HS II c©Gabriel Nagy

Hilbert Spaces II:

Applications to Measure and Integration Theory

Notes from the Functional Analysis Course (Fall 07 - Spring 08)

This section contains several fundamental results, which are proved using Hilbert spacetechniques. Since it is very likely that the reader has seen these results in the Real Analysiscourse, our presentation will be sketchy (in the form of Exercises), for most results.

Convention. Throughout this section K will be one of the fields R or C, and (X,A, µ)will denote a fixed measure space. The material presented in this section relies on resultsfrom BS IV, so the reader is urged to review that section.

A. The Radon-Nikodim Theorems

The results from this sub-section deal with the following feature.

Definition. Given a second measure ν on A, we say that ν has the Radon-Nikodimproperty with respect to µ, if there exists an A-measurable function g on X, such that

ν(A) =

∫X

gχA dµ, (1)

for all (or just some) A ∈ A. Using this terminology, a “Radon-Nikodim-type” theorem isone that has the Radon-Nikodim property as its conclusion. In other words, one looks forsome simple conditions on ν, which force (1).

Comment. The above definition is a bit imprecise due to the possible technical diffi-culties concerning the right-hand side of (1) and the fact that ultimately one also wants tocover the case when ν is a K-valued measure. There are three instances, when one wants tomake the above terminology a bit more precise, as follows.

A. If ν is an “honest” (i.e. [0,∞]-valued) measure, we want g to be non-negative (allowingg even to take value ∞), in which case the right-hand side of (1) is understood as thepositive µ-integral, so it can also take value ∞. In this case, we could require (1) tohold either for all A ∈ A, or for all A that belong to one of the following collections:

Aµfin = {A ∈ A : µ(A) <∞},

Aµσ-fin = {A ∈ A : there exists (An)∞n=1An ⊂ Aµ

fin, such that A =∞⋃

n=1

An}.

B. If ν is K-valued, we require g to belong to L1K(X,A, µ), so that the right-hand side of

(1) is always finite. As above, we may want to limit (1) to either Aµfin or Aµ

σ-fin.

1

Theorem 1 (“Easy” Radon-Nikodim Theorem). Assume µ is finite and ν is a C-valuedmeasure on A, such that

|ν(A)| ≤ Cµ(A), ∀A ∈ A, (2)

for some constant C ≥ 0. Then there exists g ∈ L1(X,A, µ), such that

ν(A) =

∫X

gχA dµ, ∀A ∈ A. (3)

Moreover:

(i) any g ∈ L1K(X,A, µ), that satisfies (3), also satisfies the inequality

|g| ≤ C, µ-a.e. (4)

(ii) the function g is essentially unique, in the sense that if g′ ∈ L1(X,A, µ) is anotherfunction that satisfies (3), then g = g′, µ-a.e., so in effect g = g′, in L1

K(X,A, µ);

(iii) if ν is R-valued, then g can be taken in L1R(X,A, µ);

(iv) if ν is an “honest” measure, then g can be taken so that Range g ⊂ [0,∞).

Sketch of Proof. For the existence part, it suffices to consider the real case K = R. Indeed,if we consider the R-valued measures ν1 = Re ν and ν2 = Re ν, then one clearly has theinequalities

|νk(A)| ≤ |ν(A)| ≤ Cµ(A), ∀A ∈ A.

Assuming that ν is R-valued, we can write its Hahn-Jordan decomposition ν = ν+ − ν−.This means that ν± are two “honest” measures given by ν+(A) = ν(A ∩ B+) and ν−(A) =−ν(A ∩B+), where (B+, B−) is a two-set A-partition of X. It is the trivial that

0 ≤ ν±(A) ≤ Cµ(A), ∀A ∈ A,

so for the existence part we can in fact assume that ν is an “honest” finite measure.Assuming that ν is “honest” and satisfies (2), it is pretty obvious that one has the

inclusionsL2(X,A, µ) ⊂ L1(X,A, µ) ⊂ L1(X,A, ν). (5)

The first inclusion is in fact continuous, and this is due to the finiteness of µ, which byHolder’s inequality implies ‖f‖1 ≤ ‖f‖2 · ‖χX‖2 = ‖f‖2

õ(X). The second inclusion in (5)

follows from (2), which yields∫

X|f | dν ≤ C

∫X|f | dµ, for all measurable functions f . Using

the inclusions (5), we can then define the map φ : L2(X,A, µ) → C by

φ(f) =

∫X

f dν, f ∈ L2(X,A, µ),

which will be linear and continuous, since by the previous estimates, we have:

|φ(f)| ≤∫

X

|f | dν ≤ C√µ(X) · ‖f‖2, ∀ f ∈ L2(X,A, µ).

2

Using Riesz’ Theorem, there exists1 g ∈ L2(X,A, µ) ⊂ L1(X,A, µ), such that∫X

fg dµ = φ(f) =

∫X

f dν, ∀ f ∈ L2(X,A, µ).

Of course, if we specialize to f = χA, we now get (3), and we are done.Having proved the existence part, we now focus on the other statement(s). To prove (i),

we fix g ∈ L1(X,A, µ) satisfying (3), and prove that the set A = {x ∈ X : |g(x)| > C}has µ(A) = 0. If we take a countable set S ⊂ T, which is dense in T, then one can writeA =

⋃s∈S

⋃n∈NAns, where Ans = {s ∈ X : Re[sg(x)] ≥ C + 1

n}, so in order to prove that

µ(A) = 0, it suffices to show that µ(Ans) = 0, ∀ s ∈ S, n ∈ N. Fix s ∈ S and n ∈ N, so thatby (3) we have

|ν(Ans)| = |sν(Ans)| =∣∣∣∣ ∫

X

sgχAns dµ

∣∣∣∣ ≥ ∫X

Re[sg]χAns dµ ≥

≥∫

X

[C + 1

n

]χAns dµ =

[C + 1

n

]µ(Ans).

Of course, by (2) this forces the inequality

Cµ(Ans) ≥[C + 1

n

]µ(Ans),

which by the finiteness of µ yields µ(Ans) = 0.Statements (ii)-(iv) are left to the reader.

Exercise 1.♥ Complete the proof of Theorem 1. Under the hypotheses of Theorem 1,also prove that, if g satisfies (3), then its norm in L∞ is given by

‖g‖∞ = min{C ≥ 0 : C satisfies (2)}.

Exercise 2.♥ Assume ν is a K-valued measure, and let |ν| be the variation measure,defined by

|ν|(A) = sup{ ∞∑

n=1

|ν(An)| : (An)∞n=1 ⊂ A disjoint, and∞⋃

n=1

An = A}, ∀A ∈ A.

Assume µ is finite.

(i) Prove that, if ν satisfies (2), then |ν| also satisfies (2), and furthermore, if g ∈L1(X,A, µ) satisfies (3), then

|ν|(A) =

∫X

|g| dµ, ∀A ∈ A.

(ii) Prove that, if |ν| = µ, so in particular (2) holds with C = 1, then any g that satisfies(3) also satisfies the equality: |g| = 1, µ-a.e.

1 Strictly speaking, by Riesz’ Theorem there exists h ∈ L2(X,A, µ), such that φ(f) = ( h | f ) =∫

Xhf dµ,

∀ f ∈ L2(X,A, µ), so we can set g = h.

3

Remark 1. Given a K-valued measure on A, we know that its variation measure |ν| isfinite, and furthermore one can write ν as a linear combination

ν =n∑

k=1

αkνk (6)

of (up to four) “honest” finite measures ν1, . . . , νn, all dominated by |ν| i.e. satisfyingνk ≤ Ck|ν|. This allows one to define, for every f ∈ L1

K(X,A, |ν|), its ν-integral by∫X

f dν =n∑

k=1

αk

∫X

f dνk. (7)

As it turns out, the right-hand side is independent of the presentation (6) of ν as as a linearcombination of “honest” finite measures dominated by |ν|, and this can be shown indirectly asfollows. By Exercise 2, there exists an (essentially unique)A-measurable function u : X → K,with |u| = 1, |ν|-a.e. on X, such that ν(A) =

∫XuχA d|ν|, so for a given presentation of ν

as (6), one has the equality

n∑k=1

αk

∫X

χA dνk = ν(A) =

∫X

χAu d|ν|, (8)

so instead of defining the ν-integral by (7), we can define it by∫X

f dν =

∫X

fu d|ν|, ∀ f ∈ L1K(X,A, |ν|) (9)

and show that (7) holds for all possible presentations of ν as (6). But this is pretty obvious,since by (8) the equality (7) holds for all f ∈ Lelem

K (X,A, |ν|), and LelemK (X,A, |ν|) is dense

in L1K(X,A, |ν|).Comment. The formula (9) can be generalized for signed measures, i.e. measures ν

that can take values in either (−∞,∞] or in [−∞,∞). (Of course, R-valued measures are aspecial case of signed measures.) Using the Hahn-Jordan decomposition, we can write anysigned measure ν as a difference ν+−ν− of two “honest” measures, one of them finite. Morespecifically, there exists an A-partition of X into two sets (B+, B−), such that ν+(A) =µ(A ∩ B+) and ν− = −ν(A ∩ B+). The variation measure is then given by |ν| = ν+ + ν−,so if we define the function u = χB+ − χB− , the equality (9) still holds.

Remark 2. Using the above constructions, we now see that, if µ is finite, and a C-valuedmeasure ν satisfies (2), then the unique g ∈ L1(X,A, µ) that satisfies (3) will automaticallysatisfy the following identities: ∫

X

f dν =

∫X

fg dµ, (10)∫X

f d|ν| =∫

X

f |g| dµ, (11)

for all f ∈ L1(X,A, µ). Such identities are often referred to as Change of Variable Formulas.

4

In what follows our goal will be to relax the hypotheses of Theorem 1 as much as possible.

Definition. An “honest” measure on A is said to be absolutely continuous with respectto µ, if: every µ-neglijeable set A ∈ A is also ν-neglijeable, that is, one has the implication

µ(A) = 0 ⇒ ν(A) = 0. (12)

In this case, one uses the notation ν � µ.Of course, it makes sense to consider the implication (12) also in the case when ν is

K-valued or a signed measure. In that case condition (12) is equivalent to |ν| � µ.

Comment. It is pretty clear that, for an “honest” measure ν, the existence of a constantC ≥ 0 that satisfies

ν(A) ≤ Cµ(A), ∀A ∈ A, (13)

forces ν � µ, so subsequent versions of Theorem 1, which have the hypothesis (2) replacedby ν � µ (or |ν| � µ in the K-valued case), will be generalizations of Theorem 1. As itturns out (and this will be helpful for proving such generalizations), one has some sort ofconverse of the implication “(13) ⇒ (12),” which goes as follows.

Lemma 1. If µ is finite, and ν is an “honest” measure on A, such that ν � µ, thenthere exists a sequence (νn)∞n=1 of finite “honest” measures on A, such that, for each A ∈ A,the sequence

(νn(A)

)∞n=1

⊂ [0,∞) has the following properties:

(i)(νn(A)

)∞n=1

is non-decreasing;

(ii) limn→∞ νn(A) = ν(A);

(iii) νn(A) ≤ nµ(A), ∀n ∈ N.

Sketch of Proof. The idea is to define νn = (nµ) ∧ ν. In general, for any two measures η,ν with at least one of them finite, it is possible to define two measures η ∧ ν and η ∨ ν, bymimicking max{a, b} = 1

2(a + b + |a − b|) and max{a, b} = 1

2(a + b − |a − b|), but using

the signed measure η − ν and its variation measure |η − ν| instead. We will not discussthis construction in general, but instead we are going to describe it explicitly in the casewhen η = nµ. For every n ∈ N we consider the signed measure2 λn = nµ − ν, and itsHahn-Jordan decomposition λn = λ+

n − λ−n , where λ±n are two “honest” measures (with λ+n

in fact finite) given by λ+n (A) = λn(A ∩B+

n ) and λ−n (A) = −λn(A ∩B−n ), where (B+

n , B−n ) is

a certain A-partition of X. Based on the equality λn+1 = λn + µ and the properties of theHahn-Jordan decomposition, one can prove that, for every A ∈ A one has the inequalities

λ+n (A) ≤ λ+

n+1(A) ≤ λ+n (A) + µ(A), (14)

λ−n (A) ≥ λ−n+1(A) ≥ λ−n − µ(A). (15)

Since by construction we have

nµ(A ∩B+n ) = ν(A ∩B+

n ) + λ+n (A) ≥ ν(A ∩B+

n ),

ν(A ∩B−n ) = nµ(A ∩B−

n ) + λ−n (A) ≥ nµ(A ∩B−n ),

2 Since µ is assumed to be finite, λn is [−∞,∞)-valued.

5

we can think B+n as “the region where nµ ≥ ν” and B−

n as “the region where nµ ≤ ν”(although the use of “the” is not justified, since the partition (B+

n , B−n ) is not unique!), so

we will define νn by

νn(A) = ν(A ∩B+n ) + nµ(A ∩B−

n ), A ∈ A,

and it is pretty obvious that condition (iii) is satisfied.Concerning the non-uniqueness of the partition (B+

n , B−n ), as well as property (i), it turns

out that, using (14) and (15), one can prove the µ-inclusions

B+n ⊂

µB+

n+1 and B−n+1 ⊂

µB−

n , ∀n ∈ N, (16)

where the notation “D ⊂µE” means µ(D r E) = 0. Of course, by the absolute continuity

assumption, we also have

B+n ⊂

νB+

n+1 and B−n+1 ⊂

νB−

n , ∀n ∈ N. (17)

Using (16) and (17), one can easily prove statement (i).To prove statement (ii), we first notice that, by construction, we clearly have ν(A) ≥

νn(A), ∀n ∈ N, which yields

ν(A) ≥ limn→∞

νn(A), ∀A ∈ A. (18)

To prove that the above inequality is in fact an equality, we first consider the sets B+ =⋃∞n=1B

+n and B− =

⋂∞n=1B

−n = XrB+, and we notice that, using3 (17) we have the equality

ν(A ∩B+) = limn→∞

ν(A ∩B+n ). (19)

Secondly, since by construction we have the inclusions B+n ⊂ B+ and B−

n ⊃ B−, we alsohave the inequalities

νn(A) = ν(A ∩B+n ) + nµ(A ∩B−

n ) ≥ ν(A ∩B+n ) + nµ(A ∩B−),

which by (19) yield

limn→∞

νn(A) ≥ ν(A ∩B+) + limn→∞

nµ(A ∩B−). (20)

On the one hand, if µ(A∩B−) > 0, this forces limn→∞ νn(A) = ∞, so by (18) we must haveν(A) = ∞. On the other hand, if µ(A ∩ B−) = 0, then by absolute continuity we also haveν(A∩B−) = 0, which forces ν(A) = ν(A∩B+), and then (20) yields limn→∞ ν(A) ≥ ν(A).

Exercise 3.♥ Fill in the details of the proof of Lemma 1.

3 It is well known that if a sequence (An)∞n=1 ⊂ A is ν-increasing, i.e. An ⊂ν

An+1, ∀n, then ν( ⋃∞

n=1 An

)=

limn→∞ ν(An).

6

Theorem 2 (Radon-Nikodim Theorem: The “Honest Vs. Finite” Case). Assume µ isfinite and ν is an “honest” measure measure on A, such that ν � µ. Then there exists anA-measurable function g : X → [0,∞], such that4

ν(A) =

∫X

gχA dµ, ∀A ∈ A, . (21)

Moreover:

(i) if g′ : X → [0,∞] is another A-measurable function that satisfies (21), then g = g′,µ-a.e.;

(ii) any A-measurable function g : X → [0,∞] that satisfies (21) also satisfies:∫X

f dν =

∫X

fg dµ, ∀ f ∈m+(X,A). (22)

Proof. We begin with the proof of the existence part. By Lemma 1, there exists a sequence(νn)∞n=1 of finite measures on A, such that

(a)(νn(A)

)∞n=1

⊂ [0,∞) is non-decreasing,

(b) limn→∞ νn(A) = ν(A),

(c) νn(A) ≤ nµ(A), ∀n ∈ N,

for all A ∈ A. Since the “honest” finite measures νn − νn−1, also satisfy (c), we can thenapply Theorem 1, to obtain, for every n ∈ N an A-measurable function hn : X → [0, n], suchthat

νn(A)− νn−1(A) =

∫X

hnχA dµ, ∀A ∈ A, (23)

for all n ∈ N. (Here we use the convention ν0 = 0.) We can then define the non-decreasingsequence (gn)∞n=1 of measurable A-functions by gn =

∑nk=1 hk, which will now satisfy

νn(A) =

∫X

gnχA dµ, ∀A ∈ A, (24)

for all n ∈ N. The function g : X → [0,∞], defined by g(x) = limn→∞ gn(x) is obviouslyA-measurable, and by Lebesgue’s Monotone Convergence Theorem, combined with property(b) above, the identities (24) will clearly yield the identity (21).

To prove (ii) we first notice that the equality (21) immediately yields∫X

f dν =

∫X

fg dµ, ∀ f ∈m elem+ (X,A). (25)

For an arbitrary A-measurable function f : X → [0,∞], we know that there exists anon-decreasing sequence (fn)∞n=1 of elementary non-negative A-measurable functions, with

4 In both (21) and (22) we use the positive integral, so we do not exclude the possibility of infinite values.

7

limn→∞ fn(x) = f(x), ∀x ∈ X. Since by (25), we know that∫

Xfn dν =

∫Xfng dµ, ∀n, the

desired identity (22) follows from Lebesgue’s Monotone Convergence Theorem.To prove the uniqueness property (i), we fix another A-measurable function g′ : X →

[0,∞], satisfying (21), and we show that g′ = g, µ-a.e. By symmetry, it suffices to provethat, for any two rational numbers r > s ≥ 0, the set Ars = {x ∈ X g(x) ≥ r > s ≥ g′(x)}has µ(Ars) = 0. (If all the Ars’s have µ(Ars) = 0, their union, which is the set A = {x ∈ X :g(x) > g′(x)}, will also have µ(A) = 0, which means that g ≤ g′, µ-a.e., so by reversing theroles one would also get g′ ≤ g, µ-a.e.) Fix r > s ≥ 0 and notice that, since g and g′ bothsatisfy (21) we have

ν(Ars) =

∫X

gχArs dµ ≥∫

X

rχArs dµ = rµ(Ars),

ν(Ars) =

∫X

g′χArs dµ ≤∫

X

sχArs dµ = sµ(Ars),

thus forcing rµ(Ars) ≤ sµ(Ars), which by the finiteness of µ, in turn yields µ(Ars) = 0.

Remark 3. Given ν, µ, and g as in Theorem 3, the space of ν-integrable functions canbe characterized as:

L1(X,A, ν) = {f ∈m(X,A) : |f |g ∈ L1(X,A, µ)}.

The Change of Variable Formula, which follows immediately from (22), has the form:∫X

f dν =

∫X

fg dµ, ∀ f ∈ L1(X,A, ν). (26)

The K-valued version of Theorem 2 is as follows.

Theorem 3 (Radon-Nikodim Theorem: The “K-Valued Vs. Finite” Case). Assume µis finite and ν is a C-valued measure measure on A, such that |ν| � µ. Then there existsg ∈ L1(X,A, µ), such that

ν(A) =

∫X

gχA dµ, ∀A ∈ A, (27)

Moreover:

(i) the function g is essentially unique, in the sense that if g′ ∈ L1(X,A, µ) is anotherfunction that satisfies (27), then g = g′, µ-a.e., so in effect g = g′, in L1(X,A, µ);

(ii) if ν is R-valued, then g can be taken in L1R(X,A, µ);

(iii) if ν is an “honest” finite measure, i.e. [0,∞)-valued, then g can be taken so thatRange g ⊂ [0,∞);

(iv) if g ∈ L1(X,A, µ) satisfies (27), then

(a) |ν|(A) =∫

XχA|g| dµ, ∀A ∈ A;

8

(b) for every f ∈ L1(X,A, |ν|), one has the Change of Variable Formulas:∫X

f dν =

∫X

fg dµ, (28)∫X

f d|ν| =∫

X

f |g| dµ. (29)

Sketch of Proof. We are only going to treat the existence part, leaving the rest for the readerto prove. By Theorem 2 and Remark 3, there exists an A-measurable function h : X →[0,∞], such that

|ν|(A) =

∫X

χAh dµ, ∀A ∈ A; (30)∫X

f d|ν| =∫

X

fh dµ, ∀ f ∈ L1(X,A, |ν|). (31)

Using the fact that all C-valued measures ν have finite variation measures, by (30) we get∫X

h dµ = |ν|(X) <∞,

so h is integrable. By Remark 1, one can also choose u ∈m(X,A), with |u| = 1, such that∫X

f dν =

∫X

fu d|ν|, ∀ f ∈ L1(X,A, |ν|),

so one can define the desired function as g = uh.

Exercise 4.♥ Complete the proof of Theorem 3.

Up to this point, we have only treated the case when µ is finite. When trying to generalizeTheorems 2 and 3 beyond this case, we will have to pay some price, namely we will have tolimit the Radon-Nikodim property (1) to sets A ∈ Aµ

fin. Concerning the Change of VariableFormulas (10) and (11), they will also have to be limited to some special type of measurablefunctions f .

Remark 4. If ν is “honest” and g : X → [0,∞] is an A-measurable function, satisfying(1), for all A ∈ Aµ

fin, then

(i) the identity (1) also holds for all A ∈ Aµσ-fin;

(ii) the Change of Variable Formula (10) holds for all A-measurable functions f : X →[0,∞], which have their support5 in Aµ

σ-fin.

Property (i) is due to the σ-additivity of both ν and µ, combined with the fact that all setsin A ∈ Aµ

σ-fin are countable disjoint unions of sets in A ∈ Aµfin.

Property (ii) follows from the observation that if f : X → [0,∞] is an A-measurablefunction such that supp f belongs to Aµ

σ-fin, then there exists a non-decreasing sequence

5 The support of f is defined as supp f = {x ∈ X : f(x) 6= 0}.

9

fn ∈ Lelem+ (X,A, µ), such that limn→∞ fn(x) = f(x), ∀x, and then by Lebesgue’s Monotone

Convergence Theorem one has the equalities∫X

f dν = limn→∞

∫X

fn dν and

∫X

fg dµ = limn→∞

∫X

fng dµ.

By this argument, it suffices to prove (10) only for f ∈ Lelem(X,A, µ) = span {χA : Aµfin},

which is trivial from the assumption, which yields∫X

χA dν = ν(A) =

∫X

χAg dµ, ∀A ∈ Aµfin.

Comment. To obtain the generalizations of Theorems 2 and 3, one replaces the finitenessassumption on (X,A, µ) with decomposability. Recall (see BS IV) that the measure space(X,A, µ) is said to be decomposable, if there exists an A-partition π = (Xi)i∈I of X, suchthat

(a) A = {A ⊂ A : A ∩Xi ∈ A, ∀ i ∈ I};

(b) µ(Xi) <∞, ∀ i ∈ I;

(c) for every A ∈ A with µ(A) <∞, one has:

µ(A) =∑i∈I

µ(A ∩Xi). (32)

In this case, we say that the partition π constitutes a decomposition of (X,A, µ).

Remarks 5-6. Assume (X,A, µ) is decomposable, and π = (Xi)i∈I is one of its decom-positions.

5. As pointed out in BS IV, using condition (a) we know that, if T is either one of thespaces R, C, [−∞,∞], of [0,∞], then for a function f : X → T , the following areequivalent

(i) f is A-measurable;

(ii) f∣∣Xi

is A∣∣Xi

-measurable, for each i ∈ I.

6. The equality (32) also holds for A ∈ Aµσ-fin.

Theorem 4 (Radon-Nikodim Theorem: The “Honest Vs. Decomposable” Case). As-sume (X,A, µ) is decomposable and ν is an “honest” measure measure on A, such thatν � µ. Then there exists an A-measurable function g : X → [0,∞], such that6

ν(A) =

∫X

gχA dµ, ∀A ∈ Aνfin, . (33)

Moreover:

6 As in Theorem 2, in (33), (34) and (35), we use the positive integral, so we do not exclude the possibilityof infinite values.

10

(i) if g′ : X → [0,∞] is another A-measurable function that satisfies (33), then g = g′,µ-l.a.e.;

(ii) any A-measurable function g : X → [0,∞] that satisfies (33) also satisfies:∫X

f dν =

∫X

fg dµ, (34)

for all f ∈m+(X,A), with supp f ∈ Aµσ-fin, so in particular it also satisfies

ν(A) =

∫X

gχA dµ, ∀A ∈ Aνσ-fin, . (35)

Sketch of Proof. To prove the existence part, we fix a decomposition π = (Xi)i∈I for (X,A, µ).For each i ∈ I consider the finite measure space (Xi,A

∣∣Xi, µ

∣∣Xi

) and the “honest” measure

ν∣∣Xi

, where A∣∣Xi

= {A ∩Xi : A ∈ A} and µ∣∣Xi

and ν∣∣Xi

denote the restrictions of µ and ν

to A∣∣Xi

. Since we clearly have ν∣∣Xi� µ

∣∣Xi

, by Theorem 2, for each i ∈ I, there exists an

A∣∣Xi

-measurable function gi : Xi → [0,∞], such that

ν(A ∩Xi) =

∫Xi

gχA∩Xid(µ

∣∣Xi

), ∀A ∈ A. (36)

Using Remark 5, there exists a (unique) A-measurable function g : X → [0,∞], such thatg∣∣Xi

= gi, ∀ i ∈ I. To check that g satisfies (33), we start with some A ∈ Aµfin and we use the

fact (see BS IV) that the set I(A) = {i ∈ I : µ(A∩Xi) 6= 0} is countable, and furthermore,the set A′ =

⋃i∈I(A)A ∩ Xi ⊂ A belongs to A, and satisfies µ(A r A′) = 0. By absolute

continuity, we also have the equality ν(Ar A′) = 0, so we get

ν(A) = ν(A′) =∑

i∈I(A)

ν(A ∩Xi).

Using (36) this clearly yields (33).The rest of the proof is left to the reader.

Exercise 5.♥ Complete the proof of Theorem 4.

Theorem 5 (Radon-Nikodim Theorem: The “K-Valued Vs. Decomposable” Case).Assume (X,A, µ) is decomposable and ν is a C-valued measure measure on A, such that|ν| � µ. Then there exists g ∈ L1(X,A, µ), such that

ν(A) =

∫X

gχA dµ, ∀A ∈ Aµfin, (37)

Moreover:

(i) the function g is essentially unique, in the sense that if g′ ∈ L1(X,A, µ) is anotherfunction that satisfies (37), then g = g′, µ-a.e., so in effect g = g′, in L1(X,A, µ);

(ii) if ν is R-valued, then g can be taken in L1R(X,A, µ);

11

(iii) if ν is an “honest” finite measure, i.e. [0,∞)-valued, then g can be taken so thatRange g ⊂ [0,∞);

(iv) if g ∈ L1(X,A, µ) satisfies (37), then for every function f ∈ L1(X,A, |ν|) withsupp f ∈ Aµ

σ-fin, one has the Change of Variable Formulas:∫X

f dν =

∫X

fg dµ, (38)∫X

f d|ν| =∫

X

f |g| dµ, (39)

so in particular, one also has the equalities

ν(A) =

∫X

gχA dµ, (40)

|ν|(A) =

∫X

|g|χA dµ, (41)

for all A ∈ Aνσ-fin.

Sketch of Proof. We are only going to treat the existence part, and the uniqueness statement(i), leaving the rest for the reader to prove. Exactly as in Theorem 3, it suffices to considerthe case when ν is an “honest” finite measure. By Theorem 4, there exists an A-measurablefunction g0 : X → [0,∞], such that

ν(A) =

∫X

g0χA dµ, ∀A ∈ Aµfin, (42)

Fix a decomposition π = (Xi)i∈I for (X,A, µ), and consider the index set

J = {i ∈ I : ν(Xi) 6= 0}.

Since ν is finite, the set J is countable. For any set A ∈ A, let us denote the union⋃

i∈I A∩Xi

by AJ .

Claim: For every A ∈ Aµfin, the set AJ belongs to Aµ

fin, and satisfies ν(Ar AJ) = 0.

First of all, since AJ ⊂ A and µ(A) <∞, it is clear that µ(AJ) <∞. Secondly, if we usethe notations from the proof of Theorem 4, we know that the index set

I(A) = {i ∈ I : µ(A ∩Xi) 6= 0}

is countable, and the set A′ =⋃

i∈I(A)A∩Xi ⊂ A satisfies µ(ArA′) = 0, which by absolute

continuity also gives ν(ArA′) = 0. Of course, if we consider the countable index set I(A)∪J ,the union B =

⋃i∈I(A)∪J A ∩Xi, will satisfy the inclusions A′, AJ ⊂ B ⊂ A, as well as the

equality µ(A r B) = 0, which by absolute continuity also gives ν(A r B) = 0, so in orderto prove the Claim we only need to show that ν(B r AJ) = 0. But this is trivial, since byconstruction we have

ν(B r AJ) =∑

i∈I(A)rJ

ν(A ∩Xi) ≤∑

i∈I(A)rJ

ν(Xi) = 0.

12

Using the above Claim, we see that if we consider the set XJ =⋃

i∈J Xi ∈ A, andthe function g = g0χXJ

, then, using Lebesgue’s Monotone Convergence Theorem and σ-additivity, it follows that, for every A ∈ Aµ

fin, we have∫X

gχA dµ =∑i∈J

∫X

g0χA∩Xidµ =

∑i∈J

ν(A ∩Xi) = ν(AJ) = ν(A),

thus proving (37). Again by Lebesgue’s Monotone Convergence Theorem, using (37) withA = Xi, we also have∫

X

g dµ =∑i∈J

∫X

gχXidµ =

∑i∈J

ν(Xi) = ν(XJ) <∞,

thus proving that g indeed belongs to L1(X,A, µ).To prove the uniqueness statement (i), we assume (working now in the general case, when

ν is C-valued) we have two functions g, g′ ∈ L1(X,A, µ), both satisfying (37), and let usprove that g = g′, µ-a.e. Using real and imaginary parts, we have∫

X

Re gχA dµ = Re ν(A) =

∫X

Re g′χA dµ,∫X

Im gχA dµ = Im ν(A) =

∫X

Im g′χA dµ,

for all A ∈ Aµfin, which means that the functions h1 = Re g − Re g′ and h2 = Im g − Im g′,

which both belong to L1R(X,A, µ), satisfy∫

X

h1χA dµ =

∫X

h2χA dµ = 0, ∀A ∈ Aµfin. (43)

If one considers the sets A+kn = {x ∈ X : hk(x) ≥ 1

n} and A−kn = {x ∈ X : hk(x) ≤ − 1

n},

k = 1, 2, n ∈ N, then the inequalities |hk| ≥ 1nχA+

kn∪A−kn, combined with the µ-integrability of

hk, force A±kn to belong to Aµfin, so using (43) with A±kn, and the inequalities hkχA+

kn≥ 1

nχA+

kn

and hkχA−kn≤ − 1

nχA−kn

we obtain 1nµ(A±kn) ≤ 0, thus forcing µ(A±kn) = 0, ∀n ∈ N, k = 1, 2.

This forces h1 = h2 = 0, µ-a.e., which yields g = g′, µ-a.e.

Exercise 6.♥ Complete the proof of Theorem 5.

Corollary 1 (The Radon-Nikodim Theorems: the “Any Vs. σ-finite” Case). If (X,A, µ)is σ-finite, then Theorems 2 and 3 hold word-for-word.

Proof. If (X,A, µ) is σ-finite, then it is decomposable, so we can use Theorems 4 and 5.In this case, however, we have the equality Aµ

σ-fin = A, and the notions of “µ-l.a.e.” and“µ-a.e.” are equivalent.

B. The duals of Lp-spaces

In this sub-section we apply the Radon-Nikodim Theorems to the calculation of thetopological duals of the Lp-spaces, for 1 ≤ p <∞. As customary, we declare two “numbers”

13

p, q ∈ [1,∞] Holder conjugate, if either both p and q are finite and 1p

+ 1q

= 1, or p = 1 andq = ∞, or vice versa.

Recall now that in BS IV, given Holder conjugate numbers p and q, we defined the Holderdual paring

Φ : LpK(X,A, µ)× Lq

K(X,A, µ) 3 (f, g) 7−→∫

X

fg dµ.

Moreover, we also showed that, for any g ∈ LqK(X,A, µ), the linear map

g# : LpK(X,A, µ) 3 f 7−→

∫X

fg dµ ∈ K

is norm-continuous on Lp, and furthermore, the map

# : LqK(X,A, µ) 3 g 7−→ g# ∈ Lp

K(X,A, µ)∗ (44)

is linear and isometric, if 1 < p ≤ ∞. Eventually we are going to show that this map is infact an isometric isomorphism, if 1 < p <∞.

Comment. The case p = 1 (when we use q = ∞) is a bit more complicated, and willrequire some additional restrictions on the ambient measure space (X,A, µ). As pointed outin BS IV, in this case the map (44) may fail to be isometric, in general. Its correct version(which is always isometric) has L∞,loc

K (X,A, µ) replaced with L∞,locK (X,A, µ), so instead of

(44), the appropriate map to consider is

#loc : L∞,locK (X,A, µ) 3 [g]loc 7−→ g# ∈ L1

K(X,A, µ)∗, (45)

where g#(f) =∫

Xfg dµ, g ∈ L∞,loc

K (X,A, µ), f ∈ L1K(X,A, µ). In the case when (X,A, µ)

is nowhere degenerate (which happens, for instance, in the σ-finite case), the surjectivecontraction

L∞K (X,A, µ) 3 g 7−→ [g]loc ∈ L∞,locK (X,A, µ)

is in fact an isometric isomorphism, so in this case the map (45) is essentially the same as(44) with p = 1 and q = ∞.

Notation. Given p ∈ [1,∞] and h ∈ L∞K (X,A, µ), we denote the multiplication mapLp

K(X,A, µ) 3 f 7−→ hf ∈ LpK(X,A, µ) by Mh. As it turns out, since for every A-measurable

function f : X → K, one has the inequality |hf | ≤ ‖h‖∞ · |f |, µ-a.e., the (obviously) linearoperator Mh is norm-continuous, and one has the inequality ‖Mh‖ ≤ ‖h‖∞.

Comment. In preparation for the proof or Theorem 6 below, it will be helpful to reduceeverything to the real case K = R. This is done in a manner similar to what we did in BSIII, as follows.

Definition. Let X be either one of the complex Banach spaces Lp(X,A, µ), p ∈ [1,∞],or Lloc,∞(X,A, µ), and let XR denote its real counterpart: Lp

R(X,A, µ), p ∈ [1,∞], or

Lloc,∞R (X,A, µ). A C-linear functional φ : X → C is said to be hermitian, if φ(XR) ⊂ R. The

features of hermitian functionals are similar to those discussed in BS III, as shown below.

Exercise 7.♥ Use the notations as above.

(i) Prove that, for every R-linear functional ψ : XR → R, there exists a unique hermitianC-linear functional φ : X → C, such that φ

∣∣XR

= ψ.

14

(ii) Prove that, if φ and ψ are as above, the following are equivalent:

• φ is continuous;

• ψ is continuous.

Moreover, in this case, prove the equality ‖φ‖ = ‖ψ‖.

(iii) Prove that for any C-linear continuous functional φ : X → C, there exist two hermitianC-linear continuous functionals φRe, φIm : X → C, such that φ = φRe + iφIm.

Theorem 6 (Holder duality: the finite case). Assume p ∈ [1,∞) and let q ∈ (1,∞] bethe Holder conjugate of p. If (X,A, µ) is a finite measure space, then the map (44) is anisometric linear isomorphism.

Proof. As previously discussed, we already know that the linear map (44) is isometric, sothe only property we need to prove is surjectivity. Start with a linear continuous functionalφ : Lp

K(X,A, µ) → K, and let us indicate how a function g ∈ LqK(X,A, µ) can be constructed,

such that

φ(f) =

∫X

fg dµ, ∀ f ∈ LpK(X,A, µ). (46)

Using Exercise 7, we can assume that φ is hermitian, so in fact we can limit ourselves to thereal case K = R.

Claim 1: The map ν : A 3 A 7−→ φ(χA) ∈ R defines an R-valued measure on A, whichsatisfies

|ν(A)| ≤ ‖φ‖ · µ(A)1/p, ∀A ∈ A. (47)

To check σ-additivity we start with a sequence (An)∞n=1 ⊂ A of disjoint sets, we letA =

⋃∞n=1An, and we must show that

∑∞n=1 ν(An) = ν(A). Since φ is linear and continuous,

it suffices to show that, when we work in LpK(X,A, µ), the series

∑∞n=1 χAn converges to χA,

in the norm topology, which means that, when we consider the partial sums fN =∑N

n=1 χAn ,we have limN→∞ ‖χA − fN‖p = 0. But this is pretty clear, since, on the one hand, we haveχA − fN = χBN

, where BN =⋃∞

n=N+1An, so

‖χA − fN‖pp =

∫X

|χBN|p dµ =

∫X

χBNdµ = µ(BN) =

∞∑n=N+1

µ(An),

and on the other hand, by σ-additivity and finiteness of µ we also have∑∞

n=1 µ(An) =µ(A) <∞, which means that

∞∑n=N+1

µ(An) = µ(A)−N∑

n=1

µ(An) −−−→N→∞

0.

The inequality (47) is obvious from the norm inequality |φ(χA)| ≤ ‖φ‖ · ‖χA‖p, and theequality ‖χA‖p

p =∫

XχA dµ = µ(A).

15

Using Claim 1 and the Radon-Nikodim Theorem 4 (the “K-valued Vs. finite” case), thereexists g ∈ L1

R(X,A, µ), such that ν(A) =∫

XχAg dµ, ∀A ∈ A. By linearity, we get in fact

the equality

φ(f) =

∫X

fg dµ, ∀ f ∈ LelemR (X,A, µ). (48)

To finish the proof of the Theorem, we must show that

(a) g in fact belongs to LqR(X,A, µ), and

(b) the equality (48) holds in fact for all f ∈ LpR(X,A, µ).

As it turns out, if we prove property (a), then (b) will follow automatically, since condition(a) forces the continuity of the linear functional g# : Lp

R(X,A, µ) 3 f 7−→∫

Xfg dµ ∈ R, so

the desired equality follows from (48) and the fact that LelemR (X,A, µ) is dense in L1

R(X,A, µ).Property (a) is pretty clear if p = 1, when we must show that g belongs to L∞R (X,A, µ).

Indeed, this follows immediately from the “easy” Radon-Nikodim Theorem 2, since in thecase p = 1, by Claim 1 we have |ν(A)| ≤ ‖φ‖ · µ(A), ∀A ∈ A.

For the remainder of the proof we are going to assume that p belongs to (1,∞), so theHolder conjugate q of p will also belong to the same interval, which means that the desiredproperty (a) now reads:

∫X|g|q dµ < ∞. Consider the sets X+ = {x ∈ X : g(x) ≥ 0} and

X− = X r X+, which both belong to A, and the functions g+ = gχX+ and g− = −gχX− ,which both belong to L1

+(X,A, µ), and satisfy the equality

g = g+ − g−. (49)

Remark now that, since the multiplication operators MχX±are continuous, the functionals

φ+ = φ ◦MχX+ and φ+ = −φ ◦MχX+ are continuous as well, and they satisfy the equality

φ = φ+ − φ−.

Moreover, since both multiplication operators MχX+ and MχX−map the space Lelem

R (X,A, µ)into itself, using (48) we also have

φ+(f) =

∫X

fg+ dµ and φ−(f) =

∫X

fg− dµ, ∀ f ∈ LelemR (X,A, µ). (50)

By (49) we see that it suffices to show that both g+ and g− belong to LqR(X,A, µ). By

(50), we see that everything reduces now to proving the Claim below. (Both pairs (φ+, g+)and (φ−, g−) satisfy (48). The added advantage here is the fact that both g+ and g− arenon-negative.)

Claim 2: If p ∈ (1,∞), and g ∈ L1+(X,A, µ) satisfies (48), then

∫Xgq dµ ≤ ‖φ‖q(<∞).

Since g ≥ 0, by the definition of the (positive) integral, we know that∫X

gq dµ = sup{ ∫

X

h dµ : h ∈ Lelem+ (X,A, µ), h ≤ gq

},

16

so in order to prove the Claim it suffices to prove that, for any h ∈ Lelem+ (X,A, µ), such

that h ≤ gq, one has the inequality∫

Xh dµ ≤ ‖φ‖q. Fix such an h, and observe that, since

1p

+ 1q

= 1, we can write h = h1/ph1/q. Furthermore, since h1/p is elementary, and h1/q ≤ g,we have ∫

X

h dµ =

∫X

h1/ph1/q dµ ≤∫

X

h1/pg dµ = φ(h1/p). (51)

By the norm inequality, we also have

φ(h1/p) ≤ ‖φ‖ · ‖h1/p‖p =

[ ∫X

|h1/p|p dµ]1/p

=

[ ∫X

h dµ

]1/p

, (52)

so if we denote∫

Xh dµ by J and we combine (51) with (52) we now have J ≤ ‖φ‖ · J1/p,

which yields J1/q = J1−1/p ≤ ‖φ‖, thus proving the desired inequality J ≤ ‖φ‖q.

In our approach to the generalization of Theorem 6 will use the following constructions.

Notations. Assume p ∈ [1,∞). Given a set Y ∈ A, we can regard the Banach spaceLp

K(Y,A∣∣Y, µ

∣∣Y) both as a quotient and as a closed linear sub-space of Lp

K(X,A, µ). When

viewing LpK(Y,A

∣∣Y, µ

∣∣Y) as a quotient, we employ the restriction map Qp

Y : f 7−→ f∣∣Y.

When viewing LpK(Y,A

∣∣Y, µ

∣∣Y) as a closed linear subspace, we employ the isometric linear

map TY : LpK(Y,A

∣∣Y, µ

∣∣Y) ↪→ Lp

K(X,A, µ), defined as follows. For every f ∈ LpK(Y,A

∣∣Y, µ

∣∣Y),

we define the function TY f : X → K by

(TY f)(x) =

{f(x) if x ∈ Y

0 if x 6∈ Y

The composition TY ◦ QpY : Lp

K(X,A, µ) → LpK(X,A, µ) coincides with the multiplication

operator MχY. Of course, the composition Qp

Y ◦ TY : LpK(Y,A, µ) → Lp

K(Y,A, µ) is theidentity map.

Given a linear functional φ : LpK(X,A, µ) → K, we denote the composition φ ◦MχY

byφY . We can also define the restricted map φ

∣∣Y

= φ ◦ TY : LpK(Y,A

∣∣Y, µ

∣∣Y) → K, which will

satisfy the equality φ∣∣Y◦Qp

Y = φY .

Remarks 7-10. Suppose p ∈ [1,∞), and π = (Xi)i∈I is an A-partition of X, which isµ-integrable, i.e. such that µ(A) =

∑i∈I µ(A ∩Xi), ∀A ∈ Aµ

fin.

7. By Proposition-Definition 5 from BS IV, the correspondence Dp : f 7−→ (QpXif)i∈I

establishes an isometric linear isomorphism

Dp : (LpK(X,A, µ), ‖ . ‖p) → `p(Lp

K(Xi,A∣∣Xi, µ

∣∣Xi

), ‖ . ‖p)i∈I . (53)

8. If we take q ∈ (1,∞] to be the Holder conjugate of p, then using the isometric isomor-phism (53) and Theorem 8 from BS II, one obtains an isometric linear isomorphism

Σ : LpK(X,A, µ)∗ → `q

(Lp

K(Xi,A∣∣Xi, µ

∣∣Xi

)∗)

i∈I, (54)

defined as Σ = (D−1p )∗ ◦Ψ, where (D−1

p )∗ is the transpose of the inverse of (53), and

Ψ : [`p(LpK(Xi,A

∣∣Xi, µ

∣∣Xi

), ‖ . ‖p)i∈I ]∗ → `q

(Lp

K(Xi,A∣∣Xi, µ

∣∣Xi

)∗)

i∈I

is the isometric isomorphism from Theorem 8 in BS II.

17

9. Upon writing down all the identifications from the preceding remarks, the isometricisomorphism (54) is simply given as Σ(φ) = (φ

∣∣Xi

)i∈I . Therefore, for a linear functional

φ : LpK(X,A, µ) → K, the following are equivalent:

(i) φ is continuous;

(ii) all restricted functionals φ∣∣Xi

= φ ◦ TXi: Lp

K(Xi,A∣∣Xi, µ

∣∣Xi

) → K, i ∈ I, are

continuous, and the I-tuple (‖φ∣∣Xi‖)i∈I belongs to `q(I), i.e.

•∑

i∈I ‖φ∣∣Xi‖q <∞, if p, q ∈ (1,∞),

• supi∈I ‖φ∣∣Xi‖ <∞, if p = 1 and q = ∞.

Furthermore, if this is the case, then

‖φ‖ =

[∑

i∈I ‖φ∣∣Xi‖q

]1/qif p, q ∈ (1,∞)

supi∈I ‖φ∣∣Xi‖ <∞ if p = 1 and q = ∞

(55)

10. In the preceding remark, the equivalence (i) ⇔ (ii), as well as the formulas (55)hold, if instead of the restricted functionals φ

∣∣Xi

, we use the functionals φXi= φ ◦

MχXi: Lp

K(X,A, µ) → K. This follows from the identities φXi= (φ

∣∣Xi

) ◦ QpXi

and

φ∣∣Xi

= φ ◦ TXi= φXi

◦ TXi, which show that φXi

is continuous, if and only if φ∣∣Xi

is

continuous, and furthermore one has the equality ‖φXi‖ = ‖φ

∣∣Xi‖.

Lemma 2. Assume p ∈ [1,∞), and φ : LpK(X,A, µ) → K is a K-linear continuous func-

tional. If we regard the collection Aµfin as a directed set (with “⊃” as the order relation), then

the nets (φY )Y ∈Aµfin⊂ Lp

K(X,A, µ)∗ and (‖φY ‖)Y ∈Aµfin⊂ [0,∞) have the following properties.

(i) The net (‖φY ‖)Y ∈Aµfin

is increasing, i.e. for Y1, Y2 ∈ Aµfin, one has the implication

Y1 ⊃ Y2 ⇒ ‖φY1‖ ≥ ‖φY2‖.

(ii) limY ∈Aµfin‖φY ‖ = ‖φ‖.

(iii) If 1 < p <∞, then (φY )Y ∈Aµfin

converges to φ in the norm topology, i.e.

limY ∈Aµ

fin

‖φ− φY ‖ = 0.

Proof. (i). This statement is pretty obvious, since the inclusion Y1 ⊃ Y2 implies the equalityMχY2 = MχY1

◦ MχY2, which yields φY2 = φY1 ◦ MχY2

, so by the norm inequality we get‖φY2‖ ≤ ‖φY1‖ · ‖MχY2

‖ ≤ ‖φY1‖.(ii). Since (‖φY ‖)Y ∈Aµ

finis increasing, we have limY ∈Aµ

fin‖φY ‖ = supY ∈Aµ

fin‖φY ‖, so it

suffices to prove the equalitysup

Y ∈Aµfin

‖φY ‖ = ‖φ‖. (56)

Denote the left-hand side of (56) by S. Since φY = φ ◦MχY, with MχY

a linear contraction,it follows that S ≤ ‖φ‖. To prove the other inequality, we fix for the moment some ε > 0,

18

and we use the definition of the norm of φ, combined with the density of LelemK (X,A, µ)

in LpK(X,A, µ), to find an elementary integrable function f = α1χA1 + · · · + αnχAn (with

α1, . . . , αn ∈ K and A1, . . . , An ∈ Aµfin), such that ‖f‖p ≤ 1, and

|φ(f)| ≥ ‖φ‖ − ε. (57)

Since the set Y = A1 ∪ · · · ∪ An obviously belongs to Aµfin, and we clearly have f = MχY

f ,it follows that φ(f) = φY (f), so we have |φ(f)| = |φY (f)| ≤ ‖φY ‖ · ‖f‖p ≤ ‖φY ‖, so if we goback to (57) we now get ‖φY ‖ ≥ ‖φ‖ − ε. This shows, of course, that

S ≥ supY ∈Aµ

fin

‖φY ‖ ≥ ‖φ‖ − ε,

which forces (by the fact that the above inequality holds for all ε > 0) the inequality S ≥ ‖φ‖.(iii). Assume now 1 < p <∞. (The case p = 1 is discussed in Exercise 8 below.) Given

any set Y ∈ Aµfin, it is pretty obvious that (Y,X r Y ) constitutes a µ-integrable A-partition

of X, so by Remarks 9-10 one has the equality

‖φ‖q = ‖φY ‖q + ‖φXrY ‖q. (58)

Since we also have MχY+ MχXrY

= I, the identity operator on LpK(X,A, µ), we also have

the equality φY +φXrY = φ, so if we replace φXrY with φ−φY , the equality (58) now reads:

‖φ− φY ‖q = ‖φ‖q − ‖φY ‖q, (59)

so by part (ii) we immediately get limY ∈Aµfin‖φ− φY ‖q = 0, and we are done.

Exercise 8. Use the notations from Lemma 2, and assume p = 1.

(i) Prove that the net (φY )Y ∈Aµfin

converges to φ in the w∗-topology, i.e. limY ∈AµfinφY (f) =

φ(f), ∀ f ∈ L1K(X,A, µ).

(ii) Let X be an infinite set, and consider the measure space (X,A, µ), with A = P(X)and µ is the counting measure, so that L1

K(X,P(X), µ) = `1K(X), and Aµfin = Pfin(X).

Give an example of a linear continuous functional φ : `1K(X) → K, such that ‖φ‖ = 1,but ‖φ− φY ‖ = 1, ∀Y ∈ Pfin(X), thus showing that statement (iii) from Lemma 2 isnot true, in general, for p = 1.

The next result show that, if 1 < p < ∞, then Theorem 6 holds, without the finitenesscondition. The case p = 1, in which we require additional restrictions, will be treatedseparately (see Theorem 8 and Corollary 2 below).

Theorem 7 (Riesz-Fischer). If 1 < p, q < ∞ are Holder conjugate of p, then map (44)is an isometric linear isomorphism.

Proof. Since the map (44) is linear and isometric, its range, which we denote here by Z, isnorm closed in the dual Banach space Lp

K(X,A, µ)∗. By Lemma 2, the set

W = {φY : φ ∈ LpK(X,A, µ)∗, Y ∈ Aµ

fin}

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is norm dense in LpK(X,A, µ)∗, so the proof will be finished once we show that W ⊂ Z. Fix

some K-linear continuous functional φ : LpK(X,A, µ) → K, and some set Y ∈ Aµ

fin, and letus indicate how a function g ∈ Lq

K(X,A, µ) can be constructed, so that

φY (f) =

∫X

fg dµ, ∀ f ∈ LpK(X,A, µ). (60)

Consider the linear continuous functional φ∣∣Y

= φ ◦ TY ∈ LpK(Y,A

∣∣Y, µ

∣∣Y), to which we

can apply Theorem 6 (the fact that Y ∈ Aµfin is essential here), so there exists g0 ∈

LqK(Y,A

∣∣Y, µ

∣∣Y), such that

(φ∣∣Y)(f0) =

∫X

f0g0 d(µ∣∣Y), ∀ f0 ∈ Lp

K(Y,A∣∣Y, µ

∣∣Y). (61)

If we consider the extension of g = TY g0 of g0 to the whole X, then using (61) we getφY (f) = φ(fχY ) = (φ

∣∣Y)(f

∣∣Y) =

∫Y(f

∣∣Y)g0 d(µ

∣∣Y) =

∫Xfg dµ, ∀ f ∈ Lp

K(X,A, µ).

Our next goal is now to identify the dual Banach space L1K(X,A, µ)∗. As discussed at

the beginning of this sub-section the natural candidate for this identification is the isometriclinear map (45).

Theorem 8. If the measure space (X,A, µ) is decomposable, then the map (45) is anisometric isomorphism.

Proof. We already know that (45) is isometric, so we only need to prove surjectivity.Start with a linear continuous functional φ : L1

K(X,A, µ) → K, and let us indicate howg ∈ L∞,loc

K (X,A, µ) can be constructed, so that

φ(f) =

∫X

fg dµ, ∀ f ∈ L1K(X,A, µ). (62)

Fix a decomposition π = (Xi)i∈I for (X,A, µ). For each i ∈ I, by applying Theorem 6 to thefinite measure space (Xi,A

∣∣Xi, µ

∣∣Xi

), there exists gi ∈ L∞K (Xi,A∣∣Xi, µ

∣∣Xi

) ⊂ L∞,locK (Xi,A

∣∣Xi, µ

∣∣Xi

),such that

(φ∣∣Xi

)(f) =

∫Xi

fgi d(µ∣∣Xi

), ∀ f ∈ L1K(Xi,A

∣∣Xi, µ

∣∣Xi

). (63)

Of course, gi : Xi → K can be chosen to be “honestly” bounded, i.e. such that |gi(x)| ≤‖gi‖∞, so using Remark 9, we can assume that

|gi(x)| ≤ ‖φ‖, ∀ i ∈ I, x ∈ Xi. (64)

Using the fact that π is a decomposition (see also sub-section D in BS IV), there exists abounded A-measurable function g : X → K, such that g

∣∣Xi

= gi, ∀ i ∈ I. In order to check

(62), we first observe that, since g is bounded, the linear functional g# : L1K(X,A, µ) 3 f 7−→∫

Xfg dµ ∈ K is continuous, so the desired equality (62) simply reads: φ = g#.Using Remark 8, we know that we have an isometric linear isomorphism

Σ : L1K(X,A, µ)∗ → `∞

(L1

K(Xi,A∣∣Xi, µ

∣∣Xi

)∗)i∈I ,

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so in order to check the equality φ = g#, it suffices to check that Σ(φ) = Σ(g#), which byRemark 9 is equivalent to the equalities

φ∣∣Xi

= (g#)∣∣Xi, ∀ i ∈ I. (65)

The equalities (65) follow trivially from (63) and from the equalities gi = g∣∣Xi

.

Corollary 2. If (X,A, µ) is σ-finite, then the map

# : L∞K (X,A, µ) 3 g 7−→ g# ∈ L1K(X,A, µ)∗

is an isometric linear isomorphism.

Proof. If (X,A, µ) is σ-finite, then it is also decomposable, and the quotient map L∞K (X,A, µ) 3g 7−→ [g]loc ∈ L∞,loc

K (X,A, µ) is an isometric linear isomorphism.

Remark 11. As pointed out in BS IV, for p ∈ [1,∞), the Lp-spaces do not change,if we replace (X,A, µ) with the Caratheodory completion (X,Ac, µc), that is, the naturalinclusion Lp

K(X,A, µ) ⊂ LpK(X,Ac, µc) yields an (isometric) equality

LpK(X,A, µ) = Lp

K(X,Ac, µc), 1 ≤ p <∞.

In particular, when p = 1, a “better candidate” to represent the dual Banach space L1K(X,A, µ)∗

is the Banach space L∞,locK (X,Ac, µc).

As it turns out (see BS IV), if (X,A, µ) is only assumed to be quasi-decomposable, i.e.there exists a µ-integrable partition π = (Xi)i∈I , with all the Xi’s in Aµ

fin, then π is adecomposition for (X,Ac, µc), so by Theorem 8 we obtain an isometric linear isomorphism

L∞,locK (X,Av, µc)

#loc−−→ L1K(X,Ac, µc)∗

∼−→ L1K(X,A, µ)∗.

Theorem 8 has a slight generalization, which goes as follows.

Exercise 9. Prove that if there exists an A-partition (X0, X1) of X, such that

• (X1,A∣∣X1, µ

∣∣X1

) is decomposable,

• (X0,A∣∣X0, µ

∣∣X0

) is degenerate, i.e. µ(A ∩X0) ∈ {0,∞}, ∀A ∈ A,

then (45) is again an isomorphism.

The next five Exercises explain how the dual Banach spaces m bK(X,A), L∞K (X,A, µ)∗

and L∞,locK (X,A, µ)∗ are computed. These characterizations are analogous to the one for

`∞K (S)∗, given in BS II.

Exercises 12-16. Denote by FK(A) the space of all bounded, finitely additive mapsτ : A → K.

12. For any τ ∈ FK(A), define the map |τ | : A → [0,∞] by

|τ |(A) = sup{ n∑

j=1

|τ(Aj)| : n ∈ N, (Aj)nj=1 ⊂ A disjoint,

n⋃j=1

An = A}.

Prove that |τ | is also finitely additive and bounded, thus |τ | ∈ FR(A).

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13. Prove that FK(A) becomes a vector space, when we define

• (τ1 + τ2)(A) = τ1(A) + τ2(A), τ1, τ2 ∈ FK(A),

• (ατ)(A) = ατ(A), τ ∈ FK(A), α ∈ K,

for every A ∈ A. Prove that the map

‖ . ‖var : FK(A) 3 τ 7−→ |τ |(X) ∈ [0,∞)

defines a norm on FK(A), referred to as the variation norm, and furthermore, (FK(A), ‖ . ‖var)is a Banach space.

14. Prove that the map

Φ : m elemK (X,A)× FK(A) 3 (f, τ) 7−→

∑α∈Range f

ατ(f−1({α})

)∈ K

establishes a dual pairing between m elemK (X,A) and FK(A). Prove that, for every

τ ∈ FK(A), the linear functional

τ# : m elemK (X,A) 3 f 7−→ Φ(f, τ) ∈ K

satisfies the inequality|τ#(f)| ≤ ‖τ‖var · ‖f‖sup.

Conclude that, for every τ ∈ FK(A), the functional τ# can be uniquely extended to alinear norm-continuous functional, still denoted by

τ# : m bK(X,A) → K.

(It is understood here that on m bK(X,A) one uses the Banach space norm ‖ . ‖sup.)

15. Prove that the correspondence

FK(A) 3 τ 7−→ τ# ∈m bK(X,A)∗ (66)

is an isometric linear isomorphism. Prove that the inverse of (66) is the map

T : m bK(X,A)∗ 3 φ 7−→ τφ ∈ FK(A), (67)

defined by: τφ(A) = φ(χA), A ∈ A.

16. Consider the surjective contractions

Π∞ : m bK(X,A) → L∞K (X,A, µ),

Π∞,loc : m bK(X,A) → L∞,loc

K (X,A, µ),

and the compositions of their transposes with the map (67):

Σ∞ : L∞K (X,A, µ)∗Π∗∞−−→m b

K(X,A)∗T−→ FK(A),

Σ∞,loc : L∞,locK (X,A, µ)∗

Π∗∞,loc−−−−→m bK(X,A)∗

T−→ FK(A).

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Prove that both Σ∞ and Σ∞,loc are isometric linear maps, and their ranges are:

Range Σ∞ = {τ ∈ FK(A) : τ(A) = 0, for all µ-neglijeable sets A ∈ A},Range Σ∞,loc = {τ ∈ FK(A) : τ(A) = 0, for all locally µ-neglijeable sets A ∈ A}.

Conclusion. If we denote the Ranges of Σ∞ and Σ∞,loc by F∞K (A, µ) and F∞,locK (A, µ),

respectively, then we have two isometric linear isomorphisms Σ∞ : L∞K (X,A, µ)∗∼−→ F∞K (A, µ)

and Σ∞,loc : L∞K (X,A, µ)∗∼−→ F∞,loc

K (A, µ).

Comment. In connection with Exercise 14, given τ ∈ FK(A) and f ∈ m bK(X,A), the

quantity τ#(f) is sometimes denoted by SX f dτ . This notation was first introduced insub-section D from BS II. Everything discussed there is a special case of the results fromExercises 12-16. In particular, the space FK(S) introduced in BS II coincides with FK(P(S)).

We conclude with a series of Exercises that contain an important construction in Prob-ability Theory.

Exercises 17-20. Assume a second σ-algebra B ⊂ A is given, which is equipped withthe measure µ

∣∣B.

17. Assume 1 < q <∞, and let p = q/(q− 1) be the Holder conjugate of q. Prove that forevery f ∈ Lq

K(X,A, µ), there exists a (unique µ-a.e.) function f ∈ LqK(X,B, µ

∣∣B) such

that ∫X

fg dµ =

∫X

fg, dµ, ∀ g ∈ LpK(X,B, µ

∣∣B).

Prove that the map

EqA↓B : Lq

K(X,A, µ) 3 f 7−→ f ∈ LqK(X,B, µ

∣∣B) (68)

has the following properties:

(i) EqA↓B is linear continuous, and has ‖Eq

A↓B‖ ≤ 1;

(ii) EqA↓B(f) = f , ∀ f ∈ Lq

K(X,B, µ∣∣B);

(iii) EqA↓B(h · f) = h · Eq

A↓B(f), ∀h ∈ L∞,locK (X,B, µ

∣∣B), f ∈ L

qK(X,A, µ).

(Hint: Consider the inclusion J : LpK(X,B, µ

∣∣B) ↪→ Lp

K(X,A, µ), and its transpose

J∗ : LpK(X,A, µ)∗ → Lp(X,B, µ

∣∣B)∗, and apply Theorem 7, by which the two dual

Banach spaces are identified with the corresponding Lq-spaces.)

18. Prove that, in the case q = 2(= p), the map E2A↓B coincides with the orthogonal

projection of L2K(X,A, µ) onto the closed linear subspace L2

K(X,B, µ∣∣B).

19. Assume (X,B, µ∣∣B) is decomposable. Prove that for every f ∈ L1

K(X,A, µ), there exists

a (unique µ-a.e.) function f ∈ L1K(X,B, µ

∣∣B) such that, whenever g ∈ mK(X,B) has

supp g ∈ Bµσ-fin and fg ∈ L1

K(X,A, µ), it follows that fg ∈ L1K(X,B, µ

∣∣B), and one has

the equality ∫X

fg dµ =

∫X

fg, dµ.

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Prove that the map

E1A↓B : L1

K(X,A, µ) 3 f 7−→ f ∈ L1K(X,B, µ

∣∣B) (69)

has the following properties:

(i) E1A↓B is linear continuous, and has ‖E1

A↓B‖ ≤ 1;

(ii) E1A↓B(f) = f , ∀ f ∈ L1

K(X,B, µ∣∣B);

(iii) E1A↓B(h · f) = h · E1

A↓B(f), ∀h ∈ L∞,locK (X,B, µ

∣∣B), f ∈ L

1K(X,A, µ).

(Hint: For every f ∈ L1K(X,A, µ), define the map νf : B 3 B 7−→

∫XfχB dµ ∈ K,

and show that νf is a K-valued measure on B, with νf � µ∣∣B. Use Radon-Nikodim

Theorem 5.)

20. Assume either

(a) q = 1 and (X,B, µ∣∣B) is decomposable, or

(b) 1 < q <∞ (without any additional restrictions).

Prove that the map (68) has the following properties:

(i) EqA↓B is positive, i.e. if f ∈ Lq

R(X,A, µ) is such that f ≥ 0, then EqA↓B(f) ≥ 0;

(i) EqA↓B is faithful, i.e. if f ∈ Lq

R(X,A, µ) is such that f ≥ 0 and EqA↓B(f) = 0, then

f = 0.

The maps EqA↓B : Lq

K(X,A, µ) → LqK(X,B, µ

∣∣B), discussed above, are called the conditional

expectation maps.

Comment. It is possible to construct a conditional expectation map

E∞,locA↓B : L∞,loc

K (X,A, µ) → L∞,locK (X,B, µ

∣∣B),

in a manner analogous to the construction from Exercise 17. For such a construction, how-ever, one needs some additional assumptions on the measure spaces (X,A, µ) and (X,B, µ

∣∣B),

which ensure that

(i) L∞,locK (X,B, µ

∣∣B) gets identified as a closed linear subspace in L∞,loc

K (X,A, µ);

(ii) L∞,locK (X,B, µ

∣∣B) ' L1

K(X,B, µ∣∣B)∗.

One sufficient condition for (i) and (ii) is the existence of a quasi-decomposition π = (Xi)i∈I

for (X,A, µ), such that

• all Xi belong to B,

• π constitutes a decomposition of (X,B, µ∣∣B).

The details, as well as the correct formulation of Exercise 17, are left to the reader.

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