BS IV c©Gabriel Nagy

Banach Spaces IV:

Banach Spaces of Measurable Functions

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

In this section we discuss another important class of Banach spaces arising from Measureand Integration theory.

Convention. Throughout this note K will be one of the fields R or C, and all vectorspaces are over K.

A. Review of Measure and Integration Theory

In this section we recollect some basic notions and results, which can be found in mostReal Analysis textbooks. The exposition is in the form of a sequence of Definitions andFacts.

Definitions. A measurable space is a pair (X,A) consisting of a non-empty set Xand a σ-algebra on X. Our basic example will be of the form (Ω, Bor(Ω)), where Ω isa topological Hausdorff space (most often locally compact), and Bor(Ω) is the Borel σ-algebra. Given another measurable space (Y,B), a map f : X → Y is said to be a measurabletransformation from (X,A) to (Y,B), if f−1(B) ∈ A, ∀B ∈ B. In practice, if one choosesa collection M ⊂ B, which generates B as a σ-algebra1, then a sufficient condition forf : (X,A) → (Y,B) to be a measurable transformation is: f−1(M) ∈ A, ∀M ∈M.

If we specialize to the case Y = K, then a function f : (X,A) → K is declared measurable,if it defines a measurable transformation f : (X,A) → (K, Bor(K)). In the complex case, thisis equivalent to the condition that both Re f, Im f : (X,A) → R are measurable. Since theBorel σ-algebra Bor(R) is generated by various collections of intervals, such as for instanceJ = (s,∞) : s ∈ R, measurability of f : (X,A) → R is equivalent to f−1(J) ∈ A.∀ J ∈ J . On occasion one also employs measurable functions which take values in one of theextended intervals [−∞,∞], [0,∞], or [−∞, 0], with the understanding that if T denotesany one of these spaces, one again uses the Borel σ-algebra Bor(T ). (Of course any such Tis homeomorphic to [0, 1].)

We denote by mK(X,A) the space of all K-valued measurable functions on (X,A). Itis well known that mK(X,A) is a K-algebra, when equipped with point-wise addition andmultiplication. As usually, when K = C, the subscript will be omitted from the notations.

The space of [0,∞]-valued measurable functions on (X,A) is denoted by m+(X,A).The simplest measurable functions on a measurable space (X,A) are the indicator func-

tions χA, of sets A ∈ A. (Recall that χA(x) = 1, if x ∈ A, and χA(x) = 0, if x 6∈ A.) Thenext class of measurable functions are those in spanKχA : A ∈ A ⊂ mK(X,A). These

1 This means that B is the intersection of all σ-algebra that contain M.

1

functions are referred to as elementary measurable functions, and the above mentioned spanis denoted by m elem

K (X,A). An equivalent description of this space is:

m elemK (X,A) =

f : X → K : Range f is finite, and f−1(α) ∈ A, ∀α ∈ Range f

.

This characterization allows one to extend the notion of elementary measurable functionsfor functions f : X → Y , where Y is an arbitrary set. In particular, if we let Y = [0,∞],the corresponding space of functions is denoted by m elem

+ (X,A).The following result is a very basic one in Measure Theory.

Fact 1. Let (X,A) be a measurable space, and let T one of the spaces R, [−∞,∞], orC. Define K to be either R, when T is R or [−∞,∞], or C, when T = C. For a functionf : X → T , the following are equivalent:

• f : (X,A) → T is measurable;

• there exists a sequence (fn)∞n=1 ⊂m elemK (X,A), such that

limn→∞

fn(x) = f(x), ∀x ∈ X.

In the case when T = [−∞,∞] or T = R, the functions fn can be chosen such that

infy∈X

f(y) ≤ fn(x) ≤ supy∈X

f(y), ∀x ∈ X.

Furthermore, in this case one has the following implications:

(i) if f(x) > −∞, ∀x, the sequence (fn)∞n=1 can be chosen to be non-decreasing;

(ii) if f(x) < ∞, ∀x, the sequence (fn)∞n=1 can be chosen to be non-increasing.

Definitions. Given a measurable space (X,A), an “honest” measure on A is a σ-additivemap µ : A → [0,∞], with µ(∅) = 0. The σ-additivity condition means that, whenever(An)∞n=1 ⊂ A is a sequence of disjoint sets, it follows that µ

( ⋃∞n=1 An

)=

∑∞n=1 µ(An).

When such a µ is identified, the triple (X,A, µ) is referred to as a measure space.Given a measure space (X,A, µ), we define two maps

I+,elemµ : m elem

+ (X,A) → [0,∞],

I+µ : m+(X,A) → [0,∞],

by the formulas:

I+,elemµ (f) =

∑α∈Range f

α · µ(f−1(α)

), f ∈m elem

+ (X,A),

I+µ (g) = sup

I+,elemµ (f) : f ∈m elem

+ (X,A, 0 ≤ f ≤ g.

(In the definition of I+,elemµ one uses the convention 0 ·∞ = ∞·0 = 0.) The map I+

µ is calledthe positive µ-integral.

Fact 2. Let (X,A, µ) be a measure space. The positive integral I+µ : m+(X,A) → [0,∞]

has the following properties:

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(i) I+µ is positive-linear, in the sense that

• I+µ (f + g) = I+

µ (f) + I+µ (g), ∀ f, g ∈m+(X,A);

• I+µ (αf) = αI+

µ (f), ∀ f ∈m+(X,A), α ∈ [0,∞];

in particular, I+µ is monotone, in the sense that, for f, g ∈ m+(X,A) one has the

implication f ≥ g ⇒ I+µ (f) ≥ I+

µ (g).

(ii) If f ∈m elem+ (X,A), then I+

µ (f) = I+,elemµ (f). In particular, one has:

I+µ (χA) = µ(A), ∀A ∈ A.

Another key feature of the positive integral, which is derived from its definition, is:

Fact 3 (Lebesgue’s Monotone Convergence Theorem). If (fn)∞n=1 ⊂ m+(X,A) is anon-decreasing sequence, then the function f : X → [0,∞] defined by f(x) = limn→∞ fn(x),x ∈ X, belongs to m+(X,A) and has its positive integral given by I+

µ (f) = limn→∞ I+µ (fn).

Definitions. Suppose (X,A, µ) is a measure space. A function f ∈ mK(X,A) is saidto be µ-integrable, if the function |f | ∈m+(X,A) has finite positive integral: I+

µ (|f |) < ∞.The space of µ-integrable K-valued functions is denoted by L1

K(X,A, µ). (As before, ifK = C, the subscript will be omitted.) The space of µ-integrable functions f : X → [0,∞)is denoted by L1

+(X,A, µ).

Comment. As it turns out, µ-integrability makes sense even for measurable functionsf : (X,A) → [−∞,∞]. The space of such functions is denoted by L1

R(X,A, µ). It turns out,however, that if f ∈ L1

R(X,A, µ), then the set A = x ∈ X : f(x) = ±∞ has µ(A) = 0.The role of sets of measure zero will be clarified a little later.

Fact 4. Assume (X,A, µ) is a measure space.

(i) The space L1K(X,A, µ) is a K-linear subspace in mK(X,A), which can also be described

as: L1K(X,A, µ) = spanKL1

+(X,A, µ).

(ii) A measurable function f : (X,A) belongs to L1(X,A, µ), if and only of both Re f andIm f belong to L1

R(X,A, µ).

(iii) There exists a unique K-linear map Iµ : L1K(X,A, µ) → K, such that Iµ(f) = I+

µ (f),∀ f ∈ L1

+(X,A, µ).

(iv) For an elementary function f ∈m elemK (X,A), the following are equivalent:

• f is µ-integrable;

• ν(f−1(α)

)< ∞, ∀α ∈ Range f ;

• f ∈ spanKχA : A ∈ A, µ(A) < ∞.

The linear map Iµ : L1K(X,A, µ) → K is referred to as the µ-integral. The traditional

notation is to write∫

Xf dµ instead of Iµ(f). The space of elementary integrable functions,

described in Fact 4 (iii) as

m elemK (X,A) ∩ L1

K(X,A, µ) = spanKχA : A ∈ A, µ(A) < ∞,

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will be denoted by LelemK (X,A, µ). Given A1, . . . , An ∈ A, with µ(A1), . . . , µ(An) < ∞, and

α1, . . . , αn ∈ K, the the elementary integrable function α1χA1 + · · ·+αnχAn has its µ-integralcomputed as: ∫

X

(α1χA1 + · · ·+ αnχAn) dµ = α1µ(A1) + · · ·+ αnµ(An). (1)

The formula (1) is referred to as the Elementary Integral Formula.Besides linearity, the integration map also has the continuity property:∣∣∣∣ ∫

X

f dµ

∣∣∣∣ ≤ ∫X

|f | dµ, ∀ f ∈ L1K(X,A, µ). (2)

The integral notation will also be used in the case of the positive integral, that is, iff ∈ m+(X,A), the positive integral I+

µ (f) will still be denoted by∫

Xf dµ, although this

quantity may equal∞. Using this notation, the space of integrable functions can be presentedas: L1

K(X,A, µ) =f ∈ mK(X,A) :

∫X|f | dµ < ∞

. The Elementary Integral Formula

(1) is still valid, for arbitrary A1, . . . , An ∈ A and α1, . . . , αn ∈ [0,∞].Integrability is preserved under neglijeable perturbations. In order to clarify this state-

ment, one employs the following terminology.

Definitions. Suppose (X,A, µ) is a measure space, We say that a set N ∈ A µ-neglijeable, if µ(N) = 0.

Assume one considers some “measurable” property (P), which refers to points in x. Herethe term “measurable” means that the set x ∈ X : x has property (P) (and of course itscomplement) belongs to A. We say that (P) holds almost µ-everywhere, if the set

N = x ∈ X : x does not have property (P)

is µ-neglijable. In practice, one considers some very specific properties, based on somerelation r on one of the spaces T = R, C, [−∞,∞], and two measurable functions f, g :(X,A) → T , so we write

f r g, µ-a.e.,

to indicate that f(x)r g(x), almost µ-everywhere.

Fact 5. Let (X,A, µ) be a measure space, let T be one of the spaces R, C, or [−∞,∞],and let f : (X,A) → T be a measurable function.

(i) If f = 0, µ-a.e., then f is integrable, and∫

Xf dµ = 0.

(ii) Conversely, if f is integrable, and∫

X|f | dµ = 0, then f = 0, µ-a.e..

As a consequence of the above properties, it follows that the space

NK(X,A, µ) = f ∈mK(X,A) : f = 0, µ-a.e.

is a linear subspace in L1K(X,A, µ).

Integrable functions can be produced via dominance conditions, as indicated by thefollowing simple observation.

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Fact 6. Let (X,A, µ) be a measure space, let T be one of the spaces R, C, or [−∞,∞],and let f : (X,A) → T be a measurable function. Assume g : (X,A) → [0,∞] is integrable,and |f | ≤ g, µ-a.e. Then f is integrable, and:

∣∣ ∫X

f dµ∣∣ ≤ ∫

Xg dµ.

The ultimate refinement of the above statement is:

Fact 7 (Lebesgue’s Dominated Convergence Theorem). Let (X,A, µ) be a measure space,let T be one of the spaces R, C, or [−∞,∞], and let f, fn : (X,A) → T , n ≥ 1, be measurablefunctions. Assume the following.

(i) There exists an integrable function g : (X,A) → [0,∞], such that

|fn| ≤ g, µ-a.e., ∀n ≥ 1.

In particular, all fn, n ≥ 1, are integrable.

(ii) The equality limn→∞ fn(x) = f(x) holds almost µ-everywhere, that is, there exists someµ-neglijeable set B ∈ A, such that limn→∞ fn(x) = f(x), ∀x ∈ X r B.

Then f is integrable, and∫

Xf dµ = limn→∞

∫X

fndµ.

We end our review by recalling some standard Measure Theory terminology that will beemployed a little later.

Definitions. Given a measure space (X,A, µ), and some A ∈ A,

• we say that A is µ-finite, if µ(A) < ∞;

• we say that A is µ-σ-finite, if there exists a sequence2 (An)∞n=1 ⊂ A of µ-finite sets,such that A =

⋃∞n=1 Anµ(A) < ∞.

The whole measure space (X,A, µ) is declared finite (resp. σ-finite), if the total set X isµ-finite (resp. µ-σ-finite). In this case it follows that all A ∈ A are µ-finite (resp. µ-σ-finite).

Exercises 1-2. Let (X,A, µ) be a measure space. For a measurable function f : (X,A),let Sf denote the set x ∈ X : f(x) 6= 0. (Since f is measurable, Sf belongs to A).

1.♥ Prove that if f ∈ L1K(X,A, µ), then the set Sf is µ-σ-finite.

2.♥ Prove that, if f is bounded, and Sf is µ-finite, then f ∈ L1K(X,A, µ).

B. The Lp-spaces

Throughout this sub-section we fix a measure space (X,A, µ).

Notations. For every p ∈ [1,∞) we define the space

LpK(X,A, µ) =

f ∈mK(X,A) :

∫X|f |p dµ < ∞

.

2 In fact the sequence can be chosen to consist of disjoint sets.

5

(The integral∫

X|f |p dµ stands for the positive integral I+

µ .) We also define the map Qp :LK(X,A, µ) → [0,∞) by

Qp(f) =

[ ∫X

|f |p dµ

]1/p

, ∀ f ∈ LpK(X,A, µ).

It is pretty clear that, when p = 1, we are dealing with the same space as the one introducedin sub-section A, which is already known to be a K-vector space. Furthermore, in this casethe map Q1 is obviously a seminorm on L1

K(X,A, µ). For 1 < p < ∞, the same will be true,as a result of the following result.

Theorem 1. Assume 1 < p, q < ∞ are Holder conjugate, i.e. 1p

+ 1q

= 1.

(i) LpK(X,A, µ) is a K-linear subspace in mK(X,A).

(ii) If f ∈ LpK(X,A, µ) and g ∈ Lq

K(X,A, µ), then fg ∈ L1K(X,A, µ), and furthermore one

has the inequalityQ1(fg) ≤ Qp(f) ·Qq(g). (3)

(iii) If f ∈ LpK(X,A, µ), then

Qp(f) = max ∣∣∫

Xfg dµ

∣∣ : g ∈ LqK(X,A, µ), Qq(g) ≤ 1

. (4)

(iv) Qp : LpK(X,A, µ) → [0,∞) is a seminorm on Lp

K(X,A, µ).

Proof. (i). It is pretty obvious that, if f ∈ LpK(X,A, µ), then αf ∈ Lp

K(X,A, µ), ∀α ∈ K.Assume now f1, f2 ∈ Lp

K(X,A, µ), and let us show that f1 + f2 ∈ LpK(X,A, µ). This is,

however, pretty obvious, since the inequality3 |f1 + f2|p ≤ 2p|f1|p + 2p|f2|p implies (bydominance) the integrability of |f1 + f2|p.

(ii). Fix f ∈ LpK(X,A, µ) and g ∈ Lq

K(X,A, µ). Use Fact 1, to choose two non-decreasingsequences (Fn)∞n=1 and (Gn)∞n=1 in m elem

+ (X,A), such that limn→∞ Fn(x) = |f(x)|p andlimn→∞ Gn(x) = |g(x)|q, ∀x ∈ X. Of course, since the two sequences are non-decreasing,we have the inequalities 0 ≤ F1 ≤ F2 ≤ · · · ≤ |f |p and 0 ≤ G1 ≤ G2 ≤ · · · ≤ |g|q, so all

the Fn’s and the Gn’s a integrable. If we define fn = G1/pn and gn = G

1/qn , then these will

also be elementary integrable, and will satisfy the inequalities 0 ≤ f1 ≤ f1 ≤ · · · ≤ |f |,0 ≤ g1 ≤ g1 ≤ · · · ≤ |g|, as well as limn→∞ fn(x) = |f(x)| and limn→∞ gn(x) = |g(x)|,∀x ∈ X. Since we clearly have 0 ≤ f1g1 ≤ f2g2 ≤ · · · ≤ |fg|, as well as limn→∞ fn(x)gn(x) =|f(x)g(x)|, ∀x ∈ X, by Lebesgue’s Monotone Convergence Theorem it follows that the(positive) integral

∫X|fg| dµ can be computed as∫

X

|fg| dµ = limn→∞

∫X

fngn dµ. (5)

Fix for the moment n, and let us write4 fn =∑k

j=1 αjχAjand gn =

∑kj=1 βjχAj

with all A’sµ-finite, and all α’s and β’s non-negative, so that∫

X

fngn dµ =k∑

j=1

αjβjµ(Aj) =k∑

j=1

xjyj, (6)

3 Use the inequality a + q ≤ 2 maxa, b, which implies (a + b)p ≤ 2p(ap + bp), ∀ a, b ≥ 0.4 One can write separately fn =

∑ri=1 αiχDi with D1, . . . , Dr disjoint µ-finite, and gn =

∑sm=1 βmχEm

with E1, . . . , Es disjoint µ-finite. The A’s will the intersections Di ∩ Em, which will also be µ-finite.

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where xj = αj[µ(Aj)]1/p and yj = βj[µ(Aj)]

1/q, j = 1, . . . , k. Using the (classical) HolderInequality, we have

k∑j=1

xjyj ≤[ k∑

j=1

xpj

]1/p

·[ k∑

j=1

yqj

]1/q

. (7)

Notice now that

k∑j=1

xpj =

k∑j=1

αpjµ(Aj) =

∫X

Fn dµ ≤∫

X

|f |p dµ = Qp(f)p,

k∑j=1

yqj =

k∑j=1

βqj µ(Aj) =

∫X

Gn dµ ≤∫

X

|g|q dµ = Qq(g)q,

so now if we combine (7) and (6), we obtain∫X

fngn dµ ≤ Qp(f) ·Qq(g). (8)

Since the inequality (8) holds for all n, using (5) it follows that∫X

|fg| dµ ≤ Qp(f) ·Qq(g).

Of course, this proves that fg is indeed integrable, and also satisfies (3).(iii). Fix f ∈ Lp

K(X,A, µ). Since, by (i), for every g ∈ LqK(X,A, µ) we clearly have the

inequality∣∣ ∫

Xfg dµ

∣∣ ≤ ∫X|fg| dµ ≤ Qp(f) ·Qq(g), it follows that

sup ∣∣∫

Xfg dµ

∣∣ : g ∈ LqK(X,A, µ), Qq(g) ≤ 1

≤ Qp(f),

so in order to prove (4), all we need to do is to produce some g ∈ LqK(X,A, µ), with Qq(g) ≤ 1,

such that ∣∣∣∣ ∫X

fg dµ

∣∣∣∣ = Qp(f). (9)

The case when Qp(f) = 0 is trivial (take g = 0), so we are going to assume that Qp(f) > 0.Define the function g0 : X → K by:

g0(x) =

|f(x)|p

f(x)if f(x) 6= 0

0 if f(x) = 0

Obviously, g0 is measurable, and satisfies the identity

|g0|q = |f |pq−q = |f |p,

so g0 belongs to LqK(X,A, µ), and

Qq(g0) =

[ ∫X

|f |p dµ

]1/q

= Qp(f)p/q. (10)

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Moreover, since fg0 = |f |p, we also have∣∣∣∣ ∫X

fg0 dµ

∣∣∣∣ =

∫X

|f |p dµ = Qp(f)p, (11)

so if we define g = Qq(g0)−1g0, then g also belongs to Lq

K(X,A, µ), but now has Qq(g) = 1,and by (10) and (11) it satisfies∣∣∣∣ ∫

X

fg dµ

∣∣∣∣ = Qq(g0)−1Qp(f)p = Qp(f)−p/qQp(f)p = Qp(f).

(iv). The identity

Qp(αf) = |α| ·Qp(f), ∀ f ∈ LpK(X,A, µ), α ∈ K

is trivial from the definition of Qp, so all that is left to prove is the triangle inequality

Qp(f1 + f2) ≤ Qp(f1) + Qp(f2), ∀ f1, f2 ∈ LpK(X,A, µ).

This inequality can be derived from (ii) and (iii) as follows. Fix some g ∈ LqK(X,A, µ), such

that Qq(g) ≤ 1 and Qp(f1 + f2) =∣∣ ∫

X(f1 + f2)g dµ

∣∣, and use linearity and continuity of theintegral to obtain

Qp(f1 + f2) ≤∫

X

|(f1 + f2)g| dµ ≤[ ∫

X

|f1g| dµ

]+

[ ∫X

|f2g| dµ

]=

= Q1(f1g) + Q1(f2g) ≤ Qp(f1) ·Qq(g) + Qp(f2) ·Qq(g) ≤ Qp(f1) + Qp(f2).

Definitions. Fix p ∈ [1,∞). By Fact 6, for f ∈ LpK(X,A, µ), one has the equivalence∫

X|f |p dµ = 0 ⇔ f = 0, µ-a.e., so one has the the equality

f ∈ LpK(X,A, µ) : Qp(f) = 0 = NK(X,A, µ). (12)

Define the quotient space

LpK(X,A, µ) = Lp

K(X,A, µ)/NK(X,A, µ).

In other words, LpK(X,A, µ) is the collection of equivalence classes associated with the re-

lation “=, µ-a.e.” For a function f ∈ LpK(X,A, µ) we denote by [f ] its equivalence class in

LpK(X,A, µ). So the equality [f ] = [g] is equivalent to f = g, µ-a.e. By (12) the seminorm

Qp induces a norm on LpK(X,A, µ), which we denote by ‖ . ‖p, so by definition we have

‖[f ]‖p = Qp(f).

Conventions. We are going to abuse a bit the notation, by writing “f ∈ LpK(X,A, µ)” if

f belongs to LpK(X,A, µ). (We will always have in mind the fact that this notation signifies

that f is almost uniquely determined.) Likewise, instead of writing ‖[f ]‖p, we are going towrite ‖f‖p, so that from now on Qp(f) will be replaced by ‖f‖p. With this notations, wenow have f ∈ Lp

K(X,A, µ) ⇔ |f |p ∈ L1K(X,A, µ), in which case ‖f‖p = ‖|f |p‖1.

The integration map∫

X: L1

K(X,A, µ) → K gives rise to a correctly defined map, whichwe denote by the same symbol, so that by the continuity condition we can write:∣∣∣∣ ∫

X

f dµ

∣∣∣∣ ≤ ∫X

|f | dµ = ‖f‖1, ∀ f ∈ L1K(X,A, µ). (13)

With these notations, statements (ii) and (iii) in Theorem 1 can be restated as follows.

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• If f ∈ LpK(X,A, µ) and g ∈ Lq

K(X,A, µ), then fg ∈ L1K(X,A, µ) and

‖fg‖1 ≤ ‖f‖p · ‖g‖q. (14)

In particular, using (13), one also has∣∣∣∣ ∫X

fg dµ

∣∣∣∣ ≤ ‖f‖p · ‖g‖q. (15)

• If f ∈ LpK(X,A, µ), then

‖f‖p = max∣∣∫

Xfg dµ

∣∣ : g ∈ LqK(X,A, µ), ‖g‖q ≤ 1

. (16)

The inequality (15) is referred to as the Integral Holder Inequality.Remarks 1-2. Let p, q ∈ (1,∞) be two Holder conjugate numbers.

1. One has a bilinear map

Φ : LpK(X,A, µ)× Lq

K(X,A, µ) 3 (f, g) 7−→∫

X

fg dµ ∈ K,

which satisfies the inequality

|Φ(f, g)| ≤ ‖f‖p · ‖g‖q. (17)

Using (16) it follows that Φ establishes a dual pairing (see DT I), which is referred toas the Holder dual pairing.

2. Using the Holder dual pairing, one defines, for every f ∈ LpK(X,A, µ) the linear func-

tional f# : LqK(X,A, µ) 3 g 7−→

∫X

fg dµ ∈ K. By (17), the functional f# is norm-continuous, and by (16) it follows that ‖f#‖ = ‖f‖p. Therefore, the Holder dualpairing produces an isometric linear map

# : LpK(X,A, µ) 3 f 7−→ f# ∈ Lq

K(X,A, µ)∗. (18)

Comment. It will be shown in HS II that the map (18) is in fact a linear isomorphism.In particular, from this identification Theorem 2 below would follow. One can, however,obtain the same result directly, as shown in the direct proof presented here.

Theorem 2. All spaces LpK(X,A, µ), p ∈ [1,∞), are Banach spaces.

Proof. Fix p ∈ [1,∞). In order to prove that LpK(X,A, µ) we are going to verify the Summa-

bility Test, so we start with some sequence (fn)∞n=1 in LpK(X,A, µ), such that the quantiy

S =∑∞

n=1 ‖fn‖p is finite, and we show that the series∑∞

n=1 fn is convergent in LpK(X,A, µ).

Define the function h : X → [0,∞] by h(x) =∑∞

n=1 |fn(x)|. Of course, we can writeh(x) = limn→∞ hn(x), where hn = |f1|+ · · ·+ |fn|. (Of course all h’s are measurable.)

On the one hand, using the triangle inequality in LpR(X,A, µ) we have∫

X

hpn =

∥∥|f1|+ · · ·+ |fn|∥∥p

p≤

[‖f1‖p + · · ·+ ‖fn‖p

]p ≤ Sp, ∀n. (19)

9

On the other hand, since we clearly have 0 ≤ h1(x)p ≤ h2(x)p ≤ · · · ≤ h(x)p, as well aslimn→∞ hn(x)p = h(x)p, by Lebesgue’s Monotone Convergence Theorem it follows that∫

X

hp dµ = limn→∞

∫X

hpn dµ,

so going back to (19) it follows that∫

Xhp dµ ≤ Sp < ∞, so h belongs to Lp. Since hp

is integrable, it follows that set A = x ∈ X : h(x) = ∞ ∈ A has µ(A) = 0. (Theobvious inequality tχA ≤ hp will force tµ(A) ≤ Sp, ∀ t > 0, and this can only happen whenµ(A) = 0.) For x ∈ X rA, the series

∑∞n=1 |fn(x)| is convergent (to h(x) < ∞), so the series∑∞

n=1 fn(x) is also convergent (in K). Define then the measurable function g : X → K by

g(x) =

∑∞n=1 fn(x) if x ∈ X r A

0 if x ∈ A

so that if we define the partial sums gn = f1 + · · ·+ fn, we have limn→∞ gn(x) = g(x), µ-a.e.We now wish to prove that g ∈ Lp and that limn→∞ ‖g − gn‖p = 0. To prove that g

is in Lp we notice that, since we obviously have |gn| ≤ hn ≤ h, we also get |gn|p ≤ hp,µ-a.e., and since limn→∞ |gn(x)|p = |g(x)|p, µ-a.e., we can apply Lebesgue’s DominatedConvergence Theorem, to conclude that |g|p is integrable. Likewise, for any n, k ≥ 1, wehave |gn+k − gn| ≤ hn+k − hn ≤ h− hn ≤ h, taking limk→∞ and the pth power yields

|g − gn|p ≤ hp, µ-a.e..

Since limn→∞ |g(x)−gn(x)|p = 0, µ-a.e., again by Lebesgue’s Dominated Convergence Theo-rem we conclude that limn→∞

∫X|g−gn|p dµ = 0, which means precisely that ‖g−gn‖p

p → 0,thus proving the desired conclusion.

Example 1. Start with some non-empty set X, let A = P(X), and let µ be the countingmeasure, defined as

µ(A) =

Card A if A is finite∞ if A is infinite

Then the Banach space LpK(X,A, µ) is precisely the space `p

K(X) discussed in BS I.

Example 2. Start with some non-empty set X, let A = P(X), and but let µ be thedegenerate counting measure, defined as

µ(A) =

0 if A = ∅∞ if A 6= ∅

Then LpK(X,A, µ) = 0. As this example suggests, unless some local finiteness condition is

assumed, integration theory can be extremely trivial.

Notation. Let (X,A, µ) be a measure space. It is not hard to see that LelemK (X,A, µ)

– the space of elementary integrable functions – is contained (as a linear subspace) inLp

K(X,A, µ), for every p ∈ [1,∞). Therefore one has a correctly defined linear map Tp :Lelem

K (X,A, µ) → LpK(X,A, µ). As it turns out, Ker Tp is independent of p, since it can

be described as NK(X,A, µ) ∩ LelemK (X,A, µ). This means that the spaces Range Tp are

10

all isomorphic, and for this reason all of them will be denoted by LelemK (X,A, µ), and we

will understand that this space is contained in all Lp-spaces. Using the (abusive) notationconvention introduced earlier, we can think

LelemK (X,A, µ) = spanKχA : A ∈ A, µ(A) < ∞, (20)

with the understanding that

• two functions f and g in the linear span on the right-hand of (20) are the same, iff = g, µ-a.e.

• the linear span on the right-hand side of (20) can be computed in LpK(X,A, µ), for any

p ∈ [1,∞).

Exercise 3.♥ With these conventions, prove that LelemK (X,A, µ) is dense in Lp

K(X,A, µ),in the norm topology, for every p ∈ [1,∞).

Exercise 4. Prove that, if p, q, r ∈ [1,∞) are such that 1p+ 1

q= 1

r, and if f ∈ Lp

K(X,A, µ)

and g ∈ LqK(X,A, µ), then fg ∈ Lr

K(X,A, µ), and furthermore, one has the inequality‖fg‖r ≤ ‖f‖p · ‖g‖q.

Exercise 5.♥ Assume (X,A, µ) is a finite measure space. Prove that, if p, q ∈ [1,∞) aresuch that p ≥ q, then Lp

K(X,A, µ) ⊂ LqK(X,A, µ). In particular, this gives rise to a linear

injective map J : LpK(X,A, µ) → Lq

K(X,A, µ). Prove that J is continuous.

C. The spaces m b, L∞ and Lloc,∞

The Holder dual pairing between Lp and Lq can be be extended to include the case p = 1.There are several “candidate” spaces that can be paired with L1. What we seek is a Banachspace X of measurable functions on (X,A), such that

(i) for any f ∈ L1K(X,A, µ) and any g ∈ X , the function fg is in L1

K(X,A, µ), andfurthermore one has the inequality ‖fg‖1 ≤ ‖f‖1 · ‖g‖;

(ii) the correspondence

Φ : L1K(X,A, µ)×X 3 (f, g) 7−→

∫X

fg dµ ∈ K

establishes a dual pairing. (By (i) the map Φ is automatically bilinear.)

In this sub-section we will investigate three natural choices for X , which satisfy (i).Concerning condition (ii), in the order we introduce these spaces, they will behave betterand better.

Definitions. The most obvious one is the space m bK(X,A) of all bounded measurable

functions g : (X,A) → K, which is a Banach space, when equipped with the norm ‖ . ‖sup,defined by

‖g‖sup = supx∈X

|g(x)|, g ∈m bK(X,A).

11

(Of course, m bK(X,A) is contained in the Banach space `∞K (X) of all bounded functions

g : X → K, and the norm ‖ . ‖sup is precisely the induced norm. Since the uniform limit of asequence of bounded measurable functions is again bounded and measurable, it follows thatm b

K(X,A) is norm-closed in `∞K (X), therefore a Banach space.)Of course, if f ∈ L1

K(X,A, µ) and g ∈ m bK(X,A), then there exists a constant, for

instance C = ‖g‖sup, such that |fg| ≤ C|f |. This will force fg ∈ L1K(X,A), as well as the

inequality ‖fg‖1 ≤ C‖f‖1, so one indeed has a bilinear map

Φ : L1K(X,A, µ)×m b

K(X,A) 3 (f, g) 7−→∫

X

fg dµ ∈ K, (21)

which satisfies the inequality

|Φ(f, g)| ≤ ‖f‖1 · ‖g‖sup, ∀ f ∈ L1K(X,A, µ), g ∈m b

K(X,A). (22)

Exercise 6.♥. Prove that, for any f ∈ L1K(X,A, µ), one has the equality

‖f‖1 = max|Φ(f, g)| : g ∈m b

K(X,A), ‖g‖sup ≤ 1.

Comment. In general, the map (21) does not give rise to a dual pairing. Since this mapis bilinear, it still gives rise to two linear continuous maps

# : L1K(X,A, µ) 3 f 7−→ f# ∈m b

K(X,A)∗, (23)

# : m bK(X,A) 3 g 7−→ g# ∈ L1

K(X,A, µ)∗ (24)

given by f#(g) = g#(f) =∫

Xfg dµ.

By (22) both maps (23) and (24) are contractive. By Exercise 6, the map (23) is infact isometric. The only drawback of the choice of m b as our “candidate” is the fact thatthe map (24) might fail to be injective. (Ultimately, with an adequate space X in placeof m b

K(X,A) we are going to make both (23) and (24) isometric.) The non-injectivity of(24) can be explained by the the possibile existence of non-zero functions g ∈ m b

K(X,A),with the property that g = 0, µ-a.e. Of course, for such a g we also have fg = 0, µ-a.e.,∀ f ∈ L1

K(X,A), thus g# = 0. Going back to argument that justified the inequality (22),we clearly see that if we take C to be any constant, such that |g| ≤ C, µ-a.e., we will have|fg| ≤ C|f |, µ-a.e., so the inequality (22) will hold with ‖g‖sup replaced by C. This justifiesthe following terminology.

Definition. Let (X,A, µ) be a measure space. A function g ∈ mK(X,A) is said tobe essentially µ-bounded, if there exists some C ≥ 0, such that |g| ≤ C, µ-a.e.. For such afunction, one can show that the set C ≥ 0 : |g| ≤ C, µ-a.e. has in fact a minimum point,which we denote by Q∞(g). One defines the space

L∞K (X,A, µ) = g ∈mK(X,A) : g essentially µ-bounded,

which is clearly a vector space. The map Q∞ : L∞K (X,A, µ) → [0,∞) is clearly a seminorm.As was the case with the Lp-spaces for p ∈ [1,∞), one has the equality

g ∈ L∞K (X,A, µ) : Q∞(g) = 0 = NK(X,A, µ).

12

Define the quotient space

L∞K (X,A, µ) = L∞K (X,A, µ)/NK(X,A, µ),

on which the seminorm Q∞ induces a norm, which we denote by ‖ . ‖∞. As was the case withthe Lp-spaces introduced in sub-section B, we are going to use the same notation conventions,so when we write “g ∈ L∞K (X,A, µ)” we mean that g belongs to L∞K (X,A, µ).

With these notations, we have a better “candidate” for X , specifically we can considerthe bilinear map

Φ : L1K(X,A, µ)× L∞K (X,A) 3 (f, g) 7−→

∫X

fg dµ ∈ K, (25)

which satisfies the inequality

|Φ(f, g)| ≤ ‖f‖1 · ‖g‖∞, ∀ f ∈ L1K(X,A, µ), g ∈ L∞K (X,A, µ). (26)

Remark 3. For every g ∈ L∞K (X,A, µ), there exists h ∈ m bK(X,A), such that h = g,

µ-a.e. and ‖h‖sup = Q∞(g). Indeed, by the definition of Q∞, we know that the set

A = x ∈ X : |g(x)| > Q∞(g) ∈ A

is µ-neglijeable, so if we define the function

h(x) =

g(x) if x ∈ X r A

0 if x ∈ A

then clearly h ∈ m bK(X,A), and satisfies h = g, µ-a.e., as well as ‖h‖sup ≤ Q∞(g). Of

course, we also have the inequality |g| ≤ ‖h‖sup, µ-a.e., so in fact we have the equality‖h‖∞ = Q∞(g).

From this observation, it follows that the inclusion map m bK(X,A) → L∞K (X,A, µ) gives

rise to a surjective linear contraction

Π∞ : m bK(X,A) → L∞K (X,A, µ),

which satisfies the condition

‖g‖∞ = min‖h‖sup : h ∈m b

K(X,A), Πh = g, ∀ g ∈ L∞K (X,A, µ). (27)

Proposition 1. If we consider the space

NbK(X,A, µ) = m b

K(X,A) ∩NK(X,A, µ)

and the map Π∞ defined above, then:

(i) Ker Π∞ = NbK(X,A, µ), hence Nb

K(A,A, µ) is norm-closed in (m bK(X,A), ‖ . ‖sup);

(ii) when we equip the quotient space m bK(X,A)/Nb

K(X,A, µ) with the quotient norm, theinduced map

Π∞ : m bK(X,A)/Nb

K(X,A, µ) → L∞K (X,A, µ)

is an isometric linear isomorphism.

13

In particular, L∞K (X,A, µ) is a Banach space.

Proof. Statement (i) is immediate. The fact that Π∞ is isometric follows from (27).

Comments. As Remark 3 and Proposition 1 show, in the definition of L∞K (X,A, µ),instead of starting with the space of essentially bounded functions in mK(X,A), we couldhave started already with the space m b

K(X,A) equipped with the seminorm Q∞, so that

f ∈m bK(X,A) : Q∞(f) = 0 = Nb

K(X,A, µ),

and then we can define L∞K (X,A, µ) as the quotient m bK(X,A)/Nb

K(X,A, µ).By (26), the bilinear map (25) induces, of course, two linear contractions

# : L1K(X,A, µ) 3 f 7−→ f# ∈ L∞K (X,A)∗, (28)

# : L∞K (X,A) 3 g 7−→ g# ∈ L1K(X,A, µ)∗ (29)

given by f#(g) = g#(f) =∫

Xfg dµ. As before (see Exercise 7 below), the map (28) is still

isometric. The map (29), however, may still fail to be injective (see Exercise 8 below), so wewill need one more “adjustment.”

Exercise 7.♥ Prove that the map (28) is isometric.

Exercise 8. Consider the measure space (X,A, µ), where X = 1, 2, A = P(X), andµ(∅) = 0, µ(1) = 1, µ(2) = µ(1, 2) = ∞. It is pretty clear that mK(X,A) is the2-dimensional space of all functions h : 1, 2 → K, and every such function is bounded. Letf = χ1 and g = χ2, and prove that:

(i) NK(X,A, µ) = 0, so LpK(X,A, µ) is identified with Lp

K(X,A, µ), for every p ∈ [1,∞];

(ii) L1K(X,A, µ) = Kf ;

(iii) L∞K (X,A, µ) = mK(X,A);

(iv) ‖f‖1 = ‖g‖∞ = 1, but∫

Xhg dµ = 0, ∀h ∈ L1

K(X,A, µ).

Conclude the, for this measure space, the map (29) fails to be injective, simply becauseg# = 0.

As Exercise 8 suggests, the Banach space L∞K (X,A, µ) is still an inadequate “candidate”to be paired with L1

K(X,A, µ). In preparation for our final adjustment we introduce thefollowing terminology.

Definitions. Given a measure space (X,A, µ), we say that a set A ∈ A is locallyµ-neglijeable, if µ(A ∩ F ) = 0, for all µ-finite sets F ∈ A.

For a “measurable” property (P), we say that (P) holds locally almost µ-everywhere, ifthe set

N = x ∈ X : x does not have property (P)

is locally µ-neglijeable. For this purpose we use the notation “µ-l.a.e.”For a measurable function g : (X,A) → K, we say that g is locally essentially µ-bounded,

if there exists some C ≥ 0, such that |g| ≤ C, µ-l.a.e.. Exactly as before, for such a function,

14

one can show that the set C ≥ 0 : |g| ≤ C, µ-l.a.e. has in fact a minimum point, whichwe denote by Qloc,∞(g).

Following the same process as the one for the construction of L∞ we consider the vectorspaces

Lloc,∞K (X,A, µ) = g ∈mK(X,A) : g locally essentially µ-bounded,Nloc

K (X,A, µ) = g ∈mK(X,A) : g = 0, µ-l.a.e.,

so that NlocK (X,A, µ) = g ∈ Lloc,∞

K (X,A, µ) : Qloc,∞(g) = 0, and we define the quotientspace

Lloc,∞K (X,A, µ) = Lloc,∞

K (X,A, µ)/NlocK (X,A, µ),

on which Qloc,∞ induces a norm, which we denote by ‖ . ‖loc,∞.

Remark 4. Exactly as in Remark 3, one can show that, for every g ∈ Lloc,∞K (X,A, µ),

there exists h ∈m bK(X,A), such that h = g, µ-l.a.e. and ‖h‖sup = Qloc,∞(g). In particular,

this gives rise to a surjective linear contraction

Πloc,∞ : m bK(X,A) → Lloc,∞

K (X,A, µ),

whose kernel is the space

Nloc,bK (X,A, µ) = g ∈m b

K(X,A) : Qloc,∞(g) = 0 = g ∈m bK(X,A) : g = 0, µ-l.a.e,

which is therefore closed in m bK(X,A).

Exactly as in Proposition 1, this yields an isometric linear isomorphism

Πloc,∞ : m bK(X,A)/Nloc,b

K (X,A, µ) → Lloc,∞K (X,A, µ), (30)

so in particular Lloc,∞K (X,A, µ) is a Banach space.

Remark 5. It is trivial that we have the inclusions

NK(X,A, µ) ⊂ L∞K (X,A, µ)∩ ∩

NlocK (X,A, µ) ⊂ Lloc,∞

K (X,A, µ)(31)

so taking quotients we obtain a linear map

Πloc : L∞K (X,A, µ) → Lloc,∞K (X,A, µ). (32)

Since both L∞K (X,A, µ) and Lloc∞K (X,A, µ) are is a quotients of m b

K(X,A) – by Remarks 3and 4 – it follows that the map (32) is surjective, and satisfies the equality ΠlocΠ∞ = Πloc,∞.

Going back to the inclusions (31), we remark also that, if f, g ∈ L∞K (X,A, µ) are suchthat f = g, µ-a.e., then one clearly has the equality Qloc,∞(f) = Qloc,∞(g). This means that,passing to quotients one has a correctly defined seminorm Qloc on L∞K (X,A, µ). In fact,using the map (32), this seminorm is simply given by

Qloc(g) = ‖Πloc(g)‖loc,∞, g ∈ L∞K (X,A, µ),

15

so in particular Ker Πloc = g ∈ L∞K (X,A, µ) : Qloc(g) = 0. In other words, if we denotethis kernel by N loc,∞

K (X,A, µ), the Banach space Lloc,∞K (X,A, µ) is isometrically isomorphic

to the quotient space L∞K (X,A, µ)/N loc,∞K (X,A, µ), thus giving another proof of the fact that

Lloc,∞K (X,A, µ) is a Banach space.

Notation. For a measurable function f : (X,X ) → K, the notation “f ∈ Lloc,∞K (X,A, µ)”

will be avoided. If f belongs to Lloc,∞K (X,A, µ), its image Πloc,∞(f) in Lloc,∞

K (X,A, µ) willbe denoted simply by [f ]loc. We use the same notation for the image Πloc(f) of an elementf ∈ L∞K (X,A, µ).

Exercise 9.♥ Let (X,A, µ) be a measure space. Prove that the space m elemK (X,A) is a

norm- dense linear subspace in (m bK(X,A), ‖ . ‖sup). Conclude that

(i) the space E∞K (X,A, µ) = Π∞

(m elem

K (X,A))

is norm-dense in L∞K (X,A, µ);

(ii) the space Eloc,∞K (X,A, µ) = Πloc,∞

(m elem

K (X,A))

is norm-dense in Lloc,∞K (X,A, µ).

Using our notation conventions, these two sub-spaces are

E∞K (X,A, µ) = spanKχA : A ∈ A (in L∞),

Eloc,∞K (X,A, µ) = spanK[χA]loc : A ∈ A (in Lloc,∞).

Proposition 2. If g ∈ Lloc,∞K (X,A, µ), if p ∈ [1,∞), and f ∈ Lp

K(X,A, µ), then

|fg| ≤ Qloc,∞(g)|f |, µ-a.e., (33)

so in particular fg also belongs to LpK(X,A, µ), and satisfies the inequality

Qp(fg) ≤ Qloc,∞(g) ·Qp(f). (34)

Proof. Fix p ∈ [1,∞), as well as f ∈ LpK(X,A, µ) and g ∈ Lloc,∞

K (X,A, µ). To prove (33), wemust show that the set N = x ∈ X : |f(x)g(x)| > Qloc,∞(g)|f(x)| ∈ A is µ-neglijeable.It is pretty clear that one can write N = Sf ∩ A, where Sf = x ∈ X : f(x) 6= 0 ∈ A,and A = x ∈ X : |g(x)| > Qloc,∞(g) ∈ A. On the one hand, by Exercise 1, we know thatSf is µ-σ-finite, so there exists a sequence (Fn)∞n=1 ⊂ A of disjoint µ-finite sets, such thatSf =

⋃∞n=1. On the other hand, we know that A is locally µ-neglijeable, so in particular

µ(Fn ∩ A) = 0, ∀n, so by σ-additivity we have

µ(N) = µ( ∞⋃

n=1

Fn ∩ A)

=∞∑

n=1

µ(Fn ∩ A) = 0.

The second statement and the inequality (34) now follow immediately from (33), whichimplies ∫

X

|fg|p dµ ≤ Qloc,∞(g)p

∫X

|f |p dµ.

16

Remark 6. Using Proposition 2 with p = 1, we obtain the existence of a bilinear mapΨloc : L1

K(X,A, µ)× Lloc,∞K (X,A, µ) → K, defined (implicitly) by

Φloc(f, [g]loc) =

∫X

fg dµ, ∀ f ∈ L1K(X,A, µ), g ∈ L∞K (X,A, µ),

which satisfies the inequality

|Φloc(f, v)| ≤ ‖f‖1 · ‖v‖loc,∞, ∀ f ∈ L1K(X,A, µ), v ∈ Lloc,∞

K (X,A, µ). (35)

Using (35), it follows that

(i) For every f ∈ L1K(X,A, µ), the map f# : Lloc,∞

K (X,A, µ) 3 v 7−→ Φloc(f, v) ∈ K is alinear continuous functional, and the linear map

#loc : L1K(X,A, µ) 3 f 7−→ f# ∈ Lloc,∞

K (X,A, µ)∗ (36)

is contractive.

(ii) For every v ∈ Lloc,∞K (X,A, µ), the map v# : L1

K(X,A, µ) 3 f 7−→ Φloc(f, v) ∈ K is alinear continuous functional, and the linear map

#loc : Lloc,∞K (X,A, µ) 3 v 7−→ v# ∈ L1

K(X,A, µ)∗ (37)

is contractive.

Proposition 3. With the notations as above, both maps (36) and (37) are isometric.

Proof. Everything can be traced back in the spaces L1 and Lloc,∞, where it suffices to provethe inequalities

Q1(f) ≤ sup∣∣∫

Xfk dµ

∣∣ : k ∈ Lloc,∞K (X,A, µ), Qloc,∞(k) ≤ 1

, ∀ f ∈ L1

K(X,A, µ); (38)

Qloc,∞(g) ≤ sup∣∣∫

Xhg dµ

∣∣ : k ∈ L1K(X,A, µ), Q1(h) ≤ 1

, ∀ g ∈ Lloc,∞

K (X,A, µ). (39)

(The reverse inequalities “≥” in both (38) and (39) hold trivially by (35).)To prove (38) we fix f ∈ L1

K(X,A, µ) and we define the function k : X → K by

k(x) =

|f(x)|f(x)

if f(x) 6= 0

0 if f(x) = 0

Since k is measurable, and |k(x)| ≤ 1, ∀x ∈ X, it is obvious that k belongs to Lloc,∞K (X,A, µ),

and has Qloc,∞(k) ≤ 1. Of course, by construction we have fk = |f |, so Q1(f) =∫

X|f | dµ =∫

Xfk dµ, and (38) follows immediately.

To prove (39), fix g ∈ Lloc,∞K (X,A, µ) and denote the supremum in the right-hand side

of (39) by C, so that in order to prove (39), it suffices to prove that for every ε > 0, one has

|g| ≤ C + ε, µ-l.a.e. (40)

17

Argue by contradiction, assuming the existence of some ε > 0, for which the inequality(40) fails, which means that the set A = x ∈ X : |g(x)| > C + ε ∈ A is not locallyµ-neglijeable, thus there exists some F ∈ A, such that F ⊂ A and 0 < µ(F ) < ∞. Considerthen the function h0 : X → K, defined by

h0(x) =

|g(x)|g(x)

if x ∈ F

0 if x ∈ X r F

(Of course, if x ∈ F , then |g(x)| > C + ε > 0, so g(x) 6= 0.) Notice that h0 is measurable,and |h0| = χF , so in particular it follows that h0 is integrable, and Q1(h0) =

∫X|h0| dµ =∫

XχF dµ = µ(F ) > 0. With this choice of h0 we have

h0g = |g|χF ≥ (C + ε)χF

(the inequality follows from the inclusion F ⊂ A), so integrating over X yields∫X

h0g dµ ≥ (C + ε)µ(F ). (41)

Of course, the function h = µ(F )−1h0 is also integrable, and has Q1(h) = 1. But now, if wedivide the inequality (41) by µ(F ) we obtain∫

X

hg dµ ≥ C + ε,

which is impossible, since the left-hand side is ≤ C.

Remark 7. The maps (37), (29) and (32) are linked by the identity #loc Πloc = #.This identity, combined with the next Exercise, explains how Πloc measures the injectivityfailure of #.

Exercise 10.♥ Prove that, for a measure space (X,A, µ), the following are equivalent:

(i) the map (32) is injective;

(ii) the map (32) is isometric;

(iii) every locally µ-neglijeable set A ∈ A is µ-neglijeable;

(iv) for every A ∈ A, with µ(A) > 0, there exists F ∈ A, F ⊂ A with 0 < µ(F ) < ∞.

Definitions. A measure space (X,A, µ), satisfying the above equivalent conditions, iscalled nowhere degenerate.

At the other extreme, we say that a measure space (X,A, µ) is degenerate, if µ(A) ∈0,∞, ∀A ∈ A. In this case, one has Lloc,∞

K (X,A, µ) = 0 and LpK(X,A, µ) = 0,

∀ p ∈ [1,∞), although the space L∞K (X,A, µ) might be non-trivial, as seen for instance inExample 2.

18

Remark 8. If (X,A, µ) is σ-finite, then it is nowhere degenerate. Indeed, if we writeX =

⋃nn=1 Fn, with (Fn)∞n=1 ⊂ A disjoint and µ-finite, then for every A ∈ A we have

µ(A) =∑∞

n=1 µ(A∩Fn), with µ(A∩Fn) < ∞, ∀n. Therefore, if µ(A) > 0, then at least oneof the sets A ∩ Fn will have µ(A ∩ Fn) > 0.

D*. Measure completions, partitions, and decomposability

Although this sub-section is optional (with most of the material presented in the form ofExercises), it contains some important results that are often overlooked, or entirely omittedfrom most Real Analysis textbooks. Our whole discussion is motivated by the followingobservation. Suppose one has two measure space structures (X,A, µ) and (X,B, ν) on thesame set X, such that A ⊂ B and µ(A) = ν(A), ∀A ∈ A, in which case we use the notation(X,A, µ) ⊂ (X,B, ν). Then it is pretty obvious that we have the inclusions

mK(X,A) ⊂mK(X,B), (42)

m bK(X,A) ⊂m b

K(X,B), (43)

NK(X,A, µ) ⊂ NK(X,B, ν), (44)

LpK(X,A, µ) ⊂ Lp

K(X,B, ν), p ∈ [1,∞], (45)

which give rise to linear maps

Tp : LpK(X,A, µ) → L∞K (X,B, ν). (46)

As it turns out, the maps (46) are isometric, for all p ∈ [1,∞], so using the conventions fromthe previous sub-sections, we are going to write them as inclusions:

LpK(X,A, µ) ⊂ Lp

K(X,B, ν). (47)

Although (42)-(45) may all be strict inclusions (if A ( B), we shall see that it is notimpossible for the inclusion (47) to be an equality.

Since it is not the goal here to include a comprehensive treatment of Measure Theory,most of the material is presented in the form of Exercises.

Comment. In general, given (X,A, µ) ⊂ (X,B, ν), it may be impossible to construct a“natural” map

Lloc,∞K (X,A, µ) → Lloc,∞

K (X,B, ν),

because the implication “A locally µ-neglijeable (in A)” ⇒ “A locally ν-neglijeable (in B)”may fail. ( The measure space (X,A, µ) may be degenerate, but very small, for instanceA = ∅, X, with µ(X) = ∞, but the measure space (X,B, ν) can be very large, with plentyof ν-finite sets, for instance B = P(X), with ν the counting measure, with X infinite.)

Notation. For two subsets M, N ⊂ X, we define the set M4N = (M r N)∪ (N r M),which is referred to as the symmetric difference of M and N . It is pretty clear that theindicator function of M4N is given by χM4N = |χM − χN |.

Exercises 11-13. Assume (X,A, µ) ⊂ (X,B, ν).

11.♥ Prove that the maps (46) are isometric, for all p ∈ [1,∞].

19

12.♥ Prove that the following are equivalent:

(i) there exists p ∈ [1,∞), such that LpK(X,A, µ) = Lp

K(X,B, ν);

(ii) for every p ∈ [1,∞), one has LpK(X,A, µ) = Lp

K(X,B, ν);

(iii) for every ν-finite set B ∈ B, there exists A ∈ A, such that ν(B4A) = 0.

(Hint: By Exercise 11, LpK(X,A, µ) is closed in Lp

K(X,B, ν). Prove that condition (iii)is equivalent to the equality Lelem

K (X,A, µ) = LelemK (X,B, ν), and then use Exercise 3.)

13.♥ Prove that the following are equivalent:

(i) L∞K (X,A, µ) = L∞K (X,B, ν);

(ii) for every B ∈ B, there exists A ∈ A, such that ν(B4A) = 0.

(Hint: Same as above, but show that condition (ii) is equivalent to the equalityE∞

K (X,A, µ) = E∞K (X,B, ν), and then use Exercise 9.)

Example 3. Given a measure space (X,A, µ), we define the collection

N(A, µ) = N ⊂ X : there exists a µ-neglijeable set D ∈ A, with D ⊃ N ⊂ P(X).

One can show that, for a set B ∈ P(X), the following conditions are equivalent:

(i) there exists A ∈ A, such that B4A ∈ N(A, µ);

(ii) there exist A ∈ A and N ∈ N(A, µ), such that B = A ∪N ;

(iii) there exist A ∈ A and N ∈ N(A, µ), such that B = A r N .

Moreover, in this case the sets A are µ-essentially unique, in the sense that if A1, A2 ∈ Aeach satisfy any one of the conditions (i)-(iii), then µ(A14A2) = 0, so in particular one hasthe equality µ(A1) = µ(A2). In particular, this means that, if we denote by A the collectionof all B ⊂ X that satisfy any one the above conditions, we can define a map µ : A → [0,∞]by µ(B) = µ(A), where A ∈ A is any set that satisfies one of the conditions (i)-(iii). One canshow that this way we obtain a measure space (X, A, µ) ⊃ (X,A, µ), called the completionof (X,A, µ), which will clearly (by Exercises 12 and 13) satisfy the equalities

LpK(X,A, µ) = Lp

K(X, A, µ), ∀ p ∈ [1,∞].

The terminology one uses in connection with this construction is as follows. We say thatthat a measure space (X,A, µ) is complete, if N(A, µ) ⊂ A, or equivalently A = A. As itturns out, if (X,A, µ) is not complete, then (X, A, µ) is complete (see Exercise 16 below).

Exercise 14.♥ Prove all above statements. (This material can be found in most RealAnalysis textbooks.)

Regarding the above construction, it is natural to ask, given a measure space (X,A, µ),if there exists other measure space structures (X,B, ν) ) (X, A, µ), for which the inclusions(47) are also equalities. One natural candidate is described as follows.

20

Definitions. Fix a measure space (X,A, µ), and define the map µ∗ : P(X) → [0,∞] by

µ∗(S) = infµ(A) : A ∈ A, A ⊃ S

, S ∈ P(X). (48)

One can show that µ∗ is an outer measure5 on X, which will be called the Caratheodory outermeasure associated with (X,A, µ). It is pretty obvious that µ∗(A) = µ(A), ∀A ∈ A. If weapply then the Caratheodory construction, we obtain a σ-algebra M(µ∗) of all µ∗-measurablesets, on which the restriction of µ∗ is a measure. From now one we are going to denote the σ-algebra M(µ∗) by Ac, and the measure µ∗

∣∣Ac by µc. The resulting measure space (X,Ac, µc)

is called the Caratheodory completion of (X,A, µ). One can show (see Exercise 17) that(X, A, µ) ⊂ (X,Ac, µc), and that (X,Ac, µc) is also complete, thus justifying the aboveterminology. (The inclusion A ⊂ Ac may be strict, however. See Exercise 15.)

In order to clarify the distinction between A and Ac it is useful to introduce the followingnotation. For a measure space (X,B, ν) we define the collection

Nloc(B, ν) = N ⊂ X : there exists a locally ν-neglijeable set D ∈ B, with D ⊃ N,

and we say that (X,B, ν) is locally complete, if Nloc(B, ν) ⊂ B. It is pretty obvious thatevery “locally complete” ⇒ “complete.”

With this terminology, one can show (see Exercise 17) that, for any measure space(X,A, µ), the Caratheodory completion (X,Ac, µc) is locally complete (but A may fail tobe locally complete).

Exercise 15. Let X be an infinite set, and let A = ∅, X be equipped with the measureµ defined by µ(X) = ∞. Prove that (X,A, µ) is complete, but the Caratheodory completion(X,Ac, µc) is given by Ac = P(X), and µc the degenerate counting measure µc(A) = ∞,∀A 6= ∅.

Exercise 16.♥ Let η on X be an outer measure on S, and let F(η) denote the collectionF ⊂ X : η(F ) < ∞.

(i) Prove that, for a set A ⊂ X, the following are equivalent:

• A is η-measurable, i.e. η(S) = η(S ∩ A) + η(S r A), ∀S ∈ P(X);

• η(F ) = η(F ∩ A) + η(F r A), ∀F ∈ F(η).

(ii) Prove that, if we consider the collections

M(η) = A ⊂ X : A η-measurable,N(η) = N ⊂ X : η(N) = 0,

Nloc(η) = N ⊂ X : η(N ∩ F ) = 0, ∀F ∈ F(η),

then one has the inclusions N(η) ⊂ Nloc(η) ⊂ M(η).

(iii) Assume (X,A, µ) is a measure space, and let µ∗ be the associated Caratheodory outermeasure.

5 At this point, the reader is urged to go back to BS III, sub-section ??, where all the meaningful featuresof outer measures are explained.

21

• Prove that, if A ⊂ X is such that µ(F ) = µ∗(F ∩ A) + µ∗(F r A), for all F ∈ Awith µ(F ) < ∞, then A ∈ M(µ∗).

• If N ⊂ X is such that, µ∗(N ∩ F ) = 0, for all F ∈ A with µ(F ) < ∞, thenN ∈ Nloc(µ∗).

Exercises 17-20. Let (X,A, µ) be a measure space, and let µ∗ be the associatedCaratheodory outer measure.

17.♥ Using the notations from the previous Exercise, prove that:

(i) N(A, µ) = N(A, µ) = N(Ac, µc) = N(µ∗);

(ii) Nloc(A, µ) = Nloc(A, µ) = Nloc(Ac, µc) = Nloc(µ∗).

Conclude, using Exercise 16, that

(a) both (X, A, µ) and (X,Ac, µc) are complete;

(b) (X,Ac, µc) is locally complete.

18.♥ Prove that, if a set S ∈ P(X) has µ∗(S) < ∞, then the following are equivalent: (i)S ∈ Ac; (ii) S ∈ A. Conclude that Lp

K(X,A, µ) = LpK(X,Ac, µc), ∀ p ∈ [1,∞).

19.♥ Prove that, for a set A ∈ A, the following are equivalent:

• A is locally µ-neglijeable (in A);

• A is locally µ-neglijeable (in A);

• A is locally µc-neglijeable (in Ac);

• A ∈ Nloc(µ∗).

Conclude that one has the inclusions

Nloc(X,A, µ) ⊂ Nloc(X, A, µ) ⊂ Nloc(X,Ac, µc), (49)

Lloc,∞(X,A, µ) ⊂ Lloc,∞(X, A, µ) ⊂ Lloc,∞(X,Ac, µc), , (50)

therefore, by passing to quotient spaces, there exists natural linear maps

Lloc,∞(X,A, µ)T ′−→ Lloc,∞(X, A, µ)

T ′′−→ Lloc,∞(X,Ac, µc). (51)

20.♥ Concerning the maps in (51), prove that:

(i) both T ′ and T ′′ are isometric, so from now on we can write them as inclusions

Lloc,∞(X,A, µ) ⊂ Lloc,∞(X, A, µ) ⊂ Lloc,∞(X,Ac, µc);

(ii) the first inclusion in (i) is an equality, i.e. Lloc,∞(X,A, µ) = Lloc,∞(X, A, µ).

22

Remark 9. If (X,A, µ) is σ-finite, thenAc = A. Indeed, if we write X =⋃∞

n=1 Fn, where(Fn)∞n=1 ⊂ A have µ(Fn) < ∞, ∀n, and we start with some set S ∈ Ac, then S ∩ Fn ∈ Ac

has µ∗(S ∩ Fn) ≤ µ∗(Fn) = µ(Fn) < ∞, so by Exercise 17 it follows that S ∩ Fn ∈ A, ∀n.and then S =

⋃∞n=1 S ∩ Fn will also belong to A.

Comment. To summarize what we have obtained so far, the inclusions (X,A, µ) ⊂(X, A, µ) ⊂ (X,Ac, µc) give rise to:

LpK(X,A, µ) = Lp

K(X, A, µ) = LpK(X,Ac, µc), ∀ p ∈ [1,∞); (52)

L∞K (X,A, µ) = L∞K (X, A, µ) ⊂ L∞K (X,Ac, µc); (53)

Lloc,∞K (X,A, µ) = Lloc,∞

K (X, A, µ) ⊂ Lloc,∞K (X,Ac, µc). (54)

Without any additional (significant) restrictions on (X,A, µ), one expects to have strictinclusion in both (53) and (54), as seen for instance in Exercise 15.

The reason we focus on the Caratheodory completion, rather than the (ordinary) com-pletion, is two-fold. On the one hand, if one is interested in the Lp-spaces, for p ∈ [1,∞),the equalities (52) show that one loses nothing by replacing (X,A, µ) with (X,Ac, µc). Onthe other hand, although the inclusion (54) is strict, it shows that a “better pair” (in thesense discussed in the previous section) for L1

K(X,A, µ) is the space Lloc,∞K (X,Ac, µc). The

advantage of using this space, instead of Lloc,∞K (X,A, µ) is the increased flexibility. Ulti-

mately, under some additional (but not too restrictive) conditions, it will be shown thatLloc,∞

K (X,Ac, µc) is isometrically isomorphic to the dual Banach space L1K(X,A, µ)∗.

We now turn our attention to the disassembling/reassembling problem for Lp-spaces. Inorder to formulate it we need to introduce some terminology.

Notation. Given a σ-algebra A on X, and a subset S ⊂ X, we denote by A∣∣S

thecollection A ∩ S : A ∈ A, which is a σ-algebra on S. In the case when S ∈ A, we clearlyhave the inclusion A

∣∣S⊂ A, so given a measure µ on A, its restriction to A

∣∣S

will be a

measure on A∣∣S, which we will denote by µ

∣∣S.

Exercise 21.♥ Let (X,A, µ) be a measure space, and let S ∈ A.

(i) Prove that the restriction operator RS : f 7−→ f∣∣S

maps (linearly):

(a) mK(X,A) onto mK(S,A∣∣S),

(b) m bK(X,A) onto m b

K(S,A∣∣S),

(c) NK(X,A, µ) onto NK(S,A∣∣S, µ

∣∣S),

(d) NlocK (X,A, µ) onto Nloc

K (S,A∣∣S, µ

∣∣S),

(e) Lloc,∞K (X,A, µ) onto Lloc,∞

K (S,A∣∣S, µ

∣∣S), and

(f) LpK(X,A, µ) onto Lp

K(S,A∣∣S, µ

∣∣S), for all p ∈ [1,∞].

In particular, one obtains the surjective linear maps

QpS : Lp

K(X,A, µ) → LpK(S,A

∣∣S, µ

∣∣S), p ∈ [1,∞], (55)

Qloc,∞S : Lloc,∞

K (X,A, µ) → Lloc,∞K (S,A

∣∣S, µ

∣∣S). (56)

23

(ii) Prove that the maps (55) and (56) are contractions.

Definition. Given a non-empty set X, a system (Xi)i∈I of nonempty subsets of X iscalled a partition of X, if all Xi’s are disjoint, and

⋃i∈I Xi. In the case when X comes

equipped with a σ-algebra A, and all Xi’s belong to A, we say that (Xi)i∈I is an A-partitionof X.

Exercise 22.♥ Let (X,A, µ) be a measure space, and let (Xi)i∈I be an A-partition ofX. Using the maps (55) and (56) defined in the previous Exercise, prove the followingstatements.

(i) If p ∈ [1,∞], and u ∈ LpK(X,A, µ), then the I-tuple

Dpu = (QpXi

u)i∈I ∈∏i∈I

LpK(Xi,A

∣∣Xi

, µ∣∣Xi

)

belongs to the Banach space `p(LpK(Xi,A

∣∣Xi

, µ∣∣Xi

), ‖ . ‖p)i∈I . Moreover, the map

Dp : (LpK(X,A, µ), ‖ . ‖p) → `p(Lp

K(Xi,A∣∣Xi

, µ∣∣Xi

), ‖ . ‖p)i∈I (57)

is a linear contraction.

(ii) For every u ∈ Lloc,∞K (X,A, µ), the I-tuple

Dloc,∞u = (Qloc,∞Xi

u)i∈I ∈∏i∈I

Lloc,∞K (Xi,A

∣∣Xi

, µ∥∥

Xi)

belongs to the Banach space `∞(Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I . Moreover, the map

Dloc,∞ : (Lloc,∞K (X,A, µ), ‖ . ‖loc,∞) → `∞(Lloc,∞

K (Xi,A∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I (58)

is a linear contraction.

(iii) If I is countable, then the maps (57) and (58) are isometric isomorphisms. (Note:For the map (57) with p ∈ [1,∞), as well as for the map (58), this will be generalizedshortly, so the reader should only focus on the map (57) for p = ∞.)

The maps (57) and (58) are called the disassembly maps associated with given partition.With this terminology, the disassembly/reassembly problem is to decide when the maps (57)and (58) are in fact isometric isomorphisms. In this case, their inverses (which will also beisometric) will be called the reassembly maps.

This problem is extremely complicated for (57) with p = ∞, and we are not going toaddress it at all. Of course, by Exercise 22 (iii) the disassembly/reassembly problem has apositive answer, when I is countable.

The next two results give a complete solution for (57) with p ∈ [1,∞).

Proposition 4. If (X,A, µ) is a measure space, and (Xi)i∈I is an A-partition of X,then the map (57) is surjective, for every p ∈ [1,∞).

24

Proof. Fix p ∈ [1,∞), and v = (vi)i∈I ∈∏

i∈I LpK(Xi,A

∣∣Xi

, µ∣∣Xi

), with∑

i∈I ‖vi‖pp < ∞, and

let us prove the existence of some u ∈ LpK(X,A, µ), such that Dpu = v. By summability,

the set Iv = i ∈ I : vi 6= 0 is countable, so the set A =⋃

i∈IvXi belongs to A. Choose

now, for every i ∈ Iv, a function fi ∈ LpK(Xi,A

∣∣Xi

, µ∣∣Xi

), such that [fi] = vi, and define thefunction f : X → K by

f(x) =

fi(x) if x ∈ Xi and i ∈ Iv

0 if x 6∈ A

It is pretty clear that f is measurable, and furthermore

f∣∣Xi

=

fi if i ∈ Iv

0 if i 6∈ Iv

so if we prove that f belongs to LpK(X,A, µ), then we will have the equality v = Dp[f ].

To prove that f belongs to LpK(X,A, µ), we list Iv = in : n ∈ M, with the in’s all

different, where either M = N, or M = 1, 2, . . . , N, and we define the increasing sequenceof sets (An)∞n=1 ⊂ A, by

An =

⋃n

k=1 Xik if n ∈ M

A if n 6∈ M

which clearly satisfies the equality⋃∞

n=1 An = A, so the sequence (χAn|f |p)∞n=1 of non-negativemeasurable functions on (X,A is non-decreasing, and satisfies

limn→∞

χAn(x)|f(x)|p = |f(x)|p, ∀x ∈ X.

By Lebesgue’s Monotone Convergence Theorem, it follows that∫X

|f |p dµ = limn→∞

∫X

χAn|f |p dµ. (59)

Of course, by construction, for every n ∈ N we have

∫X

χAn|f |p dµc =

∑n

k=1

∫X

χXik|f |p dµ =

∑nk=1

∫Xik

|fik |p d(µ∣∣Xik

), if n ∈ M

∑i∈Iv

∫X

χXi|f |p dµ =

∑i∈Iv

∫Xi|fi|p d(µ

∣∣Xi

), if n 6∈ M

so by (59) we get:∫X

|f |p dµ = limn→∞

∫X

χAn|f |p dµ =∑i∈Iv

∫Xi

|fi|p d(µ∣∣Xi

) =∑i∈Iv

‖vi‖pp = ‖v‖p

p < ∞.

Proposition-Definition 5. Let (X,A, µ) be a measure space. For an A-partition(Xi)i∈I , the following are equivalent:

(i) the disassembly map (57) is isometric, for every p ∈ [1,∞);

25

(i’) the disassembly map (57) is isometric, for at least one p ∈ [1,∞);

(ii) for any A ∈ A, with µ(A) < ∞, one has the equality6

µ(A) =∑i∈I

µ(A ∩Xi). (60)

An A-partition (Xi)i∈I , satisfying the above equivalent conditions, is called µ-integrable.

Proof. The implication (i) ⇒ (i′) is trivial.(i′) ⇒ (ii). Fix p ∈ [1,∞), for which the map (57) is isometric. This means that,∫

X

|f |p dµ =∑i∈I

∫X

χXi|f |p dµ, ∀ f ∈ Lp

K(X,A, µ). (61)

Fix now some set A ∈ A with µ(A) < ∞, and let us prove the equality (60). This followsimmediately from (61), applied to the characteristic function f = χA.

(ii) ⇒ (i). Assume condition (ii) holds, fix some arbitrary p ∈ [1,∞), and let us provethat the map (57) is isometric. Since Dp is norm-continuous, it suffices to prove the equality‖Dpu‖p = ‖u‖p, for all u in the dense subspace Lelem

K (X,A, µ) ⊂ LpK(X,A, µ). In turn, for

this condition it suffices to prove that:∫X

g dµ =∑i∈I

∫X

χXig dµ, ∀ g ∈ Lelem

K (X,A, µ), (62)

so basically we can assume that p = 1. Finally, by linearity, in order to prove (62) it sufficesto consider the case when g = χA, for A ∈ A with µ(A) < ∞, in which case the equality(62) reduces to (60).

Comment. Concerning the above definition, the reader is warned that the equality(60) is limited to sets of finite measure. In particular, if A ∈ A has the property thatµ(A ∩Xi) = 0, ∀ i ∈ I, it does not follow that µ(A) = 0.

Remark 10. Of course, if the index set I is countable, then every A-partition (Xi)i∈I isµ-integrable, since in this case the equality (60) holds for all A ∈ A.

In the general case (I arbitrary), if (Xi)i∈I is a µ-integrable A-partition, then for everyA ∈ A with µ(A) < ∞, the fact that the sum in the right-hand side of (60) is finite forcesthe set I(A) = i ∈ I : µ(A ∩ Xi) 6= 0 to be countable, so in particular, if one considersthe set A′ =

⋃i∈I(A) A ∩Xi ⊂ A, then A′ belongs to A, and by construction and by (60) it

follows that µ(A r A′) = 0.

Comment. The disassembly/reassembly problem for (58) is a bit more complicated, andwill require some extra conditions imposed on the partition. One of them – µ-integrability –is a very natural one to impose, especially if keep in mind the dual pairing between L1 andLloc,∞, so it is not surprising that we have the following result.

Proposition 6. Let (X,A, µ) be a measure space, and (Xi)i∈I is a µ-integrable A-partition of X, then the disassembly map (58) is isometric.

6 The right-hand side of (60) is understood in the sense of summability.

26

Proof. Fix some element u ∈ Lloc,∞K (X,A, µ), and let us show that the element

Dloc,∞u ∈ `∞(Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I

has ‖Dloc,∞u‖∞ = ‖u‖loc,∞. By Exercise 22 we already know that ‖Dloc,∞u‖∞ ≤ ‖u‖loc,∞,so all we need to to is to prove the inequality

‖Dloc,∞u‖∞ ≥ ‖u‖loc,∞. (63)

If we choose a representative f ∈ Lloc,∞K (X,A, µ) i.e. u = [f ], then the element Dloc,∞u is

the I-tuple ([fi])i∈I , where fi = f∣∣Xi

, so the inequality (63) simply reads:

supi∈I

Qiloc,∞(fi) ≥ Qloc,∞(f), (64)

where, for every i ∈ I, the superscript in Qiloc,∞ indicates that the seminorm Qloc,∞(fi) is

computed in Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

). Denote the supremum in the left-hand side of (63) by

C, so that, in order to prove the inequality (63) we must show that the set

N = x ∈ X : |f(x)| > C ∈ A

is locally µ-neglijeable7, that is, µ(F ∩N) = 0, for all F ∈ A with µ(F ) < ∞.Of course, for each i ∈ I, we have N ∩Xi = x ∈ Xi : |fi(x)| > C ≥ Qi

loc,∞(fi) ∈ A∣∣Xi

,

which forces N ∩Xi to be locally (µ∣∣Xi

)-neglijeable. In particular, if we start with some set

F ∈ A with µ(F ) < ∞, then the set F ∩Xi ∈ A∣∣Xi

has µ∣∣Xi

(F ∩Xi) < ∞, thus forcing

µ((F ∩N) ∩Xi) = µ∣∣Xi

(((F ∩Xi) ∩ (N ∩Xi)

)= 0, ∀ i ∈ I.

But now, using (60) with A = F ∩ N , we get µ(F ∩ N) =∑

i∈I µ((F ∩ N) ∩Xi) = 0, andwe are done.

Comment. In order to solve the disassembly/reassembly problem for (58), our approachwill be to try to construct the reassembly map. As it will turn out, one can constructa reassembly map Rloc,∞ : `∞(Lloc,∞

K (Xi,A∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I → Lloc,∞K (X,B, ν), where

(X,B, ν) ⊃ (X,A, µ) is a slightly larger measure space structure. This reassembly map willbe constructed in such a way that the composition Rloc,∞ Dloc,∞ agrees with the inclusion

Lloc,∞K (X,A, µ) ⊂ Lloc,∞

K (X,B, ν), and then the solution to the problem will depend onwhether this inclusion is an equality.

A very natural framework for reassembling I-tuples of measurable functions is providedby the following construction.

Definition. Suppose (Xi)i∈I is a partition of X. Assume on each Xi, i ∈ I, one is givena σ-algebra Bi. On can then form the direct sum σ-algebra∨

i∈I Bi = B ⊂ X : B ∩Xi ∈ Bi, ∀ i ∈ I.7 This will force ‖u‖loc,∞ = Qloc,∞(f) ≤ C.

27

It is pretty clear that this is the largest among all σ-algebras B that satisfy B∣∣Xi

= B, ∀ i ∈ I.

One should be aware of the fact that∨

i∈I Bi is larger than σ-alg( ⋃

i∈I Bi

)– the σ-algebra

generated by⋃

i∈I Bi. If I is countable, however, these two σ-algebras coincide. The nextExercises clarify this matter.

Exercises 23-24. Let (Xi)i∈I be a partition of X, and assume that on each Xi one isgiven a σ-algebra Bi..

23. Prove that:

(i) the map∏

i∈I Bi 3 (Bi)i∈I 7−→⋃

i∈I Bi ∈∨

i∈I Bi is a bijection;

(ii) given (Bi)i∈I ∈∏

i∈I Bi, the set⋃

i∈I Bi belongs to σ-alg( ⋃

i∈I Bi

), if and only if

there exists a countable set of indices I0 ⊂ I, such that Bi ∈ ∅, Xi, ∀ i ∈ I r I0.

Conclude that, if I is countable, then∨

i∈I Bi = σ-alg( ⋃

i∈I Bi

).

24.♥ Prove that for a measurable space (Y,D), and a map f : X → Y , the following areequivalent:

• f is measurable as a map f :(X,

∨i∈I Bi

)→ (Y,D);

• f∣∣Xi

: (Xi,Bi) → (Y,D) is measurable, for each i ∈ I.

Conclude that

(i) the map

D : mK(X,

∨i∈I Bi

)3 f 7−→

(f∣∣Xi

)i∈I∈

∏i∈I mK(Xi,Bi) (65)

is a linear isomorphism;

(ii) when restricted to the space of bounded measurable functions, the map D yieldsan isometric linear isomorphism

Db : m bK(X,

∨i∈I Bi

)3 f 7−→

(f∣∣Xi

)i∈I∈ `∞(m b

K(Xi,Bi), ‖ . ‖sup)i∈I . (66)

The maps (65) and (66) will also be called disassembly maps. Their inverses will alsobe called reassembly maps.

Definition. Assume A is a σ-algebra on X, and π = (Xi)i∈I is an A-partition. Thedirect sum σ-algebra

∨i∈I(A

∣∣Xi

), which we denote by Aπ, is called the π-enhancement of A.

It is obvious that Aπ

∣∣Xi

= A∣∣Xi

, ∀ i ∈ I, and one has the inclusion Aπ ⊃ A. In general, thisinclusion is strict, but if I is countable, then Aπ = A.

Comment. The problem with direct sum σ-algebras is that, given a measure space(X,A, µ) and an arbitrary A-partition π = (Xi)i∈I of X, it might be impossible8 to constructa measure µ on Aπ, that agrees with µ on A.

8 For instance, if one starts with the Lebesgue structure (R,A, λ) – where A is the σ-algebra of Lebesguemeasurable sets, and λ is the Lebesgue measure – and the partition π consists of singletons, then Aπ = P(R),and there exists no measure ν on P(R), that agrees with λ on A.

28

In the case when π is µ-integrable, the situation is much nicer, as shown below.

Theorem 3. Let (X,A, µ) be a measure space, let µ∗ be the associated Caratheodoryouter measure, and let (X,Ac, µc) denote the Caratheodory completion. Assume π = (Xi)i∈I

is a µ-integrable A-partition.

(i) The π-enhancement Aπ is contained in Ac, so when one restricts µc to Aπ, one obtainsa measure µπ on Aπ, such that: (X,A, µ) ⊂ (X,Aπ, µπ) ⊂ (X,Ac, µc).

(ii) The measure spaces (X,A, µ) and (X,Aπ, µπ) have identical associated Caratheodoryouter measures, therefore they have the same Caratheodory completion: (X,Ac, µc).

(iii) The partition π is both µc-integrable (when viewed as an Ac-partition) and µπ-integrable(when viewed as an Aπ-partition).

(iv) The disassembly map9

Dπloc,∞ : Lloc,∞

K (X,Aπ, µπ) → `∞(Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I (67)

is an isometric isomorphism.

(v) Using the notations from Exercise 16, for a set N ⊂ X, the following are equivalent:

(a) N ∈ Nloc(µ∗);

(b) Xi ∩N ∈ Nloc(µ∗), ∀ i ∈ I;

(c) N belongs to Ac and is locally µc-neglijeable.

Sketch of proof. Let us first prove statement (v). We already know that (a) ⇔ (c), byExercise 17. The implication (a) ⇒ (b) is trivial. Assume N ⊂ X has property (b), and let usshow that N belongs to Nloc(µ∗). By Exercise 16 it suffices to show that µ∗(N∩F ) = 0, for allF ∈ A, with µ(F ) < ∞. Fix such an F and let us first notice that, since µ∗(N∩F ) ≤ µ∗(F ) =µ(F ) < ∞, by the definition of the outer measure µ∗, there exists G ∈ A, with µ(G) < ∞,such that G ⊃ N ∩ F , so if we consider the countable set I(G) = i ∈ I : µ(Xi ∩G) > 0,then the set G′ =

⋃i∈I(G) Xi ∩ G belongs to A and satisfies µ(G r G′) = 0, which implies

that

µ∗(N ∩ F ) ≤ µ∗(G′ ∩N ∩ F ) + µ∗((G r G′) ∩N ∩ F ) ≤≤ µ∗(G′ ∩N ∩ F ) + µ∗(G r G′) = µ∗(G′ ∩N ∩ F ).

(68)

Of course, since G ⊃ N ∩ F , we have

G′ ∩N ∩ F =⋃

i∈I(G)

Xi ∩G ∩N ∩ F =⋃

i∈I(G)

Xi ∩N ∩ F,

which, by the countability of I(G), the σ-subadditivity of µ∗, and (68), gives

µ∗(N ∩ F ) ≤ µ∗(G′ ∩N ∩ F ) ≤∑

i∈I(G)

µ∗(Xi ∩N ∩ F ).

9 The map Dπloc,∞ is the map (58), with A replaced by Aπ.

29

Of course, by condition (b) we know that µ∗(Xi∩N∩F ) = 0, ∀ i ∈ I, so the above inequalitiesnow force µ∗(N ∩ F ) = 0.

(i). Start with some set A ∈ Aπ, which means that A =⋃

i∈I Ai, with Ai ∈ A andAi ⊂ Xi, ∀ i ∈ I, and let us show that A is µ∗-measurable. According to Exercise 16, itsuffices10 to show that

µ∗(F ∩ A) + µ∗(F r A) ≤ µ(F ), (69)

for every F ∈ A, with µ(F ) < ∞. Fix such an F , and define, for each i ∈ I, the setsFi = F ∩ Xi, so that the set I(F ) = i ∈ I : µ(Fi) > 0 is countable, and the setF ′ =

⋃i∈I(F ) Fi ⊂ F belongs to A, and satisfies µ(F r F ′) = 0. Remark now that, on the

one hand we have the equalities

F ′ ∩ A =⋃

i∈I(F )

(Fi ∩ Ai) and F ′ r A =⋃

i∈I(F )

(Fi r Ai), (70)

with Fi ∩Ai and Fi r Ai in Ai ⊂ A, so using the fact that I(F ) is countable, it follows thatboth F ′ ∩ A and F ′ r A belong to A, and therefore we have

µ∗(F ′ ∩ A) + µ∗(F ′ r A) = µ∗(F ′ ∩ A) + µ∗(F ′ r A) = µ(F ′) = µ(F ) (71)

On the other hand, since both (F r F ′) ∩ A and (F r F ′) r A are contained in F r F ′,which is µ-neglijeable, it follows that µ∗((F r F ′)∩A) = µ∗((F r F ′) r A) = 0, so using theproperties of outer measures we get the inequalities

µ∗(F ∩ A) ≤ µ∗(F ′ ∩ A) + µ∗((F r F ′) ∩ A) = µ∗(F ′ ∩ A),

µ∗(F r A) ≤ µ∗(F ′ r A) + (µ∗((F r F ′) r A) = µ∗(F ′ r A),

The desired inequality (69) now follows immediately, by adding these two inequalities, andusing (71).

Statements (ii) and (iii) are left to the reader.To prove statement (iv), it suffices to prove surjectivity. of Dloc,∞. (By (iii) and Propo-

sition 6, the map Dloc,∞ is isometric.) Start with some element

v = (vi)i∈I ∈ `∞(Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I ,

and let us construct an element u ∈ Lloc,∞K (X,Aπ, µπ), such that Dloc,∞u = v. First of

all, we know that any element in Lloc,∞ can be represented by a bounded measurablefunction with same norm (see Remark 4), so for every i ∈ I, one can choose some func-tion fi ∈ m b

K(Xi,A∣∣Xi

), such that vi = [fi], in Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

) and also ‖fi‖sup =

‖vi‖loc,∞. Since ‖fi‖sup ≤ ‖vi‖loc,∞ ≤ ‖v‖∞, it follows that the I-tuple (fi)i∈I belongs to`∞(m b

K(Xi,A∣∣Xi

, µ∣∣Xi

), ‖ . ‖sup)i∈I . Since by definition,∨

i∈I(A∣∣Xi

) = Aπ, by Exercise 24, it

follows that there exists f ∈m bK(X,Aπ), such that fi = f

∣∣Xi

, ∀ i ∈ I, and then the element

u = [f ] ∈ Lloc,∞K (X,Aπ, µπ) will clearly satisfy the equality Dloc,∞u = v.

10 The reverse inequality is always true, by the properties of outer measures.

30

Exercise 25. Complete the proof of Theorem 3.

Comment. Theorem 3 clarifies the reassembly/disassembly problem for (58), as follows.If π = (Xi)i∈I is µ-integrable, then using the isometric identification (67), we can think thedisassembly map (58) as an inclusion

Lloc,∞K (X,A, µ) ⊂ Lloc,∞

K (X,Aπ, µπ),

so the condition that (58) is an isomorphism is equivalent to the equality

Lloc,∞K (X,A, µ) = Lloc,∞

K (X,Aπ, µπ). (72)

Of course, again by Theorem 3, we also have the inclusion

Lloc,∞K (X,Aπ, µπ) ⊂ Lloc,∞

K (X,Ac, µc),

so a sufficient condition for (72) is the equality

Lloc,∞K (X,A, µ) = Lloc,∞

K (X,Ac, µc). (73)

Of course, one way to achieve (72) is to require the equality Aπ = A, which is automaticif I is countable. In general, one is also interested in having (73), and then the followingterminology will be helpful (see Theorems 4 and 5 below).

Definition. Suppose (X,A, µ) is a measure space. An A-partition π = (Xi)i∈I is calleda quasi-decomposition of (X,A, µ), if:

(a) π is µ-integrable;

(b) µ(Xi) < ∞, ∀ i ∈ I.

If in addition to these conditions, one also has

(c) Aπ = A,

then π is called a decomposition of (X,A, µ).

Theorem 4. Let π = (Xi)i∈I be a quasi-decomposition of the measure space (X,A, µ).

(i) For a subset S ⊂ X, the following are equivalent:

(a) S belongs to Ac;

(b) S ∩Xi belongs to Ac, for each i ∈ I;

(c) there exist A ∈ Aπ and a locally µc-neglijeable set N ∈ Ac, such that S = A r N .

(ii) The inclusion Lloc,∞K (X,Aπ, µπ) ⊂ Lloc,∞

K (X,Ac, µc) is an equality.

(iii) When viewed as an Aπ-partition, π is a decomposition of (X,Aπ, µπ).

(iii’) When viewed as an Ac-partition, π is a decomposition of (X,Ac, µc).

31

Proof. (i). Let µ∗ be the Caratheodory outer measure associated with (X,A, µ), which byTheorem 3 also coincides with the Caratheodory outer measure associated with (X,A, µ).The implications (c) ⇒ (a) ⇒ (b) are trivial. To prove the implication (b) ⇒ (c), fix S ⊂ X,satisfying condition (b), and let us indicate how the sets A and N can be constructed, so that(c) is satisfied. First of all, for every i ∈ I, the set Si = S∩Xi is µ∗-measurable (i.e. Si belongsto Ac), and has µ∗(Si) ≤ µ∗(Xi) = µ(Xi) < ∞. So in particular, there exists Ai ∈ A, suchthat Ai ⊃ Si, and µc(Si) = µ∗(Si) = µ(Ai). Replacing Ai with Ai ∩Xi, we can also assumethat Ai ⊂ Xi. Of course, by finiteness, the set Ni = Ai r Si ∈ Ac has µ∗(Ni) = µc(Ni) = 0.Define then the sets A =

⋃i∈I Ai and N =

⋃i∈I Ni. On the one hand, since A∩Xi = Ai ∈ A,

it is clear that A belongs to the direct sum σ-algebra∨

i∈I(A∣∣Xi

) = Aπ. On the other hand,

for every i ∈ I, the set Xi ∩N = Ni has µ∗(Ni) = µc(Ni) = 0, so we can use Theorem 3 toconclude that N belongs to Ac and is locally µc-neglijeable. Of course, by construction wealso have S = A r N .

(ii). Since we already have an isometric inclusion Lloc,∞K (X,Aπ, µπ) ⊂ Lloc,∞

K (X,Ac, µc),

all we need to do is to show that Lloc,∞K (X,Aπ, µπ) contains the dense subspace Eloc,∞

K (X,Ac, µc),which means that, for every S ∈ Ac, there exists A ∈ Ac, such that [χA] = [χS], inLloc,∞

K (X,Ac, µc). But this is immediate from part (i).Statement (iii) is immediate from Theorem 3. To prove statement (iii’) we again invoke

Theorem 3, which gives µc-integrability of π, so the only feature we must show is the equality(Ac)π = Ac. But this equality is trivial from statement (i).

Definition. A measure space (X,A, µ) is said to be (quasi-)decomposable, it it admits a(quasi-)decomposition. With this terminology, the reassembly/disassembly problem for (58)has the following solution.

Theorem 5. If (X,A, µ) is a decomposable measure space, then the inclusion

Lloc,∞K (X,A, µ) ⊂ Lloc,∞

K (X,Ac, µc)

is an equality. In particular, for any µ-integrable A-partition (Xi)i∈I of X, the disassem-bly map Dloc,∞ : Lloc,∞

K (X,A, µ) → `∞(Lloc,∞K (Xi,A

∣∣Xi

, µ∣∣Xi

), ‖ . ‖loc,∞)i∈I is an isometricisomorphism.

Proof. Fix a decomposition κ = (Yj)j∈J for (X,A, µ). On the one hand, by Theorem 4

we know that Lloc,∞K (X,Ac, µc) = Lloc,∞

K (X,Aκ, µκ). On the other hand, since Aκ = A, we

obviously have Lloc,∞K (X,Aκ, µκ) = Lloc,∞

K (X,A, µ), so the preceding equality also gives the

equality: Lloc,∞K (X,Ac, µc) = Lloc,∞

K (X,A, µ). The second statement follows from Theorem3, and the Comment that followed it.

Comments. Decomposability is perhaps the nicest notion in Measure Theory. Amongother things, it provides the correct framework for the celebrated Radon-Nikodim Theorem,which will be discussed in HS II, as well as the “correct” dual pairing between the L1-spacesand the Lloc,∞-spaces. The two (easy) examples below will be complemented by another one,in sub-section E.

Example 5. Every σ-finite measure space (X,A, µ) is decomposable. Indeed, if wechoose a countable A-partition π = (Xi)i∈I , with all Xi of finite measure, then obviously πis a decomposition.

32

Example 6. If (X,A, µ) is a quasi-decomposable measure space, then by Theorem 4,the Caratheodory completion (X,Ac, µc) is decomposable.

E. Decomposablity of Radon measures.

In this sub-section we specialize to the measure spaces (Ω, Bor(Ω), µ), where Ω is a locallycompact Hausdorff space, Bor(Ω) is the Borel σ-algebra, and µ is a Radon measure. Ameasure space of this form will be called a locally compact measure space. As in the previoussub-section, to such a measure space one associates the Caratheodory outer measure µ∗,and the Caratheodory completion (Ω, Bor(Ω)c, µc). In order not to overload the notationexcessively, the σ-algebra Bor(Ω)c will be simply denoted by M(µ). Likewise, the measureµc will be denoted by µM.

In preparation for the main result (Theorem 6 below), it is useful to introduce someterminology, and we do so in the Exercises below.

Exercises 26-27. Suppose Ω is a locally compact Hausdorff space, and µ is a Radonmeasure on Ω. A non-empty compact set K ⊂ Ω is said to be µ-tight, if for every non-emptycompact proper subset L ( K, one has the strict inequality µ(L) < µ(K).

26.♥ Prove that all singleton sets are µ-tight. Prove that, if a compact set K ⊂ Ω hasµ(K) > 0, then the following are equivalent:

• K is µ-tight;

• whenever D ⊂ Ω is open, and D ∩K 6= ∅, it follows that µ(D ∩K) > 0.

27.♥ Prove that, for every non-empty compact set K ⊂ Ω, there exists at least one compactµ-tight subset K0 ⊂ K, with µ(K r K0) = 0.

(Hint: If µ(K) = 0, one can take K0 to be a singleton. If µ(K) > 0 and K is notµ-tight, then the collection Λ(K) = L ( K : L non-empty, compact, and µ(L) =µ(K) is non-empty. Prove that if L1, . . . , Ln ∈ Λ(K), then L1 ∩ . . . Ln ∈ Λ(K), sousing the Finite Intersection Property, the compact set K0 =

⋂L∈Λ(K) L is non-empty.

Prove that K0 also belongs to Λ(K), by showing that µ(T ) = 0, for all compact subsetsT ⊂ K r K0. Indeed, if one starts with such a T , then T ∩

[⋂L∈Λ(K) L

]= ∅, so by

the Finite Intersection Property and by the above-mentioned property of Λ(K), thereexists L ∈ Λ(K) with T ∩ L = ∅, i.e. T ⊂ K r L, thus forcing µ(T ) ≤ µ(K r L) = 0.Since K0 is the smallest element in Λ(K), it is pretty obvious that K0 is µ-tight.)

Definition. Suppose η is an outer measure on some set X, and π = (Xi)i∈I is a partitionof X. Consider the index set I0 = i ∈ I : η(Xi) > 0. We say that a set S ⊂ Ω is η-essentially countably covered by π, if:

(a) the index set I0(S) = i ∈ I0 : S ∩Xi 6= ∅ is countable;

(b) the set S0 =⋃

i∈I0(S) S ∩Xi ⊂ S satisfies the equality η(S r S0) = 0.

Note that, by construction (without any conditions imposed on the partition π, or on theset S), one has the equality

S0 =⋃i∈I0

S ∩Xi, ∀S ⊂ X. (74)

33

Theorem 6 (Compact Decomposability of Radon Measures). Suppose Ω is a locallycompact Hausdorff space, and µ is a Radon measure on Ω.

(i) There exists a partition π = (Ki)i∈I of Ω, such that

• all Ki, i ∈ I, are compact;

• every open set D ⊂ Ω with µ(D) < ∞ is µ∗-essentially countably covered by π.

(ii) If π is any partition with the two properties stated above, then:

(a) π is a quasi-decomposition11 of (Ω, Bor(Ω), µ);

(b) when viewed as a M(µ)-partition, π constitutes a decomposition of the measurespace (Ω,M(µ), µM).

Proof. To prove statement (i), we consider the set Θ of all collections U ⊂ P(Ω), with thefollowing property:

(∗) all sets in K ∈ U are disjoint, compact, µ-tight, and have µ(K) > 0.

Clearly the empty collection belongs to Θ. One can easily see that if U ⊂ Θ is totally orderedby inclusion, then the collection

⋃U∈U U again satisfies property (∗), so using Zorn Lemma,

there exists a maximal collection U0 with property (∗). In general this does not exclude U0

from being the empty collection. Let us list U0 = (Ki)i∈I0 (therefore i 6= j ⇒ Ki ∩Kj = ∅),and let us complete the collection (Ki)i∈I0 to a partition π = (Ki)i∈I by adding all singletonsets in Ω r

[⋃i∈I0

Ki

], so that, for i ∈ I r I0, the set Ki is a singleton, and has µ(Ki) = 0.

(Of course, for i ∈ I r I0, the singleton set Ki is µ-tight and is disjoint from all sets in U0,so if µ(Ki) > 0 this will contradict the maximality of U0.) Trivially, π satisfies the firstcondition in (i), and the index set I0 (which could be empty!) is precisely the set of all i’swith µ(Ki) > 0. To prove the second condition, we fix an open set D ⊂ Ω, with µ(D) < ∞,and we must show that:

(a) the index set I0(D) = i ∈ I0 : D ∩Ki 6= ∅ is countable;

(b) the set D0 =⋃

i∈I0(D) D ∩Ki ⊂ D satisfies the equality µ∗(D r D0) = 0.

To prove (a) we notice that, since all Ki, i ∈ I0, are µ-tight and have µ(Ki) > 0, by Exercise26 it follows that, for every i ∈ I0(D), we have µ(D ∩ Ki) > 0. In particular, for everyfinite set F ⊂ I0(D), we have

∑j∈F µ(D ∩ Ki) = µ

( ⋃i∈F D ∩ Ki

)≤ µ(D), so we get∑

i∈I0(D) µ(D ∩Ki) ≤ µ(D) < ∞, which clearly forces I0(D) to be countable.

To prove (b) first notice that, by construction, the set D0 is Borel, being a countable unionof the Borel sets D∩Ki, i ∈ I0(D), so µ∗(DrD0) = µ(DrD0). To prove the µ(DrD0) = 0,we argue by contradiction, assuming that µ(D r D0) > 0. Since µ(D r D0) ≤ µ(D) < ∞,by the properties of Radon measures, we have

µ(D r D0) = supµ(L) : L ⊂ D r D0, L compact,11 Since compact sets are Borel, it is obvious that π is a Bor(Ω)-partition.

34

so in particular there exists some compact set L ⊂ D r D0, with µ(L) > 0. By Exercise27, such an L can be chosen to be µ-tight. But now we reached a contradiction, since L isdisjoint from all sets in U0, so the collection U0 ∪L will contradict the maximality of U0.

(ii). Fix π = (Ki)i∈I as in (i). To prove property (a), we must show that

µ(A) =∑i∈I

µ(A ∩Ki), (75)

for all Borel sets A, with µ(A) < ∞. Since, the Ki’s are disjoint, for every finite set F ⊂ I onehas

∑i∈F µ(Ki∩A) = µ

([ ⋃i∈F Ki

]∩A

)≤ µ(A), and this shows that

∑i∈I µ(Ki∩A) ≤ µ(A).

To prove the reverse inequality, we recall that, by the definition of Radon measures,we know that µ(A) = infµ(D) : D open, D ⊃ A, so in particular there exists an openset D ⊃ A, with µ(D) < ∞. By condition (i) the open set D is µ-essentially countablycovered. As argued before, the set D0 =

⋃i∈I0

D ∩ Ki =⋃

i∈I0(D) D ∩ Ki is Borel, and

satisfies µ(D r D0) = 0. Of course, the set A0 = A ∩D0 =⋃

i∈I0A ∩Ki is also Borel, and

because of the inclusion A r A0 ⊂ D r D0, we also have µ(A r A0) = 0, which yields

µ(A) = µ(A0). (76)

On the other hand, since we can also write A0 =⋃

i∈I(D) A ∩Ki (countable disjoint union),

by σ-additivity, combined with (76), we now have

µ(A) = µ(A0) =∑

i∈I0(D)

µ(A ∩Ki) ≤∑i∈I

µ(A ∩Ki).

By Theorem 4, property (b) immediately follows from (a).

Measure and Integration Theory is particularly nice in the context of Radon measures,due to the density property described below.

Remark 11. If Ω is a locally compact Hausdorff space, and µ is a Radon measure, thenfor every p ∈ [1,∞), the space Cc,K(Ω), of all continuous functions f : Ω → K, with compactsupport, is dense in Lp

K(Ω,M(µ), µM). Indeed, for every A ∈ M(µ), with µ(A) < ∞,and every ε > 0, there exist a compact set K ⊂ A, and an open set D ⊃ A, such thatµ(D r K) < ε, so if we choose, by Urysohn’s Lemma, a continuous function f : Ω → [0, 1],with compact support, such that f

∣∣K

= 0 and f∣∣ΩrD

= 0, then one has the inequalities

χK ≤ f, χA ≤ χD, which imply that ‖f − χA‖p ≤ ‖χD − χK‖p ≤ ε1/p, thus proving that thenorm-closure of Cc,K(Ω) in Lp

K(Ω,M(µ), µM) contains all elementary integrable functions.By Exercise 3 this norm-closure coincides with Lp

K(Ω,M(µ), µM).

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