Lecture 3 The Lebesgue Integral

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1 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.

2 Lecture 3: The Lebesgue Itegral 2 of 14 Remark 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, 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.

3 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 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

4 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 =

5 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 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

6 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 }.

7 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 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 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 f + g dµ f dµ + g dµ, for all f, g L 1.

8 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 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.

9 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 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 Let (S, S, µ be a measure space. 1. N S is said to be a ull set if µ(n = 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 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.

10 Lecture 3: The Lebesgue Itegral 10 of 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 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 +?

11 Lecture 3: The Lebesgue Itegral 11 of 14 Defiitio 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 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 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 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 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.

12 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

13 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 ]

14 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