Locally integrable functions and their indefinite integrals

Abstract

Consider a locally determined positive measure space $(\Omega ,\Sigma ,\mu )$ and a function $F:\Omega \rightarrow X$ taking values in a Banach space. When F is locally (Pettis or Bochner) integrable with respect to $\mu$, a vector measure $\nu _F$ with density F defined on a $\delta$-ring is obtained. We study the vector measure $\nu _F$ and its properties. We present the relation between the Banach spaces of integrable functions with respect to $\nu _F$ and the spaces of Dunford, Pettis or Bochner integrable functions.

1 Introduction

Let us consider a Banach space X and a $\sigma$-algebra $\Sigma$ on $\Omega$. The integral of X-valued functions with respect to a positive finite measure defined in $\Sigma$ was introduced by B. J. Pettis and S. Bochner in the thirties of the last century and then this theory has been studied in depth by several authors for instance [5] and [7]. Later, the theory of integration of scalar valued functions with respect to X-valued measures defined in $\Sigma$, which are called vector measures, begins to be developed. As one might expect the Pettis and the Bochner integrals define vector measures, these ones were studied among other by J. Diestel and J. J. Uhl in [5]. Otherwise in [6] N. Dinculeanu and J. J. Uhl considered a locally $\sigma$-finite measure defined on a $\delta$-ring $\mathcal {R}$ of subset of $\Omega$ and they introduced the concept of $\mathcal {R}$-locally Pettis or Bochner integrable function, namely a weakly (strongly) $\mu$-measurable function $F:\Omega \rightarrow X$ such that the function $\chi _BF$ is Pettis (Bochner) integrable, for all $B\in \mathcal {R}$. It turns out that these integrals define vector measures now on the $\delta$-ring $\mathcal {R}$. At the same time D. R. Lewis begins to develop the theory of integration with respect to vector measures defined on $\delta$-rings in [8]. Subsequently, other authors study in depth these kind of measures and their inherent spaces of integrable functions; for the concepts and notations used in this note one should look [3].

Given any positive measure $\mu$ on $\Sigma$ we can define the Dunford, Pettis and Bochner integrals of functions with values in X in analogous way to the case in which $\mu$ is finite. Also, we can obtain the spaces consisting of the Dunford and Pettis integrable functions which turn out to be normed spaces and the space of the Bochner integrable functions is a Banach space. On the other hand, the collection $\Sigma ^f$ consisting of those subsets in $\Sigma$ which have $\mu$ finite measure is a $\delta$-ring. When the measure $\mu$ is locally determined, it turns out that if $\mu$ is restricted to $\Sigma ^f$, a locally $\sigma$-finite measure is obtained; so makes sense to consider vector functions which are $\Sigma ^f$-locally Pettis or Bochner integrable, these functions will be simply called locally Pettis and locally Bochner integrable and the vector spaces obtained will be denoted by $\mathbb {P}(\mu , X)^{loc}$ and $\mathbb {B}(\mu , X)^{loc}$, respectively .

We begin this note recalling the basic concepts relative to the Dunford, the Pettis and the Bochner integrals with respect to a positive measure and also the main results about vector measures defined on $\delta$-rings, in Sect. 2. In Sect. 3 we study briefly the vector measure $\nu _F$ defined on $\Sigma ^f$ by the integral of a locally Pettis integrable function $F:\Omega \rightarrow X$ over each $B\in \Sigma ^f$ and a description of its corresponding semivariation is given. If additionally the function F is locally Bochner integrable we provide a characterization of the variation of the measure $\nu _F$. Finally, in Sect. 4, we present the existing connection between the integrable functions with respect to $\nu _F$ and the Dunford, Pettis or Bochner integrable functions as well as their corresponding integrals. In this way Theorems 8 and 13 established by G.F. Stefansson in [11] are generalized in two directions, namely the positive measure is no longer necessarily finite but locally determined and the function F is now locally integrable. Besides we obtain some conditions to determine whether a locally Pettis integrable function is in fact Pettis integrable.

2 Preliminary results

2.1 Bochner and Pettis integrals

Throughout the paper $\Omega$ will be a non empty set and X stands for a Banach space over $\mathbb {K}$ ($\mathbb {R}$ or $\mathbb {C}$). We denote by $X^*$ and $B_X$ its dual space and its unit ball respectively. Let us consider a $\sigma$-algebra $\Sigma$ on $\Omega$ and a positive measure $\mu :\Sigma \rightarrow [0,\infty ]$. The set $\Sigma ^f$ consists of the subsets $B\in \Sigma$ such that $\mu (B)<\infty$ and $\mathcal {N}_0(\mu )$ is the collection of $\mu$-null sets. Recall that the measure $\mu$ is said to be semi-finite if for each set $A\in \Sigma$ such that $\mu (A)>0$, there exists a subset $B\in \Sigma ^f$ satisfying that $B\subset A$ and $0<\mu (B)$. The measure $\mu$ is locally determined if it is semi-finite and $\Sigma =\{A\subset \Omega \ | \ A\cap B\in \Sigma , \ \forall \ B\in \Sigma ^f\}$.

We denote by $St(\mu ,X)$ the vector space of X-valued simple functions whose support has finite measure. An X-valued function $F:\Omega \rightarrow X$ is said to be strongly $\mu$ -measurable if there exists a sequence $\{S_n\}\subset St(\mu ,X)$, which converges pointwise to F $\mu$-a.e. and to be weakly $\mu$ -measurable if $\langle F,x^*\rangle :\Omega \rightarrow \mathbb {R}$ is $\mu$-measurable for any $x^*\in X^*$. Clearly each strongly $\mu$-measurable function is a weakly $\mu$-measurable function. We say that two functions $F,G:\Omega \rightarrow X$ are weakly equal $\mu$-a.e. if $\langle F,x^*\rangle =\langle G,x^*\rangle$ $\mu$-a.e. for all $x^*\in X^*$. We will denote by $L^0(\mu ,X)$ the vector space that consists of the equivalence classes that are obtained by identifying strongly $\mu$-measurable functions if they are equal $\mu$- a.e. and $L^0_w(\mu ,X)$ the vector spaces formed by the equivalence classes that are obtained when we identified weakly $\mu$-measurable functions if they are weakly equal $\mu$-a.e. We write $\mathbb {B}(\mu ,X)$ to indicate the Banach space of the Bochner integrable functions, namely the functions $F\in L^0(\mu ,X)$ such that $\Vert F\Vert _X\in L^1(\mu )$, with the norm defined by $\Vert F\Vert _1=\int _\Omega \Vert F\Vert _Xd\mu$. On the other hand a function $F\in L^0_w(\mu ,X)$ is Dunford integrable when $\langle F,x^*\rangle \in L^1(\mu )$, $\forall \ x^*\in X^*$, if additionally for each $A\in \Sigma$ there exists a vector $x_A\in X$ such that $\int _A \langle F,x^*\rangle d\mu =\langle x_A,x^*\rangle$; $\forall \ x^*\in X^*$, the function F is Pettis integrable and the vector $x_A$ is called the Pettis integral of F over A and it is denoted by $\mathbb {P}-\int _A F d\mu$. We write $\mathbb {D}(\mu ,X)$ and $\mathbb {P}(\mu ,X)$ for the vector spaces consisting of the Dunford and Pettis integrable functions respectively.

Lemma 1

The space $\mathbb {D}(\mu ,X)$ is a normed space with the norm given by

$$\begin{aligned} \Vert F\Vert _\mathbb {D}=\sup _{x^*\in B_{X^*}}\int _\Omega |\langle F,x^*\rangle | d\mu , \ \forall \ F\in \mathbb {D}(\mu ,X). \end{aligned}$$

Proof

Observe that to obtain the conclusion it only remains to establish that $\Vert F\Vert _\mathbb {D}<\infty$, $\forall \ F\in \mathbb {D}(\mu ,X)$. So, let us fix $F\in \mathbb {D}(\mu ,X)$ and define $T:X^*\rightarrow L^1(\mu )$ by $T(x^*)=\langle F,x^*\rangle , \ \forall \ x^*\in X^*$. Clearly T is a well defined linear operator. Now take $\{x^*_n\}\subset X^*$ and $x^*\in X^*$ such that $x^*_n\rightarrow x$. Let us assume that there exists $g\in L^1(\mu )$ satisfying that $Tx^*_n\rightarrow g$ in $L^1(\mu )$. Proceeding as in [4, p.46] we get a subsequence $\{x^*_{n_k}\}\subset \{x^*_n\}$ such that $\langle F,x^*_{n_k}\rangle = Tx^*_{n_k}\rightarrow g$ $\mu$-a.e. On the other hand $\langle F(t),x^*_n\rangle \rightarrow \langle F(t),x^*\rangle$, $\forall \ t\in \Omega$. Thus, $T(x^*)=\langle F(t),x^*\rangle =g$, $\mu$-a.e. By the Closed Graph Theorem we get that T is bounded. Therefore

$$\begin{aligned} \Vert F\Vert _\mathbb {D}=\sup _{x^*\in B_{X^*}}\int _\Omega |\langle F,x^*\rangle |d\mu =\sup _{x^*\in B_{X^*}}\Vert Tx^*\Vert _{L^1(\mu )}\le \Vert T\Vert . \end{aligned}$$

$\square$

Since $\mathbb {P}(\mu ,X)\subset \mathbb {D}(\mu ,X)$ it turns out that $\mathbb {P}(\mu ,X)$ is also a normed space with the same norm $\Vert \cdot \Vert _\mathbb {D}$ which will be denoted by $\Vert \cdot \Vert _\mathbb {P}$ in this case. It is well know that $\mathbb {B}(\mu ,X)\subset \mathbb {P}(\mu ,X)$ with $\Vert F\Vert _\mathbb {P}\le \Vert F\Vert _1$ and $\mathbb {B}-\int _A F d\mu =\mathbb {P}-\int _A Fd\mu$, $\forall \ F\in \mathbb {B}(\mu ,X)$.

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

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 is a function $\lambda :\mathcal {R}\rightarrow \mathbb {K}$ 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 $|\lambda |(A):=\sup \left\{ \sum _{j=1}^{n}|\lambda (A_j)|:\{A_j\}\in \pi _A\right\}$. 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 vector space consisting of the equivalence classes of $\lambda$-integrable functions when we identify two functions if they are equal $|\lambda |$-a.e.

Let X be a Banach space. A set function $\nu :\mathcal {R}\rightarrow X$ is a vector measure if for any collection $\{B_n\}\subset \mathcal {R}$ of pairwise disjoint sets satisfying that $\bigcup _{n=1}^\infty B_n \in \mathcal {R}$, we have that $\sum _{n=1}^\infty \nu (B_n)=\nu (\bigcup _{n=1}^\infty B_n)$. A vector measure $\nu$ is called strongly additive if $\nu (B_n)\rightarrow 0$ whenever $\{B_n\}$ is a disjoint sequence in $\mathcal {R}$. The variation of $\nu$ is the positive measure $|\nu |$ defined in $\mathcal {R}^{loc}$ by $|\nu |(A):=\sup \left\{ \sum _j \Vert \nu (A_j)\Vert _X : \{A_j\}\in \pi _A \right\}$. The semivariation of $\nu$ is the function $\Vert \nu \Vert :\mathcal {R}^{loc}\rightarrow [0,\infty ]$ given by ${\Vert \nu \Vert (A):=\sup \{|\langle \nu ,x^*\rangle |(A):x^*\in B_{X^*}\}}$, where $|\langle \nu ,x^*\rangle |$ is the variation of the scalar measure $\langle \nu ,x^*\rangle :\mathcal {R}\rightarrow \mathbb {K}$, defined by $\langle \nu ,x^*\rangle (B)=\langle \nu (B),x^*\rangle , \ \forall \ B\in \mathcal {R}$. The semivariation of $\nu$ is finite in $\mathcal {R}$ and $\Vert \nu (B)\Vert \le \Vert \nu \Vert (B)$ if $B\in \mathcal {R}$, moreover for any $A\in \mathcal {R}^{loc}$ is satisfied $\displaystyle {\Vert \nu \Vert (A)\le |\nu |(A)}$. 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 two functions $f,g\in L^0(\mathcal {R}^{loc})$ are equal $\nu$-a.e. if they are equal outside of a set in $\mathcal {N}_0(\nu )$.

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.

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 given by

$$\begin{aligned} \Vert f\Vert _\nu :=\sup \left\{ \int _\Omega |f| d|\langle \nu ,x^*\rangle |:x^*\in B_{X^*}\right\} , \end{aligned}$$

$L^1_w(\nu )$ is a Banach space.

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

$$\begin{aligned} \langle x_A,x^*\rangle =\int _A f d\langle \nu ,x^*\rangle , \ \forall \ x^*\in X^*. \end{aligned}$$

(1)

In this case the vector $x_A$ is denoted by $\int _A f d\nu$. With the norm $\Vert \cdot \Vert _\nu$ the subset of all $\nu$-integrable functions is a closed subspace of $L^1_w(\nu )$ and it will be denoted by $L^1(\nu )$. Therefore $L^1(\nu )$ is also a Banach space. 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 $S(\mathcal {R})$ is a dense subspace of $L^1(\nu )$. Finally the integral operator $I_\nu :L^1(\nu )\rightarrow X$ defined by $I_\nu (f)=\int _\Omega f d\nu$, is linear and bounded.

3 Vector valued indefinite integral

Recall that $(\Omega ,\Sigma ,\mu )$ is a positive measure space and X a Banach space. Given a vector valued function $F\in \mathbb {P}(\mu ,X)$ let us define the set function $\tilde{\nu }_F:\Sigma \rightarrow X$ by

$$\begin{aligned} \tilde{\nu }_F(A)=\mathbb {P}-\int _A F d\mu , \ \ \forall \ A\in \Sigma . \end{aligned}$$

(2)

In the case when $\mu$ is finite it is well known that $\tilde{\nu }_F$ is a vector measure [5, Thm. II. 3.5]. The next result generalizes this fact, it can be established in the similar way, using the Orlicz-Pettis Theorem [5, Cor. I.4.4].

Theorem 1

The set function $\tilde{\nu }_F$ defined on (2) is a vector measure with semivariation

$$\begin{aligned} \Vert \tilde{\nu }_F\Vert (A)=\sup _{x^*\in B_{X^*}}\int _A |\langle F,x^*\rangle |d\mu , \ \forall \ A\in \Sigma . \end{aligned}$$

(3)

Proof

Let us fix $x^*\in X^*$ and take a pairwise disjoint countable collection $\{A_n\}\in \Sigma$, then

$$ \begin{array}{lll} \displaystyle {\left\langle \mathbb {P}-\int _{\bigcup _n A_n} F d\mu ,x^*\right\rangle }&=&\displaystyle {\int _{\bigcup _n A_n} \langle F,x^*\rangle d\mu }\\ &=&\displaystyle {\sum _{n=1}^\infty \int _{A_n} \langle F,x^*\rangle d\mu =\sum _{n=1}^\infty \left\langle \mathbb {P}-\int _{A_n}Fd\mu ,x^*\right\rangle }. \end{array} $$

So, $\tilde{\nu }_F$ is weakly $\sigma$-additive, by the Orlicz-Pettis Theorem [5, Cor. I.4.4] $\tilde{\nu }_F$ is a vector measure. On the other hand since $\langle \tilde{\nu }_F, x^*\rangle (A)=\int _A \langle F,x ^*\rangle$, $\forall$ $A\in \Sigma$, $\mu$ is a positive measure and $\langle F,x^*\rangle \in L^1(\mu )$ from [10, Thm. 6.13] we get that

$$\begin{aligned} |\langle \tilde{\nu }_F,x^*\rangle |(A)=\int _A |\langle F,x^*\rangle |d\mu . \end{aligned}$$

It is follows (3). $\square$

Since $\mathbb {B}(\mu ,X)\subset \mathbb {P}(\mu ,X)$ we have the following consequence. In order to get (4) we can proceed as in [5, Thm. II.2.4 iv)].

Corollary 1

Let $F\in \mathbb {B}(\mu ,X)$. Then $\tilde{\nu }_F$ defined on (2) is a vector measure with bounded variation such that

$$\begin{aligned} |\tilde{\nu }_F|(A)=\int _A \Vert F\Vert _X d\mu , \ \forall \ A\in \Sigma . \end{aligned}$$

(4)

Hereafter we will consider a locally determined positive measure $\mu$ on $\Sigma$. Then $\Sigma ^f$ is a $\delta$-ring such that $(\Sigma ^f)^{loc}=\Sigma$. Let us denote the restriction of $\mu$ to $\Sigma ^f$ by $\lambda$. Since $\mu$ is a semi-finite and locally determined positive measure, it turns out that $\lambda$ is a scalar measure such that $|\lambda |=\mu$ ([1, Lemma 4.3]).

Now we will study a kind of vector functions that include vector measures having Pettis or Bochner functions as density functions.

Definition 1

Let $F:\Omega \rightarrow X$ be a vector function.

• a) The function F is locally Pettis integrable if F is weakly $\mu$-measurable function and $\chi _B F\in \mathbb {P}(\mu ,X)$, $\forall$ $B\in \Sigma ^f$. The collection of equivalence classes obtained by identifying locally Pettis integrable functions if they are weakly equal $\mu$-a.e. will be denoted by $\mathbb {P}(\mu , X)^{loc}$.

• b) Analogously F is locally Bochner integrable if F is strongly $\mu$-measurable function and $\chi _B F\in \mathbb {B}(\mu ,X)$, $\forall$ $B\in \Sigma ^f$. The collection of equivalence classes obtained by identifying locally Bochner integrable functions if they are equal $\mu$-a.e. will be denoted by $\mathbb {B}(\mu , X)^{loc}$.

Remark 1

Observe that $\mathbb {P}(\mu , X)^{loc}$ and $\mathbb {B}(\mu , X)^{loc}$ are vector spaces. Additionally we have that

$$\begin{aligned} \mathbb {B}(\mu ,X)\subset \mathbb {B}(\mu , X)^{loc}\subset \mathbb {P}(\mu , X)^{loc}. \end{aligned}$$

The following examples show that the containment $\mathbb {B}(\mu ,X)\subset \mathbb {B}(\mu , X)^{loc}$ and $\mathbb {P}(\mu ,X)\subset \mathbb {P}(\mu , X)^{loc}$ can be proper.

Example 1

Let us fix $x\in X$ and assume that $f:\Omega \rightarrow \mathbb {R}$ is such that $\chi _B f\in L^1(\mu )$, $\forall \ B\in \Sigma ^f$ (c.f. [9, Def. 2.14 c)]). Now define $F:\Omega \rightarrow X$ by

$$\begin{aligned} F(t):=f(t) x. \end{aligned}$$

(5)

Let us see that $\chi _BF\in \mathbb {B}(\mu ,X)$, $\forall$ $B\in \Sigma ^f$. Since $f\chi _B\in L^0(\Sigma )$, for all $B\in \Sigma ^f$, it turns out that $f\in L^0(\Sigma )$. Take $\{s_n\}\subset S(\Sigma )$ such that $s_n\rightarrow f$ and $|s_n|\le |f|$, $\forall$ $n\in \mathbb {N}$. Fix $B\in \Sigma ^f$. For each $n\in \mathbb {N}$, define $S_n:\Omega \rightarrow X$ by $S_n(t)= \chi _B s_n(t)x$. Since $\chi _Bf\in L^1(\mu )$, we have that $\chi _Bs_n\in L^1(\mu )$. And so $S_n\in St(\mu ,X)$ and $S_n(t)\rightarrow \chi _B f(t) x=\chi _BF(t)$, $\forall$ $t\in \Omega$ indicating that $\chi _B F$ is strongly $\mu$-measurable.

Further

$$\begin{aligned} \displaystyle {\int _B \Vert F\Vert _Xd\mu =\int _B |f|\Vert x\Vert _Xd\mu =\left( \int _B |f|d\mu \right) \Vert x\Vert _X }. \end{aligned}$$

Since $\chi _B f\in L^1(\mu )$, it follows that $\displaystyle \int _B \Vert F\Vert _Xd\mu <\infty$, and so $\chi _BF\in \mathbb {B}(\mu ,X)$. Thus $F\in \mathbb {B}(\mu , X)^{loc}$. Finally observe that $F\in \mathbb {B}(\mu ,X)$ if and only if $f\in L^1(\mu )$.

Example 2

Let us consider $(\mathbb {N},2^{N},\mu _0)$, where $\mu _0$ is the counting measure and $X=c_0$. Clearly $\mathbb {B}(\mu _0,c_0)^{loc}=L^0(\mu _0,c_0)$, so $\mathbb {P}(\mu _0,c_0)^{loc} =\mathbb {B}(\mu _0,c_0)^{loc}$. Now let us consider the functions $F,G:\mathbb {N}\rightarrow c_0$ defined by $F(n):=\sum _{k=1}^n 2^k e_k$ and $G(n):=\frac{1}{n} e_n$. Then $F,G\in L^0(\mu _0,c_0)$. However, if we take $x^*=\{\frac{1}{2^n}\}\in \ell ^1=c_0^*$ we have that $\langle F,x^*\rangle :\mathbb {N}\rightarrow \mathbb {R}$ is the sequence $\{n\}$ which is not integrable with respect to $\mu _0$, so F is not Pettis integrable. On the other hand, if $x^*=\{a_n\}\in \ell ^1$ we have that $\langle G,x^*\rangle =\{\frac{a_n}{n}\}$, thus $G\in \mathbb {P}(\mu _0,c_0)$ but $\Vert G\Vert _\infty = \{\frac{1}{n}\}$, thereby G is not Bochner integrable. Hence

$$\mathbb {B}(\mu _0,c_0)\subsetneq \mathbb {P}(\mu _0,c_0)\subsetneq \mathbb {P}(\mu _0,c_0)^{loc} =\mathbb {B}(\mu _0,c_0)^{loc}.$$

As a consequence of Theorem 1 we obtain the following result.

Proposition 1

Let $F\in \mathbb {P}(\mu , X)^{loc}$. Then the set function $\nu _F:\Sigma ^f\rightarrow X$ defined by

$$\begin{aligned} \nu _F(B):=\mathbb {P}-\int _B Fd \mu , \end{aligned}$$

(6)

is a vector measure such that

$$\begin{aligned} \Vert \nu _F\Vert (A)=\sup _{x^*\in B_{X^*}}\int _A |\langle F,x^*\rangle |d\mu , \ \forall \ A\in \Sigma . \end{aligned}$$

(7)

Proof

Let us show that $\nu _F$ is a vector measure. Let $\{B_n\}\subset \Sigma ^f$ be a disjoint collection such than $B:=\bigcup _{n=1}^\infty B_n\in \Sigma ^f$. Since $\chi _B F\in \mathbb {P}(\mu ,X)$, by Theorem 1 we have

$$\begin{aligned} \begin{array}{lll} \nu _F(B)&{}=&{} \displaystyle {\mathbb {P}-\int _B Fd\mu =\mathbb {P}-\int _B \chi _BFd\mu } \\ &{}=&{}\displaystyle {\sum _{n=1}^\infty \mathbb {P}-\int _{B_n}\chi _B Fd\mu =\sum _{n=1}^\infty \mathbb {P}-\int _{B_n} Fd\mu =\nu _F(B_n)}. \end{array} \end{aligned}$$

Thus $\nu _F$ is a vector measure. Now fix $x^*\in X^*$. Since $F\in \mathbb {P}(\mu , X)^{loc}$ we have $\langle F,x^*\rangle \in L^1_{loc}(\lambda )$. From [9, Thm. 2.31] we have that the variation of the scalar measure $\mu _{\langle F,x^*\rangle }:\Sigma ^f\rightarrow \mathbb {K}$ defined by $\mu _{\langle F,x^*\rangle }(B)=\int _B \langle F,x^*\rangle d\mu$, $\forall \ B\in \Sigma ^f$ is given by

$$\begin{aligned} |\mu _{\langle F,x^*\rangle }|(A)=\int _A |\langle F,x^*\rangle |d\mu , \ \forall \ A\in \Sigma . \end{aligned}$$

On the other hand notice that for each $B\in \Sigma ^f$, we have

$$\begin{aligned} \displaystyle {\langle \nu _F,x^*\rangle (B)=\left\langle \mathbb {P}-\int _B Fd\mu , x^*\right\rangle =\int _B \langle F,x^*\rangle d\mu =\mu _{\langle F,x^*\rangle }(B).} \end{aligned}$$

(8)

Therefore

$$\begin{aligned} |\langle \nu _F,x^*\rangle |(A)=\int _A |\langle F,x^*\rangle |d\mu . \end{aligned}$$

(9)

From this we get (7). $\square$

Observe that from (7) we have that if $A\in \mathcal {N}_0(\mu )$, then $\Vert \nu _F\Vert (B)=0$, $\forall$ $B\in \Sigma ^f_A$. So $\mathcal {N}_0(\mu )\subset \mathcal {N}_0(\nu _F)$. Also observe that $\Vert \nu _F\Vert (B)=\Vert \chi _BF\Vert _\mathbb {P}$, $\forall$ $B\in \Sigma ^f$. Moreover, from the Dunford integrability definition we get our next result.

Corollary 2

Let $F\in \mathbb {P}(\mu , X)^{loc}$. Then F is Dunford integrable if and only if $\nu _F$ has bounded semivariation. In particular if $F\in \mathbb {P}(\mu ,X)$, then $\Vert \nu _F\Vert (A)=\Vert \chi _A F\Vert _\mathbb {P}$.

In the case that F is locally Bochner integrable the variation of $\nu _F$ has the same characterization that when F is Bochner integrable, as we can see in what follow.

Proposition 2

If $F\in \mathbb {B}(\mu , X)^{loc}$, then $\displaystyle {|\nu _F|(A)=\int _A \Vert F\Vert _X d\mu }$, $\forall$ $A\in \Sigma$.

Proof

Take $B\in \Sigma ^f$. Notice that $\nu _F(A\cap B)= \nu _{\chi _BF}(A)$, $\forall$ $A\in \Sigma$. Since $\chi _BF\in \mathbb {B}(\mu ,X)$ we have that

$$\begin{aligned} |\nu _F|(B)=|\nu _{\chi _BF}|(B)=\int _B\Vert F\Vert _Xd\mu . \end{aligned}$$

Hence

$$\begin{aligned} |\nu _F|(A)&=\displaystyle {\sup _{B\in \Sigma ^f_A}|\nu _F|(B)} \\ &=\displaystyle {\sup _{B\in \Sigma ^f_A}\int _B\Vert F\Vert _Xd\mu =\int _A\Vert F\Vert _Xd\mu , \ \forall \ A\in \Sigma .} \ \end{aligned}$$

(10)

$\square$

Remark 2

Let us note that if $F:\Omega \rightarrow X$ is a strong $\mu$-measurable function, by the previous result we obtain that:

$F\in \mathbb {B}(\mu ,X)$ if and only if $F\in \mathbb {B}(\mu , X)^{loc}$ and $\nu _F$ has bounded variation.

Example 3

Let us return to the Example 1. It was shown there that F defined in (5) is locally Bochner integrable. In particular F is locally Pettis integrable. Let us obtain now the vector measure $\nu _F$, its variation and its semivariation. See that for each $B\in \Sigma ^f$

$$\begin{aligned} \nu _F(B)=\mathbb {B}-\int _B Fd\mu =\left( \int _B f d\mu \right) x, \ \forall \ B\in \Sigma ^f. \end{aligned}$$

Now take $A\in \Sigma$, from (7) and (10)

$$\begin{aligned} \Vert \nu _F\Vert (A)&=\displaystyle {\sup _{x^*\in B_{X^*}}\int _A|\langle F,x^*\rangle |d\mu =\sup _{x^*\in B_{X^*}}\int _A |f||\langle x,x^*\rangle |d\mu } \\ &=\displaystyle {\left( \int _A |f|d\mu \right) \sup _{x^*\in B_{X^*}}|\langle x,x^*\rangle | =\left( \int _A |f|d\mu \right) \Vert x\Vert _X}\\ &=\displaystyle {\int _A \Vert F\Vert _Xd\mu =|\nu _F|(A)}. \end{aligned}$$

Therefore $\Vert \nu _F\Vert =|\nu _F|$ in this case.

4 The space of $\nu _F$-integrable functions

When F is a locally Pettis or Bochner integrable function we have constructed the vector measure $\nu _F$ defined on the $\delta$-ring $\Sigma ^f$. In the present section we will study the spaces $L^0(\nu _F)$, $L^1_w(\nu _F)$, $L^1(\nu _F)$ and $L^1(|\nu _F)|)$ associated to this vector measure through the operator $M_F$ which to each measurable function g assigns the function gF. The following lemmas allow us to conclude that $M_F:L^0(\nu _F)\rightarrow L^0_w(\mu ,X)$ or $L^0(\mu ,X)$ is well defined. Clearly $M_F$ is a linear operator.

Lemma 2

Let $F:\Omega \rightarrow X$ be a function and $g\in L^0(\Sigma )$.

• i) If F is strongly $\mu$-measurable, then gF is strongly $\mu$-measurable.

• ii) If F is weakly $\mu$-measurable, then gF is weakly $\mu$-measurable.

Proof

• (i) Observe that if $\varphi \in S(\Sigma )$ and $S\in St(\mu ,X)$, then $\varphi S\in St(\mu ,X)$. Let us assume that F is strongly $\mu$-measurable. Take $\{\varphi _n\}\subset S(\Sigma )$ and $\{S_n\}\subset St(\mu ,\Sigma )$ such that $\varphi _n\rightarrow g$ and $S_n\rightarrow F$, $\mu$-a.e. Thus $\varphi _nS_n\in St(\mu ,X)$, $\forall \ n\in \mathbb {N}$ and $\varphi _nS_n\rightarrow gF$, $\mu$-a.e. It follows that gF is strongly $\mu$-measurable.

• (ii) By definition if F is weakly $\mu$-measurable, we have that for each $x^*\in X^*$, the function $\langle F,x^*\rangle$ is strongly $\mu$-measurable. Using (i) we obtain that $\langle gF,x^*\rangle =g\langle F,x^*\rangle \in L^0(\mu ,X)$ $\forall \ x^*\in X^*$.

Lemma 3

Let $F\in \mathbb {P}(\mu , X)^{loc}$, $\{g_n\}\subset L^0(\Sigma )$ and $g,h\in L^0(\Sigma )$.

• (i) If $g=h$, $\nu _F$-a.e., then $gF=hF$, weakly $\mu$-a.e.

• (ii) If $g_n\rightarrow g$, $\nu _F$-a.e., then $\langle g_nF, x^*\rangle \rightarrow \langle gF,x^*\rangle$, $\mu$-a.e., $\forall \ x^*\in X^*$.

Proof

• (i) Choose $N\in \mathcal {N}_0(\nu _F)$ such that $g(t)=h(t)$, $\forall \ t\in N^c$. Then $g\chi _{N^c} F=h\chi _{N^c} F$; Moreover, $\chi _NF=0$ weakly $\mu$ -a.e. implies that $g\chi _N F=h\chi _N F=0$, weakly $\mu$-a.e. Thus $gF=hF$, weakly $\mu$-a.e.

• (ii) Let $N\in \mathcal {N}_0(\nu _F)$ such that $g_n(t)\rightarrow g(t)$, $\forall \ t\in N^c$. So $g_n\chi _{N^c}\rightarrow g\chi _{N^c}F$ and $g_n\chi _NF=g\chi _NF=0$, weakly $\mu$-a.e. Then for each $x^*\in X^*$, $\langle g_n\chi _{N^c}F,x^*\rangle \rightarrow \langle g\chi _{N^c}F,x^*\rangle$ and $\langle g_n\chi _NF,x^*\rangle =\langle g\chi _NF,x^*\rangle =0$, $\mu$-a.e. Therefore $\langle g_nF,x^*\rangle \rightarrow \langle gF,x^*\rangle$, $\mu$-a.e., $\forall \ x^*\in X^*$.

$\square$

Proposition [11, Prop.8] established by G. F. Stefansson for the case that $F\in \mathbb {P}(\mu ,X)$ and $\mu$ is a finite positive measure defined on a $\sigma$-algebra is generalized in the next theorem.

Theorem 2

For $F\in \mathbb {P}(\mu , X)^{loc}$ and $g\in L^0(\Sigma )$, we have that

• (i) $g\in L^1_w(\nu _F)$ if, and only if, $gF\in \mathbb {D}(\mu ,X)$. Moreover, the restriction to $L^1_w(\nu _F)$ of the operator $M_F$ is a linear isometry from $L^1_w(\nu _F)$ into $\mathbb {D}(\mu ,X)$.

• (ii) $g\in L^1(\nu _F)$ if, and only if, $gF\in \mathbb {P}(\mu ,X)$. Moreover, $M_F:L^1(\nu _F)\rightarrow \mathbb {P}(\mu ,X)$, the restriction of the operator $M_F$, is a linear isometry such that $I_{\nu _F}=I_\mathbb {P}\circ M_F$.

Proof

Fix $x^*\in X^*$ and consider $\displaystyle {s=\sum _{j=1}^n a_j\chi _{A_j}\in S(\Sigma )}$. By hypothesis $\chi _BF\in \mathbb {P}(\mu ,X)$, $\forall \ B\in \Sigma ^f$. It follows that $sF\in \mathbb {P}(\mu ,X)$. From (9) we obtain

$$\begin{aligned} \begin{array}{lll} \displaystyle {\int _\Omega |s| d|\langle \nu _F,x^*\rangle |}&{}=&{}\displaystyle {\sum _{j=1}^n} |a_j| |\langle \nu _F,x^*\rangle |(A_j) =\displaystyle {\sum _{j=1}^n}| a_j|\int _{A_j}|\langle F,x^*\rangle | d\mu \\ &{}=&{}\displaystyle {\int _\Omega \sum _{j=1}^n}| a_j|\chi _{A_j}|\langle F,x^*\rangle | d\mu =\int _\Omega |\langle sF,x^*\rangle | d\mu . \end{array} \end{aligned}$$

(11)

Thus $s\in L^1_w(\nu _F)$ if and only if $sF\in \mathbb {D}(\mu ,X)$.

Proceeding in the same way, it follows from (8) that

$$\begin{aligned} \displaystyle {\int _\Omega s d\langle \nu _F,x^*\rangle }=\int _\Omega \langle sF,x^*\rangle d\mu . \end{aligned}$$

(12)

Now take $g\in L^0(\nu _F)^+$ and $\{s_n\}\subset S(\mathcal {R}^{loc})$ such that $0\le s_n\uparrow g$, $\nu _F$-a.e. From Lemma 3 we obtain $\langle s_nF,x^*\rangle \rightarrow \langle gF,x^*\rangle$, $\mu$-a.e. Then, $|\langle s_nF,x^*\rangle |\uparrow |\langle gF,x^*\rangle |$, $\mu$-a.e. By the Monotone Convergence Theorem and (11) it turns out that

$$\begin{aligned} \displaystyle {\int _\Omega |\langle gF,x^*\rangle |d\mu }&=\displaystyle {\lim _{n\rightarrow \infty } \int _\Omega |\langle s_nF,x^*\rangle |d\mu }\\ &=\displaystyle {\lim _{n\rightarrow \infty }\int _\Omega s_n d|\langle \nu _F,x^*\rangle |=\int _\Omega |g|d|\langle \nu _F,x^*\rangle |}, \end{aligned}$$

(13)

showing that $g\in L^1_w(\nu _F)$ if and only if $gF\in \mathbb {D}(\mu ,X)$.

By the Dominate Convergence Theorem and (8)

$$\begin{aligned} \int _\Omega \langle gF,x^*\rangle d\mu =\lim _{n\rightarrow \infty }\int _\Omega \langle s_nF,x^*\rangle d\mu =\lim _{n\rightarrow \infty }\int _\Omega s_n d\langle \nu _F,x^*\rangle =\int _\Omega g d\langle \nu _F,x^*\rangle . \end{aligned}$$

We conclude from here that $g\in L^1(\nu _F)$ if and only if $gF\in \mathbb {P}(\mu ,X)$ and

$$\begin{aligned} \int _\Omega gd\nu _F=\mathbb {P}-\int _\Omega gFd\mu . \end{aligned}$$

(14)

Since the involved sets are vector spaces and each $g\in L^0(\nu _F)$ is a linear combination of non negative functions, we obtain the first part in (i) and (ii).

Finally take $g\in L^1_w(\nu _F)$, since $|g|\ge 0$ we obtain equality (13) with a sequence $\{s_n\}\subset S(\Sigma )$ such that $0\le s_n\uparrow |g|$, $\nu _F$-a.e. Taking the supremum over $x^*\in B_{X^*}$ it turns out that $\Vert g\Vert _{\nu _F}=\Vert gF\Vert _\mathbb {D}$. That is, $M_F$ restricted to $L^1_w(\nu _F)$ is a linear isometry. Since $L^1(\nu _F)$ and $\mathbb {P}(\mu ,X)$ are subspaces of $L^1_w(\nu _F)$ and $\mathbb {D}(\mu ,X)$, respectively, we conclude that $M_F$ restricted to $L^1(\nu _F)$ is also an isometry. Moreover, from (14) it follows that $I_{\nu _F}=I_\mathbb {P}\circ M_F$. $\square$

Corollary 3

Let $F\in L_w^0(\mu ,X)$. Then $F\in \mathbb {P}(\mu ,X)$ if and only if $F\in \mathbb {P}(\mu , X)^{loc}$ and $\nu _F$ is strongly additive.

Proof

Let assume that $F\in \mathbb {P}(\mu ,X)$. Consider the vector measure $\tilde{\nu }_F:\mathcal {R}^{loc}\rightarrow X$ defined in (2). Since $\Sigma$ is a $\sigma$-algebra, it turns out that $\tilde{\nu }_F$ is strongly additive. Observe that $\nu _F$ is the restriction of $\tilde{\nu }_F$ to $\Sigma ^f$, so it follows that it is strongly additive.

Now assume that $F\in \mathbb {P}(\mu , X)^{loc}$ and that $\nu _F$ is strongly additive. From [3, Cor. 3.2] we obtain that $\chi _\Omega \in L^1(\nu _F)$. So, by the previous theorem $F=\chi _\Omega F\in \mathbb {P}(\mu ,X)$. $\square$

Corollary 4

Let $F\in \mathbb {P}(\mu , X)^{loc}$. If X does not contain any subspace isomorphic to $c_0$ and $\nu _F$ is bounded, then $F\in \mathbb {P}(\mu ,X)$.

Proof

Since X does not contain any subspace isomorphic to $c_0$ and $\nu _F$ is bounded it turns out that $\nu _F$ is strongly additive [4, p. 36]. Then by the previous corollary $F\in \mathbb {P}(\mu ,X)$. $\square$

The following result gives us the connection between the spaces $L^1(|\nu _F|)$ and $\mathbb {B}(\mu ,X)$ through the operator $M_F$ in case that $F\in \mathbb {B}(\mu , X)^{loc}$. We will show that, as it occurs when $F\in \mathbb {P}(\mu , X)^{loc}$, $M_F$ is a linear isometry in this case.

Proposition 3

Consider $F\in \mathbb {B}(\mu , X)^{loc}$ and $g\in L^0(|\nu _F|)$. Then $g\in L^1(|\nu _F|)$ if and only if $gF\in \mathbb {B}(\mu ,X)$. Moreover, $M_F:L^1(|\nu _F|)\rightarrow \mathbb {B}(\mu ,X)$ is a linear isometry such that $I_{\nu _F}(g)=I_\mathbb {B}\circ M_F(g)$, $\forall \ g\in L^1(|\nu _F|)$.

Proof

Clearly $M_F$ is a linear operator, we will see that its image is a subset of $\mathbb {B}(\mu ,X)$. By Lemma 3 we have that the restriction $M_F:L^1(|\nu _F|)\rightarrow \mathbb {B}(\mu ,X)$ is well defined.

Since the norms in $L^1(|\nu _F|)$ and $\mathbb {B}(\mu ,X)$ are different from those in $L^1(\nu _F)$ and $\mathbb {P}(\mu ,X)$, respectively, we need to establish that, under these norms, $M_F$ is also an isometry.

By hypothesis $F\in \mathbb {B}(\mu , X)^{loc}$, then from (10) it follows that $|\nu _F|(B)<\infty$, $\forall \ B\in \Sigma ^f$. So, $S(\Sigma ^f)\subset L^1(|\nu _F|)$. Further for each $s=\sum _{j=1}^n a_j\chi _{A_j}\in S(\Sigma )$ we have that

$$\begin{aligned} \displaystyle {\int _\Omega |s|d|\nu _F|}&=\displaystyle {\int _\Omega \sum _{j=1}^n|a_j||\nu _F|(A_j)=\sum _{j=1}^n|a_j|\int _{A_j}\Vert F\Vert _Xd\mu } \\ &=\displaystyle {\int _\Omega \sum _{j=1}^n|a_j|\chi _{A_j}\Vert F\Vert _Xd\mu =\int _\Omega \Vert sF\Vert _Xd\mu }. \end{aligned}$$

Therefore $s\in L^1(|\nu _F|)$ if and only if $sF\in \mathbb {B}(\mu ,X)$. Now consider $g\in L^0(|\nu _F|)$ and take $\{s_n\}\subset S(\Sigma )$ such that $0\le s_n\uparrow |g|$, $\nu _F$-a.e. Then $\Vert s_nF\Vert _X\uparrow \Vert gF\Vert _X$, $\mu$-a.e. By the Monotone Convergence Theorem

$$\begin{aligned} \int _\Omega \Vert gF\Vert _Xd\mu =\lim _{n\rightarrow \infty }\int _\Omega \Vert s_nF\Vert _Xd\mu =\lim _{n\rightarrow \infty }\int _\Omega |s_n|d |\nu _F|=\int _\Omega |g|d|\nu _F|. \end{aligned}$$

Thus we have that $gF\in \mathbb {B}(\mu ,X)$ if and only if $g\in L^1(|\nu _F|)$. Moreover, $\Vert g\Vert _{|\nu _F|}=\Vert gF\Vert _1$.

The equality between the operators follows from Proposition 2. $\square$

Example 4

Consider again the function F defined in (5). As we see in Example 1$F\in \mathbb {B}(\mu , X)^{loc}$.

Take $g\in L^1(\nu _F)$, from Proposition 2

$$\begin{aligned} \int _\Omega gd\nu _F=\mathbb {P}-\int _\Omega gFd\mu =\left( \int _\Omega gf d\mu \right) x, \end{aligned}$$

then $gf\in L^1(\mu )$. And so,

$$\begin{aligned} \int _\Omega \Vert gF\Vert _Xd\mu =\int _\Omega |gf|\Vert x\Vert _Xd\mu <\infty . \end{aligned}$$

(15)

By Lemma 2gF is strongly $\mu$-measurable. Thus we have that $gF\in \mathbb {B}(\mu ,X)$ and by Proposition 3, $g\in L^1(|\nu _F|)$. We conclude that $L^1(|\nu _F|)=L^1(\nu _F)$. And from [2, Prop. 5.4] it follows that $L^1(|\nu _F|)=L^1(\nu _F)=L^1_w(\nu _F)$ .