On the Banach lattice structure of $L^1_w$ of a vector measure on a $\delta$-ring

Abstract

We study some Banach lattice properties of the space $L_w^1(\nu )$ of weakly integrable functions with respect to a vector measure $\nu$ defined on a $\delta$-ring. Namely, we analyze order continuity, order density and Fatou type properties. We will see that the behavior of $L_w^1(\nu )$ differs from the case in which $\nu$ is defined on a $\sigma$-algebra whenever $\nu$ does not satisfy certain local $\sigma$-finiteness property.

1 Introduction

The space of integrable functions with respect to a vector measure finds applications in important problems as, for instance, the representation of abstract Banach lattices as spaces of functions and the study of the optimal domain of linear operators. Classical vector measures $\nu :\Sigma \rightarrow X$ are considered to be defined on a $\sigma$-algebra and with values in a Banach space. The spaces $L^1(\nu )$ and $L_w^1(\nu )$ of integrable and weakly integrable functions respectively have been studied in depth by many authors and their behavior is well understood, (see [7] and [25, Chapter 3]) and the references therein. However, this framework is not enough, for instance, for applications to operators on spaces which do not contain the characteristic functions of sets (see [2, 10, 11]) or Banach lattices without weak unit (see [12]). These cases require $\nu$ to be defined on a weaker structure than $\sigma$-algebra, namely, a $\delta$-ring. Bear in mind the spaces $\ell ^p(\Gamma ),\,1\le p \le \infty ,$ for an uncountable set $\Gamma .$ So, vector measures defined on a $\delta$-ring also play an important role and deserve to be studied together with their spaces of integrable functions. The integration theory with respect to these vector measures $\nu$ goes back to the late sixties (see [14, 18, 21–24]). In [9], there is an analysis of the space $L^1(\nu )$ which gives evidence of how large the difference can be between the $\delta$-ring and $\sigma$-algebra cases. Indeed, for the general case, bounded functions may be not integrable and this fact is crucial.

The aim of this paper is the study of the Banach lattice properties of the space $L_w^1(\nu ).$ The case when these spaces contain $c_0$ becomes specially relevant. This research is a part of a general project of analysis of these abstract integration structures that has already shown to be useful in applications. For instance, a general version of Komlós Theorem on the pointwise convergence of the Cesàro sums of functions have been recently obtained using spaces of vector measure integrable functions on a $\delta$-ring as main tool (see [17]). More applications in the setting of the theory of operators on Banach function spaces can be found in [2, 3]. The relevant case of the Hardy operator has been studied in [11].

More precisely, we study some properties related to order continuity (Sect. 3) and order density (Sect. 4), and some Fatou type properties (Sect. 5). We will see that many properties satisfied for this space when $\nu$ is defined on a $\sigma$-algebra remain true in general only in the case when $\nu$ satisfies certain local $\sigma$-finiteness property, which guarantees that every function in $L_w^1(\nu )$ is the $\nu$-a.e. pointwise limit of a sequence of functions in $L^1(\nu ).$ Also we revisit the representation theorems for abstract Banach lattices (Sect. 6), and we finish with an illustrative example (Sect. 7).

2 Preliminaries

2.1 Banach lattices

Let $E$ be a Banach lattice with norm $\Vert \cdot \Vert$ and order $\le .$ A closed subspace $F$ of $E$ is an ideal of $E$ if $y\in E$ with $|y| \le |x|$ for some $x\in F$ implies $y\in F.$ We say that $E$ is order continuous if for every $(x_\tau )\subset E$ downwards directed system $x_\tau \downarrow 0$ it follows that $\Vert x_\tau \Vert \downarrow 0$ and $E$ is $\sigma$ -order continuous if for every $(x_n)\subset E$ decreasing sequence $x_n\downarrow 0$ it follows that $\Vert x_n\Vert \downarrow 0.$ We denote by $E_{an}$ the order continuous part of $E,$ that is, the largest order continuous ideal in $E.$ It can be described as

$$\begin{aligned} E_{an}=\{x\in E:\, |x|\ge x_\tau \downarrow 0 \, \text{ implies} \, \Vert x_\tau \Vert \downarrow 0 \}. \end{aligned}$$

Similarly, $E_a$ will denote the $\sigma$ -order continuous part of $E,$ that is, the largest $\sigma$-order continuous ideal in $E,$ which can be described as

$$\begin{aligned} E_a=\{x\in E:\, |x|\ge x_n \downarrow 0 \, \text{ implies} \, \Vert x_n \Vert \downarrow 0 \}. \end{aligned}$$

The Banach lattice $E$ is Dedekind complete if every non empty subset which is bounded from above has a supremum and is Dedekind $\sigma$ -complete if every non empty countable subset which is bounded from above has a supremum. We say that $E$ has the Fatou property if for every $(x_\tau )\subset E$ upwards directed system $0\le x_\tau \uparrow$ such that $\sup \Vert x_\tau \Vert <\infty$ it follows that there exists $x=\sup x_\tau$ in $E$ and $\Vert x\Vert =\sup \Vert x_\tau \Vert ,$ and $E$ has the $\sigma$ -Fatou property if for every $(x_n)\subset E$ increasing sequence $0\le x_n\uparrow$ such that $\sup \Vert x_n\Vert <\infty$ it follows that there exists $x=\sup x_n$ in $E$ and $\Vert x\Vert =\sup \Vert x_n\Vert .$ An ideal $F$ in $E$ is said to be order dense if for every $0 \le x\in E$ there exists an upwards directed system $0\le x_\tau \uparrow x$ such that $(x_\tau )\subset F$ and is said to be super order dense if for every $0 \le x\in E$ there exists an increasing sequence $0\le x_n\uparrow x$ such that $(x_n)\subset F.$ A weak unit of $E$ is an element $0\le e\in E$ such that $x\wedge e=0$ implies $x=0.$ Every positive linear operator $T:E\rightarrow F$ between Banach lattices (i.e. $Tx\ge 0$ whenever $0\le x\in E$) is continuous, see [19, p. 2]. An operator $T:E\rightarrow F$ between Banach lattices is said to be an order isometry if it is a linear isometry which is also an order isomorphism, that is, $T$ is linear, one to one, onto, $\Vert Tx\Vert _F=\Vert x\Vert _E$ for all $x\in E$ and $T(x\wedge y)=Tx\wedge Ty$ for all $x, y\in E.$

Let $(\Omega ,\Sigma ,\mu )$ be a measure space (without assumptions of finiteness on $\mu$) and $L^0(\mu )$ be the space of all measurable real functions on $\Omega ,$ where functions which are equal $\mu$-a.e. are identified. Considering the $\mu$-a.e. pointwise order, we have that $L^0(\mu )$ is an Archimedean vector lattice. Note that for $f,f_n\in L^0(\mu ),$ it follows that $0\le f_n\uparrow f\,\mu$-a.e. if and only if $0\le f_n\uparrow f$ in $L^0(\mu ),$ that is, the $\mu$-a.e. pointwise supremum coincides with the lattice supremum. We will simple write $f\le g$ for $f\le g\,\mu$-a.e. By Banach function space (briefly, B.f.s.) related to $\mu$ we mean a Banach space $X\subset L^0(\mu )$ satisfying that if $|f|\le |g|\,\mu$-a.e. with $f\in L^0(\mu )$ and $g\in X$ then $f\in X$ and $\Vert f\Vert _X\le \Vert g\Vert _X.$ Every B.f.s. is a Banach lattice with the $\mu$-a.e. pointwise order, in which convergence in norm of a sequence implies $\mu$-a.e. convergence for some subsequence. Note that for $f,f_n\in X,$ it follows that $0\le f_n\uparrow f\,\mu$-a.e. if and only if $0\le f_n\uparrow f$ in $X.$

These and other issues related to Banach lattices can be found in [20] and [26].

2.2 Integration with respect to vector measures on $\delta$-rings.

Let $\mathcal R$ be a $\delta$ -ring of subsets of an abstract set $\Omega ,$ that is, a ring closed under countable intersections. We write $\mathcal R ^{loc}$ for the $\sigma$-algebra of all subsets $A$ of $\Omega$ such that $A\cap B\in \mathcal R$ for all $B\in \mathcal R .$ Note that if $\mathcal R$ is a $\sigma$-algebra then $\mathcal R ^{loc}=\mathcal R .$ Denote by $\mathcal M (\mathcal R ^{loc})$ the space of all measurable real functions on $(\Omega ,\mathcal R ^{loc}),$ by $\mathcal S (\mathcal R ^{loc})$ the space of all simple functions and by $\mathcal S (\mathcal R )$ the space of all $\mathcal R$ -simple functions (i.e. simple functions supported in $\mathcal R$).

Let $\lambda :\mathcal R \rightarrow \mathbb R$ be a countably additive measure, that is, $\sum \lambda (A_n)$ converges to $\lambda (\cup A_n)$ whenever $(A_n)$ is a sequence of pairwise disjoint sets in $\mathcal R$ with $\cup A_n\in \mathcal R .$ The variation of $\lambda$ is the countably additive measure $|\lambda |:\mathcal R ^{loc}\rightarrow [0,\infty ]$ given by

$$\begin{aligned} |\lambda |(A)=\sup \left\{ \sum |\lambda (A_i)|:\,(A_i) \text{ finite} \text{ disjoint} \text{ sequence} \text{ in} \mathcal R \cap 2^A\right\} . \end{aligned}$$

For every $A\in \mathcal R$ we have that $|\lambda |(A)<\infty .$ The space $L^1(\lambda )$ of integrable functions with respect to $\lambda$ is defined as the space $L^1(|\lambda |)$ with the usual norm. Every $\mathcal R$-simple function $\varphi =\sum _{i=1}^n\alpha _i\chi _{A_i}$ is in $L^1(\lambda )$ and the integral of $\varphi$ with respect to $\lambda$ is defined as usual by $\int \varphi \,d\lambda =\sum _{i=1}^n\alpha _i\lambda (A_i).$ Moreover, the space $\mathcal S (\mathcal R )$ is dense in $L^1(\lambda ).$ For every $f\in L^1(\lambda ),$ the integral of $f$ with respect to $\lambda$ is defined as $\int f\,d\lambda =\lim \int \varphi _n\,d\lambda$ for any sequence $(\varphi _n)\subset \mathcal S (\mathcal R )$ converging to $f$ in $L^1(\lambda ).$

Let $\nu :\mathcal R \rightarrow X$ be a vector measure with values in a real Banach space $X,$ that is, $\sum \nu (A_n)$ converges to $\nu (\cup A_n)$ in $X$ whenever $(A_n)$ is a sequence of pairwise disjoint sets in $\mathcal R$ with $\cup A_n\in \mathcal R .$ Denoting by $X^*$ the topological dual of $X$ and by $B_{X^*}$ the unit ball of $X^*,$ the semivariation of $\nu$ is the map $\Vert \nu \Vert :\mathcal R ^{loc}\rightarrow [0,\infty ]$ given by $\Vert \nu \Vert (A)=\sup \{|x^*\nu |(A) \, : \, x^*\in B_{X^*}\}$ for all $A \in \mathcal R ^{loc},$ where $|x^*\nu |$ is the variation of the measure $x^*\nu :\mathcal R \rightarrow \mathbb R .$ A set $A\in \mathcal R ^{loc}$ is $\nu$ -null if $\Vert \nu \Vert (A)=0,$ or equivalently, $\nu (B)=0$ for all $B\in \mathcal R \cap 2^A.$ A property holds $\nu$ -almost everywhere (briefly, $\nu$-a.e.) if it holds except on a $\nu$-null set. For every $\mathcal R ^{loc}$-measurable function $f:\Omega \rightarrow \mathbb R \cup \{\pm \infty \}$ we can define

$$\begin{aligned} \Vert f\Vert _\nu =\sup _{x^*\in B_{X^*}}\int |f|\,d|x^*\nu |\le \infty . \end{aligned}$$

Note that if $\Vert f\Vert _\nu <\infty$ then $|f|<\infty \,\nu$-a.e. Let $L_w^1(\nu )$ denote the space of functions in $\mathcal M (\mathcal R ^{loc})$ which are integrable with respect to $|x^*\nu |$ for all $x^*\in X^*,$ where functions which are equal $\nu$-a.e. are identified. The space $L_w^1(\nu )$ is a Banach space with the norm $\Vert \cdot \Vert _\nu .$ A function $f\in L_w^1(\nu )$ is integrable with respect to $\nu$ if for each $A\in \mathcal R ^{loc}$ there exists a vector denoted by $\int _Afd\nu \in X,$ such that

$$\begin{aligned} x^*\left(\int _Af\,d\nu \right)=\int \limits _Af\,dx^*\nu \text{ for} \text{ all} x^*\in X^*. \end{aligned}$$

Let $L^1(\nu )$ denote the space of all integrable functions with respect to $\nu .$ Then, $L^1(\nu )$ is a closed subspace of $L_w^1(\nu )$ and so it is a Banach space with the norm $\Vert \cdot \Vert _\nu .$ Moreover, $\mathcal S (\mathcal R )$ is dense in $L^1(\nu ).$ Note that for every $\mathcal R$-simple function $\varphi =\sum _{i=1}^n\alpha _i\chi _{A_i},$ we have that $\int \varphi \,d\nu =\sum _{i=1}^n\alpha _i\nu (A_i).$ From [1, Theorem 3.2], there always exists a measure $\lambda :\mathcal R \rightarrow [0,\infty ]$ with the same null sets as $\nu .$ Then, $L^1(\nu )$ and $L_w^1(\nu )$ are B.f.s. related to $|\lambda |.$ Moreover, $L^1(\nu )$ is order continuous and $L_w^1(\nu )$ has the $\sigma$-Fatou property.

For any measure $\mu :\mathcal R ^{loc}\rightarrow [0,\infty ]$ with the same null sets as $\nu ,$ since the $\mu$-a.e. pointwise order coincides with the $\nu$-a.e. one, we will denote $L^0(\nu )=L^0(\mu )$ and say B.f.s. related to $\nu$ for B.f.s. related to $\mu .$

For these and other issues related to integration with respect to vector measures defined on a $\delta$-ring, see [9, 18, 21, 22].

3 Order continuous part of $L_w^1(\nu )$

All along in this paper $\nu :\mathcal R \rightarrow X$ will be a vector measure defined on a $\delta$-ring $\mathcal R$ of subsets of an abstract set $\Omega ,$ with values in a real Banach space $X.$ Recall that measurable functions are referred to the $\sigma$-algebra $\mathcal R ^{loc}.$

Let us begin by noting that the $\sigma$-order continuous and the order continuous parts of $L_w^1(\nu )$ coincide. Indeed, $L_w^1(\nu )$ is Dedekind $\sigma$-complete as it has the $\sigma$-Fatou property (see [26, Theorem 113.1]), and so, since $\left(L_w^1(\nu )\right)_a$ is an ideal in $L_w^1(\nu ),$ it is also Dedekind $\sigma$-complete. Then, from [26, Theorem 103.6], $\left(L_w^1(\nu )\right)_a$ is order continuous and thus $\left(L_w^1(\nu )\right)_a=\left(L_w^1(\nu )\right)_{an}.$

It was noted in [6, p. 192], that in the case when $\mathcal R$ is a $\sigma$-algebra, the order continuous part of $L_w^1(\nu )$ is just $L^1(\nu ).$ This follows from the facts that $L^1(\nu )$ is order continuous and $\mathcal S (\mathcal R ^{loc})=\mathcal S (\mathcal R )\subset L^1(\nu ).$ In the general case, $\mathcal S (\mathcal R ^{loc})$ may not be in $L^1(\nu ),$ even so, we will see that $\left(L_w^1(\nu )\right)_a=L^1(\nu )$ remains true. First, let us characterize when a characteristic function of a measurable set is in $L^1(\nu ).$

Lemma 3.1

The following statements are equivalent for any $A\in \mathcal R ^{loc}.$

• (a) $\chi _A\in L^1(\nu ).$

• (b) $\Vert \nu \Vert (A_n)\rightarrow 0$ for all decreasing sequences $(A_n)\subset \mathcal R ^{loc}\cap 2^A$ with $\,\cap A_n\,\nu$-null.

• (c) $\nu (A_n)\rightarrow 0$ for all disjoint sequences $(A_n)\subset \mathcal R \cap 2^A.$

Proof

Suppose that $\chi _A\in L^1(\nu )$ and let $(A_n)\subset \mathcal R ^{loc}\cap 2^A$ be a decreasing sequence with $\cap A_n\,\nu$-null. Since $L^1(\nu )$ is order continuous and $\chi _A\!\ge \!\chi _{A_n}\!\downarrow 0,$ then $\Vert \nu \Vert (A_n)\!=\!\Vert \chi _{A_n}\Vert _\nu \rightarrow 0.$ So, (a) implies (b).

Let $(A_n)\subset \mathcal R \cap 2^A$ be a disjoint sequence. Taking $B_n=\cup _{j\ge n}A_j$ we have a decreasing sequence $(B_n)\subset \mathcal R ^{loc}\cap 2^A$ with $\cap B_n=\emptyset$ and $\Vert \nu (A_n)\Vert \le \Vert \nu \Vert (B_n).$ So, (b) implies (c).

Suppose that (c) holds and consider the vector measure $\nu _A:\mathcal R \rightarrow X$ defined by $\nu _A(B)=\nu (A\cap B)$ for all $B\in \mathcal R .$ Noting that $|x^*\nu _A|(B)=|x^*\nu |(A\cap B)$ for every $B\in \mathcal R ^{loc}$ and $x^*\in X^*,$ it can be checked that $\int |f|\,d|x^*\nu _A|=\int |f|\chi _A\,d|x^*\nu |.$ Indeed, this is trivial for simple functions, and for all measurable functions it is consequence of the monotone convergence theorem. Thus, $\Vert f\Vert _{\nu _A}=\Vert f\chi _A\Vert _\nu$ for every $f\in \mathcal M (\mathcal R ^{loc}).$ Then, $f\in L_w^1(\nu _A)$ if and only if $f\chi _A\in L_w^1(\nu ).$ Since $\mathcal S (\mathcal R )$ is dense in both $L^1(\nu )$ and $L^1(\nu _A),$ it follows that $f\in L^1(\nu _A)$ if and only if $f\chi _A\in L^1(\nu ).$ By hypothesis $\nu _A$ is strongly additive, so, from [9, Corollary 3.2.b)], we have that $\chi _\Omega \in L^1(\nu _A)$ and thus $\chi _A\in L^1(\nu ).$ $\square$

Let us prove now the announced result.

Theorem 3.2

The equality $\left(L_w^1(\nu )\right)_a=L^1(\nu )$ holds.

Proof

Obviously $L^1(\nu )\subset \left(L_w^1(\nu )\right)_a$ as $L^1(\nu )$ is order continuous. For the converse inclusion, consider first a set $A\in \mathcal R ^{loc}$ such that $\chi _A\in \left(L_w^1(\nu )\right)_a.$ For every decreasing sequence $(A_n)\subset \mathcal R ^{loc}\cap 2^A$ with $\,\cap A_n\,\nu$-null it follows that $\chi _A\ge \chi _{A_n}\downarrow 0$ and so $\Vert \nu \Vert (A_n)=\Vert \chi _{A_n}\Vert _\nu \rightarrow 0.$ Then we get $\chi _A\in L^1(\nu ),$ from Lemma 3.1.

Consider now $\varphi \in \mathcal S (\mathcal R ^{loc})$ such that $\varphi \in \left(L_w^1(\nu )\right)_a.$ Write $\varphi =\sum _{j=1}^n\alpha _j\chi _{A_j}$ with $(A_j)\subset \mathcal R ^{loc}$ being a disjoint sequence and $\alpha _j\not =0.$ Since $\chi _{A_j}\le |\frac{\varphi }{\alpha _j}|$ and $\left(L_w^1(\nu )\right)_a$ is an ideal, $\chi _{A_j}\in \left(L_w^1(\nu )\right)_a.$ Then, $\chi _{A_j}\in L^1(\nu )$ and so $\varphi \in L^1(\nu ).$

Finally, let $f\in \left(L_w^1(\nu )\right)_a$ and take a sequence $(\varphi _n)\subset \mathcal S (\mathcal R ^{loc})$ satisfying that $0\le \varphi _n\uparrow |f|\,\nu$-a.e. Note that $\varphi _n\in \left(L_w^1(\nu )\right)_a$ as $\varphi _n\le |f|,$ and so $\varphi _n\in L^1(\nu ).$ Since $|f|\ge |f|-\varphi _n\downarrow 0,$ we have that $\Vert \,|f|-\varphi _n\Vert _\nu \rightarrow 0.$ Then, as $L^1(\nu )$ is closed in $L_w^1(\nu ),$ we have that $|f|,$ and so also $f,$ is in $L^1(\nu ).$ $\square$

4 Order density of $L^1(\nu )$ in $L_w^1(\nu )$

The topic of this section is trivial for the case when $\mathcal R$ is a $\sigma$-algebra. Indeed, for each $0\le f\in L^0(\nu )$ there exists $(\varphi _n)\subset \mathcal S (\mathcal R ^{loc})$ such that $0\le \varphi _n\uparrow f\,\nu$-a.e. Since, in this case $\mathcal R ^{loc}=\mathcal R$ and $\mathcal S (\mathcal R )\subset L^1(\nu ),$ obviously we have that $L^1(\nu )$ is super order dense (and so order dense) in $L^0(\nu )$ (and so in $L_w^1(\nu )$). However, this argument fails for the general case as $\mathcal S (\mathcal R ^{loc})$ may not be contained in $L^1(\nu ).$

Example 4.1

Let $\Gamma$ be an uncountable abstract set, $\mathcal R$ the $\delta$-ring of finite subsets of $\Gamma$ and $\nu :\mathcal R \rightarrow c_0(\Gamma )$ the vector measure defined by $\nu (A)=\chi _A$ (see [9, Example 2.2]). Then, $\chi _\Gamma \in L_w^1(\nu )=\ell ^\infty (\Gamma ),$ but there is no sequence $(f_n)\subset L^1(\nu )=c_0(\Gamma )$ such that $0\le f_n\uparrow \chi _\Gamma .$ Indeed, in this case, since the only $\nu$-null set is the empty set, $\Gamma =\cup _n\text{ supp}(f_n)$ is countable.

Therefore, in general $L^1(\nu )$ is not super order dense in $L_w^1(\nu ),$ but order dense.

Theorem 4.2

The space $L^1(\nu )$ is order dense in $L_w^1(\nu ).$

Proof

Since every Banach lattice is Archimedean, by [20, Ch. 3, Theorem 22.3] it is enough to prove that $L^1(\nu )$ is quasi order dense in $L_w^1(\nu ),$ i.e. for every $0\not =f\in L_w^1(\nu )$ there exists $0\not =g\in L^1(\nu )$ such that $|g|\le |f|.$

Let $f\in L_w^1(\nu )$ with $\Vert \nu \Vert (\text{ supp}(f))>0.$ For $A_n=\{\omega \in \Omega :\, |f(\omega )|>\frac{1}{n}\},$ we have that $A_n\uparrow \text{ supp}(f)$ and so $\Vert \nu \Vert ( \text{ supp}(f))=\lim _n\Vert \nu \Vert (A_n)$ (see [22, Corollary 3.5.(e)]). Take $n$ large enough such that $\Vert \nu \Vert (A_n)>0.$ Since $\Vert \nu \Vert (A_n)=\sup _{B\in \mathcal R \cap 2^{A_n}}\Vert \nu \Vert (B)$ (see [22, Lemma 3.4.(g)]), there exists $B_n\in \mathcal R \cap 2^{A_n}$ such that $\Vert \nu \Vert (B_n)>0.$

On the other hand, take a sequence $(\psi _j)_j\subset \mathcal S (\mathcal R ^{loc})$ such that $0\le \psi _j\uparrow |f|\,\nu$-a.e. Then, there exists a $\nu$-null set $Z\in \mathcal R ^{loc}$ such that $0\le \psi _j(\omega )\uparrow f(\omega )$ for all $\omega \in \Omega \backslash Z.$ Let us consider $B_n=(\cup _jB_n\cap \text{ supp}(\psi _j)\backslash Z)\cup (B_n\cap Z).$ Since $B_n\cap \text{ supp}(\psi _j)\backslash Z\uparrow ,$ it follows that $\Vert \nu \Vert (B_n)=\Vert \nu \Vert (\cup _jB_n\cap \text{ supp}(\psi _j)\backslash Z) =\lim _j\Vert \nu \Vert (B_n\cap \text{ supp}(\psi _j)\backslash Z).$ Take $j_n$ large enough such that $\Vert \nu \Vert (B_n\cap \text{ supp}(\psi _{j_n})\backslash Z)>0$ and consider the function $g=\psi _{j_n}\chi _{B_n}\in \mathcal S (\mathcal R )\subset L^1(\nu ).$ Then, $g\not =0$ and $0\le g\le |f|.$ $\square$

Remark 4.3

Since $L^0(\nu )$ with the $\nu$-a.e. pointwise order is an Archimedean vector lattice, actually in Theorem 4.2 we have proved that $L^1(\nu )$ is order dense in $L^0(\nu ).$

Now, the natural question is when $L^1(\nu )$ is super order dense in $L_w^1(\nu ).$ It is easy to see that this happens if $\nu$ is $\sigma$ -finite, that is, $\Omega =(\cup A_n)\cup N$ with $N\in \mathcal R ^{loc}\,\nu$-null and $(A_n)$ a sequence in $\mathcal R .$ In this case, if $0\le f\in L^0(\nu )$ and $(\psi _n)\subset \mathcal S (\mathcal R ^{loc})$ is such that $0\le \psi _n\uparrow f\,\nu$-a.e., taking $\varphi _n=\psi _n\chi _{\cup _{j=1}^nA_j}\in \mathcal S (\mathcal R )$ we have that $0\le \varphi _n\uparrow f\,\nu$-a.e. Then, $L^1(\nu )$ is super order dense in $L^0(\nu )$ and so in $L_w^1(\nu ).$ However, $L^1(\nu )$ being super order dense in $L_w^1(\nu )$ does not imply that $\nu$ is $\sigma$-finite.

Example 4.4

The vector measure $\nu$ in Example 4.1 considered with values in $\ell ^1(\Gamma )$ instead of $c_0(\Gamma ),$ satisfies that $L^1(\nu )=L_w^1(\nu )=\ell ^1(\Gamma ).$ Then, obviously $L^1(\nu )$ is super order dense in $L_w^1(\nu )$ but $\nu$ is not $\sigma$-finite.

We will characterize the super order density of $L^1(\nu )$ in $L_w^1(\nu )$ by a weaker condition on $\nu$ than $\sigma$-finiteness. Namely, $\nu$ will be said to be locally $\sigma$ -finite if every set $A\in \mathcal R ^{loc}$ with $\Vert \nu \Vert (A)<\infty ,$ can be written as $A=(\cup A_n) \cup N,$ with $N\in \mathcal R ^{loc}\,\nu$-null and $(A_n)$ a sequence in $\mathcal R .$

Remark 4.5

If $\nu$ is such that $L^1(\nu )=L_w^1(\nu )$ (e.g. if $X$ does not contain any copy of $c_0,$ see [18, Theorem 5.1]), then for every $A\in \mathcal R ^{loc}$ with $\Vert \nu \Vert (A)<\infty ,$ we have that $\chi _A\in L_w^1(\nu )=L^1(\nu )$ and so, from [22, Theorem 4.9.(a)], $\nu$ is locally $\sigma$-finite.

Let us see that there are plenty of locally $\sigma$-finite vector measures which are not $\sigma$-finite.

Lemma 4.6

Suppose that $\nu$ is discrete, that is, for every $\omega \in \Omega$ it follows that $\{\omega \}\in \mathcal R$ and $\nu (\{\omega \})\not =0.$ Then,

• (a) $N\in \mathcal R ^{loc}$ is $\nu$-null if and only if $N=\emptyset .$

• (b) $\{A\subset \Omega : \, A \text{ is} \text{ finite}\}\subset \mathcal R \subset \{A\subset \Omega : \, A \text{ is} \text{ countable}\}.$

• (c) $\mathcal R ^{loc}=2^\Omega .$

• (d) $\nu$ is $\sigma$-finite if and only if $\Omega$ is countable.

Proof

(a) Suppose $N\in \mathcal R ^{loc}$ is $\nu$-null. If $\gamma \in N,$ then $\{\gamma \}\in \mathcal R \cap 2^N$ and so $\Vert \nu (\{\gamma \})\Vert \le \Vert \nu \Vert (N)=0$ which contradicts $\nu (\{\gamma \})\not =0.$ Hence, $N=\emptyset .$ The converse is obvious.

(b) If $A\subset \Omega$ is finite then $A=\cup _{\gamma \in A}\{\gamma \}$ is a finite union of sets in $\mathcal R ,$ so the first containment holds. For the second one, consider $A\in \mathcal R$ and the vector measure $\nu _A:\mathcal R ^{loc}\rightarrow X$ defined by $\nu _A(B)=\nu (A\cap B)$ for all $B\in \mathcal R ^{loc}.$ Note that $B\in \mathcal R ^{loc}$ is $\nu _A$-null if and only if $A\cap B$ is $\nu$-null, that is, $A\cap B=\emptyset .$ Since $\nu _A$ is defined on a $\sigma$-algebra we can take $x_A^*\in B_{X^*}$ such that $|x_A^*\nu _A|$ has the same null sets as $\nu _A$ (see [13, Theorem IX.2.2]). For every finite set $J\subset \Omega$ it follows that

$$\begin{aligned} \sum _{\gamma \in J}|x_A^*\nu _A|(\{\gamma \})=|x_A^*\nu _A|(J)\le \Vert \nu _A\Vert (J)\le \Vert \nu _A\Vert (\Omega )<\infty . \end{aligned}$$

Then, there exists a countable set $I\subset \Omega$ such that $|x_A^*\nu _A|(\{\gamma \})=0$ for all $\gamma \in \Omega \backslash I,$ that is, $A\cap \{\gamma \}=\emptyset$ for all $\gamma \in \Omega \backslash I.$ So, $A\subset I$ is countable.

(c) Note that $\{A\subset \Omega : \, A \text{ is} \text{ countable}\}\subset \mathcal R ^{loc},$ since if $A\subset \Omega$ is countable then $A=\cup _{\gamma \in A}\{\gamma \}$ is a countable union of sets in $\mathcal R .$ Given $A\in 2^\Omega ,$ from (b) we have that $A\cap B$ is countable, and so it is in $\mathcal R ^{loc}$ for every $B\in \mathcal R .$ Hence, $A\cap B=B\cap (A\cap B)\in \mathcal R$ for every $B\in \mathcal R ,$ that is, $A\in \mathcal R ^{loc}.$

(d) It follows from (a) and (b). $\square$

From Remark 4.5 and Lemma 4.6, every discrete vector measure on a $\delta$-ring of subsets of an uncountable set with values in a Banach space without any copy of $c_0$ is locally $\sigma$-finite, but not $\sigma$-finite. Also, there are locally $\sigma$-finite vector measures which are not $\sigma$-finite with values in a Banach space containing a copy of $c_0.$

Example 4.7

Consider the $\delta$-ring $\mathcal R =\{A\subset [0,\infty ):\, A \text{ is} \text{ finite}\}$ of subsets of $[0,\infty )$ and the vector measure $\nu :\mathcal R \rightarrow c_0$ defined by $\nu (A)=\sum _n\frac{\sharp (A\cap [n-1,n))}{2^n}e_n,$ where $(e_n)_n$ is the canonical basis of $c_0$ and $\sharp$ denotes the cardinal of a set. Note that $\nu$ is discrete, so $\nu$ is not $\sigma$-finite. It can be proved that $L_w^1(\nu )$ is the space of functions $f:[0,\infty )\rightarrow \mathbb R$ such that

$$\begin{aligned} f\chi _{[n-1,n)}\in \ell ^1\left([0,\infty )\right) \text{ for} \text{ all} n \text{ and} \sup _n\frac{1}{2^n}\big \Vert |f|\chi _{[n-1,n)}\big \Vert _{\ell ^1([0,\infty ))}<\infty , \end{aligned}$$

and $\Vert f\Vert _\nu =\sup _n\frac{1}{2^n}\big \Vert |f|\chi _{[n-1,n)}\big \Vert _{\ell ^1([0,\infty ))}$ for all $f\in L_w^1(\nu ).$ Moreover, $L^1(\nu )$ is the space of functions $f:[0,\infty )\rightarrow \mathbb R$ such that

$$\begin{aligned} f\chi _{[n-1,n)}\in \ell ^1\left([0,\infty )\right) \text{ for} \text{ all} n \text{ and} \lim _n\frac{1}{2^n}\big \Vert |f|\chi _{[n-1,n)}\big \Vert _{\ell ^1([0,\infty ))}=0. \end{aligned}$$

Note that every $f\in L_w^1(\nu )$ has countable support as $\text{ supp}(f)\cap [n-1,n)$ is countable for all $n.$ If $B\in \mathcal R ^{loc}$ is such that $\Vert \nu \Vert (B)<\infty ,$ that is $\chi _B\in L_w^1(\nu ),$ then $B$ is countable. Hence, $\nu$ is locally $\sigma$-finite.

Let us prove now that the super order density of $L^1(\nu )$ in $L_w^1(\nu )$ is characterized by the local $\sigma$-finiteness of $\nu .$

Theorem 4.8

The space $L^1(\nu )$ is super order dense in $L_w^1(\nu )$ if and only if $\nu$ is locally $\sigma$-finite.

Proof

Suppose that $L^1(\nu )$ is super order dense in $L_w^1(\nu ).$ Take $A\in \mathcal R ^{loc}$ with $\Vert \nu \Vert (A)<\infty .$ Since $0\le \chi _A\in L_w^1(\nu ),$ there exists a sequence $(f_n)\subset L^1(\nu )$ such that $0\le f_n\uparrow \chi _A\,\nu$-a.e. Then, there exists $Z\in \mathcal R ^{loc}\,\nu$-null such that $0\le f_n(\omega )\uparrow \chi _A(\omega )$ for all $\omega \in \Omega \backslash Z.$ Thus, $A\backslash Z=\cup _n\text{ supp}(f_n)\backslash Z.$

On the other hand, since each $f_n\in L^1(\nu ),$ from [22, Theorem 4.9.(a)] , there exist $(A_j^n)_j\subset \mathcal R$ and a $\nu$-null set $N_n\in \mathcal R ^{loc}$ such that $\text{ supp}(f_n)=(\cup _jA_j^n)\cup N_n.$ Then,

$$\begin{aligned} A=(\cup _n\cup _jA_j^n\backslash Z) \cup (\cup _n N_n\backslash Z) \cup (A\cap Z) \end{aligned}$$

where $A_j^n\backslash Z\in \mathcal R$ and $(\cup _n N_n\backslash Z) \cup (A\cap Z)$ is $\nu$-null.

Conversely, suppose that $\nu$ is locally $\sigma$-finite and let $0\le f\in L_w^1(\nu ).$ There exists a sequence $(\psi _n)\subset \mathcal S (\mathcal R ^{loc})$ such that $0\le \psi _n\uparrow f\,\nu$-a.e. For each $n,$ we can write $\psi _n=\sum _{j=1}^{k_n}\alpha _j^n\chi _{B_j^n}$ with $(B_j^n)_j$ pairwise disjoint and $\alpha _j^n>0.$ Then, taking $\beta _n=\min \{\alpha _1^n,...,\alpha _{k_n}^n\},$ it follows

$$\begin{aligned} \Vert \nu \Vert (\text{ supp}(\psi _n))=\Vert \chi _{\mathrm{supp}(\psi _n)}\Vert _\nu \le \frac{1}{\beta _n}\Vert \psi _n\Vert _\nu \le \frac{1}{\beta _n}\Vert f\Vert _\nu <\infty . \end{aligned}$$

So, there exist $(A_j^n)_j\subset \mathcal R$ and $Z_n\,\nu$-null such that $\text{ supp}(\psi _n)=(\cup _j A_j^n)\cup Z_n.$ Denote $\varphi _n=\psi _n\chi _{\cup _{i=1}^n\cup _{j=1}^nA_j^i}\in \mathcal S (\mathcal R ).$ For $\omega \not \in \cup _nZ_n$ we have that $\omega \in \Omega \backslash (\cup _n \text{ supp}(\psi _n))$ or $\omega \in \cup _n\cup _jA_j^n.$ In any case, $\varphi _n(\omega )=\psi _n(\omega )$ for all $n$ large enough. Then, $\varphi _n\uparrow f\,\nu$-a.e. $\square$

We have seen just before Example 4.4 that if $\nu$ is $\sigma$-finite then $L^1(\nu )$ is super order dense in $L^0(\nu ).$ The converse also holds, indeed taking $\Omega$ instead of $A$ in the proof of the local $\sigma$-finiteness of $\nu$ in Theorem 4.8, the same argument works to show $\Omega =(\cup A_n)\cup N,$ with $N\in \mathcal R ^{loc}\,\nu$-null and $(A_n) \subset \mathcal R .$

We know from [22, Theorem 4.9.(a)] that for each $f\in L^1(\nu )$ there are $(A_n)\subset \mathcal R$ and a $\nu$-null set $N\in \mathcal R ^{loc}$ such that $\text{ supp}(f)=(\cup A_n)\cup N.$ Does the same hold for functions in $L_w^1(\nu )$?

Proposition 4.9

For each $f\in L_w^1(\nu )$ there exist $N\in \mathcal R ^{loc}\,\nu$-null and $(A_n)\subset \mathcal R$ such that $\text{ supp}(f)=(\cup A_n)\cup N$ if and only if $\nu$ is locally $\sigma$-finite.

Proof

Suppose that $\nu$ is locally $\sigma$-finite and take $f\in L_w^1(\nu ).$ From the proof of Theorem 4.8, there exists a sequence $(\varphi _n)\subset \mathcal S (\mathcal R )$ such that $0\le \varphi _n\uparrow |f|\,\nu$-a.e. Let $Z\in \mathcal R ^{loc}$ be a $\nu$-null set such that $0\le \varphi _n(\omega )\uparrow |f(\omega )|$ for all $\omega \in \Omega \backslash Z.$ Then,

$$\begin{aligned} \text{ supp}(f)=(\cup \text{ supp}(\varphi _n)\backslash Z)\cup (\text{ supp}(f)\cap Z) \end{aligned}$$

where $\text{ supp}(\varphi _n)\backslash Z\in \mathcal R$ and $\text{ supp}(f)\cap Z$ is $\nu$-null. For the converse only note that if $B\in \mathcal R ^{loc}$ is such that $\Vert \nu \Vert (B)<\infty ,$ then $\chi _B\in L_w^1(\nu ).$ $\square$

Let $\{\Omega _\alpha :\,\alpha \in \Delta \}$ be a maximal family of non $\nu$-null sets in $\mathcal R$ with $\Omega _\alpha \cap \Omega _\beta \,\nu$-null for $\alpha \not =\beta$ (see the proof of [1, Theorem 3.1] for the existence of such a family). Then, $L^1(\nu )$ is the unconditional direct sum of the spaces $L^1(\nu _\alpha )$ where $\nu _\alpha :\Sigma _\alpha \rightarrow X$ is the restriction of $\nu$ to the $\sigma$-algebra $\Sigma _\alpha =\{A\in \mathcal R :\,A\subset \Omega _\alpha \}.$ More precisely, for each $f\in L^1(\nu )$ there exists a countable set $I\subset \Delta$ such that $f=\sum _{\alpha \in I}f\chi _{\Omega _\alpha }\,\nu$-a.e. and the sum converges unconditionally in $L^1(\nu ),$ see [9, Theorem 3.6]. Does a similar result hold for the space $L_w^1(\nu )$? The $\nu$-a.e. pointwise convergence of the sum for functions in $L_w^1(\nu )$ is again characterized by the local $\sigma$-finiteness of $\nu .$

Proposition 4.10

For each $f\in L_w^1(\nu )$ there exists a countable $I\subset \Delta$ such that $f=\sum _{\alpha \in I}f\chi _{\Omega _\alpha }\,\nu$-a.e. pointwise if and only if $\nu$ is locally $\sigma$-finite.

Proof

Suppose that for every $f\in L_w^1(\nu )$ there exists a countable $I\subset \Delta$ such that $f=\sum _{\alpha \in I}f\chi _{\Omega _\alpha }\,\nu$-a.e. pointwise. Then, given $B\in \mathcal R ^{loc}$ with $\Vert \nu \Vert (B)<\infty ,$ since $\chi _B\in L_w^1(\nu ),$ we can write $\chi _B=\sum _{\alpha \in I} \chi _{B\cap \Omega _{\alpha }}$ pointwise except on a $\nu$-null set $Z,$ for some countable $I\subset \Delta .$ So, $B=(\cup _{\alpha \in I} B\cap \Omega _\alpha ) \cup (B\cap Z),$ where $B\cap \Omega _\alpha \in \mathcal R$ and $B\cap Z$ is $\nu$-null.

Conversely, suppose that $\nu$ is locally $\sigma$-finite and take $f\in L_w^1(\nu ).$ From Proposition 4.9, there exist $(A_n)\subset \mathcal R$ and a $\nu$-null set $N\in \mathcal R ^{loc}$ such that $\text{ supp}(f)=(\cup A_n)\cup N.$ Since each $A_n\in \mathcal R ,$ there exists a countable set $I_n\subset \Delta$ such that $A_n\cap \Omega _\beta$ is $\nu$-null for all $\beta \in \Delta \backslash I_n$ (see the proof of [1, Theorem 3.1]). Take $I=\cup I_n$ and $Z=\text{ supp}(f)\backslash \cup _{\alpha \in I}\Omega _\alpha .$ Let us see that $Z$ is a $\nu$-null set. Given $B\in \mathcal R \cap 2^Z,$ if $\beta \in I$ we have that $B\cap \Omega _\beta =\emptyset .$ On the other hand, if $\beta \notin I,$ since $B\cap \Omega _\beta \subset \text{ supp}(f)\cap \Omega _\beta =(\cup A_n\cap \Omega _\beta )\cup (N\cap \Omega _\beta )$ where each $A_n\cap \Omega _\beta$ is $\nu$-null, we have that $B\cap \Omega _\beta$ is $\nu$-null. From the maximality of the family $\{\Omega _\alpha :\,\alpha \in \Delta \}$ it follows that $B$ is $\nu$-null. Then, $f=\sum _{\alpha \in I}f\chi _{\Omega _\alpha }$ pointwise except on $Z\cup (\cup _{\beta \in I}\cup _{\alpha \in I\backslash \{\beta \}}\Omega _\alpha \cap \Omega _\beta )$ which is a $\nu$-null set. $\square$

Since $f\chi _{\Omega _\alpha }\in L_w^1(\nu _\alpha )$ for all $\alpha \in \Delta$ whenever $f\in L_w^1(\nu ),$ in the case of $\nu$ being locally $\sigma$-finite, we can say that the space $L_w^1(\nu )$ is the $\nu$-a.e. pointwise direct sum of the spaces $L_w^1(\nu _\alpha ).$ We cannot expect that $\sum _{\alpha \in I}f\chi _{\Omega _\alpha }$ converges unconditionally to $f$ in $L_w^1(\nu )$ for a countable set $I\subset \Delta .$ Indeed, unconditional convergence of the sum in $L^1(\nu )$ is due to the order continuity of $L^1(\nu ).$ For instance, assume that $\nu$ is a discrete vector measure. Note that the maximal family $\left\{ \{\gamma \}: \gamma \in \Gamma \right\}$ of non $\nu$-null sets in $\mathcal R$ satisfies that $\{\alpha \}\cap \{\beta \}\,\nu$-null for $\alpha \not =\beta .$ We have that if $f\in L_w^1(\nu )$ is such that $\sum _n f\chi _{\{\gamma _n\}}$ converges to $f$ in norm $\Vert \cdot \Vert _\nu ,$ then $f\in L^1(\nu ).$ This is due to the fact that $\sum _{k=1}^nf\chi _{\{\gamma _k\}}=\sum _{k=1}^nf(\gamma _k)\chi _{\{\gamma _k\}}\in \mathcal S (\mathcal R )\subset L^1(\nu )$ and $L^1(\nu )$ is closed in $L_w^1(\nu ).$

5 Fatou property for $L_w^1(\nu )$

The space $L_w^1(\nu )$ always has the $\sigma$-Fatou property. Indeed, take $(f_n)\subset L_w^1(\nu )$ such that $0\le f_n\uparrow$ and $\sup \Vert f_n\Vert _\nu <\infty .$ Then there exists a $\nu$-null set $Z\in \mathcal R ^{loc}$ such that $0\le f_n(\omega )\uparrow$ for all $\omega \in \Omega \backslash Z.$ Taking the measurable function $g:\Omega \rightarrow [0,\infty ]$ defined by $g(\omega )=\sup f_n(\omega )$ if $\omega \in \Omega \backslash Z$ and $g(\omega )=0$ if $\omega \in Z,$ we have that $0\le f_n\chi _{\Omega \backslash Z}\uparrow g$ pointwise. Hence, the monotone convergence theorem, gives

$$\begin{aligned} \int g\,d|x^*\nu |=\lim _n \int f_n\chi _{\Omega \backslash Z}\,d|x^*\nu |\le \Vert x^*\Vert \sup \Vert f_n\Vert _\nu , \end{aligned}$$

for every $x^*\in X^*.$ So, $\Vert g\Vert _\nu \le \sup \Vert f_n\Vert _\nu <\infty ,$ and then $g<\infty \,\nu$-a.e. (except on a $\nu$-null set $N$). Taking $f=g\chi _{\Omega \backslash N}$ we have that $f:\Omega \rightarrow [0,\infty )$ and $\Vert f\Vert _\nu =\Vert g\Vert _\nu <\infty ,$ so $f\in L_w^1(\nu ).$ Moreover, $0\le f_n\uparrow f\,\nu$-a.e. with $\Vert f\Vert _\nu =\sup \Vert f_n\Vert _\nu ,$ as $\Vert f_n\Vert _\nu \le \Vert f\Vert _\nu \le \sup \Vert f_n\Vert _\nu$ for all $n.$ Therefore $L_w^1(\nu )$ always has the $\sigma$-Fatou property.

In the case when $\nu$ is defined on a $\sigma$-algebra, it was noted in [6, p. 191] that $L_w^1(\nu )$ is the $\sigma$ -Fatou completion of $L^1(\nu ),$ that is, the minimal B.f.s. related to $\nu$ with the $\sigma$-Fatou property and containing $L^1(\nu ).$ This fact does not hold for the general case. For instance, if $\nu$ is the vector measure defined in Example 4.1 and $\ell _0^\infty (\Gamma )$ denotes the Banach lattice of all real bounded functions on $\Gamma$ with countable support, then $L^1(\nu )\varsubsetneq \ell _0^\infty (\Gamma )\varsubsetneq L_w^1(\nu )$ where $\ell _0^\infty (\Gamma )$ has the $\sigma$-Fatou property. Note that in this case $\nu$ is not locally $\sigma$-finite, as $\chi _\Gamma \in L_w^1(\nu ).$ This is the reason for which $L_w^1(\nu )$ fails to be the $\sigma$-Fatou completion of $L^1(\nu ).$ Let us denote by $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ the $\sigma$-Fatou completion of $L^1(\nu ).$ In general we have that $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}\subset L_w^1(\nu ).$

Theorem 5.1

The $\sigma$-Fatou completion of $L^1(\nu )$ can be described as

$$\begin{aligned}{}[L^1(\nu )]_{_{\sigma \mathrm{-F}}}=\left\{ f\in L_w^1(\nu ):\, \text{ supp}(f)=(\cup A_n)\cup N \text{ with} (A_n)\subset \mathcal R \text{ and} N \, \nu \text{-null}\right\} . \end{aligned}$$

Consequently, the space $L_w^1(\nu )=[L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ if and only if $\nu$ is locally $\sigma$-finite.

Proof

Denote by $F$ the space of functions $f\in L_w^1(\nu )$ for which there exist $(A_n)\subset \mathcal R$ and a $\nu$-null set $N\in \mathcal R ^{loc}$ such that $\text{ supp}(f)=(\cup A_n)\cup N.$ Let us see that $F$ is a closed subspace of $L_w^1(\nu ).$ Given $f\in L_w^1(\nu )$ and $(f_n)\subset F$ such that $\Vert f-f_n\Vert _\nu \rightarrow 0,$ we can take a subsequence such that $f_{n_k}\rightarrow f\,\nu$-a.e. That is, there exists a $\nu$-null set $Z\in \mathcal R ^{loc}$ such that $f_{n_k}(\omega )\rightarrow f(\omega )$ for all $\omega \in \Omega \backslash Z.$ Then, $\text{ supp}(f) \backslash Z\subset \cup _k\text{ supp}(f_{n_k}).$ On the other hand, each $f_{n_k}$ satisfies that $\text{ supp}(f_{n_k})=(\cup _jA_j^k)\cup N_k$ for some $(A_j^k)_j\subset \mathcal R$ and $N_k\in \mathcal R ^{loc}\,\nu$-null. So, $\text{ supp}(f)=\cup _k\cup _jB_j^k\cup N$ where $B_j^k=A_j^k\cap \text{ supp}(f)\backslash Z\in \mathcal R$ and $N=(\cup _k N_k\cap \text{ supp}(f)\backslash Z)\cup (\text{ supp}(f)\cap Z)$ is $\nu$-null, that is, $f\in F.$ Note that if $|f|\le |g|\,\nu$-a.e. with $f\in L^0(\nu )$ and $g\in F,$ then $f\in F$ since $\text{ supp}(f)\backslash Z=(\text{ supp}(f)\backslash Z)\cap \text{ supp}(g)$ for some $\nu$-null set $Z.$ Therefore, $F$ endowed with the norm $\Vert \cdot \Vert _\nu ,$ is a B.f.s. related to $\nu ,$ which, by [22, Theorem 4.9.(a)], contains $L^1(\nu ).$ Let us see now that $F$ has the $\sigma$-Fatou property. Given $(f_n)\subset F$ such that $0\le f_n\uparrow$ and $\sup \Vert f_n\Vert _\nu <\infty ,$ since $L_w^1(\nu )$ has the $\sigma$-Fatou property, there exists $f=\sup f_n\in L_w^1(\nu )$ with $\Vert f\Vert _\nu =\sup \Vert f_n\Vert _\nu .$ Moreover, since $0\le f_n\uparrow f\,\nu$-a.e., $\text{ supp}(f)=(\cup \text{ supp}(f_n)\backslash Z)\cup (\text{ supp}(f)\cap Z)$ for some $\nu$-null set $Z\in \mathcal R ^{loc}.$ Then, it follows that $f\in F,$ as each $f_n\in F.$

Suppose that $E$ is a B.f.s. related to $\nu ,$ with the $\sigma$-Fatou property and containing $L^1(\nu ).$ Let $f\in F$ and take a sequence $(A_n)\subset \mathcal R$ and a $\nu$-null set $N\in \mathcal R ^{loc}$ such that $\text{ supp} (f)=(\cup A_n)\cup N.$ On the other hand, take a sequence $(\psi _n)\subset \mathcal S (\mathcal R ^{loc})$ such that $0\le \psi _n\uparrow |f|\,\nu$-a.e. Denoting $\varphi _n=\psi _n\chi _{\cup _{j=1}^nA_j}\in \mathcal S (\mathcal R )\subset L^1(\nu )$ we have that $0\le \varphi _n\uparrow |f|\,\nu$-a.e. Since $L^1(\nu )\subset E$ continuously (bear in mind that the inclusion is a positive operator) we have that $\sup \Vert \varphi _n\Vert _E\le C\sup \Vert \varphi _n\Vert _\nu \le C\Vert f\Vert _\nu <\infty$ for some positive constant $C.$ It follows that there exists $g=\sup \varphi _n\in E.$ Then, since $0\le \varphi _n\uparrow g\,\nu$-a.e., we have that $|f|=g\in E$ and so $f\in E.$

The consequence follows from Proposition 4.9. $\square$

Consider now the Fatou completion $[L^1(\nu )]_{_{\mathrm{F}}}$ of $L^1(\nu ),$ namely, the minimal B.f.s. related to $\nu$ with the Fatou property and containing $L^1(\nu ).$ The $\sigma$-Fatou completion $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ always exists since $L_w^1(\nu )$ has always the $\sigma$-Fatou property. However, we do not know if in general $L_w^1(\nu )$ has the Fatou property, so $[L^1(\nu )]_{_{\mathrm{F}}}$ could not exist.

Remark 5.2

In the case when $[L^1(\nu )]_{_{\mathrm{F}}}$ exists, we have that

$$\begin{aligned} L^1(\nu )\subset [L^1(\nu )]_{_{\sigma \mathrm{-F}}}\subset L_w^1(\nu )\subset [L^1(\nu )]_{_{\mathrm{F}}}. \end{aligned}$$

Indeed, given $f\in L_w^1(\nu ),$ from Remark 4.3, there exists $(f_\tau )\subset L^1(\nu )$ such that $0\le f_\tau \uparrow |f|$ in $L^0(\nu ).$ Since $L^1(\nu )\subset [L^1(\nu )]_{_{\mathrm{F}}}$ continuously, it follows that $\sup \Vert f_\tau \Vert _{[L^1(\nu )]_{_{\mathrm{F}}}}\le C\sup \Vert f_\tau \Vert _\nu \le C\Vert f\Vert _\nu <\infty$ for some constant $C>0.$ Then, there exists $g=\sup f_\tau$ in $[L^1(\nu )]_{_{\mathrm{F}}}.$ Noting that $f_\tau \le g\in L^0(\nu )$ for all $\tau ,$ we have that $|f|\le g$ and so $|f|\in [L^1(\nu )]_{_{\mathrm{F}}}.$ Hence, $f\in [L^1(\nu )]_{_{\mathrm{F}}}.$ Note that actually $|f|=g,$ since $f_\tau \le |f|\in [L^1(\nu )]_{_{\mathrm{F}}}$ for all $\tau$ and so $g\le |f|.$

Remark 5.3

If $L_w^1(\nu )$ has the Fatou property, then $[L^1(\nu )]_{_{\mathrm{F}}}$ exists and, from Remark 5.2, we have that $L_w^1(\nu )=[L^1(\nu )]_{_{\mathrm{F}}}.$

In the following result we give conditions under which $L_w^1(\nu )$ has the Fatou property. These conditions are satisfied for instance if $\nu$ takes values in a Banach space without any copy of $c_0.$

Proposition 5.4

The following statements are equivalent:

• (a) $L^1(\nu )=L_w^1(\nu ).$

• (b) $L_w^1(\nu )$ is order continuous.

• (c) $L^1(\nu )$ has the $\sigma$-Fatou property.

If (a)–(c) hold, then $L_w^1(\nu )$ has the Fatou property and

$$\begin{aligned} L^1(\nu )=[L^1(\nu )]_{_{\sigma \mathrm{-F}}}=L_w^1(\nu )=[L^1(\nu )]_{_{\mathrm{F}}}. \end{aligned}$$

Proof

The equivalence between (a) and (b) follows from Theorem 3.2. Condition (a) implies (c) as $L_w^1(\nu )$ always has the $\sigma$-Fatou property. Conversely, if $L^1(\nu )$ has the $\sigma$-Fatou property, from [26, Theorem 113.4], it follows that it actually has the Fatou property. Then, $[L^1(\nu )]_{_{\mathrm{F}}}=L^1(\nu )$ and, from Remark 5.2, we have that $L^1(\nu )=L_w^1(\nu ).$ So, (c) implies (a) and the last part of the theorem holds. $\square$

It is an open question if in general $L_w^1(\nu )$ has the Fatou property. The problem is that for an upwards directed system $0\le f_\tau \uparrow$ such that $(f_\tau )\subset L_w^1(\nu )$ with $\sup \Vert f_\tau \Vert _\nu <\infty$ the pointwise supremum $f=\sup f_\tau$ may not be measurable. Moreover, even if $f\in L_w^1(\nu )$ it can happen that $f_\tau \uparrow f$ does not hold, that is, $f$ may be not the lattice supremum of $(f_\tau ).$

Remark 5.5

If $\nu$ is $\sigma$-finite, we can take a measure of the type $|x_0^*\nu |$ (with $x_0^*\in B_{X^*}$) having the same null sets as $\nu ,$ see [9, Remark 3.4]. Then, since $L_w^1(\nu )\subset L^1(|x_0^*\nu |)$ and $L^1(|x_0^*\nu |)$ has the Fatou property, there exists $f=\sup f_\tau$ in $L^1(|x_0^*\nu |).$ By using the fact that $L^1(|x_0^*\nu |)$ is order separable (see [26, Theorem 113.4]), we can take a sequence $f_{\tau _n}\uparrow f$ in $L^1(|x_0^*\nu |)$ and prove that $f\in L_w^1(\nu ).$ Then, $L_w^1(\nu )$ has the Fatou property, see [12, Proposition 1]. Moreover, it follows that $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}=L_w^1(\nu )=[L^1(\nu )]_{_{\mathrm{F}}}$ from Theorem 5.1 and Remark 5.3.

We will give a more general condition than the $\sigma$-finiteness of $\nu$ under which $L_w^1(\nu )$ has the Fatou property. This new condition is inspired by the particular vector measure $\nu$ constructed in [12, Theorem 9] to prove that a Banach lattice $E$ with the Fatou property and such that $E_a$ is order dense in $E,$ is order isometric to a $L^1_w(\nu ).$ In this case, $L^1_w(\nu )$ has the Fatou property due to a good decomposition property satisfied by $\nu .$

Definition 5.6

A vector measure $\nu$ will be said to be $\mathcal R$ -decomposable if we can write $\Omega =(\cup _{\alpha \in \Delta }\Omega _\alpha )\cup N$ where $N\in \mathcal R ^{loc}$ is a $\nu$-null set and $\{\Omega _\alpha :\, \alpha \in \Delta \}$ is a family of pairwise disjoint sets in $\mathcal R$ satisfying that

• (i) if $A_\alpha \in \mathcal R \cap 2^{\Omega _\alpha }$ for all $\alpha \in \Delta ,$ then $\cup _{\alpha \in \Delta }A_\alpha \in \mathcal R ^{loc},$ and

• (ii) for each $x^*\in X^*,$ if $Z_\alpha \in \mathcal R \cap 2^{\Omega _\alpha }$ is $|x^*\nu |$-null for all $\alpha \in \Delta ,$ then $\cup _{\alpha \in \Delta }Z_\alpha$ is $|x^*\nu |$-null.

Note that condition (ii) implies that if $Z_\alpha \in \mathcal R \cap 2^{\Omega _\alpha }$ is $\nu$-null for all $\alpha \in \Delta ,$ then $\cup _{\alpha \in \Delta }Z_\alpha$ is $\nu$-null. Also note that $N$ can be taken to be disjoint with $\cup _{\alpha \in \Delta }\Omega _\alpha .$

Remark 5.7

There always exists a maximal family $\{\widetilde{\Omega }_\alpha :\, \alpha \in \Delta \}$ of non $\nu$-null sets in $\mathcal R$ with $\widetilde{\Omega }_\alpha \cap \widetilde{\Omega }_\beta \,\nu$-null for $\alpha \not =\beta$ (see the proof of [1, Theorem 3.1]). If this family satisfies (i) and (ii) of Definition 5.6, then by taking $\Omega _\alpha =\widetilde{\Omega }_\alpha \backslash (\cup _{\beta \in \Delta \backslash \{\alpha \}}\widetilde{\Omega }_\beta )$ we obtain a disjoint decomposition of $\Omega$ as in Definition 5.6.

There are plenty of $\mathcal R$-decomposable vector measures, for instance $\sigma$-finite vector measures and discrete vector measures are so.

Theorem 5.8

If $\nu$ is $\mathcal R$-decomposable, then $L^1_w(\nu )$ has the Fatou property.

Proof

Suppose that $\nu$ is $\mathcal R$-decomposable and take a $\nu$-null set $N\in \mathcal R ^{loc}$ and a family $\{\Omega _{\alpha }:\,\alpha \in \Delta \}$ of pairwise disjoint sets in $\mathcal R$ satisfying conditions (i) and (ii) in Definition 5.6. So we have $\Omega =(\cup _{\alpha \in \Delta }\Omega _\alpha )\cup N$ with disjoint union. For every finite set $I\subset \Delta ,$ consider $\Omega _I=\cup _{\alpha \in I} \Omega _\alpha \in \mathcal R$ and the vector measure $\nu _I:\mathcal R ^{loc}\rightarrow X$ defined by $\nu (A\cap \Omega _I)$ for all $A\in \mathcal R ^{loc}.$ Given $f\in \mathcal M (\mathcal R ^{loc}),$ by using a similar argument as in the proof of (c) implies (a) in Lemma 3.1, it follows that $f\in L_w^1(\nu _I)$ if and only if $f\chi _{\Omega _I}\in L_w^1(\nu ),$ and in this case $\Vert f\Vert _{\nu _I}=\Vert f\chi _{\Omega _I}\Vert _\nu .$ Note that, if $f\in L_w^1(\nu )$ then $f\chi _{\Omega _I}\in L_w^1(\nu )$ and so $f\in L_w^1(\nu _I).$ Also note that $L_w^1(\nu _I)$ has the Fatou property as $\nu _I$ is defined on a $\sigma$-algebra, see Remark 5.5.

Let $(f_\tau )\subset L_w^1(\nu )$ be such that $0\le f_\tau \uparrow$ and $\sup \Vert f_\tau \Vert _\nu <\infty .$ Since $L_w^1(\nu )\subset L_w^1(\nu _I)$ and every $Z\in \mathcal R ^{loc}\,\nu$-null is $\nu _I$-null (as $\Vert \nu _I\Vert (Z)=\Vert \nu \Vert (Z\cap \Omega _I)$), then $0\le f_\tau \uparrow$ in $L_w^1(\nu _I).$ Moreover, $\sup \Vert f_\tau \Vert _{\nu _I}=\sup \Vert f_\tau \chi _{\Omega _I}\Vert _\nu \le \sup \Vert f_\tau \Vert _\nu <\infty .$ By the Fatou property of $L_w^1(\nu _I),$ there exists $f^I=\sup f_\tau$ in $L_w^1(\nu _I)$ and $\Vert f^I\Vert _{\nu _I}=\sup \Vert f_\tau \Vert _{\nu _I}.$

Now we consider $I=\{\alpha \}$ for each $\alpha \in \Delta$ and construct the function $f:\Omega \rightarrow \mathbb R$ as $f(\omega )=f^{\{\alpha \}}(\omega )$ when $\omega \in \Omega _{\alpha }$ and $f(\omega )=0$ when $\omega \in N,$ which is well defined since $\Omega$ is a disjoint union of $(\Omega _{\alpha })_{\alpha \in \Delta }$ and $N.$ By (i), we have that $f^{-1}(B)=\cup _{\alpha \in \Delta }(f^{\{\alpha \}})^{-1}(B)\cap \Omega _\alpha \in \mathcal R ^{loc}$ for every Borel subset $B$ of $\mathbb R$ such that $0\notin B.$ If $0\in B,$ we put also in the union the set $N$ to get $f^{-1}(B).$ So, $f\in \mathcal M (\mathcal R ^{loc}).$

Let us see that $f\in L_w^1(\nu ).$ First note that for each finite set $I\subset \Delta$ and $\alpha \in I,$ it follows that $f^{\{\alpha \}}\chi _{\Omega _\alpha }\le f^I\chi _{\Omega _\alpha }\,\nu$-a.e. Indeed, $f_\tau \chi _{\Omega _\alpha }\uparrow f^{\{\alpha \}}\chi _{\Omega _\alpha }$ in $L_w^1(\nu _{\{\alpha \}})$ as $f_\tau \uparrow f^{\{\alpha \}}$ in $L_w^1(\nu _{\{\alpha \}}).$ Since $f_\tau \chi _{\Omega _\alpha }\le f^I\chi _{\Omega _\alpha }\,\nu _I$-a.e. (and so also $\nu _{\{\alpha \}}$-a.e. and $f^I\chi _{\Omega _\alpha }\in L_w^1(\nu _{\{\alpha \}})$ as $f^I\chi _{\Omega _\alpha }\le f^I\chi _{\Omega _I}\in L_w^1(\nu )$) we have that $f^{\{\alpha \}}\chi _{\Omega _\alpha }\le f^I\chi _{\Omega _\alpha }\,\nu _{\{\alpha \}}$-a.e. (except on a $\nu _{\{\alpha \}}$-null set $Z$) and so $\nu$-a.e. (except on the $\nu$-null set $Z\cap \Omega _\alpha$). Then, $f\chi _{\Omega _I}=\sum _{\alpha \in I}f^{\{\alpha \}}\chi _{\Omega _\alpha }\le f^I\chi _{\Omega _I}\,\nu$-a.e.

Fix $x^*\in X^*.$ For every finite set $I\subset \Delta ,$ it follows

$$\begin{aligned} \sum _{\alpha \in I}\int |f|\chi _{\Omega _\alpha } \,d|x^*\nu |&= \int |f| \chi _{\Omega _I}\,d|x^*\nu |\le \int |f^I|\chi _{\Omega _I} \,d|x^*\nu | \\&\le \Vert x^*\Vert \cdot \Vert f^I\chi _{\Omega _I} \Vert _\nu = \Vert x^*\Vert \cdot \Vert f^I \Vert _{\nu _I} \\&= \Vert x^*\Vert \cdot \sup \Vert f_\tau \Vert _{\nu _I} \le \Vert x^*\Vert \cdot \sup \Vert f_\tau \Vert _\nu <\infty . \end{aligned}$$

Then, there exists a countable set $J\subset \Delta$ such that $\int |f|\chi _{\Omega _\alpha } \,d|x^*\nu |=0$ for all $\alpha \in \Delta \backslash J$ and so $f\chi _{\Omega _\alpha }=0\,|x^*\nu |$-a.e. (except on a $|x^*\nu |$-null set $Z_\alpha \in \mathcal R ^{loc}$ which can be taken such that $Z\subset \Omega _\alpha$) for all $\alpha \in \Delta \backslash J.$ Hence, $f=\sum _{\alpha \in J} f\chi _{\Omega _\alpha }\,|x^*\nu |$-a.e. (except on the $|x^*\nu |$-null set $\cup _{\alpha \in \Delta \backslash J}Z_\alpha \cup N\in \mathcal R ^{loc}$). By the monotone convergence theorem we have that

$$\begin{aligned} \int |f|\,d|x^*\nu |=\sum _{\alpha \in J}\int |f|\chi _{\Omega _\alpha }\,d|x^*\nu | \le \Vert x^*\Vert \cdot \sup \Vert f_\tau \Vert _\nu <\infty . \end{aligned}$$

So $f\in L_w^1(\nu )$ and $\Vert f\Vert _\nu \le \sup \Vert f_\tau \Vert _\nu .$

Let us see now that $f_\tau \uparrow f$ in $L_w^1(\nu ).$ Fixing $\tau ,$ for each $\alpha \in \Delta ,$ there exists a $\nu _{\{\alpha \}}$-null set $Z_\alpha \in \mathcal R ^{loc}$ such that $f_\tau (\omega )\le f^{\{\alpha \}}(\omega )$ for all $\omega \in \Omega _\alpha \backslash Z_\alpha .$ Then, $Z=\cup _{\alpha \in \Delta }Z_\alpha \cap \Omega _\alpha$ is $\nu$-null and $f_\tau (\omega )\le f(\omega )$ for all $\omega \in \Omega \backslash (Z\cup N),$ that is, $f_\tau \le f\,\nu$-a.e. Suppose that $h\in L_w^1(\nu )$ is such that $f_\tau \le h\,\nu$-a.e. (except on a $\nu$-null set $Z\in \mathcal R ^{loc}$) and so $\nu _{\{\alpha \}}$-a.e. (except $Z$ which also is $\nu _{\{\alpha \}}$-null) for each $\tau .$ Since $h\in L_w^1(\nu _{\{\alpha \}}),$ we have that $f^{\{\alpha \}}\le h\,\nu _{\{\alpha \}}$-a.e. (except on a $\nu _{\{\alpha \}}$-null set $Z_\alpha \in \mathcal R ^{loc}$). Therefore, $f\le h\,\nu$-a.e. (except on the $\nu$-null set $(\cup _{\alpha \in \Delta }Z_\alpha \cap \Omega _\alpha )\cup N\in \mathcal R ^{loc}$). So, $f_\tau \uparrow f$ and $\Vert f\Vert _\nu =\sup \Vert f_\tau \Vert _\nu .$ $\square$

The converse of Theorem 5.8 does not hold as the next example shows.

Example 5.9

Following [16, p. 12, Definition 211E], a measure space $(X,\Sigma ,\mu )$ is decomposable (or strictly localizable) if there exists a disjoint family $\{X_\alpha :\,\alpha \in \Delta \}$ of measurable sets of finite measure such that $X=\cup _{\alpha \in \Delta }X_\alpha$ and

$$\begin{aligned} \Sigma =\{E\subset X: E\cap X_\alpha \in \Sigma \text{ for} \text{ all} \alpha \in \Delta \} \end{aligned}$$

with $\mu (E)=\sum _{\alpha \in \Delta }\mu (E\cap X_\alpha )$ for every $E\in \Sigma .$ In [16, p. 50, 216E], Fremlin constructs a measure space which is not decomposable as follows.

Let $C$ be an abstract set of cardinal greater than the cardinal of the continuum, $\mathcal K =\{K\subset 2^{\,C}:\, K \text{ is} \text{ countable}\}$ and $X$ the set of all functions $f:2^{\,C}\rightarrow \{0,1\}.$ For each $\gamma \in C,$ write $f_\gamma$ for the function in $X$ defined by $f_\gamma (A)=\chi _A(\gamma )$ for all $A\in 2^{\,C}$ and $F_{\gamma ,K}=\{f\in X:\, f_{|K}=f_{\gamma |K}\}$ for every $K\in \mathcal K .$ Consider the $\sigma$-algebra $\Sigma =\cap _{\gamma \in C}\Sigma _\gamma ,$ where

$$\begin{aligned} \Sigma _\gamma =\{E\subset X:\, \exists K\in \mathcal K \text{ with} F_{\gamma ,K}\subset E \text{ or} \exists K\in \mathcal K \text{ with} F_{\gamma ,K}\subset X\backslash E\}, \end{aligned}$$

and the measure $\mu :\Sigma \rightarrow [0,\infty ]$ defined by $\mu (E)=\sharp (\{\gamma \in C:\, f_\gamma \in E\})$ for all $E\in \Sigma ,$ where $\sharp$ denotes the cardinal of a set. Then, $(X,\Sigma ,\mu )$ is not decomposable.

Taking the $\delta$-ring $\mathcal R =\{E\in \Sigma :\, \mu (E)<\infty \},$ we will show that the measure $\widetilde{\mu }:\mathcal R \rightarrow [0,\infty )$ given by the restriction of $\mu$ to $\mathcal R$ is not $\mathcal R$-decomposable. Let us see first that

$$\begin{aligned} \mathcal R ^{loc}=\Sigma . \end{aligned}$$

(1)

If $A\in \Sigma ,$ then obviously $A\cap E\in \mathcal R$ for every $E\in \mathcal R ,$ that is $A\in \mathcal R ^{loc}.$ Conversely, suppose that $A\in \mathcal R ^{loc}.$ For a fixed $\gamma \in C,$ the set $G_{\{\gamma \}}=\{f\in X:\, f(\{\gamma \})=1\}$ is in $\Sigma$ and $\mu (G_{\{\gamma \}})=\sharp (\{\gamma \})=1$ (see [16, 216E.(c)]). So, $G_{\{\gamma \}}\in \mathcal R$ and thus $A\cap G_{\{\gamma \}}\in \mathcal R \subset \Sigma \subset \Sigma _{\gamma }.$ If there exists $K\in \mathcal K$ such that $F_{\gamma ,K}\subset A\cap G_{\{\gamma \}}\subset A,$ then $A\in \Sigma _\gamma .$ If there exists $K\in \mathcal K$ such that $F_{\gamma ,K}\subset X\backslash (A\cap G_{\{\gamma \}}),$ then, since $F_{\gamma ,K\cup \{\gamma \}}\subset F_{\gamma ,K}$ and $F_{\gamma ,K\cup \{\gamma \}}\subset G_{\{\gamma \}},$ it follows that $F_{\gamma ,K\cup \{\gamma \}}\subset X\backslash A$ and so $A\in \Sigma _\gamma .$ Therefore, $A\in \Sigma$ and (1) holds. Moreover, for $N\in \mathcal R ^{loc}$ we have that

$$\begin{aligned} N \text{ is} \widetilde{\mu }-\text{ null} \text{ if}\quad \text{ and} \quad \text{ only} \text{ if} N \text{ is} \mu -\text{ null.} \end{aligned}$$

(2)

Indeed, if $N$ is $\mu$-null, for every $E\in \mathcal R \cap 2^N$ we have that $\widetilde{\mu }(E)=\mu (E)\le \mu (N)=0$ and so $N$ is $\widetilde{\mu }$-null. Conversely, suppose that $N$ is $\widetilde{\mu }$-null. If $\mu (N)>0,$ then there exists $\gamma \in C$ such that $\mu (N\cap G_{\{\gamma \}})=1$ (see [16, 216E.(h)]), this is a contradiction as $N\cap G_{\{\gamma \}}\in \mathcal R \cap 2^N$ and so $\mu (N\cap G_{\{\gamma \}})=\widetilde{\mu }(N\cap G_{\{\gamma \}})=0.$

Suppose that $\widetilde{\mu }$ is $\mathcal R$-decomposable, that is, we can write $X=\left(\cup _{\alpha \in \Delta }X_\alpha \right)\cup N$ where $\{X_\alpha :\, \alpha \in \Delta \}$ is a family of pairwise disjoint sets in $\mathcal R$ satisfying that

• (i) if $A_\alpha \in \mathcal R \cap 2^{X_\alpha }$ for all $\alpha \in \Delta ,$ then $\cup _{\alpha \in \Delta }A_\alpha \in \mathcal R ^{loc},$

• (ii) if $Z_\alpha \in \mathcal R \cap 2^{X_\alpha }$ is $\widetilde{\mu }$-null for all $\alpha \in \Delta ,$ then $\cup _{\alpha \in \Delta }Z_\alpha$ is $\widetilde{\mu }$-null,

and $N\in \mathcal R ^{loc}$ is a $\widetilde{\mu }$-null set disjoint with each $X_\alpha .$ Then, $\{X_\alpha :\, \alpha \in \Delta \}\cup \{N\}$ is a disjoint family of sets in $\Sigma$ with $\mu (N), \mu (X_\alpha )<\infty .$ Let us see that

$$\begin{aligned} \Sigma =\{E\subset X: E\cap N\in \Sigma \text{ and} E\cap X_\alpha \in \Sigma \text{ for} \text{ all} \alpha \in \Delta \}. \end{aligned}$$

If $E\in \Sigma ,$ then obviously $E\cap X_\alpha \in \Sigma$ for all $\alpha \in \Delta$ and, by (1), $E\cap N\in \Sigma .$ Conversely, if $E\subset X$ is such that $E\cap N\in \Sigma$ and $E\cap X_\alpha \in \Sigma$ for all $\alpha \in \Delta ,$ since $E\cap X_\alpha \in \mathcal R \cap 2^{X_\alpha },$ by (i) and (1), we have that $\cup _{\alpha \in \Delta }E\cap X_\alpha \in \Sigma .$ So, $E=E\cap X=(\cup _{\alpha \in \Delta }E\cap X_\alpha )\cup (E\cap N)\in \Sigma .$ Moreover, $\mu (E)=\sum _{\alpha \in \Delta }\mu (E\cap X_\alpha )$ for every $E\in \Sigma .$ Indeed, if $\sum _{\alpha \in \Delta }\mu (E\cap X_\alpha )<\infty ,$ then $\mu (E\cap X_\alpha )=0$ for all $\alpha \in \Delta \backslash \Gamma$ for some countable $\Gamma \subset \Delta .$ Since, by (ii) and (2), $\cup _{\alpha \in \Delta \backslash \Gamma }E\cap X_\alpha$ is $\mu$-null,

$$\begin{aligned} \mu (E)=\mu (\cup _{\alpha \in \Gamma }E\cap X_\alpha )=\sum _{\alpha \in \Gamma }\mu (E\cap X_\alpha )=\sum _{\alpha \in \Delta }\mu (E\cap X_\alpha ). \end{aligned}$$

If $\sum _{\alpha \in \Delta }\mu (E\cap X_\alpha )=\infty$ then $\mu (E)=\infty ,$ as $\sup _{\mathop {J \subset \Delta }\limits _{{\mathrm{finite}}}}\sum _{\alpha \in J}\mu (E\cap X_\alpha )\le \mu (E).$ Therefore $(X,\Sigma ,\mu )$ is decomposable which is a contradiction.

So, $\widetilde{\mu }$ is not $\mathcal R$-decomposable. However, since $L^1(\widetilde{\mu })=L_w^1(\widetilde{\mu })$ as $\widetilde{\mu }$ takes values in $\mathbb R ,$ we have that $L_w^1(\widetilde{\mu })$ has the Fatou property (see Proposition 5.4).

Now we can say that there is no relation between the main properties used in this paper, $\mathcal R$-decomposability and local $\sigma$-finiteness. Indeed, the vector measure given in the example above is locally $\sigma$-finite (see Remark 4.5) but not $\mathcal R$-decomposable. However, the vector measure given in Example 4.1 is $\mathcal R$-decomposable, since it is discrete but not locally $\sigma$-finite.

6 Representation theorems for Banach lattices

It is always interesting to know when a Banach lattice is order isometric to some Banach function space. This problem has been studied using vector measures by several authors. It was proved in [5, Theorem 8] that every order continuous Banach lattice with a weak unit is order isometric to an space $L^1(\nu )$ for a vector measure $\nu$ defined on a $\sigma$-algebra. This result allows to represent any Banach lattice $E$ with the $\sigma$-Fatou property with a weak unit belonging to $E_a$ as an space $L_w^1(\nu )$ with $\nu$ defined on a $\sigma$-algebra, since in this case the order isometry between $E_a$ and $L^1(\nu )$ can be extended to $E$ and turns out to be an order isometry between $E$ and $L_w^1(\nu ),$ see [6, Theorem 2.5]. So, we have the following equivalences between classes of spaces:

$$\begin{aligned} \left\{ \begin{array}{l} E \text{ order} \text{ continuous} \text{ Banach} \\ \text{ lattice} \text{ with} \text{ a} \text{ weak} \text{ unit} \end{array} \right\} \equiv \left\{ \begin{array}{l} L^1(\nu ) \text{ with} \nu \text{ on} \text{ a} \sigma \text{-algebra} \end{array} \right\} \end{aligned}$$

and

$$\begin{aligned} \left\{ \begin{array}{l} E \text{ Banach} \text{ lattice} \text{ with} \text{ the} \\ \sigma \text{-Fatou} \text{ property} \text{ such} \text{ that} \\ E_a \text{ has} \text{ a} \text{ weak} \text{ unit} \end{array} \right\} \equiv \left\{ \begin{array}{l} L_w^1(\nu ) \text{ with} \nu \text{ on} \text{ a} \sigma \text{-algebra}\end{array} \right\} . \end{aligned}$$

(3)

For versions with $E$ being $p$-convex see [15, Proposition 2.4] and [8, Theorem 4]. If we forget about the weak unit, it was stated in [4, pp. 22–23] and proved in detail in [12, Theorem 5] that

$$\begin{aligned} \left\{ \begin{array}{l} E \text{ order} \text{ continuous} \text{ Banach} \text{ lattice} \end{array} \right\} \equiv \left\{ \begin{array}{l} L^1(\nu ) \text{ with} \nu \text{ on} \text{ a} \delta \text{-ring} \end{array} \right\} . \end{aligned}$$

Moreover, from [12, Theorem 9] and Theorems 3.2, 4.2, 5.8, we have that

$$\begin{aligned} \left\{ \begin{array}{l} E \text{ Banach} \text{ lattice} \text{ with} \text{ the} \text{ Fatou} \text{ property} \\ \text{ such} \text{ that} E_a \text{ is} \text{ order} \text{ dense} \text{ in} E\end{array} \right\} \equiv \left\{ \begin{array}{l} L_w^1(\nu ) \text{ with} \nu \text{ on} \text{ a} \delta \text{-ring} \\ \text{ being}\, \mathcal R - \text{ decomposable}\end{array} \right\} . \end{aligned}$$

Note that although the converse of Theorem 5.8 does not hold, if $L_w^1(\nu )$ has the Fatou property, by Theorems 3.2 and 4.2, there exists an $\mathcal R$-decomposable vector measure $\widetilde{\nu }$ such that $L_w^1(\nu )$ is order isometric to $L_w^1(\widetilde{\nu }).$

Now, we add another equivalence:

$$\begin{aligned} \left\{ \begin{array}{l} E \text{ Banach} \text{ lattice} \text{ with} \text{ the} \\ \sigma \text{-Fatou} \text{ property} \text{ such} \text{ that}\\ E_a \text{ is} \text{ super} \text{ order} \text{ dense} \text{ in} E \end{array} \right\} \equiv \left\{ \begin{array}{l} [L^1(\nu )]_{_{\sigma \mathrm{-F}}} \text{ with} \nu \text{ on} \text{ a} \delta \text{-ring}\end{array} \right\} . \end{aligned}$$

(4)

Indeed, since $L^1(\nu )\subset [L^1(\nu )]_{_{\sigma \mathrm{-F}}}\subset L_w^1(\nu ),$ then $\left([L^1(\nu )]_{_{\sigma \mathrm{-F}}}\right)_a\subset \left(L_w^1(\nu )\right)_a$ and so, from Theorem 3.2, we have that $\left([L^1(\nu )]_{_{\sigma \mathrm{-F}}}\right)_a=L^1(\nu )$ which is super order dense in $[L^1(\nu )]_{_{\sigma \mathrm{-F}}},$ see the last part of the proof of Theorem 5.1. Let us prove the converse containment.

Proposition 6.1

Every Banach lattice $E$ with the $\sigma$-Fatou property such that $E_a$ is super order dense in $E$ is order isometric to $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ for some vector measure $\nu$ defined on a $\delta$-ring.

Proof

Let $E$ be a Banach lattice with the $\sigma$-Fatou property such that $E_a$ is super order dense in $E$ and consider the vector measure $\nu$ defined on a $\delta$-ring such that the integration operator $I_\nu :L^1(\nu )\rightarrow E_a$ given by $I_\nu (f)=\int f\,d\nu$ for all $f\in L^1(\nu ),$ is an order isometry, see [12, Theorem 5]. Let us extend $I_{\nu }$ to $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}.$ First, consider $0\le f \in [L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ and take $(f_n)\subset L^1(\nu )$ such that $0\le f_n \uparrow f.$ This is always possible since $L^1(\nu )$ is super order dense in $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ as we have noted above. Since $I_{\nu }$ is an order isometry, the sequence $\left(I_\nu (f_n)\right)\subset E_a\subset E$ satisfies that $0\le I_\nu (f_n) \uparrow$ and $\sup \Vert I_\nu (f_n) \Vert _E = \sup \Vert f_n \Vert _\nu \le \Vert f \Vert _\nu < \infty .$ Then, as $E$ has the $\sigma$-Fatou property, there exists $e=\sup I_\nu (f_n)$ in $E$ and $\Vert e \Vert _E=\sup \Vert I_\nu (f_n) \Vert _E.$ We define $T(f)=e.$

A similar argument to the one in [6, Theorem 2.5], shows that $T$ is well defined. To be precise, take another sequence $(g_n)\subset L^1(\nu )$ such that $0 \le g_n\uparrow f$ and denote $z=\sup I_\nu (g_n).$ Let $0\le x^*\in E^*$ be fixed. Then, $x^*(e)\ge x^*\left(I_\nu (f_n)\right)=\int f_n\,dx^*\nu$ for all $n.$ Since $0\le f_n\uparrow f\,\nu$-a.e. and so $x^*\nu$-a.e., by using the monotone convergence theorem, we have that $x^*(e)\ge \int f\,dx^*\nu \ge x^*\left(I_\nu (f_n)\right)$ for all $n.$ In a similar way, $x^*(z)\ge \int f\,dx^*\nu \ge x^*\left(I_\nu (g_n)\right)$ for all $n.$ Thus, it follows that $x^*(e)\ge x^*\left(I_\nu (g_n)\right)$ and $x^*(z)\ge x^*\left(I_\nu (f_n)\right)$ for all $n.$ Since this holds for all $0\le x^*\in E^*,$ we have that $e\ge I_\nu (g_n)$ and $z\ge I_\nu (f_n)$ for all $n.$ Then, $e\ge z$ and $z\ge e,$ and so $e=z.$ So, $T$ is well defined. Moreover,

$$\begin{aligned} \Vert T(f)\Vert _E = \Vert e\Vert _E= \sup \Vert I_\nu (f_n) \Vert _E = \sup \Vert f_n \Vert _\nu =\Vert f \Vert _\nu , \end{aligned}$$

where in the last equality we have used that $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ has the $\sigma$-Fatou property. Let us see now that $T$ preserves the lattice structure, that is $T(f\wedge g)=Tf\wedge Tg$ for every $0\le f, g\in [L^1(\nu )]_{_{\sigma \mathrm{-F}}}.$ Consider sequences $(f_n), (g_n)\subset L^1(\nu )$ satisfying that $0\le f_n\uparrow f$ and $0\le g_n\uparrow g.$ Then, $Tf=\sup I_\nu (f_n)$ and $Tg=\sup I_\nu (g_n).$ Note that if $x_n\uparrow x$ and $y_n\uparrow y$ in a Banach lattice then $x_n\wedge y_n\uparrow x\wedge y,$ see for instance [20, Theorem 15.3]. Then, since $0\le f_n\wedge g_n\uparrow f\wedge g$ with $(f_n\wedge g_n)\subset L^1(\nu )$ and $I_\nu$ is an order isometry, we have that

$$\begin{aligned} T(f\wedge g)=\sup I_\nu (f_n\wedge g_n)=\sup I_\nu (f_n)\wedge I_\nu (g_n)=Tf\wedge Tg. \end{aligned}$$

For a general $f\in [L^1(\nu )]_{_{\sigma \mathrm{-F}}},$ we define $Tf=Tf^+-Tf^-$ where $f^+$ and $f^-$ are the positive and negative parts of $f$ respectively. So, $T:[L^1(\nu )]_{_{\sigma \mathrm{-F}}} \rightarrow E$ is a positive linear operator extending $I_\nu .$ For the linearity, see for instance [20, Theorem 15.2]. Moreover $T$ is an isometry. Indeed, $Tf^+\wedge Tf^-=T(f^+\wedge f^-)=0$ as $f^+\wedge f^-=0,$ and so $|Tf|=|Tf^+-Tf^-|=Tf^++Tf^-=T|f|,$ see [20, Theorem 14.4]. Then, $\Vert T(f)\Vert _E=\Vert T(|f|)\Vert _E=\Vert f\Vert _\nu$ for all $f\in [L^1(\nu )]_{_{\sigma \mathrm{-F}}}.$

Let us prove that T is onto. Let $0\le e \in E.$ Since $E_a$ is super order dense in $E,$ there exists $(e_n)\subset E_a$ such that $0\le e_n \uparrow e.$ Let $(f_n) \subset L^1(\nu )\subset [L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ be such that $e_n=I_\nu (f_n).$ Since $I_\nu ^{-1}$ is an order isometry, we have that $0\le f_n\uparrow$ and $\sup \Vert f_n \Vert _\nu =\sup \Vert e_n\Vert _E \le \Vert e\Vert _E<\infty .$ Then, by the $\sigma$-Fatou property of $[L^1(\nu )]_{_{\sigma \mathrm{-F}}},$ there exists $f=\sup f_n$ in $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}.$ From the definition of $T,$ we have that $Tf=\sup I_\nu (f_n)=\sup e_n=e.$ For a general $e\in E,$ consider $e^+$ and $e^-$ the positive and negative parts of $e.$ Let $g, h\in [L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ be such that $Tg=e^+$ and $Th=e^-.$ Then, taking $f=g-h\in [L^1(\nu )]_{_{\sigma \mathrm{-F}}}$ we have that $Tf=e.$ Note that $T^{-1}$ is positive. So, $T$ is positive, linear, one to one and onto with inverse being positive, then $T$ is an order isomorphism (see [19, p. 2]). $\square$

Note that the class of spaces in (3) is contained in the one in (4). Indeed, take a weak unit $0\le u\in E_a.$ Then $0\le e\wedge n\,u\uparrow e$ for each $0\le e\in E$ where $e\wedge n\,u\in E_a,$ and so $E_a$ is super order dense in $E.$ In this case we obtain that $[L^1(\nu )]_{_{\sigma \mathrm{-F}}}=L_w^1(\nu ),$ since $\nu$ is defined on a $\sigma$-algebra.

7 Example

We end by showing that there exist $\mathcal R$-decomposable vector measures $\nu$ which are not $\sigma$-finite nor discrete.

Let $\Gamma$ be an abstract set. For each $\gamma \in \Gamma ,$ consider a non null vector measure $\nu _\gamma :\Sigma _\gamma \rightarrow X_\gamma$ defined on a $\sigma$-algebra $\Sigma _\gamma$ of subsets of a set $\Omega _\gamma$ and with values in a Banach space $X_\gamma .$ Take the set $\Omega =\cup _{\gamma \in \Gamma }\{\gamma \}\times \Omega _\gamma$ and the $\delta$-ring $\mathcal R$ of subsets of $\Omega$ given by the sets $\cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma$ with $A_\gamma \in \Sigma _\gamma$ for all $\gamma \in \Gamma ,$ for which there exists a finite set $J\subset \Gamma$ such that $A_\gamma$ is $\nu _\gamma$-null for all $\gamma \in \Gamma \backslash J,$ see [12, p. 5]. Then,

$$\begin{aligned} \mathcal R ^{loc}=\left\{ \cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma : \, A_\gamma \in \Sigma _\gamma \text{ for} \text{ all} \gamma \in \Gamma \right\} . \end{aligned}$$

Note that a function $f:\Omega \rightarrow \mathbb R$ is $\mathcal R ^{loc}$-measurable if and only if $f(\gamma ,\cdot ):\Omega _\gamma \rightarrow \mathbb R$ is $\Sigma _\gamma$-measurable for all $\gamma \in \Gamma .$

Denote by $c_0\left(\Gamma {,}(X_\gamma )_{\gamma \in \Gamma }\right)$ the Banach space of all families $(x_\gamma )_{\gamma \in \Gamma }$ such that $x_\gamma \!\in \! X_\gamma$ for every $\gamma \in \Gamma$ and $\left(\Vert x_\gamma \Vert _{_{X_{\gamma }}}\right)_{\gamma \in \Gamma }\in c_0(\Gamma ),$ endowed with the norm $\Vert (x_\gamma )_{\gamma \in \Gamma }\Vert =\sup _{\gamma \in \Gamma }\Vert x_\gamma \Vert _{_{X_{\gamma }}}.$ Note that the topological dual $c_0\left(\Gamma ,(X_\gamma )_{\gamma \in \Gamma }\right)^*$ can be identified with the Banach space $\ell ^1\left(\Gamma ,(X_\gamma ^*)_{\gamma \in \Gamma }\right)$ of families $(x_\gamma ^*)_{\gamma \in \Gamma }$ such that $x_\gamma ^*\in X_\gamma ^*$ for every $\gamma \in \Gamma$ and $\left(\Vert x_\gamma ^*\Vert _{_{X_{\gamma }^*}}\right)_{\gamma \in \Gamma }\in \ell ^1(\Gamma ),$ endowed with the norm $\Vert (x_\gamma ^*)_{\gamma \in \Gamma }\Vert =\sum _{\gamma \in \Gamma }\Vert x_\gamma ^*\Vert _{_{X_{\gamma }}}.$ The action of any $x^*=(x_\gamma ^*)_{\gamma \in \Gamma }\in \ell ^1\left(\Gamma ,(X_\gamma ^*)_{\gamma \in \Gamma }\right)$ on $x=(x_\gamma )_{\gamma \in \Gamma }\in c_0\left(\Gamma ,(X_\gamma )_{\gamma \in \Gamma }\right)$ is given by $x^*(x)=\sum _{\gamma \in \Gamma }x_\gamma ^*(x_\gamma ).$

Consider the vector measure $\nu :\mathcal R \rightarrow c_0\left(\Gamma ,(X_\gamma )_{\gamma \in \Gamma }\right)$ given by

$$\begin{aligned} \nu \left(\cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma \right)=\left(\nu _\gamma (A_\gamma )\right)_{\gamma \in \Gamma }. \end{aligned}$$

Note that a set $A=\cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma \in \mathcal R ^{loc}$ is $\nu$-null if and only if $A_\gamma$ is $\nu _\gamma$-null for all $\gamma \in \Gamma .$ Then, it is direct to check that:

• (a) $\nu$ is $\mathcal R$-decomposable.

• (b) $\nu$ is $\sigma$-finite if and only if $\Gamma$ is countable.

• (c) $\nu$ is discrete if and only if $\nu _\gamma$ is discrete for all $\gamma \in \Gamma .$

Let us prove that $L_w^1(\nu )$ can be described as the space of functions $f\in \mathcal M (\mathcal R ^{loc})$ such that $f(\gamma ,\cdot )\in L_w^1(\nu _\gamma )$ for all $\gamma \in \Gamma$ with $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in \ell ^\infty (\Gamma ),$ and moreover, $\Vert f\Vert _\nu =\sup _{\gamma \in \Gamma }\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma }$ for all $f\in L_w^1(\nu ),$ that is,

$$\begin{aligned} L_w^1(\nu )=\ell ^\infty \left(\Gamma ,(L_w^1(\nu {_\gamma }))_{\gamma \in \Gamma }\right). \end{aligned}$$

Given $x^*=(x_\gamma ^*)_{\gamma \in \Gamma }\in \ell ^1\left(\Gamma ,(X_\gamma ^*)_{\gamma \in \Gamma }\right),$ since $|x^*\nu |(A)=\sum _{\gamma \in \Gamma }|x_\gamma ^*\nu _\gamma |(A_\gamma )\le \infty$ for every $A=\cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma \in \mathcal R ^{loc},$ we have that

$$\begin{aligned} \int |f|\,d|x^*\nu |=\sum _{\gamma \in \Gamma }\int |f(\gamma ,\cdot )|\,d|x_\gamma ^*\nu _\gamma |\le \infty , \text{ for} \text{ all} f\in \mathcal M (\mathcal R ^{loc}). \end{aligned}$$

(5)

Indeed, (5) holds for $\mathcal R ^{loc}$-simple functions, and so for a general $f$ by using the monotone convergence theorem. Let us see that if $f\in L^1(x^*\nu ),$ then

$$\begin{aligned} \int \limits _A f\,dx^*\nu =\sum _{\gamma \in \Gamma }\int \limits _{A_\gamma } f(\gamma ,\cdot )\,dx_\gamma ^*\nu _\gamma . \end{aligned}$$

(6)

In this case, by (5), $f(\gamma ,\cdot )\in L^1(x_\gamma ^*\nu _\gamma )$ for every $\gamma \in \Gamma$ and $\int |f(\gamma ,\cdot )|\,d|x_\gamma ^*\nu _\gamma |=0$ (and so $f(\gamma ,\cdot )=0$ except on a $x_\gamma ^*\nu _\gamma$-null set $Z_\gamma$) for all $\gamma \in \Gamma \backslash J$ with $J$ being some countable subset of $\Gamma .$ Then, $f\chi _A=f\chi _{\cup _{\gamma \in J}\{\gamma \}\times A_\gamma }\,\nu$-a.e. (except on the $\nu$-null set $\cup _{\gamma \in \Gamma \backslash J}\{\gamma \}\times A_\gamma \cap Z_\gamma$). By using the dominated convergence theorem, we have that

$$\begin{aligned} \int \limits _Af\,dx^*\nu =\sum _{\gamma \in J}\int \limits _{\{\gamma \}\times A_\gamma }f\,dx^*\nu . \end{aligned}$$

Noting that $\int _{\{\gamma \}\times A_\gamma }f\,dx^*\nu =\int _{A_\gamma }f(\gamma ,\cdot )\,dx_\gamma ^*\nu _\gamma$ holds for $\mathcal R ^{loc}$-simple functions and so for any $f\in L^1(x^*\nu )$ by density of the $\mathcal R ^{loc}$-simple functions in $L^1(x^*\nu ),$ we conclude that (6) holds.

Let $f\in L_w^1(\nu )$ and fix $\beta \in \Gamma .$ Given $x_\beta ^*\in X_\beta ^*,$ define the element $x^*=(x_\gamma ^*)_{\gamma \in \Gamma }$ in $\ell ^1\left(\Gamma ,(X_\gamma ^*)_{\gamma \in \Gamma }\right)$ by $x_\gamma ^*=x_\beta ^*$ if $\gamma =\beta$ and $x_\gamma ^*=0$ in other case. Then, from (5), we have that $\int |f(\beta ,\cdot )|\,d|x_\beta ^*\nu _\beta |=\int |f|\,d|x^*\nu |<\infty$ and so $f(\beta ,\cdot )\in L_w^1(\nu _\beta )$ with $\Vert f(\beta ,\cdot )\Vert _{\nu _\beta }\le \Vert f\Vert _\nu .$ Thus, $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in \ell ^\infty (\Gamma )$ and $\sup _{\gamma \in \Gamma }\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma }\le \Vert f\Vert _\nu .$

Let now $f\in \mathcal M (\mathcal R ^{loc})$ satisfying that $f(\gamma ,\cdot )\in L_w^1(\nu _\gamma )$ for every $\gamma \in \Gamma$ and $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in \ell ^\infty (\Gamma ).$ Given $x^*=(x_\gamma ^*)_{\gamma \in \Gamma }\in \ell ^1\left(\Gamma ,(X_\gamma ^*)_{\gamma \in \Gamma }\right),$ from (5), we have that

$$\begin{aligned} \int |f|\,d|x^*\nu |&= \sum _{\gamma \in \Gamma }\int |f(\gamma ,\cdot )|\,d|x_\gamma ^*\nu _\gamma |\le \sum _{\gamma \in \Gamma }\Vert x_\gamma ^*\Vert _{_{X_\gamma ^*}}\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma } \\&\le \sup _{\gamma \in \Gamma }\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma }\sum _{\gamma \in \Gamma }\Vert x_\gamma ^*\Vert _{_{X_\gamma ^*}}<\infty . \end{aligned}$$

Then, $f\in L_w^1(\nu )$ and $\Vert f\Vert _\nu \le \sup _{\gamma \in \Gamma }\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma }.$

Moreover, $L^1(\nu )$ can be described as the space of functions $f\in \mathcal M (\mathcal R ^{loc})$ such that $f(\gamma ,\cdot )\in L^1(\nu _\gamma )$ for every $\gamma \in \Gamma$ with $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in c_0(\Gamma ),$ that is,

$$\begin{aligned} L^1(\nu )=c_0\left(\Gamma ,(L_w^1(\nu {_\gamma }))_{\gamma \in \Gamma }\right). \end{aligned}$$

Indeed, if $f\in L^1(\nu )$ we can take $(\varphi _n)\subset \mathcal S (\mathcal R )$ such that $\varphi _n\rightarrow f$ in $L^1(\nu ).$ For each $\gamma \in \Gamma ,$ we have that $f(\gamma ,\cdot )\in L_w^1(\nu _\gamma )$ (as $f\in L_w^1(\nu )$) and $(\varphi _n(\gamma ,\cdot ))\subset \mathcal S (\Sigma _\gamma )\subset L^1(\nu _\gamma ).$ Then, since $\Vert f(\gamma ,\cdot )-\varphi _n(\gamma ,\cdot )\Vert _{\nu _\gamma }\le \Vert f-\varphi _n\Vert _\nu$ and $L^1(\nu _\gamma )$ is closed in $L_w^1(\nu _\gamma ),$ it follows that $f(\gamma ,\cdot )\in L^1(\nu _\gamma ).$ On the other hand, for each $n$ we can write $\varphi _n=\sum _{j=1}^m\alpha _j\chi _{A_j}$ where $\alpha _j\in \mathbb R$ and $A_j=\cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma ^j.$ Here, $A_\gamma ^j\in \Sigma _\gamma$ for all $\gamma \in \Gamma$ and satisfies that $A_\gamma ^j$ is $\nu _\gamma$-null for all $\gamma \in \Gamma \backslash J_j$ for some finite set $J_j\subset \Gamma .$ Then, $\varphi _n(\gamma ,\cdot )=\sum _{j=1}^m\alpha _j\chi _{A_\gamma ^j}=0\,\nu _\gamma$-a.e. for all $\gamma \in \Gamma \backslash \cup _{j=1}^mJ_j$ where $\cup _{j=1}^mJ_j$ is a finite set, and so $(\Vert \varphi _n(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in c_0(\Gamma ).$ Since $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in \ell ^\infty (\Gamma )$ and

$$\begin{aligned} \sup _{\gamma \in \Gamma }\big |\,\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma }-\Vert \varphi _n(\gamma ,\cdot )\Vert _{\nu _\gamma }\big | \le \sup _{\gamma \in \Gamma }\Vert f(\gamma ,\cdot )-\varphi _n(\gamma ,\cdot )\Vert _{\nu _\gamma }=\Vert f-\varphi _n\Vert _\nu , \end{aligned}$$

it follows that $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in c_0(\Gamma ).$

Conversely, suppose that $f\in \mathcal M (\mathcal R ^{loc})$ is such that $f(\gamma ,\cdot )\in L^1(\nu _\gamma )$ for all $\gamma \in \Gamma$ and $(\Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma })_{\gamma \in \Gamma }\in c_0(\Gamma ).$ In particular, $f\in L_w^1(\nu ).$ Given an element $x^*=(x_\gamma ^*)_{\gamma \in \Gamma }\in \ell ^1\left(\Gamma ,(X_\gamma ^*)_{\gamma \in \Gamma }\right)$ and $A=\cup _{\gamma \in \Gamma }\{\gamma \}\times A_\gamma \in \mathcal R ^{loc},$ we note that $\left(\int _{A_\gamma }f(\gamma ,\cdot )\,d\nu _\gamma \right)_{\gamma \in \Gamma }\in c_0\left(\Gamma ,(X_\gamma )_{\gamma \in \Gamma }\right)$ as $\Vert \int _{A_\gamma }f(\gamma ,\cdot )\,d\nu _\gamma \Vert _{X_\gamma }\le \Vert f(\gamma ,\cdot )\Vert _{\nu _\gamma }$ for each $\gamma \in \Gamma .$ Moreover, by (6),

$$\begin{aligned} x^*\left(\left(\int \limits _{A_\gamma }f(\gamma ,\cdot )\,d\nu _\gamma \right)_{\gamma \in \Gamma }\right)&= \sum _{\gamma \in \Gamma }x_\gamma ^*\left(\int \limits _{A_\gamma }f(\gamma ,\cdot )\,d\nu _\gamma \right) \\&= \sum _{\gamma \in \Gamma }\int \limits _{A_\gamma }f(\gamma ,\cdot )\,dx_\gamma ^*\nu _\gamma =\int \limits _Af\,dx^*\nu . \end{aligned}$$

So, $f\in L^1(\nu )$ and $\int _A f\,d\nu =\left(\int _{A_\gamma } f(\gamma ,\cdot )\,d\nu _\gamma \right)_{\gamma \in \Gamma }.$

Note that if $\nu$ is locally $\sigma$-finite, since $h=\sum _{\gamma \in \Gamma }\frac{1}{\Vert \nu _\gamma \Vert (\Omega _\gamma )}\chi _{\{\gamma \}\times \Omega _\gamma }\in L_w^1(\nu )$ and $\text{ supp}(h)=\Omega ,$ from Proposition 4.9, it follows that $\nu$ is $\sigma$-finite. So, in this case $\nu$ is locally $\sigma$-finite if and only if $\nu$ is $\sigma$-finite if and only if $\Gamma$ is countable.

In particular, consider a non atomic measure space $(\Theta ,\Sigma ,\mu )$ and an order continuous B.f.s. $X$ related to $\mu$ which does not contain any copy of $c_0$ and such that $\chi _\Theta \!\in X,$ for instance $X\!=\!L^p[0,1]$ related to the Lebesgue measure for $p\!\ge \!1.$ Then, $\eta :\Sigma \!\rightarrow X$ given by $\eta (A)\!=\!\chi _A$ for all $A\!\in \Sigma ,$ is a non discrete vector measure such that $L_w^1(\nu )\!=\!L^1(\nu )\!=\!X.$ Taking $\Gamma$ uncountable and $\nu _\gamma =\eta$ for all $\gamma \in \Gamma ,$ we obtain an $\mathcal R$-decomposable vector measure $\nu$ which is not $\sigma$-finite nor discrete. In this case, $L_w^1(\nu )=\ell ^\infty (\Gamma ,X)$ and $L^1(\nu )=c_0(\Gamma , X).$