Convergence for varying measures in the topological case

Abstract

In this paper convergence theorems for sequences of scalar, vector and multivalued Pettis integrable functions on a topological measure space are proved for varying measures vaguely convergent.

1 Introduction

Conditions for the convergence of sequences of measures $(m_n)_n$ and of their integrals $(\int f_n {\textrm{d}}m_n)_n$ in a measurable space $\Omega$ are of interest in many areas of pure and applied mathematics such as statistics, transportation problems, interactive partial systems, neural networks and signal processing (see, for instance, [1,2,3, 9,10,11,12, 17]). In particular, for the image reconstruction, which is a branch of signal theory, in the last years, interval-valued functions have been considered since the process of discretization of an image is affected by quantization errors [19] and its numerical approximation can be interpreted as a suitable sequence of interval-valued functions (see for instance [22, 28]).

Obviously, suitable convergence notions are needed for the varying measures, see for example [15, 16, 18, 21, 23, 24, 30] and the references therein. In a previous paper [13] we have examined the problem when the varying measures converge setwisely in an arbitrary measurable space. This type of convergence is a powerful tool since it permits to obtain strong results, for example the Vitali-Hahn-Saks Theorem or a Dominated Convergence Theorem [18].

But sometime in the applications it is difficult, at least technically, to prove that the sequence $(m_n(A))_n$ converges to m(A) for every measurable set A, unless e.g. the sequence $(m_n)_n$ is decreasing or increasing. So other types of convergence are studied, based on the structure of the topological space $\Omega$, such as the vague and the weak convergence which are, in general, weaker than the setwise. These convergences are useful, for example, from the point of view of applications on non-interactive particle systems (see [9, 23]).

In the present paper we continue the research started in [13] and we provide sufficient conditions in order to obtain Vitali’s type convergence results for a sequence of (multi)functions $(f_n)_n$ integrable with respect to a sequence $(m_n)_n$ of measures when $(m_n)_n$ converges vaguely or weakly to a finite measure m.

The known results, in literature, as far as we know, require that the topological space $\Omega$, endowed with the Borel $\sigma$-algebra is a metric space [15, 16], or a locally compact space which is also: separable and metric [18], metrizable [20] or Hausdorff second countable [30]. An interesting comparison among all these results is given in [23].

In the present paper, following the ideas of Bogachev [4], we assume that $\Omega$ is only an arbitrary locally compact Hausdorff space. The paper is organized as follows: in Sect. 2 the topological structure of the space $\Omega$ is introduced together with the convergence types considered and some of their properties. In Sect. 3 the scalar case is studied; the main result of this section is Theorem 3.4, where we obtain the convergence of the integrals $(\int f_n {\textrm{d}}m_n)_n$ over arbitrary Borel sets under suitable conditions. In Sect. 4 Theorem 3.4 is applied in order to obtain analogous results for the multivalued case, obtaining as a corollary also the vector case. In both cases the Pettis integrability of the integrands is considered. Finally, adding a condition as in [13, Theorem 3.2] we obtain a convergence result for (multi)functions in Proposition 4.4 on measurable spaces.

2 Topological case, preliminaries

Let $\Omega$ be a locally compact Hausdorff space and let $\mathcal {B}$ be its Borel $\sigma$-algebra. The symbol $\mathcal {F}(\Omega )$ indicates the class of all $\mathcal {B}$-measurable functions $f:\Omega \rightarrow \mathbb {R}$. We denote by $C(\Omega )$, $C_0(\Omega )$, $C_c(\Omega )$ and $C_b(\Omega )$ respectively the family of all continuous functions, and the subfamilies of all continuous functions that vanish at infinity, have compact support, are bounded.

Throughout, we will use Urysohn’s Lemma in the form ([29, Lemma 2.12]):

• If K is compact and $U\supset {K}$ is open in a locally compact $\Omega$, then there exists $f:\Omega \rightarrow [0,1]$, $f\in C_c(\Omega )$, such that $\chi _K\le {f}\le \chi _U$.

All the measures we will consider on $(\Omega , \mathcal {B})$ are finite and by $\mathcal {M}(\Omega )$ we denote the family of finite nonnegative measures. As usual a measure $m \in \mathcal {M}(\Omega )$ is Radon if it is inner regular in the sense of approximation by compact sets.

We recall the following definitions of convergence for measures.

Definition 2.1

Let m and $m_n$ be in $\mathcal {M}(\Omega )$. We say that

• 2.1.a) $(m_n)_n$ converges vaguely to m ($m_n \xrightarrow []{v}m$) ([18, Section 2.3]) if

  $$\begin{aligned} \int _{\Omega } g {\textrm{d}}m_n \rightarrow \int _{\Omega } g {\textrm{d}}m, \quad \text{ for } \text{ every } g \in C_0(\Omega ). \end{aligned}$$

• 2.1.b) $(m_n)_n$ converges weakly to m ($m_n \xrightarrow []{w} m$) ([18, Section 2.1]) if

  $$\begin{aligned} \int _{\Omega } g {\textrm{d}}m_n \rightarrow \int _{\Omega } g {\textrm{d}}m, \quad \text{ for } \text{ every } g \in C_b(\Omega ). \end{aligned}$$

• 2.1.c) $(m_n)_n$ converges setwisely to m ($m_n \xrightarrow []{s} m$) if   $\lim _n m_n (A) = m(A)$ for every $A \in \mathcal {B}$ ([18, Section 2.1], [16, Definition 2.3]) or, equivalently [23], if

  $$\begin{aligned} \int _{\Omega } g {\textrm{d}}m_n \rightarrow \int _{\Omega } g {\textrm{d}}m, \quad \text{ for } \text{ every } \text{ bounded } g \in \mathcal {F}(\Omega ). \end{aligned}$$

• 2.1.d) $(m_n)_n$ is uniformly absolutely continuous with respect to m if for each $\varepsilon > 0$ there exists $\delta > 0$ such that

  $$\begin{aligned} \left( E \in \mathcal {B} \quad \text{ and } \ \ m(E)<\delta \right) \,\, \Longrightarrow \,\, \sup _n m_n (E) < \varepsilon . \end{aligned}$$

  (1)

  We would like to note that the condition $m_n \le m$, for every $n \in {\mathbb N}$, implies that $(m_n)_n$ is uniformly absolutely continuous with respect to m.

Remark 2.2

As observed in [18] the setwise convergence is stronger than the vague and the weak convergence. For the converse implications we know, by [21, Lemma 4.1 (ii)], that if $(m_n)_n$ is a sequence in $\mathcal {M}(\Omega )$ with $m_n \le m$, where $m \in \mathcal {M}(\Omega )$ and $(m_n)_n$ converges vaguely to m, then $(m_n)_n$ converges setwisely to m. If m is $\overline{\mathbb {R}}$-valued this is not true in general, see for example [18, page 143]. The weak convergence is stronger than the vague convergence; as an example we can consider $m_n:=\delta _n$ (the Dirac measure at the point $x=n$) and $m:=0$. The sequence $(m_n)_n$ converges vaguely to m, but since $m_n(\mathbb {R})=1 \not \rightarrow 0 = m(\mathbb {R})$ the convergence cannot be weak.

Moreover we note that if $(m_n)_n$ converges weakly to m, then $m_n(\Omega ) \rightarrow m(\Omega )$ (it is enough to take $g=1$ in the definition).

We have

Proposition 2.3

Let $m_n$, $n \in \mathbb {N}$, and m be in $\mathcal {M}(\Omega )$, with m Radon. If $(m_n)_n$ is uniformly absolutely continuous with respect to m and $(m_n)_n$ is vaguely convergent to m, then $(m_n)_n$ is weakly convergent to m.

Proof

We fix $\varepsilon >0$ and let $f\in {C_b(\Omega )}$. We set $c:= \max \{1,\sup _{\Omega } \vert f(\omega )\vert \}$, let $\delta \in ]0, \varepsilon [$ be taken in such a way that if E is a Borel set with $m(E)<\delta$, then

$$\begin{aligned} \max \left\{ \int _E\vert f\vert \,{\textrm{d}}m, \, \sup _n m_n(E) \right\} <\varepsilon . \end{aligned}$$

Let K be a compact set such that $m(K^c)<\delta$. Then by Urysohn’s Lemma let $h:\Omega \rightarrow [0,1]$ be a continuous function with compact support such that $h(\omega )=1$ for $\omega \in {K}$. Let $g:=f\cdot {h}$. Then $g\in {C_0(\Omega )}$. We have for sufficiently large $n\in {\mathbb N}$, depending on the vaguely convergence,

$$\begin{aligned}&\Big \vert \int _{\Omega } f {\textrm{d}}m - \int _{\Omega } f {\textrm{d}}m_n \Big \vert \\ \quad \quad&\le \int _{\Omega } \vert f - g\vert {\textrm{d}}m +\int _{\Omega } \vert f - g\vert {\textrm{d}}m_n + \Big \vert \int _{\Omega } g \ d m - \int _{\Omega } g \ {\textrm{d}}m_n \Big \vert \\ \quad \quad&= \int _{K^c} \vert f\vert \cdot \vert 1-h\vert \,{\textrm{d}}m+\int _{K^c} \vert f\vert \cdot \vert 1-h\vert \,{\textrm{d}}m_n+ \Big \vert \int _{\Omega } g \ d m - \int _{\Omega } g \ {\textrm{d}}m_n \Big \vert \\ \quad \quad&\le \varepsilon (c +2). \end{aligned}$$

$\square$

For other relations among weak or vague convergence and setwise convergence see [21, Lemma 4.1]. Moreover

Proposition 2.4

Let $m_n$, $n \in \mathbb {N}$, and m be in $\mathcal {M}(\Omega )$ with m Radon. If $m_n \le m$, for every $n \in {\mathbb N}$, and $(m_n)_n$ is vaguely convergent to m, then for every $f\in {L^1(m)}$ and $A \in \mathcal {B}$

$$\begin{aligned} \lim _n \int _{A} f\,{\textrm{d}}m_n = \int _{A} f\,{\textrm{d}}m. \end{aligned}$$

(2)

In particular $(m_n)_n$ converges to m setwisely.

Proof

Let $f\in {L^1(m)}$ be fixed. Given $\varepsilon > 0$ there exists $g \in C_c(\Omega )$ such that

$$\begin{aligned} \int _{\Omega }\vert f-g\vert {\textrm{d}}m_n \le \int _{\Omega }\vert f-g\vert {\textrm{d}}m < \frac{ \varepsilon }{3}. \end{aligned}$$

(3)

Moreover, since $(m_n)_n$ is vaguely convergent to m, let $N(\varepsilon /3)$ be such that

$$\begin{aligned} \Big \vert \int _{\Omega } g \ {\textrm{d}}m - \int _{\Omega } g \ {\textrm{d}}m_n \Big \vert < \frac{\varepsilon }{3} \end{aligned}$$

(4)

for $n>N$. Therefore by (3) and (4) for $n>N$ we obtain

$$\begin{aligned}&\left| \int _{\Omega } f {\textrm{d}}m - \int _{\Omega } f {\textrm{d}}m_n \right| \\ \qquad \qquad&\le \int _{\Omega } \vert f - g\vert {\textrm{d}}m +\int _{\Omega } \vert f - g\vert {\textrm{d}}m_n + \left| \int _{\Omega } g \ {\textrm{d}}m- \int _{\Omega } g \ {\textrm{d}}m_n \right| < \varepsilon . \end{aligned}$$

Now if $A \in \mathcal {B}$, also $f\chi _A \in L^1(m)$ and (2) follows. In particular $m_n \xrightarrow []{s} m$. $\square$

Results of the previous type are contained for example in [18, Proposition 2.3] for the setwise convergence when the measures $m_n$ are equibounded by a measure $\nu$ for non negative $f \in L^1(\nu )$ or in [18, Proposition 2.4] under the additional hypothesis of separability of $\Omega$ for non negative and lower semicontinuous functions f.

We now introduce the following definition

Definition 2.5

Let $(m_n)_n$ be a sequence in $\mathcal {M}(\Omega )$. We say that:

• 2.5.a) A sequence $(f_n)_n \subset \mathcal {F}(\Omega )$ has uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$, if for every $\varepsilon >0$ there exists $\delta >0$ such that for every $n \in \mathbb {N}$

  $$\begin{aligned} \left( A \in \mathcal {B} \quad \text{ and } \ \ m_n(A)<\delta \right) \ \ \Longrightarrow \ \ \int _A\vert f_n\vert \,{\textrm{d}}m_n < \varepsilon . \end{aligned}$$

  (5)

  Analogously a function $f \in \mathcal {F}(\Omega )$ has uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$ if previous condition (5) holds for $f_n:=f$ for every $n \in \mathbb {N}$.

• 2.5.b) A sequence $(f_n)_n \subset \mathcal {F}(\Omega )$ is uniformly $(m_n)$-integrable on $\varOmega$ if

  $$\begin{aligned} \lim _{\alpha \rightarrow +\infty } \sup _n \int _{\vert f_n\vert >\alpha } \vert f_n\vert \, {\textrm{d}}m_n \ =0. \end{aligned}$$

  (6)

Remark 2.6

As we observed in [13, Proposition 2.6] if $(m_n)_n$ is a bounded sequence of measures and $(f_n)_n \subset \mathcal {F}$, then, $(f_n)_n$ is uniformly $(m_n)$-integrable on $\varOmega$ if and only if it has uniformly absolutely continuous $(m_n)$-integrals and

$$\begin{aligned} \sup _n \int _{\Omega } \vert f_n\vert \, {\textrm{d}}m_n < +\infty \,. \end{aligned}$$

(7)

3 The scalar case

Proposition 3.1

Let $(m_n)_n$ be a sequence in $\mathcal {M}(\Omega )$ which is uniformly absolutely continuous with respect to a Radon measure $m \in \mathcal {M}(\Omega )$ and vaguely convergent to m. Let $f \in C(\Omega )$ be a function which has uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$. Then,

$$\begin{aligned} \sup _n \int _{\Omega } \vert f \vert \, {\textrm{d}}m_n < +\infty \,. \end{aligned}$$

(8)

Proof

Let $\varepsilon >0$ be fixed and let $\sigma =\sigma (\varepsilon )$ be that of the uniform absolutely continuous $(m_n)$-integrability of f as in formula (5) (with $f_n=f$ for each $n \in \mathbb {N}$). Moreover let $\delta = \delta (\sigma ) >0$ be that of the uniform absolute continuity of $(m_n)_n$ with respect to m, as in formula (1).

Since m is Radon, there is a compact set K such that $m(\Omega {\setminus } K)< \delta$. By Urysohn’s Lemma there exists a continuous function $h:\Omega \rightarrow [0,1]$ with compact support such that $h(\omega )=1$ for $\omega \in {K}$. Let $g:=\vert f\vert \cdot {h}$. Then $g\in {C_0(\Omega )}$. Hence

$$\begin{aligned} \int _{\Omega }\vert f\vert \,{\textrm{d}}m_n \le \int _{K}\vert f\vert \,{\textrm{d}}m_n+\int _{\Omega \setminus K} \vert f\vert \, {\textrm{d}}m_n \le \int _{\Omega }g\,{\textrm{d}}m_n+\varepsilon . \end{aligned}$$

Since $(m_n)_n$ converges vaguely to m, then

$$\begin{aligned} \int _{\Omega }g\,{\textrm{d}}m_n\longrightarrow \int _{\Omega }g\,{\textrm{d}}m<+\infty . \end{aligned}$$

Hence

$$\begin{aligned} \sup _n\int _{K}\vert f\vert \,{\textrm{d}}m_n\le \sup _n \int _{\Omega }g\,{\textrm{d}}m_n<+\infty . \end{aligned}$$

$\square$

Proposition 3.2

Let $(m_n)_n$ be a sequence in $\mathcal {M}(\Omega )$ which is uniformly absolutely continuous with respect to a Radon measure $m \in \mathcal {M}(\Omega )$ and vaguely convergent to m. Moreover let $f \in C(\Omega )$ be a function which has uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$. Then $f \in L^1(m)$ and

$$\begin{aligned} \lim _n \int _{\Omega } f\,{\textrm{d}}m_n = \int _{\Omega } f\,{\textrm{d}}m. \end{aligned}$$

(9)

Proof

By Proposition 3.1$\sup _n \int _{\Omega } \vert f\vert {\textrm{d}}m_n < +\infty \,$. We denote by $(g_k)_k$ an increasing sequence of functions in $C_b(\Omega )$ such that $0\le {g_k}\uparrow \vert f\vert$, m a.e.

By Proposition 2.3$(m_n)_n$ is also weakly convergent to m. Now fix $k \in \mathbb {N}$. Let $N_1(k,1)$ be such that if $n>N_1$

$$\begin{aligned} \int _{\Omega } g_k \, {\textrm{d}}m -1 < \int _{\Omega } g_k \, {\textrm{d}}m_n . \end{aligned}$$

(10)

By Proposition 3.1 we infer

$$\begin{aligned} \int _{\Omega }g_k {\textrm{d}}m - 1< \int _{\Omega } g_k {\textrm{d}}m_n \le \sup _n \int _{\Omega } \vert f\vert {\textrm{d}}m_n < \infty \,. \end{aligned}$$

(11)

So, by the Monotone Convergence Theorem applied to the sequence $(g_k)_k$ we obtain $f \in L^1(m)$.

We are showing now that (9) holds. We fix $\sigma > 0$. Since $f \in L^1(m)$ there exists a positive $\delta _0$ such that for every $A \in \mathcal {B}$ with $m(A) < \delta _0$ then

$$\begin{aligned} \int _A \vert f\vert {\textrm{d}}m < \sigma . \end{aligned}$$

(12)

Moreover let $\varepsilon (\sigma )>0$ be that of the uniform absolutely continuous $(m_n)$-integrability of f in $\varOmega$ (with $f_n=f$ for each $n \in \mathbb {N}$) and $\delta = \delta (\varepsilon ) \in ] 0, \min \{ \varepsilon , \, \delta _0\}[$ be that of the absolute continuity of $(m_n)_n$ with respect to m.

So, if $m(A) < \delta$ then $\sup _n m_n (A)<\varepsilon$ and

$$\begin{aligned} \sup _n \int _{A} \vert f\vert {\textrm{d}}m_n < \sigma . \end{aligned}$$

(13)

By Urysohn’s Lemma one can find a compact set K with $m(K^c) < \delta$ and a function $h:\Omega \rightarrow [0,1]$ in $C_c(\Omega )$ and equal to 1 on K. So $g:=f\cdot {h} \in C_c(\Omega )$.

Since the sequence $(m_n)_n$ is vaguely convergent to m, there is $N_2(\sigma ) > N_1$ such that for $n>N_2$

$$\begin{aligned} \left| \int _{\Omega } g \, {\text {d}}m_n - \int _{\Omega } g \, {\text {d}}m \right| < \sigma .\end{aligned}$$

(14)

Then by (13), (14) and (12), for $n>N_2$, we have

$$\begin{aligned}&\left| \int _{\Omega } f\,{\textrm{d}}m_n - \int _{\Omega } f\,{\textrm{d}}m\right| \\ {}&\le \left| \int _{\Omega } (f-g)\,{\textrm{d}}m_n \right| + \left| \int _{\Omega } g\,{\textrm{d}}m_n - \int _{\Omega } g\,{\textrm{d}}m\right| + \left| \int _{\Omega } (g-f)\,{\textrm{d}}m \right| \\&\le \left| \int _{\Omega } f(1-h)\,{\textrm{d}}m_n \right| + \left| \int _{\Omega } g\,{\textrm{d}}m_n - \int _{\Omega } g\,{\textrm{d}}m\right| + \left| \int _{\Omega } f(1-h)\,{\textrm{d}}m \right| \\&\le \int _{K^c} \vert f\vert \,{\textrm{d}}m_n + \sigma + \int _{K^c} \vert f\vert \,{\textrm{d}}m < 3 \sigma \end{aligned}$$

and the thesis follows. $\square$

Now our aim is to obtain a limit result

$$\begin{aligned} \lim _n \int _{A} f_n{\textrm{d}}m_n = \int _{A} f{\textrm{d}}m, \quad \text{ for } \text{ every } A \in \mathcal {B}. \end{aligned}$$

(15)

For the scalar case, using a Portmanteau’s characterization of the vague convergence in metric spaces (see for example [20]), sufficient conditions when $A=\Omega$, are given

• in locally compact second countable and Hausdorff spaces, ([30, Theorems 3.3 and 3.5]), by Serfozo, for the vague and weak convergence respectively, when the sequence $(f_n)_n$ converges continuously to f. Under a domination condition in the first result while, in the second, the uniform $(m_n)$-integrability of the sequence $(f_n)_n$, with $f_n \ge 0$ for every $n \in \mathbb {N}$, is required;

• in locally compact separable metric spaces [18] by Hernandez-Lerma and Lasserre, obtaining a Fatou result and asking for the convergence of the sequence of measures an inequality of the $\liminf$ of the $m_n$ on each Borelian set;

• in metric spaces [15, 16], where the authors obtained a dominated convergence result for sequences of equicontinuous functions $(f_n)_n$ satisfying the uniform $(m_n)$-integrability.

In Theorem 3.4, taking into account Remarks 2.2 and 2.6, we extend [30, Theorem 3.5], obtaining a sufficient condition when the convergence is vague, the functions $f_n$ are real valued and using the uniformly absolutely continuous $(m_n)$-integrability of the sequence $(f_n)_n$. Later, in Sect. 4, we will also extend it to the vector and multivalued cases making use of the Pettis integrability.

We assume only that $\Omega$ is a locally compact Hausdorff space and then, in our setting, $\Omega$ is a Tychonoff space, i.e. a completely regular Hausdorff space ([14, Theorem 3.3.1]). So we are able to use the following Portmanteau’s characterization of the vague convergence for positive measures given in [4].

Theorem 3.3

([4, Corollary 8.1.8 and Remark 8.1.11]) Let $\Omega$ be an arbitrary completely regular space and let m and $m_n$, $n \in {\mathbb N}$, be measures in $\mathcal {M}(\Omega )$ with m Radon and assume that $\lim _{n} m_n(\Omega )=m(\Omega )$. Then the following are equivalent:

• 3.3.i) $(m_n)_n$ is vaguely convergent to m;

• 3.3.i) for any closed set $F \subset \Omega$, $\limsup _n m_n(F) \le m(F)$.

So we have

Theorem 3.4

Let m and $m_n$, $n \in {\mathbb N}$, be measures in $\mathcal {M}(\Omega )$, with m Radon. Let $f, f_n \in \mathcal {F}(\Omega )$. Suppose that

• (3.4.i) $f_n(t) \rightarrow f(t)$, m-a.e.;

• (3.4.ii) $f \in C(\Omega )$;

• (3.4.iii) $(f_n)_n$ and f have uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$;

• (3.4.iv) $(m_n)_n$ is vaguely convergent to m and uniformly absolutely continuous with respect to m.

Then $f \in L^1(m)$ and

$$\begin{aligned} \lim _n \int _{A} f_n{\textrm{d}}m_n = \int _{A} f{\textrm{d}}m \quad \text{ for } \text{ every } A \in \mathcal {B}. \end{aligned}$$

(16)

Proof

By Proposition 3.2 the function $f \in L^1(m)$. We proceed by steps.

• Step 1 We prove (16) for $A = \Omega$. Fix $\varepsilon >0$ and let $\delta := \min \left\{ \frac{\varepsilon }{6}, \delta (\frac{\varepsilon }{6}), \delta _f(\frac{\varepsilon }{6}) \right\} >0$ where $\delta _f(\frac{\varepsilon }{6})$ is that of the absolute continuity of $\int _{.} \vert f\vert {\textrm{d}}m$, and by (3.4.iii) $\delta (\frac{\varepsilon }{6})$ is that of (5) for both $(f_n)_n$ and f with respect to $(m_n)_n$. By the hypothesis (3.4.iv) let $0<\delta _0 <\delta$ be such that

  $$\begin{aligned} \left( E \in \mathcal {B} \quad \text{ and } \ \ m(E)<\delta _0 \right) \,\, \Longrightarrow \,\, \sup _n m_n (E) < \delta . \end{aligned}$$

  (17)

  By the Egoroff’s Theorem, we can find a compact set K such that $f_n \rightarrow f$ uniformly on K and $m(K^c) < \delta _0$. We observe that by condition 3.4.iv) and by Proposition 2.3$(m_n)_n$ weakly converges to m and then $\lim _nm_n(\Omega )=m(\Omega )$. So by Theorem 3.3, let $N_0 \in {\mathbb N}$ be such that

  $$\begin{aligned} m_n(K) < m(K) + 1 , \end{aligned}$$

  (18)

  for every $n>N_0$. Moreover, since the convergence is uniform on K, let $N_1 > N_0 \in {\mathbb N}$ be such that

  $$\begin{aligned} \vert f_n(t)-f(t)\vert < \frac{ \varepsilon }{6(m(K)+1)} , \end{aligned}$$

  (19)

  for every $t \in K$ and $n>N_1$. Then, for all $n>N_1$,

  $$\begin{aligned} \int _{K}\vert f_n-f\vert {\textrm{d}}m < \frac{ \varepsilon }{6}. \end{aligned}$$

  (20)

  Therefore by (18) and (19) we obtain, for every for $n> N_1$,

  $$\begin{aligned} \int _{K}\vert f_n-f\vert {\textrm{d}}m_n \le \frac{ \varepsilon }{6(m(K)+1)} \cdot m_n(K) < \frac{ \varepsilon }{6}. \end{aligned}$$

  (21)

  Since $m(K^c) < \delta _0$ by (17) it follows that $m_n (K^c) < \delta$ for every $n \in \mathbb {N}$. Moreover, by hypothesis 3.4.iii) and by the choice of $\delta$, we have that

  $$\begin{aligned} \max \, \left\{ \int _{K^c} \vert f\vert {\textrm{d}}m, \,\, \int _{K^c} \vert f_n\vert {\textrm{d}}m_n, \,\, \int _{K^c} \vert f\vert {\textrm{d}}m_n \right\} < \frac{ \varepsilon }{6}. \end{aligned}$$

  (22)

  By Urysohn’s Lemma let $h:\Omega \rightarrow [0,1]$ be a continuous function with compact support equal to 1 on K. Then $g:=f\cdot {h} \in C_c(\Omega )$ and by (22) we have

  $$\begin{aligned} \max \, \left\{ \int _{K^c}\vert f-g\vert {\textrm{d}}m_n, \,\, \int _{K^c}\vert f-g\vert {\textrm{d}}m \right\} < \frac{ \varepsilon }{6}. \end{aligned}$$

  (23)

  Moreover, since $(m_n)_n$ is vaguely convergent to m, let $n > N_2 \ge N_1$ be such that

  $$\begin{aligned} \Big \vert \int _{\Omega } g \ {\textrm{d}}m- \int _{\Omega } g \ {\textrm{d}}m_n \Big \vert < \frac{ \varepsilon }{6}. \end{aligned}$$

  (24)

  Therefore by (21)–(24) for $n > N_2$ we obtain

  $$\begin{aligned}&\left| \int _{\Omega } f {\textrm{d}}m - \int _{\Omega } f_n {\textrm{d}}m_n \right| \le \left| \int _{\Omega } (f_n -f) {\textrm{d}}m_n \right| + \left| \int _{\Omega } f {\textrm{d}}m - \int _{\Omega } f {\textrm{d}}m_n \right| \nonumber \\ {}&\le \left| \int _{K} (f_n -f) {\textrm{d}}m_n \right| + \int _{K^c} \vert f_n\vert {\textrm{d}}m_n + \int _{K^c} \vert f\vert {\textrm{d}}m_n + \left| \int _{\Omega } f {\textrm{d}}m - \int _{\Omega } f {\textrm{d}}m_n \right| \nonumber \\ {}&\le \frac{ \varepsilon }{2}+\left| \int _{\Omega } f {\textrm{d}}m - \int _{\Omega } f {\textrm{d}}m_n \right| \nonumber \\ {}&\le \frac{ \varepsilon }{2} + \int _{K^c} \vert f - g\vert {\textrm{d}}m +\int _{K^c} \vert f - g\vert {\textrm{d}}m_n + \left| \int _{\Omega } g \ {\textrm{d}}m- \int _{\Omega } g \ {\textrm{d}}m_n \right| < \varepsilon \end{aligned}$$

  (25)

  so (16) follows for $A=\Omega$.

• Step 2 Now we are proving that (16) is valid for an arbitrary compact set K. Let once again, $\varepsilon > 0$ be fixed. By (3.4.iii) and (3.4.iv) there exist $\delta _1,\delta _2>0$ such that:

  $j_1$):

  if $m_n(E)<\delta _2$, then $\displaystyle {\int _E \vert f_n\vert \,{\textrm{d}}m_n<\varepsilon }$ for every $n \in \mathbb {N}$;

  $j_2$):

  if $m(E)<\delta _1$, then $m_n(E)<\delta _2$ for every $n \in \mathbb {N}$;

  $j_3$):

  if $m(E)<\delta _1$, then $\displaystyle {\int _E \vert f\vert \,{\textrm{d}}m<\varepsilon }$.

  Let now $U\supset {K}$ be an open set such that $m(U\setminus {K})<\delta _1$. Then let $g:\varOmega \rightarrow [0,1]$ be continuous and such that $g=1$ on K and zero on $U^c$. Observe that the sequence $(f_n g)_n$ and the function fg satisfy all the hypotheses of Theorem 3.4 so, for the Step 1, we have

  $$\begin{aligned} \lim _n \int _{U} f_n g\, {\textrm{d}}m_n = \lim _n \int _{\Omega } f_n g\, {\textrm{d}}m_n = \int _{\Omega } fg \, {\textrm{d}}m = \int _{U} fg \, {\textrm{d}}m. \end{aligned}$$

  Then, by the previous inequalities and for n sufficiently large, we have

  $$\begin{aligned} \Big \vert \int _K f_n\,{\textrm{d}}m_n-\int _K f\,{\textrm{d}}m \Big \vert&= \Big \vert \int _K f_n g \,{\textrm{d}}m_n - \int _K fg \,{\textrm{d}}m + \big ( \int _{K^c} f_n g \,{\textrm{d}}m_n- \int _{K^c} f g\,{\textrm{d}}m \big ) + \\ {}&\quad - \left( \int _{K^c} f_n g\, {\textrm{d}}m_n - \int _{K^c} f g \,{\textrm{d}}m \right) \Big \vert \\ {}&\le \left| \int _{\varOmega }f_ng\,{\textrm{d}}m_n -\int _{\varOmega }fg\,{\textrm{d}}m\right| + \left| \int _{K^c}f_ng\,{\textrm{d}}m_n-\int _{K^c}fg\,{\textrm{d}}m\right| \\ {}&= \left| \int _{\varOmega }f_ng\,{\textrm{d}}m_n-\int _{\varOmega }fg\,{\textrm{d}}m\right| + \left| \int _{U\setminus {K}}f_ng\,{\textrm{d}}m_n-\int _{U\setminus {K}}fg\,{\textrm{d}}m\right| \\ {}&< \left| \int _{\varOmega }f_ng\,{\textrm{d}}m_n-\int _{\varOmega }fg\,{\textrm{d}}m\right| + \int _{U\setminus {K}}\vert f_n\vert \,{\textrm{d}}m_n + \int _{U\setminus {K}}\vert f\vert \,{\textrm{d}}m <3\varepsilon . \end{aligned}$$

• Step 3 Let now B a Borelian set and let $\varepsilon > 0$, $\delta _1,\delta _2>0$ as in Step 2. Let $C_1$ be a compact set with $C_1 \subset B$ such that $m(B \setminus C_1) <\delta _1$. So

  $$\begin{aligned}&\left| \int _B f_n\,{\textrm{d}}m_n-\int _B f\,{\textrm{d}}m\right| \le \left| \int _{C_1 }f_n \,{\textrm{d}}m_n -\int _{C_1} f \,{\textrm{d}}m\right| + \\ {}&\quad + \left| \int _{B \setminus C_1}f_n \,{\textrm{d}}m_n-\int _{B \setminus C_1}f \,{\textrm{d}}m\right| \\ {}&\le \left| \int _{C_1}f_n \,{\textrm{d}}m_n-\int _{C_1}f \,{\textrm{d}}m\right| + \int _{B\setminus {C_1}}\vert f_n\vert \,{\textrm{d}}m_n + \int _{B\setminus {C_1}}\vert f\vert \,{\textrm{d}}m. \end{aligned}$$

  So the assertion follows from $j_1)-j_3$) and the compact case in Step 2 and this ends the proof.

$\square$

Corollary 3.5

Let m and $m_n$, $n \in {\mathbb N}$, be measures in $\mathcal {M}(\Omega )$, with m Radon. If $(m_n)_n$ is vaguely convergent to m and uniformly absolutely continuous with respect to m, then $(m_n)_n$ converges setwisely to m.

Proof

It is a consequence of Theorem 3.4 if we assume $f_n=f\equiv 1$ for every $n \in \mathbb {N}$. $\square$

Remark 3.6

• 3.6.a)] We observe that under the hypotheses of Theorem 3.4, if $f \in C(\Omega )$, then also $f^{\pm }$ are in $C(\Omega )$ and $f_n^{\pm }(t) \rightarrow f^{\pm }(t)$ m-a.e. as $n \rightarrow \infty$. In fact

  $$\begin{aligned}&\Big \vert \, \vert f_n\vert - \vert f\vert \, \Big \vert \le \big \vert f_n - f \big \vert \\ {}&2 f_n^+ = f_n + \vert f_n \vert \rightarrow f + \vert f \vert = 2f^+; \\ {}&2 f_n^- = \vert f_n\vert - f_n \rightarrow \vert f\vert - f = 2f^-. \end{aligned}$$

  Moreover also $(f_n^{\pm })_n$ and $f^{\pm }$ satisfy condition 3.4.iii) since

  $$\begin{aligned} f_n^{\pm } \le \vert f_n\vert \quad \text{ and } \quad f^{\pm } \le \vert f\vert . \end{aligned}$$

  Therefore in the hypotheses of Theorem 3.4 we get also

  $$\begin{aligned} \lim _n \int _{A} f^{\pm }_n{\textrm{d}}m_n = \int _{A} f^{\pm }{\textrm{d}}m, \quad \text{ for } \text{ every } A \in \mathcal {B}. \end{aligned}$$

• 3.6.b) Theorem 3.4 is still valid if we replace condition 3.4.i) with

  • 3.6.i’) $f_n$ converges in m-measure to f.

  In fact, by 3.4.i’), there exists a subsequence of $(f_{n_k})_k$ which converges m-a.e. to f. Then Theorem 3.4 is true for such subsequence. So this implies that the result of this theorem, equality (16), is still valid for the initial sequence (with convergence in m-measure) because if, absurdly, a subsequence existed in which it is not valid, there would be a contradiction.

A simple consequence of the Theorem 3.4 is the following

Theorem 3.7

Let m and $m_n$, $n \in {\mathbb N}$, be measures in $\mathcal {M}(\Omega )$, with m Radon. Let $f, f_n \in \mathcal {F}(\Omega )$. Suppose that

• (3.7.i) $f_n(t) \rightarrow f(t)$, m-a.e.;

• (3.7.ii) $f \in C_b(\Omega )$;

• (3.7.iii) $(f_n)_n$ has uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$;

• (3.7.iv) $(m_n)_n$ is vaguely convergent to m and uniformly absolutely continuous with respect to m.

Then

$$\begin{aligned} \lim _n \int _{A} f_n{\textrm{d}}m_n = \int _{A} f{\textrm{d}}m, \quad \text{ for } \text{ every } A \in \mathcal {B}. \end{aligned}$$

Proof

The assertion follows from Theorem 3.4 since f has uniformly absolutely continuous $(m_n)$-integrals, in fact it is enough to take the pair $(\varepsilon , \delta (\varepsilon /M))$ where $M > \sup _{t \in \Omega } \vert f(t)\vert$. $\square$

4 The multivalued and the vector cases

4.1 The multivalued case

Let X be a Banach space with dual $X^*$ and let $B_{X^*}$ be the unit ball of $X^*$. The symbol cwk(X) denotes the family of all weakly compact and convex subsets of X. For every $C \in cwk(X)$ the support function of $\, C$ is denoted by $s( \cdot , C)$ and defined on $X^*$ by $s(x^*, C) = \sup \{ \langle x^*,x \rangle :\ x \in C\}$. Recall that X is said to be weakly compact generated (briefly WCG) if it possesses a weakly compact subset K whose linear span is dense in X.

A map $\Gamma : \Omega \rightarrow cwk(X)$ is called a multifunction. A space $Y \subset X$ m-determines a multifunction $\Gamma$ if $s(x^*, \Gamma )=0$ m a.e. for every $x^* \in Y^{\perp }$, where the exceptional sets depend on $x^*$.

A multifunction $\Gamma$ is said to be

• scalarly measurable if $t\rightarrow s(x^*,\Gamma (t))$ is measurable, for every $x^*\in X^*$;

• scalarly m-integrable if $t\rightarrow s(x^*,\Gamma (t))$ is m-integrable, for every $x^*\in X^*$, where $m \in \mathcal {M}(\Omega )$;

• scalarly continuous if for every $x^*\in X^*$, $t\rightarrow s(x^*,\Gamma (t))$ is continuous.

A multifunction $\Gamma : \Omega \rightarrow cwk(X)$ is said to be Pettis integrable in cwk(X) with respect to a measure m (or shortly Pettis m-integrable) if $\Gamma$ is scalarly m-integrable and for every measurable set A, there exists $M_{\Gamma }(A) \in cwk(X)$ such that

$$\begin{aligned} s(x^*,M_{\Gamma }(A))=\int _A s\big (x^*,\Gamma \big )\, {\textrm{d}}m \qquad \text{ for } \text{ all } \; x^*\in X^*. \end{aligned}$$

We set $\displaystyle {\int }_A \Gamma {\textrm{d}}m:= M_{\Gamma }(A)$.

For the properties of Pettis m-integrability in the multivalued case we refer to [5,6,7,8, 26, 27], while for the vector case we refer to [25]. If $\Gamma$ is single-valued we obtain the classical definition of Pettis integral for vector function.

Given a sequence of multifunctions we introduce now some definitions of uniformly absolutely continuous scalar integrability using Definition 2.5.

Definition 4.1

For every $n \in {\mathbb N}$, let $m_n$ be a measure in $\mathcal {M}(\Omega )$ and let $\Gamma _n:\varOmega \rightarrow cwk(X)$ be a multifunction which is scalarly $m_n$-integrable. We say that the sequence $(\Gamma _n)_n$ has uniformly absolutely continuous scalar $(m_n)$-integrals on $\varOmega$ if for every $\varepsilon >0$ there exists $\delta >0$ such that, for every $n\in {\mathbb N}$ and $A \in \mathcal {B}$, it is

$$\begin{aligned} m_n(A)<\delta \ \Rightarrow \ \sup \left\{ \int _A\vert s(x^*, \Gamma _n)\vert {\textrm{d}}m_n :\parallel x^*\parallel \le 1\right\} < \varepsilon .\, \end{aligned}$$

(26)

Analogously a multifunction $\Gamma$ has uniformly absolutely continuous scalar $(m_n)$-integrals on $\varOmega$ if previous condition (26) holds for $\Gamma _n:=\Gamma$ for every $n \in \mathbb {N}$. Moreover we say that $\Gamma$ has uniformly absolutely continuous scalar m-integrals on $\varOmega$ if, in formula (26), it is $\Gamma _n:=\Gamma$ and $m_n=m$ for every $n \in \mathbb {N}$. In this case we have, for every $A \in \mathcal {B}$,

$$\begin{aligned} m(A)<\delta \ \Rightarrow \ \sup \left\{ \int _A\vert s(x^*, \Gamma )\vert {\textrm{d}}m :\parallel x^*\parallel \le 1\right\} < \varepsilon . \end{aligned}$$

(27)

Theorem 4.2

Let $\Gamma , \Gamma _n$, $n \in {\mathbb N}$, be scalarly measurable multifunctions. Moreover let $m, \ m_n$, $n \in {\mathbb N}$, be measures in ${\mathcal M}(\Omega )$ and let m be Radon. Suppose that

• (4.2.j) $(\Gamma _n )_n$ and $\Gamma$ have uniformly absolutely continuous scalar $(m_n)$-integrals on $\varOmega$;

• (4.2.jj)) $s(x^*, \Gamma _n) \rightarrow s(x^*, \Gamma )$ m-a.e. for each $x^* \in X^*$;

• (4.2.jjj) $\Gamma$ is scalar continuous;

• (4.2.jv) $(m_n)_n$ is vaguely convergent to m and uniformly absolutely continuous with respect to m;

• (4.2.v) each multifunction $\Gamma _n$ is Pettis $m_n$-integrable.

Then the multifunction $\Gamma$ is Pettis m-integrable in cwk(X) and

$$\begin{aligned} \lim _n s\Big ( x^*, \int _{A} \Gamma _n\, {\textrm{d}}m_n \Big )= s\Big ( x^*, \int _{A} \Gamma \, {\textrm{d}}m \Big ), \end{aligned}$$

(28)

for every $x^* \in X^*$ and for every $A \in \mathcal {B}$.

Proof

Let $x^* \in X^*$ be fixed. Then the sequence of functions $\big (s(x^*, \Gamma _n)\big )_n$ and the function $s(x^*, \Gamma )$ defined on $\Omega$ satisfy the assumptions of Theorem 3.4. So, for each $A \in \mathcal {B}$

$$\begin{aligned} \lim _n \int _{A} s(x^*, \Gamma _n)\, {\textrm{d}}m_n = \int _{A}s (x^*, \Gamma )\, {\textrm{d}}m. \end{aligned}$$

(29)

In order to prove that $\Gamma$ is Pettis m-integrable, following [26, Theorem 2.5], it is enough to show that the sublinear operator $T_{\Gamma }: X^* \rightarrow L^1(m)$, defined as $T_{\Gamma }(x^*)=s(x^*, \Gamma )$ is weakly compact (step $C_w$) and that $\Gamma$ is determined by a WCG space $Y \subset X$ (step D).

$C_w$):

First of all we prove that the operator $T_{\Gamma }$ is bounded. By (4.2.jjj) $\Gamma$ is scalar m-integrable. Therefore $\Gamma$ is Dunford-integrable in $cw^*k(X^{**})$, where $X^{**}$ is endowed with the $w^*$-topology, and for every $A \in \mathcal {B}$ let $M^D_{\Gamma } (A) \in cw^*k(X^{**})$ be such that

$$\begin{aligned} s(x^*, M^D_{\Gamma } (A)) =\int _As(x^*,\Gamma ) {\textrm{d}}m < +\infty , \end{aligned}$$

(30)

for every $x^* \in X^*$. So $s(x^*,M^D_{\Gamma } ( \cdot ))$ is a scalar measure and

$$\begin{aligned} \int _{\Omega } \vert s(x^*,\Gamma ) \vert {\textrm{d}}m \le 2 \sup _{A\in \mathcal {B} }\left| \int _A s(x^*,\Gamma ) {\textrm{d}}m \right| < +\infty . \end{aligned}$$

(31)

Hence, the set $\displaystyle {\bigcup _{A \in \mathcal {B}} }M_{\Gamma }^D(A) \subset X^{**}$ is bounded, by the Banach- Steinhaus Theorem, and

$$\begin{aligned} \sup _{\parallel x^*\parallel \le 1} \int _{\Omega } \vert s(x^*,\Gamma ) \vert {\textrm{d}}m \le 2 \sup \left\{ \parallel x \parallel : x \in \bigcup _{A \in \mathcal {B}} M_{\Gamma }^D(A) \right\} <+ \infty . \end{aligned}$$

Since the set $\big \{s(x^*,\Gamma ): \parallel x^*\parallel \le 1 \big \}$ is bounded in $L^1(m)$, the operator $T_{\Gamma }$ is bounded.

In order to obtain the weak compactness of the operator $T_{\Gamma }$ it is enough to prove that $\Gamma$ has absolutely continuous scalar m-integrals on $\varOmega$. Let $x^* \in B_{X^*}$ be fixed. Now fix $\varepsilon >0$ and let $\sigma (\varepsilon ) >0$ satisfy (4.2.j). Moreover let $\delta (\sigma ) >0$ verify (4.2.jv).

Let $E \in \mathcal {B}$ be such that $m(E) < \delta$ and set

$$\begin{aligned} E^+=\{ t \in E: s(x^*, \Gamma (t)) \ge 0\} \quad E^-=\{ t \in E: s(x^*, \Gamma (t)) <0\}. \end{aligned}$$

By (29) let now $N_{x^*} \in \mathbb {N}$ be an integer such that for every $n \ge N_{x^*}$

$$\begin{aligned} \left| \int _{E^{\pm }} s(x^*, \Gamma ){\textrm{d}}m \right| < \left| \int _{E^{\pm }} s(x^*, \Gamma _{n}){\textrm{d}}m_{n} \right| + \frac{\varepsilon }{2}. \end{aligned}$$

So, for every $n \ge N_{x^*}$,

$$\begin{aligned} \int _E \vert s(x^*, \Gamma )\vert {\textrm{d}}m&= \int _{E^+} s(x^*, \Gamma ){\textrm{d}}m + \left| \int _{E^-} s(x^*, \Gamma ){\textrm{d}}m \right| \\ {}&< \left| \int _{E^+} s(x^*, \Gamma _{n}){\textrm{d}}m_{n} \right| + \left| \int _{E^-} s(x^*, \Gamma _{n}){\textrm{d}}m_{n} \right| +\varepsilon . \end{aligned}$$

Since, by (4.2.jv), it is in particular $m_{n}(E) < \sigma$ for every $n \ge N_{x^*}$, we get

$$\begin{aligned} \int _E \vert s(x^*, \Gamma )\vert {\textrm{d}}m \le \int _E \vert s(x^*, \Gamma _{n})\vert {\textrm{d}}m_{n} + \varepsilon < 2 \varepsilon \end{aligned}$$

so $\Gamma$ has uniformly absolutely continuous scalar m-integral on $\varOmega$.

D):

We have to show the existence of a WCG subspace of X which determines $\Gamma$. Since $\Gamma _n$ is Pettis $m_n$-integrable, for every $n \in {\mathbb N}$, let $Y_n \subseteq X$ be a WCG space generated by a set $W_n \in cwk( B_{X^*})$ which $m_n$-determines $\Gamma _n$, by [26, Theorem 2.5].

We may suppose, without loss of generality that each $W_n$ is absolutely convex, by Krein-Smulian’s Theorem. Let Y be the WCG space generated by $W: =\sum 2^{-n} W_n$. We want to prove that $\Gamma$ is m-determined by Y.

If $y^*\in {Y^\perp }$, then $y^*\in {Y_n^\perp }$ for each n, hence $s(y^*,\Gamma _n)=0$ $m_n$-a.e. Applying (29) with $A=\Omega ^+:=\{t: s(y^*, \Gamma (t)) \ge 0\}$ ($A = \Omega ^-:=\{t: s(y^*, \Gamma (t)) < 0\}$) we get

$$\begin{aligned} \int _{\Omega ^{\pm }}s (y^*, \Gamma )\, {\textrm{d}}m=\lim _n \int _{\Omega ^{\pm }} s(y^*, \Gamma _n)\, {\textrm{d}}m_n =0. \end{aligned}$$

Therefore $s(y^*,\Gamma (t))=0$ m-a.e. on the set $\Omega$. Thus, Y m-determines the multifunction $\Gamma$ and the Pettis m-integrability of $\Gamma$ follows. Moreover (28) follows from (29). $\square$

As an immediate consequence of the previous theorem we have a result for the vector case:

Corollary 4.3

Let $g, g_n:\Omega \rightarrow X$, $n \in {\mathbb N}$, be scalarly measurable functions. Moreover let $m, \ m_n$, $n \in {\mathbb N}$, be measures in ${\mathcal M}(\Omega )$ and let m be Radon. Suppose that

• (4.3.j) $(g_n )_n$ and g have scalarly uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$;

• (4.3.jj)) $g_n \rightarrow g$ scalarly m-a.e. where the null set depends on $x^* \in X^*$;

• (4.3.jjj) g is scalar continuous;

• (4.3.jv) $(m_n)_n$ is vaguely convergent to m and uniformly absolutely continuous with respect to m;

• (4.3.v) each $g_n$ is Pettis $m_n$-integrable.

Then g is Pettis m-integrable in X and

$$\begin{aligned} \lim _n x^*\left( \int _{A} g_n\, {\textrm{d}}m_n \right) = x^* \left( \int _{A} g\, {\textrm{d}}m \right) , \end{aligned}$$

for every $x^* \in X^*$ and $A \in \mathcal {B}$.

We conclude with the following result that holds in a general measure space without any topology on the space $\Omega$.

Proposition 4.4

Let $\Omega$ be a measurable space on a $\sigma$-algebra $\mathcal {A}$ and let $\Gamma , \Gamma _n$, $n \in {\mathbb N}$, be scalarly measurable multifunctions. Moreover let $m, \ m_n$, $n \in {\mathbb N}$, be measures in ${\mathcal M}(\Omega )$. Suppose that

• (4.4.j) $(\Gamma _n )_n$ have scalarly uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$;

• (4.4.jj) $\Gamma$ is scalarly m-integrable;

• (4.4.jjj) $(m_n)_n$ is uniformly absolutely continuous with respect to m;

• (4.4.jv) each multifunction $\Gamma _n$ is Pettis $m_n$-integrable.

• (4.4.v) for every $A \in \mathcal {A}$ and for every $x^* \in X^*$ it is

  $$\begin{aligned} \lim _n \int _{A} s(x^*, \Gamma _n) \, {\textrm{d}}m_n = \int _A s(x^*, \Gamma ) \, {\textrm{d}}m. \end{aligned}$$

Then the multifunction $\Gamma$ is Pettis m-integrable.

Proof

The weak compactness of the sublinear operator $T_{\Gamma }: X^* \rightarrow L^1(m)$, defined as $T_{\Gamma }(x^*)=s(x^*, \Gamma )$ can be proved as Theorem 4.2, taking into account hypotheses (4.4.jj), (4.4.jjj) and (4.4.v).

Moreover, the proof that $\Gamma$ is determined by a WCG space $Y \subset X$ follows as in Theorem 4.2, taking into account hypotheses (4.4.j), (4.4.jjj), (4.4.jv) and (4.4.v). Therefore $\Gamma$ is Pettis m-integrable. $\square$

At this point it is worth to observe that a similar result has been proved in [13, Theorem 3.2] under the hypothesis of the setwise convergence of the measures. Here instead of the setwise convergence we assume the uniform absolute continuity of $(m_n)_n$ with respect to m.

For the vector case, as before, we have:

Corollary 4.5

Let $\Omega$ be a measurable space on a $\sigma$-algebra $\mathcal {A}$ and let $g, g_n:\varOmega \rightarrow X$, $n \in {\mathbb N}$, be scalarly measurable functions. Moreover let $m, \ m_n$, $n \in {\mathbb N}$, be measures in ${\mathcal M}(\Omega )$. Suppose that

• (5.5.j) $(g_n )_n$ have scalarly uniformly absolutely continuous $(m_n)$-integrals on $\varOmega$;

• (5.5.jj) g is scalar m-integrable;

• (5.5.jjj) $(m_n)_n$ is uniformly absolutely continuous with respect to m;

• (5.5.jv) each function $g_n$ is Pettis $m_n$-integrable;

• (5.5.v) for every $A \in \mathcal {A}$ and for every $x^* \in X^*$ it is

  $$\begin{aligned} \lim _n \int _{A} x^*g_n \, {\textrm{d}}m_n = \int _A x^*g \, {\textrm{d}}m. \end{aligned}$$

Then the function g is Pettis m-integrable.

5 Conclusion

Some limit theorems for the sequences $\left( \int f_n\,{\textrm{d}}m_n \right) _n$ are presented for vector and multivalued Pettis integrable functions when the sequence $(m_n)_n$ vaguely converges to a measure m. The results are obtained thanks to a limit result obtained for the scalar case (Theorem 3.4).