RADON-NIKODYM THEOREMS

D. CANDELORO and A. VOLCIC

1 Introduction

Suppose λ is the Lebesgue measure on the real line and f is an integrablefunction. Then the measure ν defined for all the Lebesgue measurable sets

ν(E) =∫

E

f dλ

is called the indefinite integral of f with respect to λ. It is obvious from thedefinition of the integral that if λ(E) = 0, then ν(E) = 0 for any Lebesgue-measurable set E. This is expressed saying that ν is absolutely continuous withrespect to λ and will be denoted by ν λ.

It is not difficult to check (see [25] Sec. 30, Theorem B) that under theassumption that ν is a finite measure, this condition is equivalent to anotherone which can be given in ε - δ terms, which will be denoted by ν ε λ: givenany ε > 0, there exists a δ > 0 such that if λ(E) < δ, then ν(E) < ε, for anyLebesgue-measurable set E.

We always have that ν ε λ implies that ν λ, but the converse implicationdoes not hold in general, as it can be easily seen considering for instance themeasure ν(E) =

∫E|x| dλ(x).

In the σ-additive case, absolute continuity of a measure ν with respect toanother measure µ implies, under mild conditions on µ and ν, that ν is theintegral measure of some function f ∈ L1(µ) : such function is called Radon-Nikodym derivative of ν with respect to µ, and it is denoted by dν

dµ .

Nikodym ([41]) was the first to prove this result in a quite general setting. Ifµ and ν are two finite measures on the same σ-algebra, and if ν µ, then thereexists a Radon-Nikodym derivative dν

dµ , i.e. a µ- integrable function f such that

ν(E) =∫

E

f dµ

holds, for any measurable set E.Earlier, Radon, [43], proved (using Vitali bases) the same implication under

the assumption that the maximal element of the σ-algebra is a measurable subsetof the n-dimensional Euclidean space. For the real line, the corresponding resultgoes back to Lebesgue.

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However the situation is not so nice, either when the measures are just finitelyadditive, or when they take values in a Banach space of infinite dimension, andit gets even worse if both of the above cases occur. We shall outline a survey ofthe results known so far, involving the existence of a Radon-Nikodym derivative,in some appropriate sense, and with respect to suitable types of integrals. Inthis first section, we deal essentially with a general Radon-Nikodym theoremfor non-negative finitely additive scalar measures, from which the basic ideas offurther results can be drawn. In the second section, we turn to the σ-additivemeasures, investigating conditions which permit that even unbounded measureshave Radon-Nikodym derivatives. In the third section we face the finitely addi-tive case for scalar measures, indicating how the existence of a Radon-Nikodymderivative dν/dµ is strictly related to some geometrical properties of the rangeof the vector valued measure (µ, ν). The fourth section deals with the mostbeautiful (and difficult) results concerning those Banach spaces possessing theso-called Radon-Nikodym Property (RNP): the geometric concepts met in thethird section here become essential tools for describing such Banach spaces.The fifth section is devoted to finitely additive measures taking values in Ba-nach spaces, and to the research of weaker types of derivatives. Finally, the sixthsection concerns a number of recent results, for locally convex-valued measures,for multimeasures, for Riesz space-valued measures, and finally also a brief out-line of the so-called fuzzy integration: from this survey one can find that, thoughhidden by the abstract settings and the different kinds of integrals involved, theleading ideas of the first section are always at work.

We end the Chapter with an appendix devoted to some decomposition the-orems for measures, which are relevant for the Radon-Nikodym theorem.

We do not give all proofs, some of them being too long and technical: whenit is possible, we give an outline of the main steps, trying to clarify the basicideas. Also, we realize that it would be almost impossible to present here adetailed account of all the contributions given to this problem along almost100 years, so we simply chose among the results in our knowledge, sometimessimplifying settings, in order to reach a sufficient variety within a relativelyconcise treatment.

The general result, from which we start, is due to Greco ([23]). Though theintegration theory and the results by Greco involve more general set functions,we shall give the proof only for the finitely additive case. It should be noted thatthe idea to relate the Radon-Nikodym theorem to the existence of a “scaled”Hahn decomposition goes back, for the σ-additive case, to J. L. Kelley , [30].

We need some definitions (see also [17]).

Definition 1.1 Let (Ω, Σ) be any measure space, where Σ is a σ-algebra ofsubsets of Ω, and let µ be any monotone nonnegative set function on Σ, µ(∅) = 0.We say that a measurable function f : Ω → IR+ is integrable with respect to µif the following integral is finite:

2

∫Ω

f dµ :=∫ ∞

0

µ(x : f(x) > t)dt.

Since x : f(x) > t is non-increasing in t, the second integral has to beunderstood in the Riemann sense.

If this is the case, for each set E ∈ Σ, we set :

fµ(E) :=∫

E

fdµ :=∫

Ω

f 1Edµ.

The set function fµ is often called the integral measure of f , with respectto µ.

The integral defined above is called Choquet integral, and coincides withthe usual integral, in case f is nonnegative and µ is a σ-additive nonnegativemeasure, or a finitely additive one.

Theorem 1.2 Assume that µ and ν are two monotone nonnegative set func-tions, defined on the σ-algebra Σ, such that µ(∅) = ν(∅) = 0, satisfying:

ν(S) = µ(S) = 0 ⇒ µ(A ∪ S) = µ(A) (1)

for all A ∈ Σ.Then, the following conditions are equivalent:a) there exists a measurable non-negative function f : Ω → IR such that

ν(E) =∫

E

f dµ (2)

for all E ∈ Σ.b) there exists a (decreasing) family of sets Arr>0 in Σ, satisfying:

b1) ν(A)− ν(B) ≥ rµ(A)− rµ(B), A, B ∈ Σ, B ⊂ A ⊂ Ar, r > 0;b2) ν(E)− ν(E ∩Ar) ≤ r(µ(E)− µ(E ∩Ar)), E ∈ Σ , r > 0;b3) lim ν(Ar) = 0, as r → +∞.

It is clear that (1) is satisfied, as soon as µ and ν are additive. Moreover, incase ν and µ are finitely additive, conditions b1) and b2) above are equivalentrespectively to b′1) and b′2) below:

b′1) ν(E) ≥ rµ(E), for all E ⊂ Ar, r > 0;b′2) ν(F ) ≤ rµ(F ), for all F ⊂ Ω \Ar, r > 0.Moreover, from b′1) we deduce that lim µ(Ar) = 0 (since ν(Ar) ≤ ν(Ω) <

∞), hence, in case ν ε µ, b3) is satisfied, too.

Proof. As already mentioned, we shall give the proof just in the case offinitely additive, non-negative measures, such that ν ε µ.

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We first assume a) and prove b). Set : Ar:=x∈ Ω : f(x) > r and fixE ⊂ Ar, E ∈ Σ. As ν(E)=

∫f 1E dµ, we get ν(E) ≥ r

∫1E dµ = rµ(E), so

b’1) is proved.If F is fixed, F ∈ Σ , F ⊂ Ac

r, then ν(F ) =∫

f1F dµ ≤ rµ(F ), so b′2) holds.As already observed, b3) is satisfied because of the absolute continuity, so thefirst implication is completely proved.

Now, we assume that b) holds, and construct a suitable derivative f .For all x ∈ Ω, set: f(x) := supr > 0 : x ∈ Ar. To see that f is measurable,

fix any t ∈]0, +∞[, and observe that x : f(x) > t = ∪Ar : r ∈ D, r > t,where D denotes the set of all diadic positive numbers, i.e. D = h

2k , h and kpositive integers.

Now, for each element h2k ∈ D, set Bk

h := Ah/2k\A(h+1)/2k .We see that ν(Bk

h) ≥ h2k µ(Bk

h), and also ν(Bkh) ≤ h+1

2k µ(Bkh). For each

positive integer k, set

fk :=h

2k

k2k∑h=1

1Bkh

.

We can easily see that∫

Efdµ = limk

∫E

fkdµ, for all E ∈ Σ , and moreover

∫E

fkdµ ≤k2k∑h=1

h

2k

2k

hν(Bk

h ∩ E) ≤ ν(E)

for all k and E. Therefore, we get:∫E

fdµ ≤ ν(E) ,

for all E. On the other hand, for each k, we have:

∫fkdµ =

∫h + 1

2k(

k2k∑h=1

1Bkh

)dµ−∫

12k

(k2k∑h=1

1Bkh)dµ ≥

k2k∑h=1

ν(Bkh)− µ(Ω)/2k .

As∑k2k

h=1 ν(Bkh) = ν(A1/2k) − ν(Ak), from b3) we deduce that limk

∫fkdµ ≥

limk ν(A1/2k) from which we also get limk

∫fkdµ ≥ ν(Ω), because ν(A1/2k) ≤

2−kµ(A1/2k) ≤ 2−kµ(Ω).So far, we have seen that the finitely additive measures ν and fµ are in

this relation: ν(E) ≥ fµ(E), for all E ∈ Σ, and ν(Ω) ≤ fµ(Ω). From this itfollows immediately that the two measures agree on Σ (simply considering thecomplements of the involved sets), and therefore the theorem is proved.

Corollary 1.3 If µ and ν are finite and countably additive, and ν µ (orequivalently ν ε µ), then there exists a Radon-Nikodym derivative dν

dµ .

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Proof. Conditions b′1) and b′2) mean that there exists a Hahn decompositionfor the measure ν − rµ, for all r > 0: this is always the case, for σ-additivemeasures, and this concludes the proof.

The importance of Theorem 1.2 rests not only on its generality, but alsoon the relative simplicity of the involved conditions, from which other similarcriteria have been obtained. We shall see many of them in the sequel, dealingwith finitely additive measures, with Banach- and nuclear-valued measures, andalso with different kinds of integrals. Our dissertation however now focuses theσ-additive case for not necessarily bounded measures.

2 The σ-additive case.

In this chapter, we mainly deal with countably additive, possibly unboundedmeasures. In general, the Radon-Nikodym theorem fails to hold, for unboundedmeasures, as the following example shows.

Example 2.1 Let Ω = [0, 1], let 0 ≤ i < j ≤ 1 and consider the Hausdorff i-and j-dimensional measures Hi and Hj . Then Hj Hi but dHj/dHi does notexist.

The reason for the failure of the Radon-Nykodym theorem in the previousexample has been described already by S. Saks, [47], who considered the casei = 0 and j = 1 (see also Volcic, [55]). We will make some additional commentson this example after Theorem 2.12.

However we will see that, under suitable conditions, Radon-Nikodym deriva-tives do exist, even if both measures, µ and ν, range over [0, +∞]. Such con-ditions involve properties of the so-called measure algebra, which we are nowgoing to introduce.

As usual, (Ω, Σ, µ) denotes a measure space, where Σ is a σ-algebra and µ isany non-negative σ-additive measure, taking values in [0, +∞]. We shall assumethat the Caratheodory extension has already been done, and so Σ is actually theσ-algebra of all µ-measurable sets. In particular, all subsets of µ-null sets belongto Σ. We recall the definition of the outer measure, µ∗ : P(Ω) → [0, +∞] :

µ∗(E) = inf∞∑

n=1

µ(Fn) : Fn ∈ Σ, E ⊂ ∪Fn .

Definition 2.2 Given two subsets A and B of Ω, we say that A and B areequivalent, if µ∗(A∆B) = 0, and write: A ≈ B or A = B µ− a.e.

In a similar fashion, if f and g are real functions defined on Ω, we say thatf and g are equivalent (and write f ≈ g , or f = g a.e.) if

µ∗(x ∈ Ω : f(x) 6= g(x)) = 0 .

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In the quotient P(Ω)/ ≈ we can introduce a partial order, as follows:

[A] [B]

if and only if µ∗(B\A) = 0.Of course, P(Ω)/ ≈ contains Σ/ ≈, which is called the measure algebra. One

can easily see that P(Ω)/ ≈ and Σ/ ≈ are σ-complete lattices.

We will see that completeness of Σ/ ≈ is important for our purposes, howeverin general the two lattices are not complete. We have the following facts (moredetails and related results can be found in a series of papers by Volcic [54, 55,56]):

Under CH, if µ is the usual Lebesgue measure on Ω := [0, 1], then P(Ω)/ ≈is not complete.

There exist measures such that Σ/ ≈ is not complete (see [25], Section 31,Exercise 9).

Definition 2.3 Given any measure space (Ω, Σ, µ), we say that µ is semifiniteif

µ(E) = supµ(A) : A ⊂ E, A ∈ Σ, µ(A) < ∞ ,

for all sets E ∈ Σ.

Such measures are called essential measures by N. Bourbaki ([7]).

Proposition 2.4 Suppose ν µ, suppose that ν is semifinite and supposemoreover that ν(E) = 0 whenever µ(E) < ∞. Then dν/dµ exists only if ν isidentically zero.

Proof. Suppose dν/dµ exists and let E be any set of finite and positive νmeasure. Then

x ∈ Ω : dν/dµ > 0 ∩ E = ∪nx ∈ Ω : dν/dµ >1n ∩ E .

Each set x ∈ Ω : dν/dµ > 1n ∩ E has finite µ measure and since ν(E) > 0, at

least one of them, An say, has positive measure µ, a contradiction. This showsthat if ν(E) < ∞, then ν(E) = 0 and from the semifiniteness of the measure wededuce that ν ≡ 0.

Definition 2.5 A semifinite measure µ on (Ω, Σ) is said to be strictly localiz-able, if there exists a family of sets Eαα∈A, such that

a) 0 < µ(Eα) < ∞b) Eα ∩ Eβ = ∅ when α 6= βc) µ(E ∩ Eα) = 0,∀α ⇒ µ(E) = 0.

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Theorem 2.6 If µ(Ω) < ∞ (or more in general if µ is strictly localizable), thenΣ/ ≈ is complete.

Proof. We will limit the proof to the case of a finite measure. Given anyfamily G of measurable sets, we shall show that there exists a measurable setG0 such that:

(1) µ∗(G\G0) = 0 for all G ∈ G, and(2) µ∗(G0\G1) = 0, for all G1 ∈ Σ, such that µ∗(G\G1) = 0 for all G ∈ G.Without loss of generality, we may assume that G is a σ-ideal. Now, set:

α := supµ(G) : G ∈ G .

As µ is finite, α < ∞. Now, let (Gn)n∈IN be any sequence in G, such thatµ(Gn) > α − 1

n , and set G0 := ∪Gn. As G is a σ-ideal, G0 ∈ G. So, we onlyhave to prove that G0 satisfies (1) above. If this is not the case, then thereexists H ∈ G , satisfying µ(H\G0) > 0. This implies that H ∪ G0 ∈ G, andµ(H ∪G0) > α, a contradiction.

Definition 2.7 A semifinite measure µ is said to be Maharam (or also localiz-able) if Σ/ ≈ is complete. (see [33]).

Remark 2.8 According to Theorem 1, strictly localizable measures are Ma-haram. It was a long standing open question whether Maharam measures arestrictly localizable. The problem has been stated explicitly for the first time in[31], Problem 1, and is relevant in lifting theory. The question has been solvedin the negative by D. Fremlin ([22]).

Definition 2.9 Given two measures µ and ν on Σ we say that µ and ν arestrongly comparable if µ and ν are the Caratheodory extensions of µ|S and ofν|S respectively, where S is the ideal of all sets E ∈ Σ, such that both measuresare finite on E.

The following definition is related to the previous one.

Definition 2.10 Given two measures µ and ν on Σ we say that µ and ν areweakly comparable if µ and ν are the Caratheodory extensions of µ|Σ and ofν|Σ respectively, where Σ is the intersection of the two σ-algebras of µ- andν-measurable sets (in the Caratheodory sense).

The measures Hi and Hj are not strongly comparable, if i 6= j, but they areweakly comparable.

In 1951, I.E. Segal proved the following version of the Radon-Nikodym the-orem ([48]).

Theorem 2.11 Given a σ-additive measure µ, the following properties areequivalent:

(1) µ is Maharam;(2) for any σ-additive measure ν, strongly comparable with µ and such

that ν µ, there exists a Radon-Nikodym derivative dν/dµ. Such function isunique, up to equivalence.

7

Proof. We only prove the implication (1) ⇒ (2). So, assume that µ isMaharam, and ν is any strongly comparable, σ-additive measure, ν µ. Wedenote by S the ideal of all sets E ∈ Σ, such that µ(E) + ν(E) < ∞. For everyelement E ∈ S there exists a Radon-Nikodym derivative of ν/E with respect toµ/E : we denote by fE such derivative, and define fE on Ec as the null function.Now, for every positive real number t, set:

At := ∨At(E) : E ∈ S ,

where At(E) = x ∈ Ω : fE(x) > t. (The supremum exists as µ is Maharam).It is clear that (At) is a decreasing family in Σ/ ≈. Now, for all x ∈ Ω we define:

f(x) := supr > 0 : x ∈ Ar .

As in the proof of 1.2, one can prove that f is measurable. Let us show thatf is the required derivative. In view of the strong comparability, it’s enough tocheck that ∫

E

fdµ = ν(E)

for all E ∈ S. So, fix any element E ∈ S, and compute:∫E

f dµ =∫ ∞

0

µ(x ∈ E : f(x) > t)dt (3)

Now, µ(x ∈ E : f(x) > t) = supµ(E ∩ At+ 1n

) : n ∈ IN. For everypositive number r, we claim that

E ∩Ar ≈ Ar(E). (4)

Indeed, Ar(E) ⊂ E by definition, and Ar(E) ⊂ Ar up to a null-set, henceAr(E) ⊂ E ∩ Ar up to a null-set; conversely, E ∩ Ar = ∨(E ∩ Ar(F )), wherethe supremum is taken as F ∈ S. Now, E ∩Ar(F ) = Ar(E ∩ F ) because of theessential uniqueness of fE , hence E ∩Ar ⊂ Ar(E) a.e. and (4) is proved.

From (4), we get immediately

supµ(E ∩At+ 1n

) : n ∈ IN = supµ(At+ 1n

(E) : n ∈ IN = µ(At(E))

for all positive numbers t, and so from (3) we deduce∫E

fdµ =∫ ∞

0

µ(x ∈ Ω : f(x) > t)dt =∫ ∞

0

µ(At(E))dt =

=∫

fEdµ = ν(E) ,

by definition of fE . This concludes the proof.

For weakly comparable measures, we have the following result.

8

Theorem 2.12 Given a σ-additive measure µ on (Ω, Σ), the following proper-ties are equivalent:

(1) µ is Maharam;(2) for any σ-additive measure ν, weakly comparable with µ and such that

ν µ, there exists a decomposition Ω1, Ω2 of Ω, Ωi ∈ Σ, for i = 1, 2, suchthat:

for any E ∈ Σ and E ⊂ Ω1, µ(E) < ∞ implies that ν(E) = 0;if we denote by ν2 the restriction of ν to Σ∩Ω2, the Radon-Nikodym deriva-

tive dν2dµ exists.

The decomposition Ω1, Ω2 and the function dν2dµ are unique, up to equivalence.

The proof follows from the combination of Theorem 2.11 and the decompo-sition theorem A.4 from the Appendix.

Remark 2.13 The previous Theorem explains why Radon-Nikodym theoremfails when µ is the counting measure H0 on [0, 1], and ν is the Lebesgue measure,in spite of the fact that µ is localizable. The measures µ and ν are in this casejust weakly and not strongly comparable, and ν µ. Theorem 3 applies herewith Ω1 = Ω.

We conclude this Section presenting two interesting variants of the Radon-Nikodym theorem.

We need first two definitions.

Definition 2.14 Given a measure space (Ω, Σ, µ), we say that µ admits amonotone differentiation, if there exists a mapping d

dµ , defined on the fam-ily N of all the measures which are strictly comparable with µ and absolutelycontinuous with respect to µ, such that if νi ∈ N , for i = 1, 2, and ν1 ≤ ν2, then

dν1

dµ(x) ≤ dν2

dµ(x) ,

for any x ∈ Ω.

Definition 2.15 Given a measure space (Ω, Σ, µ), we say that µ admits a lineardifferentiation, if there exists a mapping d

dµ , defined on the family N of all themeasures which are strictly comparable with µ and absolutely continuous withrespect to µ, such that if νi ∈ N , for i = 1, 2, and a1, a2 are two real numbers,then

d(a1ν1 + a2ν2)dµ

(x) = a1dν1

dµ(x) + a2

dν2

dµ(x) ,

for any x ∈ Ω.

The following result is due to D. Kolzow ([31], Theorems 4, 7 and 8).

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Theorem 2.16 Given a measure space (Ω, Σ, µ), the following properties areequivalent:i) µ admits a monotone differentiation;ii) µ admits a linear differentiation;iii) µ is strictly localizable.

For the long and sophisticated proof we refer to Kolzow’s monograph, whichalso illustrates in great detail the strong relations between the monotone andlinear versions of the Radon-Nikodym theorem and lifting theory.

The Radon-Nikodym theorem for the Daniell integral is discussed in thecorresponding Chapter, written by M. D. Carrillo.

3 The finitely additive case

When the measures (even one of them) are just finitely additive, the Radon-Nikodym theorem does not hold, in general: (we remark that, in this case,absolute continuity is to be intended in the ε− δ sense).

It is appropriate to begin with the oldest result concerning the Radon-Nikodym theorem for finitely additive measures, attributed to Bochner ([6]).

Theorem 3.1 If µ and ν are two finitely additive measures on Σ, such thatν ε µ, then for any η > 0 there exists a µ-integrable function fη, such that∣∣∣∣ν(E)−

∫E

fη dµ

∣∣∣∣ < η ,

for any E ∈ Σ.

In order to give an example, we introduce a definition.

Definition 3.2 Let µ : Σ → IR+0 be a finitely additive measure, and let I

denote the ideal of null sets. If I is a σ-ideal, we say that µ has the property σ.

One can easily see that, if f ≥ 0 and µ has property σ, then∫fdµ = 0

implies that f = 0 a.e..Similarly, one can deduce that, whenever µ has property σ, and f and g are

two µ-integrable functions, then f = g µ-a.e. as soon as∫

Afdµ =

∫A

gdµ forall A ∈ Σ (This entails uniqueness of the Radon-Nikodym derivatives, up to a.e.equivalence).

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Example 3.3 ([9]) Let ∆ be the family of all subsets D ⊂ [0, 1], such that thefollowing limit exists:

limε→0+

λ(D ∩ [0, ε[)ε

= δ(D) .

Though ∆ is not an algebra, the function δ can be extended (using the axiomof choice) to the whole σ-algebra Σ of the Lebesgue measurable sets, in such away that

lim infε→0+

λ(B ∩ [0, ε[)ε

≤ δ(B) ≤ lim supε→0+

λ(B ∩ [0, ε[)ε

,

for all sets B ∈ Σ, and so that δ is finitely additive on Σ. Obviously, if λ(B) = 0then δ(B) = 0.

Now set µ := λ+δ. It is clear that µ(B) = 0 if and only if λ(B) = 0, hence µenjoys property σ. Moreover, λ µ also in the (ε− δ) sense, however one caneasily see that dλ/dµ does not exist. In fact, such a function f should satisfy

λ(B ∩ [ε, 1]) =∫

B∩[ε,1]

f dµ =∫

B∩[ε,1]

f dλ +∫

B∩[ε,1]

f dδ =∫

B∩[ε,1]

f dλ ,

for all ε > 0, and all B ∈ Σ, i.e. f = 1 a.e. in [ε, 1] for all ε > 0. Thus, by theproperty σ, it should follow that f = 1 µ -a.e., a contradiction.

The property σ has a double face: on one hand, it looks like a strengtheningof finite additivity, in the sense of σ-additivity. On the other hand, it preventsin some sense that a finitely additive measure might behave like a σ-additiveone, at least with respect to the Radon-Nikodym property. This statement isclarified by the next theorem.

Theorem 3.4 Let µ : Σ → IR+0 be any finitely additive measure, defined on the

σ−algebra Σ, and enjoying property σ. The following are equivalent:1) µ is countably additive;2) for every finitely additive measure ν : Σ → IR+

0 , which is (ε−δ) absolutelycontinuous with respect to µ, there exists dν/dµ.

Proof. Of course, since 1) and the other assumptions imply that ν is alsoσ-additive, just 2) ⇒ 1) needs to be proved. Let (An)n∈IN be any increasingsequence of sets in Σ, and denote by A their union. Set: ν(B) := lim µ(An∩B),for all B ∈ Σ. It is clear that ν satisfies the condition b), hence there existsf = dν/dµ.

Now, for every B ∈ Σ, we have ν(B) = ν(B ∩ A) =∫

Bf1Adµ, so f can be

replaced by f 1A. On the other hand, it is clear that f 1An = 1An µ-a.e., hencef = 1A µ-a.e.

As a consequence, we have limn→∞ µ(An) = ν(A) =∫

Afdµ =

∫A

1dµ =µ(A). As the sequence (An)n∈IN was arbitrary, µ turns out to be σ-additive.

In the spirit of the Remark preceding Theorem 3.4, we can give an antitheticexample.

11

Example 3.5 Let Ω be any infinite set, and let I denote the ideal of all finitesubsets of Ω. According to the Axiom of Choice, there exists a maximal idealI∗, including I. Define for A ∈ P(Ω):

θ(A) :=

0 if A ∈ I∗

1 otherwise .

According to a well-known theorem by Ulam ([53]), θ is a finitely additivemeasure, lacking the property σ. However, if ν : ℘(Ω) → IR+

0 is any finitelyadditive measure, (0-0) absolutely continuous with respect to θ, then the onlypossibility is that ν = kθ, for some non-negative constant k. In this case,obviously, k = dν/dθ.

To find a characterization of the Radon-Nikodym property, we have tointroduce the concept of completeness. In a recent paper by A.Basile andK.P.S. Bhaskara Rao, ([1]), an excellent presentation is given, mainly concern-ing finitely additive measures defined on algebras. We shall adapt here one oftheir results, dealing with the Radon-Nikodym property.

Definition 3.6 Let µ : Σ → IR+0 be any finitely additive measure, on a σ-

algebra Σ. If A and B are elements of Σ, we set: dµ(A,B) = µ(A∆B). It isclear that ( Σ, dµ) is a semimetrizable space. We say that (Σ, µ) is complete,if the quotient Σ / ≈ is complete as a metric space, where ≈ is the naturalequivalence relation in Σ.

The characterization given in [1] can be stated as follows (we recall that inour assumptions Σ is a σ-field, and contains all subsets of sets of measure 0).

Theorem 3.7 Let µ : Σ → IR+0 be any finitely additive measure, on a σ-algebra

Σ. The following are equivalent:1) (Σ, dµ) is complete.2) For every increasing sequence of sets (Fn)n∈IN in Σ, there exists a se-

quence of µ-null sets (Hn)n∈IN in Σ, such that

limn→∞

µ(Fn) = limn→∞

µ(Fn \Hn) = µ(∪n∈IN (Fn \Hn))

From Theorem 3.7 one can easily deduce that if ν ε µ and (Σ, dµ) iscomplete, then (Σ, dν) is complete, too.

Corollary 3.8 If (Σ, dµ) is complete, and ν is any signed finitely additivebounded measure such that ν ε µ, then ν admits a Hahn decomposition.

Proof. We only have to show that there exists a set P ∈ Σ, such thatν(P ) = supA∈Σ ν(A).

As ν is bounded, the number S := supA∈Σ ν(A) is finite, and there exists asequence (An)n∈IN in Σ, such that S = limn→∞ ν(An). Setting now

12

ν+(A) = supE∈Σ

ν(A ∩ E), and ν−(A) = − infE∈Σ

ν(A ∩ E), ∀A ∈ Σ ,

we readily see that ν+(An∆Am) → 0 and ν−(An∆Am) → 0 as both m andn diverge. Now, ν+and ν− are both ε − δ absolutely continuous with respectto µ, and so is |ν| := ν+ + ν−, hence there exists a set P ∈ Σ, such thatlimn→∞ |ν|(An∆P ) = 0. The set P has the required property.

Corollary 3.9 The following are equivalent:1) (Σ, dµ) is complete;2) For every non-negative finitely additive measure ν ε µ, there exists

dν/dµ.

Proof. We only prove the implication 1) ⇒ 2). As ν ε−δ µ, we see thatthe signed measure ν − rµ is ε− δ absolutely continuous with respect to µ, andtherefore it has a Hahn decomposition. Due to Corollary 1.3, this implies theexistence of dν/dµ.

The Corollary above has been proved in [1].A different approach to the problem consists in finding additional conditions

on the two measures µ and ν, besides absolute continuity, which ensure theexistence of dν/dµ.

One of the first consequences of Theorem 1.2 and Corollary 1.3 is the fol-lowing (see also [2],[11], [12]):

Theorem 3.10 Let µ, ν be two non-negative finitely additive measures, definedon the same σ-algebra Σ, and such that ν ε µ. If the range of the pair (µ, ν)is a closed subset of IR2, then there exists dν/dµ.

Proof. In view of 1.3, it is enough to show that, for every real number r > 0,there exists a Hahn decomposition for ν−rµ. But it is clear from the assumptionon the range of (µ, ν) that the range of ν − rµ is a closed subset of IR, hencethere exists an element A ∈ Σ in which ν − rµ attains its maximum: so (A,Ac)is a Hahn decomposition for ν − rµ, and we are done.

Of course, closedness of the range of (µ, ν) is just a sufficient condition. In[2] and [11] necessary and sufficient conditions are given on the range of (µ, ν)in order that dν/dµ exists. In both papers, the condition involves the so-calledexposed points of the range, according with the following definition.

Definition 3.11 Let R be any convex, bounded subset of IRn. Let us denoteby R its closure. We say that a point Q ∈ IRn, Q ∈∂R, is an exposed point forR if for every hyperplane H, supporting R at Q, we have R ∩H = Q.

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Theorem 3.12 Let (µ, ν) be a pair of non-negative finitely additive measures,defined on a σ-algebra Σ and such that ν ε µ , and let R denote the range of(µ, ν). Then the following are equivalent:

1) There exists dν/dµ.2) The convex hull of R contains its exposed points.

The Theorem above has been proved in [2].In [11] a similar theorem is stated, for the case of continuous measures.A measure µ : Σ → IR+

0 is said to be continuous if for every ε > 0 it is possibleto decompose Ω into a finite number of subsets A1, A2, ..., An belonging to Σ,such that µ(Ai) < ε for all i.

If µ and ν are both continuous, finitely additive and non-negative, it is well-known that the range of (µ, ν) is a bounded convex subset of the plane (see also[10], hence condition 2) of 3.12 can be expressed in a simpler way. However, in[11] the following result has been proved, which gives a full description of thosebounded convex sets R ⊂ IR2 that are the range of some pair (µ, ν) for whichdν/dµ exists.

Theorem 3.13 Let R be any bounded convex subset of [0,∞[2. Then the fol-lowing properties are equivalent:

1) There exists a pair (µ, ν) of nonnegative continuous finitely additive mea-sures, defined on a suitable σ-algebra, such that there exists dν/dµ, and suchthat R is the range of (µ, ν).

2) R contains its exposed points, and for every segment L ⊂ ∂R, at least a(possibly degenerate) closed subsegment I ⊂ L is contained in R. (see Fig.5aand 5b, at the end of this chapter).

Usually, if (µ, ν) is a pair of non-negative continuous finitely additive mea-sures, the range looks like in Fig.1, or in Fig. 3, when ν ε µ. In Fig.2 aparticular situation is shown, where ν is obtained by adding to µ some measurewhich is singular with respect to µ.

We observe that the range is always symmetric with respect to the midpoint(µ(Ω)/2, ν(Ω)/2): hence in Fig.5a and 5b just half of ∂R has been drawn.

4 The Banach-valued case

When dealing with finite-dimensional bounded measures, (including complexones), it is not difficult to apply the results quoted in the previous sections.For instance, a bounded signed measure always admits a Jordan decomposition(even in the finitely additive case), so results concerning Radon-Nikodym deriva-tives can be easily deduced from the corresponding results for the positive andnegative parts. Similarly, a complex measure can be studied by investigatingseparately its real and imaginary parts.

14

A really different situation occurs with infinite-dimensional measures, as weshall see in this section. The research on this subject led, mainly in the seventies,to a variety of quite interesting results, relating the Radon-Nikodym property totopological and geometric properties of the range space. Here, we shall confineourselves to the Banach-valued σ- additive measures, and with respect to theBochner integral. (As to absolute continuity, for the σ-additive case, the (ε− δ)definition is still equivalent to the (0-0) one). Maybe the richest survey on thissubject is [18], so we refer the reader to that paper, in order to find the proofsmissing here, and an exhaustive historical account.

We start, recalling the definition of Bochner integral. ([5], [19], [20]).

Definition 4.1 Given a finite measure space (Ω, Σ, µ), where Σ is a σ-algebra,and a Banach space X, a function f : Ω → X is said to be simple if it is of the

form: f =N

Σi=1

xi1Ai, where xi ∈ X and Ai ∈ Σ for all i = 1, ..., N . If this is the

case, then the integral of f is the element:∫fdµ :=

N

Σi=1

xiµ(Ai) ∈ X.

In the same framework, if (fn)n∈IN is any sequence of simple functions, wesay that fn converges in measure to some function f, if the following holds:

limn→∞

µ∗(ω ∈ Ω : ||fn(ω)− f(ω)|| > ε) = 0

for all ε > 0, (here, µ∗(A) is the outer measure defined in section 1).In case f : Ω → X is the limit in measure of some sequence (fn)n∈IN of

simple functions, then f is said to be measurable (also, strongly measurable).We say that f : Ω → X is (Bochner)-integrable, if there exists a sequence

(fn)n∈IN of simple functions, converging in measure to f , and such thatlim(n,m)→∞

∫||fn − fm||dµ = 0.

It can be proved that, if f : Ω → X is integrable, then the sequence(∫

fndµ)n∈IN is convergent in X, and the limit is independent of the partic-ular sequence (fn)n∈IN . Of course, the limit lim

∫fndµ is called the integral of

f.Also, as in the real-valued case, one can prove that convergence in measure

implies convergence a.e. for some subsequence, hence a strongly measurablefunction is essentially separably valued, i.e. there exists a separable subspaceY ⊂ X, such that f(ω) ∈ Y for all but a µ-null set of elements ω. Moreover,a strongly measurable function is also weakly measurable, i.e. < x∗, f > ismeasurable for all x∗ belonging to X∗.

Like in the scalar case, it can be proved that integrability of f implies in-tegrability of ||f ||, and that ||

∫fdµ|| ≤

∫||f ||dµ. When f is integrable with

respect to µ , we also say that f is in L1X(µ). The latter is a Banach space,

with the norm ||f || =∫||f ||dµ. Moreover, if f is integrable, then f1E is also

integrable, for all E ∈ Σ, and the integral becomes a function of E: we setfµ(E) :=

∫E

fdµ :=∫

f 1E dµ.One can see that fµ is an X−valued measure, which is separably valued,

whose variation is ||f ||µ. The variation of an X -valued measure ν is denotedby |ν| and is defined as:

15

|ν|(E) = sup∑||ν(Ei)||,

where the supremum is taken over all finite partitions of E: one can also regard|ν| as the least upper bound of | < x∗, ν > | : x∗ ∈ X∗, ||x∗|| = 1, in the latticeof measures ).

Moreover, fµ µ .The converse question leads, of course, to the problem of the existence of

a Radon-Nikodym derivative. The problem can be stated in various differentways:

1) Given a measure ν : Σ → X, which is absolutely continuous with respectto a scalar positive measure µ, does there exist a Bochner-integrable functionf , such that ν = fµ?

2) Given a measure ν : Σ → X, with finite variation, |ν|, does there exist afunction f , Bochner integrable with respect to |ν|, and such that ν = f |ν|?

Of course, a necessary condition for an affirmative answer to 1), is that νhas finite variation. Moreover, if |ν| is finite and ν µ, then it is clear that|ν| µ: hence, an affirmative answer to 2) would yield an affirmative answer to1). Conversely, assuming the existence of dν/dµ, one can decompose µ into thesum µ1 + µ2, where µ1 |ν| and µ2 ⊥ |ν|. (See Theorem A4). Hence |ν| µ1,dν/dµ = dν/dµ1, and dν/d|ν| = (dν/dµ1)(dµ1/d|ν|).

In conclusion, we can see that 1) and 2) are essentially equivalent.Moreover, in [15], Chatterji remarks that in 1) and 2) above the σ -algebra

Σ can be equivalently replaced by the Borel σ -algebra on the unit interval, and(up to irrelevant renorming) the measure µ by the usual Lebesgue measure.

Now, we give a couple of examples showing that the answer is not alwaysaffirmative.

Example 4.2 Let X be the L1space over the unit interval in IR. Also, let([0, 1], Σ, λ) be the measure space, where Σ denotes the Borel σ-field, and λ theLebesgue measure. For all A ∈ Σ, set ν(A) = 1A, i.e. the characteristic functionof the set A.

It is easy to see that ν is an X-valued countably additive measure, definedon Σ, and that |ν| = λ. However, if a Radon-Nikodym derivative f := dν/dλexisted, we would have:∫

Ag(x)dx =< g, 1A >=

∫A

< f(x), g > dx

for all A ∈ Σ, and all g ∈ L∞ (≡ (L1)∗). Therefore, for every fixed g ∈ L∞,we would have:

g(x) =< f(x), g > a.e.

Hence, there would exist a λ−null set N ∈ Σ such that, for all α, β in[0, 1] ∩Q, α < β, we would have

< f(x), 1[α,β] >= 1

16

for all x ∈ N c ∩ [α, β]. So, if we fix x /∈ N , and choose (αn)n∈IN , (βn)n∈IN insuch a way that αn < x < βn, αn, βn ∈ Q, and βn − αn → 0, we get:

limn→∞

< f(x), 1[αn,βn] >= 1

which is impossible, since f(x) ∈ L1.

Example 4.3 The next example concerns the same measure space, ([0, 1], Σ, λ), but with a different range: here, X is c0, the space of all real sequences,vanishing at infinity. Let us define:

µ(A) := (µ1(A), ..., µn(A), ...),

where

µn(A) =∫

Asin(2πnx)dx

for all A ∈ Σ, and n ∈ IN .As |µn(A)| ≤ λ(A) for all A and all n, it is clear that µ has bounded variation,

and |µ|([0, 1]) ≤ 1. Obviously, µ λ, however dλ/dµ does not exist. Indeed,such a derivative f , f = (f1, ..., fn, ...), should satisfy:

µn(A) =∫

Asin(2πnx)dx =

∫A

fn(x)dx

for all A and n. Thus, it would be fn(x) = sin(2πnx) a.e., which isimpossible, because sin(2πnx) is not vanishing, as n →∞.

The first affirmative answer to the Radon-Nikodym problem in infinite-dimensional spaces is due to Birkhoff.

Theorem 4.4 ([3]) If X is a Hilbert space, then every σ−additive measureµ : Σ → X with bounded variation admits a Radon-Nikodym derivative withrespect to |µ|.

We do not give the proof now, because we will see later some generalizationsof this result.

Remark 4.5 What happens if we try to adapt 4.4 to the example given in 4.2?One can always think of 1A as an element of L2, and of ν as an L2 -valuedσ−additive measure, with ν λ. However, by the same argument, one canstill show that dν/dλ does not exist. What is wrong? A simple calculationshows that µ does not have bounded variation, when considered with values inL2. This remark also shows the importance for a measure of having boundedvariation. (However, more can be said on this example, see Remark 4.18). Fromnow on, we shall write BV to mean ”bounded variation”.

17

Definition 4.6 We say that a Banach space X has the Radon-Nikodym Prop-erty (RNP) if 1) or 2) above holds, i.e. every countably additive X-valuedBV measure µ, defined on the measure space (Ω, Σ), admits a Radon-Nikodymderivative with respect to |µ|.

Hence, according to 4.4, all Hilbert spaces have the RNP.Other spaces with RNP can be derived from another important sufficient

condition, due to Dunford and Pettis, [21].

Theorem 4.7 If X is a Banach space, whose dual space X∗ is separable, thenX∗has RNP.

Proof (Sketch) Let ν : Σ → X∗ be any BV measure, with ν µ. For anyx ∈ X, and any A ∈ Σ, set νx(A) =< ν(A), x > . It is clear that νx µ. Ofcourse, there exists dνx/dµ. The idea now is to set fx = dνx/dµ, and to showthat fx defines, as x varies in X, a linear continuous functional f ; however onemust take into account that fx is defined a.e., hence a ”good” function f mustbe defined carefully, and this can be done, if X∗ is separable.

For example, the space L1 is not a dual space, because we know from 4.2that it has not RNP, however l1 is a separable dual, hence it enjoys RNP.

We will see in the sequel other classes of spaces with that property, but wewill introduce first a very important definition.

Definition 4.8 Given a bounded non-empty subset B of a Banach space X,we say that B is dentable if for every ε > 0 there exists a point xε ∈ B suchthat xε does not belong to the closed convex hull of B\B(xε, ε). In case it ispossible to choose the same x for all ε, then x is said to be a denting point of B.

Dentable sets are closely related to the existence of Radon-Nikodym deriva-tives, as Rieffel showed in [44, 45] .

We first remark that B is dentable whenever every countable subset of Bhas this property. This was proved by Maynard ([36]), and can be seen asfollows: assume that B is not dentable; then there exists ε > 0 such that x ∈co(B\B(x, ε)),∀x ∈ B. So, choose any element x ∈ B, and points y1, ..., yn ∈ B,such that ||yi−x|| > ε for all i, and some convex combination z1 of these belongsto B(x, ε/2). For the initial point x, and each of these points, say yj , one canchoose a finite subset Fj ⊂ B, such that ||y− yj || > ε, for any y ∈ Fj , and suchthat some convex combination z2 of Fj belongs to B(yj , ε/4). Now, the unionof the sets Fj , together with x and the points yj , gives a finite subset E1 ⊂ B.For each element s ∈ E1 the same construction is possible, giving rise to a finitesubset E2. We continue by induction. The union of all the sets Ej is then acountable subset of B, which is not dentable.

Now, let us state a lemma, where the construction of a Radon-Nikodymderivative is shown, under suitable assumptions.

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Lemma 4.9 Let (Ω, Σ, µ) be any finite measure space, and ν : Σ → X be anyBV measure, |ν| µ. Then dν/dµ exists, whenever the following conditionholds:

(a) For every ε > 0 there exist sequences (xεn)n∈IN in X, and (Hε

n)n∈IN inΣ, such that the sets Hε

n are pairwise disjoint, have positive measure, µ(Ω) =∑µ(Hε

n), and ν(A)

µ(A) : A∈ Σ, A ⊂ Hεn ⊂ B(xε

n, ε)

Proof Assume that (a) holds, and fix ε > 0. Consequently, a sequence(xε

n)n∈IN exists in X, and a corresponding sequence (Hεn)n∈IN in Σ, according

to (a). Now choose nε large enough, so that µ(Ω\ ∪j≤nε

Hεj ) < ε, and set

fε :=∑

j≤nε

xεj1Hε

j.

It is easy (though tedious) to see that (fε) is Cauchy in L1X(µ), hence con-

verges to some function f in this space. Now, since∫

Efdµ = lim

∫E

fεdµ asε → 0, for all E ∈ Σ, it is enough to compare ν(E) with

∫E

fεdµ. Of course,up to ε, we can assume E ⊂ ∪

j≤nε

Hεj , and so ν(E) =

∑j≤nε

ν(E ∩ Hεj ), while∫

Efεdµ =

∑j≤nε

xεjµ(E ∩Hε

j ).

Now, as E ∩Hεj ⊂ Hε

j , we have ||ν(E ∩Hεj )− xε

jµ(E ∩Hεj )|| ≤ εµ(E ∩Hε

j )by (a), therefore ||ν(E)−

∫E

fεdµ|| ≤ εµ(Ω), and this proves the lemma.

Theorem 4.10 (Rieffel). Let (Ω, Σ, µ) be any finite measure space, and letν : Σ → X be any BV measure, with |ν| µ. The following are equivalent:

1. There exists dν/dµ.2. For each set E ∈ Σ, µ(E) > 0, there exists D ⊂ E,D ∈ Σ, µ(D) > 0, such

that ν(A)

µ(A) : A ⊂ D,A ∈ Σ, µ(A) > 0 is dentable.

Proof (Sketch) 1.⇒ 2. We first observe that the result is obvious, if dν/dµis simple: for every set E ∈ Σ, having positive measure µ, there exists D ⊂ E,with D ∈ Σ, and µ(D) > 0, such that dν/dµ is constant in D, and hence ν(A)

µ(A) : A ⊂ D,A ∈ Σ, µ(A) > 0 is a singleton. In the general case, letf denote the derivative dν/dµ, and let (sn)n∈IN be any sequence of simplefunctions, converging to f in the L1−norm, and also almost uniformly. For anyset E ∈ Σ, such that µ(E) > 0, there exists a set D ⊂ E, with D ∈ Σ, suchthat µ(D) > 0, and sn → f uniformly on D. Now, it is easy to reduce to theprevious argument.

2.⇒ 1. We shall use Lemma 4.9. To prove (a), we proceed as follows: first,we show that, for every set E ∈ Σ, such that µ(E) > 0, and for any ε > 0, thereexists D ⊂ E, with D ∈ Σ, and µ(D) > 0, such that the set

A(D) = ν(A)µ(A)

: A ⊂ D,A ∈ Σ, µ(A) > 0

is contained in B(x, ε) for some element x ∈ X.

19

Once we have proved this, it is possible to replace E with Dc, and iteratethe procedure, constructing a disjoint sequence as in (a).

So fix E ∈ Σ, with µ(E) > 0, and let ε > 0. We know that there existsE0 ⊂ E, with E0 ∈ Σ, and µ(E0) > 0, such that A(E0) is dentable. Let x be anelement of A(E0), such that x /∈ co(A(E0)\B(x, ε)). Write x = ν(D0)/µ(D0),for suitable D0 ⊂ E0. If A(D0) ⊂ B(x, ε), we have finished. Otherwise, thereexists E1 ⊂ D0, with µ(E1) > 0, such that || ν(E1)

µ(E1)− x|| > ε. As ν(E1)

µ(E1)∈ A(D0),

we also have ν(E1)µ(E1)

∈ co(A(D0)\B(x, ε)).Let us denote by k1 the smallest positive integer for which such an element

E1 exists, satisfying µ(E1) ≥ 1k1

, and choose in this way E1. Set D1 = D0\E1.

It is impossible that µ(D1) = 0, because this would imply x = ν(D0)µ(D0)

= ν(E1)µ(E1)

, acontradiction.

Now, if A(D1) ⊂ B(x, ε) , we are done. Otherwise, proceeding in thisway, we can produce a disjoint sequence (En)n∈IN in Σ, an increasing sequence(kn)n∈IN in IN , such that µ(En) ≤ 1

kn, and ν(En)

µ(En) ∈ co(A(D0)\B(x, ε)). Ofcourse, we have kn ↑ ∞. Setting E∞ = ∪En, and D = D0\E∞, we can deducethat D is the desired set, i.e. µ(D) > 0 and A(D) ⊂ B(x, ε). Indeed, if it wereµ(D) = 0, we would have

x =ν(D0)µ(D0)

=ν(E∞)µ(E∞)

=∑

ν(En)µ(E∞)

=∑ ν(En)

µ(En)µ(En)µ(E∞)

∈ co(A(E0)\B(x, ε)),

which is impossible.Now, to show that A(D) ⊂ B(x, ε), fix D′ ⊂ D, with D′ ∈ Σ and µ(D′) > 0.

If ν(D′)µ(D′) /∈ B(x, ε), then ν(D′)

µ(D′) ∈ co(A(E0)\B(x, ε)), because D′ ⊂ D0\n∪

i=1Ei for

all n. But then, by the choice of (kn)n∈IN , we must have µ(D′) < 1kn

for all n,and hence µ(D′) = 0, a contradiction.

Let us see some consequences of Theorem 4.10.For instance, we can deduce that if X has the RNP then every subspace of

X has the same property. Indeed, the construction of Lemma 4.9 shows that,if ν takes values in a subspace Y ⊂ X, then the derivative can be constructedas a limit of Y -valued simple functions.

Another consequence is that X enjoys RNP as soon as every bounded subsetof X is dentable. However, the converse is also true, as Huff in [26], and Davisand Phelps, in [16], independently proved.

Theorem 4.11 Let X be any Banach space. The following are equivalent.1) X enjoys RNP.2) every bounded subset of X is dentable.

We refer the reader to [18] for a proof of the implication 1) ⇒ 2).

Another consequence is that Radon-Nikodym Property is separably deter-mined, i.e. a Banach space X possesses RNP if and only if every separable

20

subspace Y ⊂ X has RNP. This is now an easy consequence of 4.11, and the re-mark following Definition 4.8: a set is dentable as soon as its countable subsetsare dentable.

As weakly compact sets are dentable, it is now clear that every reflexiveBanach space has the RNP.

An interesting consequence of 4.7 and 4.11 is that X enjoys RNP as soonas each of its separable subspaces are duals. In [49], Stegall showed a strongerresult:

Theorem 4.12 The following are equivalent, for any Banach space X:1) X∗ has RNP;2) every separable subspace of X has separable dual;3) every separable subspace of X is embedded in some separable dual.

A consequence of 4.12 is that the dual of a separable Banach space possessesRNP if and only if it is separable.

Some of the previous results are also included in a ”martingale-type result”,obtained by Chatterji in [15]. For the sake of completeness, we give some defi-nitions.

Definition 4.13 Let (Ω, Σ, P ) be any probability space, i.e. any measure space,with P (Ω) = 1. Given any Bochner-integrable function f : Ω → X, (here, Xis any Banach space), and given any sub-σ -algebra Σ0 ⊂ Σ, the conditionalexpectation of the function f to Σ0 is the Bochner-integrable function (definedP -a.e.), denoted by E(f |Σ0), which has the following two properties:

1) E(f |Σ0) is strongly Σ0-measurable;2)

∫F

fdP =∫

FE(f |Σ0)dP , for any F ∈ Σ0.

The existence of E(f |Σ0) is independent of the RNP, and is proved first directly,for simple functions f , and then by approximation, in the general case.

Definition 4.14 Let X, and (Ω, Σ, P ) be as above. An X−valued martingale isa sequence (fn, Σn)n∈IN of Bochner-integrable functions fn and sub-σ-algebrasΣn ⊂ Σ, such that:

1) Σn ⊂ Σn+1, ∀n ∈ IN ;2) fn = E(fn+1|Σn),∀n ∈ IN.

The martingale (fn, Σn)n∈IN is said to be convergent if there exists a Bochner-integrable function f∞ : Ω → X , such that fn = E(f∞|Σn) for all n ∈ IN.

A typical way to obtain martingales is the following: assume Ω = [0, 1], Σis the Borel σ-algebra, and λ the Lebesgue measure. Construct a sequence ofdecompositions (Dn)n∈IN of the unit interval, first splitting [0, 1] into two sub-intervals of the same lenght, thus obtaining D1, and then, by induction, dividingeach interval from Dn−1 into two disjoint sub-intervals of the same lenght in

21

order to get Dn. Let Σn be the (σ−) algebra generated by Dn. Now, if ν is anymeasure defined on Σ, set: fn(x) = ν(In(x))/λ(In(x)), where In(x) is the uniqueinterval of Dn containing x. It is clear that fn is constant on each interval of Dn,hence it is Σn−measurable. It is also easy to see that

∫In

fn+1dλ =∫

Infndλ (=

ν(In)), which implies that (fn, Σn)n∈IN is a martingale.If this martingale is convergent, and if ν is σ-additive, then the function f∞

satisfies the identity: ∫F

f∞dλ = ν(F ),

for any F ∈ ∪Σn, and therefore for any F ∈ Σ, i.e. we have f∞ = dν/dλ(and of course ν λ).

In case ν is real-valued, a classical condition for the convergence of a mar-tingale is particularly simple: as soon as sup

n∈IN

∫Ω|fn|dλ < ∞, the functions fn

are uniformly integrable, and pointwise converging to f∞.When the martingale (or the measure ν) takes values in a general Banach

space X, this is no longer true, in general. Chatterji’s result clarifies the situa-tion.

Theorem 4.15 ([15]) Let (Ω, Σ, P ) be any probability space, and assume thatX is a Banach space. The following are equivalent:

1) X has the RNP with respect to (Ω, Σ, P )2) Every X-valued martingale (fn, Σn)n∈IN , satisfying supn∈IN

∫Ω|fn|dP <

∞, is convergent, in the sense that fn converges P-a.e. and strongly in X to aBochner-integrable function f∞ : Ω → X, such that E(f∞|Σn) = fn, ∀n ∈ IN

We shall not give the proof here: for the implication 2) ⇒ 1), the construc-tion above gives an idea of the main steps. The converse is less elementary. As aconsequence of this result, one can easily deduce, besides the already mentionedclasses of reflexive spaces, and separable dual spaces, another class of Banachspaces with the RNP, i.e. the weakly complete spaces with separable dual.

We conclude this section with an analytic result, due to Moedomo and Uhl([38]), connecting ”weak” derivatives with ”strong” ones. We need a furtherdefinition.

Definition 4.16 Let (Ω, Σ, µ) be any finite measure space, and X any Banachspace. Given a strongly measurable function f : Ω → X (see Definition 4.1),we say that f is Pettis integrable if for every set A ∈ Σ there exists an elementJ(A) ∈ X, such that

< x∗, J(A) >=∫

A

< x∗, f > dµ

for any x∗ ∈ X∗. The element J(A) is called the Pettis Integral of f in A,and is denoted by (P )

∫A

fdµ.

22

Theorem 4.17 ([38])Let ν : Σ → X be any σ-additive measure, ν µ. The following are equiva-

lent:1) There exists a Pettis integrable function f : Ω → X such that

(P )∫

A

fdµ = ν(A)

for all A ∈ Σ. (f is the Pettis derivative of ν).2) For every ε > 0 there exists a set Hε ∈ Σ, such that µ(Hc

ε) < ε, and theset ν(A)

µ(A) : A ⊂ Hε, µ(A) > 0 is relatively compact.3) For every A ∈ Σ, with µ(A) > 0, there exists B ∈ Σ, B ⊂ A, with

µ(B) > 0, such that ν(E)

µ(E) : E ⊂ B, µ(E) > 0 is relatively compact.(In items 2) and 3) above, ”relatively compact” can be equivalently replaced by”weakly relatively compact”).

Moreover, f turns out to be Bochner integrable, if and only if ν is BV.

We shall not give a proof of 4.17, however we emphasize again the fact thatweak compactness of bounded sets ensures they are dentable, so at least partof this theorem is a consequence of 4.10.

Remark 4.18 In some sense, theorem 4.17 tells us that bounded variation isnot an essential requirement, for ν, to be an integral measure: if properties 2)or 3) of 4.17 is satisfied, at least ν is the Pettis integral of some function f , withrespect to µ. For example, if X is reflexive, the only requirement is that 2) or3) above hold, with ”relatively compact” replaced by ”bounded”; however, if νis not BV, this is not authomatic, even for Hilbert spaces: if we consider theRemark 4.5, the example outlined there deals with an L2−valued measure ν,absolutely continuous with respect to the Lebesgue measure λ, lacking even aPettis derivative: indeed, the ”averages” ν(A)

λ(A) fail to be bounded, as soon asλ(A) decreases to 0.

5 Finitely additive Banach-valued measures

Of course, the problems concerning the existence of a Radon-Nikodym derivativeare still harder, when one allows also finitely additive measures into consider-ation. As we already observed, even for real-valued finitely additive measures,there are examples in which the derivative does not exist, hence it makes nosense to look for spaces with a property which would take the place of RNP.

This is clarified by the next theorem, which is a Banach-valued version of theapproximate Radon-Nikodym-Bochner theorem (3.1). We need first a definition.

23

Definition 5.1 If X is a Banach space, we shall say that it has the “approxi-mate finitely additive Radon-Nikodym property” (AFARNP), if for any measurespace (Ω, Σ, µ), where µ is a finitely additive X-valued BV measure, and for anyη > 0, there exists a µ-integrable function fη (which may be taken simple), suchthat

‖µ(E)−∫

E

fη dµ‖ < η ,

for any E ∈ Σ.

The following result has been communicated to us by Anna Martellotti.

Theorem 5.2 A Banach space X has AFARNP if and only if it has RNP.

Proof. Assume X has RNP and let µ be a BV finitely additive measureon (Ω, Σ), with values in X. Since µ is BV, it follows from [10] that it admitsa Stone extension, i.e. there exists a countably additive measure µ on Gδ, theBaire σ-algebra of the Stone space S associated to the quotient of Σ with respectto N , the family of all |µ|-null sets.

It is also known that µ is BV and that

|µ| = |µ| .

By RNP there exists a Bochner-integrable function f : S → X such that

µ(F ) =∫

F

f d|µ| ,

for any F ∈ Gδ.A density argument shows that, given η > 0, there exists a Gδ-simple function

g : S → X such that ∫S

‖g − f‖d |µ| < η .

Let g =∑n

i=1 xi1Gi , with Gi pairwise disjoint. Since Gi ∈ Gδ, there existpairwise disjoint sets E1, . . . , En in Σ, such that h([Ei]) = Gi, where h is thenatural embedding of Σ/N in S.

Put now γ =∑n

i=1 xi1Ei , and let ν = γ|µ|. It is easy to check that ν =∫g d|µ| and from [10] we have

|µ− ν| =∫‖f − g‖ d|µ| ,

and moreover µ− ν = µ − ν and |µ − ν|(Ω) = ^|µ− ν|(S) , so we can concludethat |µ− ν|(Ω) < η.

Conversely, assume X has AFARNP. Let µ : Σ → X be a countably additiveBV measure. For each η > 0 there exists fη : Ω → X such that

|µ− fη|µ| | < η .

24

Consider ηn = 12n and let fn = fηn

. We want to show that (fn)n∈IN is a Cauchysequence in L1(|µ|). Indeed, for any n and m, if we put νn = fn|µ|, we have

‖fn − fm‖1 =∫

Ω

‖fn − fm‖X d|µ| = |νn − νm| .

Since the total variation is a norm in the space of all countably additive BVmeasures on (Ω, Σ), we have

|νn − νm| ≤ |νn − µ|+ |µ− νm| ,

hence ‖fn − fm‖1 < ηn + ηm, and so (fn)n∈IN is Cauchy.Since |µ| is countably additive, L1(|µ|) is complete and therefore there exists

f ∈ L1(|µ|) such that fn → f (in L1(|µ|)). By definition, (fn)n∈IN is a definingsequence for f , namely

∫fn dµ →

∫f dµ. From the definition of fn it follows

now the conclusion, namely that µ = f d|µ|.

From the second part of the proof above, it follows also that an approximateRadon-Nikodym-Bochner theorem for Banach-valued σ-additive measures doesnot hold, unless the range space has RNP.

From now on, we shall limit ourselves to look for conditions, necessary and/orsufficient, to ensure the existence of some kind of weaker Radon-Nikodym deriva-tives.

To be able to do that, besides the Bochner and Pettis-type integral, we shallalso deal with the so-called Gel’fand integral, and later on with the Bartle-Dunford-Schwartz integral.

Definition 5.3 Let µ : Σ → IR+0 be a finitely additive measure, and let X be

any Banach space. Given a function f : Ω → X∗∗, we say that f is Gel’fandintegrable with respect to µ, if :

a) < f, x∗ > is µ−integrable, for all x∗ ∈ X∗.b) For every A ∈ Σ there exists an element JA ∈ X such that

< x∗, JA >=∫

A< f, x∗ > dµ,

for any x∗ ∈ X∗.c) The function A → JA is a µ-continuous finitely additive measure.d) For every ε > 0 there exists H ⊂ Ω, such that µ(Hc) < ε, and

supω∈H

||f(ω)|| < ∞.

When a finitely additive measure ν : Σ → X is given, with ν µ, (through-out this Section we use always the ε − δ definition of absolute continuity), wesay that ν has a Gel’fand-type derivative, if there exists a Gel’fand integrablefunction f : Ω → X∗∗, for which JA = ν(A) for all A ∈ Σ, i.e.∫

A

< f, x∗ > dµ =< x∗, ν(A) >,

for any A ∈ Σ, and x∗ ∈ X∗.

25

An example of such a derivative can be seen in 4.3, where the function ftakes values in l∞, while the space X is c0.

Another interesting situation is described in Example 4.2, where the exis-tence of a Gel’fand derivative is a consequence of the Lifting Theorem, [40], [33],[27], [28], in particular for the Lebesgue measure. Let us denote by φ the liftingmap. For any ω ∈ Ω = [0, 1], f(ω) can be defined as the element of L∗∞, whichassociates φ(h)(ω) to every h ∈ L∞. So we see that < f, h > turns out to be sim-ply the function φ(h), which is bounded and measurable, for each h ∈ L∞ ≡ X∗,hence a) is satisfied, and

∫A

< f, h > dλ =∫

Ah dλ =< h, 1A >=< h, ν(A) >,

for any A ∈ Σ and h ∈ X∗. This proves b) , c), and the fact that f is theGel’fand derivative dν/dλ.

Dealing with finitely additive measures, the existence of Gel’fand derivativescan be derived from the following theorems, which include the example in 4.3as a particular case.

Theorem 5.4 ([12]) Let Y be any separable Banach space, and assume that(Ω, Σ, µ) is any finitely additive measure space, with µ non-negative (and finite).Let’s suppose that, to every y ∈ Y a bounded finitely additive scalar measureT (y) is associated, such that there exists dT (y)/dµ, and moreover |T (y)|(A) ≤||y||µ(A) for every A ∈ Σ and y ∈ Y , where |T (y)| denotes as usual the totalvariation of T (y). Then, there exists a function f : Ω → Y ∗ such that:

1) f(·)(y) = dT (y)/dµ for every y ∈ Y ;

The proof of Theorem 5.4 is in some sense reminiscent of the sketch of proofgiven for Theorem 4.11. We skip the details, because they are too technical.

A direct consequence is the existence of bounded Gel’fand derivatives, undersome conditions.

Theorem 5.5 ([12]) Assume that X is any Banach space, with separable dual,and let (Ω, Σ, µ) be as above. Suppose that ν : Σ → X is a finitely additivemeasure, such that d < x∗, ν > /dµ exists, for every x∗ ∈ X∗. Assumingmoreover that the set

S := ν(A)µ(A) : A ∈ Σ, µ(A) > 0

is bounded, then there exists a bounded Gel’fand type derivative dν/dµ.

Proof It is sufficient to apply 5.4, setting Y := X∗, and T (x∗) =< x∗, ν >.

Corollary 5.6 ([12]) Let X, (Ω, Σ, µ), ν be as above. For the existence of aGel’fand derivative dν/dµ it is necessary and sufficient that the following twoconditions hold:

i) d < x∗, ν > /dµ exists, for every x∗ ∈ X∗, andii) for every ε > 0 there exists H ∈ Σ, µ(Hc) < ε, such that the set

S(H) := ν(A)µ(A) : A ∈ Σ, A ⊂ H,µ(A) > 0

is bounded.

26

Proof The necessity is clear, by definition of Gel’fand integrability. To seethe converse, let us take an increasing sequence of sets (Hn)n∈IN , such thatµ(Hc

n) ↓ 0, and with S(Hn) bounded for each n: then, applying 5.5 to Hn onegets a Gel’fand derivative fn on Hn, and pasting together the functions fn givesthe required derivative.

It is interesting to compare this result with the Moedomo-Uhl one, quotedin Theorem 4.17: in some sense, if compactness for the average sets is replacedby boundedness, then Gel’fand derivatives take the place of Pettis derivatives.

The requirement concerning the existence of d < x∗, ν > /dµ in the previ-ous theorems may look too restrictive; however, we can see that it is satisfiedunder rather mild conditions. One of these has been outlined in [12], where thefollowing well-known theorem by Rybakov is used ([46]).

Let us recall that a measure µ on Σ is said to be s-bounded, if for everydisjoint sequence of sets En ∈ Σ, we have limn µ(En) = 0.

Theorem 5.7 ([46]) Given an s-bounded finitely additive measure ν : Σ → X(here, Σ may be just an algebra), there always exists some element x∗ ∈ X∗,such that ν | < x∗, ν > |.

Definition 5.8 If ν : Σ → X is s-bounded and x∗ ∈ X∗ is such that ν | <x∗, ν > |, then the measure | < x∗, ν > | is called a Rybakov control for ν. Wecan (and do) assume that ‖x∗‖ = 1.

Theorem 5.9 ([12]) Let X be any Banach space, with separable dual, and let(Ω, Σ) be any measurable space. Assume that ν : Σ → X is any s-bounded finitelyadditive measure, and let µ be any Rybakov control for ν. If ν has weakly closedrange, then a Gel’fand derivative dν/dµ exists, if and only if condition ii) ofCorollary 5.6 holds.

Proof Thanks to 5.6, it only remains to show that d < x∗, ν > /dµ exists,as soon as x∗ ∈ X∗. Let us denote by y∗ the element of X∗, such that µ =| < y∗, ν > |. As the range of ν is weakly closed, we see that the measure< sx∗ − ry∗, ν > = s < x∗, ν > −r < y∗, ν > has closed range, hence it admitsa Hahn decomposition, for all real numbers s and r, and all elements x∗ ∈ X∗.Now, an easy consequence of Corollary 1.3 gives the desired derivative.

A more general formulation can be found in the next statement.

Theorem 5.10 ([12]) Assume that X∗ has the RNP, and let µ : Σ → X beany s-bounded measure, with separable weakly closed range. Then ii) of 5.6 isa necessary and sufficient condition for the existence of a Gel’fand derivativedν/dµ, where µ is any Rybakov control for ν.

Proof If Y denotes the separable subspace of X, generated by the range ofν, from 4.12 we find that Y ∗ is separable, hence the conclusion follows from theprevious results.

27

In order to obtain Bochner (i.e. strong) derivatives in the finitely additivecase, different conditions have been found, as Hagood in [24] showed. The defi-nition of Bochner integrable function in the finitely additive setting is formallythe same as in the σ-additive one (see also [20]). For the sake of simplicity,we shall give here just a particular form of Hagood’s result, also related to aprevious work by Maynard, [37].

Theorem 5.11 Let (Ω, Σ, µ) be any non-negative finite and finitely additivemeasure space, and let ν : Σ → X be any Banach-valued finitely additive mea-sure, ν µ. Then the following are equivalent:

1) there exists dν/dµ in the Bochner sense;2) for every ε > 0 and δ > 0, there exist C ∈ Σ, µ(C) > 0, and α ∈]0, 1[

such that:2i) µ(Cc) < δ,

2ii) the set S(C) := ν(A)µ(A) : A ∈ Σ, A ⊂ C, µ(A) > 0 is bounded,

2iii) for all E ⊂ C,E ∈ Σ, µ(E) > 0, there exists F ⊂ E,F ∈ Σ, such thatµ(F ) > αµ(E) and diam(S(F )) < ε.

Proof First, let us prove that 1) ⇒ 2). We denote by f the derivativedν/dµ, and fix ε, δ > 0. By strong measurability of f , there exists a simple

function g =n∑

i=1

βi1Bi , such that the sets Bi form a finite partition of Ω, and

µ∗(ω ∈ Ω : ||f(ω) − g(ω)|| > ε/4) < δ. Let us choose a set H ∈ Σ, suchthat ω ∈ Ω : ||f(ω) − g(ω)|| > ε/4 ⊂ H, and µ(H) < δ. Set now: C = Hc.Then, 2i) is satisfied. Moreover, if A ∈ Σ, A⊂ C, we have ||f(ω)− g(ω)|| ≤ ε/4,for all ω ∈ A, hence ||ν(A)|| ≤

∫A||g||dµ + ε

4µ(A) ≤ Mµ(A), where M =max||βi||, ε/4. So, also 2ii) is satisfied. Finally, choose α = 1

2n : if E is anyfixed element of Σ contained in C, denote by B∗ one of the elements Bi suchthat µ(E ∩ B∗) > 1

2nµ(E), and put F = B∗ ∩ E. So, µ(F ) > αµ(E). Now, letus prove that the diameter of S(F ) is less than ε. Let β∗ denote the value of gin B∗, and choose any set A ∈ Σ, with A ⊂ F, and µ(A) > 0. We get

ν(A)µ(A)

=

∫A

fdµ

µ(A)=

β∗µ(A)µ(A)

+

∫A

(f − g)dµ

µ(A),

from which we deduce that || ν(A)µ(A) − β∗|| < ε/2, and therefore diam(S(F )) < ε.

We now turn to the converse, i.e. 2) ⇒ 1). Fix ε > 0, and δ = ε. Then,let C and α be the corresponding elements, obtained from 2). Now, apply 2iii)to E = C: we find a set F1 ∈ Σ, with F1 ⊂ C, and µ(F1) > αµ(C), andsuch that diam(S(F1)) < ε. If µ(F1) = µ(C) we are finished; otherwise, applyagain 2iii) to E = C\F1, thus finding F2 ⊂ C\F1, with F2 ∈ Σ, such thatµ(F2) > αµ(C\F1), and diam(S(F2)) < ε. So, F1 and F2 are disjoint membersof Σ, both satisfying diam(S(Fi)) < ε. By an exhaustion argument, it is possibleto get a (finite or countable) family (Fn)n∈IN of subsets of C of positive measure,each satisfying diam(S(Fn)) < ∞, and such that

∑µ(Fn) = µ(C). Now define

28

fε =∑

βn1Fn, where βn is, for each n, any element of S(Fn). It is now easy to

show that fε is Bochner-integrable, (indeed, fε is bounded, from 2ii), and

||ν(A)−∫

A

fεdµ|| < ε,

for any A ∈ Σ, A ⊂ C.A routine argument now gives an increasing sequence (Ck)k∈IN , with µ(Cc

k) ↓0, and a corresponding convergent sequence (fk)k∈IN , whose limit is the requiredderivative.

Lastly, we turn to the so-called Bartle-Dunford-Schwartz integral, which al-lows to integrate a scalar function with respect to a vector-valued measure. Weshall refer to the paper [34], which in turn was inspired at Musia l’s paper [39].

Actually, in [34] locally convex-valued measures were considered, but herewe shall limit ourselves to the special case of Banach-valued ones, for simplicity.

For the rest of this section, X will denote a Banach space, µ, ν : Σ → X willbe s-bounded finitely additive measures.

Definition 5.12 We denote by P the space of all measurable functions f : Ω →IR, such that f ∈ L1(| < x∗, µ > |), for any x∗ ∈ X∗.

Given a function f ∈ P, we say that f is µ -integrable if for every A ∈ Σthere exists an element υ(A) ∈ X such that

< x∗, υ(A) >=∫

Afd(x∗µ)

for every x∗ ∈ X∗. We then set: υ(A) =∫

Afdµ.

In general, when f is µ-integrable, the function υ is a finitely additivemeasure, but it satisfies a much stronger condition than absolute continuitywith respect to µ. This is evident even in two-dimensional spaces: one canchoose ([0, 1],B, λ) as measure space, X := IR2, and then define: µ(A) =(∫

Axdx, λ(A)), υ(A) = (λ(A),

∫A

xdx). It is clear that µ and υ are both equiv-alent to λ, hence υ µ, but there is no function f : [0, 1] → IR such thatυ(A) =

∫A

fdµ: indeed, such a function should satisfy:

λ(A) =∫

Axf(x)dx, and

∫A

xdx =∫

Af(x)dx

for any A ∈ Σ, thus f(x) = x a.e. and xf(x) = 1 a.e., which is clearlyimpossible.

The key property is introduced in the following definition.

Definition 5.13 We say that ν is scalarly uniformly absolutely continuous withrespect to µ, and write: ν <<< µ, if for every ε > 0 there exists δ > 0 suchthat for every x∗ ∈ X∗ and every A ∈ Σ one has the implication:

| < x∗, µ > |(A) < δ ⇒ | < x∗, ν > |(A) < ε.

29

We say that ν is scalarly dominated by µ if there exists a positive numberM > 0 such that

| < x∗, ν > |(A) ≤ M | < x∗, µ > |(A)

holds, for any x∗ ∈ X∗ and any A ∈ Σ.

It is easy to see that, in case f : Ω → IR is bounded and µ-integrable,the measure υ, defined in 5.12 above, is scalarly dominated by µ, and satisfiesυ <<< µ.

The Radon-Nikodym theorem stated in [34] asserts that, under certain con-ditions, depending essentially on the finite additivity of the involved measures,scalar domination is equivalent to scalar uniform absolute continuity, and thateach of the two conditions is also sufficient for the existence of a bounded deriva-tive dν/dµ. The conditions we shall mention here are somewhat stronger, forthe sake of simplicity.

Theorem 5.14 ([34]) Let λ be any Rybakov control for µ. Assume that forevery x∗ ∈ X∗, the set

S x∗:= <x∗,µ>(A)

λ(A) : A ∈ Σ, λ(A) > 0is bounded. Assume also that the ranges of the measures µ and ν are closed.

Then the following are equivalent:1) ν <<< µ;2) ν is scalarly dominated by µ;3) there exists a bounded µ-integrable function f : Ω → IR, such that

< x∗, ν > (A) =∫

A

fd < x∗, µ >,

for any A ∈ Σ, and x∗ ∈ X∗.

6 Further results

In this chapter, we shall see some recent results, concerning different situations:one of them concerns Radon-Nikodym derivatives for measures taking their val-ues in locally convex spaces, and more particularly in nuclear spaces; anothersituation concerns multivalued measures, taking values in Banach or locallyconvex spaces; yet a different result concerns Riesz space-valued measures; fi-nally, we shall mention some results concerning Radon-Nikodym derivatives fora different kind of integral, the so-called Sugeno integral.

As usual, (Ω, Σ, µ) denotes a measure space, where Σ is a σ-algebra, and µis any non-negative, finitely additive measure, taking values in [0, +∞[.

30

We start with some results concerning (finitely additive) measures, takingvalues in a locally convex Hausdorff space X. Some definitions are needed, inorder to put useful conditions on X. Throughout this section, X∗ denotes thestrong dual space of X.

Definition 6.1 Let X be any locally convex space, and let B denote anybounded, non-empty subset of X. B is said to be bipolar if B = B00, ac-cording with the duality < X, X∗ > . Given any bipolar set B ⊂ X, wedenote by XB the subspace of X which can be absorbed by B (i.e. thespace of those elements x ∈ X such that rx ∈ B for some positive scalar r); thus the Minkowski functional pB of B can be defined on XB (we recall thatpB(x) = infr > 0 : x

r ∈ B, for x ∈ XB). We can endow XB with the semi-norm pB and then consider the normed space XB/ ≈, where the equivalencerelation is defined by: x ≈ x′ ⇐⇒ pB(x−x′) = 0. We shall denote by X(B) thecompletion of such normed space. We say that X satisfies the property (SP) ifX(B) is separable, for every bipolar set B ⊂ X.

Obviously, a separable locally convex space X satisfies (SP).

Definition 6.2 We say that a locally convex space X has the property (SP) ′

if X∗(B0) is separable, for all bounded subsets B ⊂ X. If X∗ is separable, thenX has (SP)′.

When ν takes values in a space X with property (SP)′, then a Radon-Nikodym theorem in the Pettis sense has been proved in [8]. The proof is toolong and technical to be presented here.

Theorem 6.3 Suppose X has property (SP)′, and ν : Σ → X is any µ-continuous finitely additive measure, satisfying the following two conditions:

1) there exists d < x∗, ν > /dµ , for any x∗ ∈ X∗;2) the set S := ν(A)

µ(A) : A ∈ Σ, µ(A) > 0 is weakly relativelycompact in X.

Then there exists a bounded weakly measurable function g : Ω → X, suchthat: ∫

A

< x∗, g(·) > dµ =< x∗, ν(A) >,

for any x∗ ∈ X∗ and all A ∈ Σ.

This theorem is applied in [8] to the case of dual-nuclear spaces. In order tostate other results, we need some more definitions (see also [42]).

Definition 6.4 Let X and Y denote two locally convex Hausdorff spaces, andlet φ : X → Y be any continuous linear map. We say that φ is nuclear if thereexist:

a) an element (λn)n∈IN ∈ l1,

31

b) an equibounded sequence (x∗n)n∈IN in X∗,c) a bounded sequence (yn)n∈IN in Y ,

such that

φ(x) =∑

λnyn < x∗n, x >

for any x ∈ X.

Definition 6.5 A locally convex Hausdorff space X is said to be nuclear ifevery linear continuous map φ : X → Y is nuclear, for every Banach space Y.

This definition is not the original one, due to Grothendieck, but we have chosenthis equivalent property, because we think it is easier to work with.

Nuclear spaces enjoy very interesting properties: for instance, it follows fromthe classical definition of a nuclear space that it is the projective limit of Hilbertspaces. Another important property is that every bounded set in a nuclear spaceis pre-compact. A useful condition for nuclear spaces is quasi-completeness, i.e.every closed bounded subset is complete. Thus, if a nuclear space X is quasi-complete, all closed bounded subsets of X are compact (hence, X is Montel).If the strong dual X∗ is nuclear, then the locally convex space X is said to bedual-nuclear. It turns out easily that a quasi-complete dual-nuclear space X issemi-reflexive, i.e. the canonical embedding c : X → X∗∗ is onto.

From all these remarks, one easily realizes that many interesting results canbe found on measures taking values in spaces of this kind. In [8] there are listedsome, for measures taking values in a dual-nuclear space. We just recall thoseresults, which are strictly connected with Radon-Nikodym derivatives.

Theorem 6.6 ([8]) Let X be a quasi-complete and dual-nuclear space. Thenevery bounded finitely additive measure ν : Σ → X is s-bounded, and admits aRybakov control.

Theorem 6.7 ([8]) Let X be as above, and assume that ν : Σ → X is anybounded finitely additive measure, such that:

1) < x∗, ν > µ , and there exists d<x∗,ν>dµ in L∞, for all x∗ ∈ X∗

2) The set S := ν(A)µ(A) : A ∈ Σ, µ(A) > 0 is bounded.

Then there exists a Bochner-type derivative dνdµ .

We remark here that ”Bochner” above means that the function dνdµ is the limit

in measure of a sequence of simple functions, and the integral is the limit of thecorresponding integrals. This looks like a strong conclusion, and it is worthmentioning how it is derived: from the assumptions we see that the mappingT : X∗ → L∞, defined as T (x∗) = d<x∗,ν>

dµ , is nuclear. (For more details werefer the reader to [8], where also completeness of L∞(µ) is proved). Thereforeone can write:

T (x∗) =∑

λn < x∗, en > yn ∀x∗ ∈ X∗,

32

where (λn)n∈IN ∈ l1, (en)n∈IN is a bounded sequence in X∗∗ = X, and(yn)n∈IN is a bounded sequence in L∞. Choosing a bounded representative fn

from the class of yn, and putting gn(ω) := λnenfn(ω) for all ω ∈ Ω , the series∑gn converges strongly to the desired derivative.

An even stronger result holds for σ-additive measures.

Theorem 6.8 ([8]). Let X be as above, and assume that µ is σ -additive. Ifν : Σ → X is any µ-continuous measure, there exists a Bochner-type derivativedν/dµ.

We now turn to multimeasures, according with the definitions of [14] and[35]: we will just recall the notations and the most relevant results in thatsetting.

Definition 6.9 Given a Hausdorff locally convex space X, the family of allnon-empty, closed, convex, bounded subsets of X will be denoted by Cc(X); ifQ denotes the set of all continuous seminorms on X, for every p ∈ Q and everypair of elements A,B ∈ Cc(X), we set:

ep(A,B) = supx∈A

infy∈B

p(x− y),

anddp(A, B) = maxep(A,B), ep(B, A).

For any A ∈ Cc(X), we set also: hp(A) = dp(A, 0).

As dp is a pseudo-distance, the family hp : p ∈ Q defines a topology on Cc(X),which is compatible with the following addition : A+B := a + b : a ∈ A, b ∈ B.

We remark that the same topology arises if Q is replaced by some absorbingsubset Q0. Also, if X is complete, so is Cc(X) with this topology. We shallassume, from now on, that X is complete. Given an absorbing subset Q0 ⊂ Q,a subset B ⊂ Cc(X) is Q0− uniformly bounded if supp∈Q0

supA∈B hp(A) < ∞.Any finitely additive measure ν : Σ → Cc(X) is said to be a multimeasure

in X.

Definition 6.10 Given a multimeasure ν : Σ → Cc(X), for every p ∈ Q thep-variation of ν is defined for all E ∈ Σ as:

νp(E) = sup(Ai)∈P (E)

∑i∈I

hp(ν(Ai)),

where P (E) denotes the family of all finite partitions of E into sets Ai ∈ Σ.We say that ν has bounded variation if νp(Ω) < ∞, ∀p ∈ Q.

Definition 6.11 Given a multimeasure ν : Σ → Cc(X), and a finitely additivemeasure µ : Σ → IR+

0 , we say that ν is absolutely continuous w.r.t. µ (ν µ)if for every ε > 0 and every p ∈ Q there exists a δ(ε, p) > 0 such that µ(E) < δimplies νp(E) < ε.

33

We now turn to integration of multifunctions.

Definition 6.12 Let µ : Ω → IR+0 be any finitely additive measure. A map

F : Ω → Cc(X) is simple if it can be written as:

F =n∑

i=1

1AiCi,

where the Ci’s are elements of Cc(X), and the Ai’s are disjoint elements ofΣ.

The integral of F with respect to µ is defined as:∫Fdµ =

n∑i=1

µ(Ai)Ci.

As each Ci is convex, the definition of integral does not depend on therepresentation of F .

In case F is not simple, F is said to be totally measurable if there exists asequence (Fn)n∈IN of simple multifunctions such that

(a1) the function dp(Fn, F ) is measurable for every n ∈ IN and everyp ∈ Q.

(a2) the sequence (dp(Fn, F )) µ-converges to 0 for every p ∈ Q.Moreover, we say that F is µ-integrable if there exists a sequence (Fn)n∈IN

of simple functions, satisfying (a1) and (a2) above, such that(a3) lim

m,n→∞

∫dp(Fn, Fm)dµ = 0 for all p ∈ Q.

Such a sequence will be called a defining sequence for F .In case Q0 is any absorbing subset of Q, and the conditions (a2) and (a3)

above hold uniformly with respect to p ∈ Q0, then F will be said to be stronglyintegrable (with respect to Q0, if confusion can arise).

In case F is integrable, and (Fn)n∈IN is any defining sequence for F , thenfor every E ∈ Σ the sequence (

∫E

Fndµ)n∈IN is Cauchy in Cc(X), and henceconvergent to some element of Cc(X), which will be called the integral of F , anddenoted by

∫E

Fdµ. Such integral does not depend on the defining sequence(see [35] for details).

Proposition 6.13 If F : Ω → Cc(X) is µ − integrable, then E →∫

EFdµ

defines a multimeasure, which is absolutely continuous w.r.t. µ.

In order to state a Radon-Nikodym theorem, we need one more definition.

Definition 6.14 Given a multimeasure ν : Σ → Cc(X), we say that ν isbounded if

supA∈Σ

hp(ν(A)) = Mp < ∞

for every p ∈ Q.

34

If this is the case, set: Qν := pMp

: p ∈ Q,Mp > 0.If µ : Σ → IR+

0 is finitely additive, for every set E ∈ Σ, and every ε > 0, weput:

S(E) :=

ν(F )µ(F )

: F ∈ Σ, µ(F ) > 0

;

Sp(E, ε) := C ∈ Cc(X) : dp(ν(F ), Cµ(F )) ≤ εµ(F ) ∀F ∈ Σ, F ⊂ E;

S(E, ε) := ∩p∈Qν

Sp(E, ε).

Finally we can state one of the Radon-Nikodym theorems from [35]. Theproof is based upon the technique of Hagood, see theorem 5.11.

Theorem 6.15 ([35]) Let ν : Σ → Cc(X) be a multimeasure, ν µ, where µis any non-negative finitely additive measure on Σ. Assume that

1) S(Ω) is Qν-uniformly bounded;2) for every ε > 0 and every E ∈ Σ, µ(E) > 0, there exists a sequence

of pairwise disjoint subsets (En)n∈IN of E, with En ∈ Σ, such that µ(E) =∑n µ(En), and such that S(En, ε) 6= ∅ for any n.Then there exists a Qν-strongly measurable multifunction G : Ω → Cc(X),

with Qν-uniformly bounded range, such that:∫E

Gdµ = ν(E) ,

for all E ∈ Σ.

A quite different problem arises when the measures are both vector-valued.An example was given in the last part of Section 5, with Banach-valued mea-sures; but sometimes it is interesting to consider a richer structure, and toassume that the vector measure takes values in a Riesz space: this is conve-nient, when the linear spaces have a natural order structure, and the ”order”convergence is not topological, like in the space L0, with the a.e. convergence.

We shall present here a very simplified version of the results, and refer to [4]for further details or generalizations.

Assume that R is an order-complete Riesz space (here, order-complete meansthat every non-empty subset F ⊂ R which is bounded from above has a leastupper bound in R). Order-completeness allows to define sup, inf, lim sup andlim inf.

We first mention the so-called Riemann-type integral for bounded functionswith values in a Riesz space.

35

Definition 6.16 Given a bounded function f : [a, b] → R, we say that f isRiemann-integrable (with respect to Lebesgue measure) if

sups∈Sf

∫s(x)dx = inf

t∈Tf

∫t(x)dx

where Sf (Tf , respectively) is the class of all R-valued step functions s ≤ f(t ≥ f, respectively), and the elementary integral of a step function is definedin the obvious way.

This integral is well-defined, and is a linear monotone R-valued functional.Moreover, one can see (by usual techniques) that monotonic functions are inte-grable.

Definition 6.17 Given a positive finitely additive measure µ : Σ → R , we saythat µ is σ-additive if, for every decreasing sequence (En) in Σ, En ↓ E impliesinfµ(En) : n ∈ IN = µ(E).

Given a positive finitely additive measure µ : Σ → R, and a bounded mea-surable function f : Ω → IR+

0 , we say that f is integrable with respect to µ ifthe following Riemann-type integral is finite:∫

Ω

fdµ :=∫ ∞

0

µ(x ∈ Ω : f(x) > t)dt

This means that the set ∫ M

0µ(x ∈ Ω : f(x) > t)dt : M > 0 is bounded

in R, and∫Ω

fdµ is its least upper bound; we observe that the function g(t) =µ(x ∈ Ω : f(x) > t) is monotone decreasing, and hence Riemann integrablein every interval [0, M ].

If f is integrable, then f1E is integrable too, for every E ∈ Σ, and E →∫f1Edµ :=

∫E

fdµ defines a measure ν, which is absolutely continuous withrespect to µ, in the sense that lim sup ν(An) = 0 as soon as (An)n∈IN is anysequence from Σ, such that lim sup µ(An) = 0 .

The Radon-Nikodym theorem for Riesz space-valued measures can be for-mulated exactly like theorem 1.2, with b1) and b2) replaced respectively by b’1)and b’2), and b3) replaced by absolute continuity.

Theorem 6.18 Let µ and ν be positive finitely additive measures defined on Σand taking values in R, with ν µ. Then the following are equivalent:

a) There exists a measurable and µ-integrable function f : Ω → [0,∞[, suchthat ∫

E

fdµ = ν(E)

for all E ∈ Σ.b) there exists a family of sets (Ar) in Σ, for r > 0, such that, for any r > 0 :

b1) ν(E) ≥ rµ(E), for any E ∈ Σ, E ⊂ Ar;b2) ν(E) ≤ rµ(E), for any E ∈ Σ, E ⊂ Ac

r.

36

Though the formulation (and the proof) of this theorem is quite similar totheorem 1.2, the consequences are not so strong: for the same example givenbefore def. 5.13 shows that even for σ-additive measures, absolute continuity isnot sufficient to ensure the existence of the derivative.

A different approach to the idea of integration was presented by Sugeno, in[50]. We shall give here quite a short outline of the involved concepts, and alsoa very simple Radon-Nikodym theorem for this integral.

Definition 6.19 Let (Ω, Σ) be any measurable space. A fuzzy measure on thisspace is a mapping g : Σ → [0, +∞[, such that:

1) g(∅) = 0;2) g(A) ≤ g(B), whenever A,B ∈ Σ, A ⊂ B;3) if (An)n∈IN is any monotone sequence in Σ, then g(lim Fn) =

lim g(Fn).If g is a fuzzy measure on (Ω, Σ), then the triple (Ω, Σ, g) is called a fuzzy

measure space.

Definition 6.20 Let (Ω, Σ, g) be a fuzzy measure space, and let h : Ω → IR+0

be any measurable function. For each A ∈ Σ, define:

(S)∫

A

h dg := supα>0

α ∧ g(A ∩ Fα),

where Fα := ω ∈ Ω : h(ω) ≥ α.The number (S)

∫A

h dg is called the Sugeno integral of h on the set A.

(Notice that the definition of Sugeno’s integral is reminiscent of the Choquetintegral, one just replaces ∧ with ordinary multiplication, and ∨ with addition).

From this definition, it follows immediately that(S)

∫A

h dg ≤ g(A), ∀A ∈ Σ.

We list a number of results, concerning this integral.

Theorem 6.21 ([13]). Let (Ω, Σ, g) be any fuzzy measure space, and let h:Ω → IR+

0 be any measurable function.1) If h is a constant function, h(x) = α, then

(S)∫

A

h dg = α ∧ g(A), for any A ∈ Σ.

2) If h and h′ : Ω → IR+0 are measurable functions, with h’≤ h, then

(S)∫

A

h′ dg ≤ (S)∫

A

h dg, for any A ∈ Σ.

3) If h = 1A, for some A ∈ Σ, then

(S)∫

Ω

h dg = g(A).

37

4) If (hn)n∈IN is any monotone sequence of measurable functions, such thathn → h, then

lim (S)∫

A

hn dg = (S)∫

A

h dg,

for any A ∈ Σ.

From part 4) of the previous theorem, it follows that A → (S)∫

Ah dg is

a fuzzy measure, not greater than g. A useful characterization of the Sugenointegral is the following:

Theorem 6.22 ([13]) Given a measurable function h : Ω → IR+0 , define, for

α > 0 :

Aα := ω ∈ Ω : h(ω) > α,

Fα := ω ∈ Ω : h(ω) ≥ α.

If (Bα) is any decreasing family in Σ, for α ≥ 0, such that Aα ⊂ Bα ⊂ Fα,then

(S)∫

A

h dg := supα>0

α ∧ g(A ∩Bα)

for all A ∈ Σ.Moreover, if B denotes the set of all α ∈ IR+

0 , such that α ≤ g(Fα), then

(S)∫

Ω

h dg = max B

This result allows us to state a Radon-Nikodym theorem.

Theorem 6.23 ([13]) Let g and γ be two fuzzy measures on (Ω, Σ) , with γ ≤ g.The following are equivalent:

a) There exists a measurable function h : Ω → IR+0 , such that

γ(A) = (S)∫

A

h dg

for any A ∈ Σ.b) There exists a decreasing family (Bα) in Σ, with α ∈ IR+

0 , such that:b1) γ(A ∩Bα) ≥ α ∧ g(A ∩Bα), andb2) γ(A) ≤ g(A ∩Bγ(A)),

for all A ∈ Σ, and all α ∈ IR+0 .

For a richer treatment of Sugeno integral, and its properties, we refer thereader [50] and [51].

Appendix. Singularity and decomposition theorems.

38

In this appendix we will present some known facts about singularity of mea-sures, the Lebesgue decomposition theorem and some related decompositiontheorems, which are relevant in connection with the Radon-Nikodym theorem.

We assume that all the measures have been already completed in the Caratheodorysense.

Definition A.1 We say that a measure µ is degenerate, if it takes no finite,non-zero values, i.e., its range is contained in 0,∞.

In connection to this definition, let us first mention the following result, dueto N. Y. Luther ([32]).

Theorem A.2 Let µ be a measure on the σ-algebra Σ. Then there exists aunique decomposition µ = µ1 + µ2, where µ1 is semifinite and µ2 is degenerate,with the property that if µ′ = µ′1 + µ′2, with µ′1 semifinite and µ′2 degenerate,then µ1 ≤ µ′1 and µ2 ≤ µ′2.

This result and the fact that degenerate measures are not very interestingfrom the point of view of their representation as integrals, justify the restrictionto semifinite measures in Section 2 of this chapter. We will make the sameassumption in the Appendix.

If µ is a measure on (Ω, Σ), we say that µ is concentrated on A ∈ Σ, ifµ(Ω \A) = 0.

Definition A.3 Suppose µ and ν are two measures on Σ, and suppose thereexists a pair of disjoint sets A and B in Σ, such that µ is concentrated on Aand ν is concentrated on B. Then we say that ν and µ are mutually singular(or that ν is singular with respect to µ) and write

ν ⊥ µ . (5)

Obviously, this relation between measures is symmetric, so (5) holds if andonly if µ ⊥ ν.

The following properties are quite obvious:If ν1 ⊥ µ and ν2 ⊥ µ, then ν1 + ν2 ⊥ µ.If ν1 µ and ν2 ⊥ µ, then ν1 ⊥ ν2.If ν µ and ν ⊥ µ, then ν ≡ 0.

Theorem A.4 If µ and ν are two measures on (Ω, Σ) and ν is σ-finite, thenthere exist ν0 µ and ν1 ⊥ µ such that ν = ν0 + ν1. The decomposition isunique.

Let us sketch the proof. Suppose first that ν(Ω) < ∞, and let α = supν(B) :B ∈ Σ , µ(B) = 0. It is easy to prove, using arguments introduced in Section2, that there exists C ∈ Σ, such that ν(C) = α and µ(C) = 0. Define nowν1(E) = ν(E ∩ C) and ν0 = ν(E \ C). It is easy to check that ν0 µ andν1 ⊥ µ.

If ν is σ-finite, we consider a measurable decomposition An of Ω such thatµ(An) < ∞ for all n, and apply the argument above to each An.

39

If ν is not σ-finite, the conclusion of the previous theorem may fail. If forinstance µ is the Lebesgue measure on [0, 1] and ν is the counting measure, sucha decomposition does not exist.

We need now two definitions. The first is due to R. A. Johnson ([29]), thesecond one is taken from [55].

Definition A.5 We say that ν is S-singular with respect to µ, denotedνSµ, if given E ∈ Σ, there exists F ∈ Σ, F ⊂ E, such that ν(E) = ν(F ) andµ(F ) = 0.

Definition A.6 We say that ν is quasi-singular (Q-singular) with respectto µ, denoted νQµ, if there exists A ∈ Σ such that ν is concentrated on A andmoreover if ν(E) < ∞ then µ(E ∩A) = 0.

Singularity implies Q-singularity and the latter implies S-singularity. Thethree concepts are not on the other hand equivalent. The counting measure on[0, 1] is Q-singular with respect to the Lebesgue measure, but it is not singular.

To show that S-singularity does not imply Q-singularity we need a moreelaborate example. Let us consider the measure space defined in [25], Exercise31.9, which provides a non Maharam measure µ as sum of two measures ν andσ. As noted by R. A. Johnson, νSσ and σSν, but it is not ν ⊥ σ. On the otherhand it can not be νQσ (or σQν) because of the following proposition ([55]).

Proposition A.7 If νQµ and µSν, then ν ⊥ µ.

R. A. Johnson proved the following decomposition theorem ([29]).

Theorem A.8 If µ and ν are two measures on (Ω, Σ), then there existν0 µ and ν1Sµ such that ν = ν0 + ν1. Always ν1 is unique. Moreover, ifν = ν′0 + ν1 is another such decomposition, ν0 ≤ ν′0.

The advantage of the previous theorem, compared to Theorem A.2 is that noassumption is required on µ and ν (not even semifiniteness). The disadvantagelies in the weakness of the concept of S-singularity.

Q-singularity stays in between and shares some good and bad aspects ofboth concepts. A reasonable assumption (specially if we deal with the Radon-Nikodym theorem) provides the following decomposition theorem (Theorem 2.1in [55]).

Theorem A.9 If µ+ν is Maharam, then Ω admits a (µ+ν)-unique decompo-sition Ωi, i = 1, 2, 3, such that if we put νi(E) = ν(E∩Ωi) and µi(E) = µ(E∩Ωi)for i = 1, 2, 3, then

ν1Qµ , (6)

µ2Qν , (7)

ν3 µ3 ν3 . (8)

Moreover, ν3 and µ3 are strongly comparable.

40

The set Ω1 is defined as the least upper bound in the measure algebra asso-ciated to (Ω, Σ, µ + ν) of the family of sets

E : E ∈ Σ, ν(E) < ∞, µ(E) = 0 .

Similarly we define Ω2. The major difficulty in the proof is to show that Ωi, fori = 1, 2 are both µ- and ν-measurable.

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