈Eξ′(28)e′(∫Ω‍fdm)=∫Ω‍fd(e′∘m)forf∈B(Σ)
(see [14, Lemma 5]). Conversely, let T:B(Σ)→E be a (∥·∥,ξ)-continuous linear operator, and let m(A)=T(1A) for A∈Σ. Then, m:Σ→E is a ξ-bounded vector measure, called the representing measure of T and Tm(f)=T(f) for f∈B(Σ) (see [14, Definition 2]).Recall that a measurem:Σ→E is said to be ξ-countably additive if m(An)→0 for ξ whenever An↓∅, (An)⊂Σ.The following characterization ofσ-order continuous operators from B(Σ) into a quasicomplete lcHs (E,ξ) displays the close connection between the Mackey topology τ(B(Σ),ca(Σ)) on B(Σ) and E-valued ξ-countably additive measures.Proposition 20.
Assume that(E,ξ) is a quasicomplete lcHs. Then for a ξ-bounded measure m:Σ→E the following statements are equivalent:(i)
Tmis(τ(B(Σ),ca(Σ)),τ(E,Eξ′))-continuous.(ii)
Tmis(τ(B(Σ),ca(Σ)),ξ)-continuous.(iii)
Tmis sequentially(τ(B(Σ),ca(Σ)),ξ)-continuous.(iv)
Tmisσ-order continuous.(v)
Tm(fn)→0forξwhenever(fn)is a uniformly bounded sequence inB(Σ)such thatfn(ω)→0for allω∈Ω.(vi)
Tm(un)→0forξfor each sequence(un)inB(Σ)+such thatun(ω)→0for allω∈Ω.(vii)
misξ-countably additive.Proof.
(i)⇒(ii)⇒(iii) It is obvious.
(iii)⇒(iv) Assume that Tm is sequentially (τ(B(Σ),ca(Σ)),ξ)-continuous, and let (fn) be a sequence in B(Σ) such that |fn|→(ο)0 in Bℝ(Σ). Then, by Theorem 16, fn→0 for τ(B(Σ),ca(Σ)). Hence, Tm(fn)→0 for ξ, that is, T is σ-order continuous.
(iv)⇒(v) Assume that Tm is σ-order continuous, and assume that (fn) is a uniformly bounded sequence in B(Σ) such that fn(ω)→0 for all ω∈Ω. Then |fn|→(ο)0 in Bℝ(Σ) (see the proof of Proposition 12). Hence, Tm(fn)→0 for ξ, as desired.
(v)⇒(vi) It is obvious.
(vi)⇒(vii) Assume that (vi) holds, and let An↓∅, (An)⊂Σ. Then, 1An(ω)↓0 for all ω∈Ω, and it follows that μ(An)=Tm(1An)→0 for ξ. This means that m is ξ-countably additive.
(vii)⇒(i) Assume that m is ξ-countably additive. Then e′∘m∈ca(Σ) for each e′∈Eξ′, and in view of (28) and Proposition 12 we get e′∘Tm∈B(Σ)c* for each e′∈Eξ′. Hence, Tm is (σ(B(Σ),B(Σ)c*),σ(E,Eξ′))-continuous (see [7, Theorem  9.26]). It follows that Tm is (τ(B(Σ),B(Σ)c*),τ(E,Eξ′))-continuous (see [7, Example 11, page 149]).



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