Vector Valued Integrals

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Vector Valued Integrals

Sankha Subhra Basu



Introduction

Sometimes it is desirable to be able to integrate functions f that are definedon some measure space X (with a real or complex measure µ) and whosevalues lie in some topological vector space V . A compelling application of the theory of integration of vector-valued functions is a theory of holomor-phic vector-valued functions, with well-known application to the resolvents of operators on Hilbert and Banach spaces. Another sort of application of un-derstanding holomorphic vector-valued functions is to generalized functions ,studying parametrized families of distributions. Many distributions whichare not classical functions appear naturally as residues or analytic continu-ations of families of classical functions with a complex parameter. A goodtheory of integration allows a natural treatment of convolutions of distribu-tions with test functions, and related operations.The first problem is to associate with these data in X that deserves to becalled

X

f dµ,

i.e., which has at least some of the properties that integrals usually have.Many approaches to vector-valued integration have been studied in greatdetail. Some of these can be classified as ‘strong’ in the sense that, theintegrals are defined somewhat directly as limits of sums. On the other hand

we have the approach of Gelfand and Pettis, which can be termed as ‘weak’in the sense that it defines an integral by the essential property that anyreasonable integral should have.

Strong Integrals

Strong integrals are generalized Lebesgue integrals for functions whose valueslie in a Banach space or more generally in a topological vector space. Some of the well-known notions are due to Bochner, Birkhoff and McShane for Banach

space valued functions. We choose the integral due Salomon Bochner herefor a flavor of the strong notion of integration.

1



The Bochner Integral

Let (X, Σ, µ) be a measure space and V a Banach space, the ground fieldbeing K (where K is either R or C). For simplicity, in most of the followingdiscussion, we will have K = R. The Bochner integral is defined in much thesame way as the Lebesgue integral. As in the case of the regular Lebesgueintegral, a function φ : X → V that assumes only a finite number of values,say v1, v2, . . . , vn, is called a simple function if E i = φ−1(vi) ∈ Σ for each i.Then the formula,

φ =n

i=1

χEi

vi

is called the standard representation of φ. If µ(E i) < ∞ for each non-zerovi, then φ is called a simple function with finite support . The integral of such a V -valued simple function with finite support is the vector

X

φ dµ of V defined via the formula

X

φ dµ =ni=1

µ(E i)vi

A simple function with finite support, φ with standard representation, φ =

n

i=1χEi

vi may have other representations

φ =m

j=1

χF j

w j

where, F j ∈ Σ with µ(F j) < ∞ whenever w j = 0. Here, the w j ’s are notnecessarily distinct and F j’s are also not necessarily pairwise disjoint.We now state a lemma about a property of σ-algebras, without proof.

Lemma. Let Σ be a σ-algebra of subsets of a set X and let A1, . . . , An belongto Σ. Then there exists a finite collection {C 1, . . . , C k} of pairwise disjoint

members of Σ such that:

1. Each C i is a subset of some A j; and

2. Each A j is a union of a subfamily of the collection {C 1, . . . , C k}.

2



Theorem. If φ is a simple function with finite support and φ = m

j=1χF j

w j

is any other representation, then X

φ dµ =m

j=1

µ(F j)w j.

Proof. Let φ = n

i=1χEi

vi be the standard representation of φ. We firstassume that the F j are pairwise disjoint. Since neither the function φ northe sum Σm

j=1µ(F j)w j changes by deleting the terms with w j = 0, we can

assume that w j = 0 for each j. In such a case, we have n

i=1E i =

m

j=1F j.

Moreover, we note that µ(E i ∩ F j)vi = µ(E i ∩ F j)w j for all i and j . Indeed, if

E i ∩ F j = ∅, the equality is obvious and if x ∈ E i ∩ F j, then vi = w j = φ(x).Furthermore, since the E i’s and the F j’s are pairwise disjoint, E i ∩ F j ’s arealso pairwise disjoint for all i and j. Therefore,

X

φ dµ =ni=1

µ(E i)vi

=m

j=1

ni=1

µ(E i ∩ F j)vi

=m

j=1

n

i=1

µ(E i ∩ F j)w j

=m

j=1

µ(F j)w j.

Now, we consider the general case. By the above lemma, there exist pairwisedisjoint sets C 1, . . . , C k ∈ Σ such that each C i is included in some F j andF j = ∪{C i | C i ⊂ F j}. For each i and j let δ ji = 1 if C i ⊂ F j and δ ji = 0 if C i ⊂ F j. Clearly,

χF j =

k

i=1

δ

j

i χC i and µ(F j) =

k

i=1

δ

j

i µ(C i).

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Consequently,

φ =m

j=1

χF j

w j

=m

j=1

ki=1

δ ji χC i

w j

=ki=1

χC i

m j=1

δ ji w j

.

So, by the preceding case, we have

X

φ dµ =ki=1

µ(C i)

m j=1

δ ji w j

=m

j=1

ki=1

δ ji µ(C i)

w j

=m

j=1

µ(F j)w j ,

and the proof is finished.

The above theorem shows that the definition of the integral of a simplefunction with finite support is well-defined. Let the vector space of all simplefunctions with finite support be denoted by S . The following theorem showsthat the integral is a linear operator from S to V .

Theorem. Let φ, ψ : X → V be simple functions with finite support. Leta, b ∈ K. Then,

X

(aφ + bψ) dµ = a

X

φ dµ + b

X

ψ dµ.

Proof. Let

φ(x) =ni=1

χEi

(x)vi

and ψ(x) =m

j=1

χF j

(x)w j

4



in the standard representation. The sets {E i}ni=1 are mutually disjoint and

vi(i = 1, . . . , n) are distict members of V . Similarly, the sets {F j}m j=1 aremutually disjoint and w j( j = 1, . . . , m) are distinct members of V . Letv0 = w0 = 0, E 0 = {x ∈ V | φ(x) = 0}, and F 0 = {x ∈ V | ψ(x) = 0}.Then,

φ(x) =ni=0

χEi

(x)vi

and ψ(x) =m

j=0

χF j

(x)w j

For each i = 0, 1, . . . , n; j = 0, 1, . . . , m, let

Ai,j = E i ∩ F j .

We note that,ni=0

Ai,j = F j andm

j=0

Ai,j = E i

Thus,

φ(x) =ni=0

χEi

(x)vi

=ni=0

m j=0

χAi,j

(x)vi

and ψ(x) =m

j=0

χF j

(x)w j

=m

j=0

ni=0

χAi,j

(x)w j

Moreover,

(aφ + bψ)(x) =

n

i=0

m

j=0

χAi,j (x)(avi + bw j)

We note that Ai,j ∈ Σ, Ai,j ∩ Ai,j = ∅ provided i = i or j = j , and

5



µ(Ai,j) < ∞ provided i = 0 or j = 0. So,

X

(aφ + bψ)(x) =ni=0

m j=0

µ(Ai,j)(avi + bw j)

= a

ni=0

m j=0

µ(Ai,j)vi + b

m j=0

ni=0

µ(Ai,j)w j

= a

ni=0

µ(E i) + b

m j=0

µ(F j)

= a X

φ(x) dµ + b X

ψ(x) dµ

If f : X → V is a vector function, thenf denotes the nonnegative real

functionf : X → R defined by

f (x) =

f (x) for each x ∈ X . We callf

the norm function of f .

If φ ∈ S and E ∈ Σ, then E

φ dµ is the integral of φ over E and is definedby

E

φ dµ = X

φχE

dµ.

Lemma. If φ ∈ S has standard representation φ = n

i=1χEi

vi, then the

norm functionφ of φ is a real simple function with finite support having

standard representationφ =

n

i=1

viχ

Ei. Moreover,

X

φ dµ =

ni=1

viµ(E i) and

X

φ dµ ≤

X

φ dµ.

Proof. Clear.

Let f

: X

→ V

be a vector function. We say that the function f

isstrongly measurable (or simply measurable ) if there exists a sequence of simplefunctions {φn} such that limn→∞

f (x) − φn(x) = 0 for almost all x ∈ X .

Lemma. If f : X → V is strongly measurable, then the real functionf is

also measurable.

6



Proof. The inequality f (x) − φn(x) ≤ f (x) − φn(x) implies thatφn(x) →

f (x) for almost every x ∈ X . We know that for any measurable

space (X, Σ, a real function f : X → R is Σ-measurable if and only if thereexists a sequence {f n} of Σ-simple functions satisfying f n(x) → f (x) for eachx ∈ X . Thus using the previous lemma, we have

f is measurable.

Let M(X, V ) denote the collection of all strongly measurable functionsfrom X to V that is,

M(X, V ) = {f ∈ V X | f is strongly measurable}.

Lemma. The collection M(X, V ) of all strongly measurable functions fromX to V is a vector subspace of V X containing all the simple functions of finite support. That is, we have the following vector subspace inclusions:

S ⊂ M(X, V ) ⊂ V X .

Proof. Straightforward.

Lemma. Let f : X → V be a strongly measurable function. Suppose thatfor two sequences {φn} amd {ψn} of simple functions with finite support, thereal functions f − φn and f − ψn are Lebesgue measurable for each n

and

limn→∞

X

f − φn

dµ = limn→∞

X

f − ψn

dµ = 0.

Then for each E ∈ Σ we have,

limn→∞

E

φn dµ = limn→∞

E

ψn dµ,

where the last two limits are taken with respect to the norm topology on V .

Proof. Suppose, {φn} and {ψn} are two sequences of simple functions withfinite support which satisfy the stated property. We choose and fix E ∈ Σ.

7



From

E

φn dµ −

E

φm dµ =

E

(φn − φm) dµ

≤

X

φn − φm

dµ

≤

X

f − φn

dµ +

X

f − φm

dµ,

we see that limn,m→∞

E

φn dµ − E

φm dµ = 0, which shows that the se-

quence { E

φn dµ} is a Cauchy sequence in V , so it converges in V . Similarly,the sequence {

E

ψn dµ} converges in V . Now the inequality

E

φn dµ −

E

ψn dµ ≤

X

f − φn

dµ +

X

f − ψn

dµ

easily implies limn→∞

E


E

ψn dµ, as claimed.

A function f : X → V is called Bochner integrable if there exists asequence φn of integrable simple functions with finite support such that thereal function

f − φn

is Lebesgue integrable for each n and

limn→∞

X f − φ

nV

dµ = 0.

In this case, for each E ∈ Σ, the Bochner integral of f over E is definedby

E

f dµ = limn→∞

E

φn dµ,

where the limit here is the limit of the sequence of vectors { E

φn dµ} in thenorm topology in V .

By the previous lemma, the Bochner integral is well-defined, in the sensethat it does not depend on the particular sequence of simple functions (withfinite support) used to approximate f .

Theorem. If f and g are two Bochner integrable functions and α, β ∈ K,then αf + βg is also Bochner integrable and

X

(αf + βg) dµ = α

X

f dµ + β

X

g dµ.

8



Proof. Since f and g are Bochner integrable, there exist sequences of simple

functions with finite support {φn} and {ψn} such that the real functionsf − φn

andg − ψn

are Lebesgue integrable for each n and

limn→∞

X

f − φn

dµ = 0

limn→∞

X

g − ψn

dµ = 0

Now, S = the set of all simple functions with finite support is a vector space,αφn+βψn is a simple function with finite support for each n. So, {αφn+βψn}is a sequence of simple functions with finite support. We note that for eachn,

X

(αf + βg) − (αφn + βψn) dµ =

X

α(f − φn) + β (g − ψn) dµ

≤

X

|α|f − φn

+ |β |g − ψn

dµ

= |α|

X

f − φn

dµ + |β |

X

g − ψn

dµ

< ∞

Thus,(αf +βg)−(αφn+βψn)

is Lebesgue integrable for each n. Moreover,

limn→∞

X

(αf + βg) − (αφn + βψn) dµ

≤|α| limn→∞

X

f − φn

dµ + |β | limn→∞

X

g − ψn

dµ

Since each of the above limits converge to zero,

limn→∞

X

(αf + βg) − (αφn + βψn)

dµ → ∞

Thus, (αf + βg) is Bochner integrable as required. Then using the fact thatthe Bochner integral is a linear operator from S to V , and continuity of

9



addition and scalar multiplication in V , we have X

(αf + βg) dµ = limn→∞

X

(αφn + βψn) dµ

= limn→∞

α

X

φn dµ + β

X

ψn dµ

= limn→∞

α

X

φn dµ + limn→∞

β

X

ψn dµ

= α limn→∞

X

φn dµ + β limn→∞

X

ψn dµ

= α X

f dµ + β X

g dµ,

as was desired.

Corollary. The collection of all Bochner integrable functions form a vectorsubspace of M(X, V ) and the Bochner integral acts as a linear operator fromthis space into V .

Proof. Straightforward consequence of the above theorem.

Weak Measurability

Suppose, (X, Σ) is a measurable space and V is a Banach space over a fieldK (K = R or C). A function f : X → V is called weakly measurable if, forevery continuous linear functional g : V → K , the function

g ◦ f : X → K : x → g(f (x))

is measurable with respect to Σ and the usual Borel σ-algebra on K . A func-tion measurable in the usual sense is sometimes called ‘strongly’ measurableto match this terminology of weak measurability.We now present a theorem and a corollary to it without proof.

Theorem. (Pettis Measurability Theorem ) A function f : X → V is measur-able (strongly measurable) if and only if f is weakly measurable and almosteverywhere separable valued, i.e., there is a set N ⊂ X, µ(N ) = 0 such thatthe set

{f (t) | t ∈ X \ N } ⊂ V

is separable under the subspace topology derived from V .

10



Corollary. 1 If V is a separable Banach space, then the strong and weak

notions of measurability are the same. In other words, any function, f : X →V is weakly measurable if and only if it is strongly measurable.

Corollary. 2 A function f : X → V is measurable (strongly measurable)if and only if it can be approximated uniformly, except possibly on a set of measure zero, by countably valued functions.

Theorem. Let (X, Σ, µ) be a finite measure space, and let f : X → V bea strongly measurable function. Then, f is Bochner integrable if and only if its norm function

f

is Lebesgue integrable that is, X

f

dµ < ∞.

Proof. Since, f is Bochner integrable, there exists a sequence {φn} of simplefunctions with finite support such that

limn→∞

X

f − φn

dµ = 0

Then, X

f dµ ≤

X

f − φn

dµ +

X

φn

dµ < ∞

for sufficiently large n.Coversely, suppose f (and consequently f ) is measurable and X f dµ <

∞. With the help of Corollary 2 above, we can choose a sequence of countablyvalued measurable functions {f n} such that

f − f n ≤ 1

n for each positive

integer n. Sincef n

≤f + 1

n almost everywhere and µ is finite, then

X

f n dµ < ∞. For each positive integer n, we can write

f n =∞

m=1

χEn,m

vn,m,

where E n,i ∩ E n,j = ∅ for i = j, E n,m ∈ Σ, vn,m ∈ V . For each n, we choose pn so large that

∪∞

m=pn+1E n,m

f n dµ < µ(X )

n

and set gn = pn

m=1χEn,m

vn,m. Then each gn is a simple function and

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X

f − gn dµ ≤

X

f − f n dµ +

X

f n − gn dµ

≤ µ(X )

n +

µ(X )

n

= 2µ(X )

n .

Therefore, f is Bochner integrable, as was to be proven.

Theorem. (Vector Dominated Convergence Theorem ) Let f : X → V bestrongly measurable and let a sequence {f n} of Bochner integrable functionssatisfy f (x) − f n(x) → 0 for almost all x ∈ X . If there exists a realnonnegative Lebesgue integrable function g : X → R such that for each n,we have

f n ≤ g almost everywhere, then f is Bochner integrable and

E

f dµ = limn→∞

E

f n dµ.

for each E ∈ Σ.

Proof. Clearly,f ≤ g almost everywhere, and

f − f n is measurable

for each n. Fromf − f n

≤ 2g almost everywhere, we see thatf − f n

is Lebesgue integrable for each n. Moreover, from f − f n → 0 almost

everywhere, and the Lebesgue Dominated Convergence Theorem, we get X

f − f n dµ → 0. Next for each n we choose a simple function (with

finite support) φn with X

f n − φn

dµ < 1

n, and note that

X

f − φn

dµ ≤

X

f − f n dµ +

X

f n − φn

dµ → 0.

This implies that f is Bochner integrable and that

E

f dµ = limn→∞

E


E

f n dµ

for each E ∈ Σ.

12



Weak Integrals

As for strong integrals, there have been many definitions of weak integralsdue to Gelfand, Dunford, Pettis and still others. We choose here the Gelfand-Pettis integral, which combines the ideas of the Gelfand and the Pettis inte-grals, for discussion of the weak integrals.

The Gelfand-Pettis Integral

Let (X, Σ, µ) be a measure space, V be a real or complex topological vectorspace and V ∗, the dual space of V . Suppose f is a function from X into V .A Gelfand-Pettis integral of f would be a vector I f ∈ V so that,

Λ(I f ) =

X

Λ(f ) dµ for all Λ ∈ V ∗.

If it exists and is unique, this vector I f is denoted by

I f =

X

f dµ

By contrast to construction of the strong integrals, this definition bankson the irreducible mimimum property that no reasonable notion of integral

should lack. So, this definition is called a weak integral. We shall provebelow that a Bochner integrable function is in fact Gelfand-Pettis integrable.

Uniqueness

Uniqueness of the Gelfand-Pettis integral of a function is immediate whenV ∗ separates points, as for locally convex spaces by Hahn-Banach theorem.In particular, since every Banach space is locally convex, the GP integral of any Banach space valued function is unique.

ExistenceIf V is a vector space and E ⊂ V , the convex hull of E is the intersection of allconvex subsets of V which contain E and is denoted by co(E ). Equivalently,co(E ) is the set of all finite convex combinations of members of E .The closed convex hull of E , written co(E ), is the closure of co(E ).

13



We will prove the existence of the GP integral only in the rather special

case in which X is compact and f is continuous. In that case, f (X ) iscompact, and the only other requirement that will be imposed is that theclosed convex hull of f (X ) should be compact. This last requirement isautomatically satisfied for Banach spaces.

Theorem. Suppose

(a) X is a topological vector space on which X ∗ separates points, and

(b) µ is a Borel probability measure on a compact Hausdorff space X .

If f : X → V is continuous and co(f (X )) is compact in V , then the GP

integral of f

I f =

X

f dµ (1)

exits. Moreover, I f ∈ co(f (X )).

Proof. Let V be a real vector space. Suppose, H = co(f (X )). We have toprove that there exists I f ∈ V such that

Λ(I f ) =

X

Λ(f ) dµ (2)

for every Λ ∈ V ∗

. Let L = {Λ1, . . . , Λn} be a finite subset of V ∗

. Let E L bethe set of all y ∈ H that satisfy (2) above for every Λ ∈ L. Each E L is closed(by the continuity of Λ) and is therefore compact, since H is compact. If no E L is empty, the collection of all E L has the finite intersection property.The intersection of E L is therefore not empty, and any y in it satisfies (2) forevery Λ ∈ V ∗. It is therefore enough to prove E L = ∅.We can regard L = (Λ1, . . . , Λn) as a mapping from X into Rn. Let K =L(f (X )). We define

mi =

X

Λi(f ) dµ (1 ≤ i ≤ n). (3)

We claim that the point m = (m1, . . . , mn) lies in the convex hull of K .Now, if t = (t1, . . . , tn) ∈ Rn is not in this hull, then there are real numbersc1, . . . , cn such that

ni=1

ciui <

ni=1

citi (4)

14



if u = (u1, . . . , un) ∈ K . Hence,

ni=1

ciΛ(f (x)) <

ni=1

citi (x ∈ X ). (5)

Since µ is a probability measure, integration of the left hand side of (5)gives

ni=1

cimi <

ni=1

citi.

Thus t = m. This shows that m lies in the convex hull of K . Since K =L(f (X )) and L is linear, it follows that m = Ly for some y in the convexhull H of f (X ). For this y, we have

Λi(y) = mi =

X

Λi(f ) dµ (1 ≤ i ≤ n). (6)

Hence y ∈ E L. This completes the proof.

Theorem. The collection of all Gelfand-Pettis (GP in short) integrable func-tions is a vector space and the GP integral is linear. That is, if f and g areGP integrable over a set E ∈ Σ, and α, β ∈ K, then αf + βg is also GP

integrable over E and

E

(αf + βg) dµ = α

E

f dµ + β

E

g dµ

Proof. Since f and g are GP integrable, there exist unique vectors I f and I gin V such that

Λ(I f ) =

X

Λ(f ) dµ

and Λ(I g) =

X

Λ(g) dµ

for all Λ ∈ V

∗

. Moreover, I f is the GP integral of f and I g is the GP integralof g over X . Now, we note that (αI f + βI g) is an element of V and for any

15



Λ ∈ V ∗,

Λ(αI f + βI g) = αΛ(I f ) + β Λ(I g)

= α

X

Λ(f ) dµ + β

X

Λ(g) dµ

=

X

αΛ(f ) dµ +

X

β Λ(g) dµ

=

X

Λ(αf ) dµ +

X

Λ(βg) dµ

=

X

Λ(αf ) + Λ(βg)

dµ

= X

Λ(αf + βg) dµ,

which proves that (αf + βg) is GP integrable and

αI f + βI g =

X

(αf + βg) dµ,

as was desired.

Let G P be the class of GP integrable functions from X to V . The abovetheorem proves that GP is a vector space over K and the GP integral is a

linear operator from G P to V .

Relationship between the above notions of in-

tegrability

Since, any notion of an integral should satisfy the irreducible minimum prop-erty through which the Gelfand-Pettis integral is defined, any strongly inte-grable function should also be weakly integrable. So, any strong notion of integrability should =⇒ Gelfand-Pettis integrability.

Lemma. Suppose (X, Σ, µ) is a mesaure space, V a real Banach space andφ : X → V a simple function with finite support. Then φ is Pettis integrableand the integrals coincide.

16



Proof. Let φ = n

i=1χEi

vi be the standard representation of φ. Then the

Bochner integral of φ, I φ ∈ V is

I φ =

X

φ dµ =ni=1

µ(E i)vi.

Then for any Λ ∈ V ∗,

Λ(φ) = Λ

ni=1

χEi

vi

=

ni=1

χEi

Λ(vi).

is a real simple function with finite support. Moreover,

Λ(I φ) = Λ

ni=1

µ(E i)vi

=ni=1

µ(E i)Λ(vi)

So, for any Λ ∈ V ∗ X

Λ(φ) dµ =ni=1

µ(E i)Λ(vi)

= Λ(I φ)

This proves that φ is GP integrable and the integrals coincide.

Theorem. Suppose V is a Banach space. If a function f : X → V is Bochnerintegrable, then f is GP integrable and the two integrals coincide.

Proof. Since f is Bochner integrable, there exists a sequence of simple func-tions with finite support {φn} such that the real function

f −φn

is Lebesgueintegrable for each n and

limn→∞

X

f − φn dµ = 0,

and the Bochner integral of f , I f ∈ V is

I f = limn→∞

X

φn dµ.

17



Then for any Λ ∈ V ∗, using the continuity of Λ, we have

Λ(I f ) = Λ

limn→∞

X

φn dµ

= limn→∞

Λ

X

φn dµ

= limn→∞

X

Λ(φn) dµ,

using the above lemma. Hence, f is GP integrable and I f is also the GP

integral of f , as required.

Thus, we have Bochner =⇒ Gelfand-Pettis. The reverse arrow does nothold in general.

Example of a GP integrable function which is not Bochner inte-

grable

Let (X, Σ, µ) be the measure space with X = [0, 1] and µ = the Lebesguemeasure on [0, 1]. Let V be the real Banach space c0. We consider thefunction f : X → V defined by

f (x) = {nχI n

(x)}

where, I n is the interval

1

n + 1, 1

n

, χ

I n is the characteristic function of I n

and {nχI n

(x)} is a sequence of non-negative integers for any x ∈ X . Clearly,a sequence {nχ

I n(x)} has atmost one non-zero term and hence is a member

of c0. So, f is well-defined.

Then f is GP integrable and the GP integral of f is the vector

1

n + 1

∈

c0. But we note that

f =∞

n=1

nχI n

,

which is not Lebesgue integrable. Since µ is a finite measure on X , thismeans that f is not Bochner integrable.

18



References

(1) W. Rudin: Functional Analysis , Second Edition, Springer (1999)

(2) Charalambos D. Aliprantis, Kim C. Border: Infinite Dimensional Anal-

ysis - A Hitchhiker’s Guide , Studies in Economic Theory, Springer-Verlag (1994)

(3) J. Diestel, J. J. Uhl, Jr.: Vector Measures , Mathematical Surveys, Num-ber 15, AMS (1977)

(4) Raymond A. Ryan: Introduction to Tensor Products of Banach Spaces,

Springer (2002)(5) B. J. Pettis: On Integration in Vector Spaces , Trans. of AMS, Vol. 44,

No. 2 (Sep. ,1938), pp. 277-304

(6) Garrett Birkhoff: Integration of Functions with Values in a Banach

Space , Trans. of AMS, Vol. 38, No. 2 (Sep. ,1935), pp. 357-378

(7) Jose Rodrıguez: Some Examples in Vector Integration , Bull. Aust. Math. Soc.(2009)

(8) Paul Garrett: Vector-Valued Integrals , http://www.math.umn.edu/

~garrett/ (2008)

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Vector Valued Integrals