Measure Theory and Integration

CHAPTER 1

One fundamental result of measure theory which we will encounter in this Chapter is that not every set can be measured. More specifically, if we wish to define the measure, or size of subsets of the real line in such a way that the measure of an interval is the length of the interval then, either we find that this measure does not extend all subsets of the real line or some desirable property of the measure must be sacrificed. An example of a desirable property that might need to be sacrificed is this: the (extended) measure of the union of two disjoint subsets might not be the sum of the measures. The Banach-Tarski paradox dramatically illustrates the difficulty of measuring general sets; it is stated in Section ???????? without proof. It is with this in mind that our discussion of measure and integration begins, in this Section, by making precise what constitutes a measurable set and a measurable function and then an integrable function.

1. Algebras of Sets Let Ω be a set and (Ω) denote the power set of Ω which is the set of all subsets of Ω. TheP reader is referred to Chapter 0, Subsection 1 if any notations or set theoretic notions are unfamiliar. Notation: N0 is the set of nonnegative integers; Q is the set of rational numbers; Z is the set of all integers. Suppose Ω and Ω are two sets and f : Ω Ω is a function with 1 2 1 → 2 domain Ω1 and range in Ω2 (but f need not be onto). For any B Ω2, denote by ⊆ 1 f − (B)= x Ω : f(x) B . { ∈ 1 ∈ } the inverse image of the set B under f. Of course f need not be 1 one-to-one (injective) and so f − need not exist as a function. Some of the elementary properties of the inverse image are worthy of note: If B,B′ Ω then ⊆ 2 1 1 1 f − (B B′) = f − (B) f − (B′) 1 ∩ 1 ∩ 1 f − (B B′) = f − (B) f − (B′) 1∪ c 1 c∪ f − (B ) = f − (B) 25 26 1.MEASURETHEORYANDINTEGRATION

The verification is asked for in the Exercises at the end of the Section. Definition: If A Ω then define the characteristic function χ on ⊆ A Ω by χA(x)=1if x A and χA(x)=0if x Ω A. ∈ 1 ∈ − Exercise: Determine χA− (B) for an arbitrary B R. There are 4 cases, if = A = Ω. ⊆ ∅ Exercise: (De Morgan’s Laws) If Aα : α is a collection of subsets of a set Ω then the operation {of complementation∈ I} has the properties. C

( α Aα) = α (Aα) C ∪ ∈I ∩ ∈I C ( α Aα) = α (Aα) C ∩ ∈I ∪ ∈I C If = N is countable then we can write n NAn = n NBn where I c c ∪ c∈ ∪ c∈ B1 = A1, B2 = A2 A1, ... Bn = An A1 A2 ... An 1. The Bn are then disjoint. ∩ ∩ ∩ ∩ ∩ − Deﬁnition: Let An, n N be a countable collection of sets in Ω. The limit supremum of A ∈ is { n} lim sup An = n N k n Ak n ∩ ∈ ∪ ≥ whereas the limit inﬁmum of A is { n} lim inf An = n N k n Ak. n ∪ ∈ ∩ ≥

Intuitively lim supn An is all points that are in infinitely many of the An whereas liminfn An is all those points that are in all except possibly finitely many of the An. If A is a decreasing sequence of sets, A A A ... then n 1 ⊇ 2 ⊇ 3 ⊇ lim sup An = n NAn = liminf An n ∩ ∈ n Definition: A collection of subsets of Ω is an algebra (field) if F a. Ω . b. If A∈ F then Ac . ∈ F ∈ F c. If Aα : α and if is finite then α . { ∈I}⊆F I ∪ ∈I ∈ F Condition c in the definition could equivalently be replaced by either of the following Conditions.

c′. If A, B then A B . ∈ F ∪ ∈ F c′′. If A, B then A B . ∈ F ∩ ∈ F because Condition c′ is equivalent to c; and Condition c′′ in conjunction with Condition b is equivalent to Conditions c′ and b, as can be seen by observing that (A B)= A B. Deﬁnition: AC collection∩ Cof subsets∪C of Ω is a σ-algebra (or σ-ﬁeld) if is an algebra and F F 1. ALGEBRAS OF SETS 27

d. If A : α and if is countable, or ﬁnite, then { α ∈I}⊆F I α Aα . ∪ ∈I ∈ F Of course condition d implies condition c in the deﬁnition of an algebra. Conditions b and d together are equivalent to b and d′ where

d′. If A : α and if is countable, or finite, then { α ∈I}⊆F I α Aα . ∩ ∈I ∈ F Example 1.1. (i) The power set (Ω) of a set Ω is a σ- algebra. P (ii) , Ω is a σ-algebra. (iii) {∅, A,} Ac, Ω is a σ-algebra, if A is any subset of Ω. {∅ } Example (i) is the largest σ-algebra of subsets of Ω; Example (ii) is the smallest and Example (iii) is the smallest σ-algebra that contains the set A. Example 1.2. Consider the set of all real intervals of the form [a,b) along with all intervals of the form [a, ) and ( ,b) where a and b are real numbers. Define to be the∞ set of all−∞ finite unions of disjoint intervals of this form alongF with the null set and R itself. We claim that is an algebra. Certainly . If A then we must check that AcFis in and that is obvious∅ if ∈A F= or A∈= FR and so F ∅ we may suppose A = 1 j n[aj,bj) where a1 ≤ ∞ c −∞ and bn < then A = ( ,a1) [b1,a2) ... [bn 1,an) [bn, ). ∞c −∞ ∪ ∪ ∪ − ∪ ∞ Therefore A ; the cases where a1 = or b1 = or both are handled similarly.∈ F Therefore is closed under−∞ taking complements.∞ It remains to check thatF is closed under finite unions and it suffices to show that if A, B F then A B . If A = or ∈ F ∪ ∈ F ∅ A = R then this is obvious and so we suppose A = 1 j n[aj,bj) where ∪ ≤ ≤ a1 < b1 < a2 < b2 < a3 <...

Proposition 1.4. Let α : α be a non empty collection of σ-algebras (resp. algebras) on{F a given∈ set I}Ω indexed by a set . Then I the intersection α α is also a σ-algebra (resp. algebra). ∩ ∈I F Proof. We need only check that α α satisfies the three properties of σ-algebras The set Ω is certainly∩ ∈I inF all σ-algebras and hence c in α α. If A α α then A is in every α. Finally if (An)n N ∩ ∈I F ∈ ∩ ∈I F F ∈ is a sequence of sets in α α then n NAn is in every α and this completes the proof in the∩ case∈I F of σ-algebras∪ ∈ and the proofF for algebras is analogous. Suppose now that (Ω) is any collection of subsets of a set Ω. Then there is a smallestS ⊆Pσ-algebra which contains which we shall denote this σ( ). To see this observe that the set of allSσ-algebras that contain is nonS empty because it contains (Ω) and so the intersection of all σ-algebrasS that contain is a σ-algebra,P by the Proposition and it certainly contains and soS it must be the minimal such σ-algebra. Definition: If S is a σ-algebra then a subset with the property that = σF( ) is said to be a system of generatorsS ⊆ F of . Consider nowF the caseS that Ω = Rn, where n 1. Similar consider-F n ≥ ations apply if Ω is any topological space. On R x =(x1,...,xn), we 2 1/2 ∋ define the norm x =( 1 j n xj ) and the corresponding (usual) topology where| a| base for≤ the≤ | neighborhoods| of a point x Rn is n ∈ Vr(x) : r > 0 where Vr(x) = y R : x y < r is an open ball{ of radius r}centered at x. Then{ ∈ a set U |is said− | to be} open in Rn, if, for every x U there is r > 0 so that Vr(x) U. The set of all open sets is denoted∈ and we define the Borel⊆subsets of Rn to be = σ( ), that is theO smallest σ-algebra that contains . We shall sometimesB O write (Rn) for when wishing to emphasizeO that it is the Borel σ-algebra onB Rn. InB general, if Ω is a topological space and is the set of all open subsets of Ω then the Borel σ-algebra on ΩO is σ( ). ORemark: If denotes the set of all closed subsets of Rn then σ( )= . For recallC that a set is closed if and only if its complement is open.C SinceB contains the open sets and is closed under the operation of taking complements,B it follows that and hence σ( ). Conversely the open sets are contained inB⊇Cσ( ): σ( ).B Therefore ⊇ C = σ( ) σ( ) and so = σ( ). ThereforeC O ⊆ isC a system of BgeneratorsO of⊆ byC the definitionB ofC and is aO second system of generators. B B C Example: The set of all intervals of the form = ( ,a): a R is a system of generatorsS of the Borel σ-algebraS of{ R−∞: = σ( ∈) Indeed,} since consists solely of open sets it is obviousB that B σ( S). S B ⊇ S 1. ALGEBRAS OF SETS 29

It therefore suﬃces to show that any open set is contained in σ( ). Certainly if a

We shall have occasion to use the Monotone Class Lemma when verifying uniqueness of measures. Exercises:

(1) Show that, if B,B′ Ω then ⊆ 2 1 1 1 f − (B B′) = f − (B) f − (B′) 1 ∩ 1 ∩ 1 f − (B B′) = f − (B) f − (B′) 1∪ c 1 c∪ f − (B ) = f − (B)

(2) Show that, in general, liminfn An lim supn An (3) Recall the deﬁnition of the limit supremum⊆ and limit inﬁmum of a sequence of real numbers an.

lim sup an = inf sup ak : k n : n 1 and n N { { ≥ } ≥ } ∈ lim inf an = sup inf ak : k n : n 1 n N ∈ { { ≥ } ≥ } where the “value” (resp. ) is possible if the sequence is unbounded above∞ (resp. below).−∞ (See page two of Ash’s book Measure, Integration, and Functional Analysis.) How

is limsup χAn related to χB where B = limsup An? What

happens if we replace limsup χAn by liminf χAn ? (4) Suppose that A = x R3 : x (0, 0, ( 1)n/n) < 1 . Find n { ∈ | − − | } lim supn An and liminfn An. Compare this with Exercise 3, page 3 of Ash’s book. (5) Show that of Example 1.2 is not a σ-algebra. F (6) Suppose that E Ω is an arbitrary subset of a set Ω and is a collection of subsets⊆ of Ω. Deﬁne E = E A : A S . Show that ∩S { ∩ ∈ S} a. If 0 is an algebra of subsets of Ω then E 0 is also an algebraF of subsets of E. ∩ F b. σ(E ) = E σ( ). That is the smallest σ-algebra in (E)∩S containing∩ ES is the intersection of E with the smallestP σ-algebra in∩S(Ω) containing . c. Conclude further that,P if is a σ-algebraS of subsets of Ω, then E is a σ-algebraF of subsets of E. (Reference: Halmos’s∩ F Measure Theory) 32 1.MEASURETHEORYANDINTEGRATION

2. Measurability It is an objective of this Chapter to define the integral of a function. If f is a nonnegative function defined on an interval in the real line then the integral should be able to tell us the area under the graph of f just as the Riemann integral does in calculus. It will tell us much more but even in this limited context we shall discover that f must be restricted: the notion of area under the graph of f will not make sense for every f and we will be forced to restrict the class of functions considered. We introduce in this Section the concept of a “measurable” function. We shall see that f must be measurable if we are to make sense of the notion of area under the graph. Of course this difficulty arises already in the case of Riemann integration: the Riemann integral is not defined for arbitrary functions f. We shall clarify this remark and discuss the Riemann integral and its relation to the “Lebesgue integral,” introduced here, at the end of this Chapter. Definition: Let Ω be a set and be a σ-algebra of subsets of Ω. The we shall refer to (Ω, ) as a measurableF space. The sets in will be referred to as measurableF sets. F Definition: Suppose (Ω , ) and (Ω , ) are two measurable 1 F1 2 F2 spaces. Then a mapping f : Ω1 Ω2 is measurable, with respect 1 → to 1 and 2 if f − (B) 1 whenever B 2. We shall sometimes sayFf is measurableF function∈ F from (Ω , )∈ to F (Ω , ) to clarify the 1 F1 2 F2 choice of σ-algebras 1 and 2. Unless otherwise specified a function n F F f : Ω1 R is said to be measurable or Borel measurable if f is measurable→ from (Ω , ) to (Rn, ). 1 F1 B An elementary example of a measurable function is given by the characteristic function χA of a set A . Then χA is measurable as a mapping from (Ω, )to(R, ) if and only if A , that is if and only if A is a measurableF set. FurtherB examples of measurable∈ F functions will be apparent once some elementary properties have been established.

Lemma 2.1. Suppose that (Ω , ), (Ω , ) and (Ω , ) are 1 F1 2 F2 3 F3 three measurable spaces and that f : Ω1 Ω2 and g : Ω2 Ω3 are two mappings. If f and g are both measurable→ then the composed→ function g f is measurable from (Ω , ) to (Ω , ). ◦ 1 F1 3 F3

Proof: We must show that, for an arbitrary set C 3, (g 1 ∈ F ◦ f)− (C) . It suﬃces to show that ∈ F1

1 1 1 (2.1) (g f)− (C)= f − (g− (C)) ◦ 2.MEASURABILITY 33

1 1 1 because g− (C) 2 because g is measurable and so f − (g− (C)) 1 because f is measurable.∈ F The veriﬁcation of (2.1) is a straightforwar∈ Fd exercise in checking the equality of sets. 2

Proposition 2.2. Suppose that f : Ω1 Ω2 is a function and and are σ-algebras on Ω and Ω respectively.→ Suppose further F1 F2 1 2 that is a system of generators of 2, which is to say 2 = σ( ). Then S F F 1 S f is measurable from (Ω1, 1) to (Ω2, 2) if and only if f − (B) 1 for every B . F F ∈ F ∈S 1 Proof: It is obvious that, if f is measurable then f − (B) 1 for 1 ∈ F every B simply because 2. Conversely, suppose f − (B) 1 ∈S S ⊆ F 1 ∈ F for every B . Let = B : f − (B) so that . We ∈S T { ∈ F2 ∈ F1} T ⊇S shall show that is a σ-algebra which implies that = 2 and so, by the deﬁnition ofT , f is measurable. This will completeT theF proof. We check thereforeT that is a σ-algebra. Certainly , because 1 T ∅∈T1 f − ( ) = 1. We check next that if B then f − (B) 1 ∅ ∅1 ∈c F 1 c ∈ T ∈ F and so f − (B ) = f − (B) 1 because 1 is closed under taking 1 ∈c F 1 c F complements. (Recall f − (B ) = f − (B) by an Exercise.) We check 1 ﬁnally that if (Bn)n N is a sequence of sets in so that f − (Bn) for each n N then∈ T ∈ T ∈ 1 1 f − ( n NBn)= n Nf − (Bn) 1 ∪ ∈ ∪ ∈ ∈ F because 1 is a σ-algebra. This shows that is closed under countable unions andF is therefore a σ-algebra and theT proof is complete. 2 Remark: This set in the proof constitute “good” sets and the argument that there areT many good sets is an instance of what Ash calls the good sets principle in his text page 5. Remark: The above proof uses a special case of the following observation: If f is a mapping f : Ω1 Ω2 and if (Bα)α is a collection of subsets of Ω indexed by a set then→ ∈I 2 I 1 1 f − ( α Bα) = α f − (Bα) ∈I ∈I 1 ∪ ∪ 1 f − ( α Bα) = α f − (Bα). ∩ ∈I ∩ ∈I On the other hand if (Aα)α is a collection of subsets of Ω1 then ∈I f( α Aα)= α f(Aα) ∪ ∈I ∪ ∈I but it may happen that f( α Aα) = α f(Aα). ∩ ∈I ∩ ∈I Corollary 2.3. A continuous function f : Rm Rn is Borel measurable. → Of course the understood σ-algebras here are the Borel σ-algebra (Rm) and (Rn). B B 34 1.MEASURETHEORYANDINTEGRATION

Proof: Since the (Rn) is generated by the open sets, it suffices 1 B m to show that f − (U) (R ) for an arbitrary open set U. The result will then follow from∈B the Proposition. However f is continuous means 1 1 m that f − (U) is open whenever U is and so f − (U) (R ). 2 ∈B The same reasoning applies in a more general setting. Corollary 2.4. If f is a continuous function from one topological space (X, Σ) to another (Y, ) then f is measurable from (X, (X)) to (Y, (Y )) where (X) is theT Borel σ-algebra which is generatedB by the openB sets Σ of XB and similarly (Y ) is generated by the open sets . B T Proposition 2.5. Suppose that (Ω, ) is a measurable space F and f1, f2, ... fm are m Borel measurable real valued functions, fj : (Ω, ) (R, ), for 1 j m. Suppose that g : (Rm, (Rm)) (R, F(R→)) is measurable.B ≤ Then≤ B → B φ(x)= g(f1(x),f2(x),...,fm(x)) defines a measurable function φ : (Ω, ) (R, (R)). F → B We shall set aside the proof of the Proposition until later and consider its consequences. Corollary 2.6. If f : (Ω, ) (R, ), j = 1, 2 are Borel j F → B measurable functions, then f1 + f2 is also Borel measurable. Proof. This is an application of the Proposition with g : R2 R → defined by g(x1,x2) = x1 + x2. Because g is continuous it is Borel measurable. Corollary 2.7. If f : (Ω, ) (R, ), j = 1, 2 are Borel j F → B measurable functions, then the product f1f2 is also Borel measurable. In particular, if k is a real constant then kf2 is Borel measurable. Proof. In this case g(x,y) = xy. The special case follows by defining f1(x)= k which makes f1 measurable because it is continuous.

Corollary 2.8. If f : (Ω, ) (R, ), is a Borel measurable function then f is also Borel measurable.F → B | | Proof. In this case g(x)= x . | | Corollary 2.9. If fj : (Ω, ) (R, ), j = 1, 2 are Borel measurable functions, and if f (x) =F 0→for allBx Ω then the ratio 2 ∈ f1/f2 is also Borel measurable. 2.MEASURABILITY 35

Proof. In this case we define x/y if y =0 g(x,y)= 0 if y =0 so that g is defined on all of R2 but of course it is not continuous. We shall check however that g is measurable, as a mapping from 2 2 1 2 (R , (R )) to (R, ). It suffices to show that g− ( ,a) (R ) B B 1 −∞ ∈ B for any real a by Proposition 2.2. If a< 0 then g− ( ,a)= (x,y) R2 : x/y < a is easily seen to be open in R2 and−∞ hence in {(R2).∈ If } 1 2 B a 0 then g− ( ,a) = (x,y) R : x/y < a (x,y) : y = 0 which≥ is the union−∞ of an open{ and a∈ closed set and is} therefore ∪ { in (R2).} This proves that g is measurable and so the above Proposition appliesB which completes the proof. Exercise: Show that if f : (Ω, ) (R, ), j = 1, 2 are Borel j F → B measurable functions then max f1,f2 and min f1,f2 are also measurable. Remark: max a,b ={ ( a b} + a + b)/{2 for any} reals a and b. { } | − | It remains to establish the Proposition. Proof of the Proposition: It suffices to show that the mapping, f say, defined by f(x)=(f1(x),f2(x),...fm(x)) is measurable as a mapping from (Ω, ) to (Rm, (Rm) because the composition of measurable functionsF is measurable.B To verify f is measurable, m it suffices to show that, for an arbitrary a = (a1,a2,...,am) R , 1 ∈ f − ( x1 < a1,x2 < a2,...,xm < am ) by Proposition 2.2. How- ever { } ∈ F 1 1 f − ( x1

f(x)= λjχAj (x) 1 j m ≤≤ The set of all simple functions is denotes (Ω, ) or when the context is clear. S F S Observe that a simple function is measurable because it is the sum of measurable functions. Also a simple function takes on only finitely many values, λj, 1 j m and possibly 0. Conversely a measurable function that takes≤ on finitely≤ many values is simple. To see this, sim- 1 ply define, for each λ in the image of f, A = f − ( λ ). The set j j { j} S 36 1.MEASURETHEORYANDINTEGRATION is closed under addition, scalar multiplication and multiplication and therefore forms a linear algebra over R of functions. (Recall that a linear algebra is a vector space with a multiplication operation (f,g) fg that is associative and distributes over addition from the left and→ right and, for any scalar α, α(fg)=(αf)g = f(αg). Reference: Naimark’s Normed Algebras 7.) It is further worth noting that the representation § f(x)= 1 j m λjχAj (x) in the definition of f is not unique. ≤ ≤ ∈S Proposition 2.11. Let fn, n N be a sequence of Borel measurable real valued functions defined on∈ a measurable space (Ω, ). Sup- pose that F

lim fn(x)= f(x) exists in R for every x Ω. n N ∈ ∈ Then the function f : (Ω, ) R, so defined, is measurable. F → This result says, briefly, that the pointwise limit of measurable functions is measurable. Proof: We will show that 1 1 (2.2) f − (( ,a]) = p N n N j nfj− (( ,a +1/p]) −∞ ∩ ∈ ∪ ∈ ∩ ≥ −∞ for every a R The set of all sets of the form ( ,a] generate the Borel σ-algebra∈ because any open set is generated.−∞ Therefore, we will have shown fB is measurable by Proposition 2.2. 1 To verify (2.2), let x f − (( ,a]) so that f(x) a. Then, ∈ −∞ ≤ for any p N there is n so that fj(x) a + 1/p for all j n: 1 ∈ ≤ 1 ≥ x fj− (( ,a +1/p]) for all j n or x j nfj− (( ,a +1/p]). ∈ −∞ ≥ ∈ ∩ ≥ −∞ But p was arbitrary and so x belongs to the right side of (2.2) and this 1 shows that f − (( ,a]) is a subset or equal to the right side of (2.2). Conversely suppose−∞ that x belongs to the right hand side of (2.2). Then, for every p N, there exists n N so that fj(x) a +1/p for all j n. Taking∈ limits in this last expression∈ as j ≤ we see that f(x) ≥ a +1/p. Since p is arbitrary f(x) a and this→ shows ∞ the right ≤ 1 ≤ side of (2.2) is in f − (( ,a]) which verifies (2.2). This proves that f is measurable. −∞ 2. The Extended Reals R: We introduce the extended real line R = [ , ], sometimes referred to as the two point compactification of −∞ ∞ the real line, and this is just R with two points adjoined: R = R . This will be a convenience when±∞ discussing convergence.∪ {∞}∪{−∞} We introduce the following topology on R. The neighborhoods of (resp. ) are those sets which contain an interval of the form (a, ∞] for some−∞a R (resp. [ ,a)) and the neighborhoods of x R are the∞ usual: those∈ sets that contain∞ an interval of the form y : ∈y x <δ { | − | } 2.MEASURABILITY 37 for some δ > 0. Of course the topology that R inherits as a subset of R is its usual topology. As a consequence of these definitions we see that R is homeomorphic to the compact interval [ 1, 1] with the (usual) topology it inherits as a subset of R. Indeed − x if x R √x2+1 ∈ (2.3) Φ(x)= 1 if x =  1 if x = ∞  − −∞ is continuous from R to [ 1, 1] with inverse − x if 1 0 be given. We shall construct a simple function gǫ so that gǫ f and f(x) gǫ(x) < ǫ for all x Ω. Since f is bounded we have≤ 0 |f −M for| some positive constant∈ M. We ≤ ≤ choose a0 =0

gǫ(x)= ajχAj 0 j m ≤≤ Then gǫ f

Remark: It is not true in general that the convergence is uniform in this more general setting. Indeed, if f is unbounded then it cannot be approximated uniformly by bounded functions. For suppose fn be a sequence of functions, uniformly convergent to a function f and suppose each f is bounded (with bound depending on n). If f f < 1 then n | − n| f fn < 1 which says that f is bounded. In the setting of the Theorem|| |−| || above, simple functions| are| bounded and so we cannot expect uniform convergence. Proof of Theorem 2.15: For each p N, define h = min f,p ∈ p { } so that hp is a bounded, nonnegative measurable function and so it is possible to choose a simple function g 0 so that h 1 g p ≥ p − p ≤ p ≤ h by Proposition 2.13. Then we define f = g , f = max f ,g , p 1 1 2 { 1 2} ... fp = max fp 1,gp so that fp is simple, nonegative and increasing. We also see{ by− induction} on p, f h f. To check the pointwise p ≤ p ≤ convergence, suppose x0 Ω. If f(x0) < then we may choose p0 N ∈ ∞ ∈1 so that p0 f(x0). Then for p p0, f(x0)= hp(x0) so that f(x0) p = 1 ≥ ≥ − hp(x0) gp fp(x0) f(x0) and so f(x0) = limp N fp(x0). On the − p ≤ ≤ ≤ ∈ other hand, if f(x ) = then h (x ) = p and g (x ) p (1/p) so 0 ∞ p 0 p 0 ≥ − that fp(x0) gp(x0) p (1/p). Therefore limp fp(x0)= = f(x0). 2 ≥ ≥ − ∞ Proof of Theorem 2.16 The proof of Proposition 2.14 applies here except the uniform convergence there must be replaced by pointwise convergence. Let gn be a sequence of nonnegative simple functions convergent to f + = max f, 0 as guaranteed by the preceding Theo- { } rem. Similarly let hn be simple functions convergent to f − and define + fn = gn hn. Then fn = gn hn gn + hn f + f − = f . Moreover,− if f(x) 0 then| | h |(x)=0− and| ≤ ≤ | | ≥ n + lim fn(x) = lim gn(x)= f (x)= f(x) n N n N ∈ ∈ Similarly f(x) < 0, then gn(x)=0 and

lim fn(x) = lim hn(x)= f −(x)= f(x) n N n N ∈ ∈ − − Note that limn gn(x) = and limn hn(y) = are both possible, but only if x = y. ∞ ∞ 2 The set of simple functions is an algebra as we have seen. Moreover if φ : R R is Borel measurable then φ f is simple whenever f is. → ◦ 1 In particular f is simple if f is and max f,g = 2 (f + g + f g ) is simple whenever| | f and g are. Similarly min{ f,g} is simple. | − | { } 40 1.MEASURETHEORYANDINTEGRATION

3. Measures and Premeasures A measure, or more generally a set function is a mapping from a σ-algebra into R = [ , ]. Arithmetic on R is defined as follows. Significantly −∞is not∞defined but ∞ − ∞ + a = if a = ∞ ∞ −∞ + a = if a = −∞ −∞ ∞ a = if a> 0 ∞ ∞

Definition: A measure on a measurable space (Ω, ) is a map- F ping of to R = [0, ] such that F + ∞ (1) ( )=0 ∅ (2) ( α Aα)= α (Aα) ∪ ∈I ∈I whenever A : α is a countable collection of sets in which { α ∈ F ∈ I} F are pairwise disjoint which means that Aα Aβ = whenever α = β. A triple (Ω, ,) where is a measure∩ on the∅ measurable space (Ω, )isa measureF space or sometimes measured space to distinguish it fromF a measurable space. In the special case that (Ω) < then is said to be a finite measure and if (Ω) = 1 then is said∞ to be a probability measure and (Ω, ,) as a probability space. Remark: If x : α Fis a set of nonnegative numbers then { α ∈ I} α xα = supF α F xα where the supremum is taken over all finite subsets∈I F of and the∈ sum could be . It is not necessary for this I ∞ definition that be countable but of course if xα > 0 : α is uncountableI then the sum is necessarily .{ (reference: General∈ I} Topology by Bourbaki, Chapter IV, 4.3 and ∞7.1.) A somewhat more primitive concept§ than a§ measure is a premeasure Definition: Let 0 be an algebra of subsets of a set Ω. A premeasure is a mappingF of to R = [0, ] such that 0 + ∞ (1) ( )=0 ∅ (2) ( α Aα)= α (Aα) ∪ ∈I ∈I whenever A : α is a countable collection of sets in { α ∈ F0 ∈ I} F0 which are pairwise disjoint and provided α Aα is in 0. Of course a measure is a premeasure and∪ ∈I a premeasureF is a measure if its domain is a σ-algebra. Property 2 for measures and premeasures is referred to as countable additivity. If a set function 0 is countably additive on an algebra 0 then F (A)= (A)+ ( ) forany A 0 0 ∅ ∈ F0 α N ∈ 3.MEASURESANDPREMEASURES 41

Therefore if 0(A) < we must have 0( ) = 0 that is, property 2 of premeasures implies∞ property 1. Therefore∅ property 1 could equally well be replaced by 0(A) < for some A . A measure or premeasure∞ must be ﬁnitely∈ Fadditive (that is could be taken to be ﬁnite in property 2) because, by property 2 I

0( α Aα) = 0([ α Aα] [ n N ]) ∪ ∈I ∪ ∈I ∪ ∪ ∈ ∅ = (A )+ ( )= (A ) 0 α 0 ∅ 0 α α n N α ∈I ∈ ∈I An elementary consequence of this is that, whenever A B and A, B then (A) (B), because (B) = (A)+ (B⊆ A) (A∈) F 0 ≤ 0 0 0 − ≥ 0 (Of course, R is ordered so that a if a R.) As a consequence, ∞ ≥ ∈ when we define a measure or premeasure 0 to be finite if 0(Ω) < , we assure that indeed (A) < for every A. ∞ 0 ∞ Example 3.1. Discrete Measures: Let Ω be any set and = (Ω) (the power set of Ω). For every x Ω assign a nonnegative FweightP 0 p and define ∈ ≤ x ≤ ∞ (A)= px. x A ∈ Property 1 of measures is easily checked and property 2 is left as an exercise to the reader. In the special case that px =1 for every x then is called the counting measure on Ω. Example 3.2. Let Ω be any nonvoid set and = (Ω) be the power set and let ( )=0 and (A)= if Ω A =F .P Then is a measure. ∅ ∞ ⊇ ∅ More interesting examples will be constructed by starting with a distribution function which we now define. Definition: A real valued function F defined on R is a distribution function if (1) F is increasing which means, whenever x

Theorem 3.5. Suppose that 0 is a premeasure on an algebra of subsets of a set Ω. Then F0 (1) If B A and A, B then 0(B) 0(A). (2) If A⊆ : α is a∈ countable F collection≤ of sets in and if { α ∈ I} F0 α Aα 0 then ∪ ∈I ∈ F 0( α Aα) 0(Aα) ∪ ∈I ≤ α ∈I (3) If A A A ... is an increasing sequence of sets in 1 ⊆ 2 ⊆ 3 ⊆ F0 and if n NAα 0 then ∪ ∈ ∈ F lim 0(An)= 0( n NAn) n N ∈ ∈ ∪ (4) If B B B ... is a decreasing sequence of sets in 1 ⊇ 2 ⊇ 3 ⊇ F0 such that n NBn is also in 0 and if 0(Bk) < for some k N then∩ ∈ F ∞ ∈ lim 0(Bn)= 0( n NBn) n N ∈ ∈ ∩ Proof. Part 1 was discussed early in this section. For Part 2 we recall from 1.1 that α Aα can be written as the union of disjoints § ∪ ∈I sets Bn . Indeed if α : N is an enumeration of and if with ∈ F c →c I I Bn = Aα(n) Aα(n 1) ... Aα(1). ∩ − ∩ ∩

0( α Aα)= 0( n NBn)= 0(Bn) 0(Aα(n)) ∪ ∈I ∪ ∈ ≤ n N n N ∈ ∈ where the last inequality follows because Bn Aα(n). Of course the order of summation is irrelevant since the terms⊆ are nonnegative. 3.MEASURESANDPREMEASURES 43

Consider Part 3. We observe that, if (A )= for some k, then 0 k ∞ 0( n NAn) 0(Ak)= and so there is nothing to check and so we ∪ ∈ ≥ ∞ can suppose that 0(Ak) is ﬁnite for every k. We have, by the additivity property for measures,

0( n NAn) = 0(A1)+ 0(A2 A1)+ ... + 0(Ap Ap 1)+ ... ∪ ∈ − − − = lim 0(A1)+ 0(A2 A1)+ ... + 0(Ap Ap 1) p N − ∈ − − = lim 0(Ap) p N ∈ We shall suppose that (B ) < ; the general case is very similar. 0 1 ∞ Let B = n NBn. Then B1 Bn is an increasing sequence of sets whose union is B∩ ∈ B and so by− Part 3 1 − 0(B1) 0(B)= 0(B1 B) = 0( n N(B1 Bn) − − ∪ ∈ − = lim 0(B1 Bn) n N ∈ − = 0(B1) lim 0(Bn) n N − ∈ This implies Part 4 because (B ) < . 0 1 ∞ Remark: In Part 4 of the preceding Theorem, the hypothesis 0(Bk) < for some k cannot be dispensed with. For example if is the counting∞ measure on the rational numbers Q then ( 1/n, 1/n)= − for every n but ( n N( 1/n, 1/n)) = ( 0 )=1. ∞ The Theorem is of∩ ∈ course− true for measures{ } in place of premeasures in which case would be replaced by a σ-algebra say and so F0 F the α Aα and n NBn are automatically in and so the statement simpliﬁes∪ ∈I slightly∩ in∈ this case. F There is a partial converse to parts 3 and 4 of the Theorem that is helpful for checking the properties of premeasures.

Proposition 3.6. Suppose that 0 is an algebra of subsets of a set Ω and : R is a mappingF so that 0 F0 → + (1) ( )=0 0 ∅ (2) ( α Aα)= α (Aα) ∪ ∈I ∈I whenever A : α is a ﬁnite collection of sets in which are α 0 0 pairwise disjoint.{ ∈ F Suppose∈ I} in addition that either one of thFe following two conditions is valid. a. Whenever A A A ... is an increasing sequence of 1 ⊆ 2 ⊆ 3 ⊆ sets in 0 such that the limit A = α Aα is also in 0 then F ∪ ∈I F limn N 0(An)= 0(A) b. Whenever∈ B B B ... is a decreasing sequence of sets 1 ⊇ 2 ⊇ 3 ⊇ in 0 such that the limit n NBn = then limn N 0(Bn)=0 F ∩ ∈ ∅ ∈ 44 1.MEASURETHEORYANDINTEGRATION

Then, in either case, 0 is a premeasure.

Proof. Let C1, C2, C3 ...be a sequence of pairwise disjoint sets in 0 and such that C = p NCp is also in 0. Define An = 1 p nCp. F ∪ ∈ F ∪ ≤ ≤ Assume Condition a. It implies limn N 0(An) = 0(C) because ∈ An C and C 0. By finite additivity, 0(An)= 1 p n 0(Cp) so thatր ∈ F ≤ ≤ 0(Cp) = lim 0(Cp)= 0(C) n N p N ∈ 1 p n ∈ ≤≤ which verifies 0 is countably additive and hence a premeasure under Condition a. Assume Condition b. We define B = C A so that B and n − n n ց ∅ therefore Condition b implies limn 0(Bn) = 0 We have

0(C)= 0(An)+ 0(Bn)= 0(Cp)+ 0(Bn) 1 p n ≤≤ Now if we take the limit in n N and we have (C) = (C ) ∈ 0 n 0 p which says 0 is countably additive. We recall from Example 3.3 that each distribution function F defines a set function so that ((a,b]) = F (b) F (a) which is a finitely 0 0 − additive set function on the set 0 of all finite disjoint unions of intervals of the form (a,b]and( ,bF]) and (a, ). Of course (( ,b]) = −∞ ∞ 0 −∞ limn N = 0(( n,b]) = limn N F (b) F ( n), with the value of pos- ∈ − ∈ − − ∞ sible. In general, if A 0 then 0(A) = limn N 0(A ( n,n]). We ∈ F ∈ ∩ − are now ready to show that 0 is a premeasure.

Lemma 3.7. Let F be any distribution function and let 0 be the corresponding finitely additive set function defined on the algebra as in Example 3.3 (so that, in particular ((a,b]) = F (b) F (a)). F0 0 − Then 0 is a premeasure. Proof: Consider first the finite case where F (x) is constant outside ( N,N] for some N so that is finite. We shall use Part b of − 0 the preceding proposition and so we assume that Bp 0 forms a decreasing sequence of sets and B . Given ǫ> 0, we∈ F will show that p ց ∅ 0Bp) <ǫ for all suffiiciently large p and this will imply the countable additivity of . We claim that there is another sequence C with 0 p ∈ F0 compact closures C B and such that (B C ) < ǫ/2p This is p ⊆ p 0 p − p possible because for each subinterval (a,b] of Bp we have 0((a,b]) = F (b) F (a) = lima′ a+ F (b) F (a′) = lima′ a+ ((a′b]) by the right continuity− of F . We→ may also− suppose that →C ( N,N] for all p. p ⊆ − Since p NCp p NBp = it follows that, for some n, 1 p nCp = ∩ ∈ ⊆∩ ∈ ∅ ∩ ≤ ≤ ∅ 3.MEASURESANDPREMEASURES 45 by a compactness argument. Consequently

0(Bn) = 0(Bn 1 p nCp) −∩ ≤ ≤ = 0( 1 p nBn Cp) ∪ ≤ ≤ − p 0( 1 p nBp Cp) ǫ/2 <ǫ ≤ ∪ ≤ ≤ − ≤ 1 p n ≤≤ Since (Bp)p N is a decreasing sequence and ǫ > 0 was arbitrary this ∈ says 0(Bp) 0 as p and this verifies 0 is countably additive, and so a premeasure,ց in→ the ∞ case F (x) is constant outside ( N,N] for some N. − Consider now the general case. For each N N let F (x) = F (x) ∈ N if x N and FN (x) = F ( N) if x N and FN (x) = F (N) if x | |N ≤. Then F is a distribution− function≤ − and the corresponding ≥ N finitely additive set function N is in fact a premeasure by the first part of the proof and (A) = (A) if A and A ( N,N]. N 0 ∈ F0 ⊆ − Moreover, for any A 0, limN N N (A) = (A) by the discussion ∈ F ∈ preceding the statement of this result. Let (Ap)p N be a sequence of ∈ pairwise disjoint sets such that A = p NAp is in 0. Then, simply because is finitely additive we have,∪ for∈ any n NF 0 ∈

0(A) 0( 1 p nAp)= 0(Ap) so that 0(A) 0(Ap) ≥ ∪ ≤ ≤ ≥ 1 p n p N ≤≤ ∈ Conversely, by the special case

0(A)= lim N (A)= lim N (Ap) lim 0(Ap)= 0(Ap) N N N N ≤ N N ∈ ∈ p N ∈ p N p N ∈ ∈ ∈ because (B) (B) for any B . Combining these two in- N ≤ 0 ∈ F0 equalities we have 0(A) = p N 0(Ap) and this verifies the 0 is a premeasure and completes the proof.∈ 2 Thus every distribution function corresponds to a premeasure (and indeed, we shall see, to a measure). The following result provides a partial converse. Definition: We shall say that is a Lebesgue-Stieltjes measure if is a measure on the Borel subsets of R and (K) < if K is a compact subset of R B ∞ Proposition 3.8. Suppose that is a Lebesgue-Stieltjes measure and C is any real constant. Define F (x)= ((0,x]) + C if x 0 and F (x) = C ((x, 0]) if x < 0. (Here (0, 0] = by convention.)≥ Then F is a distribution− function and ((a,b]) = F (b)∅ F (a) whenever a

Therefore every Lebesgue-Stieltjes measure corresponds to a distribution function which is unique up to an additive constant. Proof: Since (A) (B) whenever A B are sets in , ((0,x]) is increasing x 0 and≤((x, 0]) is decreasing⊆ for x 0 andB so F is increasing on R≥. ≤ Suppose a 0. Then ≥ lim F (x)= C + ( x>a(0,x]) = C + ((0,a]) = F (a) x a+ → ∩ so that F is right continuous on [0, ). Next suppose a< 0. Then ∞ lim F (x)= C ( x>a(x, 0]) = C ((a, 0]) = F (a) x a+ → − ∪ − so that in fact F is right continuous everwhere and is therefore a distribution function. Next check that F (b) F (a) = ((a,b]) whenever a

4. Extensions and Measures: In this Section we show that certain premeasures extend to measures and that a measure may be “completed.” As an application we are able to construct Lebesgue Stieltjes measures. Recall the definition: Definition: A real valued function F defined on R is a distribution function if (1) F is increasing which means, whenever x

Lemma 4.1. Suppose that 0 is a premeasure defined on an algebra 0 and (Ap)p N and (Bp)p N are two increasing sequences of sets F ∈ ∈ in 0 with limits, A = p NAp and B = p NBp. If A B then F ∪ ∈ ∪ ∈ ⊆ limp N (Ap) limp N (Bp) ∈ ≤ ∈ Proof: Because is a premeasure we have for any q N 0 ∈ 0(Aq) = lim 0(Aq Bp) lim 0(Bp) p N p N ∈ ∩ ≤ ∈ Taking the limit in q gives the result. 2 Define therefore to consist of all A Ω such that there exists an G ⊆ increasing sequence (Ap)p N 0, so that Ap A and define also a set function on by ∈ ∈ F ր 0 G 0(A) = lim 0(Ap). p N ∈

The preceding Lemma assures that 0 is well deﬁned because it does not depend on the particular sequence (Ap)p N 0 which approximates A. We see further that extends because∈ ∈ F any A can be 0 0 ∈ F0 written as the limit of a constant sequence. An = A.

Lemma 4.2. Suppose that 0 is a ﬁnite premeasure on an algebra 0 and 0 is its extension to as described above. Then is closed underF ﬁnite intersections and countableG unions and G a. If G , G are in then 1 2 G (G G )+ (G G )= (G )+ (G ) 0 1 ∪ 2 0 1 ∩ 2 0 1 0 2 48 1.MEASURETHEORYANDINTEGRATION

b. If (Gp)p N is an increasing sequence in then G = p NGn is in and∈ G ∪ ∈ G lim 0(Gn)= 0(G) p N ∈ c. If G is a sequence in then p G 0( p NGp) 0(Gp) ∪ ∈ ≤ p N ∈ Proof: Suppose that Ap, Bp are increasing sequences of sets in 0 and Ap G1 and Bp G2 so that G1,G2 . It follows that FA B Gր G and A ր B G G so that∈ G is closed under p ∩ p ր 1 ∩ 2 p ∪ p ր 1 ∪ 2 G ﬁnite unions and intersections. Moreover since 0 is a premeasure we can take the limit in p N in the relation (A B )+ (A B )= ∈ 0 p ∪ p 0 p ∩ p 0(Ap)+ 0(Bp) to verify Part a. To verify Part b, suppose that Gp is an increasing sequence in , G G. For each p, suppose A is an increasing sequence in G p ր pq F0 convergent to Gp = q NApq. ∪ ∈ A A ...A ... G 11 12 1q ր 1 A21 A22 ...A2q ... G2 . . . ր...... Ap1 Ap2 ...Apq ... Gp . . . ր......

Deﬁne Bq = 1 p qApq. Then Bq is an increasing sequence in 0 and ∪ ≤ ≤ F it is convergent to G so that G and so 0(G) = limq N 0(Bq). Certainly (G) (G ) for any∈p Gby the preceding Lemma.∈ On the 0 ≥ 0 p other hand 0(Gp) 0(Bp) so that limp N 0(Gp) limp N 0(Bp) = ≥ ∈ ≥ ∈ 0(G). This establishes Part b. To verify Part c, it suﬃces to check that, for any n N ∈ 0( 1 p nGp) 0(Gp) ∪ ≤ ≤ ≤ 1 p n ≤≤ because we may take the limit in n N and applying Part b above. In ∈ the case n = 2 we have 0(G1 G2) 0(G1)+ 0(G2) by Part a. The case of general n follows by induction.∪ ≤ The proof is complete. 2.

Thus extension 0 is a countably additive set function and 0 is closed under countable unions, but A does not imply AGc ⊇ F. Also unions of sets in may not be expressible∈ G as unions of disjoint∈ G sets in . G G Deﬁnition: A mapping ∗ on the power set (Ω) of a set Ω to R+ is said to be an outer measure if P

(1) ∗( )=0 ∅ 4.EXTENSIONSANDMEASURES: 49

(2) If A B Ω then ∗(A) ∗(B). ⊆ ⊆ ≤ (3) If Aα is a countable collection of sets in Ω then ∈I

∗( α Aα) ∗(Aα) ∪ ∈I ≤ α ∈I Frequently an outer measure will not be a measure because it is not additive but it is possible that a restriction to a σ-algebra of subsets of Ω may indeed be a measure. We now suppose that is a finite premeasure on an algebra 0 F0 and that 0 is the extension of 0 to the of the preceding Lemma. Later in this Section we weaken the assumptionG of finite to “σ-finite” to be defined below. Define

(4.1) ∗(A) = inf (G): G A, G { 0 ⊇ ∈ G} for any A Ω. Alternatively, in terms of itself, we deﬁne ⊆ 0

λ(A) = inf 0(Ap): Ap 0, p NAp A ∈ F ∪ ∈ ⊇ p N ∈ We claim ∗(A) = λ(A). For suppose that A , p N is a sequence in p ∈ 0 and p NAp A, Deﬁne G = p NAp so that G and 0(G) = F ∪ ∈ ⊇ ∪ ∈ ∈ G limn 0( 1 p nAp) p N 0(Ap). Consequently ∗(A) λ(A). Con- ∪ ≤ ≤ ≤ ∈ ≤ versely suppose G so that there is a sequence Ap 0 so that A G and we can∈ form G a sequence B which is∈ pairwise F dis- p ր p ∈ F0 joint and p NBp = G so that 0(G)= p N 0(Bp). This shows that ∪ ∈ ∈ ∗(A) λ(A) and complete the veriﬁcation that λ = ∗. ≥ Now let us check that ∗ is indeed an outer measure and record some of its properties.

Lemma 4.3. Suppose that 0 is a finite premeasure on an algebra of subsets of a set Ω and that ∗ is defined as indicated above; (see F0 (4.1)). Then ∗ is an outer measure and agrees with 0 on 0 and 0 on . Moreover F G (1) ∗(A B)+∗(A B) ∗(A)+∗(B) for any sets A, B Ω. ∪ ∩ ≤ c ⊆ In particular ∗(A)+ ∗(A ) 0(Ω). (2) If A , p N is an increasing≥ sequence of sets in Ω then p ∈ limp N ∗(Ap)= ∗( p NAp). ∈ ∪ ∈ Proof: It is clear that ∗(G) = (G) if G , by the definition 0 ∈ G (4.1) of ∗ and because (G) (G′) if G G′ and G′ . Similarly 0 ≤ 0 ⊆ ∈G one sees that, if A B then ∗(A) ∗(B), by the definition (4.1) of ⊆ ⊆ ∗. 50 1.MEASURETHEORYANDINTEGRATION

Therefore, to complete the check that ∗ is an outer measure, we suppose that Ap, p N is a sequence of subsets. of Ω. We choose ∈ p G , G A so that (G ) ∗(A )+2− ǫ for each p N. Then p ∈G p ⊇ p 0 p ≤ p ∈ p ∗( p NAp) 0( p NGp) 0(Gp) ∗(Ap)+ ǫ/2 ∪ ∈ ≤ ∪ ∈ ≤ ≤ p N p N ∈ ∈ where we have applied Part c of the preceding Lemma. Therefore ∗ is an outer measure. Check next Part 1. We suppose that A, B Ω and ǫ> 0 is given. ⊆ Choose G ,G A G and B G and (G ) < ∗(A)+ ǫ and 1 2 ∈ G ⊆ 1 ⊆ 2 0 1 0(G2) <∗(B)+ ǫ. Then

∗(A)+∗(B) > (G )+ (G ) 2ǫ = (G G )+ (G G ) 2ǫ 0 1 0 2 − 0 1 ∪ 2 0 1 ∩ 2 − by Part a of the preceding Lemma. Since G1 G2 A B and G G A B and G G and G G ∪ , we⊇ have∪ 1 ∩ 2 ⊇ ∩ 1 ∪ 2 ∈G 1 ∪ 2 ∈G ∗(A)+ ∗(B) > (G )+ (G ) 2ǫ ∗(A B)+ ∗(A B) 2ǫ 0 1 0 2 − ≥ ∪ ∩ − and, since ǫ> 0 was arbitrary this implies Part 1. It remains to check Part 2. Certainly ∗(An) ∗( p NAp) for any ≤ ∪ ∈ n N so that limp N ∗(Ap) ∗( p NAp). To check the converse, ∈ ∈ ≤ ∪ ∈ suppose that ǫ > 0 is arbitrary, and that, for each p N, Gp ∈p ∈ G is chosen so that G A and (G ) ∗(A )+2− ǫ. We shall p ⊇ p 0 p ≤ p show that there is a increasing sequence Hp so that Hp Ap q ∈ G ⊇ 0(Hp) ∗(Ap) + 1 q p 2− ǫ. The existence of such a sequence would imply≤ that ≤ ≤ q ∗( p NAp) 0( p NHp) = lim 0(Hp) lim ∗(Ap)+ 2− ǫ. ∪ ∈ ≤ ∪ ∈ p N ≤ p N ∈ ∈ 1 q p ≤≤ so that ∗( p NAp) limp N ∗(Ap)+ ǫ and since ǫ is arbitrary this ∪ ∈ ≤ ∈ would complete the proof. To construct the sequence Hp we proceed by induction on n supposing H1 H2 ... Hn have been constructed ⊆ ⊆ ⊆ q in and Hp Ap and 0(Hp) ∗(Ap)+ 1 q p 2− ǫ for 1 p n. G ⊇ ≤ ≤ ≤ ≤ ≤ Deﬁne Hn+1 = Hn Gn+1 so that Hn+1 and ∪ ∈G (H ) = (H G ) 0 n+1 0 n ∪ n+1 = (H )+ (G ) (H G ) 0 n 0 n+1 − 0 n ∩ n+1 p n+1 ∗(A )+ 2− ǫ + ∗(A )+ ǫ/2 ∗(A ) ≤ n n+1 − n 1 p n ≤≤ p since An Hn Gn+1. Thus 0(Hn+1) ∗(An+1)+ 1 p n+1 2− ǫ and that⊆ completes∩ the construction and≤ therefore the proof≤.≤ 2 Thus every ﬁnite premeasure 0 extends to an outer measure ∗. One of the less desirable features of ∗ is that there may exist sets A ⊆ 4.EXTENSIONSANDMEASURES: 51

c Ω so that ∗(A)+ ∗(A ) >0(Ω) (but of course 0 ∗(A) 0(Ω)). However if we omit such sets then we have a measure≤ as the≤ following Theorem asserts.

Theorem 4.4. Suppose that 0 is a finite premeasure defined on an algebra 0 of subsets of a set Ω and that ∗ is the outer measure defined by (4.1).F Define c = A Ω: ∗(A)+ ∗(A )= (Ω) F { ⊆ 0 } Then is a σ-algebra, and indeed and the restriction F F ⊇ F0 F ⊇G of ∗ to is a measure that extends and . F 0 0 Proof: It is immediately clear that and that A implies c ∅ ∈ F ∈ F A . Also 0 because ∗ = 0 on 0 and 0 is finitely additive. ∈Suppose F thatF ⊆A, F B . By Part 1F of the preceding Lemma we have ∈ F

(4.2) ∗(A B)+ ∗(A B) ∗(A)+ ∗(B) c ∪c c ∩ c ≤ c c ∗(A B )+ ∗(A B ) ∗(A )+ ∗(B ). ∪ ∩ ≤ We add the two relations and recall that A, B . ∈ F c c c c ∗(A B)+ ∗(A B)+ ∗(A B )+ ∗(A B ) 2∗(Ω) ∪ ∩ c ∪ ∩ c ≤ [∗(A B)+ ∗((A B) )]+[∗(A B)+ ∗((A B) )] 2∗(Ω) ∪ ∪ ∩ ∩ ≤ by rearranging. But we also know by Part 1 of the preceding lemma c that ∗(A B) + ∗((A B) ) ∗(Ω) and ∗(A B) + ∗((A c ∪ ∪ ≥ ∩ ∩ B) ) ∗(Ω) and so the above bound proves we must have equality ≥ c c ∗(A B)+∗((A B) )= ∗(Ω) and ∗(A B)+∗((A B) )= ∗(Ω) which∪ says that ∪is closed under union and∩ intersection.∩ Thus is an algebra. We canF say more because we must have equality in equaF tion (4.2) and so, in the case A B = , we have ∗(A B)= ∗(A)+∗(B) ∩ ∅ ∪ so that ∗ is ﬁnitely additive on . Suppose now that A , p NFis an increasing sequence in ; we p ∈ F want to show that A = p NAp is also in . By the preceding Lemma ∪ ∈ F c ∗(A) = limp N ∗(Ap) and so ∗(Ap)+∗(Ap)= ∗(Ω), for all p implies ∈ c c ∗(A)+limp N ∗(Ap)= ∗(Ω). But Ap is a decreasing sequence of sets c c ∈ c Ap A and so this implies ∗(A)+ ∗(A ) ∗(Ω). But by Part 1 of ⊇ c ≤ the previous Lemma ∗(A)+ ∗(A ) ∗(Ω) and so we have A . ≥ ∈ F This proves that is a σ-algebra and since limp N ∗(Ap)= ∗(A), by F ∈ the preceding Lemma, ∗ is countably additive on and therefore a measure when restricted to . Finally we recall thatF consists of the limits of all increasing sequencesF in and of courseG these limits are F0 in any σ-algebra that contains 0 and so and ∗ agrees with 0 by the preceding Lemma. F F⊇G 2 52 1.MEASURETHEORYANDINTEGRATION

The next step is to generalize to not necessarily ﬁnite measures.

Definition: A premeasure 0 on an algebra 0 of subsets of a set Ω is said to be σ-finite if there is a sequence setsF A so that p ∈ F0 0(Ap) < and Ω = p NAp. Thus a∞ finite premeasure∪ ∈ is σ-finite and a Lebesgue-Stieltjes measure is also σ-finite since it is bounded on every bounded interval. Theorem 4.5. Carathéodory Extension Theorem: Sup- pose that is a σ-finite premeasure defined on an algebra of sub- 0 F0 sets of a set Ω. Then 0 has an unique extension to a measure on the smallest σ-algebra, σ( ) that contains . F0 F0 Remark: The hypothesis that 0 be σ-finite is only necessary to prove the uniqueness of the extension of 0 and its not required to prove existence. See Problem 3, page 22 of Ash’s book. Proof: Let E be a sequence of disjoint subsets in of finite p F0 0 measure such that Ω = p NEp. Then Ep 0 is an algebra of subsets of E (by an Exercise∪ in∈ 1.1) and (A)=∩ F (A E ) defines p § p 0 ∩ p a premeasure on Ep 0 for each p N (check this). Since p is a finite premeasure, it extends∩F as to a measure∈ on a σ-algebra that contains Ep 0 and hence σ(Ep 0)= Ep σ( 0). Denote the extension by as∩ F well. Define on σ∩( F ) by ∩ F p F0 (A)= (A E ) p ∩ p p N ∈ Then extends because if A then 0 ∈ F0 (A)= (A E )= (A) 0 ∩ p 0 p N ∈ because 0 is countably additive. Moreover if Aq is a sequence of disjoint subsets in σ( ) then F0 ( q NAq) = 0( q NAq Ep) ∪ ∈ ∪ ∈ ∩ p N ∈ = (A E ) 0 q ∩ p p N q N ∈ ∈ = (A E )= (A E )= (A ) 0 q ∩ p p q ∩ p q q N p N q N q N ∈ ∈ ∈ ∈ because the order of summation does not matter if all terms are nonnegative. Therefore is σ-additive and therefore a measure on σ( ). F0 We have shown that 0 has an extension defined on σ( 0) and it remains to show is unique. Suppose therefore that ν isF another 4.EXTENSIONSANDMEASURES: 53 measure defined on σ( 0) so that 0(A)= ν(A) for all A 0. Define, for fixed p , =FA σ( ): (A E )= ν(A E∈) F. Then ∈N M { ∩ F0 ∩ p ∩ p } M contains 0; moreover it is a monotone class because the limit of any increasingF or decreasing sequence of sets in is also in . By the Monotone Class Lemma the smallest monotoneM class thatM contains F0 is σ( 0) and so (A Ep)= ν(A Ep) for every A σ( 0). But p N was arbitraryF and so∩ ∩ ∈ F ∈ (A)= (E A)= (E A)= ν(E A)= ν(A) p ∩ p ∩ p ∩ p p if A σ( 0) by countable additivity and so = ν which verifies uniqueness.∈ F 2. Corollary 4.6. If F is a distribution function defined on R then there is one and only measure on the Borel subsets (R) so that ((a,b]) = F (b) F (a) for every half closed interval (a,b]B. − Proof: This follows from the Carathéodory Extension Theorem by way of Lemma 3.7. Exercise Construct a set which is unbounded (and measurable) but has finite Lebesgue measure. Example: Consider the rationals Q and define an algebra 0 as follows. All intervals of the form (a,b] where a 0 there is B 0 so that (A∆B∈) F<ǫ. ∞ ∈ F Proof: In the case that is finite we recall the construction of Lemma 4.2 of and of (G) = sup (A ) : A G . Recall G ⊇ F0 0 { 0 p p ր } further that if A then (A) = ∗(A) = inf (G) : G ,G ∈ F { 0 ∈ G ⊇ A . It follows that A∆Ap =(A Ap) (Ap A) (G Ap) (G A) has} arbitrarily small measure provided− ∪ G − is⊆ chosen− appropriately∪ − ∈ G and then Ap 0 approximates G closely enough. This proves the result in the case∈ F is finite. In the general case let E be a sequence of disjoint subsets in p F0 of finite 0 measure such that Ω = p NEp. Define p by p(A) = (A E ) so that is a finite measure∪ on∈ E . By the first part of the ∩ p p p proof, if A , and (A) < and ǫ> 0, then there exist sets Ap Ep ∈ F ∞ p ⊆ with Ap 0 with p((A Ep)∆Ap) < ǫ2− . Consider B = p NAp. Then ∈ F ∩ ∪ ∈ (A∆B)= ((A∆B) E )= ((A E )∆A ) E ) <ǫ ∩ p ∩ p p ∩ p p N p N ∈ ∈ Here we have used the observations that B Ep = Ap and that (C∆D) E =(C E)∆(D E), for any sets C,D∩and E ∩ Of course∩ B may∩ not itself be in but it can be approximated by F0 BN = 1 p N Ap which is in 0 and BN B We have ∪ ≤ ≤ F ր lim (A∆BN ) = lim ((A BN ) (BN A)) N N →∞ →∞ − ∪ − = lim (A BN )+ (BN A)= (A∆B) < ǫ. N →∞ − − by Theorem 3.5 and because (A) < . Therefore for N large enough ∞ we have (A∆BN ) < 2ǫ say. Since ǫ > 0 is arbitray the proof is complete. 2 Exercise: Show that, for any Borel subset A of the real line with finite Lebesgue measure and for any ǫ> 0, there is a compact (that is closed and bounded) set K so that the Lebesgue measure of A∆K is at most ǫ. Definition 4.8. A measure space (Ω, ,) is said to be complete (or is said to be complete) if, wheneverF A has measure (A)=0, then every subset B A is also in . ∈ F ⊆ F Therefore is complete if every subset of a set of measure zero is a set of measure zero. Suppose now that (Ω, ,) is any measure space. Define F = N Ω : there exists M ,M N and (M)=0 . N { ⊆ ∈ F ⊇ } 4.EXTENSIONSANDMEASURES: 55

Define further to consist of all sets of the form A N where A and N . F ∪ ∈ F Exercise∈N Show that = A∆N : A and N F { ∈ F ∈ N } is equal to . Suggestion: Begin by showing that, for N M, F ⊆ A N = (A M)∆(M (A N) ∪ − ∩ ∪ A∆N = (A M) [M (A∆N)] − ∪ ∩ Theorem 4.9. Suppose that (Ω, ,) is a measure space and F is defined as above. Define on by (A N) = (A), for all AF N so that where A Fand N ∪. Then is a σ- ∪ ∈ F ∈ F ∈ N F algebra, is well defined and a measure that extends and (Ω, , ) is a complete measure space. F Proof: It is clear that contains . Moreover if A N is a F ∅ p ∪ p sequence in (so that AP and Np ) then p Ap Np = F ∈ F ∈ N ∪ ∈N ∪ ( p Ap) ( p Np) is in also. Suppose next that A N with∪ ∈NA ∪ and∪ ∈NN andF suppose that N M where M∪ ∈and F ∈ F ∈N ⊆ ∈ F (M) = 0. To show that contains the complement of A N we note that (A N)c = Ac FN c = (Ac M c) Ac (N c M c∪). Now N c M c = N∪c M M and∩ so Ac (∩N c M∪c) ∩ and− so (A N)c − ∩ ⊆ ∩ − ∈N ∪ is in and we have is a σ-algebra. CheckF next that Fis well defined. Suppose therefore that A N = 1 ∪ 1 A2 N2 where A1,A2 and N1,N2 . Then A1 = (A1 A2) (A ∪ A ). and (A ∈A F) = 0 because∈A N A (A N )∩ A =∪ 1 − 2 1 − 2 1 − 2 ⊆ 1 ∪ 1 − 2 (A2 N2) A2 N2. Therefore (A1) = (A1 A2); and by a symmetric∪ − argument⊆ (A ) = (A A ). This shows∩ (A N ) = 2 1 ∩ 2 1 ∪ 1 (A1) = (A2) = (A2 N2) so that is well defined. Obviously extends ∪ Check next that is countably additive. Suppose that A N is a p ∪ p sequence of pairwise disjoint sets in (so that AP and Np ) then F ∈ F ∈N

( p Ap Np) = (( p Ap) ( p Np)) ∪ ∈N ∪ ∪ ∈N ∪ ∪ ∈N = ( p Ap)= (Ap)= (Ap Np) ∪ ∈N ∪ p p ∈N ∈N because A forms a sequence of pairwise disjoint sets in . Therefore p F is a measure. To check that it is complete, suppose that A N where N and A and (A N)= (A) = 0. Then A∪ N ∈ F and so any∈N subset of∈A F N is also∪ in and so has zero ∪measure.∈N Thus is complete. ∪ N 2 56 1.MEASURETHEORYANDINTEGRATION

We have constructed Lebesgue-Stieltjes measure on the Borel sets (R) corresponding to a distribution function F and of course Lebesgue Bmeasure corresponds to the case F (x)= x. Each of these measures has a (unique) completion; and usually “Lebesgue Stieltjes measure” refers to this completion. Historical Note: Constantin Carath´eodory (September 13, 1873 to February 2, 1950) was born in Berlin of Greek parents and grew up in Brussels. He is known for his work in the calculus of variations, conformal representations and the foundations of thermodynamics be- sides measure theory; he was also an accomplished linguist, speaking six modern and several ancient languages. He joined the faculty of the ill-fated Greek University of Smyrna (in present day Izmir, Turkey) in 1920 and stayed until the Greek expulsion in 1922. He was Professor of Mathematics at the University of Munich 1924-1950. 5.THEINTEGRAL 57

5. The Integral In this section we introduce the integral and derive some of its elementary properties. Suppose that (Ω, ,) is a measure space and = (Ω, ) is the F S S F set of all simple functions; see Definition 2.10. Define 0 to be the subset of which consists of all those f so that ( x S Ω: f(x) = 0 ) < .S (Observe that x Ω : f(x) ∈S= 0 is measurable{ ∈ because f } ∞ { ∈ } is.) Consequently f = 1 j n λjχAj for some scalars λj and Aj ≤ ≤ ∈ F with (A ) < . We define the integral of f with respect to as j ∞

f d = λj(Aj) 1 j n ≤≤ Since the representation of f is not unique it is necessary to check that the integral is well deﬁned. Suppose therefore that f 0 and so f takes on ﬁnitely many non zero distinct values f : 1 ∈Si n . If A = x Ω : f(x) = f { i ≤ ≤ } i { ∈ i} then f = 1 i n fiχAi . We want to show that if there is any other ≤ ≤ representation f = 1 j m λjχBj then ≤ ≤ fi(Ai)= λj(Bj) 1 i n 1 j m ≤ ≤ ≤≤

It is convenient to deﬁne f0 = 0 and A0 = ( 1 j mBj) ( 1 i nAi). Then, for arbitrary i, 0 i n ∪ ≤ ≤ − ∪ ≤ ≤ ≤ ≤

fiχAi = λjχBj Ai ∩ 1 j m ≤≤

Since the Ai are disjoint, if one shows that fi(Ai)= 1 j m λj(Bj ≤ ≤ ∩ A ) then that suﬃces (for we can sum over i.) We also note that i 1 j mBj Ai = Ai. For i > 0 this is because fi = 0 and for i = 0 ∪ ≤ ≤ ∩ it is by the choice of A0. We now subdivide Ai into disjoint subsets as follows. For each subset I of 1 j m let { ≤ ≤ } c CI = [( j I Bj) ( j /I Bj)] Ai = ( j I Bj) j /I Bj Ai ∩ ∈ − ∪ ∈ ∩ ∩ ∈ ∩ ∩ ∈ ∩

If J is another such subset of 1 j m then C J is disjoint from { ≤ ≤ } m CI . Some of the CI may be empty but their union (over all 2 1 choices of I which are nonempty) is A . Suppose now that x −C . i ∈ I Then we must have fi = j I λj. But we also have Ai = I CI where ∈ ∪ the sum is over all 2m 1 nonvoid subsets I of 1 j m . Take the − { ≤ ≤ } intersection with Bj to get Bj Ai = I Bj CI = I jCI where now ∩ ∪ ∩ ∩ ∋ 58 1.MEASURETHEORYANDINTEGRATION the intersection is over all subsets I which contain j. Therefore λ (B A ) = λ (C ) j j ∩ i j I 1 j m 1 j m I j ≤≤ ≤≤ ∋ = λj(CI )= fi(CI )= fi(Ai) I j I I ∈ where the interchange of summation is justiﬁed by checking that index set (j,I):1 j m,I j is the same set as (j,I): I = , j I . { ≤ ≤ ∋ } { ∅ ∈ } Theorem 5.1. Suppose f,g (Ω, ,) and λ is a real. Then ∈S0 F a. Ω λf d = λ Ω f d; b. f + gd = f d + gd; Ω Ω Ω c. f d 0 if f 0; Ω ≥ ≥ d. f d gd if f g; Ω ≤ Ω ≤ e. f d f d. | Ω |≤ Ω | | Proof. The proof of Parts a and c are left as elementary exercises.

For Part b we have f = 1 i n fiχAi and g = 1 j m gjχBj . Then ≤ ≤ ≤ ≤ f + g = 1 k m+n hkχCk where hk = fk and Ck = Ak, if 1 k n ≤ ≤ ≤ ≤ and hk = gk n and Ck = Bk n, if n < k n + m. Therefore − − ≤

f + gd = hk(Ck) Ω 1 k n+m ≤ ≤ = fk(Ak)+ gk n(Bk n) − − 1 k n n

= fk(Ak)+ gj(Bj)= f d + gd 1 k n 0

Deﬁnition: Let fp be a sequence of measurable functions deﬁned on a measure space (Ω, ,) and let A and f be a measurable F ∈ F function. Then fp is said to converge in measure on A to f if, for every ǫ> 0,

lim ( x A : fp(x) f(x) >ǫ )=0. p N ∈ { ∈ | − | } 5.THEINTEGRAL 59

If (Ω, ,) is a probability space and A = Ωthen fp is said to converge in probabilityF .

Proposition 5.2. Suppose that fp is a sequence of real valued, measurable functions defined on a measure space (Ω, ,) and convergent pointwise to f. Then f is measurable and, forF every A of ∈ F finite measure, fp converges in measure to f on A. Proof. We have already seen in Proposition 2.11 that the pointwise limit of a sequence of real valued, measurable functions is measurable. We may suppose that f = 0 because otherwise we replace fp by f f. For ǫ> 0, we claim that p − m N p m x A : fp(x) ǫ = A ∪ ∈ ∩ ≥ { ∈ | |≤ } In fact, for any x A, limp fp(x) = 0 implies that there is some m for which f (x) ǫ ∈for every p m, that is there is m so that x A | p | ≤ ≥ ∈ m where Am = p m x A : fp(x) ǫ . Since x A was arbitrary this verifies the claim.∩ ≥ { Therefore∈ | A |≤A is} a decreasing∈ sequence of sets of − m finite measure converging to the empty set and so limm N (A Am)=0 by Theorem 3.5. But ∈ − A A = x A : f (x) > ǫ, for some p m − m { ∈ | p | ≥ } and the result follows. Thus pointwise convergence implies convergence in measure on sets of finite measure. Exercise: Construct a sequence of measurable functions on a probability space which converges to 0 in probability but does not converge pointwise.

Theorem 5.3. Let fp be a decreasing sequence of nonnegative functions in (Ω, ,) which converges pointwise to 0. Then S0 F

lim fp d =0. p N ∈ Ω This result will allow us to show later that the integral can be extended to a wider class of functions.

Proof. We have f1 f2 ... fp ... 0. Since f1 is finite valued, there is a constant≥ M≥ > 0 so≥ that≥f ≥M and hence f M 1 ≤ p ≤ for all p. Also, the set A = x : f1(x) > 0 has finite measure and f (x) = 0, for every p N if x{ Ac. Suppose} the ǫ> 0. Define p ∈ ∈ A = x : f (x) >ǫ p { p } Then A is a decreasing sequence of sets of finite measure; indeed A p p ⊆ A. We know from the preceding Proposition that limp N (Ap)=0. ∈ 60 1.MEASURETHEORYANDINTEGRATION

For any p we have

fp d = χA Ap fp d + χAp fp d − Ω Ω Ω

χA Ap ǫd + χAp M d ≤ − Ω Ω ǫ(A A )+ M(A ) ǫ(A)+ M(A ) ≤ − p p ≤ p Since limp N (Ap) = 0, it follows that, provided p is large enough, ∈ Ω fp d ǫ((A)+M), for example. Since ǫ> 0 was arbitrary Ω fp d must converge≤ to zero. Corollary 5.4. Let fp be an increasing sequence of functions in 0(Ω, ,) which converges pointwise to a function f also in 0. ThenS F S

lim fp d = f d p N ∈ Ω Ω Proof. Apply the Theorem to the sequence f f . − p We are now prepared to extend the integral to a broader family of functions. It will be convenient to focus first on nonnegative func- + + tions. Let 0 = 0 (Ω, ,) be the subset of 0(Ω, ,) consisting S S F + S F of nonnegative functions. Thus 0 is the set of all nonegative simple functions f such that x : f(x) =0S has finite measure. Define further +(Ω, ,)= + by{ } D F D + = f : Ω R : f = sup H for some countable H + . D { → + ⊆S0 } Later we shall see that for f +, then we can extend the concept of integral as ∈D ∗ f d = sup hd : h +,h f { ∈S0 ≤ } Ω Ω but first we explore the properties of +. A cautionary observation is that elements of + may take on theD value unlike those in +. D ∞ S0 Proposition 5.5. Properties of +(Ω, ,) + + D F (1) 0 . S ⊆D + (2) A function f : Ω R+ is in if and only if f is the → D + pointwise limit of an increasing sequence gp 0 . + ∈S (3) A function f : Ω R+ is in if and only if f is nonnegative and measurable and→ x Ω:Df(x) > 0 is the union of at most countably many sets{ of∈ finite measure.} + (4) Suppose f : Ω R+ is measurable and f g where g . Then f +. → ≤ ∈ D (5) If f,g ∈D+ then f + g +. ∈D ∈D 5.THEINTEGRAL 61

(6) If f,g + then fg +. Here it is understood the 0=0, that is∈D if f(x)= and∈Dg(x)=0 then fg(x)=0. ∞ (7) If f + and λ ∞ 0 then λf +. (8) If f,g∈D + then≥sup f,g and∈Dinf f,g are also in +. ∈D { } { } D + + Proof: If f 0 then f because f = sup H where H = f , ∈S ∈D + + { } that is H consists of only one element. Thus 0 which is Part (a). S ⊆ D Suppose f + so that f = sup H where H is a countable set, and suppose H∈= D h : p N is an enumeration of H. Deﬁne { p ∈ } g1 = h1, g2 = sup g1,h2 , and in general gp = sup gp 1,hp for p 2 so that g is an increasing{ } sequence in +. Since h{ −g }f we have≥ p S0 p ≤ p ≤ f = sup gp . Conversely if f is the pointwise limit of an increasing { } + + sequence gp in 0 then f because we can choose H = gp . If f + thenS f 0 and∈D f is the pointwise limit of an increasing{ } ∈D ≥ + sequence gp of functions in 0 . As the limit of measurable functions f is measurable by PropositionS 2.11. Moreover

x Ω: f(x) > 0 = p N x Ω: gp(x) > 0 { ∈ } ∪ ∈ { ∈ } where x Ω : gp(x) > 0 is an increasing sequence of sets of finite measure.{ ∈ } Conversely suppose that f is a nonnegative measurable function and x Ω : f(x) > 0 = p NAp where Ap is a sequence of sets { ∈ } ∪ ∈ of finite measure. We may assume that Ap is an increasing sequence. Because f 0 is measurable there is an increasing sequence gp of nonnegative≥ functions which converges pointwise to f by Theorem∈ S + 2.15. The sequence χAp gp belongs to 0 , is increasing and converges pointwise to f because, if x / A Sthen f(x)=0= g (x) and if ∈ ∪p p p x Aq, for some q then x Ap if p q and f(x) = limp gp(x) = ∈ ∈ ≥+ limp χAp (x)gp(x). This shows that f and verifies Part 3. If f is measurable and 0 f g∈Dwhere g + then x : f(x) > 0 x : g(x) > 0 . It follows≤ that≤ x : f(x) ∈> D0 can be{ written as the} ⊆ countable { union} of sets of finite{ measure because} x : g(x) > 0 can, Therefore by Part 3. f +. { } Let f and g be in + ∈Dso that, by Part 2, there are increasing D + sequences fp and gp, p N in 0 so that f = limp fp and g = limp gp. ∈ S + Consequently f + g = limp fp + gp and this implies that f + g , again by Part 2. This verifies Part (5). The proof of Part (6) is similar∈ D because fg = limp fpgp. (Here we use the convention 0 = 0.) This verifies Part 6. ∞ Exercise: Verify Properties 7 and 8 of +. 2 D 62 1.MEASURETHEORYANDINTEGRATION

For f in +, deﬁne = g + : g f and D Sf { ∈S0 ≤ } ∗ f d = sup gd : g { ∈Sf } Ω Ω Of course f is a very large set but we could actually use a much smaller set as theS following result shows. Theorem 5.6. If f + then there is an increasing sequence g in + so that f = sup g ∈. DMoreover for any such sequence, p S0 p p ∗ ∗ f d = sup gp d Ω p Ω Proof. We have already seen that there exists an increasing se- + + quence gp in 0 so that f = sup gp: this is Property 2 of . For any p, S D ∗ g d sup hd : h p ≤ ∈Sf Ω Ω simply because g . We may take the supremum over p to get p ∈Sf ∗ sup g d : p N sup hd : h = f d p ∈ ≤ ∈Sf Ω Ω Ω To prove the reverse inequality, we shall show that if h then ∈ Sf Ω hd supp Ω gp d. Taking the supremum over all such h will complete≤ the argument.{ Observe that h = sup inf h,g so that h is p { p} the pointwise limit of an increasing sequence in +. By Corollary 5.4, S0

hd = lim inf h,gp d lim gp d p N { } ≤ p N Ω ∈ Ω ∈ Ω This completes the proof. + + + Notation: We have 0 (Property 1 of ). Indeed any h + is the supremum ofS the⊆ constant D sequence h D= h and ∈S0 p ∗ hd = sup hp d = hd Ω p Ω Ω ∗ Therefore the two integrals Ω hd = Ω hd coincide on the intersection + of their domains. Later we shall construct an extension of the S0 integral on S0 and that integral (which will be ﬁnite valued on its do- ∗ + main) will be denoted Ω. The extension Ω deﬁned on D is distinct in that it may take on the value . ∞ Proposition 5.7. If f,g +(Ω, ,) and λ> 0. Then ∈D F ∗ ∗ a. λf d = λ f d; Ω Ω 5.THEINTEGRAL 63

∗ ∗ ∗ b. f + gd = f d + gd; Ω Ω Ω c. f d gd if f g. ≤ ≤ Ω Ω Proof. + Suppose that f and f = supp fp where fp is an + ∈ D increasing sequences in 0 . Then, for λ > 0. λfp is an increasing sequence in + convergentS pointwise to λf so that λf + and S0 ∈D ∗ ∗ λf d = sup λfp d = λ sup fp d = λ f d Ω p Ω p Ω Ω by the Theorem 5.1. This proves Part a because ∗ = . + Suppose now that g also and g = sup gp where gp is an ∈ D p increasing sequence in +. Then f + g = sup f + g and S0 p p ∗ f + gd = sup fp + gp d = sup fp d + gp d Ω p Ω p Ω Ω

= sup fp d + sup gp d p Ω p Ω ∗ ∗ = f d + gd Ω Ω again by Theorem 5.1 and this proves Part b. If we further assume that f g then f so that ≤ p ∈Sg ∗ fp d sup hd = gd Ω ≤ h g Ω Ω ∈S Taking the supremum over p gives Part c. Historical Notes. Henri Léon Lebesgue 1875-1941 was born in Beauvais France and received a teaching diploma from the Ecole´ Nor- male Supérieure in Paris in 1897. From 1899-1902 he was a professor at the Lycée Centrale at Nancy where he wrote several papers including his “Sur une généralisation de l’intégrale définie” in which he intro- duces his novel approach to integration on the real line. Lebesgue’s doctoral dissertation, Intégrale, longueur, aire , presented to the Fac- ulty of Science in Paris in 1902, and the 130 page work was published in the Annali di Matematica in the same year. contains a chapter on Borel measure and measure theory and ends with a discussion of the Plateau problem. 64 1.MEASURETHEORYANDINTEGRATION

6. Convergence of Integrals In this section we extend the integral to real valued functions and show that the integral of a sequence of pointwise convergent functions will converge in many circumstances.

Theorem 6.1. Monotone Convergence Theorem: Let fp be an increasing sequence in + and f = sup f . Then f + and D p p ∈D

f d = sup fp d Ω Ω Proof. (p) + For each p there is an increasing sequence hn in 0 so (p) S that fp = supn hn .

(1) (1) (1) h1 h2 ... hp ... f1 (2) (2) (2) ր h1 h2 ... hp ... f2 . . . . . ր ...... (p) (p) (p) h1 h2 ... hp ... fp . . . . . ր ...... We deﬁne g = sup h(1),h(2),...,h(p) p { p p p } so that, g + for all p N and p ∈S0 ∈ (1) gp is an increasing sequence; (2) g f ; p ≤ p (3) f = limp gp. Assuming these three for the moment we can now derive the result. We + + have f as the pointwise limit of a sequence in 0 and gp fp f so that∈D S ≤ ≤ ∗ ∗ g d f d f d p ≤ p ≤ Ω Ω Ω ∗ ∗ If we take the supremum over p, we see sup Ω fp d = Ω f d. (n) Check the three properties. The sequence gp increases because hp (n) increases as p does if n is ﬁxed. We also have g f because hp f p ≤ p ≤ n and fn fp if n p. Finally we show that, for an arbitrary x Ω, f(x) =≤ sup g (x).≤ Certainly g f f and so we need only show∈ p p p ≤ p ≤ that if λ λ. Choose p so large (p) that fp(x) > λ. Choose n N so large that hn (x) > λ and n p. (p) ∈ ≥ Then g (x) hn (x) >λ. This completes the proof. n ≥ 6. CONVERGENCE OF INTEGRALS 65

Example It is not true that, if f is a decreasing sequence in + n D which is pointwise convergent to f then Ω fn d converges to Ω f d. c For example fn to be the characteristic function of [ n,n] converges pointwise to 0 but f d = . However, if we further− suppose that R n ∞ f d < for some n then the integrals converge as can be seen by Ω n0 ∞ 0 applying the Monotone Convergence theorem to the sequence fn0 fn which is in + by property 4 of Proposition 5.5 +. − D D + Corollary 6.2. If gn, n N is a sequence in then f = + ∈ D n N gn is also in and ∈ D gn d = gn d Ω n N n N Ω ∈ ∈ Proof. + Define fk = 1 n k gn so that fk and apply the ≤ ≤ ∈ D Theorem. Because Ω fk d = 1 n k Ω gn d the result follows. ≤ ≤ We shall need the concept of liminfn an and limsupn an in the case that an is a sequence in R. For such a sequence, define bk = infn k an for each k N; b is well defined and an increasing sequence and≥ its ∈ k limit is liminfn an. Similarly, if ck = supn n ak then ck is a decreasing ≥ sequence with limit limsupn an.

Lemma 6.3. Suppose that fp : Ω R, p N is a sequence of real valued functions in +. Then lim inf→f ∈+. D p p ∈D

lim inf fp d lim inf fp d p ≥ p Ω Ω Proof. For each n N, deﬁne gn = infp n fp. Then gn is measur- ∈ ≥ able as the limit of the measurable sequence hN = infp:N p n fp. and g is in + by Property 4 of Proposition 5.5 because g ≥f≥. The se- n D n ≤ n quence gn increases to liminfp fp which, by the Monotone Convergence Theorem is in +, and D

lim inf fp d = lim gn d. p N n N Ω ∈ ∈ Ω On the other hand g f for all p n so that n ≤ p ≥ g d f d for all p n n ≤ p ≥ Ω If we take the inﬁmum over p and then the limit over n then

lim inf fp d lim inf fpd p N ≤ p N Ω ∈ ∈ Ω which proves the result. 66 1.MEASURETHEORYANDINTEGRATION

Theorem 6.4. Dominated Convergence Theorem: Pre- + liminary Version: Let fp be a sequence of functions in (Ω, ,) which converges pointwise to a function f. Suppose that gD +Fand N ∈ D + Ω gd < , g is ﬁnite valued and fp g, for all p . Then f , and f ∞f + and ≤ ∈ ∈D | p − |∈D

lim fp f d =0 p N | − | ∈ Ω Later we will allow fp to be extended real valued. Proof: Since f = liminfp fp we have, by the preceding Lemma f +. It is also clear that f f is + because it is nonnegative, ∈ D | − p| D measurable and f fp f + fp 2g so that Property 4 of Propo- sition 5.5 applies.| Consequently− |≤| | | we|≤ may as well assume that f = 0, for we simply replace fp by f fp . | − | + We further know that g fp , again by Property 4 of Propo- sition 5.5. By the preceding− Lemma∈D

lim inf g fp d lim inf g fp d = gd p − ≥ p − Ω Ω Ω

Additivity of the integral implies Ω g fp d = Ω gd Ω fp d so that − − gd lim sup f d gd − p ≥ Ω p Ω Ω and since gd < , this implies so that limsup f d 0. On Ω ∞ p Ω p ≤ the other hand f d 0 so that lim f d = 0 and that is desired Ω p p Ω p conclusion. ≥ 2 We shall now extend the definition of integral to apply to functions which may be negative. For such functions f we encounter a technical problem in computing the integral because the integral of f will contain a positive contribution from the integration over the the set f > 0 plus a negative contribution for the integration over the set {f < 0 }. If the two contributions happen to be both infinite then this{ leads to} the indeterminant and we shall say that f is not integrable. It is therefore useful∞ to − keep ∞ track of the sets where f is not finite valued more carefully than was needed when discussing functions in +; we hope the sets where f is infinite are “small” and so we introduceD the concept of “almost everywhere.”| | To be precise, we suppose that (Ω, ,) is a measure space and introduce the terminology that a prop- ertyF holds -almost everywhere or almost everywhere or a.e. if there is A so that (A) = 0 and the property does indeed hold for all ∈ F x / A. For example we shall say that a measurable function f : Ω R ∈ → 6. CONVERGENCE OF INTEGRALS 67 is finite valued -almost everywhere if ( x : f(x) = ) = 0. Fre- quently almost everywhere is an abbreviation{ | for Lebesgue| ∞} almost everywhere where the understood measure is Lebesgue measure. In the case that is a probability measure we say “almost surely” for “almost everywhere.” Proposition . + 6.5 If h (Ω, ,) and Ω hd < then h is finite valued -almost everywhere.∈D If f,gF + and if ( x∞: f(x) = g(x) )=0 then ∈D { } f d = gd Ω Ω Observe that the Proposition makes sense because h = and f = g = f g =0 are measurable. { ∞} { } { − }

Proof. The set x : h(x)= = p NAp where Ap = x : h(x) > p . Therefore { ∞} ∩ ∈ { } > hd pχ d = p(A ) ∞ ≥ Ap p Ω Ω from which we see the real sequence (Ap) converges to 0. On the + other hand limp (Ap) = ( x : h(x) = ). Therefore h is ﬁnite valued almost everywhere.{ ∞} ∈ D Now suppose f,g + and ( x : f(x) = g(x) ) = 0. Then ∈D { }

f d = χ f=g f d = χ f=g gd = gd { } { } Ω Ω Ω Ω which completes the proof. Remark In Theorem 6.4 above it is possible to dispense with the hypothesis that the sequence fp and the functions f and g be real valued. We can allow them to be extended real valued, that is taking R values in +. This is because the hypothesis Ω gd < assures that there is a measurable set A such that (Ac)=0 and g ∞is finite valued on A, by the preceding result. Consequently fp g and f are also finite values on A and we may apply Theorem 6.4 to≤ finite valued functions χAfp, χAf and χAg in place of fp, f and g respectively. We are able to conclude that χ f is in + and f χ f + and A D p − A |∈D

lim fp f d =0. p N | − | ∈ Ω That is all conclusions remain valid but with f replaced by χAf. If is complete then we can dispense with χA because then f itself will be measurable and hence in +. D 68 1.MEASURETHEORYANDINTEGRATION

Definition. A function f : Ω R is said to be integrable, and we write f 1(Ω, ,) or briefly→f 1 if ∈L F ∈L (1) f is measurable (2) f + and f d < . | |∈D | | ∞ Ω Observe that f is measurable implies f is measurable but not | | conversely. For example if A / and if f = χA χAc then f is not measurable but f = 1 is∈ measurable. F Observe− further that + includes some functions| | which are not in 1 and conversely. We nowD 1 L summarize some properties of . Recall that 0 denotes the set of simple functions f such that xL: f(x) =0 has finiteS measure. { } Remark. If f 1 then f may take on the values but, by the preceding result,∈ L x : f(x) = has measure zero.±∞ It will be convenient to add two{ functions| | in ∞}1 and for this purpose we may defined as any number in R atL all or alternatively we could say is∞ undefined. − ∞ In the latter case however the domain of functions ∞−∞in 1 would not be Ω which is a (minor) inconvenience and so we shall defineL = 0 in this restricted context. ∞ − ∞ Proposition 6.6. Properties of 1(Ω, ,): L F (1) 1 is a vector space. (See the above Remark.) (2) LIf f 1 then f 1. Conversely, if f 1 and f is measurable∈ L then |f |∈L1. | |∈L (3) If f 1 and if g ∈Lis a measurable function which is bounded (that∈ is, L for some constant K > 0, g(x) < K for all x Ω) then fg 1. | | ∈ 1 ∈L (4) 0 . (5) SIf f,g⊆L 1 then max f,g and min f,g also belong to 1. ∈L { } { } L (6) If fp is a sequence of real valued measurable functions which converges to f pointwise where f is real valued and if there is g + with gd < and so that f g for every p, ∈ D Ω ∞ | p| ≤ then f,f 1 and p ∈L

lim fp f d =0 p N | − | ∈ Ω Proof. Suppose that f,g 1. By the Remark preceding the statement of this result, f + g∈is L an extended real valued function deﬁned on all of Ω. Moreover f and g are measurable and so f + g is measurable. Also f , g + and so f + g +. Since f + g is clearly measurable| and| | |∈Df + g f +| g| we| |∈D have f + g | +;| see the Properties of +. Finally| |≤|f +| g d| | f d| + |∈Dg d < D Ω | | ≤ Ω | | Ω | | ∞ and so f + g 1. Next suppose that c R and f 1. Then ∈ L ∈ ∈ L 6. CONVERGENCE OF INTEGRALS 69 cf is measurable, and cf = c f + because f +. Also | | | || |∈D1 | |∈D 1 Ω cf d = c Ω f d < . So cf and this checks that is a vector| | space.| | | | ∞ ∈L L If f 1 then we wish to check that f 1. Because f is measurable,∈ Lf is measurable. Also we know | f|∈L+ and f d < | | | |∈D Ω | | because f 1 and so f 1. Conversely if f is measurable then ∞f 1 implies∈Lf 1 by| similar|∈L reasoning. | |∈LNext we suppose∈L the f 1 and g is a bounded measurable function with g K for some K∈L > 0. Then fg is measurable and fg = f g ≤K f and K f + because f +. Therefore fg | | +, |by|| the|≤ properties| | of | |∈D+. Finally fg| d|∈D K f d <| |∈Dand so D Ω | | ≤ Ω | | ∞ fg 1. ∈LThe proof the max f,g and min f,g belong to 1 whenever f,g 1 is a straighforward{ } exercise. { } L ∈L Finally we suppose fp is a sequence of real valued measurable functions which converges to f pointwise where f is real valued and that + g with Ω gd < and so that fp g for every p. Then f is ∈ D ∞ | | ≤ + measurable by Proposition 2.11. Moreover fp g where g im- + | |≤ + ∈D plies fp ; since f = limp fp g we have f . Therefore | |∈D1 | | | | ≤ + | |∈D f,fp and so fp f is a sequence in convergent pointwise to 0 and∈L so | − | D

lim fp f d =0 p N | − | ∈ Ω by Theorem 6.4. Integration of 1 Functions L We now define the integral of an 1 function. Recall that every measurable function is the pointwiseL limit of a sequence of simple functions by Theorem 2.16. If f 1 then f is measurable and we claim ∈ L that f is the pointwise limit of a sequence in 0. Let fn be a sequence of functions in convergent pointwise to Sf. Because f +, we know that x : fS > 0 = x : f = 0 can be written| |∈D as the union of at most{ countably| | } many{ sets of finite} measure or explicitly x : f =0 = A where each A has finite measure and we may as- { } ∪n n n sume the An for an increasing sequence. Then the sequence sn = χAn fn is in and converges pointwise to f which verifies our claim. More- S0 over sn f . Now let us suppose that sn is any sequence in 0 that converges| |≤| pointwise| to f and moreover that s g for some gS + | n|≤ ∈D with Ω gd < . (But g need not be f ) Since sn is simple we can integrate it and∞ we observe that | | s d s d s s d s f d+ f s d | n − m |≤ | n − m| ≤ | n − | | − m| Ω Ω Ω Ω Ω 70 1.MEASURETHEORYANDINTEGRATION

Of course s f d converges to 0 by Part 6 of the preceding Ω | n − | Proposition. Therefore we see that s d forms a Cauchy real valued Ω n sequence and therefore converges. We define × f d = lim sn d n N Ω ∈ Ω We observe that if there were another sequences ñ in 0 convergent to f and with s˜ h, for some h + with hd, thenS the sequence | n|≤ ∈D Ω lim s˜ d would have the same limit as lim s d for consider n Ω n n Ω n the concatenated sequence s1, s˜1,s2, s˜2,s3,... which, by the above argument, must be Cauchy. (The concatenated sequence is dominated × by max g,h .) If s 0 then we see Ω sd = Ω sd because s is { } ∈ S + the limit of the constant sequence (sn = s). If f then f is the +∈ D pointwise limit of an increasing sequence sn in 0 0. Therefore, if we further assume that fd< then S ⊆S Ω ∞ × f d = lim sn d = f d n N Ω ∈ Ω Ω × and so the integrals Ω and Ω are the same on the intersection of their domains and so we shall use the notation × = Ω Ω so that is now defined on + 1. Ω D ∪L Proposition 6.7. The integral is a linear functional defined on 1(Ω, ,). Moreover if f g and f,g 1 then L F ≤ ∈L f d gd ≤ and f d f d | |≤ | | 1 Remark: If f then f+ =( f + f)/2 and f =( f f)/2 are + ∈L | | − | |− both in and f = f+ f and D − −

f d = f+ d f d − − Ω Ω Ω by this Proposition. Thus an equivalent method of extending the integral from + to 1 would be to use the right hand side of the above equation asD the deﬁnitionL of the left side. 1 Proof: Suppose the f,g . Then there exist sequences sp,tp such that s converges pointwise∈L to f and t converges pointwise to∈ S0 p p 6. CONVERGENCE OF INTEGRALS 71 g and sp f and tp g . Then the sequence sp + tp is in 0 and converges| |≤| to f +| g which| |≤| we| know is in the linear space 1. MoreoverS s + t s + t f + g and of course f + g L+. Therefore | p p|≤| p| | p|≤| | | | | | | |∈D

f + gd = lim sp + tp d = lim sp d + lim tp d p p p Ω Ω Ω Ω = f d + g d. Ω Ω This veriﬁes that the integral is additive. To complete the proof of linearity we must show that Ω λf d = λ Ω f d. The argument is left as an exercise. Suppose now that f g and f,g 1. If we show g f d 0 ≤ ∈L Ω − ≥ then gd f d by the linearity just established. Therefore we Ω Ω may as well assume≥ that f = 0. Then g = g + and consequently gd 0. | |∈D Ω ≥ Finally we observe that f f f so that f d f d. 2 −| |≤ ≤| | | Ω |≤ Ω | | Exercise: Suppose f,g 1. Show that ∈L ( x : f(x) = g(x) )=0 { } if and only if

f d = g d, for every A ∈ F A A Theorem 6.8. Lebesgue’s Dominated Convergence The- orem: Suppose that fp is a sequence of measurable extended real valued functions defined on a measure space (Ω, ,) that converges pointwise F 1 -almost everywhere and let f = limp fp. Suppose that g and f g for all p. Then ∈ L | p|≤ (1) f 1; ∈L (2) lim fp f d =0; p | − | Ω (3) lim fp d = f d p Ω Ω Proof: There is a set A so that (Ac) = 0 so that, g is finite ∈ F valued on A and fp converges pointwise to f on all of A. Consequently χAfp is finite valued and converges pointwise to χAf which is also finite 1 valued. By Property 6 of applied to χAfp we see that χAfp,χAf 1. and L ∈ L lim χA fp f d =0 p | − | Ω 72 1.MEASURETHEORYANDINTEGRATION

Therefore

lim χAfp d χAf d lim χA fp f d =0 p | − |≤ p | − | Ω Ω Ω + by the Proposition two previous. We know g = g and so fp + + | |∈D | | and f are also in by Property 4 of . Therefore fp and f are themselves| | in 1 andD f d = χ f Dd and f d = χ f d L Ω p Ω A p Ω Ω A by the previous result. Similarly we see that χ f f d = f Ω A p Ω p f d and so the proof is complete. | − | | −2 | Since the pointwise limit of measurable functions is measurable, we see that the χAf is measurable if A is the set of convergence and so it is only necessary to assure that (1 χA)f is measurable which would be the case if f were zero on Ac for− example or if were complete since Ac has measure zero. F Example: Consider the function x if 0 x 1 f(x)= 2 x if 1 ≤ 0 and n is so large that nx > 2 then n ≤ fn(x) = 0. On the other hand

f (x) dx =1, for all n N n ∈ R (by Riemann integration if necessary). In this example each fn is continuous. It is possible to construct an example using simple functions. 2

Corollary 6.9. Suppose fn is a sequence of measurable functions on (Ω, ,) which converges almost everywhere to a measurable F function f. Suppose that g 0 is also measurable and fn g and for some p, 0

Theorem 6.10. Fatou’s Lemma Suppose that fn, n N and g are in 1(Ω, ,) and g 0. ∈ L F ≥ 6. CONVERGENCE OF INTEGRALS 73

1 a. If fn g and lim infn Ω fn d < then lim infn fn and ≥ − ∞ ∈ L lim inf fn d lim inf fn d n N ≥ n N ∈ Ω Ω ∈ b. If f g and lim sup f d > then lim sup f 1 n ≤ n Ω n −∞ n n ∈L lim sup fn d lim sup fn d n N Ω ≤ Ω n N ∈ ∈ 1 Proof: Let hn = infp n fp + g 0. Then hn because it is ≥ ≥ ∈ L measurable, as the limit of measurable functions and because0 hn f + g f + g. Also f + g h whenever p n so that ≤ ≤ n ≤| n| p ≥ n ≥

inf fp + gd hn d p n ≥ ≥ Ω Ω Since hn is an increasing sequence, the Monotone Convergence Theorem applies and we may take the limit as n : → ∞

lim inf fp d + gd lim hn d = lim inf fn + gd p N ≥ n n N ∈ Ω Ω Ω Ω ∈ 1 By assumption liminfp N Ω fp d is ﬁnite and so liminfn N fn + g 1 ∈ 1 ∈ ∈L but of course g and so liminfn N fn . Cancelling Ω gd from both sides gives∈L Part a. ∈ ∈L For Part b, we apply Part a to the sequence fn and observe that − 2 lim infn( fn)= lim supn fn. The Dominated− − Convergence Theorem can be derived from Fatou’s Lemma by a straightforward argument: see page 49 of Ash’s book. + Exercise: Suppose that f D and Ω f d = 0. Show that ( f > 0 )=0. ∈ { } 74 1.MEASURETHEORYANDINTEGRATION

7. L1-Completeness Definition: A real valued mapping ρ defined on a vector space X is a semi-norm if, whenever x,y X and λ is a scalar, ∈ a. ρ(0) = 0 b. ρ(λx)= λ ρ(x) c. ρ(x + y) | |ρ(x)+ ρ(y) ≤ From this definition it follows that 0 = ρ(0) = ρ(x + ( x)) ρ(x)+ ρ( x)=2ρ(x). Thus ρ takes on only non negative values.− If,≤ in addition− ρ(x)=0 if and only if x = 0 the ρ is said to be a norm. Example: We are primarily interested in ρ defined on 1(Ω, ,) by L F ρ(f)= f d | | Ω It is clear that ρ so defined is a semi-norm but it seldom is a norm. Definition A sequence xk in X is said to be Cauchy with respect to a seminorm ρ if, for every ǫ> 0 there is K = K(ǫ) in N so that ρ(x x ) <ǫ if k,ℓ K k − ℓ ≥ A sequence xk is convergent if there exists x so that, for every ǫ> 0, there is N N(ǫ) in N so that ρ(xn x) < ǫ for every n>N. If X is a linear space equipped with a seminorm− ρ then (X,ρ) is said to be complete if every Cauchy sequence is convergent. Exercise If xk is a Cauchy sequence and if there is a subsequence xk(n) that is convergent to x then the original sequence xk converges to x. (Recall that a subsequence of xk is a sequence yn = xk(n) for some increasing function k : N N.) → Lemma 7.1. (X,ρ) is complete if and only if every sequence xk such that ρ(x x ) < converges. k k+1 − k ∞ Proof. The condition ρ(x x ) < implies that the se- k k+1 − k ∞ quence xk is Cauchy because if ℓ > k then ρ(xℓ xk) k j<ℓ ρ(xj+1 − ≤ ≤ − x ). Therefore if (X,ρ) is complete then any sequence x for which j k k ρ(xk+1 xk) < must converge . Conversely if yk is a Cauchy − ∞ n sequence then, for each n choose k(n) so large that ρ(yk yℓ) < 2− if k,ℓ k(n). We may assume k(n + 1) > k(n). Then the− subsequence ≥ n x = y satisfies ρ(x x ) = ρ(y y ) < 2− and so n k(n) n+1 − n k(n+1) − k(n) k ρ(xk+1 xk) < which means it must converge. By the above exercise, a Cauchy− sequence∞ with a convergent subsequence converges. 2 7. L1-COMPLETENESS 75

Theorem 7.2. 1(Ω, ,) is complete with the seminorm L F ρ(f)= f d | | Ω Proof. Suppose that f is a sequence in 1 such that n L ρ(f f ) < . n+1 − n ∞ n N ∈ We wish to show f converges within 1. Define n L h = f + f f + f f + f f + ... | 1| | 2 − 1| | 3 − 2| | 4 − 3| Then h 0 is measurable and h + by Corollary 6.2 to the Monotone Convergence≥ Theorem and because∈D f f + . Moreover | n+1 − n|∈D hd = ρ(f )+ ρ(f f )+ ρ(f f )+ ... 1 2 − 1 3 − 2 Ω and so, by our hypotheses Ω hd < . Consequently A = x Ω : h(x) < has full measure: (Ac) =∞ 0. It follows that the series{ ∈ ∞} f (x)+(f (x) f (x))+(f (x) f (x))+(f (x) f (x))+ ..., x A 1 2 − 1 3 − 2 4 − 3 ∈ is absolutely convergent on A. Define f(x)= f (x)+(f (x) f (x))+ 1 2 − 1 (f3(x) f2(x)) + ... if x A and so f(x) = limn fn(x). We extend f to all of− Ω by defining f(∈x)=0if x / A. Then f is measurable by Proposition 2.11. Also f h because,∈ if x A | |≤ ∈ f(x) = f (x)+ f (x) f (x) | | | 1 n+1 − n | n N ∈ f (x) + f (x) f (x) = h(x) ≤ | 1 | | n+1 − n | n N ∈ whereas if x / A then f(x)=0 h(x) and so f and f are in 1. It remains∈ to show that lim ≤ρ(f f )=0.| But| L n − n ρ(f f ) = ρ(χ (f f )) − n A − n = ρ(χ [(f f )+(f f )+ ...]) A n+1 − n n+2 − n+1 ρ(f f )+ ρ(f f )+ ρ(f f )+ ... ≤ n+1 − n n+2 − n+1 n+3 − n+2 and the last expression goes to 0 as n goes to infinity because it is the tail of a convergent series. Therefore our Cauchy sequence converges to f and so 1 is complete L Introduce = f 1(Ω, ,): ρ(f)=0 N { ∈L F } where, of course ρ(f)= Ω f d. Then is all those functions which our integral cannot distinguish| | from 0. ObserveN that f if and only ∈N 76 1.MEASURETHEORYANDINTEGRATION if ( x : f(x) =0 ) = 0. It is further clear that is a linear subspace of {1 because, for} example ρ(f + g) ρ(f)+ ρN(g). We define f g forLf,g 1 if f g . It is clear that≤ the relation defined this way∼ is an equivalence∈L − relation∈N on 1, that is L (1) (Reflexive) For any f 1 f f. (2) (Symmetric) If f g ∈Lthen g ∼f. (3) (Transitive) If f ∼g and g ∼h then f h. ∼ ∼ ∼ This equivalence relation breaks 1 into equivalence classes and the set of all equivalence classes is denotedL L1 = 1/ L N or, when clarification is necessary, L1 = L1(Ω, ,) . We shall denote F the equivalence class that contains g 1 byg ˙ so thatg ˙ = f˙ if and only if g f . The mapping g ∈g˙ Lis denoted φ: φ(g)=g ˙ and φ : 1 −L1 is∈ surjective. N → LWe→ define addition on L1 by φ(f)+ φ(g)= φ(f + g). This requires that we check that, if f f1 and g g1 then f + g f1 + g1 to assure that the addition is∼ well defined∼ and this is of course∼ clear since is a linear space. Similarly scalar multiplication can be defined byN cφ(g) = φ(cg) where c is a scalar (real) and that is well defined because g g1 implies cg cg1. It is a routine matter to check that these definitions∼ make L1 a∼ linear space and φ is linear. Let us further define g˙ = ρ(g)= g d 1 | | Ω It is of course necessary to check that 1 is well defined. To do so observe that ρ(f) ρ(g) ρ(f g) because ρ(f)= ρ(g +(f g)) ρ(g)+ρ(f g)| so that− ρ(f|≤) ρ(g) − ρ(f g). Interchanging f and− g, we≤ find ρ(g) −ρ(f) ρ(g f)=− ρ(f≤ g) so− that ρ(f) ρ(g) ρ(f g). It follows− that f≤ g −implies ρ(f−) = ρ(g) and| this− assure|≤ that − ∼ 1 is well defined. It follows almost immediately that 1 is in fact a 1 seminorm because 0 1 = 0 because 0 L is the equivalence class corresponding to 0 1 (which is ) and∈ ρ(0) = 0. Also φ(cg) = ∈ L N 1 ρ(cg)= c ρ(g)= c g 1 if c is a real constant. Finally φ(f + g) 1 = ρ(f + g)| | ρ(f)+|ρ|(g)= φ(f) + φ(g) . In fact we have is a ≤ 1 1 1 norm because g˙ 1 = 0 implies ρ(g) = 0 so that g and sog ˙ = 0. 1 ∈ N Therefore (L , 1) is a normed linear space. 1 We would further like to conclude that (L , 1) is a Banach space. Recall that a Banach space is a normed linear space that is 1 complete. Suppose therefore thatg ˙n is a Cauchy sequence in L . Then g is a Cauchy sequence in ( 1,ρ) and as such has a limit g 1 say: n L ∈L 7. L1-COMPLETENESS 77 limn gn = g. It follows that limn g˙n =g ˙. because g˙n g˙ 1 = ρ(gn g) and lim ρ(g g) = 0. Therefore we have proved the− following result.− n n − Theorem 7.3. (L1, ) is a Banach space. 1 If t L1 then t = φ(g) for some g 1 and we define ∈ ∈L td = gd Ω Ω This is of course well defined because if t = φ(f) also then f g and so ρ(f)= ρ(g) as we saw above and ∼

gd = f d Ω Ω It is customary to include in our integration theory, functions that are defined -almost everywhere. Suppose therefore that f : A R where A and (Ac) = 0. We extend f as → ∈ F f(x) if x A F (x)= 0 if x∈ / A ∈ If we further assume that F 1 then we say that f 1 and define ∈L ∈L 1 Ω f d = Ω F d. Consequently f also defines an element φ(F ) of L . Finally we remark on the indeterminant expression . For functions in f,g 1 which are extended real valued but∞ may − ∞ take on the values only∈L on measurable sets of measure 0 the question of how to define±∞f + g at those points x where f(x) = and g(x) = becomes clear. Since those points make up a set±∞ of measure 0 it ∓∞does not make any difference for the purpose of integration theory. In particular any convention will not alter the value of Ω f + gd nor of φ(f +g) L1. Therefore we can define to be any extended real value or to∈ be undefined because it occurs∞−∞ only on sets of measure 0. Further Reading: Lp spaces It is further possible to introduce the spaces p, 1

Theorem 7.4. Suppose that fn, n N is an increasing sequence 1 ∈ in L (Ω, ),) and fn 0 Then f = sup fn = limn f is measurable and F ≥

f d = lim fn d n N Ω ∈ Ω where the limit on the right may be real or + . ∞ It should be remarked that fn, n N is an increasing sequence 1 ∈ 1 in L means that, if gn is a sequence in so the ρ(gn) = fn (that is g is any representative of f in 1) thenL 0 g g -almost n n L ≤ n ≤ n+1 everywhere so that lim gn(x) = sup gn(x)= f(x) exists for -almost all x but may be + . Indeed f +. ∞ ∈D 8. Lp SPACES 79

8. Lp Spaces We recall that, if (Ω, ,) is a measure space and 1 0 ≤ p q Proof. We use the fact that the exponential function ex is convex. (This follows, for example, because the second derivative of ex is positive. See the Section, Convex Functions.)

p q p q log a log b p q log a + log b e e a b ab = elog a+log b = e p q + = + ≤ p q p q

Exercise (An alternative derivation of Young’s Inequality. Refer- ence: S. Lang Real Analysis, page 311) Show that, t 1 t1/p + , for t 1 ≤ p q ≥ For example one could compare the t derivatives of two sides of the inequality. Derive Young’s inequality. 80 1.MEASURETHEORYANDINTEGRATION

Theorem 8.2. (Young’s Inequality with ǫ:) For 1 0 ∞ ∞ p q bq ab ǫap + for all a,b > 0 ≤ q(ǫp)q/p

Proof. Young’s Inequality can be applied to b ab = (ǫp)1/pa) (ǫp)1/p . Theorem 8.3. H¨older’s Inequality Suppose 1 < p,q < , 1 1 ∞ + =1 and f and g belong to +(Ω, ,). Then p q D F 1/p 1/q ∗ ∗ ∗ fgd f p d gq d ≤ Ω Ω Ω It is understood that 0 = 0 here. In particular if f Lp() and g Lq() then fg L1( ∞) and ∈ ∈ ∈ fg f g 1 ≤ p q 1 1 Moreover if f L and g L∞ then fg L () and ∈ ∈ ∈ fg 1 f 1 g ≤ ∞ Proof. Suppose f,g +() so that f,g 0. If f p d = , ∈ D ≥ Ω ∞ there is nothing to verify except in the case g q = 0 and in this case g = 0 -almost everywhere and so this case is clear (since 0 = 0) P ∞ and so we may assume f L (), that is f p < . Similarly, we may assume g < and∈ so we may in fact assume ∞ that f =1= q ∞ p g q, by normalizing, if necessary. (Normalizing in this context, means 1 replacing f(x) by f˜(x)= f − f(x) so that f˜ = 1 and applying the p p special case to f˜ andg ˜ whereg ˜ is deﬁned analogously.) By Young’s Inequality we have f(x)p g(x)q f(x)g(x) + ≤ p q Integrating, we have p q ∗ f g fgd p + q =1= f g ≤ p q p q Ω which shows fg 1. The case when f,g are not necessarily nonnegative follows because∈L f , g D+() and completes the argument in | | | | ∈ 8. Lp SPACES 81

1 the case 1 < p,q < . We suppose now that f , g ∞. If c> g then ∞ ∈ L ∈ L ∞ ∗ fg d c f d = c f | | ≤ | | 1 Ω Ω and taking the inﬁmum over all such c we see that fg 1() and ∈ L fg 1 f 1 g which completes the proof. ≤ ∞ Theorem 8.4. Minkowski’s Inequality Suppose f and g belong to +(Ω, ,) and >p 1. Then D F ∞ ≥ 1/p 1/p 1/p ∗ ∗ ∗ (f + g)p d f p d + gp d ≤ Ω Ω Ω p p In particular if f,g L () (resp. f,g L∞()) then f + g L ∈ ∈ ∈ (resp. f + g L∞) and ∈ f + g p f p + g p ( resp. f + g f + g ) ≤ ∞ ≤ ∞ ∞ Proof. The case p = 1 is trivial and so we suppose >p> 1. By H¨older’s Inequality we have ∞

∗ p ∗ p 1 (f + g) d = (f + g) − (f + g) d Ω Ω 1/q 1/p 1/p ∗ (p 1)q ∗ p ∗ p (f + g) − d f d + g d ≤ Ω Ω Ω Since (p 1)q = p the first inequality follows. If f,g p then f + g +(−) and so applying the first inequality we have∈ L f + g | |∈D p ≤ f p + g p. The inequality f + g f + g is verified by a straightforward argument. ∞ ≤ ∞ ∞ It follows from Minkowski’s Inequality that p is a linear space and L p p satisfies the triangle inequality. Indeed p is a semi-norm () and indeed it induces a norm on Lp = p/ since f if andL only L N ∈ N if f p = 0. (For recall is the set of all measurable functions which are zero almost everywhere.)N

p Theorem 8.5. () with the semi-norm p is complete if 1 p . L ≤ ≤ ∞ Proof. The case p = is somewhat odd and we consider it sep- arately. Suppose therefore∞ that f , n N is a Cauchy sequence in n ∈ ∞. We may suppose that each f is bounded everywhere because, in L n any case, the set where fn > fn has measure zero and so we may | | ∞ choose fn(x) to be 0 for all x for which fn(x) > fn . By the same | | ∞ 82 1.MEASURETHEORYANDINTEGRATION reasoning we can specify fn(x) fm(x) fn fm for all x: re- | − | ≤ − ∞ define fn(x) = 0 on the complementary set which has measure 0. This assures that fn is defined everywhere and is uniformly Cauchy everywhere and so f(x) = limn N = f(x) exists. In fact, since for any ǫ> 0 ∈ there is an integer N = N(ǫ) so that fn(x) fm(x) fn fm <ǫ for all m,n N and all x. Taking| the− limit in| ≤m −N, we∞ have f (x) f(x)≥ ǫ for all x and this assures that f(x) <∈ f (x) + ǫ | n − |≤ | | | n | ≤ fn + ǫ (which assures f ∞ and f fn ǫ which implies ∞ ∈ L − ∞ ≤ the completeness of ∞(). The proof of theL case 1 p < parallels that of the case p = 1, Theorem 7.2. Suppose that≤f , n∞ N is a Cauchy sequence in p n ∈ L and moreover that n fn+1 fn p < . We shall show that fn is convergent to a limit in p; this− allows ∞ us to conclude completeness, just as in the proof of TheoremL 7.2. Consider 1/p ∗ ∞ f + f f p d | 1 | n+1 − n| | Ω n=1 N 1/p ∗ p = sup f1 + fn+1 fn d N | | − | | Ω n=1 1/p 1/p ∗ ∞ ∗ f p d + f f p d ≤ | 1| | n+1 − n| Ω n=1 Ω by Minkowski’s inequality and the Monotone Convergence Theorem 6.1. Because f f < this shows that f + ∞ f f n n+1− np ∞ 1 n=1 | n+1− n| must be finite almost everywhere and so f1 + n∞=1 fn+1 fn converges N − absolutely, that is f = f + f f converges almost every- N 1 n=1 n+1 − n where. We define f = limN fN on this set of convergence and f to ∞ ∞ be 0 on the complement which has zero measure. Then f p < ∞ ∞ because f < f1 + ∞ fn+1 fn . Moreover | ∞| | | n=1 | − | f fN p = fk+1 fk p fk+1 fk p ∞ − − ≤ − k=N k=N and this shows that the convergence of fN to f is in fact in the p semi-norm. ∞ . We recall the definition of almost uniform convergence given prior to the proof of Egoroff’s Theorem.

p Corollary 8.6. Every convergent sequence in ( (), p) has a subsequence that converges almost uniformly, 1 pL . ≤ ≤ ∞ 8. Lp SPACES 83

Proof. The result is obvious for p = since a sequence conver- ∞ gent in ∞() converges uniformly on a set of all but measure zero. L Suppose therefore that 1 p < Let fk be a sequence, conver- p ≤ ∞ gent in () to a limit f and that fk(n) is a subsequence so that L 2n f f < 2− Define k(n) − p n A = x : f (x) f(x) 2− n { | k(n) − |≥ } 2np p p np Then 2− > f f f f d 2− (A ) so that k(n) p An k(n) n np − ≥ | − | ≥ (An) < 2− . If we define Bm = n>mAn then we have (Bm) mp ∪ c c ≤ (A ) 2− . On the complB = A we see that f n>m n ≤ m ∩n>m n k(n) converges uniformly by the choice of An. Theorem 8.7. (Lp Dominated Convergence Theorem) p Suppose that 1 p < and fn, n N is a sequence in L (Ω, ,) which converges≤ pointwise∞ almost everywhere∈ to a function f andF sup- p pose further that there exists g 0 and g L () so that fn g for all n. Then f Lp() and f converges≥ to∈f in Lp(). | |≤ ∈ n Proof. Define g = sup f f : ℓ,m n n {| ℓ − m| ≥ } p Then gn converges pointwise to zero and gn 2g. It follows that gn gp converges to 0 in L1() by the dominated≤ convergence theorem.| | ≤ p From this it follows that fn is a Cauchy sequence in L and so it converges in Lp by completeness to some function f˜ Lp() and f˜ = f almost everywhere by the preceding Corollary. ∈

Let E be a normed linear space over K (where K = R or C) with norm .A linear functional is a continuous linear mapping from f : E K. A linear mapping f : E K is continuous if and only if f is→ bounded, which means that sup→f(x) : x 1 is ﬁnite; one can show this equivalence by a straightforward{| | argument. ≤ } The set of all linear functionals on E is itself a linear space over K and is denoted L(E, K) or even move simply by E′ and is referred to as the dual space of E. We deﬁne f = sup f(x) : x 1 where the notation {| | ≤ } now depends on the context (acting on E or E′). This makes E′ a normed linear space and E′ is complete, by a standard argument using the completeness of K. d d Example (R )′ is (isomorphic to) R . Indeed for any 1 d< a linear functional can be represented by a 1 d matrix (with≤ respect∞ to some basis) which acts on elements of Rd ×(d 1 matrices) by matrix multiplication. × 84 1.MEASURETHEORYANDINTEGRATION

Example Suppose that g Lq(Ω, ,) and define G by ∈ F G(f)= fg d, for all f Lp() ∈ Ω where 1 q and 1/p + 1/q = 1. Then Hölder’s inequality ≤ ≤ ∞ p implies G(f) g q f p which says that G is defined on all of L () and bounded| | and ≤ since G is obviously linear, G is in the dual space: p G L ()′ and G g q. Let us show that in fact G = g q at least∈ in the case that ≤1 0 so that the open ball Br(y) of radius r and center∈y does not intersect F and moreover the set 8. Lp SPACES 85

A = x Ω : f(x) Br(y) has ﬁnite positive measure. This is because{ the∈ complement∈ of F }can be covered by countably many open balls which do not intersect F . Consider 1 1 1 f d y = f yd f y d

absolute continuity, 111 convex hull, 123 absorbing, 123–128 hull, 137 almost uniform convergence, 82, 92 countably additive, 107 balanced, 124 dual space, 175 Banach space, 158 barrel, 165 equicontinuous, 14, 167–174 barreled, 165 equilibrated, 123–128, 137–138 basis, 139 bicontinuous, 130 filtering, 139 bornological, 160 first category bounded, 160 category, 166 mapping, 160 Fréchet space, 166 set, 148–150 Fréchet space, 157–159 space, 157 Cauchy, 156–158 function Cauchy in measure, 92 characteristic, 26 Cauchy with respect to d, 157 closed Hölder’s inequality, 80 radially, 124 Hahn-Banach Theorem, 175–178 closure, 132 Hermitian form, 144 compact, 148 homeomorphism, 130 locally, 142 inductively ordered, 23 quasi-compact, 148 intersection, 7 complement, 8 complete, 158–159 locally compact, 142 complete measure, 54 continuous, 146, 159, 160 meagre absolutely, 111 meagre, 166 uniformly, 146 nonmeagre, 166 bicontinuous, 130 metric, 152–154 converge, 156 metric space, 152 convergence metrizable, 153–155 almost uniform, 82 Minkowski’s inequality, 81 almost uniform, 92 convergence in measure, 92 norm, 125 convex, 122–128, 137–138 nowhere dense, 166 201 202 INDEX open cover, 148 vector space, 121–122 partial order, 23 weak topology, 162 weak* topology, 162 quasi-compact, 148 Young’s inequality, 80 radially closed, 124–128 Radon-Nikodym derivative, 113 Zorn’s Lemma, 23, 176–177 Radon-Nikodym Theorem, 113 second category category, 166 semi-norm, 125–128 set difference, 8 empty, 7 power, 7, 25 simple topology, 167 space Baire, 166 metric, 152 vector, 122 vector , 121 adjoint, 175 Banach, 158 dual, 175 Fréchet, 159, 166 subspace, 122 topological vector, 129, 133, 164 span spanned, 122 spanning set, 122 subcover, 148 subspace, 122 support, 143 support function, 175 symmetric difference, 8 topological vector space, 129, 133, 164 topology bounded convergence, 160–162, 167 simple, 161, 167 weak, 162 weak*, 162 totally ordered, 23 uniformly continuous, 146 union, 7