Lecture 3: The Lebesgue Itegral 1 of 14 Course: Theory of Probability I Term: Fall 2013 Istructor: Gorda Zitkovic Lecture 3 The Lebesgue Itegral The costructio of the itegral Uless expressly specified otherwise, we pick ad fix a measure space (S, S, µ ad assume that all fuctios uder cosideratio are defied there. Defiitio 3.1 (Simple fuctios. A fuctio f L 0 (S, S, µ is said to be simple if it takes oly a fiite umber of values. The collectio of all simple fuctios is deoted by L Simp,0 (more precisely by L Simp,0 (S, S, µ ad the family of o-egative simple fuctios by L Simp,0 +. Clearly, a simple fuctio f : S R admits a (ot ecessarily uique represetatio f = α k 1 Ak, (3.1 k=1 for α 1,..., α R ad A 1,..., A S. Such a represetatio is called the simple-fuctio represetatio of f. Whe the sets A k, k = 1,..., are itervals i R, the graph of the simple fuctio f looks like a collectio of steps (of heights α 1,..., α. For that reaso, the simple fuctios are sometimes referred to as step fuctios. The Lebesgue itegral is very easy to defie for o-egative simple fuctios ad this defiitio allows for further geeralizatios 1 : Defiitio 3.2 (Lebesgue itegratio for simple fuctios. For f L Simp,0 + we defie the (Lebesgue itegral f dµ of f with respect to µ by f dµ = α k µ(a k [0, ], k=1 where f = k=1 α k1 Ak is a simple-fuctio represetatio of f, 1 I fact, the progressio of evets you will see i this sectio is typical for measure theory: you start with idicator fuctios, move o to o-egative simple fuctios, the to geeral oegative measurable fuctios, ad fially to (ot-ecessarily-o-egative measurable fuctios. This approach is so commo, that it has a ame - the Stadard Machie. Problem 3.1. Show that the Lebesgue itegral is well-defied for simple fuctios, i.e., that the value of the expressio k=1 α kµ(a k does ot deped o the choice of the simple-fuctio represetatio of f.
Lecture 3: The Lebesgue Itegral 2 of 14 Remark 3.3. 1. It is importat to ote that f dµ ca equal + eve if f ever takes the value +. It is eough to pick f = 1 A where µ(a = + - ideed, the f dµ = 1µ(A =, but f oly takes values i the set {0, 1}. This is oe of the reasos we start with o-egative fuctios. Otherwise, we would eed to deal with the (usolvable problem of computig. O the other had, such examples caot be costructed whe µ is a fiite measure. Ideed, it is easy to show that whe µ(s <, we have f dµ < for all f L Simp,0 +. 2. Oe ca thik of the (simple Lebesgue itegral as a geeralizatio of the otio of (fiite additivity of measures. Ideed, if the simplefuctio represetatio of f is give by f = k=1 1 A k, for pairwise disjoit A 1,..., A, the the equality of the values of the itegrals for two represetatios f = 1 k=1 A k ad f = k=1 1 A k is a simple restatemet of fiite additivity. Whe A 1,..., A are ot disjoit, the the fiite additivity gives way to fiite subadditivity µ( k=1 A k µ(a k, k=1 but the itegral f dµ takes ito accout those x which are covered by more tha oe A k, k = 1,...,. Take, for example, = 2 ad A 1 A 2 = C. The f = 1 A1 + 1 A2 = 1 A1 \C + 21 C + 1 A2 \C, ad so f dµ = µ(a 1 \ C + µ(a 2 \ C + 2µ(C = µ(a 1 + µ(a 2 + µ(c. It is easy to see that L Simp,0 + is a covex coe, i.e., that it is closed uder fiite liear combiatios with o-egative coefficiets. The itegral map f f dµ preserves this structure: Problem 3.2. For f 1, f 2 L Simp,0 + ad α 1, α 2 0 we have 1. if f 1 (x f 2 (x for all x S the f 1 dµ f 2 dµ, ad 2. (α1 f 1 + α 2 f 2 dµ = α 1 f1 dµ + α 2 f2 dµ. o-egative measurable fuctios. I fact, at o extra cost we ca cosider a slightly larger set cosistig of all measurable [0, ]-valued fuctios which we deote by L 0 +( [0, ]. Havig defied the itegral for f L Simp,0 +, we tur to geeral 2 2 Eve though there is o obvious advatage at this poit of itegratig a fuctio which takes the value +, it will become clear soo how coveiet it really is.
Lecture 3: The Lebesgue Itegral 3 of 14 Defiitio 3.4 (Lebesgue itegral for oegative fuctios. For a Note: While there is o questio that fuctio f L 0 this defiitio produces a uique umber f dµ, oe ca woder if it matches +( [0, ], we defie its Lebesgue itegral f dµ by { } the previously give defiitio of the f dµ = sup g dµ : g L Simp,0 Lebesgue itegral for simple fuctios. +, g(x f (x, x S [0, ]. A simple argumet based o the mootoicity property of part 1. of Problem 3.2 ca be used to show that this is, ideed, the case. Problem 3.3. Show that f dµ = if there exists a measurable set A with µ(a > 0 such that f (x = for x A. O the other had, show that f dµ = 0 for f of the form, x A, f (x = 1 A (x = 0, x = A, wheever µ(a = 0. Fially, we are ready to defie the itegral for geeral measurable fuctios. Each f L 0 ca be writte as a differece of two fuctios i L 0 + i may ways. There exists a decompositio which is, i a sese, miimal. We defie Note: Relate this to our covetio that 0 = 0 = 0. f + = max( f, 0, f = max( f, 0, so that f = f + f (ad both f + ad f are measurable. The miimality we metioed above is reflected i the fact that for each x S, at most oe of f + ad f is o-zero. Defiitio 3.5 (Itegrable fuctios. A fuctio f L 0 is said to be itegrable if f + dµ < ad f dµ <. The collectio of all itegrable fuctios i L 0 is deoted by L 1. The family of itegrable fuctios is tailor-made for the followig defiitio: Defiitio 3.6 (The Lebesgue itegral. For f L 1, we defie the Lebesgue itegral f dµ of f by f dµ = f + dµ f dµ. Remark 3.7. 1. We have see so far two cases i which a itegral for a fuctio f L 0 ca be defied: whe f 0 or whe f L 1. It is possible to combie the two ad defie the Lebesgue itegral for all fuctios f L 0 with f L 1. The set of all such fuctios is deoted by
Lecture 3: The Lebesgue Itegral 4 of 14 L 0 1 ad we set f dµ = f + dµ f dµ (, ], for f L 0 1. Note that o problems of the form arise here, ad also ote that, like L 0 +, L0 1 is oly a covex coe, ad ot a vector space. While the otatio L 0 ad L 1 is quite stadard, the oe we use for L 0 1 is ot. 2. For A S ad f L 0 1 we usually write A f dµ for f 1 A dµ. Problem 3.4. Show that the Lebesgue itegral remais a mootoe operatio i L 0 1. More precisely, show that if f L 0 1 ad g L 0 are such that g(x f (x, for all x S, the g L 0 1 ad g dµ f dµ. First properties of the itegral The wider the geerality to which a defiitio applies, the harder it is to prove theorems about it. Liearity of the itegral is a trivial matter for fuctios i L Simp,0 +, but you will see how much we eed to work to get it for L 0 +. I fact, it seems that the easiest route towards liearity is through two importat results: a approximatio theorem ad a covergece theorem. Before that, we eed to pick some low-hagig fruit: Problem 3.5. Show that for f 1, f 2 L 0 +( [0, ] ad α [0, ] we have 1. if f 1 (x f 2 (x for all x S the f 1 dµ f 2 dµ. 2. α f dµ = α f dµ. Theorem 3.8 (Mootoe covergece theorem. Let { f } N be a sequece i L 0 +( [0, ] with the property that f 1 (x f 2 (x... for all x S. The lim f dµ = f dµ, where f (x = lim f (x L 0 +( [0, ], for x S. Proof. The (mootoicity property (1 of Problem 3.5 above implies that the sequece f dµ is o-decreasig ad that f dµ f dµ. Therefore, lim f dµ f dµ. To show the opposite iequality, we deal with the case f dµ < ad pick ε > 0 ad g L Simp,0 + with g(x f (x, for all x S ad g dµ f dµ ε (the case f dµ =
Lecture 3: The Lebesgue Itegral 5 of 14 is similar ad left to the reader. For 0 < c < 1, defie the (measurable sets {A } N by A = { f cg}, N. By the icrease of the sequece { f } N, the sets {A } N also icrease. Moreover, sice the fuctio cg satisfies cg(x g(x f (x for all x S ad cg(x < f (x whe f (x > 0, the icreasig covergece f f implies that A = S. By o-egativity of f ad mootoicity, ad so f dµ sup f 1 A dµ c f dµ c sup g1 A dµ, g1 A dµ. Let g = k i=1 α i1 Bi be a simple-fuctio represetatio of g. The g1 A dµ = k α i 1 Bi A dµ = i=1 k α i µ(b i A. i=1 Sice A S, we have A B i B i, i = 1,..., k, ad the cotiuity of measure implies that µ(a B i µ(b i. Therefore, g1 A dµ k α i µ(b i = i=1 g dµ. Cosequetly, lim f dµ = sup f dµ c g dµ, for all c (0, 1, ad the proof is completed whe we let c 1. Remark 3.9. 1. The mootoe covergece theorem is really about the robustess of the Lebesgue itegral. Its stability with respect to limitig operatios is oe of the reasos why it is a de-facto idustry stadard. 2. The mootoicity coditio i the mootoe covergece theorem caot be dropped. Take, for example S = [0, 1], S = B([0, 1], ad µ = λ (the Lebesgue measure, ad defie f = 1 (0, 1 ], for N. The f (0 = 0 for all N ad f (x = 0 for > 1 x either case f (x 0. O the other had f dλ = λ ((0, 1 ] = 1, ad x > 0. I
Lecture 3: The Lebesgue Itegral 6 of 14 so that lim f dλ = 1 > 0 = lim f dλ. We will see later that the while the equality of the limit of the itegrals ad the itegral of the limit will ot hold i geeral, they will always be ordered i a specific way, if the fuctios { f } N are o-egative (that will be the cotet of Fatou s lemma below. Propositio 3.10 (Approximatio by simple fuctios. For each f L 0 ( + [0, ] there exists a sequece {g } N L Simp,0 + such that 1. g (x g +1 (x, for all N ad all x S, 2. g (x f (x for all x S, 3. f (x = lim g (x, for all x S, ad 4. the covergece g f is uiform o each set of the form { f M}, M > 0, ad, i particular, o the whole S if f is bouded. Proof. For N, let A k, k = 1,..., 2 be a collectio of subsets of S give by A k = { k 1 2 f < 2 k } = f 1( [ k 1 k 2, 2, k = 1,..., 2. Note that the sets A k, k = 1,..., 2 are disjoit ad that the measurability of f implies that A k S for k = 1,..., 2. Defie the fuctio g L Simp,0 + by 2 k 1 g = 2 1 A k + 1 { f }. k=1 The statemets 2., 3., ad 4. follow immediately from the followig three simple observatios: g (x f (x for all x S, g (x = if f (x =, ad g (x > f (x 2 whe f (x <. Fially, we leave it to the reader to check the simple fact that {g } N is o-decreasig. Problem 3.6. Show, by meas of a example, that the sequece {g } N would ot ecessarily be mootoe if we defied it i the followig way: g = 2 k=1 k 1 1 { f [ k 1, k + 1 } { f }.
Lecture 3: The Lebesgue Itegral 7 of 14 Propositio 3.11 (Liearity of the itegral for o-egative fuctios. For f 1, f 2 L 0 +( [0, ] ad α1, α 2 0 we have (α 1 f 1 + α 2 f 2 dµ = α 1 f 1 dµ + α 2 f 2 dµ. Proof. Thaks to Problem 3.5 it is eough to prove the statemet for α 1 = α 2 = 1. Let {g} 1 N ad {g} 2 N be sequeces i L Simp,0 + which approximate f 1 ad f 2 i the sese of Propositio 3.10. The sequece {g } N give by g = g 1 + g, 2 N, has the followig properties: g L Simp,0 + for N, g (x is a odecreasig sequece for each x S, g (x f 1 (x + f 2 (x, for all x S. Therefore, we ca apply the liearity of itegratio for the simple fuctios ad the mootoe covergece theorem (Theorem 3.8 to coclude that ( ( f 1 + f 2 dµ = lim (g 1 + g 2 dµ = lim g 1 dµ + = f 1 dµ + f 2 dµ. g 2 dµ Corollary 3.12 (Coutable additivity of the itegral. Let { f } N be a sequece i L 0 +( [0, ]. The f dµ = N N f dµ. Proof. Apply the mootoe covergece theorem to the partial sums g = f 1 + + f, ad use liearity of itegratio. Oce we have established a battery of properties for o-egative fuctios, a extesio to L 1 is ot hard. We leave it to the reader to prove all the statemets i the followig problem: Problem 3.7. The family L 1 of itegrable fuctios has the followig properties: 1. f L 1 iff f dµ <, 2. L 1 is a vector space, 3. f dµ f dµ, for f L 1. 4. f + g dµ f dµ + g dµ, for all f, g L 1.
Lecture 3: The Lebesgue Itegral 8 of 14 We coclude the preset sectio with two results, which, together with the mootoe covergece theorem, play the cetral role i the Lebesgue itegratio theory. Theorem 3.13 (Fatou s lemma. Let { f } N be a sequece i L 0 +( [0, ]. Note: The stregth of Fatou s lemma The comes from the fact that, apart from oegativity, it requires o special proper- lim if f dµ lim if f dµ. ties for the sequece { f } N. Its coclusio is ot as strog as that of the mootoe covergece theorem, but it Proof. Set g (x = if k f k (x, so that g L 0 +( [0, ] ad g (x is a proves to be very useful i various settigs because it o-decreasig sequece for each x S. The mootoe covergece gives a upper boud (amely lim if f dµ o the itegral theorem ad the fact that lim if f (x = sup g (x = lim g (x, for of the o-egative fuctio lim if f. all x S, imply that g dµ lim if dµ. O the other had, g (x f k (x for all k, ad so g dµ if k f k dµ. Therefore, lim g dµ lim if k f k dµ = lim if f k dµ. Remark 3.14. 1. The iequality i the Fatou s lemma does ot have to be equality, eve if the limit lim f (x exists for all x S. You ca use the sequece { f } N of Remark 3.9 to see that. 2. Like the mootoe covergece theorem, Fatou s lemma requires that all fuctio { f } N be o-egative. This requiremet is ecessary - to see that, simply cosider the sequece { f } N, where { f } N is the sequece of Remark 3.9 above. Theorem 3.15 (Domiated covergece theorem. Let { f } N be a Note: The domiated covergece theorem combies the lack of mootoicity requiremets of Fatou s lemma ad sequece i L 0 with the property that there exists g L 1 such that f (x g(x, for all x X ad all N. If f (x = lim f (x for all x S, the the strog coclusio of the mootoe f L 1 ad covergece theorem. The price to be f dµ = lim f dµ. Proof. The coditio f (x g(x, for all x X ad all N implies that g(x 0, for all x S. Sice f + g, f g ad g L 1, we immediately have f L 1, for all N. The limitig fuctio paid is the uiform boudedess requiremet. There is a way to relax this requiremet a little bit (usig the cocept of uiform itegrability, but ot too much. Still, it is a uexpectedly useful theorem.
Lecture 3: The Lebesgue Itegral 9 of 14 f iherits the same properties f + g ad f g from { f } N so f L 1, too. Clearly g(x + f (x 0 for all N ad all x S, so we ca apply Fatou s lemma to get g dµ + lim if f dµ = lim if (g + f dµ lim if(g + f dµ = (g + f dµ = g dµ + f dµ. I the same way (sice g(x f (x 0, for all x S, as well, we have g dµ lim sup f dµ = lim if (g f dµ lim if(g f dµ = (g f dµ = g dµ f dµ. Therefore lim sup f dµ f dµ lim if f dµ, ad, cosequetly, f dµ = lim f dµ. Remark 3.16. Null sets A importat property - iherited directly from the uderlyig measure - is that it is blid to sets of measure zero. To make this statemet precise, we eed to itroduce some laguage: Defiitio 3.17. Let (S, S, µ be a measure space. 1. N S is said to be a ull set if µ(n = 0. 2. A fuctio f : S R is called a ull fuctio if there exists a ull set N such that f (x = 0 for x N c. 3. Two fuctios f, g are said to be equal almost everywhere - deoted by f = g, a.e. - if f g is a ull fuctio, i.e., if there exists a ull set N such that f (x = g(x for all x N c. Remark 3.18. 1. I additio to almost-everywhere equality, oe ca talk about the almost-everywhere versio of ay relatio betwee fuctios which ca be defied o poits. For example, we write f g, a.e. if f (x g(x for all x S, except, maybe, for x i some ull set N.
Lecture 3: The Lebesgue Itegral 10 of 14 2. Oe ca also defie the a.e. equality of sets: we say that A = B, a.e., for A, B S if 1 A = 1 B, a.e. It is ot hard to show (do it! that A = B a.e., if ad oly if µ(a B = 0 (Remember that deotes the symmetric differece: A B = (A \ B (B \ A. 3. Whe a property (equality of fuctios, e.g. holds almost everywhere, the set where it fails to hold is ot ecessarily ull. Ideed, there is o guaratee that it is measurable at all. What is true is that it is cotaied i a measurable (ad ull set. Ay such (measurable ull set is ofte referred to as the exceptioal set. Problem 3.8. Prove the followig statemets: 1. The almost-everywhere equality is a equivalece relatio betwee fuctios. 2. The family {A S : µ(a = 0 or µ(a c = 0} is a σ-algebra (the so-called µ-trivial σ-algebra. The blidess property of the Lebesgue itegral we referred to above ca ow be stated formally: Propositio 3.19. Suppose that f = g, a.e,. for some f, g L 0 +. The f dµ = g dµ. Proof. Let N be a exceptioal set for f = g, a.e., i.e., f = g o N c ad µ(n = 0. The f 1 N c = g1 N c, ad so f 1 N c dµ = g1 N c dµ. O the other had f 1 N 1 N ad 1 N dµ = 0, so, by mootoicity, f 1N dµ = 0. Similarly g1 N dµ = 0. It remais to use the additivity of itegratio to coclude that f dµ = = f 1 N c dµ + g1 N c dµ + f 1 N dµ g1 N dµ = g dµ. A statemet which ca be see as a coverse of Propositio 3.19 also holds: Problem 3.9. If f L 0 + ad f dµ = 0, show that f = 0, a.e. The mootoe covergece theorem ad the domiated covergece theorem both require the sequece { f } N fuctios to coverge for each x S. A slightly weaker otio of covergece is required, though: Hit: What is the egatio of the statemet f = 0. a.e. for f L 0 +?
Lecture 3: The Lebesgue Itegral 11 of 14 Defiitio 3.20. A sequece of fuctios { f } N is said to coverge almost everywhere to the fuctio f, if there exists a ull set N such that f (x f (x for all x N c. Remark 3.21. If we wat to emphasize that f (x f (x for all x S, we say that { f } N coverges to f everywhere. Propositio 3.22 (Mootoe (almost-everywhere covergece theorem. Let { f } N be a sequece i L 0 +( [0, ] with the property that f f +1 a.e., for all N. The lim f dµ = f dµ, if f L 0 + ad f f, a.e. Proof. There are + 1 a.e.-statemets we eed to deal with: oe for each N i f f +1, a.e., ad a extra oe whe we assume that f f, a.e. Each of them comes with a exceptioal set; more precisely, let {A } N be such that f (x f +1 (x for x A c ad let B be such that f (x f (x for x B c. Defie A S by A = ( A B ad ote that A is a ull set. Moreover, cosider the fuctios f, { f } N defied by f = f 1 A c, f = f 1 A c. Thaks to the defiitio of the set A, f (x f +1 (x, for all N ad x S; hece f f, everywhere. Therefore, the mootoe covergece theorem (Theorem 3.8 ca be used to coclude that f dµ f dµ. Fially, Propositio 3.19 implies that f dµ = f dµ for N ad f dµ = f dµ. Problem 3.10. State ad prove a versio of the domiated covergece theorem where the almost-everywhere covergece is used. Is it ecessary for all { f } N to be domiated by g for all x S, or oly almost everywhere? Remark 3.23. There is a subtlety that eeds to be poited out. If a sequece { f } N of measurable fuctios coverges to the fuctio f everywhere, the f is ecessarily a measurable fuctio (see Propositio 1.23. However, if f f oly almost everywhere, there is o guaratee that f is measurable. There is, however, always a measurable fuctio which is equal to f almost everywhere; you ca take lim if f, for example.
Lecture 3: The Lebesgue Itegral 12 of 14 Additioal Problems Problem 3.11 (The mootoe-class theorem. Prove the followig re- Hit: Use Theorems 3.10 ad 2.12 sult, kow as the mootoe-class theorem (remember that a a meas that a is a o-decreasig sequece ad a a Let H be a class of bouded fuctios from S ito R satisfyig the followig coditios 1. H is a vector space, 2. the costat fuctio 1 is i H, ad 3. if { f } N is a sequece of o-egative fuctios i H such that f (x f (x, for all x S ad f is bouded, the f H. The, if H cotais the idicator 1 A of every set A i some π-system P, the H ecessarily cotais every bouded σ(p-measurable fuctio o S. Problem 3.12 (A form of cotiuity for Lebesgue itegratio. Let (S, S, µ be a measure space, ad suppose that f L 1. Show that for each ε > 0 there exists δ > 0 such that if A S ad µ(a < δ, the A f dµ < ε. Problem 3.13 (Sums as itegrals. I the measure space (N, 2 N, µ, let µ be the coutig measure. 1. For a fuctio f : N [0, ], show that f dµ = f (. =1 2. Use the mootoe covergece theorem to show the followig special case of Fubii s theorem k=1 =1 a k = =1 k=1 a k, wheever {a k : k, N} is a double sequece i [0, ]. 3. Show that f : N R is i L 1 if ad oly if the series coverges absolutely. f (, =1
Lecture 3: The Lebesgue Itegral 13 of 14 Problem 3.14 (A criterio for itegrability. Let (S, S, µ be a fiite measure space. For f L 0 +, show that f L1 if ad oly if µ({ f } <. N Problem 3.15 (A limit of itegrals. Let (S, S, µ be a measure space, Hit: Prove ad use the iequality ad suppose f L 1 log(1 + x α αx, valid for x 0 ad + is such that f dµ = c > 0. Show that the limit α 1. lim ( log 1 + ( f / α dµ exists i [0, ] for each α > 0 ad compute its value. Problem 3.16 (Itegrals coverge but the fuctios do t.... Costruct a sequece { f } N of cotiuous fuctios f : [0, 1] [0, 1] such that f dµ 0, but the sequece { f (x} N is diverget for each x [0, 1]. Problem 3.17 (... or they do, but are ot domiated. Costruct a sequece { f } N of cotiuous fuctios f : [0, 1] [0, such that f dµ 0, ad f (x 0 for all x, but f L 1, where f (x = sup f (x. Problem 3.18 (Fuctios measurable i the geerated σ-algebra. Let S = be a set ad let f : S R be a fuctio. Prove that a fuctio g : S R is measurable with respect to the pair (σ( f, B(R if ad oly if there exists a Borel fuctio h : R R such that g = h f. Problem 3.19 (A chage-of-variables formula. Let (S, S, µ ad (T, T, ν be two measurable spaces, ad let F : S T be a measurable fuctio with the property that ν = F µ (i.e., ν is the push-forward of µ through F. Show that for every f L 0 + (T, T or f L1 (T, T, we have f dν = ( f F dµ. Problem 3.20 (The Riema Itegral. A fiite collectio = {t 0,..., t }, where a = t 0 < t 1 < < t = b ad N, is called a partitio of the iterval [a, b]. The set of all partitios of [a, b] is deoted by P([a, b]. For a bouded fuctio f : [a, b] R ad = {t 0,..., t } P([a, b], we defie its upper ad lower Darboux sums U( f, ad L( f, by ( U( f, = sup f (t (t k t k 1 k=1 t (t k 1,t k ]
Lecture 3: The Lebesgue Itegral 14 of 14 ad ( L( f, = if f (t (t k t k 1. k=1 t (t k 1,t k ] A fuctio f : [a, b] R is said to be Riema itegrable if it is bouded ad sup L( f, = if U( f,. P([a,b] P([a,b] I that case the commo value of the supremum ad the ifimum above is called the Riema itegral of the fuctio f - deoted by (R b a f (x dx. 1. Suppose that a bouded Borel-measurable fuctio f : [a, b] R is Riema-itegrable. Show that b [a,b] f dλ = (R f (x dx. 2. Fid a example of a bouded a Borel-measurable fuctio f : [a, b] R which is ot Riema-itegrable. 3. Show that every cotiuous fuctio is Riema itegrable. 4. It ca be show that for a bouded Borel-measurable fuctio f : [a, b] R the followig criterio holds (ad you ca use it without proof: f is Riema itegrable if ad oly if there exists a Borel set D [a, b] with λ(d = 0 such that f is cotiuous at x, for each x [a, b] \ D. Show that all mootoe fuctios are Riema-itegrable, f g is Riema itegrable if f : [c, d] R is Riema itegrable ad g : [a, b] [c, d] is cotiuous, products of Riema-itegrable fuctios are Riema-itegrable. 5. Let ([a, b], B([a, b], λ deote the completio of ([a, b], B([a, b], λ. Show that ay Riema-itegrable fuctio o [a, b] is B([a, b] - measurable. Hit: Pick a sequece { } N i P([a, b] so that +1 ad U( f, L( f, 0. Usig those partitios ad the fuctio f, defie two sequeces of Borel-measurable fuctios { f } N ad { f } N so that f f, f f, f f f, ad ( f f dλ = 0. Coclude that f agrees with a Borel measurable fuctio o a complemet of a subset of the set { f = f } which has Lebesgue measure 0. a