On vector measures with values in $\ell_\infty$

We study some aspects of countably additive vector measures with values in $\ell_\infty$ and the Banach lattices of real-valued functions that are integrable with respect to such a vector measure. On the one hand, we prove that if $W \subseteq \ell_\infty^*$ is a total set not containing sets equivalent to the canonical basis of $\ell_1(\mathfrak{c})$, then there is a non-countably additive $\ell_\infty$-valued map $\nu$ defined on a $\sigma$-algebra such that the composition $x^* \circ \nu$ is countably additive for every $x^*\in W$. On the other hand, we show that a Banach lattice $E$ is separable whenever it admits a countable positively norming set and both $E$ and $E^*$ are order continuous. As a consequence, if $\nu$ is a countably additive vector measure defined on a $\sigma$-algebra and taking values in a separable Banach space, then the space $L_1(\nu)$ is separable whenever $L_1(\nu)^*$ is order continuous.


Introduction
Given a σ-algebra Σ and a Banach space X, we denote by ca(Σ, X) the set of all countably additive X-valued measures defined on Σ; when X is the real field, this set is simply denoted by ca(Σ).The topological dual of X is denoted by X * .The Orlicz-Pettis theorem states that a map ν : Σ → X belongs to ca(Σ, X) if (and only if) the composition x * • ν belongs to ca(Σ) for every x * ∈ X * (see, e.g., [10,p. 22,Corollary 4]).It is natural to wonder whether testing on a "big" subset, instead of all X * , is enough for countable additivity.For instance, one might consider a total subset of X * , that is, a set W ⊆ X * satisfying x * ∈W ker x * = {0}.In general, this does not work.A typical example is given by the map ν : P(N) → ℓ ∞ defined on the power set of N by ν(A) := χ A (the characteristic function of A) for all A ⊆ N. Indeed, ν is not countably additive, while the composition π n • ν : P(N) → R is countably additive for every n ∈ N, where π n ∈ ℓ * ∞ is the nth-coordinate functional given by π n (x) := x(n) for all x ∈ ℓ ∞ .
Thanks to the injectivity of ℓ ∞ , such an example can be easily carried over to any Banach space containing subspaces isomorphic to ℓ ∞ .Actually, the existence of such subspaces is the only obstacle, as the following result of Diestel and Faires [9] (cf.[10,p. 23,Corollary 7]) shows: Theorem 1.1 (Diestel-Faires).Let X be a Banach space.
(i) Suppose X contains a subspace isomorphic to ℓ ∞ .Then there exist a total set W ⊆ X * and a map ν : P(N) → X such that x * • ν ∈ ca(P(N)) for every x * ∈ W and ν ∈ ca(P(N), X). (ii) Suppose X does not contain subspaces isomorphic to ℓ ∞ .Let W ⊆ X * be a total set.If ν : Σ → X is a map defined on a σ-algebra Σ such that x * • ν ∈ ca(Σ) for every x * ∈ W , then ν ∈ ca(Σ, X).
The following concept (going back to [32, Appendice II]) arises naturally: Definition 1.2.Let X be a Banach space.A set W ⊆ X * is said to have the Orlicz-Thomas (OT) property if, for every map ν : Σ → X defined on a σ-algebra Σ, we have ν ∈ ca(Σ, X) whenever x * • ν ∈ ca(Σ) for all x * ∈ W .
Any set having the OT property is total (Proposition 3.1).With this language, the Diestel-Faires Theorem 1.1 says that every total subset of the dual of a Banach space X has the OT property if and only if X contains no subspace isomorphic to ℓ ∞ .Without additional assumptions on X, it is easy to check that any normdense subset of X * has the OT property as explicitly given in [22,Lemma 3.1]; alternatively, we can apply the Vitali-Hahn-Saks-Nikodým theorem (see, e.g., [10,p. 24,Corollary 10]) together with the Orlicz-Pettis theorem.Given any total set W ⊆ X * , a classical result of Dieudonné and Grothendieck (see, e.g., [10,p. 16,Corollary 3]) states that an X-valued map defined on a σ-algebra is bounded and finitely additive if (and only if) x * • ν is bounded and finitely additive for every x * ∈ W .This result and the Rainwater-Simons theorem (see, e.g., [12,Theorem 3.134]) ensure that any James boundary of X has the OT property, see [13, Proposition 2.9] (cf.[22,Remark 3.2

(i)]).
The previous discussion makes clear that the particular case of ℓ ∞ is the most interesting one as the OT property is concerned.In this paper we study the OT property in ℓ ∞ and some aspects of the Banach lattices of real-valued functions which are integrable with respect to countably additive ℓ ∞ -valued measures.The paper is organized as follows.
In Section 2 we introduce some terminology and preliminary facts.
In Section 3 we focus on the OT property.We begin with some basic results in arbitrary Banach spaces, including an application to the factorization of vector measures and their integration operators (Proposition 3.7).The core of this section is devoted to studying the OT property in ℓ ∞ .Our main result here (Theorem 3.15) states that any subset of ℓ * ∞ having the OT property contains a copy of the usual basis of ℓ 1 (c), where c stands for the cardinality of the continuum (i.e., c = |R|).Some consequences of this result are given.Let us mention that the existence of copies of the usual basis of ℓ 1 (c) inside James boundaries (which always have the OT property) has been studied thoroughly (see [16,17] and the references therein).
In Section 4 we study some structural properties of the Banach lattice L 1 (ν) of real-valued functions which are integrable with respect to a given ν ∈ ca(Σ, X), where Σ is a σ-algebra and X is a Banach space.These spaces play an important role in Banach lattice and operator theory.It is known that any order continuous Banach lattice having a weak order unit is lattice-isometric to a space of the form L 1 (ν).Suppose that ν has separable range or equivalently that ν takes its values in ℓ ∞ .Does it follow that L 1 (ν) is separable?The answer is in the negative, in general, even if ν is a finite measure (see Subsection 2.4).However, the answer is in the affirmative with an additional assumption that L 1 (ν) * is order continuous as asserted in Theorem 4.3.We shall instead prove a more general fact: Theorem 4.6, which asserts that if E is a Banach lattice admitting a countable positively norming set and both E and E * are order continuous, then E is separable.In particular, if µ is a finite measure for which L 1 (µ) is non-separable and 1 < p < ∞, then L p (µ) is not isomorphic to L 1 (ν) for any ν with separable range (Example 4.16).

Terminology and preliminaries
Given a set S, we denote by P(S) its power set, that is, the set of all subsets of S. The cardinality of S is denoted by |S|.The density character of a topological space (T, T), denoted by dens(T, T) or simply dens(T ), is the minimal cardinality of a T-dense subset of T .
2.1.Banach spaces.All Banach spaces considered in this paper are real.An operator is a continuous linear map between Banach spaces.Let X be a Banach space.The norm of X is denoted by • X or simply • .The closed unit ball of X is B X := {x ∈ X : x ≤ 1}.The weak (resp., weak * ) topology on X (resp., X * ) is denoted by w (resp., w * ).By a subspace of X we mean a norm-closed linear subspace.In almost all cases we will deal with norm-closed linear subspaces, so we prefer to use such an abridged terminology; when norm-closedness is not assumed, we use the term linear subspace unless otherwise stated.By a projection from X onto a subspace Y ⊆ X we mean an operator P : X → X such that P (X) = Y and P is the identity when restricted to Y .The convex hull and linear span of a set D ⊆ X are denoted by co(D) and span(D), respectively, and their closures are denoted by co(D) and span(D).A set B ⊆ B X * is said to be norming if there is a constant c > 0 such that x ≤ c sup x * ∈B |x * (x)| for every x ∈ X.A set B ⊆ B X * is said to be a James boundary of X if for every x ∈ X there is x * ∈ B such that x = x * (x).If X is a Banach lattice, then its positive cone is X + = {x ∈ X : x ≥ 0}.Given a compact Hausdorff topological space K, we denote by C(K) the Banach space of all real-valued continuous functions on K, equipped with the supremum norm.
2.2.Banach function spaces.Let (Ω, Σ, µ) be a finite measure space.To define Banach function spaces, we consider linear subspaces, not necessarily norm-closed, of L 1 (µ).A Banach space E is said to be a Banach function space (or a Köthe function space) over (Ω, Σ, µ) if the following conditions hold: In this case, E is a Banach lattice when endowed with the µ-a.e.order and the inclusion map E → L 1 (µ) is an operator.The Köthe dual of E is Any g ∈ E ′ gives raise to a functional ϕ g ∈ E * defined by ϕ g (f ) := Ω f g dµ for all f ∈ E. It is known that E is order continuous if and only if E * = {ϕ g : g ∈ E ′ } (see, e.g., [21, p. 29]).
2.3.L 1 of a vector measure.Let (Ω, Σ) be a measurable space, let X be a Banach space and let ν ∈ ca(Σ, X).A set A ∈ Σ is said to be ν-null if ν(B) = 0 for every B ∈ Σ with B ⊆ A. The family of all ν-null sets is denoted by N (ν).By a Rybakov control measure of ν we mean a finite measure of the form µ = |x * • ν| (the variation of x * • ν) for some x * ∈ X * such that N (µ) = N (ν) (see, e.g., [10,p. 268,Theorem 2] for a proof of the existence of Rybakov control measures).A Σ-measurable function f : By identifying functions which coincide ν-a.e., the set L 1 (ν) of all (equivalence classes of) ν-integrable functions is a Banach space with the norm The integration operator is the (norm one) operator I ν : L 1 (ν) → X defined by If µ is any Rybakov control measure of ν, then L 1 (ν) is a Banach function space over (Ω, Σ, µ).As a Banach lattice, L 1 (ν) is order continuous and has a weak order unit (the function χ Ω ).Conversely, any order continuous Banach lattice having a weak order unit is lattice-isometric to the L 1 space of a countably additive vector measure defined on a σ-algebra.Indeed, on the one hand, such a Banach lattice is lattice-isometric to a Banach function space E over some finite measure space (Ω, Σ, µ) (see, e.g., [21,Theorem 1.b.14]).On the other hand, thanks to the order continuity of E, the map ν : Σ → E given by ν(A) := χ A for all A ∈ Σ is countably additive and one has E = L 1 (ν) (see [5,Theorem 8], [11, Proposition 2.4(vi)]).We refer the reader to [25,Chapter 3] for more information on the L 1 space of a vector measure.
2.4.The usual measure on {0, 1} I .Let I be a non-empty set.For each i ∈ I we denote by π i : {0, 1} I → {0, 1} the ith-coordinate projection.Let Σ I be the σ-algebra on {0, 1} I generated by all the sets of the form i∈F π −1 i (w(i)), where F ⊆ I is finite and w ∈ {0, 1} F .The usual product probability measure on {0, 1} I , denoted by λ I , is defined on Σ I and satisfies λ I (π 2 for all i ∈ I.For simplicity, we just call λ I the usual measure on {0, 1} I .We have dens(L 1 (λ I )) = |I| if I is infinite.In particular, L 1 (λ I ) is not separable whenever I is uncountable.We refer the reader to [15, §254] for more information on infinite product measures and the usual measure on {0, 1} I .2.5.Measure algebras.Let (Ω, Σ, µ) be a probability space.We consider the equivalence relation on Σ defined by A ∼ B if and only if µ(A△B) = 0.The set of equivalence classes, denoted by Σ/N (µ), becomes a measure algebra when equipped with the usual Boolean algebra operations and the functional defined by µ • (A • ) := µ(A) for all A ∈ Σ, where A • ∈ Σ/N (µ) denotes the equivalence class of A. Given another probability space (Ω 0 , Σ 0 , µ 0 ), the measure algebras of µ and µ 0 are said to be isomorphic if there is a Boolean algebra isomorphism In this case, there is a lattice isometry Φ : For more information on measure algebras, see [14].
3. The Orlicz-Thomas property 3.1.The OT property in arbitrary Banach spaces.Throughout this subsection X is a Banach space.We begin with an observation: Proof.If W is not total, then there is x ∈ X \ {0} such that x * (x) = 0 for all x * ∈ W .Let ξ : P(N) → [0, 1] be a finitely additive measure which is not countably additive.Then there is a disjoint sequence (A n ) n∈N in P(N) such that the sequence (ξ( m>n A m )) n∈N does not converge to 0. Define ν : P(N) → X by ν(A) := ξ(A)x for all A ⊆ N.For each x * ∈ W we have (x * • ν)(A) = 0 for all A ⊆ N, hence x * • ν ∈ ca(P(N)).Since ν( m>n A m ) = ξ( m>n A m ) x for every n ∈ N, we have ν ∈ ca(P(N), X).
The following proposition is straightforward.As usual, ω 1 denotes the first uncountable ordinal.Given any set D ⊆ X * , we denote by S 1 (D) ⊆ X * the set of all limits of w * -convergent sequences contained in D. For any ordinal α ≤ ω 1 , we define S α (D) by transfinite induction as follows: Then S ω1 (D) is the smallest w * -sequentially closed subset of X * containing D. In general, we have The following statements are equivalent: (i) W has the OT property.
(ii) W • has the OT property.
(iii) S 1 (W ) has the OT property.
(iv) S ω1 (W ) has the OT property.
(v) W w has the OT property.
• , the set co(W ) • has the OT property.By the equivalence (i)⇔(ii) applied to co(W ), this set has the OT property and so does W (by Proposition 3.2).
The following proposition gives an operator theoretic reformulation of the OT property.Given a σ-algebra Σ, the set ca(Σ) is a Banach space when equipped with the variation norm.It is known that a set H ⊆ ca(Σ) is relatively weakly compact if and only if it is bounded and there is a non-negative µ 0 ∈ ca(Σ) such that H is uniformly µ 0 -continuous, i.e., for every ε > 0 there is δ > 0 such that sup µ∈H |µ(A)| ≤ ε for every A ∈ Σ satisfying µ 0 (A) ≤ δ (see, e.g., [7,p. 92,Theorem 13]).Proposition 3.4.Let W ⊆ X * be a subspace and let ν : Σ → X be a map defined on a σ-algebra Σ such that x * • ν ∈ ca(Σ) for all x * ∈ W . Then the map is an operator and the following statements hold: Proof.A routine application of the Closed Graph Theorem ensures that T is an operator.
(ii) Since B W is absolutely convex and norming, we have for some c > 0, by the Hahn-Banach separation theorem.Fix a non-negative µ 0 ∈ ca(Σ) such that T (B W ) is uniformly µ 0 -continuous.Observe that ν is finitely additive because W is total (bear in mind that B W is norming and so total) and x * • ν is finitely additive for every x * ∈ W .To prove that ν is countably additive it suffices to check that it is µ 0 -continuous.Fix ε > 0. Choose δ > 0 such that |x * (ν(A))| ≤ cε for every A ∈ Σ with µ 0 (A) ≤ δ and for every x * ∈ B W .
Clearly, the previous inequality is also valid for all x * ∈ B W w * and then (3.1) implies that Therefore, ν is µ 0 -continuous and so it is countably additive.Proposition 3.5.Let T : X → Y be an operator between Banach spaces and let ν : Σ → X be a map defined on a σ-algebra Σ.The following statements hold: Proof.(i) follows from the w * -compactness of B Y * and the w * -to-w * continuity of T * .(ii) is immediate.(iii) follows at once from the Orlicz-Pettis theorem.(iv) is a consequence of (iii).
To prove (v), note that the injectivity of T * * is equivalent (via the Hahn-Banach separation theorem) to the norm denseness of T * (Y * ) in X * .From the Orlicz-Pettis theorem and Proposition 3. Typical examples of non-isomorphic embeddings having injective biadjoints are the operators associated to the Davis-Figiel-Johnson-Pe lczyński factorization (see, e.g., [2, Theorem 5.37]).The following simple example shows that the conclusion of Proposition 3.5(iv) can fail for arbitrary injective operators.
Example 3.6.Let ν : P(N) → ℓ ∞ be the finitely additive measure defined by ν(A) := χ A for all A ⊆ N and let T : ℓ ∞ → ℓ 1 be the injective operator defined by T (( Let T : X → Y be an injective operator between Banach spaces, let (Ω, Σ) be a measurable space and suppose that the integration operator I ν of ν ∈ ca(Σ, Y ) factors as for some operator S. Define ν(A) := S(χ A ) for all A ∈ Σ.In [22,Theorem 3.7] it was proved that ν ∈ ca(Σ, X) satisfies: Then there is ν ∈ ca(Σ, X) such that: Proof.Observe that T is injective (by Propositions 3.1 and 3.5(ii)).For each A ∈ Σ we have ν(A) = I ν (χ A ) ∈ T (X), so there is a unique ν(A) ∈ X such that The so-defined map ν : Σ → X satisfies ν = T • ν and belongs to ca(Σ, X) because T * (B Y * ) has the OT property and T * (y To prove the reverse inclusion

has the OT property and the indefinite integral
Indeed, let (s (n) k ) k∈N be a sequence of Σ-simple functions which are uniformly convergent to This verifies (3.2) as T is injective.Now (3.2) together with countable additivity of η imply that lim increasing with union Ω).Since this holds for an arbitrarily fixed A ∈ Σ and since lim n→∞ f n = f pointwise on Ω, it follows from a result by Lewis (see, e.g., [25,Theorem 3.5]) that f ∈ L 1 (ν).Therefore we have proved L 1 (ν) ⊆ L 1 (ν) and hence, (c) holds.The Closed Graph Theorem can be used to show that both inclusions L 1 (ν) ⊆ L 1 (ν) and L 1 (ν) ⊆ L 1 (ν) are continuous, so that the norms of L 1 (ν) and L 1 (ν) are equivalent.
We finish this subsection with two results showing that the study of countable additivity of vector measures in arbitrary Banach spaces can be reduced somehow to the ℓ ∞ -valued case.
Let ν : Σ → X be a map defined on a σ-algebra Σ and define ν W := i W • ν.Let us consider the following statements: To prove (iii)⇔(iv), observe first that for each x * ∈ W we have x * • ν = e x * • ν W , where e x * ∈ ℓ 1 (W ) is the vector defined by e x * (z * ) = 0 for all z * ∈ W \ {x * } and is a James boundary of C(W ) and so it has the OT property, as we already mentioned in the Introduction.Finally, observe that (iv) is equivalent to saying that γ • ν W ∈ ca(Σ) for every γ ∈ Γ.
(b) This follows at once from Proposition 3.5(iv-v).
Observe that if W ⊆ B X * is norming, then the operator i W of Proposition 3.8 is an isomorphic embedding.Proposition 3.9.Let W ⊆ B X * be a norming set and let ν : Σ → X be a map defined on a σ-algebra Σ such that x * •ν ∈ ca(Σ) for every x * ∈ W .If ν ∈ ca(Σ, X), then there is a countable set W 0 ⊆ W such that ν W0 ∈ ca(Σ, ℓ ∞ (W 0 )), and we have a commutative diagram where P W0 is the operator defined by P W0 (u) := u| W0 for all u ∈ ℓ ∞ (W ).
Proof.Observe that ν is finitely additive.Since ν is not countably additive, we can take a sequence (A i ) i∈N of pairwise disjoint elements of Σ such that the sequence (ν( i>n A i )) n∈N does not converge to 0. Write B n := i>n A i for all n ∈ N. Since W is norming, there is a constant k > 0 such that for each n ∈ N there is The last statement is immediate.

3.2.
The OT property in ℓ ∞ .As noted in the Introduction, the finitely additive map ν : P(N) → ℓ ∞ defined by ν(A) := χ A for A ⊆ N is not countably additive while π n • ν ∈ ca(P(N)) for each coordinate functional π n ∈ ℓ * ∞ .This example has been used to see that the set {π n : n ∈ N} ⊆ ℓ * ∞ fails to have the OT property.To provide further examples of the same nature, we shall first determine the form of general ℓ ∞ -valued countably additive measures in Proposition 3.12 below.The proof uses a couple of results which shall also be needed later.The first one goes back to Bartle, Dunford and Schwartz [4] (cf. [10, p. 14, Corollary 7]) while the second one is due to Rosenthal [28] (cf.[10, p. 252, Theorem 13]).Theorem 3.10 (Bartle-Dunford-Schwartz). Let Σ be a σ-algebra, let X be a Banach space and let ν ∈ ca(Σ, X).Then the range of ν, that is, the set Theorem 3.11 (Rosenthal).Any weakly compact subset of ℓ ∞ is norm-separable.Proposition 3.12.Let ν : Σ → ℓ ∞ be a map defined on a σ-algebra Σ such that {π n • ν : n ∈ N} ⊆ ca(Σ).The following conditions are equivalent: Proof.The implications (i)⇒(ii) and (ii)⇒(iii) follow from Theorems 3.10 and 3.11, respectively.
To prove (iii)⇒(i), note that (iii) implies that ν takes values in a separable subspace X of ℓ ∞ .Since X contains no subspace isomorphic to ℓ ∞ and the set of restrictions {π n | X : n ∈ N} ⊆ X * is total (for X), the Diestel-Faires Theorem 1.1(ii) implies that ν is countably additive.
Motivated by condition (vi) in Proposition 3.12 above, we present the following: According to the proof of Proposition 3.12 and the Dunford-Pettis theorem (see, e.g., [7, p. 93]), the map ν is countably additive if and only if (f n ) n∈N is uniformly λ-integrable, that is, lim This holds when there is g ∈ L 1 [0, 1] such that |f n | ≤ g for all n ∈ N.For further criteria for uniform integrability, see [8], for example.
It is easy to check that the norm-bounded sequence (nχ [0,1/n) ) n∈N in L 1 [0, 1] is not uniformly λ-integrable.The same holds for the sequence (f n ) n∈N defined by f n (t) := It turns out that the existence of a norm-separable subset of X * having the OT property prevents the Banach space X from containing subspaces isomorphic to ℓ ∞ .This is asserted in Corollary 3.22 below, whose proof requires our main result: ∞ be a set not containing sets equivalent to the canonical basis of ℓ 1 (c).Then W fails the OT property.
Given a Banach space X and a non-empty set I, recall that a set {x i : i ∈ I} ⊆ X is said to be equivalent to the canonical basis of ℓ 1 (I) if it is bounded and there is a constant c > 0 such that for every (a i ) i∈I ∈ ℓ 1 (I).In this case, span({x i : i ∈ I}) is isomorphic to ℓ 1 (I).
The proof of Theorem 3.15 requires some known facts and uses the following result of Talagrand (see [31,Théorème 4]).Recall that the cofinality (denoted by cf (κ)) of a cardinal κ is the smallest cardinal κ ′ such that κ is the union of κ ′ many sets of cardinality < κ.Both cf (ω 1 ) and cf (c) are uncountable (see, e.g., [18,Corollary 5.12]).Theorem 3.16 (Talagrand).Let I be a set such that |I| has uncountable cofinality.Let X be a Banach space and let D ⊆ X be a set such that X = span(D).If X contains a subspace isomorphic to ℓ 1 (I), then D contains a set which is equivalent to the canonical basis of ℓ 1 (I).
The following result is an application of [28, Lemma 1.1]: Lemma 3.17.Let X be the ℓ 1 -sum of a family of Banach spaces {X i : i ∈ I} and, for each i ∈ I, let π i : X → X i be the canonical projection.Let W ⊆ X be a subspace.If the set {i ∈ I : π i (W ) = {0}} contains a set J such that |J| has uncountable cofinality, then W contains a subspace isomorphic to ℓ 1 (J).
Proof.For each i ∈ J we fix For each i ∈ J, let φ i ∈ ℓ 1 (J) * be the ith-coordinate functional.Since J k is contained in the set we have |J ′ | = |J|.We can now apply [28, Lemma 1.1] to conclude that W contains a subspace isomorphic to ℓ 1 (J).
Another ingredient for proving Theorem 3.15 is Theorem 3.18 below.It can be proved as [19,Proposition 4] (which can also be found in [1, Theorem 2.5.4]),bearing in mind that there is an almost disjoint family A of infinite subsets of N with |A| = c (see, e.g., the proof of [1, Lemma 2.5.3]).Theorem 3.18 (Kalton).Let T : ℓ ∞ → ℓ ∞ (I) be an operator, where I is a nonempty set with |I| < c.If T vanishes on c 0 , then there is an infinite set A ⊆ N such that T vanishes on the subspace Let K be a compact Hausdorff topological space.By Riesz's representation theorem, the dual C(K) * is the Banach space of all real-valued regular Borel measures on K with the variation norm.The subset of C(K) * consisting of all regular Borel probability measures on K is denoted by P (K).Given µ ∈ P (K), any ξ ∈ C(K) * can be written (in a unique way) as ξ = f dµ + ξ ′ for some f ∈ L 1 (µ) and some ξ ′ ∈ C(K) * which is singular with respect to µ; here f dµ ∈ C(K) * is given by (f dµ)(B) = B f dµ for every Borel set B ⊆ K and, as usual, we write f = dξ dµ .The Banach spaces ℓ ∞ and C(βN) are isometrically isomorphic, where βN denotes the Stone-Čech compactification of N with the discrete topology.Recall that βN is the set of all ultrafilters on N, which is a compact Hausdorff topological space such that the family A := {U ∈ βN : A ∈ U}, for A ⊆ N, forms a basis of clopen sets.Each n ∈ N is identified with the ultrafilter {A ⊆ N : n ∈ A} ∈ βN.
We are now ready to prove the main result of this section: Proof of Theorem 3.15.Let R : C(βN) → ℓ ∞ be the isometric isomorphism satisfying R(χ A ) = χ A for all A ⊆ N. Let µ 0 ∈ P (βN) be the regular Borel probability measure on βN satisfying µ 0 ({n}) = 2 −n for all n ∈ N. Observe that for each f ∈ L 1 (µ 0 ) the series of real numbers n∈N f (n)2 −n is absolutely convergent and we have Zorn's lemma ensures the existence of a set ∆ ⊆ P (βN) containing µ 0 and consisting of mutually singular elements of P (βN) such that ∆ is maximal (with respect to the inclusion) among all subsets of P (βN) satisfying those properties.Then for any ξ ∈ C(βN) * we have (see, e.g., the proof of [1, Proposition 4.3.8(iii)]).By Theorem 3.16 we can assume without loss of generality that W is a subspace of ℓ * ∞ .The conclusion is obvious if W = {0}, so we assume that W = {0}.For each µ ∈ ∆, let π µ : Z → L 1 (µ) be the canonical projection.Since W does not contain subspaces isomorphic to ℓ 1 (c), the same holds for the subspace (S • R * )(W ) ⊆ Z and so the set is non-empty and has cardinality |∆ 0 | < c (by Lemma 3.17).Let T : ℓ ∞ → ℓ ∞ (∆ 0 ) be the operator defined by for every µ ∈ ∆ 0 and for every x ∈ ℓ ∞ .Every µ ∈ ∆ \ {µ 0 } is singular with respect to µ 0 and hence, µ( A) = 0 for every finite set A ⊆ N. Bearing in mind that c 0 = span({χ A : A ⊆ N finite}) ⊆ ℓ ∞ , we deduce that T (x) = 0 for every x ∈ c 0 .The fact that |∆ 0 | < c allows us to apply Theorem 3.18 to get an infinite set A ⊆ N such that T vanishes on βN R −1 (x) dµ = 0 for every x ∈ Z A and for every µ ∈ ∆ 0 \ {µ 0 }.
Define a finitely additive map ν : P(A) → ℓ ∞ by ν(B) := χ B for all B ⊆ A. Note that ν is not countably additive, because A is infinite and ν({n}) = 1 for every n ∈ A. To finish the proof we will show that W fails the OT property by checking that ϕ • ν ∈ ca(P(A)) for arbitrarily ϕ ∈ W . Observe first that R * (ϕ) for every B ⊆ A, where the series n∈A dR * (ϕ) The converse of Theorem 3.15 fails to hold, in general.An example follows: Example 3.19.Let 2N − 1 (resp., 2N) be the set of all odd (resp., even) natural numbers.With the notation of Theorem 3.18, let W ⊆ ℓ * ∞ be the norm-closure of (Z 2N−1 ) ⊥ + ℓ 1 .Then: (i) W is total and contains a subspace isometric to ℓ 1 (2 c ); and (ii) W fails the OT property.Indeed, W is total because it contains ℓ 1 .Let P : ℓ ∞ → Z 2N be the canonical projection and define Φ : Then Φ is an isometric embedding and therefore W contains a subspace isometric to Z * 2N .Since Z 2N is isometric to C(βN), its dual contains a subspace isometric to ℓ 1 (|βN|), and the same holds for W . Now, bear in mind that |βN| = 2 c (see, e.g., [33, 19.13(d)]) to get (i).In order to show that W fails the OT property, define ν : Clearly, ν is not countably additive.However, we claim that ϕ • ν ∈ ca(P(N)) for every ϕ ∈ W . Indeed, by the Vitali-Hahn-Saks-Nikodým theorem (see, e.g., [10, p. 24, Corollary 10]), it suffices to check it whenever ϕ ∈ (Z 2N−1 ) ⊥ ∪ {e n : n ∈ N}, where {e n : n ∈ N} is the canonical basis of ℓ 1 .On the one hand, we have ϕ • ν = 0 (hence it is countably additive) whenever ϕ ∈ (Z 2N−1 ) ⊥ .On the other hand, for each n ∈ N the composition e n • ν is countably additive, because (e n • ν)(A) = χ A∩(2N−1) (n) for all A ⊆ N.This establishes the claim and hence, W fails the OT property.
Corollary 3.20.Let X be a Banach space such that there is a subset of X * having the OT property but not containing sets equivalent to the canonical basis of ℓ 1 (c).Then X does not contain subspaces isomorphic to ℓ ∞ .
Proof.Let W ⊆ X * be a set having the OT property such that W does not contain sets equivalent to the canonical basis of ℓ 1 (c).Given any subspace Y ⊆ X, the set W | Y := {x * | Y : x * ∈ W } ⊆ Y * has the OT property and does not contain sets equivalent to the canonical basis of ℓ 1 (c).By Theorem 3.15, Y cannot be isomorphic to ℓ ∞ .
Corollary 3.22.Let X be a Banach space.The following statements are equivalent: (i) X does not contain subspaces isomorphic to ℓ ∞ and X * is w * -separable.
(ii) There is a countable subset of X * having the OT property.
(iii) There is a norm-separable subset of X * having the OT property.
Proof.(i)⇒(ii) follows from the Diestel-Faires Theorem 1.1(ii), bearing in mind that the w * -separability of X * is equivalent to the existence of a countable total subset of X * .The implication (ii)⇒(i) is a consequence of Corollary 3.20 and Proposition 3.1.The equivalence (ii)⇔(iii) follows from Proposition 3.3.

L 1 spaces of vector measures with separable range
Let Σ be a σ-algebra, let X be a Banach space and let ν ∈ ca(Σ, X).By Theorems 3.10 and 3.11, the set (the range of ν) is separable whenever X = ℓ ∞ and so, in this case, ν can be seen as an element of ca(Σ, Y ) for some separable subspace Y ⊆ ℓ ∞ .Conversely, if X is separable, then it is isometric to a subspace of ℓ ∞ and so ν can be seen as an element of ca(Σ, ℓ ∞ ).As a consequence, we get: Proposition 4.1.Let E be a Banach function space over a finite measure space (Ω, Σ, µ).The following statements are equivalent: Moreover, both statements hold if E is order continuous and separable.
In general, the separability of R(ν) does not imply that the space L 1 (ν) is separable, as witnessed by the space L 1 (λ I ) of the usual measure on {0, 1} I for any uncountable set I (see Subsection 2.4).
The following example provides a vector measure ν such that R(ν) is separable, span(R(ν)) is infinite-dimensional and L 1 (ν) is neither separable nor latticeisomorphic to any AL-space.
We will obtain Theorem 4.3 as a consequence of a more general approach dealing with the concept of positively norming set introduced in [30]: Lemma 4.5.Let (Ω, Σ) be a measurable space, let X be a separable Banach space and let ν ∈ ca(Σ, X).Then L 1 (ν) admits a countable positively norming set.
Proof.Let µ be a Rybakov control measure of ν, so that L 1 (ν) is a Banach function space over (Ω, Σ, µ).Since X is separable, B X * is w * -separable and so we can take a w * -dense sequence (x * n ) n∈N in B X * .For each n ∈ N, the measure |x * n • ν| is µcontinuous and we consider its Radon-Nikodým derivative g n := The proof of Theorem 4.6 requires some previous work.Proposition 4.10 below presents a special case when E is a Banach function space and will be used to prove Theorem 4.6.Lemma 4.7.Let E be a Banach function space over a finite measure space (Ω, Σ, µ).
(ii) Since E is order continuous, S is dense in E (see, e.g., [25,Remark 2.6]).So, it suffices to check that {χ A : A ∈ Σ} is separable as a subset of E. Now, since L 1 (µ) is separable, there is a sequence (A n ) n∈N in Σ such that inf n∈N µ(A n △A) = 0 for every A ∈ Σ.
Given a finite measure space (Ω, Σ, µ) and A ∈ Σ \ N (µ), we define Σ A := {B ∈ Σ : B ⊆ A} and µ A (B) := µ(A) −1 µ(B) for every B ∈ Σ A , so that µ A is a probability measure on the measurable space (A, Σ A ). Lemma 4.8.Let (Ω, Σ, µ) be a finite measure space such that L 1 (µ) is not separable and let (g n ) n∈N be a sequence in L 1 (µ).Then there exist A ∈ Σ \ N (µ) and a sequence (A m ) m∈N of pairwise disjoint sets of for every f ∈ span({χ Am : m ∈ N}) and for every n ∈ N.
For each n ∈ N we have Φ(g n | A ) ∈ L 1 (λ I ) and so there exist a countable set [15, 254Q]).Then the set I ′ := n∈N I n is countable and for each n ∈ N we have Note that the set J := I \ I ′ is uncountable and, in particular, infinite.So, we can find a sequence (B m ) m∈N of pairwise disjoint elements of Σ J \ N (λ J ). Define Claim.For every h ∈ span({χ Cm : m ∈ N}) and for every n ∈ N we have (4.4) Indeed, note that h = h • ρ IJ for some h ∈ span({χ Bm : m ∈ N}) and then Fubini's theorem yields (4.5) Since the function ρ II ′ is Σ I -to-Σ I ′ measurable and λ I ′ (A) = λ I (ρ −1 II ′ (A)) for every A ∈ Σ I ′ (see, e.g., [15, 254O]), we have Similarly, or by direct computation, we also have The proof is finished.
We shall make use of the following well known characterization of Banach lattices with order continuous dual (see, e.g., [ for every f ∈ span({χ Am : m ∈ N}) and for every n ∈ N.
Define f m := µ(A m ) −1 χ Am ∈ E for all m ∈ N. We can use (4.11) to prove that the disjoint sequence (f m ) m∈N in E + is equivalent to the canonical basis of ℓ 1 .This contradicts the order continuity of E * (see Theorem 4.9).
Recall that an unconditional Schauder decomposition of a Banach space X is a family {X i : i ∈ I} of subspaces of X such that each x ∈ X can be written in a unique way as x = i∈I x i , where x i ∈ X i for all i ∈ I, the series being unconditionally convergent.In this case, for each i ∈ I one has a projection P i from X onto X i in such a way that x = i∈I P i (x) for all x ∈ X. Lemma 4.11.Let X be a Banach space and let {X i : i ∈ I} be an unconditional Schauder decomposition of X.For each i ∈ I, let P i be the associated projection from X onto X i .If X * contains no subspace isomorphic to ℓ ∞ , then for every x * ∈ X * the set {i ∈ I : x * • P i = 0} is countable.
Proof.For each J ⊆ I, let Q J be the projection from X onto span( i∈J X i ) defined by Q J (x) := i∈J P i (x) for all x ∈ X. Fix x * ∈ X * and define ν : P(I) → X * by ν(J) := x * • Q J for all J ⊆ I.
We identify X as a subspace of X * * in the canonical way.Note that x•ν ∈ ca(P(I)) for every x ∈ X, because the series of real numbers i∈I x * (P i (x)) is absolutely convergent and (x • ν)(J) = x * (Q J (x)) = i∈J x * (P i (x)) for all J ⊆ I.
Since X * contains no subspace isomorphic to ℓ ∞ and X is a total subset of X * * , we can apply the Diestel-Faires Theorem 1.1(ii) to conclude that ν ∈ ca(P(I), X * ).
We claim that for every ε > 0 the set I ε := {i ∈ I : x * • P i ≥ ε} is finite.Indeed, if not, then there is a sequence (i n ) n∈N of distinct elements of I ε .However, the countable additivity of ν implies that the series n∈N ν({i n }) = n∈N x * • P in is unconditionally convergent in X * which is impossible because x * • P in ≥ ε for all n ∈ N. Therefore, the set {i ∈ I : x * • P i = 0} = n∈N I 1/n is countable.Lemma 4.12.Let E be a Banach lattice admitting a countable positively norming set.Then E admits a strictly positive functional, i.e., there is ϕ ∈ (E * ) + such that ϕ(x) > 0 whenever x ∈ E + \ {0}.
Proof.Take a sequence (ϕ n ) n∈N in B E * ∩ (E * ) + and a constant c > 0 such that x E ≤ c sup n∈N ϕ n (|x|) for every x ∈ E. Now, it is clear that the functional ϕ := n∈N 2 −n ϕ n satisfies the required property.
We have gathered all the tools needed to prove the main result of this section: Proof of Theorem 4.6.Since E is order continuous, it admits an unconditional Schauder decomposition {E i : i ∈ I} consisting of pairwise disjoint bands, each having a weak order unit (see, e.g., [21, Proposition 1.a.9]).For each i ∈ I, let P i be the associated projection from E onto E i .
Fix i ∈ I. Since E i is order continuous and has a weak order unit, it is latticeisometric to a Banach function space over a finite measure space (see, e.g., [21,Proposition 1.b.14]).Since E * is order continuous, the same holds for E * i (bear in mind the equivalence (i)⇔(iii) in Theorem 4.9).Moreover, E i admits a countable positively norming set (consider the restriction to E i of a countable positively norming set for E).Then, E i is separable by Proposition 4.10.
Fix ϕ ∈ (E * ) + such that ϕ(x) > 0 whenever x ∈ E + \ {0} (see Lemma 4.12).For each i ∈ I we have E i = {0} and so ϕ • P i = 0. Since E * is order continuous, it contains no subspace isomorphic to ℓ ∞ (see Theorem 4.9) and then Lemma 4.11 implies that I is countable.From the separability of each E i it follows that E is separable.
Example 4.13.The conclusion of Theorem 4.6 can fail if the order continuity of E is dropped.For instance, the non-separable Banach lattice ℓ ∞ admits a countable positively norming set (the coordinate functionals form a norming set) and ℓ * ∞ is an AL-space, hence it is order continuous (see, e.g., [2,p. 194 and Theorem 4.23]).

Proposition 3 . 2 .
Let W ⊆ X * .The following statements are equivalent: (i) W has the OT property.(ii) co(W ) has the OT property.(iii) span(W ) has the OT property.
3 it follows that T * (Y * ) has the OT property.Since span(T * (B Y * )) = T * (Y * ), we can apply Proposition 3.2 to conclude that T * (B Y * ) has the OT property.

(4. 7 )
{0,1} I h dλ I = {0,1} J h dλ J .By putting together (4.5), (4.6) and (4.7) we get (4.4), as claimed.Finally, let (A m ) m∈N be a sequence of pairwise disjoint elements of Σ A such that Φ(χ Am | A ) = χ Cm for all m ∈ N. Then µ(A m ) = µ(A)λ I (C m ) > 0 for all m ∈ N. Given any f ∈ span({χ Am : m ∈ N}), we have Φ(f | A ) ∈ span({χ Cm : m ∈ N}) and for each n ∈ N we have A f g n dµ A (4.1) see Subsections 2.2 and 2.3).Since (x * n ) n∈N is w * -dense in B X * , for each f ∈ L 1 (ν) we can apply [27, Lemma 2.2] to get Theorem 4.6.Let E be a Banach lattice such that both E and E * are order continuous.If E admits a countable positively norming set, then E is separable.
2, Theorem 4.69]): Theorem 4.9.Let E be a Banach lattice.The following statements are equivalent: (i) E * is order continuous.(ii)There is no disjoint sequence in E + which is equivalent to the canonical basis of ℓ 1 .(iii)E does not contain sublattices which are lattice isomorphic to ℓ 1 .(iv)E * does not contain subspaces isomorphic to ℓ ∞ .Proposition 4.10.Let E be a Banach function space over a finite measure space (Ω, Σ, µ) such that both E and E * are order continuous.If E admits a countable positively norming set, then E is separable.Proof.Fix a sequence (φ n ) n∈N in B E * ∩ (E * ) + and a constant c > 0 such that Since E is order continuous, we can identify E * with E ′ , hence for each n ∈ N we have φ n = ϕ gn for some g n ∈ E ′ (see Subsection 2.2).Suppose, by contradiction, that E is not separable.By Lemma 4.7(ii), the space L 1 (µ) is not separable.Then we can apply Lemma 4.8 to (g n ) n∈N (as a sequence in L 1 n∈N φ n (|f |) for every f ∈ E.