Banach Spaces IV: Banach Spaces of Measurable Functions

BS IV c Gabriel Nagy

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

In this section we discuss another important class of Banach spaces arising from Measure and Integration theory. Convention. Throughout this note K will be one of the ﬁelds R or C, and all vector spaces 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 most Real Analysis textbooks. The exposition is in the form of a sequence of Definitions and Facts. Definitions. A measurable space is a pair (X, A) consisting of a non-empty set X and a σ-algebra on X. Our basic example will be of the form (Ω, Bor(Ω)), where Ω is a 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 measurable transformation from (X, A) to (Y, B), if f −1(B) ∈ A, ∀ B ∈ B. In practice, if one chooses a collection M ⊂ B, which generates B as a σ-algebra1, then a sufficient condition for f :(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, this is equivalent to the condition that both Re f, Im f :(X, A) → R are measurable. Since the Borel σ-algebra Bor(R) is generated by various collections of intervals, such as for instance J = {(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 the extended intervals [−∞, ∞], [0, ∞], or [−∞, 0], with the understanding that if T denotes any one of these spaces, one again uses the Borel σ-algebra Bor(T ). (Of course any such T is homeomorphic to [0, 1].)

We denote by m K(X, A) the space of all K-valued measurable functions on (X, A). It is well known that m K(X, A) is a K-algebra, when equipped with point-wise addition and multiplication. 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 functions χA, of sets A ∈ A. (Recall that χA(x) = 1, if x ∈ A, and χA(x) = 0, if x 6∈ A.) The

next class of measurable functions are those in spanK{χA : A ∈ A} ⊂ m K(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 span is denoted by elem(X, A). An equivalent description of this space is: m K elem(X, A) = f : X → : Range f is ﬁnite, and f −1({α}) ∈ A, ∀ α ∈ Range f . m K K This characterization allows one to extend the notion of elementary measurable functions for functions f : X → Y , where Y is an arbitrary set. In particular, if we let Y = [0, ∞], elem the corresponding space of functions is denoted by m + (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, [−∞, ∞], or C. Deﬁne K to be either R, when T is R or [−∞, ∞], or C, when T = C. For a function f : X → T , the following are equivalent: • f :(X, A) → T is measurable; • there exists a sequence (f )∞ ⊂ elem(X, A), such that n n=1 m K

lim fn(x) = f(x), ∀ x ∈ X. n→∞

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

inf f(y) ≤ fn(x) ≤ sup f(y), ∀ x ∈ X. y∈X y∈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-additive map µ : A → [0, ∞], with µ(∅) = 0. The σ-additivity condition means that, whenever ∞ S∞ P∞ (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

+,elem elem Iµ : m + (X, A) → [0, ∞], + Iµ : m +(X, A) → [0, ∞], by the formulas:

+,elem X −1 elem Iµ (f) = α · µ f ({α}) , f ∈ m + (X, A), α∈Range f + +,elem elem Iµ (g) = sup Iµ (f): f ∈ m + (X, A, 0 ≤ f ≤ g .

+,elem + (In the deﬁnition of Iµ one uses the convention 0 · ∞ = ∞ · 0 = 0.) The map Iµ is called the positive µ-integral. + Fact 2. Let (X, A, µ) be a measure space. The positive integral Iµ : m +(X, A) → [0, ∞] has the following properties:

2 + (i) Iµ is positive-linear, in the sense that

+ + + • Iµ (f + g) = Iµ (f) + Iµ (g), ∀ f, g ∈ m +(X, A); + + • Iµ (αf) = αIµ (f), ∀ f ∈ m +(X, A), α ∈ [0, ∞];

+ in particular, Iµ is monotone, in the sense that, for f, g ∈ m +(X, A) one has the + + implication f ≥ g ⇒ Iµ (f) ≥ Iµ (g).

elem + +,elem (ii) If f ∈ m + (X, A), then Iµ (f) = Iµ (f). In particular, one has:

+ Iµ (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 a non-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 Iµ (f) = limn→∞ Iµ (fn).

Definitions. Suppose (X, A, µ) is a measure space. A function f ∈ m K(X, A) is said + to be µ-integrable, if the function |f| ∈ m +(X, A) has finite positive integral: Iµ (|f|) < ∞. The space of µ-integrable -valued functions is denoted by L1 (X, A, µ). (As before, if K K K = C, the subscript will be omitted.) The space of µ-integrable functions f : X → [0, ∞) 1 is denoted by L+(X, A, µ). Comment. As it turns out, µ-integrability makes sense even for measurable functions f :(X, A) → [−∞, ∞]. The space of such functions is denoted by L1 (X, A, µ). It turns out, however, that if f ∈ L1 (X, A, µ), then the set A = {x ∈ X : f(x)R = ±∞} has µ(A) = 0. The role of sets of measureR zero will be clarified a little later. Fact 4. Assume (X, A, µ) is a measure space. (i) The space L1 (X, A, µ) is a -linear subspace in (X, A), which can also be described K K m K as: L1 (X, A, µ) = span L1 (X, A, µ). K K + (ii) A measurable function f :(X, A) belongs to L1(X, A, µ), if and only of both Re f and Im f belong to L1 (X, A, µ). R (iii) There exists a unique -linear map I : L1 (X, A, µ) → , such that I (f) = I+(f), K µ K K µ µ 1 ∀ f ∈ L+(X, A, µ). (iv) For an elementary function f ∈ elem(X, A), the following are equivalent: m K • f is µ-integrable; • νf −1({α}) < ∞, ∀ α ∈ Range f;

• f ∈ spanK{χA : A ∈ A, µ(A) < ∞}. 1 The linear map Iµ : L (X, A, µ) → K is referred to as the µ-integral. The traditional R K notation is to write X f dµ instead of Iµ(f). The space of elementary integrable functions, described in Fact 4 (iii) as

elem(X, A) ∩ L1 (X, A, µ) = {χ : A ∈ A, µ(A) < ∞}, m K K spanK A

3 will be denoted by Lelem(X, A, µ). Given A ,...,A ∈ A, with µ(A ), . . . , µ(A ) < ∞, and K 1 n 1 n α1, . . . , αn ∈ K, the the elementary integrable function α1χA1 +···+αnχAn has its µ-integral computed as: Z

(α1χA1 + ··· + αnχAn ) dµ = α1µ(A1) + ··· + αnµ(An). (1) X The formula (1) is referred to as the Elementary Integral Formula. Besides linearity, the integration map also has the continuity property: Z Z f dµ ≤ |f| dµ, ∀ f ∈ L1 (X, A, µ). (2) K X X The integral notation will also be used in the case of the positive integral, that is, if + R f ∈ m +(X, A), the positive integral Iµ (f) will still be denoted by X f dµ, although this quantity may equal ∞. Using this notation, the space of integrable functions can be presented as: L1 (X, A, µ) = f ∈ (X, A): R |f| dµ < ∞ . The Elementary Integral Formula K m K X (1) is still valid, for arbitrary A1,...,An ∈ A and α1, . . . , αn ∈ [0, ∞]. Integrability is preserved under neglijeable perturbations. In order to clarify this statement, 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. Here the term “measurable” means that the set {x ∈ X : x has property (P) } (and of course its complement) 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 some relation 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. R (i) If f = 0, µ-a.e., then f is integrable, and X f dµ = 0. R (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 ∈ m K(X, A): f = 0, µ-a.e.} is a linear subspace in L1 (X, A, µ). K Integrable functions can be produced via dominance conditions, as indicated by the following simple observation.

4 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, R R and |f| ≤ g, µ-a.e. Then f is integrable, and: X f dµ ≤ X g 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 measurable functions. 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 some µ-neglijeable set B ∈ A, such that limn→∞ fn(x) = f(x), ∀ x ∈ X r B. R R Then f is integrable, and X f dµ = limn→∞ X fndµ. We end our review by recalling some standard Measure Theory terminology that will be employed 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) < ∞;

2 ∞ • we say that A is µ-σ-finite, if there exists a sequence (An)n=1 ⊂ A of µ-finite sets, S∞ 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 is µ-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 ∈ L1 (X, A, µ), then the set S is µ-σ-ﬁnite. K f 2♥. Prove that, if f is bounded, and S is µ-ﬁnite, then f ∈ L1 (X, A, µ). f K

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

Lp (X, A, µ) = f ∈ (X, A): R |f|p dµ < ∞ . K m K X 2 In fact the sequence can be chosen to consist of disjoint sets.

5 R p + (The integral X |f| dµ stands for the positive integral Iµ .) We also define the map Qp : LK(X, A, µ) → [0, ∞) by Z 1/p Q (f) = |f|p dµ , ∀ f ∈ Lp (X, A, µ). p K X It is pretty clear that, when p = 1, we are dealing with the same space as the one introduced in sub-section A, which is already known to be a K-vector space. Furthermore, in this case the map Q is obviously a seminorm on L1 (X, A, µ). For 1 < p < ∞, the same will be true, 1 K as a result of the following result. 1 1 Theorem 1. Assume 1 < p, q < ∞ are Hölderconjugate, i.e. p + q = 1. (i) Lp (X, A, µ) is a -linear subspace in (X, A). K K m K (ii) If f ∈ Lp (X, A, µ) and g ∈ Lq (X, A, µ), then fg ∈ L1 (X, A, µ), and furthermore one K K K has the inequality Q1(fg) ≤ Qp(f) · Qq(g). (3) (iii) If f ∈ Lp (X, A, µ), then K Q (f) = max R fg dµ : g ∈ Lq (X, A, µ),Q (g) ≤ 1 . (4) p X K q (iv) Q : Lp (X, A, µ) → [0, ∞) is a seminorm on Lp (X, A, µ). p K K p p Proof. (i). It is pretty obvious that, if f ∈ LK(X, A, µ), then αf ∈ LK(X, A, µ), ∀ α ∈ K. p p Assume now f1, f2 ∈ LK(X, A, µ), and let us show that f1 + f2 ∈ LK(X, A, µ). This is, 3 p p p p p however, pretty obvious, since the inequality |f1 + f2| ≤ 2 |f1| + 2 |f2| implies (by p dominance) the integrability of |f1 + f2| . (ii). Fix f ∈ Lp (X, A, µ) and g ∈ Lq (X, A, µ). Use Fact 1, to choose two non-decreasing K K ∞ ∞ elem p sequences (Fn)n=1 and (Gn)n=1 in m + (X, A), such that limn→∞ Fn(x) = |f(x)| and q limn→∞ Gn(x) = |g(x)| , ∀ x ∈ X. Of course, since the two sequences are non-decreasing, p q we have the inequalities 0 ≤ F1 ≤ F2 ≤ · · · ≤ |f| and 0 ≤ G1 ≤ G2 ≤ · · · ≤ |g| , so all 1/p 1/q the Fn’s and the Gn’s a integrable. If we define fn = Gn and gn = Gn , 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 R (positive) integral X |fg| dµ can be computed as Z Z |fg| dµ = lim fngn dµ. (5) X n→∞ X 4 Pk Pk Fix for the moment n, and let us write fn = j=1 αjχAj and gn = j=1 βjχAj with all A’s µ-finite, and all α’s and β’s non-negative, so that k k Z X X fngn dµ = αjβjµ(Aj) = xjyj, (6) X j=1 j=1

3 Use the inequality a + q ≤ 2 max{a, b}, which implies (a + b)p ≤ 2p(ap + bp), ∀ a, b ≥ 0. 4 Pr Ps One can write separately fn = i=1 αiχDi with D1,...,Dr disjoint µ-finite, and gn = m=1 βmχEm with E1,...,Es disjoint µ-finite. The A’s will the intersections Di ∩ Em, which will also be µ-finite.

6 1/p 1/q where xj = αj[µ(Aj)] and yj = βj[µ(Aj)] , j = 1, . . . , k. Using the (classical) H¨older Inequality, we have k k 1/p k 1/q X X p X q xjyj ≤ xj · yj . (7) j=1 j=1 j=1 Notice now that k k Z Z X p X p p p xj = αj µ(Aj) = Fn dµ ≤ |f| dµ = Qp(f) , j=1 j=1 X X k k Z Z X q X q q q yj = βj µ(Aj) = Gn dµ ≤ |g| dµ = Qq(g) , j=1 j=1 X X so now if we combine (7) and (6), we obtain Z fngn dµ ≤ Qp(f) · Qq(g). (8) X Since the inequality (8) holds for all n, using (5) it follows that Z |fg| dµ ≤ Qp(f) · Qq(g). X Of course, this proves that fg is indeed integrable, and also satisﬁes (3). (iii). Fix f ∈ Lp (X, A, µ). Since, by (i), for every g ∈ Lq (X, A, µ) we clearly have the R K R K inequality X fg dµ ≤ X |fg| dµ ≤ Qp(f) · Qq(g), it follows that sup R fg dµ : g ∈ Lq (X, A, µ),Q (g) ≤ 1 ≤ Q (f), X K q p so in order to prove (4), all we need to do is to produce some g ∈ Lq (X, A, µ), with Q (g) ≤ 1, K q such that Z

fg dµ = Qp(f). (9) X

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:  |f(x)|p  if f(x) 6= 0  f(x) g0(x) =   0 if f(x) = 0

Obviously, g0 is measurable, and satisﬁes the identity

q pq−q p |g0| = |f| = |f| , so g belongs to Lq (X, A, µ), and 0 K Z 1/q p p/q Qq(g0) = |f| dµ = Qp(f) . (10) X

7 p Moreover, since fg0 = |f| , we also have Z Z p p fg0 dµ = |f| dµ = Qp(f) , (11) X X so if we deﬁne g = Q (g )−1g , then g also belongs to Lq (X, A, µ), but now has Q (g) = 1, q 0 0 K q and by (10) and (11) it satisﬁes Z −1 p −p/q p fg dµ = Qq(g0) Qp(f) = Qp(f) Qp(f) = Qp(f). X (iv). The identity Q (αf) = |α| · Q (f), ∀ f ∈ Lp (X, A, µ), α ∈ p p K K

is trivial from the deﬁnition of Qp, so all that is left to prove is the triangle inequality Q (f + f ) ≤ Q (f ) + Q (f ), ∀ f , f ∈ Lp (X, A, µ). p 1 2 p 1 p 2 1 2 K This inequality can be derived from (ii) and (iii) as follows. Fix some g ∈ Lq (X, A, µ), such R K that Qq(g) ≤ 1 and Qp(f1 + f2) = X (f1 + f2)g dµ , and use linearity and continuity of the integral to obtain Z Z Z Qp(f1 + f2) ≤ |(f1 + f2)g| dµ ≤ |f1g| dµ + |f2g| dµ = X X X

= 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 ∈ Lp (X, A, µ), one has the equivalence K R p X |f| dµ = 0 ⇔ f = 0, µ-a.e., so one has the the equality {f ∈ Lp (X, A, µ): Q (f) = 0} = N (X, A, µ). (12) K p K Define the quotient space Lp (X, A, µ) = Lp (X, A, µ)/N (X, A, µ). K K K In other words, Lp (X, A, µ) is the collection of equivalence classes associated with the re- K lation “=, µ-a.e.” For a function f ∈ Lp (X, A, µ) we denote by [f] its equivalence class in K Lp (X, A, µ). So the equality [f] = [g] is equivalent to f = g, µ-a.e. By (12) the seminorm K Q induces a norm on Lp (X, A, µ), which we denote by k . k , so by definition we have p K p k[f]kp = Qp(f). Conventions. We are going to abuse a bit the notation, by writing “f ∈ Lp (X, A, µ)” if K f belongs to Lp (X, A, µ). (We will always have in mind the fact that this notation signifies K that f is almost uniquely determined.) Likewise, instead of writing k[f]kp, we are going to write kfkp, so that from now on Qp(f) will be replaced by kfkp. With this notations, we now have f ∈ Lp (X, A, µ) ⇔ |f|p ∈ L1 (X, A, µ), in which case kfk = k|f|pk . K K p 1 The integration map R : L1 (X, A, µ) → gives rise to a correctly defined map, which X K K we denote by the same symbol, so that by the continuity condition we can write: Z Z f dµ ≤ |f| dµ = kfk , ∀ f ∈ L1 (X, A, µ). (13) 1 K X X With these notations, statements (ii) and (iii) in Theorem 1 can be restated as follows.

8 • If f ∈ Lp (X, A, µ) and g ∈ Lq (X, A, µ), then fg ∈ L1 (X, A, µ) and K K K

kfgk1 ≤ kfkp · kgkq. (14)

In particular, using (13), one also has Z

fg dµ ≤ kfkp · kgkq. (15) X

• If f ∈ Lp (X, A, µ), then K kfk = max R fg dµ : g ∈ Lq (X, A, µ), kgk ≤ 1 . (16) p X K q

The inequality (15) is referred to as the Integral HölderInequality. Remarks 1-2. Let p, q ∈ (1, ∞) be two Hölder conjugate numbers. 1. One has a bilinear map Z Φ: Lp (X, A, µ) × Lq (X, A, µ) 3 (f, g) 7−→ fg dµ ∈ , K K K X which satisfies the inequality

|Φ(f, g)| ≤ kfkp · kgkq. (17)

Using (16) it follows that Φ establishes a dual pairing (see DT I), which is referred to as the H¨olderdual pairing.

2. Using the Hölder dual pairing, one defines, for every f ∈ Lp (X, A, µ) the linear func- K tional f # : Lq (X, A, µ) 3 g 7−→ R fg dµ ∈ . By (17), the functional f # is norm- K X K # continuous, and by (16) it follows that kf k = kfkp. Therefore, the Hölder dual pairing produces an isometric linear map

#: Lp (X, A, µ) 3 f 7−→ f # ∈ Lq (X, A, µ)∗. (18) K K

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 Lp (X, A, µ), p ∈ [1, ∞), are Banach spaces. K Proof. Fix p ∈ [1, ∞). In order to prove that Lp (X, A, µ) we are going to verify the Summa- K ∞ p bility Test, so we start with some sequence (fn)n=1 in L (X, A, µ), such that the quantiy P∞ P∞K p S = n=1 kfnkp is finite, and we show that the series n=1 fn is convergent in L (X, A, µ). P∞ K Define the function h : X → [0, ∞] by h(x) = n=1 |fn(x)|. Of course, we can write h(x) = limn→∞ hn(x), where hn = |f1| + ··· + |fn|. (Of course all h’s are measurable.) On the one hand, using the triangle inequality in Lp (X, A, µ) we have R Z p p p p hn = |f1| + ··· + |fn| p ≤ kf1kp + ··· + kfnkp ≤ S , ∀ n. (19) X

9 p p p On the other hand, since we clearly have 0 ≤ h1(x) ≤ h2(x) ≤ · · · ≤ h(x) , as well as p p limn→∞ hn(x) = h(x) , by Lebesgue’s Monotone Convergence Theorem it follows that Z Z p p h dµ = lim hn dµ, X n→∞ X

R p p p p so going back to (19) it follows that X h dµ ≤ S < ∞, so h belongs to L . Since h is integrable, it follows that set A = {x ∈ X : h(x) = ∞} ∈ A has µ(A) = 0. (The p p obvious inequality tχA ≤ h will force tµ(A) ≤ S , ∀ t > 0, and this can only happen when P∞ µ(A) = 0.) For x ∈ X rA, the series n=1 |fn(x)| is convergent (to h(x) < ∞), so the series P∞ n=1 fn(x) is also convergent (in K). Deﬁne then the measurable function g : X → K by P∞ f (x) if x ∈ X A g(x) = n=1 n r 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. p We now wish to prove that g ∈ L and that limn→∞ kg − gnkp = 0. To prove that g p p p is in L we notice that, since we obviously have |gn| ≤ hn ≤ h, we also get |gn| ≤ h , p p µ-a.e., and since limn→∞ |gn(x)| = |g(x)| , µ-a.e., we can apply Lebesgue’s Dominated Convergence Theorem, to conclude that |g|p is integrable. Likewise, for any n, k ≥ 1, we th have |gn+k − gn| ≤ hn+k − hn ≤ h − hn ≤ h, taking limk→∞ and the p power yields

p p |g − gn| ≤ h , µ-a.e..

p Since limn→∞ |g(x)−gn(x)| = 0, µ-a.e., again by Lebesgue’s Dominated Convergence Theo- R p p rem we conclude that limn→∞ X |g − gn| dµ = 0, which means precisely that kg − gnkp → 0, thus proving the desired conclusion. Example 1. Start with some non-empty set X, let A = P(X), and let µ be the counting measure, defined as Card A if A is finite µ(A) = ∞ if A is infinite Then the Banach space Lp (X, A, µ) is precisely the space `p (X) discussed in BS I. K K Example 2. Start with some non-empty set X, let A = P(X), and but let µ be the degenerate counting measure, defined as

0 if A = µ(A) = ∅ ∞ if A 6= ∅ Then Lp (X, A, µ) = {0}. As this example suggests, unless some local ﬁniteness condition is K assumed, integration theory can be extremely trivial. Notation. Let (X, A, µ) be a measure space. It is not hard to see that Lelem(X, A, µ) K – the space of elementary integrable functions – is contained (as a linear subspace) in Lp (X, A, µ), for every p ∈ [1, ∞). Therefore one has a correctly deﬁned linear map T : K p Lelem(X, A, µ) → Lp (X, A, µ). As it turns out, Ker T is independent of p, since it can K K p be described as N (X, A, µ) ∩ Lelem(X, A, µ). This means that the spaces Range T are K K p

10 all isomorphic, and for this reason all of them will be denoted by Lelem(X, A, µ), and we K will understand that this space is contained in all Lp-spaces. Using the (abusive) notation convention introduced earlier, we can think

Lelem(X, A, µ) = {χ : A ∈ A, µ(A) < ∞}, (20) K spanK A with the understanding that

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

• the linear span on the right-hand side of (20) can be computed in Lp (X, A, µ), for any K p ∈ [1, ∞).

Exercise 3♥. With these conventions, prove that Lelem(X, A, µ) is dense in Lp (X, A, µ), K K in the norm topology, for every p ∈ [1, ∞). Exercise 4. Prove that, if p, q, r ∈ [1, ∞) are such that 1 + 1 = 1 , and if f ∈ Lp (X, A, µ) p q r K and g ∈ Lq (X, A, µ), then fg ∈ Lr (X, A, µ), and furthermore, one has the inequality K K kfgkr ≤ kfkp · kgkq. Exercise 5♥. Assume (X, A, µ) is a ﬁnite measure space. Prove that, if p, q ∈ [1, ∞) are such that p ≥ q, then Lp (X, A, µ) ⊂ Lq (X, A, µ). In particular, this gives rise to a linear K K injective map J : Lp (X, A, µ) ,→ Lq (X, A, µ). Prove that J is continuous. K K C. The spaces m b, L∞ and Lloc,∞

The H¨olderdual 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 Banach space X of measurable functions on (X, A), such that

(i) for any f ∈ L1 (X, A, µ) and any g ∈ X , the function fg is in L1 (X, A, µ), and K K furthermore one has the inequality kfgk1 ≤ kfk1 · kgk; (ii) the correspondence Z Φ: L1 (X, A, µ) × X 3 (f, g) 7−→ fg dµ ∈ K K X 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 better and better. Deﬁnitions. The most obvious one is the space b (X, A) of all bounded measurable m K functions g :(X, A) → K, which is a Banach space, when equipped with the norm k . ksup, deﬁned by kgk = sup |g(x)|, g ∈ b (X, A). sup m K x∈X

11 (Of course, b (X, A) is contained in the Banach space `∞(X) of all bounded functions m K K g : X → K, and the norm k . ksup is precisely the induced norm. Since the uniform limit of a sequence of bounded measurable functions is again bounded and measurable, it follows that b (X, A) is norm-closed in `∞(X), therefore a Banach space.) m K K Of course, if f ∈ L1 (X, A, µ) and g ∈ b (X, A), then there exists a constant, for K m K instance C = kgk , such that |fg| ≤ C|f|. This will force fg ∈ L1 (X, A), as well as the sup K inequality kfgk1 ≤ Ckfk1, so one indeed has a bilinear map Z Φ: L1 (X, A, µ) × b (X, A) 3 (f, g) 7−→ fg dµ ∈ , (21) K m K K X which satisﬁes the inequality

|Φ(f, g)| ≤ kfk · kgk , ∀ f ∈ L1 (X, A, µ), g ∈ b (X, A). (22) 1 sup K m K Exercise 6♥.. Prove that, for any f ∈ L1 (X, A, µ), one has the equality K kfk = max |Φ(f, g)| : g ∈ b (X, A), kgk ≤ 1 . 1 m K sup

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

#: L1 (X, A, µ) 3 f 7−→ f # ∈ b (X, A)∗, (23) K m K #: b (X, A) 3 g 7−→ g# ∈ L1 (X, A, µ)∗ (24) m K K # # R given by f (g) = g (f) = X fg dµ. By (22) both maps (23) and (24) are contractive. By Exercise 6, the map (23) is in fact isometric. The only drawback of the choice of m b as our “candidate” is the fact that the map (24) might fail to be injective. (Ultimately, with an adequate space X in place of b (X, A) we are going to make both (23) and (24) isometric.) The non-injectivity of m K (24) can be explained by the the possibile existence of non-zero functions g ∈ b (X, A), m K 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 (X, A), thus g# = 0. Going back to argument that justiﬁed the inequality (22), K 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 kgksup replaced by C. This justiﬁes the following terminology.

Deﬁnition. Let (X, A, µ) be a measure space. A function g ∈ m K(X, A) is said to be essentially µ-bounded, if there exists some C ≥ 0, such that |g| ≤ C, µ-a.e.. For such a function, 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∞(X, A, µ) = {g ∈ (X, A): g essentially µ-bounded}, K m K which is clearly a vector space. The map Q : L∞(X, A, µ) → [0, ∞) is clearly a seminorm. ∞ K As was the case with the Lp-spaces for p ∈ [1, ∞), one has the equality

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

12 Deﬁne the quotient space

L∞(X, A, µ) = L∞(X, A, µ)/N (X, A, µ), K K K on which the seminorm Q∞ induces a norm, which we denote by k . k∞. As was the case with the Lp-spaces introduced in sub-section B, we are going to use the same notation conventions, so when we write “g ∈ L∞(X, A, µ)” we mean that g belongs to L∞(X, A, µ). K K With these notations, we have a better “candidate” for X , speciﬁcally we can consider the bilinear map Z Φ: L1 (X, A, µ) × L∞(X, A) 3 (f, g) 7−→ fg dµ ∈ , (25) K K K X which satisﬁes the inequality

|Φ(f, g)| ≤ kfk · kgk , ∀ f ∈ L1 (X, A, µ), g ∈ L∞(X, A, µ). (26) 1 ∞ K K Remark 3. For every g ∈ L∞(X, A, µ), there exists h ∈ b (X, A), such that h = g, K m K µ-a.e. and khksup = 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 g(x) if x ∈ X A h(x) = r 0 if x ∈ A

then clearly h ∈ b (X, A), and satisﬁes h = g, µ-a.e., as well as khk ≤ Q (g). Of m K sup ∞ course, we also have the inequality |g| ≤ khksup, µ-a.e., so in fact we have the equality khk∞ = Q∞(g). From this observation, it follows that the inclusion map b (X, A) ,→ L∞(X, A, µ) gives m K K rise to a surjective linear contraction

Π : b (X, A) → L∞(X, A, µ), ∞ m K K which satisﬁes the condition

kgk = min khk : h ∈ b (X, A), Πh = g , ∀ g ∈ L∞(X, A, µ). (27) ∞ sup m K K Proposition 1. If we consider the space

Nb (X, A, µ) = b (X, A) ∩ N (X, A, µ) K m K K

and the map Π∞ deﬁned above, then: (i) Ker Π = Nb (X, A, µ), hence Nb (A, A, µ) is norm-closed in ( b (X, A), k . k ); ∞ K K m K sup (ii) when we equip the quotient space b (X, A)/Nb (X, A, µ) with the quotient norm, the m K K induced map Πˆ : b (X, A)/Nb (X, A, µ) → L∞(X, A, µ) ∞ m K K K is an isometric linear isomorphism.

13 In particular, L∞(X, A, µ) is a Banach space. K ˆ 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∞(X, A, µ), K instead of starting with the space of essentially bounded functions in m K(X, A), we could have started already with the space b (X, A) equipped with the seminorm Q , so that m K ∞ {f ∈ b (X, A): Q (f) = 0} = Nb (X, A, µ), m K ∞ K and then we can deﬁne L∞(X, A, µ) as the quotient b (X, A)/Nb (X, A, µ). K m K K By (26), the bilinear map (25) induces, of course, two linear contractions

#: L1 (X, A, µ) 3 f 7−→ f # ∈ L∞(X, A)∗, (28) K K #: L∞(X, A) 3 g 7−→ g# ∈ L1 (X, A, µ)∗ (29) K K

# # R given by f (g) = g (f) = X fg 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 we will 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), and

µ(∅) = 0, µ({1}) = 1, µ({2}) = µ({1, 2}) = ∞. It is pretty clear that m K(X, A) is the 2-dimensional space of all functions h : {1, 2} → K, and every such function is bounded. Let f = χ{1} and g = χ{2}, and prove that: (i) N (X, A, µ) = {0}, so Lp (X, A, µ) is identified with Lp (X, A, µ), for every p ∈ [1, ∞]; K K K (ii) L1 (X, A, µ) = f; K K (iii) L∞(X, A, µ) = (X, A); K m K (iv) kfk = kgk = 1, but R hg dµ = 0, ∀ h ∈ L1 (X, A, µ). 1 ∞ X K Conclude the, for this measure space, the map (29) fails to be injective, simply because g# = 0. As Exercise 8 suggests, the Banach space L∞(X, A, µ) is still an inadequate “candidate” K to be paired with L1 (X, A, µ). In preparation for our final adjustment we introduce the K following terminology. Definitions. Given a measure space (X, A, µ), we say that a set A ∈ A is locally µ-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, if the 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, which we denote by Qloc,∞(g). Following the same process as the one for the construction of L∞ we consider the vector spaces

Lloc,∞(X, A, µ) = {g ∈ (X, A): g locally essentially µ-bounded}, K m K Nloc(X, A, µ) = {g ∈ (X, A): g = 0, µ-l.a.e.}, K m K so that Nloc(X, A, µ) = {g ∈ Lloc,∞(X, A, µ): Q (g) = 0}, and we deﬁne the quotient K K loc,∞ space Lloc,∞(X, A, µ) = Lloc,∞(X, A, µ)/Nloc(X, A, µ), K K K

on which Qloc,∞ induces a norm, which we denote by k . kloc,∞. Remark 4. Exactly as in Remark 3, one can show that, for every g ∈ Lloc,∞(X, A, µ), K there exists h ∈ b (X, A), such that h = g, µ-l.a.e. and khk = Q (g). In particular, m K sup loc,∞ this gives rise to a surjective linear contraction

Π : b (X, A) → Lloc,∞(X, A, µ), loc,∞ m K K whose kernel is the space

Nloc,b(X, A, µ) = {g ∈ b (X, A): Q (g) = 0} = {g ∈ b (X, A): g = 0, µ-l.a.e}, K m K loc,∞ m K which is therefore closed in b (X, A). m K Exactly as in Proposition 1, this yields an isometric linear isomorphism

Πˆ : b (X, A)/Nloc,b(X, A, µ) → Lloc,∞(X, A, µ), (30) loc,∞ m K K K so in particular Lloc,∞(X, A, µ) is a Banach space. K Remark 5. It is trivial that we have the inclusions N (X, A, µ) ⊂ L∞(X, A, µ) K K ∩ ∩ (31) Nloc(X, A, µ) ⊂ Lloc,∞(X, A, µ) K K so taking quotients we obtain a linear map

Π : L∞(X, A, µ) → Lloc,∞(X, A, µ). (32) loc K K Since both L∞(X, A, µ) and Lloc∞(X, A, µ) are is a quotients of b (X, A) – by Remarks 3 K K m K and 4 – it follows that the map (32) is surjective, and satisﬁes the equality Πloc ◦Π∞ = Πloc,∞. Going back to the inclusions (31), we remark also that, if f, g ∈ L∞(X, A, µ) are such K that 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 Q on L∞(X, A, µ). In fact, loc K using the map (32), this seminorm is simply given by

Q (g) = kΠ (g)k , g ∈ L∞(X, A, µ), loc loc loc,∞ K

15 so in particular Ker Π = {g ∈ L∞(X, A, µ): Q (g) = 0}. In other words, if we denote loc K loc this kernel by N loc,∞(X, A, µ), the Banach space Lloc,∞(X, A, µ) is isometrically isomorphic K K to the quotient space L∞(X, A, µ)/N loc,∞(X, A, µ), thus giving another proof of the fact that K K Lloc,∞(X, A, µ) is a Banach space. K Notation. For a measurable function f :(X, X ) → , the notation “f ∈ Lloc,∞(X, A, µ)” K K will be avoided. If f belongs to Lloc,∞(X, A, µ), its image Π (f) in Lloc,∞(X, A, µ) will K loc,∞ K be denoted simply by [f]loc. We use the same notation for the image Πloc(f) of an element f ∈ L∞(X, A, µ). K Exercise 9♥. Let (X, A, µ) be a measure space. Prove that the space elem(X, A) is a m K norm- dense linear subspace in ( b (X, A), k . k ). Conclude that m K sup (i) the space E∞(X, A, µ) = Π elem(X, A) is norm-dense in L∞(X, A, µ); K ∞ m K K (ii) the space Eloc,∞(X, A, µ) = Π elem(X, A) is norm-dense in Lloc,∞(X, A, µ). K loc,∞ m K K Using our notation conventions, these two sub-spaces are

E∞(X, A, µ) = {χ : A ∈ A} (in L∞), K spanK A Eloc,∞(X, A, µ) = span {[χ ] : A ∈ A} (in Lloc,∞). K K A loc Proposition 2. If g ∈ Lloc,∞(X, A, µ), if p ∈ [1, ∞), and f ∈ Lp (X, A, µ), then K K

|fg| ≤ Qloc,∞(g)|f|, µ-a.e., (33)

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

Qp(fg) ≤ Qloc,∞(g) · Qp(f). (34)

Proof. Fix p ∈ [1, ∞), as well as f ∈ Lp (X, A, µ) and g ∈ Lloc,∞(X, A, µ). To prove (33), we K K must 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 that ∞ Sf is µ-σ-ﬁnite, so there exists a sequence (Fn)n=1 ⊂ A of disjoint µ-ﬁnite sets, such that S∞ Sf = 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

∞ ∞ [ X µ(N) = µ Fn ∩ A = µ(Fn ∩ A) = 0. n=1 n=1 The second statement and the inequality (34) now follow immediately from (33), which implies Z Z p p p |fg| dµ ≤ Qloc,∞(g) |f| dµ. X X

16 Remark 6. Using Proposition 2 with p = 1, we obtain the existence of a bilinear map Ψ : L1 (X, A, µ) × Lloc,∞(X, A, µ) → , deﬁned (implicitly) by loc K K K Z Φ (f, [g] ) = fg dµ, ∀ f ∈ L1 (X, A, µ), g ∈ L∞(X, A, µ), loc loc K K X which satisﬁes the inequality

|Φ (f, v)| ≤ kfk · kvk , ∀ f ∈ L1 (X, A, µ), v ∈ Lloc,∞(X, A, µ). (35) loc 1 loc,∞ K K Using (35), it follows that (i) For every f ∈ L1 (X, A, µ), the map f # : Lloc,∞(X, A, µ) 3 v 7−→ Φ (f, v) ∈ is a K K loc K linear continuous functional, and the linear map

# : L1 (X, A, µ) 3 f 7−→ f # ∈ Lloc,∞(X, A, µ)∗ (36) loc K K is contractive.

(ii) For every v ∈ Lloc,∞(X, A, µ), the map v# : L1 (X, A, µ) 3 f 7−→ Φ (f, v) ∈ is a K K loc K linear continuous functional, and the linear map

# : Lloc,∞(X, A, µ) 3 v 7−→ v# ∈ L1 (X, A, µ)∗ (37) loc K K 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 prove the inequalities

Q (f) ≤ sup R fk dµ : k ∈ Lloc,∞(X, A, µ),Q (k) ≤ 1 , ∀ f ∈ L1 (X, A, µ); (38) 1 X K loc,∞ K Q (g) ≤ sup R hg dµ : k ∈ L1 (X, A, µ),Q (h) ≤ 1 , ∀ g ∈ Lloc,∞(X, A, µ). (39) loc,∞ X K 1 K (The reverse inequalities “≥” in both (38) and (39) hold trivially by (35).) To prove (38) we ﬁx f ∈ L1 (X, A, µ) and we deﬁne the function k : X → by K K  |f(x)|  if f(x) 6= 0  f(x) k(x) =   0 if f(x) = 0

Since k is measurable, and |k(x)| ≤ 1, ∀ x ∈ X, it is obvious that k belongs to Lloc,∞(X, A, µ), K R and has Qloc,∞(k) ≤ 1. Of course, by construction we have fk = |f|, so Q1(f) = |f| dµ = R X X fk dµ, and (38) follows immediately. To prove (39), ﬁx g ∈ Lloc,∞(X, A, µ) and denote the supremum in the right-hand side K 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 locally µ-neglijeable, thus there exists some F ∈ A, such that F ⊂ A and 0 < µ(F ) < ∞. Consider then the function h0 : X → K, deﬁned by  |g(x)|  if x ∈ F  g(x) h0(x) =   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, R and |h0| = χF , so in particular it follows that h0 is integrable, and Q1(h0) = |h0| dµ = R X X χF 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 Z h0g dµ ≥ (C + ε)µ(F ). (41) X

−1 Of course, the function h = µ(F ) h0 is also integrable, and has Q1(h) = 1. But now, if we divide the inequality (41) by µ(F ) we obtain Z hg dµ ≥ C + ε, X 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 injectivity failure 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, is called 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,∞(X, A, µ) = {0} and Lp (X, A, µ) = {0}, K K ∀ p ∈ [1, ∞), although the space L∞(X, A, µ) might be non-trivial, as seen for instance in K Example 2.

18 Remark 8. If (X, A, µ) is σ-ﬁnite, then it is nowhere degenerate. Indeed, if we write Sn ∞ X = n=1 Fn, with (Fn)n=1 ⊂ A disjoint and µ-ﬁnite, then for every A ∈ A we have P∞ µ(A) = n=1 µ(A ∩ Fn), with µ(A ∩ Fn) < ∞, ∀ n. Therefore, if µ(A) > 0, then at least one of 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 of Exercises), it contains some important results that are often overlooked, or entirely omitted from most Real Analysis textbooks. Our whole discussion is motivated by the following observation. Suppose one has two measure space structures (X, A, µ) and (X, B, ν) on the same 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

m K(X, A) ⊂ m K(X, B), (42) b (X, A) ⊂ b (X, B), (43) m K m K

NK(X, A, µ) ⊂ NK(X, B, ν), (44) Lp (X, A, µ) ⊂ Lp (X, B, ν), p ∈ [1, ∞], (45) K K which give rise to linear maps

T : Lp (X, A, µ) → L∞(X, B, ν). (46) p K K As it turns out, the maps (46) are isometric, for all p ∈ [1, ∞], so using the conventions from the previous sub-sections, we are going to write them as inclusions:

Lp (X, A, µ) ⊂ Lp (X, B, ν). (47) K K Although (42)-(45) may all be strict inclusions (if A ( B), we shall see that it is not impossible 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,∞(X, A, µ) → Lloc,∞(X, B, ν), K K 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 instance A = {∅,X}, with µ(X) = ∞, but the measure space (X, B, ν) can be very large, with plenty of 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 the indicator 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 Lp (X, A, µ) = Lp (X, B, ν); K K (ii) for every p ∈ [1, ∞), one has Lp (X, A, µ) = Lp (X, B, ν); K K (iii) for every ν-ﬁnite set B ∈ B, there exists A ∈ A, such that ν(B4A) = 0.

(Hint: By Exercise 11, Lp (X, A, µ) is closed in Lp (X, B, ν). Prove that condition (iii) K K is equivalent to the equality Lelem(X, A, µ) = Lelem(X, B, ν), and then use Exercise 3.) K K 13♥. Prove that the following are equivalent:

(i) L∞(X, A, µ) = L∞(X, B, ν); K K (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 equality E∞(X, A, µ) = E∞(X, B, ν), and then use Exercise 9.) K K 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 ∈ A each satisfy any one of the conditions (i)-(iii), then µ(A14A2) = 0, so in particular one has ˜ the equality µ(A1) = µ(A2). In particular, this means that, if we denote by A the collection of all B ⊂ X that satisfy any one the above conditions, we can deﬁne a mapµ ˜ : A˜ → [0, ∞] byµ ˜(B) = µ(A), where A ∈ A is any set that satisﬁes one of the conditions (i)-(iii). One can show that this way we obtain a measure space (X, A˜, µ˜) ⊃ (X, A, µ), called the completion of (X, A, µ), which will clearly (by Exercises 12 and 13) satisfy the equalities

Lp (X, A, µ) = Lp (X, A˜, µ˜), ∀ p ∈ [1, ∞]. K K The terminology one uses in connection with this construction is as follows. We say that that a measure space (X, A, µ) is complete, if N(A, µ) ⊂ A, or equivalently A˜ = A. As it turns 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 Real Analysis 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) = inf µ(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 outer measure associated with (X, A, µ). It is pretty obvious that µ∗(A) = µ(A), ∀ A ∈ A. If we apply then the Caratheodory construction, we obtain a σ-algebra M(µ∗) of all µ∗-measurable sets, on which the restriction of µ∗ is a measure. From now one we are going to denote the σ- ∗ c ∗ c c c algebra M(µ ) by A , and the measure µ Ac by µ . The resulting measure space (X, A , µ ) 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 above terminology. (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 following notation. 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 that every “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 to be locally complete). Exercise 15. Let X be an inﬁnite set, and let A = {∅,X} be equipped with the measure µ 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 collection {F ⊂ 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 outer measure. 5 At this point, the reader is urged to go back to BS III, sub-section ??, where all the meaningful features of 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 ∈ A with µ(F ) < ∞, then A ∈ M(µ∗). • If N ⊂ X is such that, µ∗(N ∩ F ) = 0, for all F ∈ A with µ(F ) < ∞, then N ∈ Nloc(µ∗).

Exercises 17-20. Let (X, A, µ) be a measure space, and let µ∗ be the associated Caratheodory 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 (X, A, µ) = Lp (X, Ac, µc), ∀ p ∈ [1, ∞). K K 19♥. Prove that, for a set A ∈ A, the following are equivalent:

• A is locally µ-neglijeable (in A); • A is locallyµ ˜-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

0 00 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 0 and T 00 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 c ˜ S∞ Remark 9. If (X, A, µ) is σ-ﬁnite, then A = A. Indeed, if we write X = n=1 Fn, where ∞ c c (Fn)n=1 ⊂ A have µ(Fn) < ∞, ∀ n, and we start with some set S ∈ A , then S ∩ Fn ∈ A ∗ ∗ ˜ has µ (S ∩ Fn) ≤ µ (Fn) = µ(Fn) < ∞, so by Exercise 17 it follows that S ∩ Fn ∈ A, ∀ n. S∞ ˜ 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:

Lp (X, A, µ) = Lp (X, A˜, µ˜) = Lp (X, Ac, µc), ∀ p ∈ [1, ∞); (52) K K K L∞(X, A, µ) = L∞(X, A˜, µ˜) ⊂ L∞(X, Ac, µc); (53) K K K Lloc,∞(X, A, µ) = Lloc,∞(X, A˜, µ˜) ⊂ Lloc,∞(X, Ac, µc). (54) K K K Without any additional (signiﬁcant) restrictions on (X, A, µ), one expects to have strict inclusion in both (53) and (54), as seen for instance in Exercise 15. The reason we focus on the Caratheodory completion, rather than the (ordinary) completion, 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). On the other hand, although the inclusion (54) is strict, it shows that a “better pair” (in the sense discussed in the previous section) for L1 (X, A, µ) is the space Lloc,∞(X, Ac, µc). The K K advantage of using this space, instead of Lloc,∞(X, A, µ) is the increased ﬂexibility. Ulti- K mately, under some additional (but not too restrictive) conditions, it will be shown that Lloc,∞(X, Ac, µc) is isometrically isomorphic to the dual Banach space L1 (X, A, µ)∗. K K We now turn our attention to the disassembling/reassembling problem for Lp-spaces. In order 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 the collection {A ∩ S : A ∈ A}, which is a σ-algebra on S. In the case when S ∈ A, we clearly have the inclusion A ⊂ A, so given a measure µ on A, its restriction to A will be a S S 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) m K(X, A) onto m K(S, A S), (b) b (X, A) onto b (S, A ), m K m K S

(c) NK(X, A, µ) onto NK(S, A S, µ S), (d) Nloc(X, A, µ) onto Nloc(S, A , µ ), K K S S (e) Lloc,∞(X, A, µ) onto Lloc,∞(S, A , µ ), and K K S S (f) Lp (X, A, µ) onto Lp (S, A , µ ), for all p ∈ [1, ∞]. K K S S In particular, one obtains the surjective linear maps

Qp : Lp (X, A, µ) → Lp (S, A , µ ), p ∈ [1, ∞], (55) S K K S S Qloc,∞ : Lloc,∞(X, A, µ) → Lloc,∞(S, A , µ ). (56) S K K S S

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

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

p Y p Dpu = (Q u)i∈I ∈ L (Xi, A , µ ) Xi K Xi Xi i∈I

p p belongs to the Banach space ` (L (Xi, A , µ ), k . kp)i∈I . Moreover, the map K Xi Xi

p p p Dp :(L (X, A, µ), k . kp) → ` (L (Xi, A , µ ), k . kp)i∈I (57) K K Xi Xi is a linear contraction.

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

loc,∞ Y loc,∞ Dloc,∞u = (Q u)i∈I ∈ L (Xi, A , µ ) Xi K Xi Xi i∈I

∞ loc,∞ belongs to the Banach space ` (L (Xi, A , µ ), k . kloc,∞)i∈I . Moreover, the map K Xi Xi

loc,∞ ∞ loc,∞ Dloc,∞ :(L (X, A, µ), k . kloc,∞) → ` (L (Xi, A , µ ), k . kloc,∞)i∈I (58) K K Xi Xi 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 generalized shortly, 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 be isometric) will be called the reassembly maps. This problem is extremely complicated for (57) with p = ∞, and we are not going to address it at all. Of course, by Exercise 22 (iii) the disassembly/reassembly problem has a positive 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 Q p P p Proof. Fix p ∈ [1, ∞), and v = (vi)i∈I ∈ L (Xi, A , µ ), with kvik < ∞, and i∈I K Xi Xi i∈I p let us prove the existence of some u ∈ Lp (X, A, µ), such that D u = v. By summability, K p the set I = {i ∈ I : v 6= 0} is countable, so the set A = S X belongs to A. Choose v i i∈Iv i p now, for every i ∈ Iv, a function fi ∈ L (Xi, A , µ ), such that [fi] = vi, and deﬁne the K Xi Xi function f : X → K by f (x) if x ∈ X and i ∈ I f(x) = i i v 0 if x 6∈ A

It is pretty clear that f is measurable, and furthermore fi if i ∈ Iv f X = i 0 if i 6∈ Iv so if we prove that f belongs to Lp (X, A, µ), then we will have the equality v = D [f]. K p To prove that f belongs to Lp (X, A, µ), we list I = {i : n ∈ M}, with the i ’s all K v n n different, where either M = N, or M = {1, 2,...,N}, and we define the increasing sequence ∞ of sets (An)n=1 ⊂ A, by  Sn X if n ∈ M  k=1 ik An =  A if n 6∈ M S∞ p ∞ which clearly satisfies the equality n=1 An = A, so the sequence (χAn |f| )n=1 of non-negative measurable functions on (X, A is non-decreasing, and satisfies

p p lim χA (x)|f(x)| = |f(x)| , ∀ x ∈ X. n→∞ n By Lebesgue’s Monotone Convergence Theorem, it follows that Z Z p p |f| dµ = lim χAn |f| dµ. (59) X n→∞ X

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

 Pn R p Pn R p k=1 X χXi |f| dµ = k=1 X |fik | d(µ ), if n ∈ M  k ik Xik R p c  X χAn |f| dµ = P R p P R p  χX |f| dµ = |fi| d(µ ), if n 6∈ M i∈Iv X i i∈Iv Xi Xi so by (59) we get: Z Z Z p p X p X p p |f| dµ = lim χAn |f| dµ = |fi| d(µ ) = kvikp = kvkp < ∞. n→∞ Xi X X Xi i∈Iv i∈Iv

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 X µ(A) = µ(A ∩ Xi). (60) i∈I

An A-partition (Xi)i∈I , satisfying the above equivalent conditions, is called µ-integrable. Proof. The implication (i) ⇒ (i0) is trivial. (i0) ⇒ (ii). Fix p ∈ [1, ∞), for which the map (57) is isometric. This means that,

Z X Z |f|p dµ = χ |f|p dµ, ∀ f ∈ Lp (X, A, µ). (61) Xi K X i∈I X Fix now some set A ∈ A with µ(A) < ∞, and let us prove the equality (60). This follows immediately from (61), applied to the characteristic function f = χA. (ii) ⇒ (i). Assume condition (ii) holds, fix some arbitrary p ∈ [1, ∞), and let us prove that the map (57) is isometric. Since Dp is norm-continuous, it suffices to prove the equality kD uk = kuk , for all u in the dense subspace Lelem(X, A, µ) ⊂ Lp (X, A, µ). In turn, for p p p K K this condition it suffices to prove that: Z X Z g dµ = χ g dµ, ∀ g ∈ Lelem(X, A, µ), (62) Xi K X i∈I X so basically we can assume that p = 1. Finally, by linearity, in order to prove (62) it suffices to 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 that µ(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 is µ-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 every A ∈ A with µ(A) < ∞, the fact that the sum in the right-hand side of (60) is ﬁnite forces the set I(A) = {i ∈ I : µ(A ∩ Xi) 6= 0} to be countable, so in particular, if one considers 0 S 0 the 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 A0) = 0. Comment. The disassembly/reassembly problem for (58) is a bit more complicated, and will 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 and Lloc,∞, 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,∞(X, A, µ), and let us show that the element K

∞ loc,∞ Dloc,∞u ∈ ` (L (Xi, A , µ ), k . kloc,∞)i∈I K Xi Xi

has kDloc,∞uk∞ = kukloc,∞. By Exercise 22 we already know that kDloc,∞uk∞ ≤ kukloc,∞, so all we need to to is to prove the inequality

kDloc,∞uk∞ ≥ kukloc,∞. (63)

loc,∞ If we choose a representative f ∈ L (X, A, µ) i.e. u = [f], then the element Dloc,∞u is K the I-tuple ([fi])i∈I , where fi = f , so the inequality (63) simply reads: Xi

i sup Qloc,∞(fi) ≥ Qloc,∞(f), (64) i∈I

i where, for every i ∈ I, the superscript in Qloc,∞ indicates that the seminorm Qloc,∞(fi) is loc,∞ computed in L (Xi, A , µ ). Denote the supremum in the left-hand side of (63) by K Xi Xi 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 ) < ∞. i Of course, for each i ∈ I, we have N ∩ Xi = {x ∈ Xi : |fi(x)| > C ≥ Qloc,∞(fi)} ∈ A , Xi which forces N ∩ Xi to be locally (µ )-neglijeable. In particular, if we start with some set Xi F ∈ A with µ(F ) < ∞, then the set F ∩ Xi ∈ A has µ (F ∩ Xi) < ∞, thus forcing Xi Xi µ((F ∩ N) ∩ Xi) = µ ((F ∩ Xi) ∩ (N ∩ Xi) = 0, ∀ i ∈ I. Xi P But now, using (60) with A = F ∩ N, we get µ(F ∩ N) = i∈I µ((F ∩ N) ∩ Xi) = 0, and we are done. Comment. In order to solve the disassembly/reassembly problem for (58), our approach will be to try to construct the reassembly map. As it will turn out, one can construct ∞ loc,∞ loc,∞ a reassembly map Rloc,∞ : ` (L (Xi, A , µ ), k . kloc,∞)i∈I → L (X, B, ν), where K Xi Xi K (X, B, ν) ⊃ (X, A, µ) is a slightly larger measure space structure. This reassembly map will be constructed in such a way that the composition Rloc,∞ ◦ Dloc,∞ agrees with the inclusion Lloc,∞(X, A, µ) ⊂ Lloc,∞(X, B, ν), and then the solution to the problem will depend on K K whether this inclusion is an equality. A very natural framework for reassembling I-tuples of measurable functions is provided by the following construction.

Deﬁnition. Suppose (Xi)i∈I is a partition of X. Assume on each Xi, i ∈ I, one is given a σ-algebra Bi. On can then form the direct sum σ-algebra W i∈I Bi = {B ⊂ X : B ∩ Xi ∈ Bi, ∀ i ∈ I}.

7 This will force kukloc,∞ = Qloc,∞(f) ≤ C.

It is pretty clear that this is the largest among all σ-algebras B that satisfy B X = B, ∀ i ∈ I. W S i One should be aware of the fact that Bi is larger than σ-alg Bi – the σ-algebra S i∈I i∈I generated by i∈I Bi. If I is countable, however, these two σ-algebras coincide. The next Exercises clarify this matter.

Exercises 23-24. Let (Xi)i∈I be a partition of X, and assume that on each Xi one is given a σ-algebra Bi.. 23. Prove that: Q S W (i) the map i∈I Bi 3 (Bi)i∈I 7−→ i∈I Bi ∈ i∈I Bi is a bijection; Q S S (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. W S 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 are equivalent: W • f is measurable as a map f : X, i∈I Bi → (Y, D);

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

(i) the map W Q D : m X, Bi 3 f 7−→ f ∈ m (Xi, Bi) (65) K i∈I Xi i∈I i∈I K is a linear isomorphism; (ii) when restricted to the space of bounded measurable functions, the map D yields an isometric linear isomorphism

b b W ∞ b D : m X, Bi 3 f 7−→ f ∈ ` (m (Xi, Bi), k . ksup)i∈I . (66) K i∈I Xi i∈I K

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

Deﬁnition. Assume A is a σ-algebra on X, and π = (Xi)i∈I is an A-partition. The W direct sum σ-algebra i∈I (A ), which we denote by Aπ, is called the π-enhancement of A. Xi It is obvious that Aπ = A , ∀ i ∈ I, and one has the inclusion Aπ ⊃ A. In general, this Xi Xi inclusion is strict, but if I is countable, then Aπ = A. Comment. The problem with direct sum σ-algebras is that, given a measure space 8 (X, A, µ) and an arbitrary A-partition π = (Xi)i∈I of X, it might be impossible to construct a 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 Lebesgue measurable 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 Caratheodory c c outer measure, and let (X, A , µ ) denote the Caratheodory completion. Assume π = (Xi)i∈I is a µ-integrable A-partition.

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

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

c c (iii) The partition π is both µ -integrable (when viewed as an A -partition) and µπ-integrable (when viewed as an Aπ-partition). (iv) The disassembly map9

π loc,∞ ∞ loc,∞ D : L (X, Aπ, µπ) → ` (L (Xi, A , µ ), k . kloc,∞)i∈I (67) loc,∞ K K Xi Xi is an isometric isomorphism.

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

(a) N ∈ Nloc(µ∗); loc ∗ (b) Xi ∩ N ∈ N (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), by Exercise 17. The implication (a) ⇒ (b) is trivial. Assume N ⊂ X has property (b), and let us show that N belongs to Nloc(µ∗). By Exercise 16 it suffices to show that µ∗(N ∩F ) = 0, for all F ∈ 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}, 0 S 0 then the set G = i∈I(G) Xi ∩ G belongs to A and satisfies µ(G r G ) = 0, which implies that ∗ ∗ 0 ∗ 0 µ (N ∩ F ) ≤ µ (G ∩ N ∩ F ) + µ ((G r G ) ∩ N ∩ F ) ≤ ∗ 0 ∗ 0 ∗ 0 (68) ≤ µ (G ∩ N ∩ F ) + µ (G r G ) = µ (G ∩ N ∩ F ). Of course, since G ⊃ N ∩ F , we have

0 [ [ G ∩ N ∩ F = Xi ∩ G ∩ N ∩ F = Xi ∩ N ∩ F, i∈I(G) i∈I(G) which, by the countability of I(G), the σ-subadditivity of µ∗, and (68), gives

∗ ∗ 0 X ∗ µ (N ∩ F ) ≤ µ (G ∩ N ∩ F ) ≤ µ (Xi ∩ N ∩ F ). i∈I(G)

9 π The map Dloc,∞ 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 inequalities now force µ∗(N ∩ F ) = 0. S (i). Start with some set A ∈ Aπ, which means that A = i∈I Ai, with Ai ∈ A and ∗ Ai ⊂ Xi, ∀ i ∈ I, and let us show that A is µ -measurable. According to Exercise 16, it suffices10 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 sets Fi = F ∩ Xi, so that the set I(F ) = {i ∈ I : µ(Fi) > 0} is countable, and the set 0 S 0 F = 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

0 [ 0 [ F ∩ A = (Fi ∩ Ai) and F r A = (Fi r Ai), (70) i∈I(F ) i∈I(F ) with Fi ∩ Ai and Fi r Ai in Ai ⊂ A, so using the fact that I(F ) is countable, it follows that both F 0 ∩ A and F 0 r A belong to A, and therefore we have

∗ 0 ∗ 0 ∗ 0 ∗ 0 0 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 0) ∩ A and (F r F 0) r A are contained in F r F 0, which is µ-neglijeable, it follows that µ∗((F r F 0) ∩ A) = µ∗((F r F 0) r A) = 0, so using the properties of outer measures we get the inequalities

∗ ∗ 0 ∗ 0 ∗ 0 µ (F ∩ A) ≤ µ (F ∩ A) + µ ((F r F ) ∩ A) = µ (F ∩ A), ∗ ∗ 0 ∗ 0 ∗ 0 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, and using (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

∞ loc,∞ v = (vi)i∈I ∈ ` (L (Xi, A , µ ), k . kloc,∞)i∈I , K Xi Xi

and let us construct an element u ∈ Lloc,∞(X, A , µ ), such that D u = v. First of K π π loc,∞ all, we know that any element in Lloc,∞ can be represented by a bounded measurable function with same norm (see Remark 4), so for every i ∈ I, one can choose some func- b loc,∞ tion fi ∈ m (Xi, A ), such that vi = [fi], in L (Xi, A , µ ) and also kfiksup = K Xi K Xi Xi kvikloc,∞. Since kfiksup ≤ kvikloc,∞ ≤ kvk∞, it follows that the I-tuple (fi)i∈I belongs to ∞ b W ` (m (Xi, A , µ ), k . ksup)i∈I . Since by deﬁnition, (A ) = Aπ, by Exercise 24, it K Xi Xi i∈I Xi b follows that there exists f ∈ m (X, Aπ), such that fi = f , ∀ i ∈ I, and then the element K Xi u = [f] ∈ Lloc,∞(X, A , µ ) will clearly satisfy the equality D u = v. K π π loc,∞ 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 the disassembly map (58) as an inclusion

Lloc,∞(X, A, µ) ⊂ Lloc,∞(X, A , µ ), K K π π so the condition that (58) is an isomorphism is equivalent to the equality

Lloc,∞(X, A, µ) = Lloc,∞(X, A , µ ). (72) K K π π Of course, again by Theorem 3, we also have the inclusion

Lloc,∞(X, A , µ ) ⊂ Lloc,∞(X, Ac, µc), K π π K so a suﬃcient condition for (72) is the equality

Lloc,∞(X, A, µ) = Lloc,∞(X, Ac, µc). (73) K K

Of course, one way to achieve (72) is to require the equality Aπ = A, which is automatic if I is countable. In general, one is also interested in having (73), and then the following terminology 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 called a 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

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; c (b) S ∩ Xi belongs to A , for each i ∈ I; c c (c) there exist A ∈ Aπ and a locally µ -neglijeable set N ∈ A , such that S = A r N. (ii) The inclusion Lloc,∞(X, A , µ ) ⊂ Lloc,∞(X, Ac, µc) is an equality. K π π K

(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 by Theorem 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 belongs c ∗ ∗ to A ), and has µ (Si) ≤ µ (Xi) = µ(Xi) < ∞. So in particular, there exists Ai ∈ A, such c ∗ that Ai ⊃ Si, and µ (Si) = µ (Si) = µ(Ai). Replacing Ai with Ai ∩ Xi, we can also assume c ∗ c that Ai ⊂ Xi. Of course, by finiteness, the set Ni = Ai r Si ∈ A has µ (Ni) = µ (Ni) = 0. S S Define then the sets A = Ai and N = Ni. On the one hand, since A∩Xi = Ai ∈ A, i∈I i∈I W it is clear that A belongs to the direct sum σ-algebra (A ) = Aπ. On the other hand, i∈I Xi ∗ c for every i ∈ I, the set Xi ∩ N = Ni has µ (Ni) = µ (Ni) = 0, so we can use Theorem 3 to conclude that N belongs to Ac and is locally µc-neglijeable. Of course, by construction we also have S = A r N. (ii). Since we already have an isometric inclusion Lloc,∞(X, A , µ ) ⊂ Lloc,∞(X, Ac, µc), K π π K all we need to do is to show that Lloc,∞(X, A , µ ) contains the dense subspace Eloc,∞(X, Ac, µc), K π π K c c which means that, for every S ∈ A , there exists A ∈ A , such that [χA] = [χS], in Lloc,∞(X, Ac, µc). But this is immediate from part (i). K 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 c c (A )π = A . 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,∞(X, A, µ) ⊂ Lloc,∞(X, Ac, µc) K K

is an equality. In particular, for any µ-integrable A-partition (Xi)i∈I of X, the disassem- loc,∞ ∞ loc,∞ bly map Dloc,∞ : L (X, A, µ) → ` (L (Xi, A , µ ), k . kloc,∞)i∈I is an isometric K K Xi Xi isomorphism.

Proof. Fix a decomposition κ = (Yj)j∈J for (X, A, µ). On the one hand, by Theorem 4 we know that Lloc,∞(X, Ac, µc) = Lloc,∞(X, A , µ ). On the other hand, since A = A, we K K κ κ κ obviously have Lloc,∞(X, A , µ ) = Lloc,∞(X, A, µ), so the preceding equality also gives the K κ κ K equality: Lloc,∞(X, Ac, µc) = Lloc,∞(X, A, µ). The second statement follows from Theorem K K 3, and the Comment that followed it. Comments. Decomposability is perhaps the nicest notion in Measure Theory. Among other 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-spaces and 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 we choose a countable A-partition π = (Xi)i∈I , with all Xi of ﬁnite measure, then obviously π is 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 locally compact Hausdorﬀ space, Bor(Ω) is the Borel σ-algebra, and µ is a Radon measure. A measure space of this form will be called a locally compact measure space. As in the previous sub-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 notation excessively, the σ-algebra Bor(Ω)c will be simply denoted by M(µ). Likewise, the measure c µ will be denoted by µM. In preparation for the main result (Theorem 6 below), it is useful to introduce some terminology, and we do so in the Exercises below. Exercises 26-27. Suppose Ω is a locally compact Hausdorﬀ space, and µ is a Radon measure on Ω. A non-empty compact set K ⊂ Ω is said to be µ-tight, if for every non-empty compact 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 has µ(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 compact µ-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 not µ-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), so T using 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 subsets T T ⊂ 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), there exists 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 partition of 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 S = S S ∩ X ⊂ S satisﬁes the equality η(S S ) = 0. 0 i∈I0(S) i r 0 Note that, by construction (without any conditions imposed on the partition π, or on the set S), one has the equality [ S0 = S ∩ Xi, ∀ S ⊂ X. (74)

i∈I0

33 Theorem 6 (Compact Decomposability of Radon Measures). Suppose Ω is a locally compact 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 measure space (Ω, M(µ), µM). Proof. To prove statement (i), we consider the set Θ of all collections U ⊂ P(Ω), with the following 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 ordered S by 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 singleton sets in Ω S K , so that, for i ∈ I I , the set K is a singleton, and has µ(K ) = 0. r i∈I0 i r 0 i i (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 first condition in (i), and the index set I0 (which could be empty!) is precisely the set of all i’s with 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 D = S D ∩ K ⊂ D satisﬁes the equality µ∗(D D ) = 0. 0 i∈I0(D) i r 0

To prove (a) we notice that, since all Ki, i ∈ I0, are µ-tight and have µ(Ki) > 0, by Exercise 26 it follows that, for every i ∈ I0(D), we have µ(D ∩ Ki) > 0. In particular, for every P S ﬁnite set F ⊂ I0(D), we have j∈F µ(D ∩ Ki) = µ i∈F D ∩ Ki ≤ µ(D), so we get P µ(D ∩ K ) ≤ µ(D) < ∞, which clearly forces I (D) to be countable. i∈I0(D) i 0 To prove (b) ﬁrst notice that, by construction, the set D0 is Borel, being a countable union ∗ of 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) = sup{µ(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 Exercise 27, such an L can be chosen to be µ-tight. But now we reached a contradiction, since L is disjoint 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 X µ(A) = µ(A ∩ Ki), (75) i∈I for all Borel sets A, with µ(A) < ∞. Since, the Ki’s are disjoint, for every finite set F ⊂ I one P S P has 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) = inf{µ(D): D open, D ⊃ A}, so in particular there exists an open set D ⊃ A, with µ(D) < ∞. By condition (i) the open set D is µ-essentially countably covered. As argued before, the set D = S D ∩ K = S D ∩ K is Borel, and 0 i∈I0 i i∈I0(D) i satisfies µ(D D ) = 0. Of course, the set A = A ∩ D = S A ∩ K is also Borel, and r 0 0 0 i∈I0 i 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) S 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 X X µ(A) = µ(A0) = µ(A ∩ Ki) ≤ µ(A ∩ Ki).

i∈I0(D) i∈I

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, then

for every p ∈ [1, ∞), the space Cc,K(Ω), of all continuous functions f :Ω → K, with compact support, is dense in Lp (Ω, M(µ), µ ). Indeed, for every A ∈ M(µ), with µ(A) < ∞, K M and every ε > 0, there exist a compact set K ⊂ A, and an open set D ⊃ A, such that µ(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 = 0 and f = 0, then one has the inequalities K ΩrD 1/p χK ≤ f, χA ≤ χD, which imply that kf − χAkp ≤ kχD − χK kp ≤ ε , thus proving that the norm-closure of C (Ω) in Lp (Ω, M(µ), µ ) contains all elementary integrable functions. c,K K M By Exercise 3 this norm-closure coincides with Lp (Ω, M(µ), µ ). K M