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California State University, FresnoMATH 271REAL VARIABLESFall 2013Instructor : Stefaan Delcroix

Chapter 1Preliminary MaterialIn this chapter, we review some material that will be used to develop measure theory. Note that anymaterial from Math 171 or Math 172 may be needed.1.1 Unions, Intersections and ComplementsLet X be a fixed set.We denote by P(X) the set of all subsets of X. P(X) is sometimes called the power set of X. Allsets in this section will be subsets of X.Let A, B ⊆ X. The intersection of A and B (notation: A ∩ B) and the union of A and B (notation:A ∪ B) are defined asA ∩ B = {x ∈ X : x ∈ A and x ∈ B}A ∪ B = {x ∈ X : x ∈ A or x ∈ B}We can generalize this to an arbitrary family of subsets. Let I be an index set and A i ⊆ X for alli ∈ I. Then∩ i∈I A i = {x ∈ X : x ∈ A i for all i ∈ I}∪ i∈I A i = {x ∈ X : x ∈ A i for some i ∈ I}The Distributive Laws give relations between intersections and unions:For arbitrary collections, we getA ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)A ∩ (∪ i∈I B i ) = ∪ i∈I (A ∩ B i )A ∪ (∩ i∈I B i ) = ∩ i∈I (A ∪ B i )Let A ⊆ X. The complement of A (in X) (notation: Ã) is defined asà = {x ∈ X : x /∈ A}We easily get that˜∅ = X , ˜X = ∅ andà = A1

Let A, B ⊆ X. Then the set difference A \ B (also called the relative complement of B in A) isdefined asA \ B = A ∩ ˜B = {x ∈ X : x ∈ A and x /∈ B}De Morgan’s Laws give relations between complements and unions or intersections:For arbitrary collections, we getà ∩ B = à ∪ ˜B and à ∪ B = à ∩ ˜B˜∩ i∈I A i = ∪ i∈I à i and ˜∪ i∈I A i = ∩ i∈I à iA collection of sets {A i : i ∈ I} is called a disjoint collection if the sets are pairwise disjoint:A i ∩ A j = ∅ for all i, j ∈ I with i ≠ j.1.2 FunctionsLet X and Y be sets.We denote a function f with domain X and function values in Y by f : X → Y .For A ⊆ X, we define f(A) = {f(a) : a ∈ A}. So we can extend f to a function f : P(X) → P(Y ).For B ⊆ Y , we define f −1 (B) = {x ∈ X : f(x) ∈ B}. So f −1 : P(Y ) → P(X). If f is one-to-one andY is the range of f (so f is onto), then f −1 sends singletons to singletons and we have the standarddefinition of f −1 .The following properties hold (where A i ⊆ X and B i ⊆ Y for all i ∈ I and B ⊆ Y ):f (∪ i∈I A i ) = ∪ i∈I f(A i )f (∩ i∈I A i ) ⊆ ∩ i∈I f(A i )f −1 (∪ i∈I B i ) = ∪ i∈I f −1 (B i )f −1 (∩ i∈I B i ) = ∩ i∈I f −1 (B i )( )f −1 ˜B = f ˜−1(B)Note that in the last property, the complements are taken with respect to different sets (Y on theleft and X on the right).1.3 Countable and Uncountable setsDefinition 1.1(a) A set S is countable if either S is finite or if there exists a bijection between N and S.(b) A set S is uncountable if S is not countable.✄2

An infinite set S is countable if we can enumerate the set:S = s 1 , s 2 , s 3 , . . .where every element of S shows up exactly once in the enumeration.Example: Z is countable sinceis an enumeration of Z.0, 1, −1, 2, −2, 3, −3, 4, −4, . . .✄Note that we can relax the bijection requirement under certain circumstances (every element of Smust show up at least once in the enumeration but not exactly once).Lemma 1.2 Let S be a non-empty set. Then S is countable if and only if there exists an onto mapθ : N → S.Proof :Suppose first that S is countable. If S is finite, say S = {s 1 , . . . , s n }, then the map{ k if k < nθ : N → S : k →n if k ≥ nis onto. If S is infinite, there exists a bijection between S and N and hence there exists an onto mapθ : N → S.Suppose next that there exists an onto map θ : N → S. If S is finite then clearly S is countable. Soassume that S is infinite. We define a map φ : N → S as follows:• φ(1) = θ(1)• For n ≥ 1, we put φ(n+1) = θ(m) where m is the smallest value with θ(m) /∈ {φ(1), . . . , φ(n)}.Note that φ is well-defined since S is infinite. One easily proves that φ is a bijection. Hence S iscountable.✷Remark: In Lemma 1.2, we can replace N be any infinite countable set.✄Lemma 1.3 Let S be a non-empty set. Then S is countable if and only if there exists a one-to-onemap θ : S → N.Proof :Suppose first that S is countable. If S is finite, say S = {s 1 , . . . , s n }, then the mapθ : S → N : s k → kis one-to-one. If S is infinite, there exists a bijection between S and N and hence there exists aone-to-one map θ : S → N.Suppose next that there exists a one-to-one map θ : S → N. Fix x ∈ S. We define a map φ : N → Sas follows:• φ(n) = x if θ −1 (n) = ∅• φ(n) = θ −1 (n) if θ −1 (n) ≠ ∅3

Note that φ is well-defined since θ is one-to-one. One easily proves that φ is onto. Hence S iscountable by Lemma 1.2.✷Remark: In Lemma 1.3, we can replace N be any infinite countable set.✄Corollary 1.4 A subset of a countable set is countable.Proof : Let ∅ ̸= A ⊆ B with B countable. By Lemma 1.3, there exists a one-to-one map θ : B → N.Then the restriction of θ to A is still one-to-one. Hence A is countable by Lemma 1.3. ✷Proposition 1.5 The cartesian product of a finite number of countable sets is countable.Proof :table:First, we prove that N × N is countable. We can list the elements of N × N in an infinite(1, 1) (1, 2) (1, 3) (1, 4) · · ·(2, 1) (2, 2) (2, 3) (2, 4) · · ·(3, 1) (3, 2) (3, 3) (3, 4) · · ·(4, 1) (4, 2) (4, 3) (4, 4) · · ·.. . . . ..It is pretty clear that we can enumerate all the elements in this table. Hence N × N is countable.Next, we prove the proposition. Let A 1 , . . . , A n be non-empty countable sets. We prove by inductionon n that A 1 ×· · ·×A n is countable. This is obvious for n = 1. So let n ≥ 2. Then A = A 1 ×· · ·×A n−1is non-empty and countable (this is obvious for n = 2; use induction for n ≥ 3). By Lemma 1.2,there exist onto maps θ : N → A and φ : N → A n . Hence the mapψ : N × N → A × A n : (x, y) → (θ(x), φ(y))is onto. Since N × N is infinite and countable, there exists a bijection π : N → N × N. Thenψ ◦ π : N → A × A n is onto. So A 1 × · · · × A n = A × A n is countable by Lemma 1.2.✷Corollary 1.6 Q is countable.Proof : The map θ : Z × N → Q : (m, n) → m is clearly onto. Since Z×N is countable by Propositionn1.5, we get that Q is countable by Lemma 1.2. ✷Corollary 1.7 The countable union of countable sets is countable.Proof : Let I be a non-empty countable index set and let A i be a non-empty countable set for alli ∈ I. By Lemma 1.2, there exists an onto map θ : N → I and for each i ∈ I, there exists an ontomap φ i : N → A i . Then the mapψ : N × N → ∪ i∈I A i : (m, n) → φ θ(m) (n)is onto. Since N × N is infinite and countable by Proposition 1.5, we get that ∪ i∈I A i is countable byLemma 1.2.✷Note that there exist uncountable sets. It is fairly easy to prove that P(N) is uncountable. Later, wewill show (using measure theory) that any non-degenerate interval of real numbers is uncountable.Using Cantor’s Diagonalization process, one can show that the set of all infinite sequences from {0, 1}is uncountable. This result can in turn be used to prove that the interval [0, 1] is uncountable.4

1.4 Axiom of ChoiceIn this section, we mention the Axiom of Choice. This axiom is really an axiom (it seems like it is‘always true’ but there are some equivalent forms that are not so ‘obviously true’).Axiom of Choice: Let I be an index set and A i a non-empty set for all i ∈ I. Thenthere exists a function f : I → ∪ i∈I A i with f(i) ∈ A i for all i ∈ I.Another formulation of the Axiom of Choice is:Let I be an index set and {A i : i ∈ I} a disjoint collection of non-empty sets. Then thereexists a set containing exactly one element of A i for each i ∈ I.In this class, we accept the Axiom of Choice. Note that we do not need the Axiom of Choice if I iscountable.1.5 The Real Numbers1.5.1 Axioms for the Real NumbersThe real numbers (notation: R) are a complete, ordered field. The satisfy the field axioms, the orderaxioms and the completeness axiom.• Field AxiomsA set F with two binary operations + and · is a field ifF1. ∀x, y ∈ F : x + y ∈ FF2. ∀x, y ∈ F : x + y = y + xF3. ∀x, y, z ∈ F : (x + y) + z = x + (y + z)F4. ∃0 ∈ F : ∀x ∈ F : x + 0 = xF5. ∀x ∈ F : ∃y ∈ F : x + y = 0F6. ∀x, y ∈ F : xy ∈ FF7. ∀x, y ∈ F : xy = yxF8. ∀x, y, z ∈ F : (xy)z = x(yz)F9. ∃1 ∈ F : ∀x ∈ F : x · 1 = xF10. ∀x ∈ F \ {0} : ∃z ∈ F : xz = 1F11. ∀x, y, z ∈ F : x(y + z) = xy + xzThe elements 0 (from Axiom F4)) and 1 (from (Axiom F9) are unique. The element y (fromAxiom F5) is unique and is denoted by −x. The element z (from Axiom F10) is unique and isdenoted by x −1 or 1 x .Note that there are plenty of fields: Q, R, C, Z p with p prime, etc.• Order AxiomsLet (F, +, ·) be a field. Then F is an ordered field if there exists a subset P of F satisfying:5

O1. ∀x, y ∈ P : x + y ∈ PO2. ∀x, y ∈ P : xy ∈ PO3. ∀x ∈ P : −x /∈ PO4. ∀x ∈ F : x = 0 or x ∈ P or − x ∈ PIt is easy to see that 0 /∈ P and 1 ∈ P . This implies that an ordered field has characteristic 0:1 + 1 + · · · + 1 is never 0. Hence an ordered field contains isomorphic copies of N, Z and Q.R and Q are examples of ordered fields. The set P is the set of all positive numbers (under theusual order of real numbers). Note that C is not an ordered field.In an ordered field, we can introduce an order: we define x < y if y − x ∈ P ; we define x ≤ yif x = y or if x < y.• Completeness AxiomLet F be an ordered field. Let S be a subset of F. An element m ∈ F is an upper bound for Sif s ≤ m for all s ∈ S. An element m ∈ F is a least upper bound for S if m is an upper boundfor S and if m ≤ b for any upper bound b for S.Note that a set S has at most one least upper bound.A least upper bound for S is also called the supremum of S (notation: sup S).Now we can state the Completeness Axiom:An ordered field F is complete ifC1. Every non-empty subset of F which has an upper bound has a least upper bound.It turns out that the real numbers are a complete, ordered field. This is not at all obvious (doesthere exist a complete, ordered field). Standard proofs use either Dedekind cuts or equivalenceclasses of rational Cauchy sequences.Another fact that is not obvious: any complete ordered field is isomorphic to the real numbers.1.5.2 Archimedean Property and Density of the RationalsThe real numbers have the Archimedean Property (also called the Axiom of Eudoxus).Proposition 1.8 (Archimedes) Let x ∈ R. Then there exists n ∈ Z with x < n.Proof : If x < 0 we can put n = 0. So we may assume that x ≥ 0. Put S = {k ∈ Z : k ≤ x}. Notethat S is a non-empty subset of R (since 0 ∈ S) and is bounded above (by x). By the CompletenessAxiom, S has a supremum, say p = sup S. Since p − 1 2 < p, it follows that p − 1 2is not an upperbound for S. Hence q > p − 1 2 for some q ∈ S. Then q + 1 > p − 1 2 + 1 = p + 1 2since p = sup S. Since q + 1 ∈ Z, it must be that q + 1 > x.> p. So q + 1 /∈ S✷The Archimedean Property allows us to prove that the rational numbers are dense in the real numbers(that is, between any two real numbers is a rational number).6

Proposition 1.9 Between any two real numbers there exists a rational number.Proof : Let x, y ∈ R with x < y. Since y − x > 0, it follows from the Archimedean Property thatthere exists n ∈ N with n > 1 . Hence ny − nx > 1. Consider the set S = {s ∈ Z : s ≥ yn}.y − xNote that S is not empty by the Archimedean Property. Let m be the smallest element in S (notethat m exists since S is bounded below). Then m − 1 /∈ S and so m − 1 < yn. Hencenx = ny − (ny − nx) < ny − 1 ≤ m − 1We get that nx < m − 1 < yn and so x < m − 1n< y. ✷1.6 Supremum and Infimum of a Set of Real NumbersDefinition 1.10 Let S be a non-empty set of real numbers.(a) The supremum of S (notation: sup S) is the least upper bound for S. So sup S ∈ R ∪ {+∞}.(b) The infimum of S (notation: inf S) is the greatest lower bound for S. So inf S ∈ {−∞} ∪ R.✄These definitions are equivalent to the following:• sup S = α ∈ R ⇐⇒{ (1) ∀s ∈ S : s ≤ α(2) ∀ε > 0 : ∃s ∈ S : α − ε < s• sup S = +∞ ⇐⇒ ∀M ∈ R : ∃s ∈ S : M ≤ s• inf S = α ∈ R ⇐⇒{ (1) ∀s ∈ S : α ≤ s(2) ∀ε > 0 : ∃s ∈ S : s < α + ε• inf S = −∞ ⇐⇒ ∀M ∈ R : ∃s ∈ S : s ≤ MWe mention one property of suprema/infima that might not be obvious.Proposition 1.11 Let S be a non-empty set of real numbers. Put −S = {−s : s ∈ S}. Theninf(−S) = − sup S and sup(−S) = − inf S.Proof : We show that inf(−S) = − sup S. Then sup(−S) = − inf(−(−S)) = − inf S.Put sup S = α. Suppose first that α = +∞. Let M ∈ R. Then −M ≤ s for some s ∈ S. Hence−s ≤ M. So inf(−S) = −∞. Suppose next that α ∈ R. Then s ≤ α for all s ∈ S. Thus−α ≤ −s for all s ∈ S. Let ε > 0. Then α − ε < s for some s ∈ S. Hence −s < −α + ε. Soinf(−S) = −α = − sup S.✷7

1.7 Sequences of Real Numbers1.7.1 Limit of a SequenceDefinition 1.12 Let {a n } n≥1 be a sequence of real numbers.(a) The sequence {a n } n≥1 converges to α ∈ R (notation: limn→∞a n = α) if(b) The sequence {a n } n≥1 is a Cauchy sequence if∀ε > 0 : ∃N ∈ N : ∀n ≥ N : |a n − α| < ε∀ε > 0 : ∃N ∈ N : ∀m, n ≥ N : |a n − a m | < ε✄Over the real numbers, these concepts are equivalent:Theorem 1.13 A sequence of real numbers converges to a real number if and only if it is a Cauchysequence.We also want to deal with sequences converging to +∞ or −∞.Definition 1.14 Let {a n } n≥1 be a sequence of real numbers.(a) The sequence {a n } n≥1 converges to +∞ (notation: limn→∞a n = +∞) if∀M ∈ R : ∃N ∈ N : ∀n ≥ N : a n > M(b) The sequence {a n } n≥1 converges to −∞ (notation: limn→∞a n = −∞) if∀M ∈ R : ∃N ∈ N : ∀n ≥ N : a n < MDefinition 1.15 The extended real numbers (notation: R) is the set R ∪ {−∞, +∞}.✄✄We mention some important properties of limits:Proposition 1.16 Let {a n } n≥1 and {b n } n≥1 ) be sequences of real numbers that converge to extendedreal numbers. Then the following holds:1. If a n ≤ b n for all n large enough then limn→∞a n ≤ limn→∞b n .2. If α, β ∈ R then lim (αa n + βb n ) = α lim a n + β lim b n provided the right-hand side is definedn→∞ n→∞ n→∞(note that 0 · (±∞) = 0 here; undefined would be (+∞) + (−∞)).8

1.7.2 Limit Superior and Limit Inferior of a SequenceNot every sequence of extended real numbers has a limit.numbers has a limit superior and a limit inferior.Definition 1.17 Let {x n } n≥1 be a sequence of extended real numbers.But every sequence of extended real(a) The limit inferior of the sequence {x n } n≥1 (notation : lim x n ) is the extended real numberinf x k.k≥nsupn≥1(b) The limit superior of the sequence {x n } n≥1 (notation : lim x n ) is the extended real numberinf x k .✄supn≥1 k≥nRemark : Let {x n } n≥1 be a sequence of extended real numbers. For n ≥ 1, putA n = inf x k = inf{x n , x n+1 , . . .} and B n = sup x k = sup{x n , x n+1 , . . .}k≥nThen {A n } n≥1 is an increasing sequence of extended real numbers. So {A n≥1 } converges to sup A n .n≥1Hencelim x n = sup A n = lim A nn≥1 n→∞Similarly, {B n } n≥1 is a decreasing sequence of extended real numbers. So {B n≥1 } converges to infn≥1 B n.Hencek≥nlim x n = infn≥1 B n = limn→∞B nThere are equivalent forms of these definitions, depending on whether the limit superior/inferior isfinite or infinite.Proposition 1.18 Let {x n } n≥1 be a sequence of extended real numbers. Then the following holds :(a) lim x n = +∞ ⇐⇒ ∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≥ M(b) lim x n = −∞ ⇐⇒ ∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k ≤ M{(1) ∀ε > 0 : ∃n ∈ N : ∀k ≥ n : α − ε ≤ x(c) lim x n := α ∈ R ⇐⇒k(2) ∀ε > 0, ∀n ∈ N : ∃k ≥ n : x k ≤ α + ε(d) lim x n = −∞ ⇐⇒ ∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≤ M(e) lim x n = +∞ ⇐⇒ ∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k ≥ M{(1) ∀ε > 0 : ∃n ∈ N : ∀k ≥ n : x(f) lim x n := α ∈ R ⇐⇒k ≤ α + ε(2) ∀ε > 0, ∀n ∈ N : ∃k ≥ n : α − ε ≤ x k✄9

Proof : We prove the statements about the limit inferior. The statements for the limit superior areproven similarly.For n ≥ 1, put A n = inf x k = inf{x n , x n+1 , . . .}. Then lim x n = sup A n .k≥n(a) Suppose first that lim x n = +∞. Let M ∈ R. Since sup n≥1 A n = +∞, there exists n ∈ N withM ≤ A n = inf k≥n x k . Hence M ≤ A n ≤ x k for all k ≥ n.Suppose next that the given property hold. Let M ∈ R. Then there exists n ∈ N with x k ≥ M forall k ≥ n. Hence A n = inf k≥n x k ≥ M. We proved:So lim x n = sup n≥1 A n = +∞.∀M ∈ R : ∃n ∈ N : A n ≥ M(b) Suppose first that lim x n = −∞. Then sup n≥1 A n = −∞. Hence A n = −∞ for all n ≥ 1. LetM ∈ R and let n ≥ 1. Since A n = inf k≥n x k = −∞, there exists k ≥ n with x k ≤ M.Suppose next that the given property hold. Let M ∈ R and let n ≥ 1. Then there exists k ≥ n withx k ≤ M. Hence A n = inf k≥n x k ≤ M. Since this is true for all n ∈ N, we get that sup n≥1 A n ≤ M.Since this is true for all M ∈ R, we have that lim x n = sup n≥1 A n = −∞.(c) Suppose first that lim x n = α ∈ R. Let ε > 0. Since sup n≥1 A n = α, there exists n ≥ 1 such thatα − ε < A n = inf k≥n x k . Hence α − ε ≤ x k for all k ≥ n and (1) holds. Let ε > 0 and let n ≥ 1. ThenA n ≤ sup n≥1 A n = α. If x k > α + ε for all k ≥ n then A n = inf k≥n x k ≥ α + ε > α, a contradiction.Hence there exists k ≥ n with x k ≤ α + ε and (2) holds.Suppose next that properties (1) and (2) hold. Let n ≥ 1 and let ε > 0. By (2), there existsk ≥ n with x k ≤ α + ε. Hence A n = inf k≥n x k ≤ α + ε. Since this is true for all ε > 0, we havethat A n ≤ α. Let ε > 0. By (1), there exists n ≥ 1 such that α − ε ≤ x k for all k ≥ n. HenceA n = inf k≥n x k ≥ α − ε. We proved:n≥1(i)(ii)∀n ≥ 1 : A n ≤ α∀ε > 0 : ∃n ≥ 1 : α − ε ≤ A nHence lim x n = sup n≥1 A n = α.✷We mention several properties of the limit superior and limit inferior of a sequence.Proposition 1.19 Let {x n } n≥1 be a sequence of extended real numbers. Then the following holds :(a) lim x n ≤ lim x n(b) lim x n = lim x n := α ∈ R if and only if the sequence {x n } n≥1 converges to α.(c) lim (−x n ) = −lim x n and lim (−x n ) = −lim x n .Proof :(a) For n ≥ 1, put A n = inf x k and B n = sup x k . Thenk≥nk≥nA n ≤ B n for all n ≥ 1Taking the limit on both sides as n goes to ∞, we getlim x n = limn→∞A n ≤ limn→∞B n = lim x n10

(b) Suppose first that lim x n = lim x n := α. If α = −∞ then∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≤ Mby using the formulation for the limit superior. Hence {x n } n≥1 converges to −∞. If α = +∞ then∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≥ Mby using the formulation for the limit inferior. Hence {x n } n≥1 converges to +∞. So we may assumethat α ∈ R. Pick ε > 0. Using the formulation for the limit inferior, we have∃n 1 ∈ N : ∀k ≥ n 1 : α − ε ≤ x kUsing the formulation for the limit superior, we have∃n 2 ∈ N : ∀k ≥ n 2 : x k ≤ α + εPut n = max{n 1 , n 2 }. Pick k ≥ n. Then α − ε ≤ x k ≤ α + ε and so |x n − α| ≤ ε. We proved:∀ε > 0 : ∃n ∈ N : ∀k ≥ n : |x k − α| ≤ εHence {x n } n≥1 converges to α.Suppose next that the sequence {x n } n≥1 converges to α. If α = −∞ then∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≤ MHence lim x n = −∞ and so lim x n = −∞ since lim x n ≤ lim x n . If α = +∞ then∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≥ MHence lim x n = +∞ and so lim x n = +∞ since lim x n ≥ lim x n . So we may assume that α ∈ R. Letε > 0. Then there exists n ≥ 1 such that |x k − α| < ε for all k ≥ n. Hence α − ε < x k < α + εfor all k ≥ n. Moreover, if m ≥ 1 is given, then α − ε < x k < α + ε for all k ≥ max{m, n}. Solim x n = α = lim x n by the formulations for the limit inferior and the limit superior.(c) We prove that lim (−x n ) = −lim x n . Then lim (−x n ) = −lim (−(−x n )) = −lim x n .Using that inf(−S) = − sup S, we easily get thatlim (−x n ) = lim inf (−xn→∞k) = lim (− sup x k ) = − limk≥n n→∞k≥nsupn→∞ k≥nx k = −lim x nWe conclude by proving some properties of the limit inferior/superior of two sequences.Proposition 1.20 Let {x n } n≥1 and {y n } n≥1 be two sequences of real numbers. Then the followingholds:(a) If x n ≤ y n for all n big enough then lim x n ≤ lim y n and lim x n ≤ lim y n .(b) lim x n + lim y n ≤ lim (x n + y n ) ≤ lim x n + lim y n ≤ lim (x n + y n ) ≤ lim x n + lim y n unless a sumis undefined (like (+∞) + (−∞)).Proof : (a) We prove that lim x n ≤ lim y n . The proof of lim x n ≤ lim y n is similar. By assumption,there exists n 1 ∈ N with x n ≤ y n for all n ≥ n 1 .We consider three cases.✷11

1. lim y n = +∞Then clearly lim x n ≤ lim y n .2. lim y n = −∞Let M ∈ R. Then we have∃n 2 ∈ N : ∀k ≥ n 2 : y k ≤ MPut n = max{n 1 , n 2 }. Let k ≥ n. Then x k ≤ y k ≤ M. We proved:Hence lim x n = −∞ = lim y n .3. lim y n = β ∈ R∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k ≤ MIf lim x n = −∞ then clearly lim x n ≤ lim y n . So we may assume that lim x n ≠ −∞. Let ε > 0.Since lim y n = β, we have:∃n 2 ∈ N : ∀k ≥ n 2 : y k < β + εHence x k ≤ y k < β + ε for all k ≥ max{n 1 , n 2 }. So lim x n ≠ +∞ (indeed, we proved: ∃M ∈R, ∃n ∈ N : ∀k ≥ n : x k < M which is the negation of ∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k ≥ M).Hence lim x n = α ∈ R. Then α − ε < x k for some k ≥ max{n 1 , n 2 }. For this k, we haveα − ε < x k ≤ y k < β + εSo we provedTaking the limit as ε → 0 + , we get∀ε > 0 : α − ε < β + εlim x n = α ≤ β = lim y n(b) First, we prove that lim (x n + y n ) ≤ lim x n + lim y n . The proof of lim x n + lim y n ≤ lim (x n + y n )is similar.We consider three cases.1. lim x n = +∞ or lim y n = +∞Then lim x n + lim y n = +∞. Hence lim (x n + y n ) ≤ lim x n + lim y n .2. lim x n = −∞ or lim y n = −∞.Then lim x n + lim y n = −∞. Assume that lim x n = −∞. We consider two subcases.2a. lim y n = −∞Let M ∈ R. Since lim x n = −∞, we haveSince lim y n = −∞, we have∃n 1 ∈ N : ∀k ≥ n 1 : x k ≤ M 2∃n 2 ∈ N : ∀k ≥ n 2 : y k ≤ M 212

Put n = max{n 1 , n 2 }. Let k ≥ n. Then x k + y k ≤ M 2 + M 2= M. We proved:∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k + y k ≤ MHence lim (x n + y n ) = −∞ = lim x n + lim y n .2b. lim y n = β ∈ RLet M ∈ R. Since lim x n = −∞, we have∃n 1 ∈ N : ∀k ≥ n 1 : x k ≤ M − β − 1Since lim y n = β,∃n 2 ∈ N : ∀k ≥ n 1 : y k ≤ β + 1Put n = max{n 1 , n 2 }. Let k ≥ n. Then x k + y k ≤ M − β − 1 + β + 1 = M. We proved:∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k + y k ≤ MHence lim (x n + y n ) = −∞ = lim x n + lim y n .3. lim x n = α ∈ R and lim y n = β ∈ RIf lim (x n + y n ) = −∞ then clearly lim (x n + y n ) ≤ lim x n + lim y n . So we may assume thatlim (x n + y n ) ≠ −∞. Let ε > 0. Since lim x n = α, we have∃n 1 ∈ N : ∀k ≥ n 1 : x k < α + εSince lim y n = β, we have∃n 2 ∈ N : ∀k ≥ n 1 : y k < β + εPut n = max{n 1 , n 2 }. Let k ≥ n. Then x k + y k < α + β + 2ε. Hence lim (x n + y n ) ≠ +∞(indeed, we proved: ∃M ∈ R, ∃n ∈ N : ∀k ≥ n : x k + y k < M which is the negation of∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k + y k ≥ M). So lim (x n + y n ) = γ ∈ R. Then γ − ε < x k + y k forsome k ≥ n. For this k, we getSo we proved:Taking the limit as ε → 0 + , we getγ − ε < x k + y k < α + β + 2ε∀ε > 0 : γ − ε < α + β + 2εlim (x n + y n ) = γ ≤ α + β = lim x n + lim y nNext, we prove that lim x n + lim y n ≤ lim (x n + y n ). The proof of lim (x n + y n ) ≤ lim x n + lim y n issimilar.We consider three cases.1. lim y n = −∞Then lim x n + lim y n = −∞. Hence lim x n + lim y n ≤ lim (x n + y n ).13

2. lim y n = +∞Then lim x n + lim y n = +∞. We consider two subcases.2a. lim x n = +∞Let M ∈ R. Let n ∈ N. Since lim x n = +∞, we haveSince lim y n = +∞, we get∃n 1 ∈ N : ∀k ≥ n 1 : x k > M 2∃k ≥ max{n, n 1 } : y k > M 2For this k, we find thatx k + y k > M 2 + M 2 = MSo we proved:∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k + y k > MHence lim (x n + y n ) = +∞ = lim x n + lim y n .2b. lim x n = α ∈ RLet M ∈ R. Let n ∈ N. Since lim x n = α, we haveSince lim y n = +∞, we getFor this k, we findSo we proved:∃n 1 ∈ N : ∀k ≥ n 1 : x k > α − 1∃k ≥ max{n, n 1 } : y k > M − α + 1x k + y k > α − 1 + M − α + 1 = M∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k + y k > MHence lim (x n + y n ) = +∞ = lim x n + lim y n .3. lim y n = β ∈ RWe consider three subcases.3a. lim x n = −∞Then lim x n + lim y n = −∞. Hence lim x n + lim y n ≤ lim (x n + y n ).3b. lim x n = +∞Then lim x n + lim y n = +∞. Let M ∈ R. Let n ∈ N. Since lim x n = +∞, we haveSince lim y n = β, we have∃n 1 ∈ N : ∀k ≥ n 1 : x k > M − β + 1∃k ≥ max{n 1 , n} : y k > β − 114

For this k, we findSo we proved:x k + y k > M − β + 1 + β − 1 = M∀M ∈ R, ∀n ∈ N : ∃k ≥ n : x k + y k > MHence lim (x n + y n ) = +∞ = lim x n + lim y n .3c. lim x n = α ∈ RIf lim (x n + y n ) = +∞, then clearly lim x n + lim y n ≤ lim (x n + y N ). Hence we may assumethat lim (x n + y n ) ≠ +∞.Let n ∈ N. Since lim x n = α, we get thatSince lim y n = β, we getSo for this k, we findSo we proved:which is the negation of∃n 1 ∈ N : ∀k ≥ n 1 : x k > α − 1∃k ≥ max{n, n 1 } : y k > β − 1x k + y k > α + β − 2∃M ∈ R : ∀n ∈ N : ∃k ≥ n : x k + y k > M∀M ∈ R : ∃n ∈ N : ∀k ≥ n : x k + y k ≤ MSo lim (x n + y n ) ≠ −∞.Hence we may assume that lim (x n + y n ) = γ ∈ R. Let ε > 0. Since lim (x n + y n ) = γ, wehave∃n 1 ∈ N : ∀k ≥ n 1 : x k + y k < γ + εSince lim x n = α, we get thatSince lim y n = β, we getFor this k, we findWe proved:Taking the limit as ε → 0 + , we get∃n 2 ∈ N : ∀k ≥ n 2 : x k > α − ε∃k ≥ max{n 1 , n 2 } : y k > β − εα − ε + β − ε < x k + y k < γ + ε∀ε > 0 : α + β − 2ε < γ + εlim x n + lim y n = α + β ≤ γ = lim (x n + y n )✷15

1.8 Open and Closed Sets of Real NumbersDefinition 1.21 A subset O ⊆ R is open if∀x ∈ O : ∃r > 0 : (x − r, x + r) ⊆ O✄One can easily show that ∅, R, (a, b), (a, +∞), (−∞, b) are open sets where a, b ∈ R.The next couple of propositions deal with intersections and unions of open sets.Proposition 1.22 The intersection of finitely many open sets is open.Proof : Let {O i : 1 ≤ i ≤ n} be a finite collection of open sets. Let x ∈ ∩ 1≤i≤n O i . Then x ∈ O i for1 ≤ i ≤ n. Since O i is open, we get that for 1 ≤ i ≤ n, there exists r i > 0 with (x − r i , x + r i ) ⊆ O i .Put r = min{r 1 , r 2 , . . . , r n }. Then r > 0 and (x − r, x + r) ⊆ (x − r i , x + r i ) ⊆ O i for 1 ≤ i ≤ n.Hence (x − r, x + r) ⊆ ∩ 1≤i≤n O i . So ∩ 1≤i≤n O i is open.✷Remark: The intersection of infinitely many open sets does not need to be open.✄Proposition 1.23 The union of arbitrarily many open sets is open.Proof : Let {O i : i ∈ I} be an arbitrary collection of open sets. Let x ∈ ∪ i∈I O i . Then x ∈ O jfor some j ∈ I. Since O j is open, we get that there exists r > 0 with (x − r, x + r) ⊆ O j . So(x − r, x + r) ⊆ O j ⊆ ∪ i∈I O i . Hence ∪ i∈I O i is open. ✷It turns out that we can completely describe open sets of real numbers in terms of open intervals.Note that these intervals could be infinite.Theorem 1.24 Every open set of real numbers is the union of a countable number of pairwisedisjoint open intervals.Proof : Let O be a non-empty open set of real numbers.First, we associate with an open interval (a x , b x ) with each x ∈ O. So let x ∈ O. We definea x = inf{y ∈ R : y ≤ x and (y, x) ⊆ O} and b x = sup{z ∈ R : x ≤ z and (x, z) ⊆ O}Note that these sets are not empty: since x ∈ O and O is open, we have that (x − r, x + r) ⊆ Ofor some r > 0; hence (x − r, x) ⊆ O and (x, x + r) ⊆ O). Also note that a x ∈ R ∪ {−∞} andb x ∈ R ∪ {+∞}. We prove two properties of the interval (a x , b x ).Claim 1: (a x , b x ) ⊆ OWe show that (x, b x ) ⊆ O. Let w ∈ (x, b x ). Since b x is a supremum, there exists z ≥ x suchthat (x, z) ⊆ O and w < z. Hence w ∈ (x, z) ⊆ O. So (x, b x ) ⊆ O.Similarly, we show that (a x , x) ⊆ O. Hence (a x , b x ) ⊆ O.16

Claim 2: a x , b x /∈ OSuppose that b x ∈ O. Then b x ∈ R. Since O is open, we get that (b x − r, b x + r) ⊆ O for somer > 0. By Claim 1, we have that (x, b x ) ⊆ O. Hence (x, b x + r) ⊆ (x, b x ) ∪ (b x − r, b x + r) ⊆ O.So b x + r ∈ {z ∈ R : x ≤ z and (x, z) ⊆ O}, a contradiction since b x = sup{z ∈ R : x ≤z and (x, z) ⊆ O}. Thus b x /∈ O.Similarly, we show that a x /∈ O.Next, we prove the proposition. Consider the collection of open intervals {(a x , b x ) : x ∈ O}.Clearly O = ∪ x∈O (a x , b x ). Indeed, (a x , b x ) ⊆ O for all x ∈ O and so ∪ x∈O (a x , b x ) ⊆ O. Let y ∈ O.Then y ∈ (a y , b y ) ⊆ ∪ x∈O (a x , b x ) and so O ⊆ ∪ x∈O (a x , b x ). Hence O = ∪ x∈O (a x , b x ).We show that this collection is a disjoint collection. Suppose that (a x , b x ) ∩ (a y , b y ) ≠ ∅ for somex, y ∈ O, say z ∈ (a x , b x ) ∩ (a y , b y ). Then a y < z < b x . Since a y /∈ O and (a x , b x ) ⊆ O and a y < b x ,we get that a y ≤ a x . Similarly, we find that a x ≤ a y and so a x = a y . Similarly, we get that b x = b y .Hence (a x , b x ) = (a y , b y ).Finally, we show that this collection is countable. Note that O is uncountable but we will end upwith the same interval (a x , b x ) for different values of x (in fact, any element in (a x , b x ) will lead tothe same interval). For each interval (a x , b x ), pick a rational number in (a x , b x ). Since the intervalsare pairwise disjoint, different intervals will give us different rational numbers. Hence we have aone-to-one map from the collection of intervals to the rational numbers. Since the rational numbersare countable, it follows that the collection of intervals is countable.✷Definition 1.25 Let E be a set of real numbers and x ∈ R. Then x is a point of closure of E if∀δ > 0 : ∃y ∈ E : |x − y| < δWe denote the set of of closure points of E by E and call it the closure of E.✄Proposition 1.26 The following holds:(a) Let E ⊆ R and x ∈ R. Then x ∈ E if and only if O ∩ E ≠ ∅ for every open set O containingx.(b) E ⊆ E for all E ⊆ R.(c) E = E for all E ⊆ R.(d) Let A, B ∈ R with A ⊆ B. Then A ⊆ B.Definition 1.27 Let F ⊆ R. Then F is closed if F = F .✄The next proposition gives a relation between open sets and closed sets.Proposition 1.28 Let E ⊆ R. Then E is open if and only ifẼ is closed.Proof : Put F = Ẽ.Suppose first that E is open. Let x ∈ F . If x ∈ E, then O := (x − r, x + r) ⊆ E for some r > 0and so O ∩ F = O ∩ Ẽ = ∅, a contradiction since x ∈ F and O is an open set containing x. Hencex /∈ E. So x ∈ Ẽ = F . Thus F ⊆ F . Hence F = F and F is closed.17

Suppose next that F is closed. Let x ∈ E. Then x /∈ Ẽ = F = F and so∃δ > 0 : ∀y ∈ F : |x − y| ≥ δLet z ∈ (x − δ, x + δ). If z ∈ F , then |x − z| ≥ δ, a contradiction since z ∈ (x − δ, x + δ). Hencez /∈ F = Ẽ. So z ∈ E. Thus (x − δ, x + δ) ⊆ E and E is open.✷Remark: Since Ẽ = E, we get that E is closed if and only if Ẽ is open.✄The next couple of propositions deal with intersections and unions of closed sets.Proposition 1.29 The intersection of arbitrarily many closed sets is closed.Proof :Let {F i : i ∈ I} be an arbitrary collection of closed sets. Then ˜F i is open for all i ∈ I byProposition 1.28. Using De Morgan’s Law for intersections, we get that ˜∩ i∈I F i = ∪ i∈I˜Fi is open byProposition 1.23. So ∩ i∈I F i is closed by Proposition 1.28.✷Proposition 1.30 The union of finitely many closed sets is closed.Proof : Let {F i : 1 ≤ i ≤ n} be a finite collection of closed sets. Then ˜F i is open for 1 ≤ i ≤ n byProposition 1.28. Using De Morgan’s Law for unions, we get that ∪ 1≤i≤n˜F i = ∩ 1≤i≤n˜Fi is open byProposition 1.22. So ∪ 1≤i≤n F i is closed by Proposition 1.28.✷Remark: The union of infinitely many closed sets does not need to be closed.✄The proof of the next theorem uses the famous Heine-Borel Theorem. Recall from Math 172 thatthere are three equivalent definitions of a compact set E (E is closed and bounded; every sequencein E has a subsequence that converges to some element in E; every open covering of E has a finitesubcovering).Theorem 1.31 [Nested Set Theorem] Let F 1 ⊇ F 2 ⊇ F 3 ⊇ · · · be a descending sequence of nonemptyclosed sets such that F 1 is bounded. Then ∩ ∞ n=1F n ≠ ∅.Proof :Suppose that ∩ ∞ n=1F n = ∅. Then using De Morgan’s Law for intersections, we get that∪ ∞ n=1˜F n = ˜ ∩ ∞ n=1F n = ˜∅ = RNote that ˜F n is open for all n ≥ 1. Hence {˜F n : n ≥ 1} is an open covering of F 1 . Since F 1 iscompact, it follows from the Heine-Borel Theorem that this open covering has a finite subcovering,say {F n1 , . . . , F nk }. Put m = max{n 1 , . . . , n k }. Since {F n } n≥1 is a descending sequence of sets, itfollows from De Morgan’s Law for intersections thatF 1 ⊆ ∪ k i=1 ˜F ni ⊆ ∪ m n=1˜F n = ˜ ∩ m n=1F n = ˜F mSo F 1 ⊆ ˜F m . Since F m ⊆ F 1 , we get that ˜F 1 ⊆ ˜F m . Hence R = F 1 ∪ ˜F 1 ⊆ ˜F m . So R = ˜F m and thus∅ = ˜R = ˜Fm = F m , a contradiction.Hence ∩ ∞ n=1F n ≠ ∅.Remark: This theorem is false if we drop the condition that F 1 is bounded or that the F n ’s areclosed.✄✷18

1.9 Continuous FunctionsDefinition 1.32 Let E ⊆ R and f : E → R a function.(a) Let a ∈ E. Then f is continuous at a if∀ε > 0 : ∃δ > 0 : ∀x ∈ E : |x − a| < δ =⇒ |f(x) − f(a)| < ε(b) f is continuous on E if f is continuous at a for all a ∈ E.✄There is a relation between continuous functions and open sets.Theorem 1.33 Let f : E → R be a function. Then f is continuous on E if and only if for everyopen set O, there exists an open set O ∗ with f −1 (O) = E ∩ O ∗ .Proof : Suppose first that for every open set O, there exists an open set O ∗ with f −1 (O) = E ∩ O ∗ .Let a ∈ E. Let ε > 0. Put O = (f(a) − ε, f(a) + ε). Since O is open, we have that f −1 (O) = E ∩ O ∗for some open set O ∗ . Note that a ∈ f −1 (O) since f(a) ∈ O. Hence a ∈ O ∗ . So (a−δ, a+δ) ⊆ O ∗ forsome δ > 0 since O ∗ is open. Let x ∈ E with |x−a| < δ. Then x ∈ E∩(a−δ, a+δ) ⊆ E∩O ∗ = f −1 (O).Thus f(x) ∈ O = (f(a) − ε, f(a) + ε). So |f(x) − f(a)| < ε. We proved:∀ε > 0 : ∃δ > 0 : ∀x ∈ E : |x − a| < δ =⇒ |f(x) − f(a)| < εHence f is continuous at a. Since a was arbitrary, we get that f is continuous on E.Suppose next that f is continuous on E. Let O be an open set. We associate with every element off −1 (O) an open set. So let a ∈ f −1 (O). Then f(a) ∈ O. Hence (f(a) − ε a , f(a) + ε a ) ⊆ O for someε a > 0 since O is open. Using the continuity of f on E, we get∃δ a > 0 : ∀x ∈ E : |x − a| < δ a=⇒ |f(x) − f(a)| < ε aPut O ∗ = ∪ a∈f −1 (O)(a − δ a , a + δ a ). Then O ∗ is open by Proposition 1.23.Let b ∈ f −1 (O). Then b ∈ E and b ∈ (b − δ b , b + δ b ). Henceb ∈ E ∩ (b − δ b , b + δ b ) ⊆ E ∩ ( ∪ a∈f −1 (O)(a − δ a , a + δ a ) ) = E ∩ O ∗So f −1 (O) ⊆ E ∩ O ∗ .Let x ∈ E ∩ O ∗ . Then x ∈ E and x ∈ (a − δ a , a + δ a ) for some a ∈ f −1 (O). So |x − a| < δ a and thus|f(x)−f(a)| < ε a . Hence f(x) ∈ (f(a)−ε a , f(a)+ε a ) ⊆ O. So x ∈ f −1 (O). Thus E ∩O ∗ ⊆ f −1 (O).Hence f −1 (O) = E ∩ O ∗ .✷19

Chapter 2Measurable SetsIn this chapter, we will generalize the concept ‘length of an interval’.2.1 Outer Measure of a SetWe want to generalize the concept ‘length of an interval’ to more general sets: what is the ‘length’ ofof any kind of subset of real numbers. We use the terminology measure of a set instead of ‘length’.We want this measure to have certain properties.Does there exist a function µ : P(R) → [0, +∞] such that1. If I is an interval then µ(I) = l(I) where l(I) is the length of I.2. µ is translation invariant: µ(E + x) = µ(E) for all E ⊆ R and all x ∈ R.3. µ is countable additive: If {E n } n≥1 is a sequence of disjoint sets of real numbers then µ (∪ ∞ n=1E n ) =∑ ∞n=1 µ(E n).It turns out that such a function µ does not exist! In this course, we will relax the third condition:countable additivity will only hold for a specific class of sets (called the measurable sets).Definition 2.1 Let A ⊆ R.(a) A countable cover of open intervals for A (abbreviation: ccoi for A) is a countable collection{I n : n ∈ N} of open intervals with A ⊆ ∪ ∞ n=1I n .(b) The outer measure of A (notation: m ∗ (A)) is defined as{ ∞}∑m ∗ (A) = inf l(I n ) : {I n : n ∈ N} is a ccoi for An=1✄We list some immediate but important properties of the outer measure.Proposition 2.2 The outer measure has the following properties:(a) 0 ≤ m ∗ (A) ≤ +∞ for all A ⊆ R.20

(b) [Monotonicity] If A ⊆ B ⊆ R then m ∗ (A) ≤ m ∗ (B).(c) m ∗ (∅) = 0(d) If A is countable then m ∗ (A) = 0.Proof : (a) Obvious.(b) This follows from the fact that a ccoi for B is also a ccoi for A if A ⊆ B.(c) Note that {(0, ε)} is a ccoi for ∅ for all ε > 0. Hence 0 ≤ m ∗ (∅) ≤ l((0, ε)) = ε for all ε > 0. Som ∗ (∅) = 0.{((d) Let a 1 , a 2 , a 3 , . . . be an enumeration of A. Let ε > 0. Then a n −ε2 , a n+1 n + ε ) }: n ∈ N is2 n+1a ccoi for A. Hence0 ≤ m ∗ (A) ≤∞∑n=1(l a n −ε2 , a n+1 n + ε )=2 n+1∞∑n=1ε2 n = εSo m ∗ (A) = 0.✷Recall that we would like the outer measure to have certain properties. We show that the outermeasure of an interval is indeed the length of the interval (this is by no means obvious), that theouter measure is translation invariant and that the outer measure is countable subadditive (this isthe closest we can get to countable additive).Proposition 2.3 Let I be an interval. Then m ∗ (I) = l(I).Proof : Suppose first that I is compact, say I = [a, b].Note that {(a−ε, b+ε)} is a ccoi for [a, b] for all ε > 0. Hence m ∗ ([a, b]) ≤ l((a−ε, b+ε)) = b−a+2εfor all ε > 0. So m ∗ ([a, b]) ≤ b − a.Let {I n : n ∈ N} be a ccoi of [a, b]. We want to show that ∑ ∞n=1 l(I n) ≥ b − a. Hence we may assumethat I n is bounded for all n ∈ N. Since [a, b] is compact, it follows from the Heine-Borel Theoremthat this ccoi has a finite subcovering for [a, b], say {I 1 , . . . , I m }.We will construct a finite list of bounded intervals {(a 1 , b 1 ), . . . , (a k , b k )} such that• a 1 < a < b 1• a i+1 < b i < b i+1 for i = 1, 2, . . . , k − 1• b ≤ b kPut I j = (a j , b j ) for 1 ≤ j ≤ m. Since a ∈ [a, b] ⊆ ∪ m j=1I j , we have that a ∈ I j for some 1 ≤ j ≤ m,say a ∈ I 1 = (a 1 , b 1 ). If b ≤ b 1 , we stop. So assume that b 1 < b. Then a < b 1 < b and sob 1 ∈ I j = (a j , b j ) for some 1 ≤ j ≤ m. Clearly j ≥ 2 (since b 1 < b j ), say j = 2: b 1 ∈ I 2 = (a 2 , b 2 ). Ifb ≤ b 2 , we stop. So we may assume that b 2 < b. Then a < b 1 < b 2 < b and so b 2 ∈ I j = (a j , b j ) forsome 1 ≤ j ≤ m. Clearly j ≥ 3 (since b 1 < b 2 < b j ), say j = 3: b 2 ∈ I 3 = (a 3 , b 3 ). Continuing thisway, this process must stop, say after k steps, since we have a finite subcovering. Then b ≤ b k andwe have constructed our desired list.Then b i − a i+1 > 0 for i = 1, . . . , k − 1. Hence we get that21

Sok∑l(I k ) = (b 1 − a 1 ) + (b 2 − a 2 ) + · · · + (b k − a k )j=1= b k + (b k−1 − a k ) + (b k−2 − a k−1 ) + · · · + (b 1 − a 2 ) − a 1> b k − a 1> b − a∞∑l(I n ) ≥n=1k∑l(I k ) > b − aSince this is true for any ccoi {I n } n≥1 of [a, b], we find that m ∗ ([a, b]) ≥ b − a.Hence m ∗ ([a, b]) = b − a = l([a, b]).Next, suppose that I is a bounded interval, say with endpoints a and b. Then I ⊆ [a, b] and son=1m ∗ (I) ≤ m ∗ ([a, b]) = b − aby monotonicity. For 0 < ε < b−a , we have that [a + ε, b − ε] ⊆ I and so2m ∗ (I) ≥ m ∗ ([a + ε, b − ε]) = b − a − 2εby monotonicity. Since this is true for all 0 < ε < b−a , we get that2Hence m ∗ (I) = b − a = l(I).m ∗ (I) ≥ b − aFinally, suppose that I is an unbounded interval, say I is unbounded above. Let a ∈ I. Then for alln ∈ N, we have that [a, a + n] ⊆ I and so m ∗ (I) ≥ m ∗ ([a, a + n]) = n by monotonicity. Since this istrue for all n ∈ N, we get that m ∗ (I) = +∞ = l(I).✷Corollary 2.4 Any non-degenerate interval of real numbers is uncountable.Proof : Let I be a non-degenerate interval. Then l(I) ≠ 0. By Proposition 2.3, we have thatm ∗ (I) = l(I) > 0. It follows from Porposition 2.2(d) that I is not countable.✷Proposition 2.5 The outer measure is translation invariant: m ∗ (A + y) = m ∗ (A) for all A ⊆ Rand for all y ∈ R.Proof : Let A ⊆ R and y ∈ R. Note that if I is an open interval, then I + y is an open intervaland l(I + y) = l(y). It follows that if {I n } n≥1 is a ccoi of A then {I n + y} n≥1 is a ccoi of A + y and∞∑∞∑l(I n ) = l(I n + y). Hence m ∗ (A) ≥ m ∗ (A + y).n=1n=1Using this general inequality, we get m ∗ (A) = m ∗ ((A + y) + (−y) ≤ m ∗ (A + y).Thus m ∗ (A + y) = m ∗ (A).Proposition 2.6 (Countable Subadditivity) Let A n ⊆ R for n ∈ N. Then( ∞)∪m ∗ A n ≤n=1∞∑m ∗ (A n )n=1✷22

Proof :This is obvious if∞∑m ∗ (A n ) = +∞. Hence we may assume thatn=1particular, m ∗ (A n ) ≠ +∞ for all n ≥ 1.Let ε > 0. For n ≥ 1, there exists a ccoi {I n k } k≥1 of A n with∞∑m ∗ (A n ) ≠ +∞. Inn=1∞∑k=1l(I n k ) < m ∗ (A n ) + ε2 nsince m ∗ (A n ) is an infimum. By Proposition 1.5, {Ik n : k, n ∈ N} is a ccoi of ∪∞ n=1A n . Hence we get( ∞) (∪∞∑∞∑ ∞)(∑∞∑ (m ∗ A n ≤ l(Ik n ) = l(Ik n ) < m ∗ (A n ) + ε ) ∞)∑= m ∗ (A2 n n ) + εn=1k,n=1k=1n=1k=1n=1( ∞)∪∞∑Since this is true for all ε > 0, we find that m ∗ A n ≤ m ∗ (A n ).n=1n=1✷Remarks:(a) This result holds (of course) for a finite number of sets:( n)∪m ∗ A k ≤k=1n∑m ∗ (A k )k=1We can use countable subadditivity with A k = ∅ for all k > n since m ∗ (∅) = 0.(b) Note that we can not improve this proposition to have equality even if {A n } n≥1 is a disjointcollection. Using the Axiom of Choice, we can construct sets A, B ⊆ R with A ∩ B = ∅ andm ∗ (A ∪ B) < m ∗ (A) + m ∗ (B).✄2.2 σ-AlgebrasIn order to establish countable additivity for some special disjoint collections, we will use a structurecalled σ-algebras, which are special case of algebras.Definition 2.7 Let X be a set and A a collection of subsets of X (so A ⊆ P(X)).(a) A is an algebra if(1) ∀A ∈ A : Ã ∈ A(2) ∀A, B ∈ A : A ∪ B ∈ A(b) A is a σ-algebra if(1) ∀A ∈ A : Ã ∈ A∞∪(2) ∀A 1 , A 2 , . . . ∈ A : A n ∈ An=1✄23

Remarks:(a) A σ-algebra is clearly an algebra but not every algebra is a σ-algebra.(b) For any set X, there are at least two algebras: {∅, X} and P(X). Note that P(X) is actuallya σ-algebra.(c) Let A be an algebra and A, B ∈ A. Then Ã, ˜B ∈ A by (1). So à ∪ ˜B ∈ A by (2). Again by(1), A ∋ ˜Ã ∪ ˜B = ˜Ã ∩ ˜B = A ∩ B. So A ∩ B ∈ A.Similarly, if A is a σ-algebra and A n ∈ A for all n ∈ N, then ∩ ∞ n=1A n ∈ A.(d) Let A be an algebra. Pick A ∈ A. ThenX = ˜∅ ∈ A by (1).à ∈ A by (1) and so ∅ = A ∩ à ∈ A by (c). Hence(e) If we replace unions with intersections in conditions (2), we get equivalent definitions.✄The next proposition shows that every collection of subsets of X generates an algebra and a σ-algebra.Proposition 2.8 Let X be a set and C ⊆ P(X). Then the following holds:(a) There exists a smallest algebra on X containing C.(b) There exists a smallest σ-algebra on X containing C.Proof : We prove (a). The proof of (b) is similar.Let A be the intersection of all the algebras containing C. Note that A is well-defined since P(X) isan algebra containing C. Moreover, C ⊆ A.We show that A is an algebra. Let A ∈ A. If B is an algebra containing C then A ∈ B and so à ∈ B.Hence à ∈ A. Let A, B ∈ A. If B is an algebra containing C then A, B ∈ B and so A ∪ B ∈ B.Hence A ∪ B ∈ A.Finally, we show that A is the smallest algebra containing C. Let B be an algebra containing C.Since A is the intersection of all the algebras containing C, we have that A ⊆ B.✷Remark: This algebra is called the algebra generated by C and is sometimes denoted by ⟨C⟩. Theσ-algebra generated by C is denoted by ⟨C⟩ σ .✄We finish this section with a technical lemma that will be used later on in the proof of countableadditivity of measurable sets.Lemma 2.9 Let A be an algebra on X and {A n : n ∈ N} a countable collection of elements ofA. Then there exists a countable disjoint collection {B n : n ∈ N} of elements of A with ∪ ∞ n=1A n =∪ ∞ n=1B n .Proof :Put A 0 = ∅ ∈ A. For n ≥ 1, we putB n = A n \ (A 0 ∪ A 1 ∪ A 2 ∪ · · · ∪ A n−1 ) = A n ∩ ∪ ˜n−1k=0 A kNote that B n ∈ A for all n ≥ 1 since A is an algebra.24

Next, we show that the collection {B n : n ∈ N} is a disjoint collection. Let 1 ≤ m < n. ThenB n = A n ∩ ∪ ˜n−1k=0 A k = A n ∩ Ã0 ∩ Ã1 ∩ · · · ∩ Ã n−1 ⊆ Ãm and B m = A m ∩ ∪ ˜m−1k=0 A k ⊆ A mHence B n ∩ B m = ∅.Finally, we show that ∪ ∞ n=1A n = ∪ ∞ n=1B n . Since B n ⊆ A n for all n ≥ 1, we have that ∪ ∞ n=1B n ⊆∪ ∞ n=1A n . Let x ∈ ∪ ∞ n=1A n . Then x ∈ A n for some n ≥ 1. Let m be the smallest index with x ∈ A m .Then m ≥ 1 since A 0 = ∅ and x /∈ A i for 0 ≤ i ≤ m − 1. Hence x /∈ ∪ m−1i=0 A i. SoThus ∪ ∞ n=1A n ⊆ ∪ ∞ n=1B n .It follows that ∪ ∞ n=1A n = ∪ ∞ n=1B n .2.3 Measurable Setsx ∈ A m ∩ ∪ ˜m−1i=0 A i = B mIn this section, we define a class of sets for which the outer measure is countable additive.Definition 2.10 Let E ⊆ R. Then E is measurable ifRemarks:m ∗ (A) = m ∗ (A ∩ E) + m ∗ (A ∩ Ẽ) for all A ⊆ R ✄(a) Let A, E ⊆ R. Then A = A ∩ R = A ∩ (E ∪ Ẽ) = (A ∩ E) ∪ (A ∩ Ẽ). So by countablesubadditivity, we have thatm ∗ (A) ≤ m ∗ (A ∩ E) + m ∗ (A ∩ Ẽ)Hence we get that E is measurable if and only if✷m ∗ (A ∩ E) + m ∗ (A ∩ Ẽ) ≤ m∗ (A)for all A ⊆ RThis inequality is clearly satisfied if m ∗ (A) = +∞. So we finally get that E is measurable ifand only ifm ∗ (A ∩ E) + m ∗ (A ∩ Ẽ) ≤ m∗ (A) for all A ⊆ R with m ∗ (A) ≠ +∞(b) It is not at all clear that the outer measure is countable additive on measurable sets. But thedefinition of measurable sets does have something to do with disjoint sets (note that A ∩ Eand A ∩ Ẽ are disjoint). Recall that we mentioned there exist disjoint sets A, B ⊆ R withm ∗ (A ∪ B) < m ∗ (A) + m ∗ (B). That is not possible if A or B are measurable. Indeed, supposeB is measurable. Since A and B are disjoint, we get thatandSosince B is measurable.(A ∪ B) ∩ B = (A ∩ B) ∪ (B ∩ B) = ∅ ∪ B = B(A ∪ B) ∩ ˜B = (A ∩ ˜B) ∪ (B ∩ ˜B) = A ∪ ∅ = Am ∗ (A ∪ B) = m ∗ ((A ∪ B) ∩ B) + m ∗ ((A ∪ B) ∩ ˜B) = m ∗ (B) + m ∗ (A)25

(c) ∅ and R are clearly measurable.(d) Since the definition of measurable set is symmetric in E and Ẽ, we get that E is measurable ifand only if Ẽ is measurable.✄Definition 2.11 The Lebesgue measure of a measurable set E is the outer measure of E and isdenoted by m(E).✄Remark: The notation m(E) implies that the set E is measurable. If we do not know if a certainset is measurable, we use the m ∗ notation (for outer measure).✄The next proposition gives us a family of measurable sets.Proposition 2.12 Let E ⊆ R with m ∗ (E) = 0. Then E is measurable.Proof :Let A ⊆ R. Then A ∩ E ⊆ E and A ∩ Ẽ ⊆ A. Hencem ∗ (A ∩ E) + m ∗ (A ∩ Ẽ) ≤ m∗ (E) + m ∗ (A) = m ∗ (A)by monotonicity. So E is measurable.✷Before establishing countable additivity, we show that the collection of all measurable sets forms aσ-algebra.Lemma 2.13 The union of finitely many measurable sets is measurable.Proof : Using induction, it is enough to prove this result for two measurable sets. So let E 1 and E 2be measurable sets. Let A ⊆ R. Note thatandA ∩ ˜ E 1 ∪ E 2 = A ∩ Ẽ1 ∩ Ẽ2A ∩ (E 1 ∪ E 2 ) = (A ∩ E 1 ) ∪ (A ∩ E 2 ) = (A ∩ E 1 ) ∪ (A ∩ Ẽ1 ∩ E 2 )Hence using countable subadditivity, the measurability of E 2 and the measurability of E 1 , we getm ∗ (A ∩ (E 1 ∪ E 2 )) + m ∗ (A ∩ ˜ E 1 ∪ E 2 ) = m ∗ ((A ∩ E 1 ) ∪ (A ∩ Ẽ1 ∩ E 2 )) + m ∗ (A ∩ Ẽ1 ∩ Ẽ2)≤ m ∗ (A ∩ E 1 ) + m ∗ (A ∩ Ẽ1 ∩ E 2 ) + m ∗ (A ∩ Ẽ1 ∩ Ẽ2)= m ∗ (A ∩ E 1 ) + m ∗ (A ∩ Ẽ1)= m ∗ (A)So E 1 ∪ E 2 is measurable.✷Lemma 2.14 Let {E 1 , . . . , E n } be a finite disjoint collection of measurable sets. Thenfor all A ⊆ R.m ∗ (A ∩( n∪·i=1E i))=n∑m ∗ (A ∩ E i )i=126

Proof : Let A ⊆ R. We use induction on n. The result is obvious for n = 1. So assume that n ≥ 2.Since E i ∩ E n = ∅ for 1 ≤ i ≤ n − 1, we have that(A ∩ (∪· ni=1E i )) ∩ E n = A ∩ ((∪· ni=1E i ) ∩ E n ) = A ∩ (∪ n i=1(E i ∩ E n )) = A ∩ E nand(A ∩ (∪· ni=1E i )) ∩ Ẽn = A ∩((∪· ni=1E)i ) ∩ Ẽn = A ∩ (∪ n i=1(E i ∩ Ẽn)) = A ∩ ( ∪· n−1i=1 E )iUsing the measurability of E n and induction, we findm ∗ (A ∩ (∪· ni=1E i )) = m ∗ ((A ∩ (∪· ni=1E i )) ∩ E n ) + m ∗ ((A ∩ (∪· ni=1E i )) ∩ Ẽn)= m ∗ (A ∩ E n ) + m ∗ (A ∩ ( ∪· n−1i=1 E i))= m ∗ (A ∩ E n ) + ∑ n−1i=1 m∗ (A ∩ E i )= ∑ ni=1 m∗ (A ∩ E i ) ✷Proposition 2.15 The union of countably many measurable sets is measurable.Proof : Let {E n } n≥1 be a countable collection of measurable sets. Note that we already proved thatthe collection of all measurable sets is an algebra (this follows from Lemma 2.13 and the fact that Eis measurable if and only if Ẽ is measurable). Since we want to show that ∪∞ n=1E n is measurable, itfollows from Lemma 2.9 that we may assume that {E n } n≥1 is a disjoint collection.Put E = ∪· ∞i=1E i . For n ≥ 1, put F n = ∪· ni=1E i . Note that F n is measurable for all n ≥ 1 by Lemma2.13.Let A ⊆ R. Let n ≥ 1. Since F n ⊆ E, we get that Ẽ ⊆ ˜F n and soby monotonicity. By Lemma 2.14, we havem ∗ (A ∩ Ẽ) ≤ m∗ (A ∩ ˜F n )m ∗ (A ∩ F n ) = m ∗ (A ∩ (∪· ni=1E i )) =n∑m ∗ (A ∩ E i )i=1Since F n is measurable, we getn∑m ∗ (A ∩ Ẽ) + m ∗ (A ∩ E i ) ≤ m ∗ (A ∩ ˜F n ) + m ∗ (A ∩ F n ) = m ∗ (A)i=1Son∑m ∗ (A ∩ Ẽ) + m ∗ (A ∩ E i ) ≤ m ∗ (A) for all n ≥ 1Taking the limit as n → ∞, we find∞m ∗ (A ∩ Ẽ) + ∑m ∗ (A ∩ E i ) ≤ m ∗ (A)i=1Butm ∗ (A ∩ E) = m ∗ (A ∩ (∪· ∞i=1E∞∑i )) = m ∗ (∪ ∞ i=1(A ∩ E i )) ≤ m ∗ (A ∩ E i )i=1i=127

y countable subadditivity. Hence∞m ∗ (A ∩ Ẽ) + m∗ (A ∩ E) ≤ m ∗ (A ∩ Ẽ) + ∑m ∗ (A ∩ E i ) ≤ m ∗ (A)i=1So E = ∪ ∞ i=1E i is measurable.✷Theorem 2.16 The collection of all measurable sets forms a σ-algebra.Proof : This follows from Proposition 2.15 and the fact that E is measurable if and only if Ẽ ismeasurable.✷Now we can prove that countable additivity holds for disjoint measurable sets.Proposition 2.17 (Countable Additivity) Let {E n } n≥1 be a countable disjoint collection of measurablesets. Then( ∞)∞∑m∪· E n = m(E n )n=1n=1Proof :By countable subadditivity, we havem( ∞∪·i=1E i)≤∞∑m(E i )i=1Let n ≥ 1. By Lemma 2.14 (with ‘A = R’) and monotonicity, we get( ∞) ( n)m∪· E i ≥ m∪· E i =i=1i=1n∑m(E i )i=1So(n∑∞)m(E i ) ≤ m∪· E i for all n ≥ 1i=1i=1Taking the limit as n → ∞, we find(∞∑∞)m(E i ) ≤ m∪· E ii=1i=1Hence m( ∞∪·i=1E i)=∞∑m(E i ).i=1✷The following formula can be quite useful.Proposition 2.18 (Excision) Let A, B be sets with B ⊆ A. Suppose that B is measurable andm(B) ≠ +∞. Thenm ∗ (A \ B) = m ∗ (A) − m(B)28

Proof :Since B is measurable, we have thatm ∗ (A) = m ∗ (A ∩ B) + m ∗ (A ∩ ˜B) = m(B) + m ∗ (A \ B)Since m(B) ≠ +∞, we can subtract m(B) from both sides.✷Our next goal is to show that a large family of sets (which includes the open sets and the closed sets)are measurable.Definition 2.19 Let A ⊆ R.collection of all open sets.Then A is a Borel set if A is in the σ-algebra generated by the✄Remark: Clearly, every open set if a Borel set. Since σ-algebra are closed under complements, itfollows from Proposition 1.28 that every closed set is also a Borel set.✄Lemma 2.20 The interval (a, +∞) is measurable for all a ∈ R.Proof : Let a ∈ R. Let A ⊆ R. Let {I n } n≥1 be a ccoi of A. For n ≥ 1, put In 1 = I n ∩ (a, +∞) andIn 2 = I n ∩ (a, ˜ +∞) = I n ∩ (−∞, a]. We easily get thatandA ∩A ∩ (a, +∞) ⊆ (∪ ∞ n=1I n ) ∩ (a, +∞) = ∪ ∞ n=1(I n ∩ (a, +∞)) = ∪ ∞ n=1I 1 n˜ (a, +∞) = A ∩ (−∞, a] ⊆ (∪ ∞ n=1I n ) ∩ (−∞, a] = ∪ ∞ n=1(I n ∩ (−∞, a]) = ∪ ∞ n=1I 2 nIt follows from countable subadditivity thatm ∗ (A ∩ (a, +∞)) + m(A ∗ ∩ (a, ˜)+∞) ≤∞∑m ∗ (In) 1 +n=1∞∑m ∗ (In) 2 =n=1∞∑(m ∗ (In) 1 + m ∗ (In))2But I 1 n and I 2 n are intervals and l(I 1 n) + l(I 2 n) = l(I n ) for all n ≥ 1. Indeed, let n ≥ 1. We considerfive cases for I n :n=1I n I 1 n = I n ∩ (a, +∞) I 2 n = I n ∩ (−∞, a](−∞, +∞) (a, +∞) (−∞, a](−∞, d){ ∅ if d ≤ a(a, d) if d > a{ (−∞, d] if d ≤ a(−∞, a] if d > a(c, +∞)(c, d){ (a, +∞) if c < a(c, ∞) if c ≥ a⎧⎨ ∅ if d ≤ a(a, d) if c < a < d⎩(c, d) if a ≤ c{ (c, a] if c < a∅ if c ≥ a⎧⎨ (c, d) if d ≤ a(c, a] if c < a < d⎩∅ if a ≤ c∅ ∅ ∅29

Hence we getm ∗ (A ∩ (a, +∞)) + m(A ∗ ∩ (a, ˜) ∞∑+∞) ≤ (m ∗ (In) 1 + m ∗ (In)) 2 =SoThusn=1m ∗ (A ∩ (a, +∞)) + m(A ∗ ∩ (a, ˜)+∞) ≤Hence (a, +∞) is measurable.Theorem 2.21 Every Borel set is measurable.∞∑l(I n )n=1∞∑(l(In) 1 + l(In)) 2 =n=1m ∗ (A ∩ (a, +∞)) + m(A ∗ ∩ (a, ˜)+∞) ≤ m ∗ (A)for every ccoi {I n } n≥1 of A∞∑l(I n )Proof : Let M be the collection of all measurable sets. Note that M is a σ-algebra.First, we prove that every open interval is measurable. Note that ∅, I = (−∞, +∞) and I = (a, +∞)for a ∈ R are measurable ( (by Lemma 2.20). So (−∞, a] = (a, ˜ +∞) ∈ M for all a ∈ R. Hence(−∞, b) = ∩ ∞ n=1 −∞, b −1n]∈ M for all b ∈ R. Thus (a, b) = (−∞, b) ∩ (a, +∞) ∈ M for alla, b ∈ R with a < b.Next, we show that every open set is measurable. Let O be an open set. Then O is a countableunion of open intervals by Theorem 1.24. Since every open interval is measurable, we get that O ismeasurable.Finally, we show that every Borel set is measurable. Let B be the collection of all Borel sets. ThenB is the smallest σ-algebra containing all the open sets. Since M is a σ-algebra containing everyopen set, we have that B ⊆ M. So every Borel set is measurable.✷n=1✷Remark: Not every measurable set is a Borel set.✄We finish this section with a couple more properties of measurable sets.Proposition 2.22 Let E ⊆ R be a measurable set. Then E + x is measurable for all x ∈ R.Proof :We use the following properties:• A ∩ (B + x) = ((A − x) ∩ B) + x for all A, B ⊆ R and all x ∈ R• Ã + x = Ã + x for all A ⊆ R and all x ∈ RLet x ∈ R. Let A ⊆ R. Thenm ∗ (A ∩ Ẽ + x) = m ∗ (A ∩ (Ẽ + x)) = m∗ (((A − x) ∩ Ẽ) + x)Using Proposition 2.5 (namely the outer measure is translation invariant), the measurability of Eand again Proposition 2.5, we getm ∗ (A ∩ (E + x)) + m ∗ (A ∩ Ẽ + x) = m ∗ (((A − x) ∩ E) + x) + m ∗ (((A − x) ∩ Ẽ) + x)= m ∗ ((A − x) ∩ E) + m ∗ ((A − x) ∩ Ẽ)= m ∗ (A − x)= m ∗ (A)So E + x is measurable.✷30

Proposition 2.23 (Continuity of Measure) Let {E n } n≥1 be a sequence of measurable sets. Thenthe following holds:( ∞)∪(a) If E 1 ⊆ E 2 ⊆ E 3 ⊆ · · · then m E n = lim m(E n ).n→∞n=1(b) If E 1 ⊇ E 2 ⊇ E 3 ⊇ · · · and m(E 1 ) ≠ +∞ then m( ∞∩n=1E n)= limn→∞m(E n ).Proof : (a) Note that {m(E n )} n≥1 is an increasing sequence of extended real numbers (by monotonicity)and hence converges to an extended real number.Put E = ∪ ∞ n=1E n .Suppose first that m(E k ) = +∞ for some k ≥ 1. Since E k ⊆ E, we have that +∞ = m(E k ) ≤ m(E)by monotonicity. Hence lim m(E n ) = +∞ = m(E).n→∞Suppose next that m(E k ) ≠ +∞ for all k ≥ 1.Put E 0 = ∅. Then ∪ k≥1 E k = ∪· ∞k=1 (E k \ E k−1 ). Indeed, let x ∈ ∪ ∞ k=1 E k. Then x ∈ E k for somek ≥ 1. Let n be minimal with x ∈ E n . Then n ≥ 1, x ∈ E n and x /∈ E n−1 . So x ∈ E n \ E n−1 ⊆∪ ∞ k=1 (E k \ E k−1 ). Hence ∪ ∞ k=1 E k ⊆ ∪ ∞ k=1 (E k \ E k−1 ). Clearly, ∪ ∞ k=1 (E k \ E k−1 ) ⊆ ∪ ∞ k=1 E k and so∪ ∞ k=1 E k = ∪ ∞ k=1 (E k \ E k−1 ). Finally, suppose (E k \ E k−1 ) ∩ (E l \ E l−1 ) ≠ ∅ for some 1 ≤ k < l, sayx ∈ (E k \ E k−1 ) ∩ (E l \ E l−1 ). Then x /∈ E l−1 , a contradiction since x ∈ E k ⊆ E k+1 ⊆ · · · ⊆ E l−1 . So∪ ∞ k=1 (E k \ E k−1 ) is a disjoint union.Using countable additivity and excision (note that m(E k ) ≠ +∞ for all k ≥ 1), we get)m(E) = m( ∞∪·=k=1+∞∑k=1(E k \ E k−1 )m(E k \ E k−1 )( n∑= lim (m(E k ) − m(E k−1 ))n→∞k=1= limn→∞(m(E n ) − m(E 0 ))= limn→∞m(E n )(b) Note that {m(E n )} n≥1 is a decreasing sequence of positive real numbers (by monotonicity andthe fact that m(E 1 ) ≠ +∞) and hence converges to a real number.Put E = ∩ ∞ k=1 E k. Note that ∪· ∞k=1 (E k\E k+1 ) = E 1 \E. Indeed, let x ∈ ∪· ∞k=1 (E k\E k+1 ). Then x ∈ E n \E n+1 for some n ≥ 1. Hence x ∈ E n ⊆ E 1 and x /∈ E n+1 ⊇ ∩ ∞ k=1 E k = E. So ∪ ∞ k=1 (E k\E k+1 ) ⊆ E 1 \E.Let x ∈ E 1 \ E. Then x ∈ E 1 and x /∈ E = ∩ ∞ k=1 E k. Hence x /∈ E k for some k ≥ 1. Let n be thesmallest index with x /∈ E n . Then n ≥ 2 and so x ∈ E n−1 . Thus x ∈ E n−1 \ E n ⊆ ∪ ∞ k=1 (E k \ E k+1 ).Hence E 1 \ E ⊆ ∪ ∞ k=1 (E k \ E k+1 ). It follows that ∪ ∞ k=1 (E k \ E k+1 ) = E 1 \ E. Similarly as in part (a),we prove that {E k \ E k+1 : k ≥ 1} is a disjoint collection.)31

Using excision, countable additivity and again excision (note that m(E k ) ≠ +∞ for all k ≥ 1 andm(E) ≠ +∞), we getm(E 1 ) − m(E) = m(E 1 \ E)= m( ∞∪·=k=1(E k \ E k+1 )∞∑m(E k \ E k−1 )k=1)( n∑= lim (m(E k ) − m(E k+1 ))n→∞k=1= lim (m(E 1 ) − m(E n+1 ))n→∞= m(E 1 ) − lim m(E n )n→∞)Since m(E 1 ) < +∞, we get that m(E) = limn→∞m(E n ).✷Remarks(a) Part (a) of this theorem remains true if we remove the condition ‘measurable’(b) Part (b) of this theorem is false if we remove the condition ‘m(E 1 ) < +∞’ or if we remove thecondition ‘measurable’.✄2.4 Non-Measurable SetsIn this section, we construct a family of non-measurable sets, using the Axiom of Choice.Definition 2.24 Let A be a non-empty set of real numbers.(a) We define the rational equivalent relation (notation: ∼) on A by x ∼ y if x − y ∈ Q. One easilyshows that ∼ is an equivalence relation on A.• A choice set for A is a set containing exactly one element of each equivalence class of ∼ on A.Note that choice sets exist by the Axiom of Choice.✄We can now construct non-measurable sets.Theorem 2.25 Let A be a bounded set of real numbers with m ∗ (A) > 0. Then any choice set for Ais non-measurable.Proof : Let C be a choice set for A. Since A is bounded, we have that A ⊆ [−N, N] for some N > 0.Let {q n } n≥1 be an enumeration of [−2N, 2N] ∩ Q. Put C n = C + q n for all n ≥ 1.We show thatA ⊆∞∪·n=1Let x ∈ A. It follows from the definition of a choice set that x ∼ c for some c ∈ C. So x − c ∈ Q.Since x, c ∈ A ⊆ [−N, N], we get that x − c ∈ [−2N, 2N]. So x − c = q n for some n ≥ 1. ThusC n32

x = c + q n ∈ C n . Hence A ⊆ ∪ ∞ n=1C n . Suppose C k ∩ C n ≠ ∅ for some k, n ≥ 1, say x ∈ C k ∩ C n .Then x = a + q k = b + q n for some a, b ∈ C. So a − b = q n − q k ∈ Q. Hence a ∼ b. Since C is a choiceset, we have that a = b. So q k = q n and k = n. Thus {C n } n≥1 is a disjoint collection.Suppose that C is measurable. Then for all n ≥ 1, we have that C n is measurable by Proposition2.22 and m(C n ) = m(C) by Proposition 2.5. Using monotonicity and countable additivity, we getm ∗ (A) ≤ m( ∞∪·n=1C n)=Since m ∗ (A) > 0, we must have that∞∑m(C n ) =n=1∞∑m(C) =n=1{ 0 if m(C) = 0+∞ if m(C) ≠ 0m( ∞∪·n=1C n)= +∞Let x ∈ ∪· ∞n=1C n . Then x ∈ C n for some n ≥ 1. Hence x = c + q n for some c ∈ C. So |x| =|c + q n | ≤ |c| + |q n | ≤ N + 2N = 3N. Thus ∪· ∞n=1C n ⊆ [−3N, 3N]. By monotonicity, we get thatm(∪· ∞n=1C n ) ≤ m([−3N, 3N]) = 6N, a contradiction since m(∪· ∞n=1C n ) = +∞.Hence C is not measurable.The existence of non-measurable sets allows us to prove that countable additivity does not hold forall sets.Proposition 2.26 There exist sets A and B such that A ∩ B = ∅ and m ∗ (A ∪· B) < m ∗ (A) + m ∗ (B).✷Proof :Let C be a non-measurable set. It follows from the definition of measurable set thatm ∗ (D ∩ C) + m ∗ (D ∩ ˜C) ≠ m ∗ (D)for some set DSince D = (D∩C)∪(D∩ ˜C), we get that m ∗ (D) ≤ m ∗ (D∩C)+m ∗ (D∩ ˜C) by countable subadditivity.Hencem ∗ (D ∩ C) + m ∗ (D ∩ ˜C) > m ∗ (D)Put A = D ∩ C and B = D ∩ ˜C. Then A ∩ B = ∅ and A ∪ B = D. So m ∗ (A) + m ∗ (B) > m ∗ (A ∪· B).✷2.5 Littlewood’s First PrincipleIn this section, we prove some equivalent definitions of a measurable set and prove Littlewood’s FirstPrinciple: every measurable set is nearly a finite union of open intervals.Note that the next proposition could be interpreted as an expansion of Littlewood’s First Principle:every measurable set is nearly an open set (or a closed set, a G δ -set or an F σ -set).Definition 2.27 Let A ⊆ R.(a) A is a G δ -set if A is the intersection of countably many open sets.(b) A is an F σ -set if A is the union of countably many closed sets.✄33

Remarks:(a) Since the family of Borel sets is a σ-algebra containing all the open sets, we get that G δ -setsand F σ -sets are Borel sets.(b) Every open set is a G δ -set but not every G δ -set is open.(c) It follows from De Morgan’s Laws that the complement of a G δ -set is an F σ -set and thecomplement of an F σ -set is a G δ -set.(d) We can define other types of Borel sets: a G δσ -set is the union of countably many G δ -sets, aF σδσ -set is the union of countably many F σδ -sets, etc. All these sets are Borel sets. One canshow that there exist Borel sets that are not of any of the types described here.✄Proposition 2.28 Let E be a set of real numbers. Then the following are equivalent:(i) E is measurable.(ii) For all ε > 0, there exists an open set O containing E with m ∗ (O \ E) < ε.(iii) There exists a G δ -set G containing E with m ∗ (G \ E) = 0.(iv) For all ε > 0, there exists a closed set F contained in E with m ∗ (E \ F) < ε.(v) There exists an F σ -set F contained in E with m ∗ (E \ F ) = 0.Proof : (i) =⇒ (ii): Let ε > 0. Suppose first that m(E) < +∞. Since the outer measure is an∞∑infimum, there exists a ccoi {I n } n≥1 of E with l(I n ) < m(E) + ε. Put O = ∪ ∞ n=1I n . Then O isn=1an open set containing E and( ∞)∑m(O \ E) = m(O) − m(E) = m(∪ ∞ n=1I n ) − m(E) ≤ l(I n ) − m(E) < m(E) + ε − m(E) = εby excision and countable subadditivity. Suppose next that m(E) = +∞. Then E is the disjointunion of countably many measurable sets of finite measure (indeed, E = ∪· k∈Z (E ∩ [k, k + 1))), sayE = ∪· ∞n=1E n where E n is measurable and m(E n ) < +∞ for all n ∈ N. By the previous case, wehave that for all n ∈ N, there exists an open set O n containing E n with m(O n \ E n ) < ε2 . Put nO = ∪ ∞ n=1O n . Then O is an open set containing ∪ ∞ n=1E n = E. Note that O \ E ⊆ ∪ ∞ n=1(O n \ E n ).Indeed, let x ∈ O \ E. Then x ∈ O = ∪ ∞ n=1O n and so x ∈ O k for some k ≥ 1. Also, x /∈ E = ∪ ∞ n=1E n .Hence x /∈ E n for all n ≥ 1. In particular, x /∈ E k and so x ∈ O k \ E k .It follows from monotonicity and countable subadditivity thatm(O \ E) ≤ m(∪ ∞ n=1(O n \ E n )) ≤n=1∞∑m(O n \ E n )

y monotonicity. Taking the limit as n → ∞, we get that 0 ≤ m ∗ (G \ E) ≤ 0.(iii) =⇒ (i): Let G be a G δ -set containing E with m ∗ (G \ E) = 0. Then G \ E is measurableby Proposition 2.12. Clearly G = (G \ E) ∪· E and so E = G \ (G \ E). Since G and G \ E aremeasurable, we get that E is measurable.(i) =⇒ (iv): Note that Ẽ is measurable. Let ε > 0. By (ii) applied to Ẽ, we get that there existsan open set O containing Ẽ with m(O \ Ẽ) < ε. Put F = Õ. Then F is a closed set contained inẼ = E. Since E \ F = E ∩ ˜F = E ∩ O = O ∩ Ẽ = O \ Ẽ, we get that m(E \ F) = m(O \ Ẽ) < ε.(iv) =⇒ (v): Similar to (ii) =⇒ (iii)(v) =⇒ (i): Similar to (iii) =⇒ (i).Proposition 2.29 (Littlewood I) Let E be a set with m ∗ (E) < +∞. Then E is measurable ifand only if for all ε > 0, there exists a finite disjoint collection of open intervals {I k } n k=1 such thatm ∗ (E \ O) + m ∗ (O \ E) < ε where O = ∪ n k=1 I k.Proof : Suppose first that E is measurable. Let ε > 0. By Proposition 2.28(ii), there exists an openset O ∗ containing E with m(O ∗ \E) < ε. By Proposition 1.24, 2 O∗ is the union of a countable disjointcollection of open intervals, say O ∗ = ∪· ∞k=1 I k. Since O ∗ = (O ∗ \ E) ∪· E, it follows from countableadditivity that m(O ∗ ) = m(O ∗ \ E) + m(E) < ε + m(E) < +∞. So by countable additivity, we get2✷∞∑l(I k ) =k=1(∞∑∞)m(I k ) = m∪· I k = m(O ∗ ) < +∞k=1k=1Hence there exists n ∈ N such that ∑ ∞k=n+1 l(I k) < ε. Put O = ∪· nk=1 I 2 k. Since O ⊆ ∪· ∞k=1 I k = O ∗ , wehave that O \E ⊆ O ∗ \E and so m(O \E) ≤ m(O ∗ \E) < ε by monotonicity. Also E \O ⊆ 2 O∗ \O =(∪· ∞k=1 I k)\(∪· nk=1 I k) = ∪· ∞k=n+1 I k and so m(E\O) ≤ m(∪· ∞k=n+1 I k) = ∑ ∞k=n+1 m(I k) = ∑ ∞k=n+1 l(I k) < ε 2by monotonicity and countable additivity. Hencem(O \ E) + m(E \ O) < ε 2 + ε 2 = εSuppose next that the given property holds. Let ε > 0. Then there exists an open set O 1 (namelythe union of a finite disjoint collection of open intervals) with m ∗ (E \ O 1 ) + m ∗ (O 1 \ E) < ε. Since2m ∗ (E \ O 1 ) < ε, there exists a ccoi {I 2 n} ∞ n=1 of E \ O 1 with ∑ ∞n=1 l(I n) < ε. Put O 2 2 = ∪ ∞ n=1I n andO = O 1 ∪ O 2 . Then O is an open set containing E andO \ E = (O 1 ∪ O 2 ) \ E = (O 1 \ E) ∪ (O 2 \ E) ⊆ (O 1 \ E) ∪ O 2Moreover, m ∗ (O 2 ) = m ∗ (∪ ∞ n=1I n ) ≤ ∑ ∞n=1 l(I n) < ε 2by countable subadditivity. We get thatm ∗ (O \ E) ≤ m ∗ (O 1 \ E) + m ∗ (O 2 ) < ε 2 + ε 2 = εby monotonicity and countable subadditivity. So E is measurable by Proposition 2.28.✷35

2.6 Cantor SetIn this section, we provide an example of an uncountable set of measure zero.Put C 0 = [0, 1]. We divide C 0 into three intervals of equal length and remove the interior of themiddle interval to obtain C 1 . So( 1C 1 = [0, 1] \3 , 2 [= 0,3)1 ] [ ] 2∪3 3 , 1We repeat this process: C 1 is the union of two closed disjoint intervals of length 1 . We divide each of3these two intervals into three intervals of equal length and remove the interior of the middle interval.So[C 2 = 0, 1 ] [ 2∪9 9 , 1 [ 2∪3]3 , 7 [ ] 8∪9]9 , 1In general, for n ≥ 0, C n is the union of 2 n disjoint closed intervals of length 1 and C3 n n+1 ⊆ C n .We define the Cantor set C as∞∩C =Proposition 2.30 The Cantor set C is an uncountable, closed set of measure zero.n=0Proof : First, we show that C is closed. For all n ≥ 0, we have that C n is closed since it is the finiteunion of closed intervals. Hence C = ∩ ∞ n=0C n is closed.Next, we show that m(C) = 0. Note that m(C n ) = 2n for all n ≥ 0 by countable additivity. Since3 nC ⊆ C n for all n ≥ 0, it follows from monotonicity thatC nm(C) ≤ m(C n ) = 2n3 n for all n ≥ 0Taking the limit as n → ∞, we find that m(C) = 0.Finally, we show that C is uncountable. Note that C ≠ ∅ since 0, 1 ∈ C. Suppose that C is countable.Let {c n } k n=1 be an enumeration of C (so 1 ≤ k ≤ +∞). Since C 1 is the union of two disjoint intervalsof length 1 3 , it must be that one of these intervals (say F 1) does not contain c 1 . Since F 1 is the unionof two disjoint intervals of length 1 9 , it must be that one of these intervals (say F 2) does not containc 2 . Continuing this way, we obtain a sequence {F n } k n=1 such that• F n is non-empty and closed for all n < k + 1• F n ⊆ C n for all n < k + 1• F n+1 ⊆ F n for all n < k• c n /∈ F n for all n < k + 1If k < +∞, then let F k+1 be one of the two closed intervals whose union is F k , let F K+2 be one ofthe two closed intervals whose union is F k+1 , etc.It follows from the Nested Set Theorem (Theorem 1.31) that ∩ ∞ n=1F n ≠ ∅. So let x ∈ ∩ ∞ n=1F n . Thenx ∈ F n for all n ≥ 1. Since F n ⊆ C n for all n ≥ 1, we get that x ∈ ∩ ∞ n=1F n ⊆ ∩ ∞ n=1C n = C. Hencex = c i for some 1 ≤ i < k + 1. But then x = c i /∈ F i , a contradiction since x ∈ F n for all n ≥ 1.So C is uncountable.✷36

Chapter 3Measurable Functions3.1 Measurable FunctionsIn this section, we generalize the concept of a continuous function and prove some general propertiesof these generalized functions.Proposition 3.1 Let D ⊆ R be measurable and f : D → R an extended real-valued function. Thenthe following are equivalent:(i) The set {x ∈ D : f(x) > α} is measurable for all α ∈ R.(ii) The set {x ∈ D : f(x) ≥ α} is measurable for all α ∈ R.(iii) The set {x ∈ D : f(x) < α} is measurable for all α ∈ R.(iv) The set {x ∈ D : f(x) ≤ α} is measurable for all α ∈ R.Proof :Recall that the collection of measurable sets is a σ-algebra.(i) =⇒ (ii) Let α ∈ R. Then{x ∈ D : f(x) ≥ α} =∞∩n=1{x ∈ D : f(x) > α − 1 }nSince {x ∈ D : f(x) > α − 1 } is measurable for all n ∈ N, we get that {x ∈ D : f(x) ≥ α} isnmeasurable.(ii) =⇒ (iii) Let α ∈ R. Then{x ∈ D : f(x) < α} = D \ {x ∈ D : f(x) ≥ α}Since D and {x ∈ D : f(x) ≥ α} are measurable, we get that {x ∈ D : f(x) < α} is measurable.(iii) =⇒ (iv) Let α ∈ R. Then{x ∈ D : f(x) ≤ α} =∞∩n=1{x ∈ D : f(x) < α + 1 }n37

Since {x ∈ D : f(x) < α + 1 } is measurable for all n ∈ N, we get that {x ∈ D : f(x) ≤ α} isnmeasurable.(iv) =⇒ (i) Let α ∈ R. Then{x ∈ D : f(x) > α} = D \ {x ∈ D : f(x) ≤ α}Since D and {x ∈ D : f(x) ≤ α} are measurable, we get that {x ∈ D : f(x) < α} is measurable.✷Remark: Note that {x ∈ D : f(x) > α} = f −1 ((α, +∞]).✄Definition 3.2 A function f : D → R is measurable if D is measurable and if f satisfies one (andhence all four) of the statements in Proposition 3.1.✄Proposition 3.3 Let f : D → R be a measurable function. Then the set {x ∈ D : f(x) = α} ismeasurable for all α ∈ R.Proof : Let α ∈ R.Suppose first that α ∈ R. Then{x ∈ D : f(x) = α} = {x ∈ D : f(x) ≥ α} ∩ {x ∈ D : f(x) ≤ α}Since {x ∈ D : f(x) ≥ α} and {x ∈ D : f(x) ≤ α} are measurable (because f is measurable), we getthat {x ∈ D : f(x) = α} is measurable.Suppose next that α /∈ R. We prove the case ‘α = +∞’ as the case ‘α = −∞’ is similar. Then{x ∈ D : f(x) = +∞} =∞∩{x ∈ D : f(x) > n}Since {x ∈ D : f(x) > n} is measurable for all n ∈ N (because f is measurable), we get that{x ∈ D : f(x) = +∞} is measurable. ✷n=1Remark: Note that this property does not imply f is measurable. There exists a measurable set Dand a function f : D → R such that {x ∈ D : f(x) = α} is measurable for all α ∈ R but f is notmeasurable.✄Proposition 3.4 Let I be an interval and f : I → R a function that is monotone on I. Then f ismeasurable.Proof : Exercise. ✷Proposition 3.5 Let D ⊆ R be measurable and f : D → R a measurable function. Then |f| ismeasurable.Proof : Exercise. ✷The next theorem gives a characterization of measurable functions similar to the characterization ofcontinuous functions (Theorem 1.33).38

Theorem 3.6 Let D ⊆ R be measurable and f : D → R a real-valued function defined on D. Thenf is measurable if and only if f −1 (O) is measurable for every open set O.Proof : Suppose first that f −1 (O) is measurable for every open set O. Let α ∈ R. Then (α, +∞) isan open set. Hence f −1 ((α, +∞)) = {x ∈ D : f(x) > α} is measurable. So f is measurable.Suppose next that f is measurable. We show that f −1 (I) is measurable for all open intervals I.Clearly f −1 (∅) = ∅ is measurable and f −1 ((−∞, +∞)) = D is measurable. For a ∈ R, we have thatf −1 ((a, +∞)) = {x ∈ D : f(x) > a} is measurable and f −1 ((−∞, a)) = {x ∈ D : f(x) < a} ismeasurable since f is measurable. Finally, if a, b ∈ R with a < b, thenf −1 ((a, b)) = {x ∈ D : a < f(x) < b} = {x ∈ D : f(x) > a} ∩ {x ∈ D : f(x) < b}is measurable since {x ∈ D : f(x) > a} and ∩{x ∈ D : f(x) < b} are measurable (because f ismeasurable).Now we can show that f −1 (O) is measurable for every open set O. Let O be an open set. ByTheorem 1.24, O is the union of countably many open intervals, say O = ∪ ∞ n=1I n . Hencef −1 (O) = f −1 (∪ ∞ n=1I n ) = ∪ ∞ n=1f −1 (I n )is measurable since f −1 (I n ) is measurable for all n ∈ N.✷This characterization allows us to prove that every continuous function on a measurable domain isa measurable function.Corollary 3.7 Let D ⊆ R be measurable and f : D → R continuous on D. Then f is measurable.Proof : Let O be an open set. Since f is continuous on D, it follows from Theorem 1.33 thatf −1 (O) = D ∩ O ∗ for some open set O ∗ . So f −1 (O) is measurable. Hence f is measurable byTheorem 3.6.✷Proposition 3.8 Let D ⊆ R be measurable and f : D → R a real-valued function defined on D.Then f is measurable if and only if f −1 (B) is measurable for every Borel set B.Proof : Exercise. ✷Proposition 3.9 Let D ⊆ R be measurable, g : D → R a measurable function, E ⊆ R a Borel setcontaining g(D) and f : E → R a function that is continuous on E. Then f ◦ g is measurable.Proof : Let O ⊆ R be open. Then (f ◦ g) −1 (O) = g −1 (f −1 (O)). Since f is continuous, it followsfrom Theorem 1.33 that f −1 (O) = E ∩ O ∗ for some open set O ∗ . So (f ◦ g) −1 (O) = g −1 (E ∩ O ∗ ).Since E ∩ O ∗ is a Borel set and g is measurable, it follows from Proposition 3.8 that g −1 (E ∩ O ∗ ) ismeasurable. So (f ◦ g) −1 (O) is measurable. Hence f ◦ g is measurable by Theorem 3.6. ✷The next proposition allows us to show that a function is measurable by looking at the restrictionto measurable subsets.39

Proposition 3.10 Let D ⊆ R be measurable and f : D → R an extended real-valued function.Suppose that D = ∪ ∞ n=1D n where D n is measurable for all n ∈ N. Then f is measurable (over D) ifand only if f| Dn is measurable (over D n ) for all n ∈ N.Proof :Suppose first that f is measurable over D. Let n ∈ N. Let α ∈ R. Then{x ∈ D n : f(x) > α} = D n ∩ {x ∈ D : f(x) > α}is measurable since f is measurable. So f| Dn is measurable over D n .Suppose next that f| Dn is measurable for all n ∈ N. Let α ∈ R. Since D = ∪ ∞ n=1D n , we get that{x ∈ D : f(x) > α} = ∪ ∞ n=1{x ∈ D n : f(x) > α}Note that for all n ∈ N, we have that {x ∈ D n : f(x) > α} is measurable since f| DnSo {x ∈ D : f(x) > α} is measurable. Hence f is measurable over D.is measurable.✷Definition 3.11 Let D ⊆ R be a set.(a) Let f k : D → R be a function for 1 ≤ k ≤ n for some n ∈ N. We define the functionsmin{f 1 , . . . , f n } and max{f 1 , . . . , f n } pointwise as follows:min{f 1 , . . . , f n } : D → R : x → min{f 1 (x), . . . , f n (x)}max{f 1 , . . . , f n } : D → R : x → max{f 1 (x), . . . , f n (x)}(b) Let f n : D → R be a function for all n ∈ N. We define the functions inf f n, sup f n , lim infn≥1 n≥1(also denoted by lim f n ) and lim supn≥1n≥1f n (also denoted by lim f n ) pointwise as follows:f ninf f n : D → R : x → inf f n(x)n≥1 n≥1sup f n : D → R : x → sup f n (x)n≥1n≥1lim infn≥1lim supn≥1f n : D → R : x → lim infn≥1f n : D → R : x → lim supn≥1f n (x)f n (x)✄Proposition 3.12 Let D ⊆ R be a measurable set and f n : D → R a measurable function forall n ∈ N. Then min{f 1 , . . . , f n }, max{f 1 , . . . , f n }, inf f n, sup f n , lim inf f n and lim sup f n aren≥1 n≥1 n≥1n≥1measurable.Proof : Let α ∈ R.Let n ∈ N and put g = min{f 1 , . . . , f n }. ThenSo min{f 1 , . . . , f n } is measurable.{x ∈ D : g(x) > α} =n∩{x ∈ D : f k (x) > α}k=140

Let n ∈ N and put g = max{f 1 , . . . , f n }. ThenSo max{f 1 , . . . , f n } is measurable.Put g = infn≥1 f n. ThenSo infn≥1 f n is measurable.Put g = sup f n . Thenn≥1So sup f n is measurable.n≥1{x ∈ D : g(x) > α} ={x ∈ D : g(x) ≥ α} ={x ∈ D : g(x) > α} =n∪{x ∈ D : f k (x) > α}k=1∞∩{x ∈ D : f n (x) ≥ α}n=1∞∪{x ∈ D : f n (x) > α}For n ≥ 1, put g n = inf f k. Then g n is measurable for all n ∈ N. Hence lim f n = sup inf f k = supk≥n n≥1 k≥n n≥1is measurable.Similarly, we get that lim f n is measurable.n=1Corollary 3.13 Let D ⊆ R be measurable and f n : D → R measurable for all n ∈ N. Suppose thatthe sequence {f n } n≥1 converges pointwise to the function f : D → R. Then f is measurable.g n✷Proof :Note that f = limn→∞f n = lim f n .✷Before we tackle the problem of sums or products of measurable functions, we introduce the conceptof ‘almost everywhere’. It will allow us to consider functions like f + g that are defined ‘almosteverywhere’.Definition 3.14 A property holds almost everywhere on a set E if there exists a subset E 0 of Ewith m(E 0 ) = 0 such that the property holds on E \ E 0 . We write that the property holds a.e. onE. ✄Proposition 3.15 Let f, g : D → R be functions such that f is measurable and g = f a.e. on D.Then g is measurable.Proof : Put D 0 = {x ∈ D : f(x) ≠ g(x)}. Then m ∗ (D 0 ) = 0 since g = f a.e. on D. If A ⊆ D 0 thenm ∗ (A) ≤ m ∗ (D 0 ) = 0 by monotonicity and so A is measurable by Proposition 2.12.Let α ∈ R. Then{x ∈ D : g(x) > α} = ({x ∈ D : f(x) > α} \ D 0 ) ∪ {x ∈ D 0 : g(x) > α}Note that {x ∈ D : f(x) > α} is measurable since f is measurable, and {x ∈ D 0 : g(x) > α} ismeasurable since it is a subset of D 0 . Hence {x ∈ D : g(x) > α} is measurable. So g is measurable.✷41

Proposition 3.16 Let D ⊆ R be measurable and f : D → R a function such that f is continuousa.e. on D. Then f is measurable.Proof : Exercise. ✷Proposition 3.17 Let f, g : D → R be real-valued measurable functions. Then the following holds:(a) λf + µg is measurable for all λ, µ ∈ R.(b) fg is measurable.(c) If g ≠ 0 on D then f gis measurable.Proof : (a) Let λ ∈ R. We show that λf is measurable. If λ = 0 then λf is continuous on D andhence measurable. If λ > 0 then for all α ∈ R, we have that{{x ∈ D : (λf)(x) > α} = x ∈ D : f(x) > α }λand so λf is measurable. If λ < 0 then for all α ∈ R, we have that{{x ∈ D : (λf)(x) > α} = x ∈ D : f(x) < α }λand so λf is measurable.Next, we show that f + g is measurable. Let α ∈ R. Then{x ∈ D : (f + g)(x) > α} = ∪ q∈Q({x ∈ D : f(x) > q} ∩ {x ∈ D : g(x) > α − q}) (∗)Indeed, if x ∈ D such that f(x) > q and g(x) > α−q for some q ∈ Q, then (f +g)(x) = f(x)+g(x) >q+α−q = α. Conversely, let x ∈ D with (f +g)(x) > α. Then f(x)+g(x) > α and so f(x) > α−g(x).By the density of the rational numbers, there exists q ∈ Q with f(x) > q > α − g(x). So f(x) > qand g(x) > α − q.Since f and g are measurable, it follows from (*) that f + g is measurable.Now we can prove (a). Let λ, µ ∈ R. Then λf and µg are measurable by the first part of the proof.Hence λf + µg is measurable by the second part of the proof.(b) First, we show that f 2 is measurable. Let α ∈ R. If α < 0 then {x ∈ D : f 2 (x) > α} = D ismeasurable. If α ≥ 0 then{x ∈ D : f 2 (x) > α} = {x ∈ D : f(x) < − √ α} ∪ {x ∈ D : f(x) > √ α}is measurable since f is measurable. So f 2 is measurable.To prove (b), note thatfg = 1 2 [(f + g)2 − f 2 − g 2 ]By (a), f + g is measurable. By what we just proved in (b), (f + g) 2 , f 2 and g 2 are measurable.Then fg is measurable by (a).42

(c) We leave this as an exercise.✷Suppose f and g are two extended-real valued functions defined on some measurable set set D. Thenf + g is not defined at the points where we get a sum of the form ∞ − ∞. However, if f and g arefinite a.e. on D, say D 0 = {x ∈ D : f(x) = ±∞ or g(x) = ±∞}, then f + g is defined on D \ D 0 . Weassume that f + g is somehow (and it does not matter how) defined on D 0 so that f + g is definedon D. Since a function defined on a set of measure zero is always measurable, we see that f + g ismeasurable on D if and only if f + g is measurable on D \ D 0 .Similarly, we have that fg is defined on D \D 0 . If fg is defined somehow on D then fg is measurableon D if and only if fg is measurable on D \ D 0 .If λ ∈ R then λf is defined on D unless λ = 0; if λ = 0 then we define λf = 0 on D.Proposition 3.18 Let f, g : D → R be extended real-valued measurable functions that are finite a.e.on D. Then the following holds:(a) λf + µg is measurable for all λ, µ ∈ R.(b) fg is measurable.(c) If g ≠ 0 on D then f gis measurable.Proof :Let D 0 = {x ∈ D : f(x) = ±∞ or g(x) = ±∞}. SinceD 0 = {x ∈ D : f(x) = ±∞} ∪ {x ∈ D : g(x) = ±∞}and f and g are finite a.e. on D, we have that m(D 0 ) = 0. Note that f| D\D0 and g| D\D0 aremeasurable real-valued functions.(a) Let λ, µ ∈ R. By Proposition 3.17, λf + µg is measurable on D \ D 0 . Since m(D 0 ) = 0, we havethat λf + µg is measurable on D 0 . Hence λf + µg is measurable on (D \ D 0 ) ∪ D 0 = D.(b) (c) Similar as above.3.2 The Cantor-Lebesgue FunctionIn this section, we introduce the Cantor-Lebesgue function and disprove several ‘conjectures’ aboutmeasurable functions and measurable sets.Recall the definition of the Cantor set C.For k ∈ N, put O k = [0, 1] \ C k . Then for k ∈ N, we have that O k ⊆ O k+1 and O k is the disjointunion of 1 + 2 + 2 2 + · · · + 2 k−1 = 2 k − 1 open intervals. Put O = ∪ ∞ k=1 O k. Then C = [0, 1] \ O. Weproved that O is dense in [0, 1].We define the Cantor-Lebesgue function φ : [0, 1] → R as follows:Let k ∈ N. Then on O k , φ is defined as the increasing function which is constant on the 2 k − 1 openintervals of O k and takes on the valuesWe extend φ to all of [0, 1] as follows:12 , 2 k 2 , 3 k 2 , . . . , 2k − 2, 2k − 1k 2 k 2 k✷43

• φ(0) = 0• φ(x) = sup{φ(t) : t ∈ O ∩ [0, x]} if 0 < x ≤ 1Note that φ is well-defined on O. Let k ∈ N. Let I 1 , I 2 , . . . , I 2 k −1 be the open intervals of O k in‘increasing’ order. Let J 1 , J 2 , . . . , J 2 k be the 2 k open intervals of O k+1 \O k in ‘increasing’ order. ThenJ 1 , I 1 , J 2 , I 2 , . . . , J 2 k −1, I 2 k −1, J 2 kare the 2 k+1 − 1 open intervals of O k+1 in ‘increasing’ order. Now let x ∈ I j for some 1 ≤ j ≤ 2 k − 1.Using the definition of φ on O k , we get that φ(x) = j . Using the definition of φ on O2 k k+1 , we getthat φ(x) =2j . So φ is well-defined on O.2 k+1Note that φ is non-decreasing on O: if x, y ∈ O with x ≤ y then x, y ∈ O k for some k ∈ N and henceφ(x) ≤ φ(y).If 0 < x ≤ 1 then O ∩ [0, x] ≠ ∅ since O is dense in [0, 1]. So the definition of φ(x) makes sense on[0, 1].The ‘old’ definition of φ on O coincides with the ‘new’ definition of φ on [0, 1]: if x ∈ O thenφ(x) = max{φ(t) : t ∈ O ∩ [0, x]} since φ is non-decreasing on O.Finally, since 0 < φ(t) < 1 for all t ∈ O, we get that 0 < φ(x) ≤ 1 for all x ∈ (0, 1].Proposition 3.19 The Cantor-Lebesgue function φ is a non-decreasing, continuous function thatmaps [0, 1] onto [0, 1]. Moreover, φ is differentiable on O and φ ′ = 0 on O.Proof : Let x, y ∈ [0, 1] with x ≤ y. If x = 0 then φ(x) = 0 ≤ φ(y). So we may assume that x > 0.Then O ∩ [0, x] ⊆ O ∩ [0, y]. Henceφ(x) = sup{φ(t) : t ∈ O ∩ [0, x]} ≤ sup{φ(t) : t ∈ O ∩ [0, y]} = φ(y)So φ is non-decreasing on [0, 1].Let x 0 ∈ [0, 1]. Suppose first that x 0 ∈ O. Then x 0 ∈ O k for some k ∈ N. Hence φ is constant on anopen interval containing x 0 . Clearly, φ is differentiable at x 0 with φ ′ (x 0 ) = 0 and φ is continuous atx 0 . Suppose next that x 0 /∈ O. Assume first that 0 < x 0 < 1. Since O is dense in [0, 1], there existk 0 ∈ N and a, b ∈ O k0 with 0 < a < x < b < 1. Let k ≥ k 0 . Then x lies between two ‘consecutive’intervals in O k . Hence there exist a k , b k ∈ O k with a k < x 0 < b k and φ(b k ) = φ(a k ) + 1 . Since φ2 kis non-decreasing, this implies that |φ(x) − φ(x 0 )| ≤ 1 if a2 k k < x < b k . So φ is continuous at x 0 .Assume finally that x 0 = 0 (the case x 0 = 1 is similar). For k ∈ N, let a k be an element in the ‘first’interval of O k . Then 0 ≤ x 0 < a k and φ(a k ) = 1 for all k ∈ N. Since φ is non-decreasing, this2 kimplies that φ is continuous at x 0 = 0.We show that φ(1) = 1. We know that φ(1) ≤ 1. Let k ∈ N and b k an element in the ‘last’ intervalof O k . Since φ is non-decreasing, we get that 2k −1= φ(b2 k k ) ≤ φ(1) ≤ 1. Considering the limit ask → +∞, we get that φ(1) = 1.Since φ(0) = 0 and φ(1) = 1 and φ is continuous and non-decreasing on [0, 1], we get that φ([0, 1]) =[0, 1] by the Intermediate Value Theorem. ✷Proposition 3.20 The function ψ : [0, 1] → R : x → φ(x) + x has the following properties:(a) ψ is strictly increasing and continuous on [0, 1].(b) ψ([0, 1]) = [0, 2].44

(c) ψ(C) is measurable and m(ψ(C)) = 1.(d) There exists a subset E of C such that E is measurable and ψ(E) is non-measurable.Proof :(a) Obvious.(b) Since ψ(0) = φ(0) + 0 = 0 + 0 = 0 and ψ(1) = φ(1) + 1 = 1 + 1 = 2 and ψ is strictly increasingon [0, 1], we have that ψ([0, 1]) ⊆ [0, 2]. Since ψ(0) = 0 and ψ(1) = 2 and ψ is continuous, it followsfrom the Intermediate Value Theorem that [0, 2] ⊆ ψ([0, 1]).(c) (d) We show that ψ(O ′ ) and ψ(F ′ ) are measurable for any open set O ′ and any closed setF ′ . Since ψ is a strictly increasing continuous function defined on an interval, it has a continuousinverse. Put g = ψ −1 : [0, 2] → [0, 1]. Let O ′ be an open set. Since g is continuous, we have thatψ(O ′ ) = g −1 (O ′ ) = [0, 2] ∩ O ∗ for some open set O ∗ by Theorem 1.33. So ψ(O ′ ) is measurable.Similarly, if F ′ is a closed set, then ψ(F ′ ) = [0, 2] ∩ F ∗ for some closed set F ∗ and so ψ(F ′ ) ismeasurable.In particular, ψ(O) and ψ(C) are measurable. Since [0, 1] = O ∪· C and ψ is one-to-one, we get that[0, 2] = ψ([0, 1]) = ψ(O) ∪· ψ(C)Since ψ(O) and ψ(C) are measurable, it follows from countable additivity thatNote that2 = m(ψ(O)) + m(ψ(C)) (∗)m(O) = m([0, 1] \ C) = m([0, 1]) − m(C) = 1 − 0 = 1by excision. Put O = ∪· ∞k=1 I k where {I k } k≥1 are the disjoint open intervals that make up O. Fixk ∈ N. Note that I k is one of the intervals of O k0 for some k 0 . Hence φ is constant on I k . Let c k bethis constant value: φ(x) = c k for all x ∈ I k . Then ψ(x) = φ(x) + x = c k + x for all x ∈ I k . Henceψ(I k ) = I k + c k . Since the outer measure is translation-invariant, we get thatm(ψ(I k )) = m(I k + c k ) = m(I k )Using that O = ∪· ∞k=1 I k, that ψ is one-to-one and countable additivity, we find that( ( ∞)) ( ∞)(∞∑∞∑∞m(ψ(O)) = m ψ∪· I k = m∪· ψ(I k ) = m(ψ(I k )) = m(I k ) = m∪·k=1k=1k=1It follows from (*) that m(ψ(C)) = 1, which proves (c).k=1k=1I k)= m(O) = 1Since ψ(C) is bounded and m(ψ(C)) > 0, it follows from Theorem 2.25 that ψ(C) contains a nonmeasurablesubset D. Put E = ψ −1 (D). Then E ⊆ C and so m ∗ (E) ≤ m(C) = 0 by monotonicity.Hence m ∗ (E) = 0 and E is measurable. But ψ(E) = D is non-measurable, which proves (d). ✷We finish this section by proving that not every measurable set is a Borel set.Proposition 3.21 There exists a measurable set that is not a Borel set.Proof : Recall that ψ −1 : [0, 2] → [0, 1] is continuous and hence measurable. By Proposition 3.20(d),there exists a measurable subset E of C and a non-measurable subset D of [0, 2] with ψ(E) = D. IfE is a Borel set, then D = ψ(E) = (ψ −1 ) −1 (E) is measurable by Proposition 3.8, a contradiction.Hence E is not a Borel set.✷45

3.3 The Simple Approximation TheoremIn this section, we prove that every measurable function is the pointwise limit of a sequence of simplefunctions.Definition 3.22(a) Let D ⊆ R. The characteristic function of D (notation : χ D ) is the function{ 1 if x ∈ Dχ D : R → R : x →0 if x /∈ D(b) Let D ⊆ R be measurable and φ : D → R a function. Then φ is a simple function if φ ismeasurable and assumes only a finite number of function values.✄Remark: Let D 1 , . . . , D n be measurable subsets of R and a 1 , . . . , a n ∈ R. Then it is quite easy ton∑check that φ = a i χ Di is a simple function (on any measurable domain). ✄i=1Definition 3.23 Let D ⊆ R and f : E → R be a function. We define the positive part of f(notation: f + ) and the negative part of f (notation: f − ) asf + = max{f, 0} and f − = max{−f, 0}✄Remarks:(a) Note that f + and f − are nonnegative measurable functions if f is measurable.(b) Note that f = f + − f − and |f| = f + + f − .✄Lemma 3.24 Let f : D → R be a measurable function and m, M ∈ R with m ≤ f ≤ M on D.Then for all ε > 0, there exist simple functions φ ε and ψ ε defined on D such thatm ≤ φ ε ≤ f ≤ ψ ε ≤ M and 0 ≤ ψ ε − φ ε < ε on DProof : Let ε > 0. Let m = y 0 < y 1 < · · · < y n−1 < y n = M be a partition of [m, M] withy k − y k−1 < ε for 1 ≤ k ≤ n. Put I k = [y k−1 , y k ) for 1 ≤ k ≤ n − 1 and I n = [y n−1 , y n ]. PutD k = f −1 (I k ) for 1 ≤ k ≤ n. Note that for 1 ≤ k ≤ n − 1, we have thatandD k = f −1 (I k ) = {x ∈ D : y k−1 ≤ f(x) < y k } = {x ∈ D : f(x) ≥ y k−1 } ∩ {x ∈ D : f(x) < y k }D n = f −1 (I n ) = {x ∈ D : y n−1 ≤ f(x) ≤ y n } = {x ∈ D : f(x) ≥ y n−1 } ∩ {x ∈ D : f(x) ≤ y n }Since f is measurable, we have that D k is measurable for 1 ≤ k ≤ n.46

Note that D = ∪· nk=1 D k since f(D) ⊆ [m, M] = ∪· nk=1 I k. We putφ ε =n∑y k−1 χ Dk and ψ ε =n∑y k χ Dkon Dk=1k=1Then φ ε and ψ ε are simple functions. Let x ∈ D. Then x ∈ D j for a unique 1 ≤ j ≤ n. SinceD = ∪· nk=1 D k, we get that φ ε (x) = y j−1 and ψ ε (x) = y j . Note that m ≤ y j−1 , y j ≤ M andy j−1 ≤ f(x) ≤ y j since x ∈ D j . Also, 0 ≤ y j − y j−1 < ε. Hencem ≤ φ ε (x) ≤ f(x) ≤ ψ ε (x) ≤ M and 0 ≤ ψ ε (x) − φ ε (x) < ε for all x ∈ D ✷Theorem 3.25 (Simple Approximation Theorem) Let f : D → R be a measurable function.Then there exists a sequence of simple functions {f n } n≥1 defined on D such that {f n } n≥1 convergespointwise to f on D and |f n | ≤ |f| on D for all n ∈ N.If f ≥ m on D for some m ∈ R, we may choose f n ≥ m on D for all n ∈ N (instead of |f n | ≤ |f| onD for all n ∈ N) and {f n } n≥1 non-decreasing.Proof : First, we prove this theorem assuming f ≥ m on D for some m ∈ R.Let n ∈ N. Put D n = {x ∈ D : f(x) ≤ n}. Then D n is measurable since f is measurable. Supposefirst that D n = ∅. Put g n = max{m, n} on D. Then g n is a simple function and m ≤ g n ≤ f onD. Suppose next that D n ≠ ∅ (note this implies that n ≥ m). Then f| Dn is a bounded measurablefunction. By Lemma 3.24, there exist simple functions a n and b n defined on D n withm ≤ a n ≤ f ≤ b n ≤ n and 0 ≤ b n − a n < 1 non D nHence0 ≤ f − a n ≤ b n − a n < 1 non D nWe define g n on D by extending a n to D:{g n : D → R : x →a n (x)n = max{m, n}if x ∈ D nif x /∈ D nThen g n is still a simple function and m ≤ g n ≤ f on D.For n ∈ N, put f n = max{g 1 , . . . , g n }. Then f n is a simple function and m ≤ f n ≤ f n+1 ≤ f on Dfor all n ∈ N. We show that the sequence {f n } n≥1 converges to f on D. Let x 0 ∈ D.Suppose first that f(x 0 ) = +∞. Then g n (x 0 ) = max{m, n} for all n ∈ N. So f n (x 0 ) = max{m, n}for all n ∈ N. Hence {f n (x 0 )} n≥1 converges to f(x 0 ).Suppose next that f(x 0 ) is finite. Let N ∈ N with f(x 0 ) < N. Then x 0 ∈ D n for all n ≥ N. So0 ≤ f(x 0 ) − g n (x 0 ) < 1 nfor all n ≥ NSince g n (x 0 ) ≤ f n (x 0 ) ≤ f(x 0 ), we get that0 ≤ f(x 0 ) − f n (x 0 ) < 1 nfor all n ≥ NAgain, we see that {f n (x 0 )} n≥1 converges to f(x 0 ).47

Next, we prove the general case of the theorem.Applying the first part of our proof to f + and f − (with m = 0), we get that there exist non-decreasingsequences of nonnegative simple functions {a n } n≥1 and {b n } n≥1 such that {a n } n≥1 converges to f +on D and {b n } n≥1 converges to f − on D. Put f n = a n − b n for all n ∈ N. Then f n is still a simplefunction for all n ∈ N and {f n } n≥1 converges to f + − f − = f on D. Moreover, for all n ∈ N, we havethat|f n | = |a n − b n | ≤ |a n | + |b n | = a n + b n ≤ f + + f − = |f| on D ✷3.4 Littlewood’s Second and Third PrincipleIn this section, we prove Littlewood’s Second Principle:Every measurable function is nearly continuous.and Littlewood’s Third Principle:Every pointwise convergent sequence of measurable functions is nearly uniformly convergent.3.4.1 Littlewood’s Third PrincipleIn this subsection, we prove one version of Littlewood’s Third Principle: Egoroff’s Theorem.Theorem 3.26 (Egoroff, Littlewood III) Let D ⊆ R be a measurable set with m(D) < +∞,f n : D → R a measurable function for all n ∈ N and f : D → R a real-valued function such that{f n } n≥1 converges to f on D. Then for all ε > 0, there exists a measurable subset A of D such thatm(A) < ε and {f n } n≥1 converges uniformly to f on D \ A.Proof : Note that f is measurable (since it is the pointwise limit of a sequence of measurablefunctions). Hence |f − f k | is defined and measurable on D since f is real-valued.First, we prove the following claim:For all ε 1 , ε 2 > 0, there exist N ∈ N and a measurable subset A of D such that m(A) < ε 1and |f − f n | < ε 2 on D \ A for all n ≥ N.Indeed, let ε 1 , ε 2 > 0. Note that {x ∈ D : |f(x) − f k (x)| < ε 2 } is measurable for all k ∈ N. HenceD n = {x ∈ D : |f(x) − f k (x)| < ε 2 for all k ≥ n} =∞∩{x ∈ D : |f(x) − f k (x)| < ε 2 }is measurable for all n ∈ N. Clearly, {D n } n≥1 is an increasing sequence and ∪ ∞ n=1D n = D (indeed,let x 0 ∈ D; since {f n (x 0 )} → f(x 0 ) ∈ R, there exists n ∈ N with |f k (x 0 ) − f(x 0 )| < ε 2 forall k ≥ n). It follows from the Continuity of the Measure that {m(D n )} n≥1 → m(D). Sincem(D) < +∞, there exists N ∈ N with |m(D n ) − m(D)| < ε 1 for all n ≥ N. In particular,0 ≤ m(D) − m(D N ) < ε 1 . Put A = D \ D N . By excision, m(A) = m(D) − m(D N ) < ε 1 . Moreover,since D N = {x ∈ D : |f(x) − f n (x)| < ε 2 for all n ≥ N}, we have that |f − f n | < ε 2 on D N = D \ Afor all n ≥ N, which proves the claim.k=n48

Next, we prove the theorem. Let ε > 0.Let n ∈ N. Using the claim with ε 1 = ε2 and ε n 2 = 1 n , there exist an N n ∈ N and a measurablesubset A n of D such that m(A n ) 0. ByEgoroff’s Theorem, there exists a measurable subset B of D such that m(B) < ε and {f 2n} n≥1converges uniformly to f on D \ B. By Proposition 2.28(i)(iv), there exists a closed subset F ofD \ B with m((D \ B) \ F ) < ε . Put A = D \ F . Then D \ A = F is closed,2m(A) = m(B ∪· ((D \ B) \ F ) = m(B) + m((D \ B) \ F ) < ε 2 + ε 2 = εand {f n } n≥1 converges uniformly to f on D \ A since D \ A = F ⊆ D \ B.✄3.4.2 Littlewood’s Second PrincipleIn this subsection, we prove one version of Littlewood’s Second Principle: Lusin’s Theorem.Lemma 3.27 Let F 1 , F 2 , . . . , F n be pairwise disjoint closed sets and g i : F i → R continuous on F ifor 1 ≤ i ≤ n. Put F = ∪· ni=1F i andThen g is continuous on F .g : F → R : x → g i (x) if x ∈ F iProof : Note that g is well-defined since the F i ’s are pairwise disjoint. Let x 0 ∈ F . Then x 0 ∈ F i fora unique 1 ≤ i ≤ n, say x 0 ∈ F 1 . Let ε > 0. Since g 1 is continuous on F 1 , we have:∃δ 1 > 0 : ∀x ∈ F 1 : |x − x 0 | < δ 1=⇒ |g 1 (x) − g 1 (x 0 )| < εBecause the F i ’s are pairwise disjoint, we have that x 0 /∈ ∪· ni=2F i . Hence x 0 ∈ ˜∪· ni=2 F i = ∩ n i=2 ˜F i ,which is open since the F i ’s are closed. So (x 0 − δ 2 , x 0 + δ 2 ) ⊆ ˜∪· ni=2 F i for some δ 2 > 0. Hence(x 0 − δ 2 , x 0 + δ 2 ) ∩ (∪· ni=2F i ) = ∅. Put δ = min{δ 1 , δ 2 }. Let x ∈ F with |x − x 0 | < δ. Since δ ≤ δ 2 ,we have that x ∈ (x 0 − δ 2 , x 0 + δ 2 ) and so x /∈ (∪· ni=2F i ). Hence x ∈ F 1 . So |g(x) − g(x 0 )| =|g 1 (x) − g 1 (x 0 )| < ε. Thus g is continuous on F . ✷Lemma 3.28 Let F ⊆ R be closed and f : F → R continuous. Then there exists a continuousfunction g : R → R with f = g on F .Proof : Exercise. ✷49

Lemma 3.29 Let D ⊆ R be measurable and f : D → R a simple function. Then for all ε > 0, thereexist a measurable subset A of D and a continuous function g : R → R such that m(A) < ε, D \ Ais closed and f = g on D \ A.Proof : Let a 1 , . . . , a n be the distinct function values of f on D. For 1 ≤ i ≤ n, put D i = {x ∈ D :f(x) = a i }. Then D i is measurable for all 1 ≤ i ≤ n and D = ∪· ni=1D i .Let ε > 0. For 1 ≤ i ≤ n, it follows from Proposition 2.28(i)(iv) that there exists a closed subsetF i ⊆ D i with m(D i \ F i ) < ε n . Put F = ∪· ni=1F i and A = ∪· ni=1(D i \ F i ). Then A ⊆ D, F = D \ A andm(A) ≤n∑m(D i \ F i ) 0, there exist a measurable subset A of D and a continuous function g : R → Rsuch that m(A) < ε and f = g on D \ A. Moreover, we can choose A such that D \ A is closed.Proof : We prove this theorem assuming m(D) < ∞ and leave the general case as an exercise.Let ε > 0. By the Simple Approximation Theorem, there exists a sequence of simple functions{f n } n≥1 , defined on D, that converges to f on D. For n ∈ N, it follows from Lemma 3.29 that thereexist a measurable subset A n ⊆ D and a continuous function g n : R → R such that m(A n )

Chapter 4The Lebesgue Integral4.1 Lebesgue Integral of Nonnegative Simple FunctionsDefinition 4.1 Let φ : E → R be a simple function.(a) Let a 1 , . . . , a n be the different function values of φ. For 1 ≤ i ≤ n, put A i = {x ∈ E | φ(x) = a i }.Note that A i is measurable for all 1 ≤ i ≤ n. Thenφ =n∑a i χ Aii=1This is called the canonical representation or canonical form of φ.(b) Suppose φ is nonnegative. The Lebesgue integral of φ over E (notation : ∫ φ) isE∫Eφ =n∑a i m(A i )i=1where φ = ∑ ni=1 a iχ Ai is the canonical representation of φ and a i m(A i ) = 0 if a i = 0. ✄Remark : It follows immediately that we don’t have to write down 0 as a function value of φ in thecanonical form of φ to calculate ∫ φ : if a E n = 0 then φ = ∑ n−1i=1 a iχ Ai and ∫ φ = ∑ n−1E i=1 a im(A i ). ✄Lemma 4.2 Let E be a measurable set, C 1 , C 2 , . . . , C m disjoint measurable subsets of E and c 1 , . . . , c mnonnegative real numbers. Put φ = ∑ mj=1 c jχ Cj . Then ∫ φ = ∑ mE i=1 c jm(C j ) where c j m(C j ) = 0 ifc j = 0.Proof : We may assume that C j ≠ ∅ for j = 1, 2, . . . , m and that E = ∪· mj=1C j . Let φ = ∑ ni=1 a iχ Aibe the canonical representation of φ. For i = 1, 2, . . . , n, put Ω i = {1 ≤ j ≤ m | c j = a i }.Since the a i ’s are distinct, we have that the Ω i ’s are disjoint. Pick 1 ≤ j ≤ m. Pick x ∈ C j .Then there exists a unique 1 ≤ i ≤ n with x ∈ A i . Hence a i = φ(x) = c j and j ∈ Ω i . So{Ω i | i = 1, 2, . . . , n} forms a partition of {1, 2, . . . , m}.Pick 1 ≤ i ≤ n. For all j ∈ Ω i and all x ∈ C j , we have that φ(x) = c j = a i and so x ∈ A i . Hence∪· j∈Ωi C j ⊆ A i . Pick x ∈ A i . Then there exists a unique 1 ≤ j ≤ m with x ∈ C j . So a i = φ(x) = c j .Hence j ∈ Ω i and A i ⊆ ∪· j∈Ωi C j .51

So A i = ∪· j∈Ωi C j . Hence m(A i ) = ∑ j∈Ω im(C j ). We now get thatm∑n∑ ∑n∑ ∑n∑ ∑c j m(C j ) = c j m(C j ) = a i m(C j ) = a i m(C j ) =j=1i=1 j∈Ω i i=1 j ∈Ω i i=1 j∈Ω in∑∫a i m(A i ) = φi=1E✷Proposition 4.3 Let φ, ψ : E → R be nonnegative simple functions. Then the following holds :(a) If φ = ∑ ni=1 a iχ Ai where A i ⊆ E is measurable and a i ≥ 0 for i = 1, 2, . . . , n, then ∫ E φ =∑ ni=1 a i m(A i ) where a i m(A i ) = 0 if a i = 0.(b) ∫ (aφ + cψ) = a ∫ φ + c ∫ ψ for all a, c ≥ 0.E E E(c) If E = E 1 ∪· E 2 where E 1 , E 2 are measurable, then ∫ E φ = ∫ E 1φ + ∫ E 2φ.(d) If m(E) = 0 then ∫ E φ = 0.(e) If φ = ψ a.e. on E then ∫ E φ = ∫ E ψ.(f) If φ ≤ ψ a.e. on E then ∫ E φ ≤ ∫ E ψ.Proof :that(b) Let φ = ∑ mi=1 a iχ Ai and ψ = ∑ nj=1 c jχ Cj be the canonical forms. By definition, we get∫ m∑∫ n∑φ = a i m(A i ) and ψ = c j m(C j )Ei=1Put B ij = A i ∩ C j for i = 1, 2, . . . , m and j = 1, 2, . . . , n. Then {B ij | 1 ≤ i ≤ m, 1 ≤ j ≤ n} forms apartition of E. Moreover, A i = ∪· nj=1B ij and so m(A i ) = ∑ nj=1 m(B ij) for i = 1, 2, . . . , m. Similarly,C j = ∪· mi=1B ij and so m(C j ) = ∑ mi=1 m(B ij) for j = 1, 2, . . . , n. Note thatφ =m∑i=1n∑a i χ Bij , ψ =j=1By Lemma 4.2, we get thatm∑i=1Ej=1n∑c j χ Bij and so aφ + cψ =j=1∫E (aφ + cψ) = ∑ m ∑ ni=1 j=1 (aa i + cc j )m(B ij )= a ∑ mi=1 a im∑i=1n∑(aa i + cc j )χ Bijj=1∑ nj=1 m(B ij) + c ∑ nj=1 c j∑ mi=1 m(B ij)= a ∑ mi=1 a im(A i ) + c ∑ nj=1 c jm(C j )= a ∫ E φ + c ∫ E ψ(a) Note that χ Ai is a nonnegative simple function on E and ∫ χ E A i= m(A i ) for i = 1, 2, . . . , n.Hence by (b), we get∫ ( n∑)n∑∫n∑a i χ Ai = a i χ Ai = a i m(A i )E i=1i=1 E i=1(c) Let φ = ∑ n ji=1 a ijχ Aij be the canonical form of φ on E j for j = 1, 2. Since the A ij ’s are pairwisedisjoint, we getn 2∑ ∑ jφ = a ij χ Aijj=1i=152

By (a), we get that∫Eφ =n 2∑ ∑ ja ij m(A ij ) =j=1i=1(d) Let φ = ∑ ni=1 a iχ Ai be the canonical form of φ on E. Since A i ⊆ E and m(E) = 0, we get thatm(A i ) = 0 for i = 1, 2, . . . , n. Hence by definition, we get that∫Eφ =2∑∫j=1n∑a i m(A i ) = 0i=1(e) Put D = {x ∈ E | φ(x) ≠ ψ(x)}. Then m(D) = 0. Note that φ| E\D= ψ| E\D. Using (c) and (d),we find ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫φ = φ + φ = φ = ψ = ψ + ψ = ψEE\DDE\D(f) Similarly as in (e), we may assume that φ ≤ ψ on E. Note that ψ − φ is a nonnegative simplefunction. Clearly, ∫ (ψ − φ) ≥ 0. Hence by (b), we haveE∫ ∫∫ ∫∫ψ = [φ + (ψ − φ)] = φ + (ψ − φ) ≥ φ✷EE4.2 Lebesgue Integral of Nonegative Measurable FunctionsEE\DEE\DE jφDefinition 4.4 : Let f : E → R be a nonnegative measurable function.(1) We define the set Φ f as follows :Φ f = {φ | φ is a nonnegative simple function on E and φ ≤ f on E}DEE(2) We define the Lebesgue integral of f over E (notation : ∫ f) asE∫E f = sup ∫φ∈Φ fE φ✄Remark : This definition of ∫ f is consistent with our ‘old’ definition if f is a nonnegativeEsimple function. ∫ Indeed, suppose that f is a nonnegative simple function. Then f ∈ Φ f and sosup φ ≥ f. On the other hand, if φ ∈ Φ f , then φ ≤ f on E and soφ∈Φ f∫E∫ φ ≤ ∫ f by PropositionE EE∫∫4.3(f). Since this is true for all φ ∈ Φ f , we get sup φ ≤ f. So sup φ = f. ✄φ∈Φ f∫E E φ∈Φ f∫E EProposition 4.5 Let f, g : E → R be nonnegative measurable functions. Then the following holds(a) ∫ D f = ∫ E (f · χ D) for all measurable D ⊆ E.(b) ∫ (λf) = λ ∫ f for all λ ≥ 0.E E(c) If E = E 1 ∪· E 2 where E 1 , E 2 are measurable, then ∫ E f = ∫ E 1f + ∫ E 2f.(d) If m(E) = 0 then ∫ E f = 0. 53

(e) If f = g a.e. on E then ∫ E f = ∫ E g.(f) If f ≤ g a.e. on E then ∫ E f ≤ ∫ E g.Proof : (a) Let D ⊆ E be measurable.{ φ(x) if x ∈ DPick φ ∈ Φ f,D . Put ψ : E → R : x →0 if x ∈ E \ D . Then ψ ∈ Φ f·χ D ,E. If φ = ∑ ni=1 a iχ Aiis the canonical form of φ over D, then ψ = ∑ ni=1 a iχ Ai over E. So∫ ∫∫ ∫φ = ψ ≤ψ = (f · χ D )DEsupψ∈Φ f·χD ,ESince this is true for all φ ∈ Φ f,D , we get∫∫f = sup φ ≤D φ∈Φ f,D∫DEEE(f · χ D ) (%)Pick φ ∈ Φ f·χD ,E. Put ψ = φ| D. Then ψ ∈ Φ f,D . If ψ = ∑ ni=1 a iχ Ai is the canonical form of ψ overD, then φ = ∑ ni=1 a iχ Ai over E. So∫ ∫∫φ = ψ ≤ sup ψ = fE D ψ∈Φ f,D∫D DSince this is true for all φ ∈ Φ f·χD ,E, we get that∫∫(f · χ D ) = supφ∈Φ f.χD ,EEE∫φ ≤From (%) and (% %), it follows that ∫ D f = ∫ E (f · χ D).Df (%%)(b) We may assume that λ > 0. Pick φ ∈ Φ f . Then λφ ∈ Φ λf . By Proposition 4.3(b), we get that∫∫ ∫(λf) = sup ψ ≥ (λφ) = λ φE ψ∈Φ λf∫E EESince this is true for all φ ∈ Φ f , we get that∫( ∫ )(λf) ≥ sup λ φφ∈Φ fEE∫= λ sup φ = λ fφ∈Φ f∫EE(∗)Since (*) is true for all λ > 0 and all nonnegative measurable functions f, we can replace λ by 1 λand f by λf to get∫ ∫ ( ) 1f =E E λ (λf) ≥ 1 ∫∫ ∫(λf) and so λ f ≥ (λf) (∗∗)λ EE EFrom (*) and (**), it follows that ∫ E (λf) = λ ∫ E f.(c) Pick φ ∈ Φ f . Then φ| Ei∈ Φ f,Ei for i = 1, 2. Using Proposition 4.3(c), we get∫ ∫ ∫∫∫ ∫ ∫φ = φ| E1+ φ| E2≤ sup ψ + sup ψ = f +E E 1 E 2 ψ∈Φ f,E1 E 1 ψ∈Φ f,E2 E 2 E 1E 2f54

Since this is true for all φ ∈ Φ f , we get∫∫ ∫f = sup φ ≤ f + f (#)E φ∈Φ f∫E E 1 E 2{φ1 (x) if x ∈ EPick φ i ∈ Φ f,Ei for i = 1, 2. Define φ : E → R : x →1. Note that this isφ 2 (x) if x ∈ E 2well-defined since E = E 1 ∪· E 2 . Then φ ∈ Φ f and φ| Ei= φ i for i = 1, 2. Using Proposition 4.3(c),we get that ∫ ∫ ∫ ∫ ∫∫φ 1 + φ 2 = φ| E1+ φ| E2= φ ≤ sup ψ = fE 1 E 2 E 1 E 2 E ψ∈Φ f∫E ESince this is true for all φ 1 ∈ Φ f,E1 and all φ 2 ∈ Φ f,E2 , we get that∫ ∫∫∫ ∫f + f = sup φ 1 + sup φ 2 ≤E 1 E 2 φ 1 ∈Φ f,E1 E 1 φ 2 ∈Φ f,E2 E 2From (#) and (##), it follows that ∫ E f = ∫ E 1f + ∫ E 2f.Ef (##)(d) Pick φ ∈ Φ f . By Proposition 4.3(d), we have that ∫ φ = 0. Since this is true for all φ ∈ Φ E f, weget that∫f = sup φ = sup 0 = 0E φ∈Φ f∫E φ∈Φ f(e) Put D = {x ∈ E | f(x) ≠ g(x)}. Then m(D) = 0. Note that f = g on E \ D. Using (c) and (d),we get that ∫ ∫ ∫ ∫ ∫ ∫ ∫ ∫f = f + f = f = g = g + g = gEE\DDE\D(f) Similarly as in (e), we may assume that f ≤ g on E. Pick φ ∈ Φ f . Since f ≤ g on E, we getthat φ ∈ Φ g . Hence∫∫g = sup ψ ≥ φE ψ∈Φ g∫E EE\DSince this is true for all φ ∈ Φ f , we have that∫∫g ≥ sup φ =φ∈Φ f∫EERemark: At this point, we can not prove linearity for the Lebesgue integral of nonnegative measurablefunctions. We will need to develop quite a bit of machinery (Fatou’s Lemma and the MonotoneConvergence Theorem) before we can prove linearity.✄4.3 Fatou’s Lemma, Monotone Convergence TheoremLemma 4.6 (Fatou’s Lemma) Let f, f n : E → R be nonnegative measurable functions for alln ≥ 1 such that f = lim f n a.e. on E. Then∫ ∫f ≤ limEEf nEfE\DDE✷55

Proof : We may assume that f = lim f n on E.Pick φ ∈ Φ f and 0 < t < 1. We will show that t ∫ φ ≤ lim ∫ f E E n. So we may assume thatφ ≠ 0. Then we can write φ = ∑ mk=1 a kχ Ak where a 1 , a 2 , . . . , a m > 0 and A 1 , A 2 , . . . , A m are disjointmeasurable subsets of E.For n ≥ 1 and 1 ≤ k ≤ m, we defineB kn = {x ∈ A k | f l (x) > ta k for all l ≥ n} =+∞∩l=n{x ∈ A k | f l (x) > ta k }Hence {B kn } n≥1 is an increasing sequence of measurable subsets of A k for 1 ≤ k ≤ m.CLAIM : A k = ∪ +∞n=1B kn for 1 ≤ k ≤ mProof : Pick 1 ≤ k ≤ m. Clearly, ∪ +∞n=1B kn ⊆ A k . Pick x ∈ A k . Note thatlim f n (x) = f(x) ≥ φ(x) = a k > ta kSuppose first that f(x) = +∞. Then lim f n (x) = +∞ and so∀M > 0 : ∃n ∈ N : ∀l ≥ n : f l (x) > MUsing this with ‘M = ta k ’, we get that there exists n ∈ N such that f l (x) > ta k for alll ≥ n. So x ∈ B kn .Suppose next that f(x) < +∞. Then lim f n (x) < +∞ and so∀ε > 0 : ∃n ∈ N : ∀l ≥ n : f l (x) > lim f n (x) − εUsing this with ‘ε = lim f n (x) − ta k ’, we get that there exists n ∈ N such that f l (x) > ta kfor all l ≥ n. Again x ∈ B kn .Hence A k ⊆ ∪ +∞n=1B kn .By the Continuity of the Measure, we get that m(A k ) = limn→∞m(B kn ) for 1 ≤ k ≤ m.Pick n ≥ 1. Put ψ n = ∑ mk=1 ta kχ Bkn . Pick x ∈ E. If x ∈ B kn for some 1 ≤ k ≤ m, then x ∈ A k andf n (x) > ta k . So ψ n (x) = ta k < f n (x). If x /∈ B kn for all 1 ≤ k ≤ m, then ψ n (x) = 0 ≤ f n (x). Henceψ n ∈ Φ fn and so by Proposition 4.3(a), we get∫E∫ ∫f n = sup ψ ≥ ψ n =ψ∈φ fn∫E E E(∑ m)ta k χ Bkn =k=1m∑ta k m(B kn )Since this is true for all n ≥ 1, we can apply the limit inferior to both sides. Using the fact thatm(A k ) = lim m(B kn ) for 1 ≤ k ≤ m and Proposition 4.3(a), we getn→∞∫limE( m)∑f n ≥ lim ta k m(B kn )k=1= limn→+∞k=1(∑ m)ta k m(B kn ) = tk=1m∑k=1✄∫a k m(A k ) = t φESince this is true for all 0 < t < 1, we can take the limit as t goes to 1 − on both sides :∫( ∫ ) ∫lim f n ≥ lim t φ = φt→1 −EEE56

Since this is true for all φ ∈ Φ f , we finally get∫∫lim f n ≥ sup φ =E φ∈Φ f∫ERemark : Another way of viewing Fatou’s Lemma : ∫ E lim f n ≤ lim ∫ E f n.Ef✷✄Theorem 4.7 Let f, f n : E → R be nonnegative measurable functions for all n ≥ 1 such that{f n } n≥1 → f a.e. on E and f n ≤ f a.e. on E for all n ≥ 1. Then {∫ f }E n → ∫ f.n≥1 EProof : Since f n ≤ f a.e. on E, we get by Proposition 4.5(f) that∫ ∫f n ≤ f for all n ≥ 1E EApplying the limit superior to both sides, we find∫ ∫lim f n ≤ f (∗)E EBy Fatou’s Lemma, we get ∫ ∫f ≤ limf nEECombining (*) and (**), we find∫ ∫lim f n ≤E(∗∗)∫f ≤ limf nEEHence lim ∫ f E n = lim ∫ f E n = ∫ f and {∫ f }E E n → ∫ f.n≥1 E✷Corollary 4.8 (Monotone Convergence Theorem) Let f, f n : E → R be nonnegative measurable{∫functions for all n ≥ 1 such that f 1 ≤ f 2 ≤ f 3 ≤ · · · and {f n } n≥1 → f a.e. on E. Thenf }E n → ∫ f.n≥1 EProof : Since {f n } n≥1 is an increasing sequence of functions converging to f a.e. on E, we have thatf n ≤ f a.e. on E for all n ≥ 1. Hence the result follows from Theorem 4.7.✷∫∫Remark : Another way of viewing the last theorem and corollary : lim f n = lim f n . ✄En→+∞ n→+∞EProposition 4.9 Let f, g, f n : E → R be nonnegative measurable functions for all n ≥ 1. Then thefollowing holds :(a) ∫ (f + g) = ∫ f + ∫ gE E E(b) ∫ (∑ +∞E n=1 f ) ∑n = +∞∫n=1E f n(c) If E 1 , E 2 , . . . are disjoint measurable sets such that E = ∪· +∞n=1E n , then ∫ E f = ∑ +∞n=1∫E nf57

Proof : (a) By the Simple Approximation Theorem, there exists a nondecreasing sequence of nonnegativesimple functions {φ n } n≥1 (resp. {ψ n } n≥1 ) that converges to f (resp. g) on E. Then{φ n + ψ n } n≥1 is a nondecreasing sequence of nonnegative simple functions that converges to f + gon E. By the Monotone Convergence Theorem, we get{∫ } ∫ {∫ } ∫ {∫ } ∫φ n → f , ψ n → g and (φ n + ψ n ) → (f + g)EEEEEEn≥1n≥1By Proposition 4.3(b), we have that ∫ (φ E n + ψ n ) = ∫ φ E n + ∫ ψ E n for all n ≥ 1. Hence{∫ } {∫ ∫ } ∫ ∫(φ n + ψ n ) = φ n + ψ n → f + gEE EE ESo ∫ E (f + g) = ∫ E f + ∫ E g.n≥1(b) For k ≥ 1, put S k = ∑ kn=1 f n. Then {S k } k≥1 is a nondecreasing sequence of nonnegativemeasurable functions that converges to ∑ +∞n=1 f n on E. By the Monotone Convergence Theorem, weget that{∫ } ∫ ( +∞)∑S k → f nBy (a), we get that ∫ S E k = ∑ k∫n=1f E n for all k ≥ 1. Hence{∫ } { k∑ ∫ }S k = f n →n=1So ∫ E(∑ +∞n=1 f ) ∑n = +∞∫n=1f E n.E(c) Note that f = ∑ +∞n=1 fχ E n. By (b) and Proposition 4.5(a), we get∫ ∫ ( +∞)∑+∞∑∫+∞∑∫f = fχ En = fχ En =EEEk≥1n=1k≥1EEn=1En=1k≥1n≥1+∞∑n=1∫n=1f nE4.4 Lebesgue Integral of Measurable FunctionsDefinition 4.10 :(a)∫Let f : E → R be a nonnegative measurable function. Then f is Lebesgue integrable over E iff is finite.E(b) Let f : E → R be a measurable function. Then f is Lebesgue integrable over E if and only iff + and f − are Lebesgue integrable over E. In that case we define the Lebesgue integral of fover E (notation : ∫ f) asE∫ ∫ ∫f = f + − f −✄EERemark : This definition of ∫ f is consistent with our ‘old’ definition if f is a nonnegative measurablefunction because f + = f and f − = 0 if f is a nonnegative measurable function.E✄EE nfn≥1✷58

Lemma 4.11 Let f, g : E → R be nonnegative measurable functions such that g is integrable overE and f ≤ g a.e. on E. Then f is integrable over E.Proof : Since f is nonnegative, we have that f − = 0 and f + = f. So ∫ f − = ∫ 0 = 0. ByE EProposition 4.5(f), we have ∫ f + = ∫ f ≤ ∫ g < +∞. Hence f is integrable over E.✷E E ELemma 4.12 Let f : E → R be measurable.integrable over E.Then f is integrable over E if and only if |f| isProof : Suppose first that f is integrable over E. Then f + and f − are integrable over E. Since|f| = f + + f − , we get that ∫ |f| = ∫ (f + + f − ) = ∫ f + + ∫ f − < +∞ by Proposition 4.9(a). SoE E E E|f| is integrable over E.Suppose next that |f| is integrable over E. Since f + , f − ≤ |f| over E, we get that f + and f − areintegrable over E by Lemma 4.11. Hence f is integrable over E.✷Lemma 4.13 (Tchebychev’s Inequality) Let f : D → R be a nonnegative measurable function.For α > 0, put D α = {x ∈ D | f(x) ≥ α}. Then m(D α ) ≤ 1 ∫f for all α > 0.αProof : Exercise. ✷Lemma 4.14 Let f : E → R be integrable over E. Then f is finite a.e. on E.Proof : Put D = {x ∈ E | f(x) = ±∞}. By Lemma 4.12, |f| is integrable over E.∫Pick n ≥ 1.Put D n = {x ∈ E | |f(x)| > n}. By Tchebychev’s Inequality, we get that m(D n ) ≤ 1 |f|. Sincen ED ⊆ D n , we get thatm(D) ≤ m(D n ) ≤ 1 ∫|f|nSince this is true for all n ≥ 1, we can take the limit as n goes to +∞ on both sides. Since |f| isintegrable over E, we find∫1m(D) ≤ lim |f| = 0✷n→+∞ n ELemma 4.15 Let f : E → R be measurable and D ⊆ E with m(D) = 0. Then f is integrable overE if and only if f is integrable over E \ D.EDProof :By Proposition 4.5(c)(d), we get that∫ ∫ ∫ ∫f ± = f ± + f ± =f ±E E\D D E\DHence ∫ E f ± is finite if and only if ∫ E\D f ± is finite.✷Remark : Let f, g : E → R be integrable over E. By Lemma 4.14, f + g is well-defined a.e. on E.So by Lemma 4.15, we can investigate if f + g is integrable over E, even though f + g might not bedefined everywhere.✄Lemma 4.16 Let h 1 , h 2 : E → R be nonnegative and integrable over E. Put h = h 1 − h 2 . Then his integrable over E and ∫ E h = ∫ E h 1 − ∫ E h 2.59

Proof : We will show that h + ≤ h 1 and h − ≤ h 2 . Pick x ∈ E. If h(x) ≤ 0, then h + (x) = 0 ≤ h 1 (x)and h 2 (x) = h 1 (x) − h(x) ≥ −h(x) = h − (x). If h(x) ≥ 0, then h − (x) = 0 ≤ h 2 (x) and h 1 (x) =h(x) + h 2 (x) ≥ h(x) = h + (x).Hence h + ≤ h 1 and h − ≤ h 2 . By Lemma 4.11, h + and h − are integrable over E. So h is integrableover E. Since h + − h − = h = h 1 − h 2 , we get that h + + h 2 = h − + h 1 . By Proposition 4.9(a), we find∫ ∫ ∫∫∫ ∫h + + h 2 = (h + + h 2 ) = (h − + h 1 ) = h − +EEEEEh 1ESo ∫ E h = ∫ E h+ − ∫ E h− = ∫ E h 1 − ∫ E h 2.✷Proposition 4.17 Let f, g : E → R be integrable over E. Then the following holds(a) f is integrable over D and ∫ D f = ∫ E (f · χ D) for all measurable D ⊆ E.(b) λf + µg is integrable over E and ∫ (λf + µg) = λ ∫ f + µ ∫ g for all λ, µ ∈ R.E E E(c) If E = ∪· +∞n=1E n where E n is measurable for all n ≥ 1, then ∫ f = ∑ +∞∫E n=1 E nf.(d) If m(E) = 0 then ∫ E f = 0.(e) If f = g a.e. on E then ∫ E f = ∫ E g.(f) If f ≤ g a.e. on E then ∫ E f ≤ ∫ E g.(g) ∣ ∣ ∫ E f∣ ∣ ≤∫E |f|.Proof :(a) Note that (f · χ D ) ± = f ± · χ D . By Proposition 4.5(a), we get that∫ ∫∫f ± = (f ± · χ D ) = (f · χ D ) ±D EEBy Proposition 4.5(f), we have that ∫ (f ± · χE D ) ≤ ∫ f ± < +∞. So f ± is integrable over D andE(f · χ D ) ± is integrable over E. Hence f is integrable over D and f · χ D is integrable over E. Weeasily get that∫ ∫ ∫ ∫∫∫f = f + − f − = (f · χ D ) + − (f · χ D ) − = (f · χ D )DDDE(b) First, we prove that λf is integrable over E and ∫ (λf) = λ ∫ f for all λ ∈ R.E EAssume first that λ ≥ 0. Then (λf) ± = λf ± . By Proposition 4.5(b), we get that ∫ ∫E (λf)± =(λf ± ) = λ ∫ f ± < +∞. So (λf) ± is integrable over E. Hence λf is integrable over E. We nowE Eget ∫ ∫ ∫∫ ∫ (∫ ∫ ) ∫(λf) = (λf) + − (λf) − = λ f + − λ f − = λ f + − f − = λ fEEEEEE EEAssume∫next∫that λ < 0. Then (λf) ± = −λf ∓ . Similarly, we get that λf is integrable over E and(λf) = λ f. ENext, we prove that f + g is integrable over E and ∫ (f + g) = ∫ f + ∫ g. We have thatE E Ef + g = (f + − f − ) + (g + − g − ) = (f + + g + ) − (f − + g − )EE60

By Prosition 4.9(a), ∫ E (f ± + g ± ) = ∫ E f ± + ∫ E g± < +∞. So f ± + g ± is integrable over E. ByLemma 4.16, f + g is integrable over E and∫E∫(f + g) =E(∫=(∫=∫=E∫(f + + g + ) −EE∫f + +∫f + −∫f +EgEEE(f − + g − )) (∫g + −) (∫f − +EE∫f − +∫g + −(c) By (a), f is integrable over E n for all n ≥ 1. Using Proposition 4.9(c), we get∫ ∫ ∫ ( +∞) (∑+∞)∑+∞∑(∫)f = f + − f − = fE E E∫E + − fn∫E − = f + − fn E n∫E − =nn=1(d) Using Proposition 4.5(d), we get∫ ∫f =EEn=1∫f + −En=1f − = 0 − 0 = 0(e) Since f = g a.e. on E, we get that f ± = g ± a.e. on E. By Proposition 4.5(e), we find∫ ∫ ∫ ∫ ∫ ∫f = f + − f − = g + − g − = gEEE(f) Note that g − f ≥ 0 a.e. on E. By Proposition 4.5(f), we have that ∫ (g − f) ≥ ∫ 0 = 0. UsingE E(b), we find∫ ∫∫ ∫ ∫g = ((g − f) + f) = (g − f) + f ≥ fEE(g) Note that |f| = f + + f − . Hence by Proposition 4.9(a), we get∫∣∫∫ ∣ ∫ ∣ ∫ ∣ ∫ ∫ ∫∣∣∣ ∣∣∣ ∣ f∣ = ∣∣∣ ∣∣∣f + − f − ≤∣ f + +∣ f − = f + + f − =EEE4.5 Lebesgue Convergence TheoremEEEEEEEEEEEg − )g − )EE∫(f + + f − ) =Theorem 4.18 Let f n , f : E → R be measurable functions and g n , g : E → R integrable over Esuch that {f n } n≥1 → f a.e. on E, {g n } n≥1 → g a.e. on E, { ∫ E g n} n≥1 → ∫ E g and |f n| ≤ g n a.e. onE for all n ≥ 1. Then f n , f are integrable over E for all n ≥ 1 and { ∫ E f n} n≥1 → ∫ E f.Proof : We may assume that the inequalities and the convergence is everywhere on E instead of a.e.on E.Pick n ≥ 1. Since |f n | ≤ g n on E, we get that |f n | is integrable over E by Lemma 4.11. So f n isintegrable over E by Lemma 4.12.+∞∑n=1E∫|f|E nf✷61

Since |f n | ≤ g n for all n ≥ 1, we can apply the limit as n goes to +∞ on both sides. We get|f| =lim |f n| ≤ lim g n = gn→+∞ n→+∞Similarly as above, f is integrable over E.Since |f n | ≤ g n , we get −g n ≤ f n ≤ g n for all n ≥ 1. Hence g n − f n ≥ 0 and f n + g n ≥ 0 for alln ≥ 1. Applying Fatou’s Lemma to {g n − f n } n≥1 and using some properties of limit inferior andlimit superior and using Proposition 4.17, we find∫g − ∫ f = ∫ (g − f) = ∫ lim (g E E E E n − f n ) ≤ lim ∫ (g E n − f n )= lim (∫ g E n − ∫ f )E n≤ lim ∫ g E n + lim ( − ∫ f )E n= ∫ E g − lim ∫ E f nSo ∫ E g − ∫ E f ≤ ∫ E g − lim ∫ E f n. Hence lim ∫ E f n ≤ ∫ E f.Applying Fatou’s Lemma to {f n +g n } n≥1 and using some properties of limit inferior and limit superiorand using Proposition 4.17, we find∫f + ∫ g = ∫ (f + g) = ∫ lim (f E E E E n + g n ) ≤ lim ∫ (f E n + g n )= lim (∫ f E n + ∫ g )E n≤ lim ∫ E f n + lim ∫ E g n= lim ∫ E f n + ∫ E gSo ∫ f + ∫ g ≤ lim ∫ f E E E n + ∫ g. Hence ∫ f ≤ lim ∫ f E E E n.So we get that∫ ∫ ∫lim f n ≤ f ≤ limf nE EEHence lim ∫ E f n = lim ∫ E f n = ∫ E f. So {∫ E f n} n≥1 → ∫ E f.✷Corollary 4.19 (Lebesgue Convergence Theorem) Let f n , f : E → R be measurable functionsand g : E → R integrable over E such that {f n } n≥1 → f a.e. on E and |f n | ≤ g a.e. on E for alln ≥ 1. Then f n , f are integrable over E for all n ≥ 1 and { ∫ E f n} n≥1 → ∫ E f.Proof : This follows immediately from the previous theorem with ‘g n = g’ for all n ≥ 1. ✷∫∫Remark : Another way of viewing the last theorem and corollary : lim f n = lim f n . ✄n→+∞ n→+∞4.6 Riemann and Lebesgue IntegralsWe start by noting the following :Let φ : E → R be a simple function and m(E) < +∞. Then φ is integrable over E.Moreover, if φ = ∑ ni=1 a iχ Ei where a i ∈ R and E i is a measurable subset of E, then∫E φ = ∑ ni=1 a im(E i ).EE62

Definition 4.20 : Let E be a measurable set with m(E) < +∞ and f : E → R a bounded function.We define the sets L f and U f as follows :L f = {φ | φ is a simple function on E and φ ≤ f on E}U f = {ψ | ψ is a simple function on E and f ≤ ψ on E}✄Remark : Let E be a measurable set with m(E) < +∞ and f : E → R a bounded function. Pickφ ∈ L f and ψ ∈ U f . Then φ ≤ f ≤ ψ on E. By Proposition 4.17(f), we have that ∫ E φ ≤ ∫ E ψ.Since this is true for all φ ∈ L f and all ψ ∈ U f , we get that sup φ ≤ inf ψ. ✄φ∈L f∫Eψ∈U f∫ETheorem 4.21 Let E be a measurable set with m(E) < +∞ and f : E → R a bounded function.Then the following are equivalent :(a) f is measurable(b) f is Lebesgue integrable over E(c) sup φ = inf ψφ∈L f∫Eψ∈U f∫E∫In this case, f = sup φ = inf ψ.E φ∈L f∫Eψ∈U f∫EProof : (a) ⇒ (b) : Since f is bounded, there exists M > 0 such that |f| ≤ M on E. So −Mχ E ≤f ≤ Mχ E on E. Since −Mχ E and Mχ E are integrable over E, we get that f is integrable over E.(b) ⇒ (a) : By definition, a function that is Lebesgue integrable over E is measurable over E.(a) ⇒ (c) : Since f is bounded, there exists M > 0 such that |f| ≤ M on E.Pick n ≥ 1. Put{E nk = x ∈ E | k M }n ≤ f(x) < (k + 1)M for k = −n, −(n − 1), . . . , n − 1, nnNote that E nk is measurable for −n ≤ k ≤ n and E = ∪· nk=−n E nk. Putφ n =Then φ n ∈ L f and ψ n ∈ U f . So∫sup φ ≥ φ n =φ∈L f∫EM E nn∑k=−nn∑k=−nk M n χ E nkand ψ n =n∑(k + 1) M n χ E nkk=−n∫k m(E nk ) and inf ψ ≤ ψ n =ψ∈U f∫EM E nn∑(k + 1) m(E nk )k=−nHence we find0 ≤ inf ψ − sup φ ≤ψ∈U f∫E φ∈L f∫E M nn∑k=−nm(E kn ) = Mm(E)n63

Since this is true for all n ≥ 1, we can take the limit as n goes to +∞ on both sides. Using the factthat m(E) is finite, we getMm(E)0 ≤ inf ψ − sup φ ≤ limψ∈U f∫E φ∈L f∫En→+∞ nHence sup φ = inf ψ.φ∈L f∫Eψ∈U f∫EPick φ ∈ L f and ψ ∈ U f . Then φ ≤ f ≤ ψ on E. Since f is integrable over E, we find ∫ ∫φ ≤f ≤ ∫ ψ. Since this is true for all φ ∈ L EE E f and all ψ ∈ U f , we get∫sup φ ≤φ∈L f∫E∫Hence sup φ = inf ψ = f.φ∈L f∫Eψ∈U f∫E EEf ≤ inf ψψ∈U f∫E(c) ⇒ (a) Note that S := sup φ = inf ψ is finite since f is bounded and m(E) is finite.φ∈L f∫Eψ∈U f∫EUsing some properties of infimum and supremum, we get that for all n ≥ 1, there exist φ n ∈ L f andψ n ∈ U f such thatS − 12n∫E< φ n and S + 12n∫E> ψ nHence∫ ∫ψ n − φ n < 1 for all n ≥ 1 (∗)nEEPut φ = sup φ n and ψ = sup ψ n . By Theorem 3.20, we have that φ and ψ are measurable. Pickn≥1n≥1x ∈ E. Then φ n (x) ≤ f(x) ≤ ψ n (x) for all n ≥ 1. Hence= 0φ(x) = sup φ n (x) ≤ f(x) ≤ inf ψ n(x) = ψ(x)n≥1n≥1So φ ≤ f ≤ ψ on E.{Put ∆ = {x ∈ E | φ(x) < ψ(x)} and ∆ k = x ∈ E | φ(x) < ψ(x) − 1 }for all n ≥ 1. Note thatk∆ = ∪ +∞k=1 ∆ k.Pick k ≥ 1. Since φ n ≤ φ ≤ ψ ≤ ψ n on E, we get that 0 ≤ ψ − φ ≤ ψ n − φ n for all n ≥ 1. UsingTchebychef’s Inequality and (*), we find∫∫m(∆ k ) ≤ k (ψ − φ) ≤ k (ψ n − φ n ) < k for all n ≥ 1nTaking the limit as n goes to +∞ on both sides, we getSince this is true for all k ≥ 1, we haveEEk0 ≤ m(∆ k ) ≤ limn→+∞ n = 0m(∆) = m ( ∪ +∞k=1 ∆ ) ∑+∞ k ≤ m(∆ k ) = 0k=164

Hence φ = ψ a.e. on E. Since φ ≤ f ≤ ψ on E, we get that f = φ a.e. on E. Since φ is measurableover E, we finally get that f is measurable by Proposition 3.21.✷Definition 4.22 : A function φ : [a, b] → R is a step function if there exista = x 0 < x 1 < · · · < x n−1 < x n = band c 1 , . . . , c n ∈ R such that φ(x) = c i if x i−1 < x < x i .✄Remark : If φ is a step function over [a, b] then φ is a simple function over [a, b] and hence Lebesgueintegrable over [a, b]. So ∫ φ makes sense (and is a real number).✄[a,b]Definition 4.23 : Let f : [a, b] → R be a bounded function.(a) We putRL f = {φ | φ is a step function on [a, b] and φ ≤ f on [a, b]}RU f = {ψ | ψ is a step function on [a, b] and f ≤ ψ on [a, b]}(b) We say that f is Riemann integrable over [a, b] ifputsup φ = inf ψ. In this case, weφ∈RL f∫[a,b]ψ∈RU f∫[a,b]∫ baf(x) dx = sup φ = inf ψφ∈RL f∫[a,b]ψ∈RU f∫[a,b]✄Remark : Let φ : [a, b] → R be a step function. Then φ is Riemann and Lebesgue integrable over∫ b ∫[a, b] and φ(x) dx = φ. This is just a special case of the next theorem. ✄a[a,b]Theorem 4.24 Let f : [a, b] → R be a bounded function. If f is Riemann integrable over [a, b], then∫ b ∫f is Lebesgue integrable over [a, b] and f(x) dx = f.a[a,b]Proof : Note that RL f ⊆ L f and RU f ⊆ U f . So we always havesup φ ≤ sup φ ≤ inf ψ ≤ inf ψφ∈RL f∫[a,b] φ∈L f∫[a,b]ψ∈U f∫[a,b]ψ∈RU f∫[a,b]Since f is Riemann integrable over [a, b], we have equalities instead of inequalities :∫ baf(x) dx = sup φ = sup φ = inf ψ = inf ψφ∈RL f∫[a,b] φ∈L f∫[a,b]ψ∈U f∫[a,b]ψ∈RU f∫[a,b]By Theorem 4.21(b)(c), we get that f is Lebesgue integrable over [a, b] and∫f = sup φ = inf ψ[a,b] φ∈L f∫[a,b]ψ∈U f∫[a,b]∫ ∫ bHence f = f(x) dx.[a,b]a✷65

Definition 4.25 : Let f : [a, b] → R be a function.(a) For all x 0 ∈ [a, b] and all δ > 0, we definem δ (x 0 ) = inf{f(x) | x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]}M δ (x 0 ) = sup{f(x) | x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]}(b) We define the lower boundary of f (notation : m) and the upper boundary of f (notation :M) as follows :m(x 0 ) = limδ→0 + m δ(x 0 ) and M(x 0 ) = limδ→0 + M δ(x 0 ) for all x 0 ∈ [a, b] ✄Remark : Let x 0 ∈ [a, b]. If 0 < δ 1 ≤ δ 2 thenSo limδ→0 + m δ(x 0 ) and limδ→0 + M δ(x 0 ) exist andm δ2 (x 0 ) ≤ m δ1 (x 0 ) ≤ f(x 0 ) ≤ M δ1 (x 0 ) ≤ M δ2 (x 0 )m(x 0 ) = lim m δ(x 0 ) = sup m δ (x 0 ) ≤ f(x 0 ) ≤ inf M δ(x 0 ) = lim M δ(x 0 ) = M(x 0 )δ→0 + δ>0 δ→0 +δ>0✄Proposition 4.26 Let f : [a, b] → R be a function and x 0 ∈ [a, b]. Then f is continuous at x 0 ifand only if m(x 0 ) = M(x 0 ).Proof :Suppose first that f is continuous at x 0 . Pick ε > 0. Then we have∃ρ > 0 : ∀x ∈ [a, b] : |x − x 0 | < ρ ⇒ |f(x) − f(x 0 )| < εPick 0 < δ < ρ. Pick x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]. Then x ∈ [a, b] and |x − x 0 | < ρ. Hence|f(x) − f(x 0 )| < ε and sof(x 0 ) − ε < f(x) < f(x 0 ) + εSince this is true for all x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b], we getf(x 0 ) ≥ m δ (x 0 ) = inf{f(x) | x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]} ≥ f(x 0 ) − εf(x 0 ) ≤ M δ (x 0 ) = sup{f(x) | x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]} ≤ f(x 0 ) + εHence |m δ (x 0 ) − f(x 0 )| < ε and |M δ (x 0 ) − f(x 0 )| < ε. Som(x 0 ) = limδ→0 + m δ(x 0 ) = f(x 0 ) = limδ→0 + M δ(x 0 ) = M(x 0 )Suppose next that m(x 0 ) = M(x 0 ).Pick ε > 0. We have :Then limδ→0 + m δ(x 0 ) = m(x 0 ) = f(x 0 ) = M(x 0 ) = limδ→0 + M δ(x 0 ).∃ρ 1 > 0 : ∀δ > 0 : 0 < δ < ρ 1 ⇒ |m δ (x 0 ) − f(x 0 )| < ε∃ρ 2 > 0 : ∀δ > 0 : 0 < δ < ρ 2 ⇒ |M δ (x 0 ) − f(x 0 )| < ε66

Let 0 < δ < min{ρ 1 , ρ 2 }. Then0 ≤ f(x 0 ) − m δ (x 0 ) < ε and 0 ≤ M δ (x 0 ) − f(x 0 ) < εPick x ∈ [a, b] with |x − x 0 | < δ. Then x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]. Sof(x 0 ) − ε < m δ (x 0 ) = inf{f(y) | y ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]} ≤ f(x)f(x) ≤ sup{f(y) | y ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]} = M δ (x 0 ) < f(x 0 ) + εHence f(x 0 ) − ε < f(x) < f(x 0 ) + ε. So |f(x) − f(x 0 )| < ε and f is continuous at x = x 0 .✷Proposition 4.27 Let f : [a, b] → R be bounded, m its lower boundary and M its upper boundary.Then the following holds :(a) m and M are measurable.(b) Let φ ∈ RL f and x 0 ∈ [a, b] such that φ is continuous at x 0 . Then φ(x 0 ) ≤ m(x 0 ).(c) Let ψ ∈ RU f and x 0 ∈ [a, b] such that ψ is continuous at x 0 . Then M(x 0 ) ≤ ψ(x 0 ).∫∫(d) sup φ = m and inf ψ = M.φ∈RL f∫[a,b] [a,b]ψ∈RU f∫[a,b] [a,b]Proof :(a) Pick α ∈ R. Put D = {x ∈ [a, b] | m(x) > α}. We will prove∀x 0 ∈ D : ∃ρ > 0 : (x 0 − ρ, x 0 + ρ) ∩ [a, b] ⊆ DPick x 0 ∈ D. Let α < β < m(x 0 ). Then we have :∃ρ > 0 : ∀δ > 0 : 0 < δ ≤ ρ ⇒ |m δ (x 0 ) − m(x 0 )| < m(x 0 ) − βSo m δ (x 0 ) > β for all 0 < δ ≤ ρ. Pick x ∈ [a, b] with |x − x 0 | < ρ. Then x ∈ (x 0 − ρ, x 0 + ρ) ∩ [a, b]and sof(x) ≥ inf{f(y) | y ∈ (x 0 − ρ, x 0 + ρ) ∩ [a, b]} = m ρ (x 0 ) > βPick 0 < δ < ρ − |x − x 0 |. Pick y ∈ (x − δ, x + δ) ∩ [a, b]. Then |y − x 0 | < ρ and so f(y) > β. Hencem δ (x) = inf{f(y) | y ∈ (x − δ, x + δ) ∩ [a, b]} ≥ β. Since this is true for all 0 < δ < ρ − |x − x 0 |, wehave m(x) = limδ→0 + m δ(x) ≥ β > α. Hence x ∈ D.So we proved the following :∀x 0 ∈ D : ∃ρ > 0 : (x 0 − ρ, x 0 + ρ) ∩ [a, b] ⊆ DHence D = [a, b] ∩ O for some open set O. So D is measurable. Hence m is measurable.Similarly, M is measurable.(b) Since φ ∈ RL f is continuous at x 0 , we have :∃ρ > 0 : ∀x ∈ (x 0 − ρ, x 0 + ρ) ∩ [a, b] : φ(x 0 ) = φ(x) ≤ f(x)Pick 0 < δ < ρ. Then φ(x 0 ) = φ(x) ≤ f(x) for all x ∈ (x − δ, x + δ) ∩ [a, b] and som δ (x 0 ) = inf{f(x) | x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]} ≥ φ(x 0 )67

Since this is true for all 0 < δ < ρ, we get(c) Similar as in (b).m(x 0 ) = limδ→0 + m δ(x 0 ) ≥ φ(x 0 )(d) Let φ ∈ RL f . Then φ is continuous a.e. on [a, b]. By (b), φ ≤ m a.e. on [a, b]. Hence∫ ∫φ ≤ m[a,b][a,b]Since this is true for all φ ∈ RL f , we get thatsup φ ≤φ∈RL f∫[a,b]∫[a,b]m(∗)Pick n ≥ 1. We partition the interval [a, b] into 2 n intervals I n1 , I n2 , . . . , I n2 nI n1 =[a, a + b − a ]2 nand I nk =We define φ n : [a, b] → R as follows :(a + (k − 1) b − a2 , a + k b − a ]n 2 n∀x ∈ [a, b] : φ n (x) = inf{f(y) | y ∈ I nk } if x ∈ I nkThen φ n ∈ RL f .Let D be the set of all the partition points of all the partitions :{D = a + k b − a}| n ≥ 1 ; 1 ≤ k ≤ 2 n − 12 nof length b − a2 n :for k = 2, 3, . . . , 2 nNote that D is countable.Pick x 0 ∈ [a, b]\D. We will show that {φ n (x 0 )} n≥1 → m(x 0 ). Pick ε > 0. Since m(x 0 ) = limδ→0 + m δ(x 0 ),there exists δ > 0 such that 0 ≤ m(x 0 ) − m δ (x 0 ) < ε. For n ≥ 1, let 1 ≤ k n ≤ 2 n with x 0 ∈ I nkn . LetN ∈ N with b − a2 N < δ. Pick n ≥ N. Then I nk n⊆ (x 0 − δ, x 0 + δ). Since x 0 /∈ D, we have that φ n iscontinuous at x 0 . So by (b) we get :m δ (x 0 ) = inf{f(x) | x ∈ (x 0 − δ, x 0 + δ) ∩ [a, b]} ≤ inf{f(x) | x ∈ I nkn } = φ n (x 0 ) ≤ m(x 0 )Hence |m(x 0 ) − φ n (x 0 )| ≤ |m(x 0 ) − m δ (x 0 )| < ε. Thus {φ n (x 0 )} n≥1 → m(x 0 ).Since this is true for all x 0 ∈ [a, b] \ D and D is countable, we have that {φ n } n≥1 → m a.e. on [a, b].Since f is bounded over [a, b], there exists λ > 0 such that −λ ≤ f(x) ≤ λ for all x ∈ [a, b]. Pickn ≥ 1 and x 0 ∈ [a, b] \ D. Let 1 ≤ k ≤ 2 n with x 0 ∈ I nk . Then by (b) we get−λ ≤ inf{f(x) | x ∈ [a, b]} ≤ inf{f(x) | x ∈ I nk } = φ n (x 0 ) ≤ m(x 0 ) ≤ f(x 0 ) ≤ λSo |φ n | ≤ λχ [a,b] a.e. on [a, b]. Note that λχ [a,b] is integrable over [a, b].By the Lebesgue Convergence Theorem, we get{∫ } ∫φ n → m[a,b] n≥1 [a,b]68

Since φ n ∈ RL f , we get∫sup φ ≥ φ n for all n ≥ 1φ∈RL f∫[a,b] [a,b]Since this is true for all n ≥ 1, we can take the limit as n goes to +∞ on both sides :∫ ∫sup φ ≥ lim φ n = m (∗∗)φ∈RL f∫[a,b]n→+∞[a,b] [a,b]It now follows from (*) and (**) that∫sup φ = mφ∈RL f∫[a,b] [a,b]∫Similarly, we prove that inf ψ = M. ✷ψ∈RU f∫[a,b] [a,b]Theorem 4.28 Let f : [a, b] → R be a function. Then f is Riemann integrable over [a, b] if andonly if f is bounded on [a, b] and m({x ∈ [a, b] | f is discontinuous at x}) = 0.Proof : Put E = {x ∈ [a, b] | f is discontinuous at x}. If f is bounded over [a, b], we have :f is Riemann integrable over [a, b] ⇐⇒ sup φ = inf ψφ∈RL f∫[a,b]ψ∈RU f∫[a,b]∫ ∫⇐⇒ m = M⇐⇒∫[a,b][a,b][a,b](M − m) = 0⇐⇒ M − m = 0 a.e. on [a, b](definition)(Proposition 4.27(d))(M − m ≥ 0 on [a, b])⇐⇒ m(E) = 0 (Proposition 4.26) ✷69

Contents1 Preliminary Material 11.1 Unions, Intersections and Complements . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.3 Countable and Uncountable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.4 Axiom of Choice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5.1 Axioms for the Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.5.2 Archimedean Property and Density of the Rationals . . . . . . . . . . . . . . . 61.6 Supremum and Infimum of a Set of Real Numbers . . . . . . . . . . . . . . . . . . . . 71.7 Sequences of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7.1 Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.7.2 Limit Superior and Limit Inferior of a Sequence . . . . . . . . . . . . . . . . . 91.8 Open and Closed Sets of Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . 161.9 Continuous Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 Measurable Sets 202.1 Outer Measure of a Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.2 σ-Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232.3 Measurable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.4 Non-Measurable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.5 Littlewood’s First Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 332.6 Cantor Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 Measurable Functions 373.1 Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373.2 The Cantor-Lebesgue Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.3 The Simple Approximation Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.4 Littlewood’s Second and Third Principle . . . . . . . . . . . . . . . . . . . . . . . . . 483.4.1 Littlewood’s Third Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483.4.2 Littlewood’s Second Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . 494 The Lebesgue Integral 514.1 Lebesgue Integral of Nonnegative Simple Functions . . . . . . . . . . . . . . . . . . . 514.2 Lebesgue Integral of Nonegative Measurable Functions . . . . . . . . . . . . . . . . . 534.3 Fatou’s Lemma, Monotone Convergence Theorem . . . . . . . . . . . . . . . . . . . . 554.4 Lebesgue Integral of Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . 584.5 Lebesgue Convergence Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 614.6 Riemann and Lebesgue Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6270