Ingegneria Energetica - Analisi e Geometria 1

Completed notes of the course

Complete course

Real and Functional Analysis * Angelo Pasquale, Michele Bucelli„ September 15, 2019 *Master of Science in Mathematical Engineering at Politecnico di Milano. Unocial notes from the lectures of Professor Maurizio Grasselli of the Mathematics Department.„With special thanks to Andrea Di Primio, Lorenzo Fiorello and Matteo Contini. 2 Contents Contents1 1 Elements of set theory5 1.1 Binary relations. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5 1.2 Equivalence classes, quotient set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .6 1.3 Finite and innite sets, cardinals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 1.4 About the axiom of choice, Part One. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .7 1.5 Cardinals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8 1.6 Ordinals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .8 1.7 About the axiom of choice, Part Two. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9 2 Measure theory and Real Analysis112.1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11 2.2algebra. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11 2.3 Topological spaces andalgebras. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .11 2.4 Measurable functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .122.4.1 Properties of measurable functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .12 2.4.2R andB(R ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13 2.4.3 Simple functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .14 2.5 Positive measures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15 2.6 Lebesgue measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .162.6.1 Construction ofandM. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17 2.6.2 Complete measures, completeness of. . . . . . . . . . . . . . . . . . . . . . . . . . . . .18 2.6.3B(Rn );L(Rn ). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19 2.6.4 Completion of a measure. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .20 2.6.5 Vitali set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21 2.7 Abstract integration. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .212.7.1 Basic properties of abstract integral. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .22 2.7.2 Denition of measures through measurable functions. . . . . . . . . . . . . . . . . . . . .24 2.7.3 Integration of real valued functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25 2.8 Almost everywhere. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .26 2.9 The metric spaceL1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .28 2.10 Lebesgue integral (LR ) vs CauchyRiemann integral (RR ). . . . . . . . . . . . . . . . . . .28 2.10.1 Improper integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 2.11 Convergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29 2.12 Integration and dierentiation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .322.12.1 Properties of the RN derivative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 2.13 1st Fundamental Theorem of Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 2.14 2nd Fundamental Theorem of Calculus. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .34 2.14.1 Absolutely continuous functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35 2.14.2 2nd Fundamental Theorem of Calculus, Part One. . . . . . . . . . . . . . . . . . . . . . .36 2.14.3 Bounded variation functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37 2.14.4 2nd Fundamental Theorem of Calculus, Part Two. . . . . . . . . . . . . . . . . . . . . . .38 2.15 Lebesgue decomposition of BV functions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .40 2.16 Product spaces, multiple integrals. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .412.16.1 Productalgebra. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41 2.16.2 Measurable functions on productalgebras. . . . . . . . . . . . . . . . . . . . . . . . .42 2.16.3 Product measures. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42 2.16.4 Integrals on product spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .42 1 CONTENTS 3 2.16.5 Completeness of . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43 3 Functional Analysis45 3.1 Vector spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45 3.2 Normed vector spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .46 3.2.1 Sequences in normed vector spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .47 3.2.2 Separability. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .48 3.3 Banach spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .483.3.1 Compactness. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .50 3.3.2 Equivalent norms. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .52 3.4Lp spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53 3.4.1 Basic inequalities inLp spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53 3.4.2 Inclusions betweenLp spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .56 3.4.3 Compactness inLp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57 3.4.4 Separability ofLp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57 3.5 Linear bounded operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .59 3.5.1 Spaces of linear operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61 3.5.2 Baire's theorem, and its consequences. . . . . . . . . . . . . . . . . . . . . . . . . . . . .63 3.5.3 Open mapping theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .66 3.5.4 Closed linear operators and graphs. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .67 3.5.5 Structure of topological dual spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .68 3.5.6 Characterization of(Lp ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .69 3.5.7 HahnBanach theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .70 3.5.8 Implications of the HahnBanach theorem. . . . . . . . . . . . . . . . . . . . . . . . . . .73 3.6 Reexivity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .743.6.1 Properties of reexive spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .75 3.7 Weak convergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .773.7.1 Properties of weak convergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .78 3.8 Weakconvergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .79 3.8.1 BanachAlaoglu. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .80 3.9 Uniformly convex spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .82 3.10 Compact (linear) operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .84 3.11 Fredholm's alternative. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .87 3.12 Hilbert spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .883.12.1 Orthogonal spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .90 3.12.2 Minimal distance theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91 3.12.3 Dual of a Hilbert space and Riesz representation theorem. . . . . . . . . . . . . . . . . .93 3.12.4 Orthogonal basis of a Hilbert space. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95 3.12.5 Fourier series, Fourier coecients. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95 3.12.6 Linear bounded operators in Hilbert spaces. . . . . . . . . . . . . . . . . . . . . . . . . .97 3.12.7 Eigenvalues and eigenvectors. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .98 3.12.8 Spectral theorem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99 3.13 Fredholm's alternative in Hilbert spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .103 4 Proof of the RadonNikodym theorem105 Appendices107 A Typewriter Sequence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .109 B Convergence implications. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .110 C A second proof of reexivity of Hilbert spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . .111 D GramSchmidt orthogonalization process. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .112 4 CONTENTS Chapter 1 Elements of set theory The purpose of this rst part is to provide a formal description of the concept of set and related ones.The rst thing we need to do is to dene what a set is. One possible approach (called  naïve set theory) is to dene a set through common language (as in the denition given by George Cantor). In particular we dene in this way the concepts of set, element and the belonging relation and assume the existence of an empty set as an axiom. An immediate consequence of this construction is that the empty set is unique, and that it is contained in every set. Since the naïve set theory relies on common language, it lacks a solid formal base. While this is enough for many purposes (including ours), it may lead to paradoxes (e.g. Bertrand Russel's barber paradox). To avoid this, one can develop a more precise and axiomatic set theory, the ZermeloFraenkel set theory (1920s). In it, the concepts of set and element are assumed as primitive. We will use the naïve set theory, assuming to have dened the usual operations and relations (union, intersection, ...) along with their usual properties (e.g. De Morgan's law). 1.1 Binary relations Binary relations are a formal generalization of the well known concept of function. Denition 1.1.1. Given two sets X; Y , non empty, a binary relation is a predicate r(x; y )in two variables x2X; y2Y. By xingxandy, the predicaterbecomes aproposition. Denition 1.1.2 (Graph) .Given a binary relation r(x; y )between two sets X; Y , its graph is the set dened byG(r) =f(x; y)2XY:r(x; y)is trueg. Note that the graph is a generalization of the graph of a function. Denition 1.1.3(Domain).dom(r) =fx2X:9y2Ys.t.(x; y)2G(r)g. Denition 1.1.4(Range).ran(r) =fy2Y:9x2Xs.t.(x; y)2G(r)g. Denition 1.1.5(Function).A binary relationris afunctionif8x2dom(r);9!y2Y: (x; y)2G(r). A function has the following properties: for anyA; BX(orY), ˆf(A[B) =f(A)[f(B) ˆf 1 (A[B) =f 1 (A)[f 1 (B) ˆf(A\B)f(A)\f(B) ˆf 1 (A[B) =f 1 (A)\f 1 (B) 5 6 CHAPTER 1. ELEMENTS OF SET THEORY Example 1.1.1.X=R2 ; A=f(x;0) :x2[0;1]g; B=f(x;1) :x2[0;1]g f:R2 !R2 ; f(x; y) = (x;0) f(A\B) =f(;) =; f(A) =f(x;0) :x2[0;1]g=A f(B) =f(x;0) :x2[0;1]g=A f(A)\f(B) =A ; This example shows how in the third of the properties the strict inclusion might hold. Denition 1.1.6(Order relation) .A binary relation rbetween X and X is said to be an order relation if it is reexive, transitive and antisymmetric, namely if:(i)8x2X;(x; x)2G(r); (ii)8x; y; z2X;(x; y);(y; z)2G(r) =)(x; z)2G(r); (iii)8x; y2X;(x; y);(y; x)2G(r) =)x=y. Denition 1.1.7 (Total and partial order) .An order relation is called total order relation if8x; y2X , either r(x; y )or r(y; x )are true. If this doesn't happen, namely if 9(x; y )2X2 such that neither r(x; y )nor r(y; x )are true, ris calledpartial order relation. Denition 1.1.8 (Equivalence relation) .A binary relation between X and X is said to be an equivalence relationif it is reexive, transitive and symmetric, namely if: (i)8x2X;(x; x)2G(r); (ii)8x; y; z2X;(x; y);(y; z)2G(r) =)(x; z)2G(r); (iii)8x; y2X;(x; y)2G(r) =)(y; x)2G(r). Example 1.1.2.X6 =; A; B2}(X); r(A; b) =AB.ris a partial order relation on}(X)2 . Example 1.1.3.X=N; r(x; y) =xdividesy.ris a partial order relation onN2 . Example 1.1.4.X=R; r(x; y) =xy.ris a total order relation onR2 . Example 1.1.5.X=fstraight lines on a planeg.r(x; y) =xky is an equivalence inX2 . Example 1.1.6.X=Z2 ; r((p; q);(p0 ; q0 )) =pq0 =qp0  is an equivalence relation inX2 . Example 1.1.7.X=Z; q2Nxed.r(m; n) =9k2Z:mn=kq is an equivalence relation inX2 . 1.2 Equivalence classes, quotient set Denition 1.2.1 (Equivalence class) .Given X6 = ;, and requivalence in X2 ,x 2X , the equivalence class whose representative isxis the setfxg=fy2X: (x; y)2G(r)g. Notice that if, given x; x0 2X , if (x; x0 )2 G (r), then fxg \ fx0 g = ;. Vice versa, if (x; x0 )2 G (r), then fxg=fx0 g. This property allows a complete characterization of an equivalence class given one of its elements. Denition 1.2.2. Given a set X6 = ;and an equivalent relation r, the quotient set is the set X=r = ffxg; x2 Xg }(X). With reference to the examples in the previous section, we have:ˆ(1.1.5)X=ris the set of possible directions in a plane; ˆ(1.1.6)X=r=Q; ˆ(1.1.7)Z=r= convergence classes (e.g.Z 5= ff0g;f1g;f2g;f3g;f4g;f5g g;f0g=f0;5;10; : : :g). Remark 1.2.1 .Not all relations are of order or of equivalence. For example, let X be the set of the points in a plane, andr(P; Q) = dist(P; Q)1;ris symmetric but not transitive. 1.3. FINITE AND INFINITE SETS, CARDINALS 7 1.3 Finite and innite sets, cardinals Denition 1.3.1(Equivalence between sets) .Two sets are equivalent (or equipotent, or have the same power) if there exists a bijection between them (namely, a function A!B whose domain is the whole A and whose range is the wholeB). Denition 1.3.2 (Innite set) .A set is innite if it is equivalent to one of its proper subsets (denition by Dedekind, XIX century). Example 1.3.1. Natural numbers are equivalent to odd numbers, which are a subset of naturals. Thus, N is innite. Obviously, a set isniteif it's not innite. Denition 1.3.3. A set is countable if it is equivalent to N or one of it's subsets. In particular, if a set is countable and innite, we will say it iscountably innite. Example 1.3.2.N;Z;Q; A = froots of algebraic polynomials with rational coecients, aka algebraic numbers g are all countably innite. In particular,Ais countable because it is the union of many countable sets. There exists uncountable sets, for exampleR(proof of this: diagonalization method by Cantor). The notion of equipotence relation induces an equivalence relation in some universe of sets ( ); namely, we can say two sets are equivalent if they have the same power. (: can't say the set of all sets, as that leads to paradox in our naïve set theory, while within ZF such set does not exist, i.e. the set of all sets is not a set) Denition 1.3.4. The equivalence class fAg (w.r.t. the equipotence relation) is called cardinality of A, and is denoted bym(A)(wheremstands for magnitude). If A is nite, m(A)is (can be identied with) the number of elements in A. In fact, natural numbers can be generated using this property. We will now show that, in some sense,m(A)behaves like a number even ifAis innite. In particular, we can introduce the following order relation between cardinalities: given two setsA; B; m(A)m(B)(= (9B 1 B:B 1 A)^(6 9A 1 A:A 1 B) This relation, a priori, is a partial order. Indeed, in the following cases, neither m(A)< m (B)nor m(B)< m (A): (1)(9B 1 B:B 1 A)^(9A 1 A:A 1 B) (2)(9A 1 A:A 1 B)^(6 9B 1 B:B 1 A) However, the following theorem holds: Theorem 1.3.1(CantorBernstein).(1) =)AB, hencem(A) =m(B). (2)never holds, as long as our theory includes the axiom of choice, along with th other ZF axioms. 1.4 About the axiom of choice, Part One The axiom of choice assumes it is possible to construct a function (the choice function) that to any non empty set associates an element belonging to that set:8M6 =;;9f:}(M)!M:f(A)2A8AM ; A6 =;(AC) This means there's a criterion to choose an element from every subset of a non empty set. We'll return on the importance of this axiom later on. 8 CHAPTER 1. ELEMENTS OF SET THEORY 1.5 CardinalsThe CantorBernstein theorem (Theorem1.3.1) implies that the order relation within cardinalities is a total order relation as long as the(AC)holds. Moreover, we can introduce sum and multiplication of cardinalities, associate and commutative as in N. Ba- sically, the set of cardinalities is a generalization of N. We call it the set of cardinal numbers (or simply cardinals ). Theorem 1.5.1 (Cantor) .Given a set A, the power set of A has a larger cardinality then A:m(}(A)) > m (A). This implies in particular that there exists no larger set.Example 1.5.1.[a; b][a; b)(a; b](a; b). Proof.LetA= [a; b]; B= (a; b); A 1= ( a+; b)A; B 1= [ a+; b]B, with >0. We have A 1 A; A 1 B , and B 1 B; B 1 A (because intervals of the same type are all equivalent through an ane mapping). Hence, thanks to Cantor-Bernstein Theorem1.3.1, AB . The other equivalence are proven in a similar way.Example 1.5.2. }(N)[0;1]. Example 1.5.3.R2 R(the bijection is the Peano curve). Proof.R}(N)}(ZnN);RR}(N)}(ZnN)}(Z)}(N)R.1.6 Ordinals Denition 1.6.1. Let (A; )ans (B; )be two totally ordered sets. A and B have the same order (or type) if 9f:A!B, bijective, such thatxy=)f(x)f(y)8x; y2A(i.e. an order preserving function). Remark 1.6.1 .Two sets with the same order are necessarily equipotent (because there exists a bijection); the opposite implication does not hold, namely: A; Bhave the same order) :A B. Remark 1.6.2 .N can be totally ordered in an uncountable number of ways, and for each of them there ex- ists a class of sets with the same order, dierent from the others. The order is a structure independent from the set. Denition 1.6.2. A totally order set A is well ordered if each non-empty subset of A has a minimum (i.e. 8EA;9x2Es.t.xy8y2E). Example 1.6.1.Nis well ordered with respect to the canonical order. Example 1.6.2.Z is not well ordered with respect to the canonical order (take for instance E = fz2Z; z < 0g Z). Example 1.6.3.Q\[0;1]is not well ordered with respect to the canonical order. Denition 1.6.3. The order, or type, of a set is called ordinal number . If it is innite, the ordinal is called transnite. For nite sets cardinals and ordinals are  the same, in the sense that they can be identied (i.e. two nite sets with the same cardinality also have the same order; vice versa, if they don't have the same cardinality, they can't have the same order) Theorem 1.6.1 (Zermelo well-ordering theorem, 1904) .If the axiom of choice( AC )holds, any set can be wel l ordered. Remark 1.6.3 .Assuming that any set can be well ordered, it's easy to prove that the axiom of choice holds (it's enough to take the minimum as choice function). In other terms( AC )and Zermelo well-ordering theorem are 1.7. ABOUT THE AXIOM OF CHOICE, PART TWO 9 equivalent. Remark1.6.4.To every cardinal, then, can be associated an ordinal by well-ordering the corresponding set.It is possible to introduce a sum and a product on ordinals, which are not commutative. It is also possible to introduce a total order relation among ordinals, as per the next theorem. Theorem 1.6.2.Given two wel l-ordered setsA; B, eitherAB, orm(A)< m(B), orm(B)< m(A). In other words, ordinals can be ordered. Example 1.6.4.! is the ordinal produced by (N; )(i.e. the equivalence class of sets with the same order as (N;)), and it is the smallest transnite ordinal. The corresponding cardinality is denoted by@ 0. Example 1.6.5. The order of the set of all countable ordinals is the rst uncountable ordinal, whose cardinality is denoted by@ 1. 1.7 About the axiom of choice, Part Two (AC )is necessary to prove many important theorems. However, is also allows some counter-intuitive facts to happen. For instance, a sphere can be decomposed into a nite number of subsets, mutually disjoint, such that, if suitably rearranged, they produce two spheres, each equal to the original one (BanachTarski, 1924) (which cannot be done if we don't assume(AC)to hold...). It was proven by Gödel that( AC )is not a theorem within ZF (i.e. cannot be proven using ZF axioms); it was also proven by Cohen that also ZF +:AC is a consistent theory (meaning that no contradictions can be deducted). Basically, both ZF+AC and ZF+:AC are consistent. A similar fact aected the so-called continuum hypothesis (CH), that say that there does not exist an uncountable set whose cardinality is less than 2@ 0 (namely, there's no set with cardinality between @ 0 and @1 = 2 @0). In other terms,Rwould be the smallest uncountable set. However, (CH) can't be proven either true or false within ZF (Gödel and Cohen). Therefore, both ZF + CH and ZF+:CH are consistent. We will use from now on the theory constructed from the set of axioms ZF+AC+CH. (Reference book:KolmogorovFomin.) 10 CHAPTER 1. ELEMENTS OF SET THEORY Chapter 2 Measure theory and Real Analysis 2.1 IntroductionLet 6= ;be an arbitrary set. How can we dene a function :}( ) ! [0;1 ]such that (A)represents, in some appropriate sense, the  measure of the setA ? Some reasonable requirements for such function are, for instance, that (;) = 0 , and (A[B ) = (A) + (B) whenA\B=;. In order to construct such function, we need to introduce a number of intermediate notions. 2.2algebra Denition 2.2.1(algebra).Given a set 6 =;, the setM}( )is aalgebraif: (i); 2M (ii)8E2M; Ec 2M (iii)8fE ig i2N M=)S i2NE i2 M Remark 2.2.1 .(ii) + (iii) =)T i2NE i2 M , through De Morgan's law. Also, from the properties, follows AnB2M8A; B2M . Basically, a  algebra is closed with respect to most reasonable operations on its sets. Denition 2.2.2 (Measurable space) .Let 6= ;be a set, and M} ( ) be a  algebra. The pair ( ;M )is calledmeasurable space. Denition 2.2.3. Let 6= ;be a set, and E } ( ) be a collection of subsets of . The intersection of all  algebras containing E is called  algebra generated by E, and it's denoted by . It is the smallest algebra containingE. : =T fM:Malgebra containingE g. Example 2.2.1.E = fE 1; : : : ; E ng  } ( ) ; E i\ E j = ; 8i6 = j, and = S n j=1E j (i.e. Ej are a partition of ). Then,has2n elements. 2.3 Topological spaces andalgebras Denition 2.3.1.Given a set 6 =;, and}( ), the pair( ; )is calledtopological spaceif: (i);; 2 (ii)A 1; : : : ; A n2 =)T n i=1A i2 (nite intersections) (iii)fA ig i2I =)S i2IA i2 (arbitrary unions) 11 12 CHAPTER 2. MEASURE THEORY AND REAL ANALYSIS The setis calledtopologyand its elements are calledopen sets. Remark2.3.1.It is possible to give an equivalent denition in terms of closed sets. Topological spaces are the most general contexts in which to dene the concepts of limit and continuity. Denition 2.3.2(Borel  algebra) .The  algebra generated by the topology of a topological space is called Borelalgebra, and its sets are calledBorel sets. The Borel  algebra is the most natural  algebra associated with a topological space. It is denoted by B( ) . 6 =;;E=fA :Aopeng }( );B( ) = Obviously, both open and closed sets are Borel sets. More precisely, a Borel set can be either a countable union of closed sets (F ) or a countable intersection of open sets ( G ). 2.4 Measurable functions Denition 2.4.1. Let f: ( ;M )! (X;  )be a function from a measurable space to a topological space. We say thatfis ameasurable functionif8A2 ; f 1 (A)2M. Remark 2.4.1 .This denition is dierent from the one given in probability theory. We do that to make sure continuous functions are measurable. Remark 2.4.2 .Let f : ( ;~ )! (X;  )be a function between topological spaces. If f is continuous, i.e. 8A2 ; f 1 (A)2~ , thenf: ( ;B( ))!(X; )is measurable (it is calledBorel-measurable). 2.4.1 Properties of measurable functions Theorem 2.4.1. Let f : ( ;M )! (X;  )be measurable and g: ( X;  )! (Y ; 0 )be B measurable. Then h =gf: ( ;M)!(Y ; 0 )is measurable. Proof. Take A20 ; then g 1 (A)2 B (X), which means either g 1 (A)2F  or g 1 (A)2G  (because a Borel set can be either a countable union of closed sets (F ) or a countable intersection of open sets ( G )). Then: ˆg 1 (A)2G = )g 1 (A) =T i2Na i= )f 1 T i2Na i
=T i2Nf 1 (a i) 2M ˆg 1 (A)2F = )g 1 (A) =S i2Nc i =)f 1 S i2Nc i
=S i2Nf 1 (c i) =S i2Nf 1 (ac i) =S i2N( f 1 (a i))c 2M Remark 2.4.3 .It is not possible to relax the hypothesis of the theorem above, namely, if gis measurable but not Bmeasurable, thenfgis not necessary measurable, even whenfis continuous. Proposition 2.4.1. Any open set ERn is a countable union of open rectangles (i.e. sets in the form R = I 1


 I n , where Ij are open intervals in R). Moreover, if n= 1 , the rectangles (in this case intervals) can be chosen to be disjoint. Theorem 2.4.2. Let u; v : ( ;M )!R be measurable, and :R2 ! (X;  )be continuous. Then h=  (u; v ) : ( ;M)!(X; )is measurable. Proof. First, observe that, if R = IJ is an open rectangle in R2 , then setting f: ( ; M )!R2 ,x 7! (u(x); v(x)) we havef 1 (R) =u 1 (I)\v 1 (J)2M(simple case). Therefore, letA2, we want to show thath 1 (A) =f 1 ( 1 (A))2M(general case). We have that  1 (A)2 B (R2 )open, because is continuous so, thanks to the previous proposition, we can write:  1 (A) =S j2NR jwhere R jare open rectangles, and nally f 1 ( 1 (A)) =S i2Nf 1 (R j) 2M. Remark 2.4.4 .A consequence of this is that given ( ;M )measurable space and two measurable functions u; v : !R , the functions u+v; uv; u+ ; u ;juj; : : : , are all measurable, being particular cases of the previous theorem. 2.4. MEASURABLE FUNCTIONS 13Later we will prove that measurability is preserved by limit operations as well, but before doing that we'll need some further observations on Borelalgebra. 2.4.2R andB(R ) Denition 2.4.2.R :=R[ f+1;1g. It is possible to partially extend usual arithmetic toR , and to dene atotal orderand atopologyon it. The topology inR can be introduced through:  :=ff(a;+1]g a2R; f(a; b)g a;b2R; f[1; b)g b2Rg () (plus all the nite unions of sets in  , and innite intersections, i.e. by taking the smallest topology containing  ). This particular construction can be useful because it is convenient to deal with function that can receive 1 as arguments. Remark 2.4.5 .If in( )we take a; b2Q we obtain the same topology, by the density of Q in R, and so generate the samealgebra. Theorem 2.4.3.B(R ) = . Proof.LetM= , we show thatB(R ) M . The inclusion B(R ) M is obvious because 8a2R; (a;+1]2 B (R )(by denition, all open sets are in Borel  algebra). We deal with B(R ) M . By Denition2.3.2, B(R ) = <  > , so we have to show that all the generators of  are in thealgebraM. ˆ(a;+1]c = [1; a]2M, ˆ[1; a) =S n2N 0 1; a1n  |{z} 2M2 M, ˆ(a; b) = [1; b)\(a;+1]2M8b > a. Hence, all the generators of  are in the  algebra M , which implies, by the denition of generated  algebra, B(R )M. Remark 2.4.6 .Obviously, even here we can further restrict the set of generators to f[a;+1]g a2Q , since Q is dense inR. Remark2.4.7.Even inR , as inR, the open sets are countable unions of open intervals and/or half lines. At this point we can introduce the following, very useful, theorem about measurable functions. Theorem 2.4.4. Let ( ;M )be a measurable space, and f n : !R be a sequence of measurable functions. Then: (1)sup n2Nf nand inf n2Nf nare measurable; (2)lim sup n!+1f nand lim inf n!+1f nare measurable. Proof.(1) Let g= supn2Nf n . We need to show that if B2 B (R ), then g 1 (B)2M . We have g 1 ((a;+1]) = S n2Nf 1 n(( a;+1])2M . Similarly, if g= inf n2Nf n , we have g 1 ((a;+1]) = T n2Nf 1 n(( a;+1])2M . Thus, thanks to Theorem2.4.3,gis measurable in both cases. (2)Seth= lim sup n!+1f n( t) = inf n0 supknf k( t)
is measurable by (1). Same goes forlim inf. 14 CHAPTER 2. MEASURE THEORY AND REAL ANALYSISThis shows how measurability is preserved by common limit operations, making it a valuable property. It also shows that it is not trivial to construct non measurable functions. Finally, a bounded geometrical model for R is that of a half a circle in which every P 2R has a matching P 2R , and inR we have the additional pointsQ1 . 2.4.3 Simple functions The importance of measurable functions is linked to them being approximable with simple functions , namely very trivial functions on which the concept of integral is naturally dened.Denition 2.4.3 (Simple function) .Given a measurable space ( ;M ), a function s: !R is a simple function if it has the form s(t) = P N j=1a j Ej (t)where a j2 R , and fE jgN j=1 is a measurable partition of (collection of nonempty disjoint subsets of whose union is ). Remark 2.4.8 .In other words a function sis simple if s( ) is nite ( s( ) = fa 1; : : : a ng ). Actually, this characterisation lacks measurability! More precisely I should say:  a measurable function sis simple if .... Remark2.4.9.Every simple function is measurable. Indeed,s 1 (a j) = E j2 M8j2 f1; : : : ; Ngby denition. Theorem 2.4.5 (Approximation of measurable functions with simple functions) .Let f: ( ;M )! [0;+1]be a positive, measurable function. Then,9fs ng n2N, sequence of simple functions, such that: (i)s n s n+18 n2N; (ii)0s n( t)f(t)8t2 8n2N; (iii)8t2 ;lim n!+1s n( t) =f(t), namelys nconverges pointwise to f; (iv)iffis bounded (9M >0 :f(t)< M8t2 ), thens nconverges uniformly to fin . Proof. Let n2N . We split [0;+1]into [0; n )and [n; +1]. We divide the interval [0; n )into k= n2nintervals (a j; b j) , each of length2 n . So(a j; b j) = j12 n ;j2 n =I jwith j2 f1; : : : ; kg. Then, let E 0 = f 1 ([n;+1]); E j = f 1 ((a j; b j)) . We have that, by the measurability 1of f,E 0; E j2 M8j . Dene nows n( t) =n E0( t) +P k j=1a j Ej( t). (practically,s n( t) =a jif f(t)2(a j; b j) ands n( t) =nisf(t)2[n;+1]) The functions nis simple and has the following properties: (i)It's easy to check that8t2 ; s n( t)s n+1( t). (ii)We show thatfs n. f(t)s n( t) =f(t)n E0( t)P k j=1j 12 n  Ej( t)8t2 . Ift2E j= f 1 (I j) , thenf(t)s n( t) =f(t)j 12 n =2 n f(t)(j1)2 n . Moreover,f(t)2I j= )j12n f(t)j. So,(j1)(j1)2n f(t)(j1)j(j1) =)02n f(t)(j1)1. Moreover, observe that, if t2 is such that f(t)< +1, then 9n 0 (t)2N such that f(t)< n8nn 0 (t). Finally,( t2 :f(t) = +1=)s n( t) =n8n2N t2 :f(t)< n8nn 0= )0f(t)s n( t)12 n 8 nn 0 (iii)Follows that:0f(t)s n( t)12 n 8 nn 0( t) =)s n( t)!f(t)8t2 . (iv) Moreover, if fis bounded, n0 can be chosen uniformely in tand sn! f uniformely. This means that we have the pointwise convergence for allt2 , i.e. the second case holds for everyt: 8t2 ; f(t)< n 0= )0f(t)s n( t)12 n 8 t:1 E0= f 1 ([n;+1]) =f 1 ([1; n)c ) = f 1 ([1; n))
c 2M 2.5. POSITIVE MEASURES 15 2.5 Positive measures Denition 2.5.1.Let ( ;M )be a measurable space. A set function :M! [0;+1]is called positive measure if: (1)9E2M:(E) > > :(  1;  1) \( 2;  2) = ; 2 E;if 1<  2( or 2<  1) ( 1;  1) \( 2;  2) = (  2;  1) 2 E;if 1<  2<  1<  2 ( 1;  1) \( 2;  2) = (  1;  2) 2 E;if 2<  1<  2<  1 (many other cases...) Basically, the intersection of two intervals is either empty or another interval. ˆ(n; n+ 2)2 E;S n2N( n; n+ 2) = . Thus, two measures such that (a; b ) = (a; b ) = ba are inevitably equal on the whole B(R): the Lebesgue measure is unique, if it exists. The same kind of arguments can be used to prove the uniqueness of the Lebesgue measure onRn . Now, we want to nd a measure on a  algebra M B (Rn )such that 8 K = Q n i=1 (a i; b i );  (K) = Q n i=1 (b i a i ), and (E + a) = (E) 8E2M; a2 Rn (where by E + awe denote the set f x + a:x2Eg ). The latter is a reasonable property of invariance of the volume by translations. The rst problem is to determinate the  algebra M , namely the set of measurable sets. Of course, the larger it is, the better. It is however impossible to construct such a measure on the whole power set of Rn , assuming (AC)and (CH) to be true. Remark 2.6.1 (By Buch & Wikipedia) .Beyond Ulam's theorem (see below), we can deduce this from Ba- nachTarski paradox : the sets into which the sphere is decomposed are non measurable. If they were (consequence of M = }(R3 )), their measure would be preserved by roto-translations and it couldn't be possible to construct two spheres. The paradox shows it is not possible to dene a notion of measurability coherent with the classical notion of volume on the one hand, and such that all sets are measurable on the other hand. All this provided (AC)holds, of course. Denition 2.6.1. An atom is a countable set of positive measure which does not contain subsets of positive measure (its measurable subsets have measure zero). Example 2.6.1. Consider the set X = f1;2; : : : ; 9;10 gand the  algebra  = }(X). Dene the measure  of a set to be its cardinality, that is, the number of elements in the set. Then, each of the singletons fig , for 2.6. LEBESGUE MEASURE 17 i= 1;2; : : : ;9;10is an atom. Denition 2.6.2.A measure is purely non atomic if singletons have measure zero (and consequently all countable sets, being countable unions of singletons, have measure zero, so that there are no atoms, hence the name). Example 2.6.2.Consider the Lebesgue measure on the real line. This measure has no atoms. Theorem 2.6.2(Ulam).The only purely non atomic measure on}(R)is the zero measure. Remark 2.6.2 .In the proof of Theorem2.6.2, both the axiom of choice and the continuum hypothesis are used. Remark 2.6.3 .The theorem implies that it is not possible to dene the Lebesgue measure on }(R), as it would have to be non atomic and yet be dierent from the zero measure: ((; )) =8; 2R(Lebesgue measure) E=fx 0g ; >0; E n= x02 n ; x 0+2 n
8n2N (E n) =2 n 1 ES n2NE n= )(E) S n2NE n
P n2N2 n = 8 >0 (E) = 0 =)is purely non atomic=)contradiction with Ulam's Theorem2.6.2. Remark 2.6.4 .Also note that if we assume ¬AC to hold, it is possible to construct a Lebesgue measure on the whole}(R). 2.6.1 Construction ofandM We start introducing a function, called outer measure , dened on all subsets of a given set with values in the extended real numbers satisfying some additional technical conditions. Denition 2.6.3. :}(R)![0;+1]is calledouter measure(orexterior measure) if  (E) = inf ES nI nX n2Nl (I n) 8E2}(R) whereI n= ( a n; b n) ; l(I n) = b n a n. By using the denition of inmum, we can interpreter this equation in terms of inequality. Let  > 0. Then 9f I ng n2Ns.t. ES nI n;P nl (I n)  (E) +. It is easy to prove that if a setEis countable, then (E) = 0. ˆFirst of all consider singletons: x02 R=)x 02 (x 0 ; x 0+ ); l((x 0 ; x 0+ )) = 2, hence:  (fx 0g ) = inf fx 0gS nI nP n2Nl (I n) inf >0l ((x 0 ; x 0+ )) = 08x 02 R. ˆ Let now E be a nite set, say E = fa 1; : : : ; a kg  R . Then for every  > 0; ES k i=1I i , where Ii= ( a i ; a i+ ). Hence, (E)2k. Since this is true for every >0;  (E) = 0. ˆNext suppose thatEis a countably innite set, sayE=fa i: i2Ng. Then takingI n= ( a n2 n +1; a n+2 n +1) , we haveES n2NI n, so that:  (E)P n2Nl (I n) =P n2N 2 n
. Since this is true for every >0, we obtain (E) = 0. Proposition 2.6.1. is monotone and countably sub-additive. Proof.ˆ Monotonicity . If EF , then any covering 2 fI jg j2N of F is also a covering of E. Namely, the set of coverings of F is also contained in the set of coverings of E. Thus, by the denition of inmum,  (E) (F), seeing as (E)is the inmum over a larger set.2 A cover of a setXis a collection of sets whose union containsXas a subset 18 CHAPTER 2. MEASURE THEORY AND REAL ANALYSIS ˆCountably sub-additivity. Consider now a familyfE ng n2N }(R), and let >0. 8n2N;9fIn jg s.t.E nS j2NIn jandP jl (In j)  (E n) +2 n , by the denition of inmum. SetE=S n2NE nS n2NS j2NIn j. Then:  (E) S nS jIn j () P j;n2Nl (In j) P n  (E n) +2 n
=P n (E n) + 8 >0 ((): by the denition of , seeingfIn jg j;n2Nas one particular covering ofS nS jIn j). Thus,  (E)P n (E n) + ;  >0 =) (E)P n (E n) :However,  can't be nitely additive. If it were, then it would be a measure (see Theorem2.5.2), and it would be non-atomic yet dierent from zero measure (see above  (f x 0g ) = 0 ), contradicting Ulam's theorem (Theorem 2.6.2). Still, it is true that ((a; b)) =ba, and (E+a) =  (E). To turn  into a measure, we will dene a convenient  algebra (called Lebesgue  algebra ) and restrict  to it. In order to do so, we give the following denition. Denition 2.6.4.A setE2}(R)satises theCaratheodory condition(C)if: 8T2}(R);  (T) (T\E) + (T\Ec ) Theorem 2.6.3. Set L(R) = fE2} (R) : Esatises (C)g. Then, L(R)is a  algebra, called Lebesgue algebra, and= jL(R)is a measure on (R;L(R)), cal led Lebesgue measure. Remark2.6.5.If 2L(R), thenj L( )is again a measure. Proposition 2.6.2.L(R) B(R). Proof. It is enough to prove that (a;+1)2L (R), because the half-lines generate B(R). Thus, we have to check the(C)condition on(a;+1). Let T2} (R), and set T 1 = T\ (a;+1),T 2 = T\ (1; a ]. If  (T) = + 1, then (C)is trivially satised; so let's assume  (T)0. Then,9fI ng n2Ns.t. TS nI n;P nl (I n)  (T) +(denition of outer measure ofT). SetI1 n= I n\ (a;+1); I2 n= I n\ (1; a+2 n ) , so thatl(I n) +2 n  l(I1 n) + l(I2 n) . We have8i2 f1;2g; T iS nIi n= ) (T i) P nl (Ii n) .  (T 1) +  (T 2) P n l(I1 n) + l(I2 n)
P n l(I n) +2 n
=P nl (I n) +   (T) + 28 >0 =) (T 1) +  (T 2)  (T) The last inequality is the Caratheodory condition for the set (a;+1): thus that set belongs to L(R), proving our thesis.2.6.2 Complete measures, completeness of  Denition 2.6.5. A measure on ( ;M )is complete if8N2M;  (N) = 0 = ) 8PN ; P2M (and so (P) = 0). Proposition 2.6.3.Lebesgue measure is complete. Proof.LetN2L(R)be such that(N) = 0, and letEN. We need to show thatE2L(R), namely thatE satises the Caratheodory condition. 8T2}(R); T\ET\NN=) (T\E) (N) = jL(R) =(N) = 0 =) (T\E) = 0 T\Ec T=) (T\Ec ) (T)  (T) (T\Ec ) = (T\Ec ) + (T\E) =)E2L(R) 2.6. LEBESGUE MEASURE 19Completeness is important to ensure that the integral on subset of sets of zero measure is still well dened and still equal to zero, as one would naturally expect. Denition 2.6.6. A measure  is  nite on ( ;M )if 9fE ng n2N2 M such that S n2NE n = and (E n) 0;9FE such that F =2L (R)(i.e. F not Lebesgue measurable). Refer to the gure here below for a more visual idea of the following proof:Proof. We will consider only the caseE= [0;1]. Let's introduce inEthe following equivalence relation: ab()ab2Q and the related quotient set E (for example   + 3 and p2 6 ). Using( AC )we can pick one element from each equivalence class, and form a setV(calledVitali set) of the chosen elements.3 All the rationals are in one class, and the irrationals are in the other classes. We can assume V E . Observe that: (i)(V+r)\(V+s) =; 8r; s2Q; r6 =s Indeed, if a+r= b+s, then ab , and they would belong to the same class, thus they wouldn't belong both toV; (ii) (V) = (V +r) 8r2Q (if we assume V2L (R)) by the invariance with respect to the translations of ; (iii)R=S r2Q( V+r) Indeed, any number x2R belongs to one of the equivalence classes, and so it can be obtained by translating its representative by a rational. Then, ifV2L(R): +1=(R) = S r2Q( V+r) =P r2Q (V+r) =P r2Q (V) =)(V)>0 Observe now that e E = S r2Q\[0;1] (V + r)is such that  e E = P r2Q\[0;1] (V) = + 1, but e E [0;2], so  e E 0, by denition of supremum, 9 n 02 N s.t. cs(t)f n (t)f (t) 8nn 0 . So t2E n8 nn 0. (note that the previous could have been impossible if we allowed c= 1 , as it might have required f n (t) = f(t), which is not necessarily true...) Thus, = S n2NE n . Moreover, E1 E 2


(due to f f n (t)gn2N being monotonically increasing). Furthermore, all the sets E n are measurable ( En = h 1 n ((c;+1)), where hn = fns, and both f n and sare measurable). Finally: Z f nd Z Enf nd cZ Ens d=c(E n) where withwe denote the measure induced on( ; m)bys(see Remark2.7.2). c(E n) %c( ) (by Thorem2.5.1, part 5) c( ) =cZ s d8s;simple function s.t.0sf csup 0sfZ s d=cZ f d8c2(0;1) By taking the limit asc!1, we have:Z f d(2) . (1) + (2) =)R f d=, which is thesis of the theorem.Remark: the hypothesis f n (t) 0in BL theorem cannot be dropped (see Exercise Book, Chapter 2, Integrals and limits, Exercise 2.2.3 for a counterexample). Theorem 2.7.2 (Additivity of the integral) .Let ( ;M;  )be a measure space, and f ; g : ! [0;+1]be two measurable functions. Then: Z ( f+g) d=Z f d+Z g d() Proof.(i)Iff ; gare simple functions,()holds trivially. (ii) Iff ; g are not simple, then (Theorem2.4.5), 9ff ng n2N; fg ng n2N , sequences of simple functions, such that f n" f ; g n" g. By (i),R ( f n+ g n) d =R f nd +R g nd . Taking the limit asn! 1, and applying monotone convergence (Theorem2.7.1) we nd the thesis.Corollary 2.7.1 (Monotone convergence for series) .Let f n : ! [0;+1]be measurable, and set f(t) = P n2Nf n( t). Then,fis measurable, andR f d=P n2NR f nd . Proof.(i) The series is the limit of a sequence of partial sums of measurable functions, hence it is measurable (see Theorem2.4.4). (ii) Set Fk = P k n=0f n , and observe that Fk+1 (t)F k (t) 8t2 ;8k2N ; then, by monotone convergence (Theorem2.7.1) and nite additivity (Theorem2.7.2) we have: lim k!+1Z F kd =Z lim k!+1F kd =Z f d Z F kd =k X n=0Z f nd 9 > > > = > > > ;= )Z f d= lim k!+1k X n=0Z f nd =X n2NZ f nd  24 CHAPTER 2. MEASURE THEORY AND REAL ANALYSISRemark 2.7.3 .Take (N; } (N);  c ). In this case, non negative measurable functions are the non negative sequences fa ng n2N. The previous corollary can be rephrased as: X k2NX n2Na kn=X n2NX k2Na kn( a kn 0;8k; n2N) Also note that on such space,R Na nd  c=P n2Na n. Theorem 2.7.3(Fatou's lemma).Letf n: ![0;+1]be measurable for al ln2N. Then: Z limn !1f nd limn !1Z f nd  Proof. First of all, the idea is to use monotone convergence theorem, so we want to create from fn (t)a monotone increasing sequence of functionsg n( t). Set g n (t) = infknf k (t); then, sup n0g n (t) = limn !1f n (t). Observe that gn+1 (t)g n (t) 8t2 ;8n2N . We can then apply monotone convergence (Theorem2.7.1), and nd: lim n!1Z g nd =Z lim n!1g nd =Z limn !1f nd  On the other hand, since8n2Ng n f n, by monotonicity of the integral, we have: Z g nd Z f nd =)limn !1Z g nd  |{z} =Z limn !1f nd  limn !1Z f nd  Remark 2.7.4 .In Fatou's lemma, the strict inequality may hold. Indeed, on (R;L (R);  ), consider the following: f n( t) =( 1n ; jtj n 0;jtj> n8 n2N 0 limn !1f n( t) = 0;butR Rf nd = 2>0;limn !1R Rf nd = 2>0. For other counterexamples see Exercise Book, Chapter 2, Integrals and limits, Exercise 2.2.4. 2.7.2 Denition of measures through measurable functions Let( ;M; )be a measure space, and: ![0;+1]be a measurable function. If we set 8E2M; (E) =Z E d() thenis a measure on( ;M). Indeed, the following theorem holds. Theorem 2.7.4. The set function dened by( )is a measure on ( ;M ), and 8f : ! [0;+1], measurable, there holds: Z f d=Z f  d() Proof. (1: is a measure) Trivially, (;) = 0 . Let fE ng n2N M be disjoint, and set E = S n2NE n . We have E=P n2N En, so: (E) =Z E d=Z  Ed =Z X n2N End ( ?) =X n2NZ  End =X n2NZ En d=X n2N (E n) (?) :thanks to monotone convergence for series, Corollary2.7.1, since Enare measurable. 2.7. ABSTRACT INTEGRATION 25(2:( )) We will prove the formula for characteristic functions only; the extension to generic functions is easily made thanks to Theorem2.4.5and monotone convergence (Theorem2.7.1). Let E2M , and  E be its characteristic function. Then: Z  Ed =Z Ed =(E) =Z E d=Z  Ed 2.7.3 Integration of real valued functions We want to extend the denition of integral to arbitrary real-valued functions (not necessarily positive).Denition 2.7.3.L1 ( ;M; ) =ff: !R;measurable, s.t.R j fjd < > :R b af difb > a 0ifa=b R b af difb < a Theorem 2.7.5 (Lebesgue dominated convergence) .Let ff ng n2N be a sequence of measurable functions f n : !R. If (i)f n( t)!f(t)8t2 , (ii)9g2 L1 ( ;M; )s.t.jf n( t)j g(t)8t2 ;8n2N. Then,(I)f n2 L1 ( ;M; ); f2 L1 ( ;M; ); (II)lim n!1R j f n fjd= 0(which implieslim n!1R f nd =R f d, by triangular inequality). Proof.(I) By (ii), we have R j f nj dR g d < +1, so that f n , being measurable, is in L1 . Moreover, since j f nj  g8n2N, we have alsojfj= lim n!1j f nj  g, which similarly impliesf2 L1 . 26 CHAPTER 2. MEASURE THEORY AND REAL ANALYSIS (II)Let n = 2 g jf n fj . n is measurable and non-negative (because 2g jf n fj  2g jf nj j fj  0). Then, by Fatou's lemma: 0Z 2 gd=Z lim n!1 nd limn !1Z  nd = 2Z g d+ limn !1Z j f n fjd =)0limn !1 Z j f n fjd () =)0 lim n!1Z j f n fjd=)lim n!1Z j f n fjd0 =)lim n!1Z j f n fjd= 0 (): because we can show thatlim( g) =lim g. Finally,lim n!1   Z f d Z f nd    lim n!1Z j ff nj d= 0, and hencelim n!1Z f nd =Z f d:Theorem 2.7.6 (Dominated convergence for series) .Let f f ng n2N L1 ( ; M;  )be such that P n2NR j f nj d < +1. Then, (i)P n2Nf n( t)converges pointwise tof(t), withf2 L1 ( ;M; ); (ii)R f d=P n2NR f nd . Proof. Consider the series g(t) = Pn2Nj f n( t)j :gis measurable by construction, and by monotone convergence for series (Corollary2.7.1) we have: R g d = P n2NR j f nj d < +1. Thus g2 L1 . Then, by dominated convergence of the sequence of partial sums, the thesis holds. Indeed, jP n k=1f k( t)j P n k=1j f k( t)j g(t)8t2 ;8n2N Hence we can apply Lebesgue dominated convergence to the sequencefF ng n2N= fP n k=1f k( t)g n2N. (A more precise proof of this theorem will be given after introducing the notion of almost everywhere in the next section, see Corollary2.8.1)Remark 2.7.11.Functions belonging toL1 ( ;M; )are calledintegrable. 2.8 Almost everywhere Zero measure sets, and the behaviour of functions on such sets, do not eect integrals, as we've seen (Proposition 2.7.5). The purpose of this section is to introduce a notion that allows to formally ignore what happens on sets of zero measure. Let( ;M; )be a complete measure space. Denition 2.8.1. A property P(t), where t2 , is said to hold almost everywhere (a.e. ) if the sets of points where it doesn't hold has measure zero, namely ifft2 ::P(t)g= 0. Denition 2.8.2. A function f: !X , where X is a topological space, is measurable almost everywhere in if9 0 such that( c 0) = 0 and8AX; f 1 (A)\ 02 M(whereAis open). Remark 2.8.1 .Denition2.8.1of almost everywhere applies only to pointwise properties. That's why Denition 2.8.2, being about measurability (which is not a pointwise property, of course), is not a particular case of Denition2.8.1. Remark 2.8.2 .Iffis measurable a.e., and 0is the set within which fis measurable, only extension of fto c 0 does not aect its measurability a.e.. Denition 2.8.3.A functionf: !Risessential ly boundedif9M0such that: (ft2 :jf(t)j> Mg) = 0. Referring to Denition2.8.1, we can say that fis essentially bounded if the property  jf(t)j M  holds almost everywhere. 2.8. ALMOST EVERYWHERE 27 Denition 2.8.4.If RN is a topological space (on top of being a measure space), then f: ! R is continuous almost everywhereif the set of discontinuities has zero measure. Remark 2.8.3 .The last denition is one particular case of Denition2.8.1, by taking P(t) = fis continuous at t (as long as we are in RN , where the concept of discontinuity point is rather straightforward, and not in abstract topological space, on which we would need a denition closer to Denition2.8.2, due to the dierent denition of continuity). Example 2.8.1. Q = ( 1; t2Q 0; t =2Q .8M > 0;  (ft2R : Q (t)> Mg ) = 0 , thus  Q is essentially bounded. Q is never continuous. However, Q= 0 a.e., and0is a continuous function, which is curious. Example 2.8.2.f (t) = ( 1; t0 0; t 0, then there exists, by continuity, an interval (; 0] on which g > 0, and thus g6= fon (; 0], which has a positive measure. Similarly, for the casesg(0) = 0org(0)0, which contradict the almost everywhere equality. Example 2.8.4.
f(t) =( tet 2 ; t2Q sint; t =2Qis essentially bounded, but not bounded, as inff(t) =1;supf(t) = +1.
f(t) =( sint; t2RnQ 1; t2Q is equal a.e. to a continuous function, but formally it is nowhere continuous.
f(t) =( sint; t2[0;1]\(RnQ) +1; t2[0;1]\Qis nowhere continuous, but it is bounded and measurable; =)fisLintegrable=) 9R 1 0f (x) d 0) =)(ft2 :f(t)Cg)1C R f d 2.9 The metric spaceL1 We introduce, inL1 ( ;M; ), the following equivalence relation:fg()f=ga.e.. We now dene L1 as the quotient set L1 = L 1 .L 1 is a vector space, with respect to the canonical operations of sum and product, whose elements are classes of equivalent functions, i.e. functions dened except at least a zero measure set. Moreover,L1 is a metric space with the distance:d 1( f ; g) =R j fgjd ; f ; g2L1 ( ;M; ). Indeed, the reason why we use the quotient set L1 instead of L1 is that within L1 ,d 1 wouldn't be a proper distance (in particular, we'd have d(f ; g ) = 0 6 =)f = g). We have, by the vanishing lemma, d 1(f ; g ) = 0 = ) jfgj= 0a.e.=)f=ga.e.). If is also a topological space (and so the notion of continuous function is well dened), each class of L1 contains either one or some continuous representative (see Example2.8.1,2.8.2,2.8.3,2.8.4). We'll see later thatL1 can be made into a normed vector space, namely thatd 1is induced from a norm. 2.10 Lebesgue integral (LR ) vs CauchyRiemann integral (RR ) The purpose of this section is to highlight the dierences and similarities of the two notions of integral on the space(R;L(R); ). Theorem 2.10.1. Let f : ( a; b )R!R be bounded. Then, f is R integrable in (a; b )i the set of discontinuities has measure zero, with respect to the Lebesgue measure. This theorem is a characterisation of Riemann-integrable functions. For example, let us consider  Q\(a;b) . This function isn't Riemann-integrable (it's nowhere continuous), but  Q\(a;b) = 0 a.e. in (a; b )so R b a Q\(a;b)d  = 0 . Theorem 2.10.2.Iff: (a; b)!Ris bounded and measurable, then it'sLintegrable. Theorem 2.10.3. Let f: (a; b )!R be bounded in (a; b )and such that the set of its discontinuities has measure zero. Then, fis both R integrable and L integrable (by the theorems above) and the two integrals coincide h Rb af (t) dt=R b af di . (For the proof, see KolmogorovFomin, pp. 309-310. The general idea is to construct a sequence of simple functions that mimic the upper and lower sums in the denition of Riemann integral). By this theorem, we can use the traditional techniques of integral calculus to computeLintegrals. Also note that there exist L integrable functions which are not bounded or not continuous. For example: 9T  (0;1)(known as generalized Cantor set) such that: (T ) =3(1 )3 2;  2(0;1) (T is measurable ) 6 9I= (a; b); IT ( T doesn't contain intervals, it has not interior points ) 2.11. CONVERGENCE 29Then,  T is measurable and bounded, thus L integrable, but it has a positive-measure set of discontinuities (namely,T itself ), so it's not Rintegrable. 2.10.1 Improper integrals Proposition 2.10.1. Let f: (a; b )!R , with a; b2R ; a < b . If fis R integrable in improper sense, and changes sign a nite number of times, thenfis alsoLintegrable, and the two integrals coincide. Proof. Since fchanges sign a nite number of times, we can assume, without loss of generality, f 0. Let (a; b) =S nE n, with E n= ( a n; b n) , such thata n# a,b n" b. Moreover, suppose fis bounded and R integrable on E n for each n2N (which is a consequence of fbeing Rintegrable on(a; b)in improper sense). Set F n = S n k=0E k; f n = f
 Fn . Observe that as n! 1; F n" (a; b )and f n" f pointwise. Thus, by monotone convergence (Theorem2.7.1), we can write lim n!1R (a;b)f nd  = R (a;b)f d . Since fis both R and L integrable on(a n; b n) we can also write: R(a;b)f nd =R (a n;b n)f d=R bn anf (t) dt Hence,R(a;b)f d= lim n!1R (a;b)f nd = lim n!1R bn anf (t) dt= |{z} improperRRR bn! b an! af (t) dt=R (a;b)f (t) dt This argument fails if the function changes sign an innite number of times. For example, the integralR +1 0 sintt
dt does exist in the improper R sense, but the function is not L integrable, because R +1 0 sin tt  d = +1.2.11 Convergence Let ( ; M;  )be a measure space, complete. Let ff ng n2N; f n : !R , and f: !R be measurable functions. Denition 2.11.1(Pointwise convergence a.e.). fna:e: !f() ft2 : lim n!1f n( t)6 =f(t)g = 0 Denition 2.11.2(Essential supremum). f: !R;ess sup t2 f (t): = inffM0s.t.(ft2 :f(t)> Mg) = 0g Note that, equivalently,ess sup t2 f (t) = inffsup t2 g (t) :g=fpointwise a.e.g. Thus, the essential supremum of a function depends only on its-a.e. equivalence class. Moreover,(ft2 :f(t)> Mg) = 0 =)f(t)Ma.e.. Denition 2.11.3(Uniform convergence a.e.). fnu:a:e: !f()lim n!1ess sup t2 j f n( t)f(t)j= 0 Denition 2.11.4(Convergence in mean). fnL 1 !f()lim n!1Z j f n fjd= 0 Denition 2.11.5(Convergence in measure). fn !f() 8 >0;lim n!1 (ft2 :jf n fj> g) = 0 Theorem 2.11.1.fnu:a:e: !f=)f na:e: !f 30 CHAPTER 2. MEASURE THEORY AND REAL ANALYSIS Remark2.11.1.The opposite implication is not true. Indeed, consider the following sequence of functions fn( t) =8 > < > :nt 0t 0 9E such that (Ec )<  andf nu:a:e: !finE. This theorem means that almost everywhere convergence  a:e ! implies uniform a.e. convergence  u:a:e ! in an arbitrary large set. Proof. Since fna:e: !f , then 9 0 s.t. ( 0) = ( ) and f n (t)!f (t) 8t2 0(by denition of a.e. convergence). Then,8k2N;8t2 09 n 0= n 0( k; t)s.t.jf n( t)f(t)j n 0. Let us now set:E k;m= ft2 0: n 0( k; t)mg. Observe thatS m2NE k;m= 0, and E k;m+1 E k;m. For every k2N, we choosem k= m k( k; )such that: (E k;mk) > ( 0) 2 k = ( )2 k = )(Ec k;mk) 12 g
= +1 8n=)f n 6!0. Remark 2.11.8 .f n !f6 =)f na:e: !f . Consider for example the typewriter sequence.  En ! 0indeed 8 >0 :(ft2[0;1] : En( t)> g) 0be such that n! 0asn! 1, and let n> 0be such thatP n2N n< +1. Consider n h such that nh+1> n h , and (ft2 : jf nh( t)f(t)j>  nhg )<  nh (we can do it because f n !f ). SetE i=S +1 h=if t2 :jf nh( t)f(t)j>  nhg .E=T i2NE i= lim sup h!+1f t2 :jf nh( t)f(t)j>  nhg . 32 CHAPTER 2. MEASURE THEORY AND REAL ANALYSISThe sequence E i is nested ( E1 E 2


), and (E i )< +1 8i , because (E i ) S+1 h=0f


g  P+1 n=0 n< +1. Then,(E i) !0asi!+1, and also(E i) !(E), thus(E) = 0. Take nowt2Ec . This means9i2Ns.t.t =2E i, therefore jf nh( t)f(t)j  nh8 hi. Thus,f nha:e: !f(becausef nh! finEc , and(E) = 0). Remark 2.11.9 .In general, f nu:a:e: !f6 =)f nL 1 !f . However, if we assume f2 L1 and ( ) < +1, we can say 9 n 02 N s.t. f(t) 1< f n (t)< f (t) + 1 8nn 0 , and for almost every t. Then, by dominated convergence (Theorem2.7.5),f nL 1 !f. Theorem 2.11.5.f nL 1 !f=)f n !f. Proof.8 >0;R j f n fjdR ft2 :jf n fj>gj f n( t)f(t)jd(ft2 :jf n fj> g) limn!1R j f n( t)f(t)jd= 0, thus(f


g)!08 >0. By this and Theorem2.11.4, we deduce that from every sequence converging in mean is possible to extract a subsequence converging pointwise. However,f n !f6 =)f nL 1 !f. Indeed, considerf n( t) =n [0 ;1n ]( t); t2R. Thus,f na:e: !0; f n !0h (ft2R:f n( t)> g)1n ! n!10i , butR f nd = 16!0. Note:A scheme summing up the dierent convergence implications is available in AppendixB. Theorem 2.11.6.f n !f ; f n !g=)f=ga.e.. Proof (by Buch).(ft2 :jf(t)g(t)j> g)(ft2 :jf(t)f n( t)j+jg(t)f n( t)j> g)  t2 : maxfjf(t)f n( t)j;jg(t)f n( t)jg>2  = t2 :jf(t)f n( t)j>2  [ t2 :jg(t)f n( t)j>2   t2 :jf(t)f n( t)j>2  + t2 :jg(t)f n( t)j>2  |{z} !0 Thus,8 >0,(ft2 :jf(t)g(t)j> g) = 0, which impliesf=ga.e..Remark 2.11.10. ˆThe convergence in mean is induced by the metricd 1(see Section2.9). ˆThere is no metric that induces the pointwise convergence. ˆUniform convergence is induced byd 1. ˆ Convergence in measure is induced by (f ; g ) = R j fgj1+ jfgjd  , provided ( ) < +1. Actually there are a few other norms inducing the convergence in measure, for instance(f ; g) =R arctan( jfgj) d. 2.12 Integration and dierentiation Denition 2.12.1. Let ( ;M;  )be a complete measure space, and ': ! [0;+1]be a measurable function. Let be the measure induced by ', namely (E) = R E' d8E2M . Then, ' is called RadonNikodym derivativeofwith respect to, and we write'=d d . 2.12. INTEGRATION AND DIFFERENTIATION 33Remark 2.12.1 .With the notation of the above denition, we have that (see Sub-Section2.7.2and Theorem 2.7.4):R Ef d=R Ef ' d8f: ![0;+1];measurable. It follows thatf2L1 ( ;M; )if '2L1 ( ;M; ). Remark2.12.2.(E) = 0 =)(E) = 0. Denition 2.12.2. Let 1;  2 be two measures on the measurable space ( ;M ). We say that  1 isabsolutely continuouswith respect to 2if: 8E2M;  2( E) = 0 =) 1( E) = 0. In this case we write 1  2. We already know that if 9 ' = dd  , then  (Remark2.12.2). It is a legitimate question whether the opposite is true, namely if it is true that=) 9'=d d . The answer lies in the following, very important theorem, whose proof will be given later on as it requires more advanced concepts (see Chapter4). Theorem 2.12.1 (RadonNikodym) .Let ( ;M )be a measurable space, and ;  two measures on it such that: (i)isnite, i.e.9fE ng n2N Ms.t.8n2N(E n) < > :G is dierentiable a.e. in[a; b] G0 2L1 ([a; b]) G(x)G(a) =R x aG0 (t) dt(CF) Proof. As observed on page36, we'll only need to prove the direct implication. Since G is AC, G is BV (Proposition2.14.5), and so it can be written as the dierence of two monotone functions V and VG (Theorem 2.14.3). Both these functions are AC in [a; b ]. Then, let us assume, without loss of generality, that G is monotone non decreasing. Then, G is dierentiable a.e. in [a; b ]and has nite derivative, so that G0 2L1 ([a; b ])(Lebesgue Theorem2.14.4). Let us set '(x) =G(x)R x aG0 (t) dt. ˆ'2AC([a; b])by Theorem2.14.1; 2.14. 2 ND FUNDAMENTAL THEOREM OF CALCULUS39 ˆ'(x)is not decreasing. Indeed, setax 2< x 1 b: '(x 1) '(x 2) = G(x 1) G(x 2) R x1 x2G 0 (t) dt0(by Theorem2.14.5); ˆ'0 (x) = 0a.e. (applying the 1st FTC Theorem2.13.2). It follows, by Proposition2.14.6, that'is constant everywhere, and equal to'(a) =G(a). Thus: G(a) =G(x)R x aG0 (t) dtand henceG(x) =G(a) +R x aG0 (t) dt.Theorem 2.14.6(Change of variable) .Let G : [a; b ]! [c; d ]be AC in [a; b ], strictly monotone. If f: [c; d ]! [0;+1]isLmeasurable, then the fol lowing formula holds: Zd cf d=Z b af (G)
jG0 jd(CV) Remark 2.14.12 .8E2L (R);  (E) = R Ej G0 ()jd denes a measure. The formula( CV )can be written as R d cf d=R b af (G) d(see also Proposition2.12.1). Remark 2.14.13 .The hypothesis of Theorem2.14.6are larger than its  classic counterpart ( G2AC( = G2 C1 ). Remark 2.14.14 .The theorem implies that f(G(
))j G0 (
)j is measurable, and that f2L1 ([c; d ]; L ([c; d ]);  )i f(G)jG0 j 2L1 ([a; b];L([a; b]); ). The Theorem2.14.2states that being absolutely continuous is a condition for calculus formula to hold, which is a really useful property. Then, we are interested in checking whether a function is AC or not. In general, checking the absolute continuity of a function is not easy. Unless one is lucky enough to know how to write the function as an integral, we need to use the denition of AC function, and that's rather inconvenient. Then, we would like to have (sucient) conditions for absolute continuity (like for instance Lipschitz continuity). One such condition is the following. Theorem 2.14.7. Let f: [a; b ]!R be dierentiable everywhere in [a; b ]. If f0 2L1 ([a; b ]), then f 2AC ([a; b ]). Proof.See Rudin, Theorem 8.2.1.Note that this condition is very strict, as it excludes jumps and discontinuities in f0 . Example 2.14.3.f(x) =( x32 sin1x ; x 2(0;1] 0; x= 0. fis dierentiable everywhere. f0 (x) =( 32 px sin1x 1p x cos1x ; x 2(0;1] 0; x= 0= )f0 2L1 ([0;1]) =)f2AC([0;1]). Remark 2.14.15 .Weakening the assumptions of the theorem is not possible. Namely, if f2 C ([a; b ]) is dierentiable almost everywhere in [a; b ], and such that f0 2L1 ([a; b ]), then fis not necessarily AC in [a; b ]. An example of this is the Cantor function. Example 2.14.4.Consider the functionf(x) =( x sin1x ; x 2(0;1] 0; x= 0. Recalling Remark2.14.9, we know that this function is not is of bounded variation on [0;1] if= 1 . We now show that it is AC, and henceBV, if >1. fis dierentiable, let's compute the derivative. f0 (x) =( x 1 sin1x 1x 2 cos1x ; x 2(0;1] 0; x= 0 Hence,jf0 (x)j x 1 +x 2 . For  2the derivative is bounded. Moreover, the integrals R 1 0x 1 dx and R 1 0x 2 dx both converge for 1<  1, hencef2AC([0;1]), thusf2BV([0;1]). Remind thatf2BV([0;1])if the total variation is nite. 40 CHAPTER 2. MEASURE THEORY AND REAL ANALYSIS Observe thatV1 0( f)R 1 0j f0 (x)jdxandfis of bounded variation on[0;1]as the rhs of the inequality is nite.Indeed, notice that f0 is continuous on (0;1]. So for 0< u < v  1, we have f(v)f (u) = R v uf0 (t)d t. Hence, jf(v)f(u)j= R v uf0 (t) dt R v uj f0 (t)jdt. This gives thatV1 ( f)R 1 j f0 (t)jdtR 1 0j f0 (t)jdt. But, as a consequence offbeing continuous on[0;1], we have: lim!0V1 ( f) =V1 0( f)which is a nite number asV1 ( f)R 1 0j f0 (t)jdt8. For a more general case, see Exercise Book, Section 2.4, Exercise 2.4.6, p. 35. 2.15 Lebesgue decomposition of BV functions Denition 2.15.1.Ajump functionis a functionf: [a; b]!Rin the form: f(x) =X xn xh n+X x0 n xh 0 n( ) where x n; x0 n2 [a; b ]are sequences of discontinuities, and h n; h0 n are such that P nj h nj < +1;P nj h0 nj < +1. Remark 2.15.1 .It's easy to check that the rst sum in( )is continuous from the left, while the second is continuous from the right. Remark 2.15.2 .A jump function is a step function ifx 1< x 2< : : : ,x0 1< x0 2< : : : (i.e. if the sequences can be or- dered in R, which is not always possible - e.g., f x ng = Q, ...), and it is a simple function if the sequences are nite. Example 2.15.1. Take f x ng n2N = Q, enumeration of the rationals and h n = 2 n . The function f(x) = P xn = > ;= )e Vis a v.s. onR;dime V @ 0( sinceVe Vis a subspace ofV). 3.2 Normed vector spacesWe now want to dene notions in a vector space that will allow us to do some calculus (basically, to dene and perform limit operations). In order to do so, we need to construct a metric space (which is the simplest structure that induces a notion of convergence). The way to do this in a coherent way with the vector space structure is to dene a norm. Denition 3.2.1 (Normed vector space) .A normed vector space is a structure (V;k
k ), where V is a vector space overR, andk
k:V!Ris a function such that: (i)8v2 V;kvk  0^ kvk = 0()v= 0 (ii)8v2 V;82R;kvk =jj
kvk (iii)8u; v2 V;ku+ vk  k uk +kvk (triangle inequality) Such function is callednorm. If a function veries (ii) and (iii), then it's calledseminorm. (V;k
k )is always a metric space w.r.t. the induced metric d(u; v ):= kuvk 8u; v2V . However, a metric space(X; d)is not always a n.v.s. (take a ball inR3 with the euclidean distance). IfV= R is a v.s. (w.r.t. standard operations) then any norm has the following form kuk = cjujfor somec > 0. Remark3.2.1.(iii)=) jkuk k vkj  k u vk u; v2 V(reverse triangle inequality). Indeed,( kxk=kxy+yk  kxyk+kyk=) kxyk  kxk kyk kyk=kyx+xk  kxyk+kxk=) kxyk  (kxk kyk). Example 3.2.1.V=RN ; x2 V=)x= (x 1; : : : ; x N) . The following are norms:
kxk p=h PN i=1j x ijpi 1p ; p2[1;+1). In particular,k
k 2is the canonical euclidean norm.
kxk 1= max i=1;:::;Nj x ij Thus, (RN ;k
k p )is a family o