Associate space with respect to a locally $\sigma$-finite measure on a $\delta$-ring and applications to spaces of integrable functions defined by a vector measure

Abstract

We show that for a locally $\sigma$-finite measure $\mu$ defined on a $\delta$-ring, the associate space theory can be developed as in the $\sigma$-finite case, and corresponding properties are obtained. Given a saturated $\sigma$-order continuous $\mu$-Banach function space E, we prove that its dual space can be identified with the associate space $E ^\times$ if, and only if, $E^\times$ has the Fatou property. Applying the theory to the spaces $L^p (\nu )$ and $L_w^p (\nu )$, where $\nu$ is a vector measure defined on a $\delta$-ring $\mathcal {R}$ and $1 \le p < \infty$, we establish results corresponding to those of the case when the vector measure is defined on a $\sigma$-algebra.

1 Introduction

Let $\Omega$ be a set and $\mathcal {R}$ a $\delta$-ring consisting of subsets of $\Omega$. Given a vector measure $\nu : \mathcal {R}\rightarrow X$, where X is a (real or complex) Banach space, we obtain the Banach space of weakly-integrable functions $L^1_w(\nu )$, which has the space of $\nu$-integrable functions $L^1 (\nu )$ as a closed subspace. With the order given by $f \ge g$ if $f \ge g$ outside a $\nu$-null set, we have that $L^1_w (\nu )$ is a $\sigma$-Fatou Banach lattice and $L^1 (\nu )$ is an order continuous Banach lattice. (See Sect. 2 for definitions.) The integration theory with respect to vector measures defined on $\delta$-rings was developed mainly by Lewis [18], Masani and Niemi [21, 22] and Delgado [8]. It extends the well known theory for vector measures defined on $\sigma$-algebras [25, Ch. 3].

If E is a real or complex order continuous Banach lattice, Curbera [5, p. 22], [10, p. 246] showed that there is a set $\Omega$, a $\delta$-ring $\mathcal {R}$ consisting of subsets of $\Omega$, and a vector measure $\nu : \mathcal {R}\rightarrow E$, such that E and $L^1 (\nu )$ are order and isometrically isomorphic . It follows that the dual spaces $L^1 (\nu )^*$ and $E^*$, both of which are Banach lattices, are also order and isometrically isomorphic. Hence by studying a dual space of the form $L^1 (\nu )^*$ we are implicitly analyzing the dual space of the most general order continuous Banach lattice.

The study of $L^1 (\nu )^*$ can be done through the associate space theory, systematically developed by Luxemburg and Zaanen [28, Ch. 15], [29, Ch. 16, Sect. 112]. This theory begins with an arbitrary positive measure space $(\Omega , \Sigma , \mu )$ and applies to a $\mu$-Banach function space ($\mu$-B.f.s. for short) E which is saturated. Then also its associate space $E^\times$ is a $\mu$-Banach function space and the operator $R: E^\times \rightarrow E^*$, defined by $R(g) := R_g$, where $R_g (f) := \int _\Omega fg d \mu$, is a linear isometry. This operator, called the canonical isometry, allows us to consider the associate space $E^\times$ as a closed subspace of the dual space $E^*$. When the involved measure $\mu$ is $\sigma$-finite and the saturated B.f.s. E is $\sigma$-order continuous, then the canonical isometry R is onto. Moreover, R also preserves the lattice structure and so we write $E^\times =E^*$. In the following we will maintain this notation to indicate that R is onto.

To apply the above theory for studying $L^1(\nu )^*$, in the first place we have to find a positive measure $\mu$ which is also a local control measure for $\nu$, that is, such that $\mu$ and $\nu$ have the same null sets. In this situation $L^1 (\nu )$ is a $\mu$-B.f.s. Assume the $\delta$-ring $\mathcal {R}$ on which the vector measure $\nu$ is defined is a $\sigma$-algebra. Then it is well known that $\nu$ has a finite local control measure $\mu$ with respect to which $L^1(\nu )$ is a saturated $\mu$-B.f.s. [25, pp. 107–108]. It follows that in this case we have $L^1 (\nu )^\times = L^1 ( \nu )^*$.

If the $\delta$-ring $\mathcal {R}$ is not a $\sigma$-algebra, we cannot proceed directly as above to obtain for $L^1 (\nu )$ similar results to those we have just mentioned, since in this situation $\nu$ may not have a $\sigma$-finite local control measure. This is a problem, since a key fact for the associate space theory to work is that when the measure $\mu$ is $\sigma$-finite, then the saturation of a $\mu$-B.f.s. E implies that of $E^\times$.

However, relying on a result of Brooks and Dinculeanu [3], it has recently been pointed out by Jiménez et al. [16] the fact that any vector measure defined on a $\delta$-ring $\mathcal {R}$, always has a local control measure $\mu$ that is also $\sigma$-finite on any set $B \in \mathcal {R}$. We will see that a measure of this kind, which we have called Brooks–Dinculeanu measure, is appropriate for our objectives.

In this paper we first study the associate space $E^\times$ of a saturated $\mu$-B.f.s., for a locally $\sigma$-finite positive measure $\mu$. Thereafter we consider a vector measure $\nu$ and apply the results to the $\mu$-Banach function spaces of p integrable functions $L^p (\nu )$ and of weakly integrable functions $L ^p_w (\nu ), 1 \le p < \infty$, where $\mu$ is a Brooks–Dinculeanu measure for $\nu$. The main question we discussed was that of the validity of the equality $L^1(\nu )^\times = L^1(\nu )^*$.

We divided our work in five sections, including this Introduction. In Sect. 2 we present the notation, definitions and basic results that we have needed.

The general theory for the associate space $E^\times$ of a saturated $\mu$-B.f.s E is given in Sect. 3. As in the well known $\sigma$-finite case, it turns out that the associate space $E^\times$ always has the $\sigma$-Fatou property. When $\mu$ is a $\sigma$-Fatou property, we show that the $\sigma$-Fatou property and the Fatou properties are equivalent for E. Based on general properties of the dual space of an arbitrary Banach lattice [29, Ch. 14, Ch. 15], we establish that if $E^\times = E^*$, then E is order continuous and $E^\times$ has the Fatou property.

In Sect. 4 we restrict our considerations to a locally $\sigma$-finite measure $\mu$. In this setting we obtain results that are well known in the $\sigma$-finite case. Particularly we show that $E^\times$ is also a saturated $\mu$-B.f.s., that $E^{\times \times } = E$ when E has the Fatou property, and that the factorization $L^1 (\mu ) = E E ^\times$ holds. When the saturated $\mu$-B.f.s. is $\sigma$-order continuous, we prove that $E^\times = E^*$ if, and only if, $E^\times$ has the Fatou property. In a forthcoming work we plan to show that this last is not always the case.

In the last section we apply our development to the spaces $L^p(\nu )$ and $L^p_w(\nu )$, $1\le p <\infty$, obtained from a vector measure $\nu$. These are Banach function spaces with respect to any Brooks–Dinculeanu measure for $\nu$. Thus we show that $L^p (\nu )^{\times \times } = L^p_w (\nu )$ and $L^p(\nu )^\times = L^p_w (\nu )^\times$ for $1 \le p < \infty$. When the vector measure $\nu$ is defined on a $\sigma$-algebra the first of these equalities was established by Curbera and Ricker [6, Prop. 2.4], [7, Prop. 1].

We give several situations where $L^1(\nu )^\times =L^1(\nu )^*$ holds, one of them being the case of a decomposable vector measure. This turns out to be important, since the vector measure $\nu$ that Calabuig et al. used to represent an order continuous Banach lattice as $L^1(\nu )$, is a decomposable vector measure [4].

Finally, we establish that $L^p(\nu )^\times =L^p(\nu )^*$ if $1<p<\infty$ and that $L^1(\nu )^\times =L^1(\nu )^*$ when $L^1(\nu )$ is reflexive. We also verify that a reflexivity criterion proven by Fernández et al. [12, Cor. 3.10] for a vector measure defined on a $\sigma$-algebra is still valid in the $\delta$-ring case.

To complete this introduction, we want to note that Okada was the first to obtain a description of $L^1(\nu )^*$ for a classical vector measure $\nu$ [24]. Later Galaz-Fontes gave a representation for $L^p(\nu )^*$ when $1 < p < \infty$ [14] and recently Mastylo and Sánchez Pérez have established representations of these kind for a dual Banach space in a more general context [23].

2 Notation and basic results

2.1 Banach lattices

Throughout this paper all vector spaces considered will be with respect to $\mathbb {K}$, where $\mathbb {K}=\mathbb {C}$, the field of complex numbers or $\mathbb {K}=\mathbb {R}$, the field of real numbers. Let X be a normed space. By $B_X$ we will indicate its unit closed ball and by $X^*$ its dual space. We will represent by $\langle \cdot ,\cdot \rangle$ the duality pairing, i.e. $\langle x,x^*\rangle :=x^*(x), \forall \ x\in X \text{ and } \ x^*\in X^*$. If Y is other normed space, to express that $X=Y$ as sets and with equal norms, we will write $X\equiv Y$.

Let X be a real vector lattice with order $\le$. For $A\subset X$ we will be denote by $A^+$ the subset of X consisting of all $f\in A$ such that $0\le f$. Given $f\in X$, we indicate by $f^+$, $f^-$ and |f| its positive part, negative part and modulus, respectively. Finally X is called Dedekind $\sigma$ -complete if every non-empty countable subset which is bounded from above has a supremum.

Let J be a directed set. A net $\{f_\tau \}_{\tau \in J}\subset X$ is an upwards directed system if for $\tau _1$ and $\tau _2$ in J, there exists $\tau _3\in J$ such that $f_{\tau _1}\le f_{\tau _3}$ and $f_{\tau _2}\le f_{\tau _3}$. In this case we will use the notation $f_\tau \uparrow$. If additionally there exists $f=\sup _\tau f_\tau \in X$ we write $f_\tau \uparrow f$. Similarly, if the sequence $\{f_n\}\subset X$ is such that $f_n\le f_{n+1}$, $\forall \ n\in \mathbb {N}$, we will indicate $f_n\uparrow$ and if additionally $f=\sup _n f_n$ then we will write $f_n\uparrow f$. Analogously, we define a downwards directed system $\{f_\tau \}$ and the notations $f_\tau \downarrow$ and $f_\tau \downarrow f$.

A real normed vector lattice X is a real normed space that is a vector lattice and whose norm $\Vert \cdot \Vert _X$ has the lattice property, that is

$$\begin{aligned} \mathrm{if } f,g\in Y\, \mathrm{satisfy },\ |f|\le |g|, \text{ then } \Vert f\Vert _X\le \Vert g\Vert _X. \end{aligned}$$

(2.1)

If in addition the space is complete, we will say that X is real Banach lattice.

Let X be a real normed vector lattice. Then X has the weak Fatou property if for each upwards directed system $\{f_\tau \}\subset X^+$ with $\sup _\tau \Vert f_\tau \Vert _X<\infty$, there exists $f=\sup _\tau f_\tau \in X$; if additionally $\Vert f\Vert _X=\sup _n \Vert f_n\Vert _X$, then X has the Fatou property. Similarly, X has the weak $\sigma$ -Fatou property, if given $\{f_n\}\subset X^+$ such that $f_n\uparrow$ and $\sup _n \Vert f_n\Vert _X<\infty$, then there exists $f=\sup _n f_n\in X$, and if additionally $\Vert f\Vert _X=\sup _n \Vert f_n\Vert _X$, then X is said to have the $\sigma$ -Fatou property. We say that X is order continuous, if for any system $\{f_\tau \}\subset X$ satisfying $f_\tau \downarrow 0$ it follows that $\Vert f_\tau \Vert _X\downarrow 0$. Analogously, X is $\sigma$ -order continuous if for any sequence $\{f_n\}\subset X$ satisfying $f_n\downarrow 0$ we have that $\Vert f_n\Vert _X\downarrow 0$.

Take a real Banach lattice X. Then in $Z:=X+iX$, the complexification of X, the modulus is defined by $|h|:=\sup \{|(\cos \theta ) f+(\sin \theta ) g|:0\le \theta <2\pi \}, \ \forall \ h:=f+ig\in Z$ [29, Ch. 14; Thm. 91.2], the norm by $\Vert h\Vert _Z=\Vert \ |h| \ \Vert _X, \ \forall \ h\in Z$, and the order is given by $f\le g$ in Z, if $f,g\in X$ and $f\le g$. In this case Z is called a complex Banach lattice, X is its real part and we write $X = Z_\mathbb {R}$. Observe that $Z^+=X^+$. We will say that a complex Banach lattice has one of the properties we introduced above if its real part has it. Henceforth we will say only Banach lattice (normed vector lattice) to refer to a complex or real Banach lattice (normed vector lattice).

Let X be a Banach lattice. An ideal Y of X is a vector subspace of X if $f\in X$ with $|f|\le |g|$ for some $g\in Y$ implies $f\in Y$.

Let $T : X \rightarrow Y$ be a linear operator between Banach lattices. Then T is said to be positive if for each $f \in X^+$ we have that $T(f) \in Y^+$. In this case $T(X_\mathbb {R})\subset Y_\mathbb {R}$ and T is bounded [1, Lemma 3.22]. We will say that T is an order isometry if T is an isometry, T is onto and both T and $T^{-1}$ are positive operators. This last condition is equivalent to

$$\begin{aligned} Tf \ge 0 \text{ if } \text{ and } \text{ only } \text{ if } f \ge 0, \quad \forall f \in X. \end{aligned}$$

In this case we have that $T(X_\mathbb {R}) = Y_\mathbb {R}$, $T(\sup \{f, g\}) = \sup \{Tf, Tg\}, \forall f, g \in X_\mathbb {R}$ and $T|f| = |Tf|, \ \forall f \in X$.

Let X be a real Banach lattice. Then its dual space $X^*$ is a Banach lattice with the order given by

$$\begin{aligned} \varphi \le \psi , \text{ if } \varphi (f)\le \psi (f), \quad \forall \ f\in X^+, \ \varphi ,\psi \in X^*. \end{aligned}$$

(2.2)

In this case the supremum and infimum are uniquely determined by

$$\begin{aligned} \begin{array}{c} \sup \{\varphi ,\psi \}(f):=\sup \{\varphi (g)+\psi (h): f=g+h, g\ge 0, \ h\ge 0 \}, \\ \inf \{\varphi ,\psi \}(f):=\inf \{\varphi (g)+\psi (h): f=g+h, g\ge 0, \ h\ge 0 \}, \end{array} \end{aligned}$$

(2.3)

for each $\varphi ,\psi \in X^*$ and $f\in X^+$ [27][Chap II, Props. 4.2, 5.5] .

Now assume that X is a complex Banach lattice. Given $\varphi \in X_\mathbb {R}^{ \ *}$, we will indicate by $\widetilde{\varphi }:X\rightarrow \mathbb {C}$ its canonical extension, that is, $\widetilde{\varphi }(x+iy)=\varphi (x)+i\varphi (y)$. If $\Phi :X\rightarrow \mathbb {C}$ is a bounded linear functional, then $\Phi$ has the form $\Phi =\widetilde{\varphi }+i\widetilde{\psi }$, where $\widetilde{\varphi }$ and $\widetilde{\psi }$ are the canonical extensions of linear functionals $\varphi ,\psi \in X_\mathbb {R}^{ \ *}$. Identifying $X_\mathbb {R}^{ \ *}$ with $\widetilde{X}^*_{ \ \mathbb {R}} \subset X^*$, we have that $X^*=X_\mathbb {R}^{ \ *}+iX_\mathbb {R}^{ \ *}$ is a Banach lattice. As can be seen in [27, §11], in this case

$$\begin{aligned} |\Phi |(f)=\sup _{|g|\le f}|\Phi (g)|, \quad \forall \ f\in X^+. \end{aligned}$$

(2.4)

2.2 $\mu$-Banach function spaces

Given a measurable space $(\Omega ,\Sigma )$ we will denote by $L^0(\Sigma )$ the space formed by the $\Sigma$-measurable functions $f:\Omega \rightarrow \mathbb {K}$. If additionally we have a positive measure $\mu$ defined on $\Sigma$, we indicate by $\mathcal {N}_0(\mu )$ the family of $\mu$ -null subsets, i. e., the sets $A\in \Sigma$ such that $\mu (A)=0$. As usual a property holds $\mu$ -almost everywhere (briefly $\mu$-a.e.) if it holds except on a $\mu$-null set. We indicate by $L^0(\mu )$ the space of equivalence classes of functions in $L^0(\Sigma )$, where two functions are identified when they are equal $\mu$-a.e.

Note that, when $\mathbb {K}=\mathbb {C}$, the space $L^0(\mu )$ is the complexification of the real space $L^0(\mu )_\mathbb {R}:=\{f\in L^0(\mu ): f \text{ take } \text{ its } \text{ values } \text{ in } \ \mathbb {R}\ \mu \text{-a.e. } \}$.

In $L^0(\mu )_\mathbb {R}$ we will always consider the $\mu$-a.e. pointwise order. Let $f\in L^0(\mu )$. So $\text{ Re }f,\text{ Im }f\in L^0(\mu )_\mathbb {R}$ and $f=\text{ Re }f+i\text{ Im }f$. Moreover,

$$\begin{aligned} {\sup _{0\le \theta <2\pi }|(\cos \theta )\text{ Re }f +(\sin \theta ) \text{ Im }f|=\sqrt{(\text{ Re }f)^2+(\text{ Im }f)^2}=|f|.} \end{aligned}$$

(2.5)

We will say that a normed space $E\subset L^0(\mu )$ is a normed function space related to $\mu$ (briefly $\mu$-n.f.s.) if E is a vector subspace of $L^0(\mu )$ such that $f\in L^0(\mu )$ with $|f|\le |g|$ for some $g\in E$, implies $f\in E$ and the lattice property (2.1) holds (with E instead of X). If additionally E is complete we will call it Banach function space related to $\mu$ (briefly $\mu$-B.f.s.). We must note that in the literature there appear other definitions with the same name, such as in [19, Def. 1.b.17] and in [2, Def. I.1.3].

Let E be a $\mu$-B.f.s. Then E is Dedekind $\sigma$-complete. So, E is $\sigma$-order continuous if and only if E is order continuous [29, 103.9]. Now if E is a $\mu$-n.f.s. with the $\sigma$-Fatou property, then E is complete [28, Ch. 15, §65, Thm. 1]. Thus, E is a $\mu$-B.f.s. with the $\sigma$ -Fatou property.

Let E be a complex $\mu$-B.f.s. Note that $E_\mathbb {R}=E\cap L^0(\mu )_\mathbb {R}$, with the $\mu$-a.e. pointwise order, is a real Banach lattice and $E=E_\mathbb {R}+iE_\mathbb {R}$. It follows from (2.5) that E is a complex Banach lattice. Moreover, $f\in E$ if and only if $(\text{ Re }f)^+,(\text{ Re }f)^-,(\text{ Im }f)^+,(\text{ Im }f)^-\in E^+$.

2.3 Integration with respect to measures defined on $\delta$-rings

Let $\Omega$ be a set. A family $\mathcal {R}$ of subsets of $\Omega$ is a $\delta$ -ring if $\mathcal {R}$ is a ring which is closed under countable intersections. From now on in this paper $\mathcal {R}$ will be a $\delta$-ring. We denote by $\mathcal {R}^{loc}$ the $\sigma$-algebra of all sets $A\subset \Omega$ such that $A\cap B\in \mathcal {R}$, $\forall \ B\in \mathcal {R}$. Given $A\in \mathcal {R}^{loc}$ we indicate by $\mathcal {R}_A$ the $\delta$-ring $\{B\subset A:B\in \mathcal {R}\}$ and by $\pi _A$ the collection of finite families of pairwise disjoint sets in $\mathcal {R}_A$. Note that if $\Omega \in \mathcal {R}$, then $\mathcal {R}$ is a $\sigma$-algebra, and in this case we have that $\mathcal {R}^{loc}=\mathcal {R}$. Moreover, for each $B\in \mathcal {R}$ it turns out that $\mathcal {R}_B$ is a $\sigma$-algebra.

A scalar measure (positive measure) is a function $\lambda :\mathcal {R}\rightarrow \mathbb {K}$ ($\lambda :\mathcal {R}\rightarrow [0,\infty ]$) satisfying that if $\{B_n\}\subset \mathcal {R}$, is a family of pairwise disjoint sets such that $\bigcup _{n=1}^{\infty }B_n\in \mathcal {R}$, then $\sum _{n=1}^{\infty } \lambda (B_n)=\lambda \left( \bigcup _{n=1}^{\infty }B_n\right)$. The variation of $\lambda$ is the countably additive measure $|\lambda |:\mathcal {R}^{loc}\rightarrow [0,\infty ]$ defined by

$$\begin{aligned} |\lambda |(A):=\sup \left\{ \sum _{j=1}^{n}|\lambda (A_j)| :\{A_j\}\in \pi _A\right\} . \end{aligned}$$

A function $f\in L^0(\mathcal {R}^{loc})$ is $\lambda$ -integrable if $f\in L^1(|\lambda |)$. We denote by $L^1(\lambda )$ the subspace of $L^0(\lambda )$ formed by the $\lambda$-integrable functions. Then $L^1(\lambda )$ with norm given by $|f|_{1,\lambda }:=\int _\Omega |f|d|\lambda |$ is a $\sigma$-order continuous $|\lambda |$-B.f.s. with the $\sigma$-Fatou property. The following result is basic in the theory; when $\lambda$ is a scalar measure it was established by Masani and Niemi [21, Lemma 2.30, Thm. 2.32], for the case of a positive measure we can proceed similarly.

Proposition 2.1

If $f\in L^0(\mathcal {R}^{loc})$, then

$$\begin{aligned} {\int _A |f|d|\lambda |=\sup _{B\in \mathcal {R}_A}\int _B|f|d|\lambda |,} \quad \forall \ A\in \mathcal {R}^{loc}. \end{aligned}$$

(2.6)

Therefore, $f\in L^1(|\lambda |)$ if and only if ${\sup _{B\in \mathcal {R}}\int _B|f|d|\lambda |<\infty }$.

Let X be a Banach space. A function $\nu :\mathcal {R}\rightarrow X$ is a vector measure if $\sum _{n=1}^\infty \nu (B_n)=\nu \left( \bigcup _{n=1}^\infty B_n\right)$, for any collection $\{B_n\}\subset \mathcal {R}$ of pairwise disjoint sets such that $\bigcup _{n=1}^\infty B_n \in \mathcal {R}$. The variation of $\nu$ is the positive measure $|\nu |$ defined in $\mathcal {R}^{loc}$ by

$$\begin{aligned} |\nu |(A):=\sup \left\{ \sum _j \Vert \nu (A_j)\Vert _X : \{A_j\}\in \pi _A \right\} . \end{aligned}$$

The semivariation of $\nu$ is the function $\Vert \nu \Vert :\mathcal {R}^{loc}\rightarrow [0,\infty ]$ given by

$$\begin{aligned} {\Vert \nu \Vert (A):=\sup \{|\langle \nu ,x^*\rangle |(A):x^*\in B_{X^*}\}}, \end{aligned}$$

where $|\langle \nu ,x^*\rangle |$ is the variation of the scalar measure $\langle \nu ,x^*\rangle :\mathcal {R}\rightarrow \mathbb {K}$, where

$$\begin{aligned} \langle \nu ,x^*\rangle (B):=\langle \nu (B),x^*\rangle , \quad \forall \ B\in \mathcal {R}. \end{aligned}$$

The semivariation of $\nu$ is finite in $\mathcal {R}$ and for any $A\in \mathcal {R}^{loc}$ satisfies $\Vert \nu \Vert (A)\le |\nu |(A)$. A set $A\in \mathcal {R}^{loc}$ is said to be $\nu$-null if $\Vert \nu \Vert (A)=0$. We will denote by $\mathcal {N}_0(\nu )$ the collection of $\nu$-null sets. It turns out that $\mathcal {N}_0(\nu )=\mathcal {N}_0(|\nu |)$. Moreover $A\in \mathcal {N}_0(\nu )$ if and only if $\nu (B)=0$, $\forall \ B\in \mathcal {R}_A$. We say that a positive measure $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ is a local control measure for $\nu$, if $\mathcal {N}_0(|\lambda |)=\mathcal {N}_0(\nu )$ [8, p. 437]. Then, $|\nu |$ is a local control measure for $\nu$. We define $L^0(\nu )$ as the space of equivalence classes of functions in $L^0(\mathcal {R}^{loc})$, where two functions are identified when they are equal $\nu$-a.e. So, $L^0(\nu )=L^0(|\lambda |)$, where $\lambda$ is any local control measure for $\nu$.

A function $f\in L^0(\mathcal {R}^{loc})$ is weakly $\nu$ -integrable, if $f\in L^1(\langle \nu ,x^*\rangle )$, for each $x^*\in X^*$. We will denote by $L^1_w(\nu )$ the subspace of $L^0(\nu )$ of all weakly $\nu$-integrable functions. With the norm ${\Vert f\Vert _\nu :=\sup \{\int _\Omega |f| d|\langle \nu ,x^*\rangle |:x^*\in B_{X^*} \}}$, $L^1_w(\nu )$ is a $|\lambda |$-B.f.s. with the $\sigma$-Fatou property, where $\lambda$ is a local control measure for $\nu$.

A function $f\in L^1_w(\nu )$ is $\nu$ -integrable, if for each $A\in \mathcal {R}^{loc}$ there exists a vector $x_A\in X$, such that ${ \langle x_A,x^*\rangle =\int _A f d\langle \nu ,x^*\rangle , \ \forall \ x^*\in X^*}$. The subset of all $\nu$-integrable functions is a closed subspace of $L^1_w(\nu )$ and it will be denoted by $L^1(\nu )$. We indicate by $S(\mathcal {R})$ the collection of simple functions in $L^0(\mathcal {R}^{loc})$ which have support in $\mathcal {R}$. It turns out that $L^1(\nu )$, with norm $\Vert \cdot \Vert _\nu$ is a $\sigma$-order continuous $\mu$-B.f.s. where $S(\mathcal {R})$ is a dense subspace.

We also notice that $L^1(\nu )=L^1_w(\nu )$ if X does not contain a copy of $c_0$ [18, Thm. 5.1].

3 Associate space

Let $(\Omega ,\Sigma ,\mu )$ be a positive measure space and let us consider a $\mu$-B.f.s. E. We will show that several of the basic results about the associate space of E when $\mu$ is a $\sigma$-finite measure, can also be established in the case that $\mu$ is not $\sigma$-finite. We will begin by just enunciating some of these results; they can be proven as in the $\sigma$-finite case [28, Ch. 15], [2, Ch. 1].

The vector space defined by $E^\times :=\{g\in L^0(\mu ): gf\in L^1(\mu ), \ \forall \ f\in E \}$ is called the associate space of E and the function

$$\begin{aligned} \Vert g\Vert _{E^\times }:=\sup \left\{ \int _\Omega |gf|d\mu : f\in B_E\right\} , \quad \forall \ g\in E^\times , \end{aligned}$$

(3.1)

is a seminorm. For each $f\in E$ and $g\in E^\times$ the Hölder inequality is satisfied:

$$\begin{aligned} {\int _\Omega |gf|d\mu \le \Vert g\Vert _{E^\times }\Vert f\Vert _E}. \end{aligned}$$

We also have that for the function $\Vert \cdot \Vert _{E^\times }$ to be a norm, it is necessary and sufficient that E be saturated, that is, for each $A\in \Sigma$ with positive measure there exists $B\in \Sigma _A$ such that $\mu (B)>0$ and $\chi _B\in E$. Next we give an equivalent condition for saturation. For this, let us first recall that a $\mu$-B.f.s. Y is order dense in $L^0(\mu )$ if for any $f\in L^0(\mu )^+$ there exists an upwards directed system $\{f_\tau \}\subset Y^+$ such that $f_\tau \uparrow f$.

Lemma 3.1

Let E a $\mu$-B.f.s. The following statements are equivalent:

• (i) The space E is saturated.

• (ii) The space E is order dense in $L^0(\mu )$.

• (iii) The seminorm $\Vert \cdot \Vert _{E^\times }$ is a norm.

Proof

The equivalence (i) $\Leftrightarrow$ (iii) is proved as in the $\sigma$-finite case [28, Ch. 15, §69, Thm. 4]. Let us prove (i) $\Leftrightarrow$ (ii).

Since E is a Banach lattice we have that E is archimedean. Then, it is enough to prove that E is saturated if and only if for each $0\ne f\in L^0(\mu )$ there exists $g\in E$ such that $0<|g|\le |f|$ [20, Thm. 22.3(vi)]. Assume that E is saturated. Consider $0\ne f\in L^0(\mu )$ and define

$$\begin{aligned} A_n:=\left\{ x\in \Omega :\frac{1}{n}\le |f(x)|\right\} , \quad \forall \ n\in \mathbb {N}. \end{aligned}$$

Let us fix $N\in \mathbb {N}$ such that $\mu (A_N)>0$. Since E is saturated there exists $B\subset A_N$ with $\mu (B)>0$ and $\chi _B\in E$. Then, $g:=\frac{1}{N}\chi _B\in E$ and $0<|g|\le |f|$.

To establish the other implication, take $A\in \Sigma$ such that $\mu (A)>0$. Hence $0\ne \chi _A\in L^0(\mu )$. Thus there exists $g\in E$ satisfying $0<|g|\le \chi _A$. As $0\ne g\in L^0(\mu )$ we can take $\varphi \in S(\Sigma )$ with $0<\varphi \le |g|$. Therefore $\varphi \in E$ and so, there exists $B\in \Sigma$ such that $B\subset \mathrm{supp}\varphi$, $\mu (B)>0$ and $\chi _B\in E$. $\square$

Henceforth we will assume that E is a saturated $\mu$-B.f.s. Then as in the $\sigma$-finite case we have:

Proposition 3.2

The space $E^\times$ is a $\mu$-B.f.s. with the $\sigma$-Fatou property.

For our next result, let us recall that a Banach lattice E is super Dedekind complete if every non-empty subset D of E which is bounded from above has a supremum and it contains an at most countable subset possessing the same supremum as D.

Proposition 3.3

If there exists a $\sigma$-order continuous $\mu$-B.f.s. F with the $\sigma$-Fatou property, such that $E^\times \subset F$, then $E^\times$ has the Fatou property.

Proof

Consider $\{g_\tau \}\subset E^\times$ such that $0\le g_\tau \uparrow$ and $\sup _\tau \Vert g_\tau \Vert _{E^\times }<\infty$. Then $\{g_\tau \}\subset F^+$ is an upwards directed system such that $\sup _\tau \Vert g_\tau \Vert _F<\infty$. Since F is $\sigma$-order continuous and has the $\sigma$-Fatou property, then F has the Fatou property and is super Dedekind complete [29, Thm. 113.4]. Hence $g:=\sup _\tau g_\tau \in F$ and there exists a sequence $\{g_{\tau _n}\}\subset \{g_\tau \}$ such that $g_{\tau _n}\uparrow g$ [20, Thm. 23.2.(iii)]. From the $\sigma$-Fatou property in $E^\times$ we obtain that $g\in E^\times$. And since $\Vert \cdot \Vert _E^\times$ is a lattice norm, $\sup _\tau \Vert g_\tau \Vert _{E^\times }\le \Vert g\Vert _{E^\times }$.

Now take $f\in B_E$. Using the monotone convergence theorem and the Hölder inequality we have

$$\begin{aligned} \int _\Omega |gf|d\mu =\sup _n\int _\Omega |g_{\tau _n} f|d\mu \le \sup _n\Vert g_{\tau _n}\Vert _{E^\times }\Vert f\Vert _E\le \sup _n\Vert g_\tau \Vert _{E^\times }. \end{aligned}$$

Thus, $\Vert g\Vert _{E^\times }\le \sup _\tau \Vert g_\tau \Vert _{E^\times }<\infty$. Hence, $E^\times$ has the Fatou property. $\square$

Since $L^1(\mu )$ is $\sigma$-order continuous and has the $\sigma$-Fatou property, we obtain:

Corollary 3.4

If $\chi _\Omega \in E$, then $E^\times$ has the Fatou property.

Next we will show that when $\mu$ is $\sigma$-finite, it turns out that the associate space always has the Fatou property. For this it is necessary to make before a brief discussion.

Let A be in $\Sigma$. We denote by $\mu _A$ the restriction of the measure $\mu$ to the $\sigma$-algebra $\Sigma _A$ formed by the measurable subsets of A. Thus $(A,\Sigma _A,\mu _A)$ is a measure space.

For each $f\in L^0(\Sigma _A)$ define the function $f^\Omega :\Omega \rightarrow \mathbb {K}$ by $f^\Omega (x)=f(x)$, if $x\in A$ and $f^\Omega (x)=0$ otherwise. Then $f^\Omega$ is a $\Sigma$-measurable function which is called canonical extension of f. Now, the set $E_A$ defined by

$$\begin{aligned} E_A:=\left\{ f\in L^0(\mu _A): f^\Omega \in E\right\} \end{aligned}$$

(3.2)

is a vector space. If $h\in E$, then $(h_A)^\Omega =h\chi _A\in E$, where $h_A$ is the restriction of h to A. Thus, $h_A\in E_A$. In $E_A$ we define the norm $\Vert \cdot \Vert _A$ by

$$\begin{aligned} \Vert f\Vert _A:=\Vert f^\Omega \Vert _E, \quad \forall \ f\in E_A. \end{aligned}$$

(3.3)

Since E is a saturated $\mu$-B.f.s., it follows that also $E_A$ is a saturated $\mu _A$-B.f.s. On the other hand if ${A=\bigcup _{n=1}^\infty A_n}$, where $A_n\in \Sigma$ and $\mu (A_n)<\infty$, $\forall \ n\in \mathbb {N}$, we obtain that $\mu _A$ is $\sigma$-finite. In this case it is well known that $E_A^\times :=(E_A)^\times$ is saturated [28, Ch. 15, §71, Thm. 4]. Furthermore $E^\times _A= (E^\times )_A$ and

$$\begin{aligned} {\Vert g\Vert _{E_A^\times }=\sup \left\{ \int _A|gf|d\mu _A:f\in B_{E_A}\right\} }=\Vert g^\Omega \Vert _{E^\times }, \quad \forall \ g\in E_A^\times . \end{aligned}$$

(3.4)

Theorem 3.5

If the measure $\mu$ is $\sigma$-finite, then $E^\times$ has the Fatou property.

Proof

We will assume that $\mu (\Omega )>0$. Let us take an upwards directed system $\{g_\tau \}_{\tau \in I}\subset E^\times$ such that $g_\tau \ge 0$, $\forall \ \tau \in I$ and $\sup _{\tau }\Vert g_\tau \Vert _{E^\times }<\infty$. Now since $\mu$ is $\sigma$-finite and E is a saturated $\mu$-B-f.s., there exist $\{\Omega _n\}\subset \Sigma$ and $N\in \mathcal {N}_0(\mu )$ such that $\Omega _n\subset \Omega _{n+1}$, $0\ne \chi _{\Omega _n}\in E$, $\forall \ n\in \mathbb {N}$ and ${\Omega =\bigcup _{n=1}^\infty \Omega _n \cup N}$ [28, Ch. 15, §67, Thm. 4].

Fix $n\in \mathbb {N}$. Let us denote by $\Sigma _n$ the $\sigma$-algebra $\Sigma _{\Omega _n}$ and by $\mu _n$ the restriction of $\mu$ to $\Sigma _n$. Thus $(\Omega _n,\Sigma _n,\mu _n)$ is a finite measure space. Then the space $E_n:=E_{\Omega _n}$ with the norm $\Vert \cdot \Vert _n:=\Vert \cdot \Vert _{\Omega _n}$ is a saturated $\mu _n$-B.f.s. such that $\chi _{\Omega _n}\in E_n$. By the above corollary we have that $E^\times _n$ has the Fatou property.

Let $g_{\tau ,n}$ be the restriction of $g_\tau$ to the set $\Omega _n$. Thus $\{g_{\tau ,n}\}_{\tau \in I}\subset E^\times _n$ is an upwards directed system with $\sup _{\tau }\Vert g_{\tau ,n}\Vert _{E^\times _n}\le \sup _{\tau }\Vert g_\tau \Vert _{E^\times }<\infty$. Therefore $g_{(n)}:=\sup _{\tau }g_{\tau ,n}\in E^\times _n$ and $\Vert g_{(n)}\Vert _{E^\times _n}=\sup _{\tau }\Vert g_{\tau ,n}\Vert _{E^\times _n}$.

Now let $g_n:=(g_{(n)})^\Omega$ be the canonical extension of $g_{(n)}$. Then $\{g_n\}\subset E^\times$ is an increasing sequence. By (3.4),

$$\begin{aligned} {\sup _n\Vert g_n\Vert _{E^\times }=\sup _n\Vert g_{(n)}\Vert _{E^\times _n}\le \sup _{\tau }\Vert g_\tau \Vert _{E^\times }}. \end{aligned}$$

(3.5)

From the $\sigma$-Fatou property in $E^\times$ and (3.5), it follows that

$$\begin{aligned} {g:=\sup _n g_n\in E^\times } \text{ and } {\Vert g\Vert _{E^\times }\le \sup _{\tau }\Vert g_\tau \Vert _{E^\times }}. \end{aligned}$$

(3.6)

Let us prove that $g=\sup _\tau g_\tau$. Fix $\tau \in I$. Since $g_\tau \chi _{\Omega _n}\le g_n\le g$ for each $n\in \mathbb {N}$, we have that $g_\tau \le g$. Suppose that $g'\in E^\times$ satisfies $g_\tau \le g'$, $\forall \ \tau \in I$. Now fix $n\in \mathbb {N}$. Then $g_\tau \chi _{\Omega _n}\le g'\chi _{\Omega _n}$, $\forall \ \tau$, so $g_n\le g'$. Therefore, $g\le g'$ and we obtain that $g=\sup _\tau g_\tau$. Finally $\Vert \cdot \Vert _E^\times$ is a lattice norm and so the conclusion follows from (3.6). $\square$

Since we are assuming that E is a saturated $\mu$-B.f.s., then $E^\times$ is a $\mu$-B.f.s. and we can consider its associate space, which is called second associate space of E and it is denoted by $E^{\times \times }$. Thus

$$\begin{aligned} E^{\times \times }:=(E^\times )^\times =\left\{ h\in L^0(\mu ): hg\in L^1(\mu )\quad \forall \ g\in E^\times \right\} \end{aligned}$$

and the seminorm $\Vert \cdot \Vert _{E^{\times \times }}:E^{\times \times }\rightarrow [0,\infty )$ is given by

$$\begin{aligned} \Vert h\Vert _{E^{\times \times }}:=\sup \left\{ \int _\Omega |hg|d\mu : g\in B_{E^\times }\right\} . \end{aligned}$$

Unlike the $\sigma$-finite case, we will see in Example 4.5 that it can happen that $E^\times$ is not saturated. Nevertheless, we have that $E\subset E^{\times \times }$ and

$$\begin{aligned} \Vert f\Vert _{E^{\times \times }}\le \Vert f\Vert _E \quad \forall \ f\in E. \end{aligned}$$

(3.7)

Corollary 3.6

Let E be a saturated $\mu$-B.f.s. If $\mu$ is $\sigma$-finite, then

• (i) $E^{\times \times }$ is a $\mu$-B.f.s. with the Fatou property.

• (ii) E has the $\sigma$-Fatou property if and only if E has the Fatou property.

Proof

Since $\mu$ is $\sigma$-finite we have that $E^\times$ is a saturated $\mu$-B.f.s. [28, Ch. 15, §71, Thm. 4]. Thus, from Theorem 3.5 we obtain (i).

Now from the $\sigma$-Fatou property in E it follows that $E\equiv E^{\times \times }$ [28, Ch. 15, §71, Thm. 1]. Therefore from (i) we have (ii). $\square$

Let us fix $g\in E^\times$. Then the function $\varphi _g:E\rightarrow \mathbb {K}$ defined by

$$\begin{aligned} {\varphi _g(f):=\int _\Omega gfd\mu }, \end{aligned}$$

(3.8)

is a linear and bounded functional such that $\Vert \varphi _g\Vert =\Vert g\Vert _{E^\times }$. Thus we consider the operator

$$\begin{aligned} R:E^\times \rightarrow E^* \text{ defined } \text{ by } R(g):=\varphi _g. \end{aligned}$$

(3.9)

Clearly R is a linear isometry, called canonical isometry. Accordingly, the associate space $E^\times$ can be identified with a certain closed subspace of $E^*$. The canonical isometry also preserves the order in the sense that

$g \ge 0$ if, and only if, $\varphi _g \ge 0$.

In the case $\mathbb {K}=\mathbb {C}$ we also have that g is real if, and only if, $\varphi _g$ is real. Therefore if R is onto, then R is an order isometry. Hence in what follows we will write $E^\times = E^*$ to mean that the canonical isometry R is onto.

Next we distinguish two necessary conditions for $E^\times =E^*$ to hold. We will need the following result, which is obtained from [29, Thm. 102.3, p. 415].

Lemma 3.7

If E is a Banach lattice, then $E^*$ has the Fatou property.

We also need to recall that a functional $\varphi \in E^*$ is $\sigma$ -order continuous whenever $f_n\downarrow 0$ implies $\varphi (f_n)\rightarrow 0$.

Proposition 3.8

Let E be a saturated $\mu$-B.f.s. If $E^*=E^\times$, then E is order continuous and $E^\times$ has the Fatou property.

Proof

Since E is a Dedekind $\sigma$-complete Banach lattice, we only have to show that E is $\sigma$-order continuous. And so, by [29, Lemma 84.1, Thm. 102.7] it is enough to establish that $\varphi$ is $\sigma$-order continuous for any $\varphi \in (E^*)^+$. So, take $\varphi \in E^*$ such that $\varphi \ge 0$ and consider $\{f_n\}\subset E$ satisfying that $f_n\downarrow 0$. Since $\varphi$ is positive, there exists $g\in (E^\times )^+$ such that

$$\begin{aligned} \varphi (f)=\int _\Omega gf d\mu , \quad \forall \ f\in E. \end{aligned}$$

Then $\{gf_n\}\subset L^1(\mu )$ is a decreasing sequence such that $0\le gf_n$. Since the space $L^1(\mu )$ is Dedekind $\sigma$-complete, there exists $0\le h=\inf _n gf_n$. Let $A:=\mathrm{supp}g$. It is clear that $h\chi _{\Omega \setminus A}=0$. Taking $\frac{0}{0}:=0$, we have that

$$\begin{aligned} \frac{h\chi _A}{g}\le f_n\chi _A\le f_n, \quad \forall \ n\in \mathbb {N}. \end{aligned}$$

It follows that $\frac{h\chi _A}{g}\in E$ and from $f_n\downarrow 0$ we have that $\frac{h\chi _A}{g}\le 0$. Hence $h\chi _A=0$ and then $gf_n\downarrow 0$. Therefore ${ \varphi (f_n)=\int _\Omega gf_n d\mu \downarrow 0}$. So $\varphi$ is $\sigma$-order continuous.

The other affirmation follows from the above lemma. $\square$

4 Locally $\sigma$-finite measure defined on a $\delta$-ring

Let us assume now that the $\sigma$-algebra $\Sigma$ that we have been considering is given as $\Sigma =\mathcal {R}^{loc}$, where $\mathcal {R}$ is a $\delta$-ring and $\mu = |\lambda |$, where $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ is a measure on $\mathcal {R}$. Noting that for any $A\in \mathcal {R}^{loc}$ with $|\lambda |(A)>0$ we can find $B\in \mathcal {R}_A$ with $\lambda (B)>0$, next we give a simple sufficient condition for E to be saturated.

Lemma 4.1

Let E be a $|\lambda |$-B.f.s. If $S(\mathcal {R})\subset E$, then E is saturated.

Remark 4.2

When $S(\mathcal {R})\subset E$, the space E is a B.f.s. with respect to $(\Omega ,\mathcal {R},\lambda )$ in the sense introduced by Delgado in [9, Def. 3.1]. Thus these class of spaces are always saturated.

Let $\nu :\mathcal {R}\rightarrow X$ be a vector measure having $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ as a local control measure. For $1\le p<\infty$, the spaces $L^p_w(\nu )$ and $L^p(\nu )$ are defined by

$$\begin{aligned} L^p_w(\nu ):=\left\{ f\in L^0(\nu ):|f|^p\in L^1_w(\nu )\right\} \quad \text{ and }\quad L^p(\nu ) :=\left\{ f\in L^0(\nu ): |f|^p\in L^1(\nu )\right\} . \end{aligned}$$

Each function in $L^p_w(\nu )$ is called weakly p -integrable with respect to $\nu$ and each function in $L^p(\nu )$ is called p -integrable with respect to $\nu$. Note that $L^p(\nu )\subset L^p_w(\nu )$. Moreover, $L^p_w(\nu )$ and $L^p(\nu )$ are $|\lambda |$-B.f.s. with norm

$$\begin{aligned} \Vert f\Vert _{p,\nu }:=\Vert |f|^p\Vert _\nu ^\frac{1}{p}=\sup _{x^*\in B_{X^*}}\left( \int _\Omega |f|^pd|\langle \nu ,x^*\rangle |\right) ^{\frac{1}{p}}, \quad \forall \ f\in L^p_w(\nu ). \end{aligned}$$

Also $S(\mathcal {R})$ is a dense subspace of $L^p(\nu )$, the space $L^p(\nu )$ is $\sigma$-order continuous and $L^p_w(\nu )$ has the $\sigma$-Fatou property [17, p. 37].

From the above lemma and Theorem 3.5 we obtain:

Proposition 4.3

Let $1\le p<\infty$. Then $L^p(\nu )$ is saturated. Thus, $L^p(\nu )^\times$, with norm $\Vert \cdot \Vert _{\nu ^\times }:=\Vert \cdot \Vert _{L^p(\nu )^\times }$, is a $|\lambda |$-B.f.s. with the $\sigma$-Fatou property. If in addition the measure $\lambda$ is $\sigma$-finite, then $L^p(\nu )^\times$ has the Fatou property.

Remark 4.4

Since $L^1(\nu )$ is always saturated with respect to any local control measure for $\nu$, by Lemma 3.1 we have that $L^1(\nu )$ is order dense in $L^0(|\lambda |)$. Using other methods, this result was established by Calabuig et al. [4, 4.2].

Example 4.5

Given a vector measure $\nu$, let us consider its variation $|\nu |$ as a local control measure. Then $L^1(\nu )^\times$ is a $|\nu |$-B.f.s. with the $\sigma$-Fatou property. It may happen that the range of $|\nu |$ is $\{0,\infty \}$. For instance, if $\Sigma$ is the Lebesgue $\sigma$-algebra on [0, 1], then the function $\nu :\Sigma \rightarrow L^2([0,1])$ defined by $\nu (A):=\chi _A$ is a vector measure whose range is $\{0,\infty \}$ [5, p. 57]. In this case $L^1(|\nu |)=\{0\}$ and so $L^1(\nu )^\times =\{g\in L^0(|\nu |): gf=0, \ \forall \ f\in L^1(\nu )\}=\{0\}$. Thus clearly the space $L^1(\nu )^\times$ is not saturated. Then in this situation the study of the associate space will not give interesting information.

As we have just seen, when the measure involved is not $\sigma$-finite the associate space is not necessarily saturated. This motivates to look for a class of measures for which this problem does not occur. In this direction, let us recall the following definition, introduced by Brooks and Dinculeanu [3, p. 162].

Definition 4.6

A measure $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ is locally $\sigma$ -finite, if for each $B\in \mathcal {R}$, there exists $\{B_n\}_{n\in \mathbb {N}}\subset \mathcal {R}$ such that $B={\bigcup _{n=1}^\infty B_n}$ and $\lambda (B_n)<\infty$, $\forall$ $n\in \mathbb {N}$.

Clearly any positive $\sigma$-finite measure on a $\sigma$-algebra is locally $\sigma$-finite.

Example 4.7

Let us consider an uncountable set $\Gamma$. Let $\mathcal {R}:=\{B\subset \Gamma : B \text{ is } \text{ finite }\}$ and $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ be the counting measure. Then $\mathcal {R}$ is a $\delta$-ring and $\lambda$ is a locally $\sigma$-finite measure which is not $\sigma$-finite.

Remark 4.8

Consider a locally $\sigma$-finite measure $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ and a $|\lambda |$-B.f.s. E. Let us take $A\in \mathcal {R}^{loc}$ and assume that

$$\begin{aligned} {A:=\bigcup _{n=1}^\infty B_n\cup N} \text{ with } \{B_n\}\subset \mathcal {R}, \ \lambda (B_n)<\infty , \quad \forall \ n\in \mathbb {N} \text{ and } N\in \mathcal {N}_0(|\lambda |). \end{aligned}$$

(4.1)

Hence $(A,(\mathcal {R}_A)^{loc},|\lambda _A|)$ is a $\sigma$-finite measure space, where $(\mathcal {R}_A)^{loc}$ is the $\sigma$-algebra related to $\mathcal {R}_A$ and $\lambda _A$ is the restriction of $\lambda$ to $\mathcal {R}_A$. It follows that the space $E_A$ defined in (3.2), with norm $\Vert \cdot \Vert _A$, is a saturated $|\lambda _A|$-B.f.s. As in this case $|\lambda _A|$ is $\sigma$-finite, we have that $E_A^\times$, with norm $\Vert \cdot \Vert _{E^\times _A}$, is a saturated $|\lambda _A|$-B.f.s.

Note that if $B\in \mathcal {R}$, then B has the form (4.1) and in this case $\mathcal {R}_B=(\mathcal {R}_B)^{loc}$. Hence $(B,\mathcal {R}_B,\lambda _B)$ is a $\sigma$-finite measure space.

We now show that the problem of having nonsaturated associate spaces does not appear when we work with a locally $\sigma$-finite measure.

Theorem 4.9

Let E be a saturated $|\lambda |$-B.f.s. If the measure $\lambda$ is locally $\sigma$-finite, then $E^\times$ is a saturated $|\lambda |$-B.f.s.

Proof

Let $A\in \mathcal {R}^{loc}$ be such that $|\lambda |(A)>0$ and consider $B\in \mathcal {R}_A$ satisfying $\lambda (B)>0$. Since $E_B$, defined in (3.2), is a saturated $\lambda _B$-B.f.s. and $\lambda _B(B)>0$ there exists $C\in \mathcal {R}_B$ with $0<\lambda _B(C)=\lambda (C)$ and $\chi _C\in E_B^\times$. Now take $f\in E$. Then $f_B\in E_B$ and ${\int _\Omega |f|\chi _Cd|\lambda |=\int _B|f_B|\chi _Cd|\lambda _B|<\infty }$. Hence $\chi _C\in E^\times$. $\square$

Hereafter, to the condition that E be a saturated $|\lambda |$-B.f.s. we will add that of $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ being always a locally $\sigma$-finite measure and sometimes we will omit it explicitly. In the $\sigma$-finite case it is well known that $E\equiv E^{\times \times }$ when E has the $\sigma$-Fatou property. By using this fact we will establish the corresponding result in the more general context we are discussing.

Theorem 4.10

If E has the Fatou property, then $E\equiv E^{\times \times }$.

Proof

As we have that $E\subset E^{\times \times }$ and $\Vert f\Vert _{E^{\times \times }}\le \Vert f\Vert _E$, $\forall \ f\in E$, it only rests to prove the another contention and the other norm inequality. For this, it is enough to establish the conclusion only for non-negative functions.

Let $0\le f\in E^{\times \times }$. Fix $B\in \mathcal {R}$. Since E has the Fatou property we have that $E_B$ is also a $\lambda _B$-B.f.s. with the Fatou property. Moreover, as established in Remark 4.8, $\lambda _B$ is a $\sigma$-finite measure. Then we obtain $E_B\equiv E^{\times \times }_B$ [28, Ch. 15, §71, Thm. 1]. Denote by $f_B$ the restriction of f to B. Hence $f_B\in E_B$ and $\Vert f_B\Vert _B=\Vert f_B\Vert _{E^{\times \times }_B}$. Noting that $(f_B)^\Omega =f\chi _B$, we have $f\chi _B\in E$ and $\Vert f\chi _B\Vert _E=\Vert f\chi _B\Vert _{E^{\times \times }}$.

On the other hand, as $\mathcal {R}$ is a directed set with the order given by $B\le C \ \text{ if, } \ B\subset C, \ \forall B,C\in \mathcal {R}$, we can consider the net $\{f\chi _B\}_{B\in \mathcal {R}}\subset E$. Then $\{f\chi _B\}_{B\in \mathcal {R}}$ is an upwards directed system and $\Vert f\chi _B\Vert _E=\Vert f\chi _B\Vert _{E^{\times \times }}\le \Vert f\Vert _{E^{\times \times }}$, $\forall \ B\in \mathcal {R}$. Since E has the Fatou property, there exists $h\in E\subset E^{\times \times }$ with ${h=\sup _{B\in \mathcal {R}} f\chi _B}$ and

$$\begin{aligned} {\Vert h\Vert _E=\sup _{B\in \mathcal {R}}\Vert f\chi _B\Vert _E=\sup _{B\in \mathcal {R}}\Vert f\chi _B\Vert _{E^{\times \times }}\le \Vert f\Vert _{E^{\times \times }}}. \end{aligned}$$

(4.2)

Assume that there exists $A\in \mathcal {R}^{loc}$ such that $h\chi _A<f\chi _A$ and $|\lambda |(A)>0$. Then for some $B\in \mathcal {R}_A$ with positive measure we have that $h\chi _B<f\chi _B$, which is a contradiction and it follows that $f\in E$. Moreover, as $f\chi _B\le f$, $\forall \ B\in \mathcal {R}$ we have that $h\le f$, $\lambda$-a.e. Therefore $h=f$, $|\lambda |$-a.e. The remaining inequality between the norms follows from (4.2). $\square$

It is well known that if $\mu$ is a $\sigma$-finite measure and E is a $\mu$-B.f.s., then we can write $L^1 (\mu ) = \{f g: f \in E, g \in E^\times \}$. Next we will show that this result remains valid when we consider a locally $\sigma$-finite measure.

Proposition 4.11

Let $\lambda$ be a locally $\sigma$-finite measure and E be a saturated $|\lambda |$-B.f.s. If $h\in L^1(\lambda )$:

• (i) then for each $\varepsilon >0$ there exist $f\in E$ and $g\in E^\times$ such that

  $$\begin{aligned} h=fg \quad \text{ and } \quad {\Vert f\Vert _E\Vert g\Vert _{E^\times }\le (1+\varepsilon )\int _\Omega |h|d|\lambda |.} \end{aligned}$$

• (ii) if in addition E has the $\sigma$-Fatou property, then there exist $f\in E$ and $g\in E^\times$ such that

  $$\begin{aligned} h=fg \quad \text{ and } \quad {\Vert f\Vert _E\Vert g\Vert _{E^\times }=\int _\Omega |h|d|\lambda |.} \end{aligned}$$

Proof

If $h=0$, the conclusion is clear. Assume that $h\ne 0$. Since $h\in L^1(\lambda )$, we have that ${A:=\mathrm{supp}h=\bigcup _{n=1}^\infty B_n\cup N}$, where $\{B_n\}\subset \mathcal {R}$ and $N\in \mathcal {N}_0(\lambda )$. As $\lambda$ is a locally $\sigma$-finite measure we can assume that $\lambda (B_n)<\infty$, $\forall \ n\in \mathbb {N}$. Then $\lambda _A$ is $\sigma$-finite and $h_A\in L^1(\lambda _A)$.

(i) Given $\varepsilon >0$, [15, Thm. 1, (ii)] there exist $\tilde{f}\in E_A$ and $\tilde{g}\in E^\times _A$ such that

$$\begin{aligned} h_A=\tilde{f}\tilde{g} \quad \text{ and } \quad {\Vert \tilde{f}\Vert _{E_A}\Vert \tilde{g}\Vert _{E^\times _A}\le (1+\varepsilon )\int _A|h_A|d|\lambda _A|.} \end{aligned}$$

(ii) Since E has the $\sigma$-Fatou property it follows that $E_A$ also has it. From [15, Thm. 1i)] we get $\tilde{f}\in E_A$ and $\tilde{g}\in E^\times _A$ such that

$$\begin{aligned} h_A=\tilde{f}\tilde{g} \quad \text{ and } \quad {\Vert \tilde{f}\Vert _{E_A}\Vert \tilde{g}\Vert _{E^\times _A}=\int _A|h_A|d|\lambda _A|.} \end{aligned}$$

As $h=h_A\chi _A$, by taking $f:=\tilde{f}^\Omega \in E$ and $g:=\tilde{g}^\Omega \in E^\times$, the conclusion follows. $\square$

In the $\sigma$-finite case we know that $E^\times =E^*$ if, and only if, E is $\sigma$-order continuous [28, Ch. 15, §72, Thm. 5]. We proved in Proposition 3.8 that if $E^\times =E^*$, then E is $\sigma$-order continuous and $E^\times$ has the Fatou property. Now we will show that the converse also is valid in our context. For this let us recall that an ideal Y of a Banach lattice X is a band whenever, for every subset D of Y possessing a supremum in X, this supremum is already in Y.

Theorem 4.12

If E is $\sigma$-order continuous, then the following properties are equivalent:

• (i) $E^\times =E^*$.

• (ii) $E^\times$ is a band of $E^*$.

• (iii) $E^\times$ has the Fatou property.

• (iv) $E^\times$ has the weak Fatou property.

Proof

The implications (i) $\Rightarrow$ (ii) and (iii) $\Rightarrow$ (iv) are clear. (ii) $\Rightarrow$ (iii) Let $\{g_\tau \}\subset E^\times$ be an upwards directed system such that $\sup _\tau \Vert g_\tau \Vert _{E^\times }<\infty$. Let us take $\varphi _\tau :=R(g_\tau )$. Then $\{\varphi _\tau \}\subset E^*$ is an upwards directed system such that $\sup _\tau \Vert \varphi _\tau \Vert <\infty$. From Lemma 3.7 we have that $E^*$ has the Fatou property. Thus there exists $\varphi \in E^*$ such that $\varphi _\tau \uparrow \varphi$ and $\Vert \varphi \Vert =\sup _\tau \Vert \varphi _\tau \Vert$. Now since $\{\varphi _\tau \}\subset R(E^\times )$ and $R(E^\times )$ is a band of $E^*$ we have that $\varphi \in R(E^\times )$. Let $g\in E^\times$ be such that $\varphi =R(g)$. Since R is an order isometry $g_\tau \uparrow g$ and $\Vert g\Vert _{E^\times }=\sup _\tau \Vert g_\tau \Vert _{E^\times }$.

(iv) $\Rightarrow$ (i) Now let us assume that $E^\times$ has the weak Fatou property. First note that since $E^*$ is a Banach lattice and R is a linear operator it is enough to represent only the positive functionals. Take $0\le \varphi \in E^*$. Consider the $\delta$-ring $\mathcal {R}(E):=\{B\in \mathcal {R}: \chi _B\in E\}$ and define $m:\mathcal {R}(E)\rightarrow [0,\infty )$ by $m(B):=\varphi (\chi _B), \ \forall \ B\in \mathcal {R}(E)$. Since E is $\sigma$-order continuous and $\varphi$ is a positive linear functional we have that m is a positive measure.

Fix $B\in \mathcal {R}(E)$. Then $\mathcal {R}_B=\mathcal {R}(E)_B$. Let us denote by $m_B$ the restriction of m to $\mathcal {R}_B$.Then $m_B$ is bounded. On the other hand as $\lambda$ is a locally $\sigma$-finite measure, its restriction $\lambda _B$ to $\mathcal {R}_B$ is a positive $\sigma$-finite measure.

Now let $A\in \mathcal {R}_B$ with $\lambda _B(A)=|\lambda |(A)=0$, then $\chi _A=0$, $\lambda$-a.e. Hence $m_B(A)=0$. By the Radon–Nikodym Theorem [26, p. 121] there exists a unique $h_B\in L^0(\mathcal {R}_B)$ such that

$$\begin{aligned} \varphi (\chi _A)=m_B(A)=\int _A h_Bd\lambda _B=\int _B h_B\chi _A d\lambda _B, \quad \forall \ A\in \mathcal {R}_B. \end{aligned}$$

Using a standard procedure it follows now that

$$\begin{aligned} \varphi (f)=\int _B h_Bfd\lambda _B, \quad \forall \ f\in E_B. \end{aligned}$$

(4.3)

Let us denote by $H_B$ the canonical extension of $h_B$, then $H_B\in E^\times$, $\varphi _B(f):=\varphi (f\chi _B)=\int _\Omega fH_Bd|\lambda |$, $\forall \ f\in E$ and, by the uniqueness of $h_B$ it turns out that $H_B\chi _C=H_{B\cap C}, \ \forall \ C\in \mathcal {R}$. Noting that $\varphi _B\le \varphi$ we have that

$$\begin{aligned} \Vert H_B\Vert _{E^\times }=\Vert \varphi _B\Vert \le \Vert \varphi \Vert <\infty . \end{aligned}$$

This shows that $\sup _{B\in \mathcal {R}(E)}\Vert H_B\Vert _{E^\times }<\infty$. It turns out that $\{H_B\}_{B\in \mathcal {R}(E)}\subset E^\times$ is an upwards directed system. Now since $E^\times$ has the weak Fatou property, there exists ${h=\sup _{B\in \mathcal {R}(E)} H_B\in E^\times }$.

Let $f\in E^+$. To prove that $\varphi (f)=\int _\Omega f h d|\lambda |$ first we will show that $H_B=h\chi _B, \ \forall \ B\in \mathcal {R}(E)$. Fix $B\in \mathcal {R}(E)$. It is clear that $H_B\le h\chi _B$. Let us assume that $H_B<h\chi _B$, so we can take $C\in \mathcal {R}_B$ with positive measure such that $H_B(t)<h\chi _B(f), \ \forall \ t\in C$. Define $k=h\chi _{\Omega \setminus C}+H_B\chi _C$. Now let $D\in \mathcal {R}(E)$, then

$$\begin{aligned} H_D=H_D\chi _{\Omega \setminus C}+H_{D\cap C}\le h\chi _{\Omega \setminus C}+H_B\chi _C=k, \end{aligned}$$

that is, k is an upper bound of $\{H_B\}_{B\in \mathcal {R}(E)}$ which contradicts that h is the supremum. We conclude that $H_B=h\chi _B$.

Since $fh\in L^1(|\lambda |)$, it follows that $A:=\mathrm{supp}fh=\bigcup _{n=1}^\infty B_n\cup N$ where $\{B_n\}$ is a disjoint family of subsets of $\mathcal {R}$ and N is a $|\lambda |$-null set. Observe that since $\lambda _B$ is $\sigma$-finite and $E_B$ is saturated, $\forall \ B\in \mathcal {R}$, by [28, Ch. 15, §67, Thm. 4] we can take $B_n\in \mathcal {R}(E)$. Since E is order continuous, it follows that $f \chi _A = \sum _{n=1}^\infty f \chi _{B_n}$ in E. Assume that $\varphi (f\chi _{\Omega \setminus A})>0$. Then $f\chi _{\Omega \setminus A}>0$, so we can choose $B\in \mathcal {R}(E)_{\Omega \setminus A}$ such that $|\lambda |(B)>0$ and $f\chi _B>0$. Take its corresponding $H_B\in E^\times$. So, $0<\varphi (f\chi _B)=\int _\Omega fH_B d|\lambda |=\int _\Omega fh\chi _B d|\lambda |$, but $B\subset \Omega \setminus \mathrm{supp}fh$, thus $\int _\Omega fh\chi _B d|\lambda |=0$, which is a contradiction. Therefore $\varphi (f\chi _{\Omega \setminus A})=0$ and so

$$\begin{aligned} \varphi (f)= & {} {\sum _{n=1}^\infty \varphi (f\chi _{B_n})=\sum _{n=1}^\infty \int _\Omega fH_Bd|\lambda |} \\= & {} {\sum _{n=1}^\infty \int _\Omega fh\chi _Bd|\lambda |=\int _\Omega \sum _{n=1}^\infty f\chi _Bhd|\lambda |=\int _\Omega fhd|\lambda |}. \end{aligned}$$

This show that $\varphi =\varphi _h\in R(L^1(\nu )^\times )$ and hence the conclusion follows. $\square$

Although the above theorem characterizes when $E^\times =E^*$, up to now we do not know if $E^\times$ always has the Fatou property. Next we present a situation where this holds. The proof follows from Theorem 4.12 and Proposition 3.3.

Corollary 4.13

Let $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ a locally $\sigma$-finite measure and E be a saturated $\lambda$-B.f.s. If E and $E^\times$ are $\sigma$-order continuous, then $E^\times =E^*$.

Using Theorem 4.12 we can give a characterization of reflexivity as follows.

Theorem 4.14

Let $\lambda$ be a locally $\sigma$-finite measure and E a saturated $\lambda$-B.f.s. If E is order continuous then the space E is reflexive if, and only if, $E=E^{\times \times }$ and $E^\times$ is $\sigma$-order continuous.

Proof

Let us denote by $R_1:E^\times \rightarrow E^*$ and $R_2:E^{\times \times }\rightarrow {E^\times }^*$ the corresponding canonical isometries. Then for the adjoint operator of $R_1$ we have $R_1^*:E^{**}\rightarrow {E^\times }^*$. First assume that $E^\times$ is $\sigma$-order continuous and $E=E^{\times \times }$. Then by Corollary 4.13 $R_1$ is onto and by hypothesis $E^{\times \times }=E$. So $E^{\times \times }$ is $\sigma$-order continuous and has the $\sigma$-Fatou property. It follows that $E^{\times \times }$ has the Fatou property and so we can apply Theorem 4.12 to obtain that $R_2$ is onto. Let us see that $R_1^*j=R_2$ where $j:E\rightarrow E^{**}$ is the canonical injection. Take $f\in E$ and $g\in E^\times$, then

$$\begin{aligned} \langle g,R_1^*j(f)\rangle =\langle R_1(g),j(f)\rangle =\langle f,R_1(g)\rangle ={\int _\Omega fgd|\lambda |}=\langle g,R_2(f)\rangle . \end{aligned}$$

Therefore j is onto, that is, E is reflexive.

Now assume that E is reflexive. To establish that $E=E^{\times \times }$ it only rests to prove that $E^{\times \times }\subset E$. Take $h\in E^{\times \times }$. Since $R_1$ is an injective linear operator with closed range, it follows that $R_1^*$ is onto [11, Thm. VI.6.2]. Hence there exists $\varphi \in E^{**}$ such that $R_1^*(\varphi )=R_2(h)$. Let $f\in E$ satisfy $j(f)=\varphi$. Thus for $g\in E^\times$ we have

$$\begin{aligned} \langle g,R_1^*(\varphi )\rangle =\langle R_1(g), \varphi \rangle =\langle R_1(g),j(f)\rangle =\langle f,R_1(g)\rangle ={\int _\Omega fgd|\lambda |}. \end{aligned}$$

Hence, ${\int _\Omega fgd|\lambda |=\langle g,R_2(h)\rangle =\int _\Omega hgd|\lambda |}$. Then $h=f$, $|\lambda |$-c.t.p. and $h \in E$.

We will now prove that $R_2$ is onto, then by Proposition 3.8 we will obtain that $E^\times$ is order continuous. Consider $\varphi \in L^1(\nu )^{**}$, then $\varphi \circ R_1^{-1}:R_1(E^\times )\rightarrow \mathbb {K}$ is linear and bounded. By the Hahn–Banach Theorem there exists $\widetilde{\varphi }\in E^{**}$ such that $\langle \psi ,\widetilde{\varphi }\rangle =\langle \psi ,\varphi \circ R_1^{-1}\rangle$, $\forall \ \varphi \in R_1(E^\times )$. Let $f\in E=E^{\times \times }$ be satisfy $j(f)=\widetilde{\varphi }$. Then for each $g\in E^\times$ we have

$$\begin{aligned} \langle g,R_2(f)\rangle= & {} \int _\Omega fgd|\lambda |=\langle f,R_1(g)\rangle = \langle R_1(g),j(f)\rangle \\= & {} \langle R_1(g),\widetilde{\varphi }\rangle =\left\langle R_1(g), \varphi \circ R_1^{-1}\right\rangle = \langle g,\varphi \rangle \end{aligned}$$

It follows that $\varphi =R_2(f)$ and we conclude that $R_2$ is onto. $\square$

Proposition 4.15

Let $\lambda$ be a locally $\sigma$-finite measure and E a saturated $|\lambda |$-B.f.s. If $E^\times$ has the weak Fatou property, then $E^\times$ is a band of $E^*$.

Proof

To prove that $R(E^\times )$ is an ideal of $E^*$ we can proceed as in implication (iv) $\Rightarrow$ (i) of Theorem 4.12 only observing that if $0\le \varphi \le \varphi _g\in R(E^\times )$, then the set function $m_\varphi :\mathcal {R}(E)\rightarrow [0,\infty )$, defined by $m_\varphi (B)=\varphi (\chi _B)$ is a positive measure. Now let $A\subset E^\times$ be a non empty set such that there exists ${\varphi :=\sup _{g\in A} \varphi _g\in E^*}$. We have to prove that $\varphi \in R(E^\times )$.

Let us note that $\mathcal {F}:=\{F\subset A: F~\mathrm{is~finite}\}$ is a directed set with the order given by $F_1\le F_2$ if $F_1\subset F_2$. For each $F\in \mathcal {F}$ define ${\varphi _F:=\max _{g\in F} \varphi _g}$. Let us fix $F_0\in \mathcal {F}$ and take $\mathcal {F}_0:=\{F\in \mathcal {F}: F_0\subset F\}$. It turns out that ${\sup _{F\in \mathcal {F}_0}\varphi _F=\varphi }$. Then $0\le \varphi _F-\varphi _{F_0}\le \varphi -\varphi _{F_0}$, $\forall \ F\in \mathcal {F}_0$. Thus $\{\varphi _F-\varphi _{F_0}\}_{\mathcal {F}_0}$ is an upwards directed system such that ${\sup _{F\in \mathcal {F}_0}\Vert \varphi _F-\varphi _{F_0}\Vert <\infty }$. Since $\{\varphi _F-\varphi _{F_0}\}_{\mathcal {F}_0}\subset R(E^\times )$ and R is an order isometry we obtain an upwards directed system $\{g_F\}_{\mathcal {F}_0}\subset {E^\times }^+$ with ${\sup _{F\in \mathcal {F}_0}\Vert g_F\Vert _{E^\times }}<\infty$. By the weak Fatou property in $E^\times$, there exists $g\in E^\times$ such that $g_F\uparrow g$. Using again that R is an order isometry we have that $\varphi _F-\varphi _{F_0}\uparrow \varphi _g$. Since ${\sup _{F\in \mathcal {F}_0}\varphi _F=\varphi }$ we have that $\varphi -\varphi _{F_0}=\varphi _g$. Therefore $\varphi \in R(E^\times )$. $\square$

The following result was established in [29, p. 418]. We obtain it as consequence of the above proposition and Theorem 3.5.

Corollary 4.16

Let E be a saturated $|\lambda |$-B.f.s. If $|\lambda |$ is $\sigma$-finite, then $E^\times$ is a band of $E^*$.

5 Brooks–Dinculeanu measure

Let $\nu :\mathcal {R}\rightarrow X$ be a vector measure defined on a $\delta$-ring. Since we are interested in providing a representation of the dual space of $L^1(\nu )$ as its associate space, it is important to know if $\nu$ has a local control measure which is locally $\sigma$-finite. Then, by Theorem 4.9, the associate space of $L^1(\nu )$, with respect to this local control measure, will be saturated. Let us distinguish this kind of measures.

Definition 5.1

A measure $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ is a Brooks–Dinculeanu measure for $\nu$, if $\lambda$ is a local control measure for $\nu$ which is locally $\sigma$-finite.

Example 5.2

• 1. Let $\nu :\Sigma \rightarrow X$ be a vector measure defined on a $\sigma$-algebra. If $\mu :\Sigma \rightarrow [0,\infty )$ is a Rybakov control measure for $\nu$, then $\mu$ is a Brooks–Dinculeanu measure for $\nu$.

• 2. Let $\nu :\mathcal {R}\rightarrow X$ be a $\sigma$-finite vector measure. Then $\nu$ has a bounded local control measure $\lambda :\mathcal {R}\rightarrow [0,\infty )$ [8, Thm. 3.3]. Hence, $\lambda$ is a Brooks–Dinculeanu for $\nu$.

Fortunately, it turns out that each vector measure defined in a $\delta$-ring has a Brooks–Dinculeanu measure. This result was established by Jiménez Fernández et al. in [16, p. 3]. Given its importance, we will state it below.

Theorem 5.3

If $\nu :\mathcal {R}\rightarrow X$ is a vector measure, then $\nu$ has a Brooks–Dinculeanu measure.

Let us define

$$\begin{aligned} \widehat{\mathcal {R}}:=\{B\in \mathcal {R}: \lambda (B)<\infty \}. \end{aligned}$$

It is clear that $\widehat{\mathcal {R}}$ is a $\delta$-ring satisfying that $\widehat{\mathcal {R}}\subset \mathcal {R}$. Moreover, it turns out that $\mathcal {R}^{loc}=\widehat{\mathcal {R}}^{loc}$. Now let us show that

$$\begin{aligned} |\langle \nu ,x^*\rangle |=|\langle \widehat{\nu },x^*\rangle |, \quad \forall \ x^*\in X^*, \end{aligned}$$

(5.1)

where $\widehat{\nu }$ is the restriction of $\nu$ to $\widehat{\mathcal {R}}$. By definition we obtain that $|\langle \widehat{\nu },x^*\rangle |\le |\langle \nu ,x^*\rangle |$. To establish the other inequality let us fix $x^*\in X^*$ and consider $A\in \mathcal {R}^{loc}$. Take $B\in \mathcal {R}_A$. Since $\lambda$ is locally $\sigma$-finite, there exists an increasing sequence, $\{B_n\}\subset \widehat{\mathcal {R}}$ such that ${B=\bigcup _{n=1}^\infty B_n}$. So,

$$\begin{aligned} {|\langle \nu ,x^*\rangle |(B)}= & {} {\sup _n|\langle \nu ,x^*\rangle |(B_n) =\sup _n|\langle \widehat{\nu },x^*\rangle |(B_n)} \\\le & {} {\sup _{C\in \widehat{\mathcal {R}}_A}|\langle \widehat{\nu },x^*\rangle |(C)=|\langle \widehat{\nu },x^*\rangle |(A).} \end{aligned}$$

It follows that $|\langle \nu ,x^*\rangle |(A)\le |\langle \widehat{\nu },x^*\rangle |(A)$. Hence we have established (5.1).

From (5.1) we have that

$$\begin{aligned} \Vert f\Vert _\nu =\Vert f\Vert _{\widehat{\nu }}, \quad \forall \ f\in L^0(\mathcal {R}^{loc}). \end{aligned}$$

(5.2)

Thus $L^1_w(\nu )\equiv L^1_w(\widehat{\nu })$. Since (5.2) is valid, from the density of $S(\mathcal {R})$ in $L^1(\nu )$ and of $S(\widehat{\mathcal {R}})$ in $L^1(\widehat{\nu })$, to prove that $L^1(\nu )\equiv L^1(\widehat{\nu })$, it is sufficient to check that $S(\mathcal {R})\subset L^1(\widehat{\nu })$ and that $S(\widehat{\mathcal {R}})\subset L^1(\nu )$. Noting that $\widehat{\mathcal {R}}\subset \mathcal {R}$ the second contention is clear. Now consider $B\in \mathcal {R}$ and take $\{B_n\}\subset \widehat{\mathcal {R}}$ satisfying that $B_n\subset B_{n+1}$ and ${B=\bigcup _{n=1}^\infty B_n}$. Then $\chi _{B_n}\rightarrow \chi _B$, moreover

$$\begin{aligned} \int _A \chi _{B_n}d\widehat{\nu }=\widehat{\nu }(B_n\cap A)=\nu (B_n\cap A)=\int _A\chi _{B_n}d\nu \rightarrow \int _A\chi _Bd\nu , \quad \forall \ A\in \mathcal {R}^{loc}. \end{aligned}$$

From [8, Prop. 2.3] we have that $\chi _B\in L^1(\widehat{\nu })$. It follows that $S(\mathcal {R})\subset L^1(\widehat{\nu })$ and $I_\nu (s)=I_{\widehat{\nu }}(s)$, $\forall \ s\in S(\mathcal {R})$. By the continuity of the integration operators we have $I_\nu =I_{\widehat{\nu }}$. Therefore, we have proven the following result and so, whenever we find it convenient we can work on the $\delta$-ring $\widehat{\mathcal {R}}$ instead of $\mathcal {R}$.

Lemma 5.4

If $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ is a Brooks–Dinculeanu measure for a given vector measure $\nu$, then

• (i) for each $x^*\in X^*$ we have that $|\langle \nu ,x^*\rangle |=|\langle \widehat{\nu },x^*\rangle |$,

• (ii) $L^1_w(\nu )\equiv L^1_w(\widehat{\nu })$, $L^1(\nu )\equiv L^1(\widehat{\nu })$ and ${\int _\Omega fd\nu =\int _\Omega fd\widehat{\nu }}$, $\forall \ f\in L^1(\nu )$.

Curbera and Ricker established that $L^p(\nu )^{\times \times }\equiv L^p_w(\nu )$ when a vector measure defined on a $\sigma$-algebra and a Rybakov control measure are considered [7, Prop. 2]. We will show that this equality remains true if we consider instead a vector measure defined on a $\delta$-ring and a Brooks–Dinculeanu measure. Before it is necessary to establish a useful characterization for the functions in $L^1_w(\nu )$.

Lemma 5.5

Let $f\in L^0(\mathcal {R}^{loc})$. Then $f\in L^1_w(\nu )$ if and only if for each $B\in \mathcal {R}$, $f\chi _B\in L^1_w(\nu )$ and ${\sup _{B\in \mathcal {R}}\Vert f\chi _B\Vert _\nu <\infty }$. In this case ${\Vert f\Vert _\nu =\sup _{B\in \mathcal {R}}\Vert f\chi _B\Vert _\nu }$.

Proof

First let us assume that $f\in L^1_w(\nu )$. Since $L^1_w(\nu )$ is a Banach lattice we have that $f\chi _B\in L^1_w(\nu )$ and $\Vert f\chi _B\Vert _\nu \le \Vert f\Vert _\nu$, $\forall$ $B\in \mathcal {R}$. Then, ${\sup _{B\in \mathcal {R}}\Vert f\chi _B\Vert _\nu \le \Vert f\Vert _\nu }<\infty$.

Now assume that $f\chi _B\in L^1_w(\nu )$, $\forall$ $B\in \mathcal {R}$ and ${M:=\sup _{B\in \mathcal {R}}\Vert f\chi _B\Vert _\nu <\infty }$. Let $x^*\in B_{X^*}$, then

$$\begin{aligned} {\sup _{B\in \mathcal {R}}\int _B |f|d|\langle x^*,\nu \rangle |\le M}. \end{aligned}$$

From Proposition 2.1 we obtain that $f\in L^1(|\langle x^*,\nu \rangle |)$, $\forall$ $x^*\in B_{X^*}$. Thus $f\in L^1_w(\nu )$ and $\Vert f\Vert _\nu \le M$. $\square$

Although most of the time we will not state it explicitly, in what follows $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ will be a Brooks–Dinculeanu measure for a given vector measure $\nu$ and we will consider $L^p_w(\nu )$ and $L^p(\nu )$ as Banach function spaces with respect to $|\lambda |$.

Theorem 5.6

Let $1\le p<\infty$. Then $L^p(\nu )^{\times \times }\equiv L^p_w(\nu )$.

Proof

First we prove that $L^p_w(\nu )\subset L^p(\nu )^{\times \times }$ and $\Vert f\Vert _{p,\nu ^{\times \times }}\le \Vert f\Vert _{p,\nu }$, $\forall \ f\in L^p_w(\nu )$. Let $\varphi \in S(\mathcal {R}^{loc})$ be such that $\varphi \in L^p_w(\nu )$ and $B\in \widehat{\mathcal {R}}$. Then $\varphi \chi _B\in S(\mathcal {R})\subset L^p(\nu )$. By the Hölder inequality, for each $g\in L^p(\nu )^\times$

$$\begin{aligned} {\int _B|g\varphi |d|\lambda |\le \Vert g\Vert _{p,\nu ^\times }\Vert \varphi \chi _B\Vert _{p,\nu }\le \Vert g\Vert _{p,\nu ^\times }\Vert \varphi \Vert _{p,\nu }}. \end{aligned}$$

From Proposition 2.1 we have

$$\begin{aligned} {\int _\Omega |g\varphi |d|\lambda |=\sup _{B\in \widehat{\mathcal {R}}}\int _B|g\varphi |d|\lambda |\le \Vert g\Vert _{p,\nu ^\times }\Vert \varphi \Vert _{p,\nu }}. \end{aligned}$$

It follows that $\varphi \in L^p(\nu )^{\times \times }$ and

$$\begin{aligned} \Vert \varphi \Vert _{p,\nu ^{\times \times }}\le \Vert \varphi \Vert _{p,\nu }. \end{aligned}$$

(5.3)

Now consider $f\in L^p_w(\nu )$ and take $\{\varphi _n\}\subset S(\mathcal {R}^{loc})$ with $0\le \varphi _n\uparrow |f|$. Then, $\{\varphi _n\}\subset L^p_w(\nu )$. By (5.3), $\Vert \varphi _n\Vert _{p,\nu ^{\times \times }}\le \Vert \varphi _n\Vert _{p,\nu }\le \Vert f\Vert _{p,\nu }$, $\forall$ $n\in \mathbb {N}$. Since $L^p(\nu )^{\times \times }$ has the $\sigma$-Fatou property it turns out that $f\in L^p(\nu )^{\times \times }$ and $\Vert f\Vert _{p,\nu ^{\times \times }}\le \Vert f\Vert _{p,\nu }$.

For the other contention let us fix $B\in \mathcal {R}$ and let $\nu _B$ be the restriction of $\nu$ to the $\sigma$-algebra $\mathcal {R}_B$. As $L^1(\nu _B)\equiv L^1(\nu )_B$, it follows that $L^p(\nu _B)\equiv L^p(\nu )_B$. Hence we obtain that $L^p(\nu _B)^{\times \times }\equiv L^p_w(\nu _B)$ [7, Prop. 2].

Take $f\in L^p(\nu )^{\times \times }$ and let us denote by $f_B$ its restriction to B, then $f_B\in L^p_w(\nu _B)$ and $\Vert f_B\Vert _{p,\nu _B}=\Vert f_B\Vert _{p,\nu ^{\times \times }_B}\le \Vert f\Vert _{p,\nu ^{\times \times }}$. And so, for each $x^*\in B_{X^*}$

$$\begin{aligned} {\int _B|f|^pd|\langle \nu ,x^*\rangle |\le \Vert f\Vert _{p,\nu ^{\times \times }}^p}. \end{aligned}$$

From the above proposition we have $|f|^p\in L^1_w(\nu )$ and $\Vert |f|^p\Vert _\nu \le \Vert f\Vert ^p_{p,\nu ^{\times \times }}$. Hence $f\in L^p_w(\nu )$ and $\Vert f\Vert _{p,\nu }\le \Vert f\Vert _{p,\nu ^{\times \times }}$. $\square$

The following result was established in [4, p. 77] by other methods, we obtain it as consequence of the above proposition and Corollary 3.6.

Corollary 5.7

Let $1\le p<\infty$. If $\nu :\mathcal {R}\rightarrow X$ is a $\sigma$-finite vector measure, then $L^p_w(\nu )$ has the Fatou property.

From Theorem 5.6 and Proposition 3.3 we obtain:

Corollary 5.8

If $L^p_w(\nu )\subset L^1(\lambda )$, $1\le p<\infty$, then $L^p_w(\nu )$ has the Fatou property.

The sufficiency in the next result was proven in [4, Prop. 5.4]. Since $L^1(\nu )$ is $\sigma$-order continuous we obtain it from Theorems 4.10 and 5.6.

Corollary 5.9

$L^1(\nu )$ has the Fatou property if, and only if, $L^1(\nu )= L^1_w(\nu )$.

As in the $\sigma$-finite case, we have the next result.

Lemma 5.10

If E and F are $\mu$-B.f.s such that $E\subset F$ and there exists $a>0$ with $\Vert f\Vert _F\le a\Vert f\Vert _E$, $\forall \ f\in E$, then $F^\times \subset E^\times$ and

$$\begin{aligned} \Vert g\Vert _{E^\times }\le a\Vert g\Vert _{F^\times }, \quad \forall \ g\in F^\times . \end{aligned}$$

Corollary 5.11

$L^p(\nu )^\times \equiv L^p_w(\nu )^\times$ and $L^p_w(\nu )^{\times \times } \equiv L^p_w (\nu )$, $1\le p<\infty$.

Proof

From the above lemma, we have $L^p_w(\nu )^\times \subset L^p(\nu )^\times$ and $\Vert g\Vert _{p,\nu ^\times }\le \Vert g\Vert _{L^p_w(\nu )^\times }$, $\forall \ g\in L^p_w(\nu )^\times$. Now consider $f\in L^p_w(\nu )$ and $g\in L^p(\nu )^\times$, from the Hölder inequality and Theorem 5.6 we have that

$$\begin{aligned} {\int _\Omega |gf|d|\lambda |\le \Vert g\Vert _{p,\nu ^\times }\Vert f\Vert _{p,\nu ^{\times \times }}=\Vert g\Vert _{p,\nu ^\times }\Vert f\Vert _{p,\nu }.} \end{aligned}$$

Hence $g\in L^p_w(\nu )^\times$ and $\Vert g\Vert _{L^p_w(\nu )^\times }\le \Vert g\Vert _{p,\nu ^\times }$.

The second equality follows from Theorem 5.6. $\square$

In Proposition 3.5 we have seen that the associate space of a $\mu$-B.f.s. has the Fatou property when $\mu$ is $\sigma$-finite. We will show that this result remains true for certain Brooks–Dinculeanu measures, introduced by Calabuig et al. [4, p. 77].

Definition 5.12

A vector measure $\nu$ is $\mathcal {R}$ -decomposable if we can write $\Omega =\bigcup _{\alpha \in \Delta }\Omega _\alpha \cup N$, where $N\in \mathcal {N}_0(\nu )$ and ${\{\Omega _\alpha \}_{\alpha \in \Delta }\subset \mathcal {R}}$ is a family of pairwise disjoint sets satisfying that

• (a) if $A_\alpha \in \mathcal {R}_{\Omega _\alpha }$, $\forall$ $\alpha \in \Delta$, then ${\bigcup _{\alpha \in \Delta }A_\alpha \in \mathcal {R}^{loc}}$, and

• (b) if $x^*\in X^*$ and $N_\alpha \in \mathcal {N}_0(\langle \nu ,x^*\rangle )$, $\forall$ $\alpha \in \Delta$, then ${\bigcup _{\alpha \in \Delta }}N_\alpha \in \mathcal {N}_0(\langle \nu ,x^*\rangle )$.

Note that if $\nu$ is an $\mathcal {R}$-decomposable vector measure and $A\in \mathcal {R}^{loc}$ is such that $A\cap \Omega _\alpha \in \mathcal {N}_0(\nu )$, $\forall \ \alpha \in \Delta$, then $A\cap \Omega _\alpha$ is $\langle \nu ,x^*\rangle$-null, $\forall \ \alpha \in \Delta$ and $\forall \ x^*\in B_{X^*}$. From b) in the above definition it follows that A is $\nu$-null.

Some examples of $\mathcal {R}$-decomposable measures are the $\sigma$-finite vector measures and the discrete vector measures [4, Lemma 4.6, p. 77]. However there are $\mathcal {R}$-decomposable measures which are neither $\sigma$-finite nor discrete [4, p. 85].

Proposition 5.13

Let $\nu :\mathcal {R}\rightarrow X$ be a vector measure, $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ be a Brooks–Dinculeanu measure for $\nu$ and E be $|\lambda |$-B.f.s. If $S(\mathcal {R})\subset E$ and $\nu$ is $\mathcal {R}$-decomposable, then $E^\times$ has the Fatou property.

Proof

Since $\nu$ is $\mathcal {R}$-decomposable, ${\Omega =\bigcup _{\alpha \in \Delta }\Omega _\alpha \cup N}$, where $N\in \mathcal {N}_0(\nu )$ and ${\{\Omega _\alpha \}_{\alpha \in \Delta }\subset \mathcal {R}}$ is a family of pairwise disjoint sets satisfying (a) and (b) in Definition 5.12. Moreover, since $\lambda$ is locally $\sigma$-finite we can consider that $\lambda (\Omega _\alpha )<\infty$, $\forall \alpha \in \Delta$. Let us note that by Lemma 4.1, E is a saturated $|\lambda |$-e.f.B.

Let $I\subset \Delta$ be a countable set and take ${\Omega _I:=\bigcup _{\alpha \in I}\Omega _\alpha }$, $\mathcal {R}_I:=\mathcal {R}_{\Omega _I}$ and $\lambda _I$ the restriction of $\lambda$ to the $\delta$-ring $\mathcal {R}_I$. Then $(\Omega _I,(\mathcal {R}_I)^{loc},|\lambda _I|)$ is a $\sigma$-finite measure space and $E_I:=E_{\Omega _I}$ is a $|\lambda _I|$-B.f.s. Then, from Theorem 3.5, we obtain that $E^\times _I$ is a $|\lambda _I|$-B.f.s. whit the Fatou property.

Now let us consider an upwards directed system $\{g_\tau \}_{\tau \in K}\subset E^\times$ such that $g_\tau \ge 0$, $\forall \ \tau \in K$ and $M:=\sup _{\tau }\Vert g_\tau \Vert _{E^\times }<\infty$. Denoting by $g_{\tau ,I}$ to the restriction of $g_\tau$ to $\Omega _I$, we have that $\{g_{\tau ,I}\}_{\tau \in K}\subset E^\times _I$ is an upwards directed system, and from (3.4), $\sup _\tau {\Vert g_{\tau ,I}\Vert _{E^\times _I}\le M<\infty }$. Since $E^\times _I$ has the Fatou property, it turns out that $g_I:=\sup _{\tau }g_{\tau ,I}\in E^\times _I$ and

$$\begin{aligned} {\Vert g_I\Vert _{E^\times _I}=\sup _{\tau }\Vert g_{\tau ,I}\Vert _{E^\times _I}\le M}. \end{aligned}$$

(5.4)

In particular, for each $\alpha \in \Delta$ exists $g_{\{\alpha \}}\in E^\times _{\{\alpha \}}$ such that ${g_{\{\alpha \}}=\sup _{\tau }g_{\tau ,\{\alpha \}}}$ and

$$\begin{aligned} {\Vert g_{\{\alpha \}}\Vert _{E^\times _{\{\alpha \}}}=\sup _{\tau }\Vert g_{\tau ,\{\alpha \}}\Vert _{E^\times _{\{\alpha \}}}}. \end{aligned}$$

Let us denote by $g^\alpha$ the canonical extension of $g_{\{\alpha \}}$ and define ${g:=\sum _{\alpha \in \Delta }g^\alpha }$. As $\mathcal {N}_0(\nu )=\mathcal {N}_0(|\lambda |)$ and from (a) in Definition 5.12 we have that $g\in L^0(|\lambda |)$.

Let us prove that $g=\sup _\tau g_\tau$. Consider $\alpha \in \Delta$ and $\tau \in K$. Since $g_\tau \chi _{\Omega _\alpha }\le g^\alpha \le g$, it follows that $g_\tau \le g$. Let us assume that $g'\in L^0(|\lambda |)$ is such that $g_\tau \le g'$. Then $g_\tau \chi _{\Omega _\alpha }\le g'\chi _{\Omega _\alpha }$. So, $g^\alpha \le g'$. Hence $g\le g'$ and $g=\sup _\tau g_\tau$.

Finally we will establish that $g\in E^\times$. Let us fix $f\in B_E$ and let $I\subset \Delta$ be a countable set. Note that the canonical extension of $g_I=\sup _{\tau }g_{\tau ,I}$ is given by ${g^I=\sum _{\alpha \in I} g^\alpha }$ and $f_I\in B_{E_I}$, where $f_I$ is the restriction of f to $\Omega _I$. Moreover, ${\int _\Omega |g^I f|d|\lambda |=\int _{\Omega _I}|g_I f_I|d|\lambda _I|}$. By using the monotone convergence theorem

$$\begin{aligned} {\sum _{\alpha \in I}\int _\Omega |g^\alpha f|d|\lambda |=\int _\Omega |g^If|d|\lambda |=\int _{\Omega _I}|g_I f_I|d|\lambda _I|\le \Vert g_I\Vert _{E^\times _I}}. \end{aligned}$$

From this and (5.4) we obtain that ${\sum _{\alpha \in I}\int _\Omega |g^\alpha f|d\lambda |\le M}$, for each finite subset I of $\Delta$. Then, there exists a countable set $J\subset \Delta$ such that ${\int _\Omega |g^\alpha f|d|\lambda |=0}$, $\forall \ \alpha \in \Delta \setminus J$. This implies that

$$\begin{aligned} {\int _\Omega |gf|d|\lambda |=\sum _{\alpha \in J}\int _\Omega |g^\alpha f|d|\lambda |=\int _\Omega |g^Jf|d|\lambda |\le M}. \end{aligned}$$

(5.5)

We conclude that $g\in E^\times$; moreover, from lattice property of the norm in $E^\times$ and from (5.5), we have that $\Vert g\Vert _{E^\times }=\sup _\tau \Vert g_\tau \Vert _{E^\times }$. $\square$

Let us consider the canonical isometry R between $L^1(\nu )^\times$ and $L^1(\nu )^*$. When $\nu$ is a $\sigma$-finite vector measure, then $\nu$ has a bounded local control measure $\lambda :\mathcal {R}\rightarrow [0,\infty )$ [8, Thm. 3.3]. Since $L^1(\nu )$ is a $\sigma$-order continuous $|\lambda |$-B.f.s., then we have $L^1(\nu )^*=L^1(\nu )^\times$ [28, Ch. 15, §72, Thm. 5]. In what follows we will present other situations where this holds.

Since $L^1(\nu )$ is an order continuous $|\lambda |$-B.f.s. from Theorem 4.12 we obtain the following result.

Corollary 5.14

The following properties are equivalent:

• (i) $L^1(\nu )^\times =L^1(\nu )^*$.

• (ii) $L^1(\nu )^\times$ is a band of $L^1(\nu )^*$.

• (iii) $L^1(\nu )^\times$ has the Fatou property.

• (iv) $L^1(\nu )^\times$ has the weak Fatou property.

The next result is a consequence of Proposition 5.13 and the previous result.

Corollary 5.15

If $\nu$ is $\mathcal {R}$-decomposable, then $L^1(\nu )^\times =L^1(\nu )^*$.

If E is a real $\sigma$-order continuous Banach lattice it is well known that there exists an $\mathcal {R}$-decomposable vector measure $\nu :\mathcal {R}\rightarrow E$ such that E is order isometric to the space $L^1(\nu )$ [10, Thm. 5]. Then from the above corollary we obtain the following result.

Corollary 5.16

If E is a $\sigma$-order continuous Banach lattice, then there exist an $\mathcal {R}$- decomposable vector measure $\nu :\mathcal {R}\rightarrow E$ and a order isometry from $L^1(\nu )^\times$ onto $E^*$. More precisely, if T is a lattice isometry from E onto $L^1(\nu )$, $\lambda :\mathcal {R}\rightarrow [0,\infty ]$ is a Brooks–Dinculeanu measure for $\nu$ and $\varphi \in E^*$, then there exists $g\in L^1(\nu )^\times$ such that

$$\begin{aligned} \varphi (f)=\int _\Omega (Tf)gd|\lambda |, \quad \forall \ f\in E. \end{aligned}$$

Proof

Note that it only rests to verify that the result mentioned before remains valid in the complex case. Since E is a $\sigma$-order continuous Banach lattice, then $E_\mathbb {R}$ is also a $\sigma$-order continuous Banach lattice. Thus, there exist an $\mathcal {R}$-decomposable vector measure $\tilde{\nu }:\mathcal {R}\rightarrow E_\mathbb {R}$ and an onto lattice isometry $S:L^1(\nu )\rightarrow E_\mathbb {R}$. Now let us define $\nu :\mathcal {R}\rightarrow E$, by $\nu (B)=\tilde{\nu }(B), \ \forall \ B\in \mathcal {R}$. It turns out that $\nu$ is an $\mathcal {R}$-decomposable vector measure and $L^1(\nu )_\mathbb {R}=L^1(\tilde{\nu })$. Let $T:L^1(\nu )\rightarrow E$ be the canonical extension of S, then T is an onto lattice isometry [25, Lemma 3.8]. $\square$

As a consequence of Theorem 4.14 we obtain the following result.

Corollary 5.17

$L^1(\nu )$ is reflexive if, and only if, $L^1(\nu )=L^1_w(\nu )$ and $L^1(\nu )^\times$ is $\sigma$-order continuous.

Now from the previous result and Corollary 4.13 we have:

Corollary 5.18

If $L^1(\nu )$ is reflexive, then $L^1(\nu )^\times =L^1(\nu )^*$.

If $1<p<\infty$, Ferrando and Rodríguez established that $L^p(\nu )^*$ is order continuous when $\nu$ is defined on a $\sigma$-algebra [13, Thm 3.1]. Using the same arguments, it follows that in our context we also have that $L^p(\nu )^\times$ is order continuous. Then from Corollary 4.13 we have the following result.

Corollary 5.19

$L^p(\nu )^\times =L^p(\nu )^*$, $1<p<\infty$.

Since $L^p(\nu )^\times$ is order continuous and $L^p(\nu )=L^p_w(\nu )$ if, and only if, $L^1(\nu )=L^1_w(\nu )$ [17, Prop. 3.1.6], the next result follows from Theorems 4.14 and 5.6. It was proven when $\nu$ is defined in a $\sigma$-algebra by Fernández et al. [12, Cor. 3.10].

Corollary 5.20

Let $1 < p < \infty$. Then $L^p(\nu )$ is reflexive if, and only if, $L^1(\nu )= L^1_w(\nu )$.

Let us fix $1<p<\infty$. Then Corollary 5.19 implies that each functional in $L^p(\nu )^*$ has the form $\varphi _g$, $g\in L^p(\nu )^\times$. So we can define $S:L^p(\nu )^{\times \times }\rightarrow L^p(\nu )^{**}$ by

$$\begin{aligned} \langle \varphi _g, S(h)\rangle :=\int _\Omega ghd|\lambda |. \end{aligned}$$

It turns out that S is a linear isometry and we will write $L^p(\nu )^{\times \times }=L^p(\nu )^{**}$ to indicate that is onto. Let $R_1:L^p(\nu )^\times \rightarrow L^p(\nu )^*$ and $R_2:L^p(\nu )^{\times \times }\rightarrow L^p(\nu )^{\times *}$ be the corresponding canonical isometries, then $S=(R_1^*)^{-1}\circ R_2$. Thus S is onto if, and only if, $R_2$ is it. So from Theorems 5.6 and 4.12 we have:

Corollary 5.21

Let $1 < p < \infty$. Then $L_w^p(\nu )$ has the Fatou property if, and only if, $L^p_w(\nu )=L^p(\nu )^{**}$.

Remark 5.22

Calabuig, Delgado, Juan and Sánchez-Pérez asked if in general $L^1_w(\nu )$ always has the Fatou property [4, pp. 77–78]. With respect to this question we have the following. Let $1<p<\infty$ and notice that $L^1_w(\nu )$ has the Fatou property if, and only if, $L^p_w(\nu )$ has it. Then from the previous result we obtain that

$L^1_w(\nu )$ has the Fatou property if, and only if, $L^p_w(\nu )=L^p(\nu )^{**}$.