Itegratio Theory: Lecture otes 203 Joha Joasso September 203 Preface These lecture otes are writte whe the course i itegratio theory is for the first time i more tha twety years, give joitly by the the two divisios Mathematics ad Mathematical Statistics. The major source is G. B. Follad: Real Aalysis, Moder Techiques ad Their Applicatios. However, the parts o probability theory are mostly take from D. Williams: Probability with Martigales. Aother source is Christer Borell s lecture otes from previous versios of this course, see www.math.chalmers.se/math/grudutb/gu/mma0/a/ 2 Itroductio This course itroduces the cocepts of measures, measurable fuctios ad Lebesgue itegrals. The itegral used i earlier math courses is the so called Riema itegral. The Lebesgue itegral will tur out to be more powerful i the sese that it allows us to defie itegrals of ot oly Riema itegrable fuctios, but also some fuctios for which the Riema itegral is ot defied. Most importatly however, is that it will allow us to rigorously prove may results for which proofs of the correspodig results i the Riema settig are usually ever see by studets at the basic ad itermediate level. Such results iclude precise coditios for whe we ca chage order of itegrals ad limits, chage order of itegratio Chalmers Uiversity of Techology Göteborg Uiversity joasso@chalmers.se
i multiple itegrals ad whe we ca use itegratio by parts. Of course, we will also prove may ew results. The cocept of measurability is a advaced oe, i the sese that a lot of people at first fid it difficult to master; it teds to feel fudametally more abstract tha thigs oe has ecoutered before. Therefore, a atural first questio is why the cocept is eeded. To aswer this, cosider the followig example. Let X = R/Z, the circle of circumferece, with additio ad multiplicatio defied modulo. Suppose we wat to itroduce the cocept of the legth of subsets of X. A atural first assumptio is that oe should be able to do this so that the legth is defied for all subsets of X. It is also extremely atural to claim that the legth l, should satisfy l( ) = 0, l(x) =, l( )A = l(a ) for all disjoit A, A 2,..., l(a + x) = l(a) for all A X ad x X. However, if we isist o defiig l for all subsets, this turs out to be impossible. Let us see why. Partitio X ito equivalece classes by sayig that x ad y are equivalet if x y is a ratioal umber. By the axiom of choice, there exists a set A cotaiig exactly oe elemet from each equivalece class. For each q Q X, let A q = A + q. The q A q = X, for sice for each x X, A cotais a elemet y equivalet to x, i.e. x A x y ad x y Q. O the other had, the A q s are disjoit, for if x A q A q2, the x = y+q = z + q 2 for two elemets y, z A. However, the y z = q 2 q Q, so y ad z are equivalet, cotradictig the costructio of A. If we could assig legths to the A q s, the these legths must be equal by the fourth coditio o l. O the other had, the legths of the A q s must sum to by the third coditio. However, these two coditios are mutually exclusive. The moral of the example is that the set A must be declared o-measurable; o legth of A ca be defied. The costructio of the example is based o the axiom of choice ad it ca be show that all costructios of o-measurable sets must rely o the axiom of choice. There are eve more absurd examples tha this oe. The famous Baach- Tarski paradox proves, usig the axiom of choice, that for ay two bouded compact sets i R 3, the oe ca be divided ito a fiite umber of parts which ca be 2
traslated ad rotated ad mirrored ad the put back together to form the other. For example: ay grai of sad ca be divided ito a umber of pieces that ca be put back together to form a ball the size of the earth! Clearly theses pieces caot have a well defied volume. Examples like these call for a theory of measures ad measurable sets. 3 Measures We are goig to cosider measures i a very geeral framework: we will cosider measures o a a abstract space X o we which we make o iitial assumptios whatsoever. As the above example revealed, it is ot always possible with meaigful measures defied o all subsets of X. Hece a cocept of what classes of subsets to defie a desired measure o, is eeded. The two last coditios o a legth measure i the above example were atural i that particular situatio, but it is easy to thik of other situatios where either of them is atural or eve meaigful. The two first coditios however, are such that they should hold for aythig that deserves to be called a measure, o matter what structure X has. Thus we keep those two coditios i mid, ad ask for classes of subsets large eough to esure that all iterestig set operatios o measurable sets results i a measurable set, but restrictive eough to make sure that o coflict with the basic assumptios arises. The aswer is σ-algebras. 3. Algebras ad σ-algebras Defiitio 3. Let A be a class of subsets of X such that (i) X A, (ii) E c A wheever E A, (iii) E F A wheever E, F A. The A is called a algebra (o X). Note that by (i) ad (ii), = X c A. Also, if E, F A, the E F = (E c F c ) c A by (ii) ad (iii). Defiitio 3.2 Let M be a class of subsets of X such that 3
(i) X M, (ii) E c M wheever E M, (iii) = E M wheever E, E 2,... M. The M is called a σ-algebra. Clearly ay σ-algebra is a algebra. As above M, ad aalogously, if E, E 2,... M, the E = ( Ec ) c M. A measure will always be defied o a σ-algebra. The smallest possible σ- algebra o ay space X is {, X}. The largest σ-algebra is P(X), the class of all subsets of X (but we have see that meaigful measures caot always be defied o this σ-algebra). If M is a σ-algebra o X, the the pair (X, M) is called a measurable space ad a set E M is called M-measurable. 3.2 Geerated σ-algebras Let C be a arbitrary class of subsets of X. We defie the σ-algebra geerated by C as the smallest σ-algebra cotaiig C, i.e. σ(c) = {F : F σ-algebra, F C}. (It is a easy exercise to show that ay itersectio of σ-algebras is a σ-algebra.) The most importat example is the Borel σ-algebra; if X is a topological space ad T is the class of ope sets, the the Borel σ-algebra, B(X), is give by B(X) = σ(t ). Sice ay ope set i R is a coutable uio of ope itervals, it follows that B(R) = σ((a, b) : a, b R). It is ow easy to see (check this!) that we also have B(R) = σ([a, b) : a, b R) = σ((a, b] : a, b R) = σ([a, b] : a, b R) = σ((, b) : b R) = σ((a, ) : a R). I itegratio theory, oe ofte works with the exteded real lie, R = [, ] ad, eve more, with the exteded positive half-lie R + = [0, ]. Here the arithmetics ivolvig the poits ad work as oe would ituitively guess, ad a subset is regarded as ope if it is either a subset of R ad ope as such, of the form [, a) or (a, ], or the whole space. It is ow straightforward to prove aalogous expressios for B(R) ad B(R + ). 4
3.3 Measures If C is a class of subsets of X ad µ 0 : C R +, the µ 0 is called a set fuctio. Let A be a algebra. If µ 0 is a set fuctio o A such that µ 0 ( ) = 0 ad E, F A, E F = implies µ 0 (E F ) = µ 0 (E) + µ 0 (F ), the µ 0 is said to be additive. If µ 0 ( ) = 0 ad µ 0 satisfies the stroger coditio that µ 0 ( E ) = µ 0(E ) wheever E, E 2,... A ad E A, the µ 0 is said to be coutably additive or a premeasure. (Stroger sice additivity follows from coutable additivity by takig E = E, E 2 = F ad E 3 = E 4 =... =.) Defiitio 3.3 Let M be a σ-algebra ad µ a set fuctio defied o M. If µ is coutably additive, the µ is said to be a measure. Let µ be a measure o the σ-algebra M. Here are a few classificatios. µ is said to be fiite if µ(x) <. µ ca be said to be a probability measure if µ(x) =. µ is said to be σ-fiite if there exist sets E, E 2,... M such that E = X ad µ(e ) < for all. µ is said to be semi-fiite if for every E M such that µ(e) =, there exists a set F E such that 0 < µ(f ) <. The trivial measure is the measure µ with µ(e) = 0 for all E M. Clearly ay probability measure is fiite, ay fiite measure is σ-fiite ad every σ-fiite measure is semi-fiite. Example. Let µ( ) = 0 ad µ(e) = for ay oempty measurable E. The µ is a measure which is ot eve semi-fiite. Example. Legth measure o [0, ] (which, to be true, we have ot defied yet) is a probability measure. Legth measure o R is σ-fiite; take e.g. E = (, ). Whe M is a σ-algebra o X ad µ is a measure o M, the triple (X, M, µ) is called a measure space. If µ(x) =, the we may also speak of (X, M, µ) as a probability space ad if we do that, we usually refer to M-measurable sets as evets. 5
Remark. Suppose that µ(x) =. The we ca choose to call µ a probability measure ad (X, M, µ) a probability space. Whether or ot we actually do that depeds o the poit of view we wat to adopt. I may situatios it is either our mai purpose to model a radom experimet or it is istructive or useful for some other reaso to thik of the poits x X as the possible outcomes of a radom experimet. If this is ot the case, we may istead prefer to just refer to µ as a fiite measure of total mass. Some geeral properties of measures follow. I all of these, it is assumed that (X, M, µ) is a measure space. Propositio 3.4 (a) E, F M, E F µ(e) µ(f ). (b) E, E 2,... M µ( E ) µ(e ), (c) If µ(x) <, the µ(e F ) = µ(e) + µ(f ) µ(e F ), (d) If µ(x) <, E, F M ad E F, the µ(f \ E) = µ(f ) µ(e). Proof. By additivity of µ, µ(f ) = µ(e) + µ(f \ E) wheever E F. This proves (d) ad sice µ(f \ E) 0, (a) follows too. For (b), let F = E ad recursively F = E \ F j, = 2, 3,.... The the F s are disjoit ad F = E, so by (a) µ( E ) = µ(f ) µ(e ). Fially (c) follows from µ(e F ) = µ(e) + µ(f \ E F ) = µ(e) + µ(f ) µ(e F ) by additivity ad (d). Propositio 3.5 (Cotiuity of measures) (a) If E E 2... ad E = E, the µ(e) = lim µ(e ). (b) If F F 2..., F = F ad µ(f ) <, the µ(f ) = lim µ(f ). 6
Proof. For (a), let A = E ad recursively A = E \ E. The E = A ad the A s are disjoit, so µ(e) = µ(a j ) = lim µ(a j ) = lim µ(e ) sice E = A j. Now (b) follows from applyig (a) to E = F \ F ad E = F \ F ad usig Propositio 3.4(d). Corollary 3.6 If µ(n ) = 0 for all, the µ( N ) = 0. Proof. Apply e.g. Propositio 3.4(b). 3.4 Almost everywhere ad completeess Let S be a propositio about poits of X ad suppose that F = {x : S(x) is false} is measurable. If µ(f ) = 0, the S is said to hold almost everywhere (with respect to µ if other measures are also uder discussio), abbreviated a.e. I case µ is a probability measure, oe ofte istead says that S holds almost surely, abbreviated a.s. If S holds a.e. ad T is aother propositio such that T (x) is true wheever S is true, the oe would clearly wat to thik of T as also holdig a.e. However this is ot so i geeral, sice eve if µ(f ) = 0, it may be the case that some subset E of F is ot measurable. If (X, M, µ) is such that E M wheever E F, F M ad µ(f ) = 0, the the measure space is said to be complete ad µ is said to be a complete measure. If µ is ot complete, the oe ca always exted the measure space, by defiig the larger σ-algebra M = {E F : E M, N M : F N, µ(n) = 0} (exercise: prove that M is a σ-algebra) ad the measure µ o M by µ(e F ) = µ(e). The (X, M, µ) is complete ad µ is called the completio of µ. 3.5 Dyki s Lemma ad the Uiqueess Theorem Dyki s Lemma will be a fudametal tool for theorem provig. It is based o the cocepts of π-systems ad d-systems. A π-system is a class I of subsets of X 7
that is closed uder fiite itersectios, i.e. E F I wheever E, F I. The defiitio of a d-system follows. Defiitio 3.7 Let D be a class of subsets of X. The D is said to be d-system if (a) X D, (b) E, F D, E F F \ E D, (c) E D, E E E D. Geerated d-systems are defied aalogously with geerated σ-algebras: d(c) = {D C : D d-system}s. (Check that ay itersectio of d-systems is a d-system.) Theorem 3.8 Let M be a class of subsets of X. The M is a σ-algebra if ad oly if it is π-system ad a d-system. Proof. The oly if-directio is obvious. The if directio follows from that X M by (a) i the defiitio of a d-system, E c = X \ E M wheever E M by (b) ad if E M, =, 2,..., the F := E j = ( Ec j) c M sice M is a π-system, so E := E j M by (c) sice F E. Sice ay σ-algebra is also a d-system, it follows that σ(c) d(c) for ay C. Dyki s Lemma provides a aswer to whe we have equality. Theorem 3.9 (Dyki s Lemma) If I is a π-system, the d(i) = σ(i). Proof. It suffices to prove that d(i) σ(i). By Theorem 3.8 it thus suffices to prove that d(i) is a π-system. I other words, it suffices to prove that D 2 := {B d(i) : B C d(i) for all C d(i)} equals d(i). The proof is doe i two similar steps. For step, defie D := {B d(i) : B C d(i) for all C I}. Sice I is a π-system, D cotais I, so if we ca show that D is a d-system, the D = d(i). Part (a) i the defiitio of a d-system obviously holds. If B, B 2 8
D ad B B 2, the for ay C I, (B 2 \B ) C = (B 2 C)\(B C) d(i) sice d(i) is a d-system. Hece part (b) holds for D. Fially if B D ad B B, the B C B C, so B D sice d(i) is a d-system. That D = d(i) meas that D 2 I, so it suffices ow to prove that D 2 is a d-system, which is ow doe i complete aalogy with step. (Check that you ca fill this i.) Our first applicatio is the followig uiqueess theorem for measures. Theorem 3.0 (Uiqueess of fiite measures) Suppose that I is a π-system ad M = σ(i). If µ ad µ 2 are two measures o M such that µ (X) = µ 2 (X) < ad µ (I) = µ 2 (I) for all I I, the µ = µ 2. Proof. By Dyki s Lemma, it suffices to prove that D := {E M : µ (E) = µ 2 (E)} is a d-system. That X D follows from the first part of the assumptio. If E, F D ad E F, the µ (F \ E) = µ (F ) µ (E) = µ 2 (F ) µ 2 (E) = µ 2 (F \ E), so F \ E D. Fially if E D ad E E, the µ (E ) = µ 2 (E ), so µ (E) = µ 2 (E) by the cotiuity of measures. Corollary 3. If two probability measures agree o I, the they are equal. 3.6 Borel-Catelli s First Lemma Defiitio 3.2 Let E, E 2,... be subsets of X. The Note that lim sup E := lim if E := m= =m m= =m E E. lim sup E = {x X : x E for ifiitely may } ad lim if E = {x X : x E for all but fiitely may }. 9
Oe sometimes writes E i.o. for lim sup E, where i.o. stads for ifiitely ofte. (There is o correspodig abbreviatio for lim if E.) Let (X, M, µ) be a measure space ad suppose that E, E 2,... M. Sice a σ-algebra is closed uder coutable itersectios ad uios, it is clear that lim sup E ad lim if E are the also measurable. Lemma 3.3 (Borel-Catelli s Lemma I) If = µ(e ) <, the µ(lim sup E ) = 0. Proof. Write F m = =m E ad F = lim sup E. The F F. Sice M = E F it follows from the cotiuity of measures (from below) ad the hypothesis that M µ(f ) = lim µ( E ) lim M M M µ(e ) = µ(e ) <. Hece the cotiuity of measures (from above) ad the hypothesis imply that µ(f ) = lim µ(f m ) µ(e ) = 0. m =m The Borel-Catelli Lemma is a importat tool, i particular i probability theory. Example. (The doublig strategy.) Assume that (X, M, P) is a probability space ad suppose that E, E 2,... are evets such that P(E ) = 2, =, 2,.... The by the Borel-Catelli Lemma, P(lim sup E ) = P(E i.o.) = 0. Oe way to describe this i words is the followig. Suppose we play a sequece of games such that at the th game we wi oe c.u. with probability 2 ad lose 2 c.u. with probability 2. Each game is fair i terms of expectatio, but by the Borel-Catelli Lemma, we will almost surely lose moey oly fiitely may times. Hece, over the whole ifiite sequece of games, we will almost surely wi a ifiite amout of moey. (I practice this strategy fails, of course, sice there are always some bouds that will set thigs up, e.g. oe ca oly play a certai umber of games i a lifetime.) 0
3.7 Carathéodory s Extesio Theorem A set fuctio µ : P(X) [0, ] is said to be a outer measure if µ ( ) = 0, µ (E) µ (F ) wheever E F, µ ( = E ) = µ (E ) for all sets E, E 2,.... If µ is a outer measure, the we say that a set A P(X) is µ -measurable if, for all E P(X), µ (E) = µ (E A) + µ (E A c ). By the defiitio of outer measure, it is immediate that the left had side is bouded by the right had side, so to prove that a give set A is µ -measurable, it suffices to show that µ (E) µ (E A) + µ (E A c ) for arbitrary E with µ (E) <. Theorem 3.4 (Carathéodory s Extesio Theorem) Let A be a algebra o X ad let µ 0 : A [0, ] be a coutably additive set fuctio. The there exists a measure µ o σ(a) such that µ(a) = µ 0 (A) for all A A. If µ 0 (X) <, the µ is the uique such measure. The uiqueess part follows immediately from Theorem 3.0. The existece part will be proved via a sequece of claims. These will also reveal some other useful facts, apart from the statemet of the theorem. Claim I. Let µ be a outer measure ad let M be the collectio of µ -measurable sets. The M is a σ-algebra. Moreover, the restrictio of µ to M is a complete measure. Proof. It is obvious that X M. From the symmetry betwee A ad A c i the defiitio of µ -measurability, it is also obvious that M is closed uder complemets. It remais to show that M is closed uder coutable uios. Suppose that A, B M ad let E be a arbitrary subset of X. The A B M sice µ (E) = µ (E A) + µ (E A c ) = µ (E A B) + µ (E A B c ) + µ (E A c B) + µ (E A c B c ) = µ (E (A B)) + µ (E (A B) c
where the last iequality follows from that A B = (A B) (A B c ) (A c B), so that the defiitio of outer measure implies that the first three terms i the middle expressio boud the first term i the last expressio, ad that (A B) c = A c B c. Moreover, if A B =, the (A B) A = A ad (A B) A c = B, so the applyig the defiitio of µ -measurability of A with E = A B gives µ (A B) = µ (A) + µ (B). I summary M is closed uder fiite uios ad µ is additive o M. Now suppose that A j M, j =, 2,... are disjoit sets. Write B = A j ad B = A j. Let E be a arbitrary subset of X. By the µ -measurability of A, µ (E B ) = µ (E B A ) + µ (E B A c ) so by iductio it follows that = µ (E A ) + µ (E B ) µ (E B ) = µ (E A j ). Above, we proved that M is closed uder fiite uios, so B M for each. Hece µ (E) = µ (E B ) + µ (E B c ) = µ (E A j ) + µ (E B c ). µ (E A j ) + µ (E B) c Lettig ad usig the defiitio of outer measure, it follows that µ (E) ( ) µ (A j ) + µ (E B c ) µ (E A j ) + µ (E B c ) = µ (E B) + µ (E B c ) µ (E). Hece all the iequalities must be equalities ad it follows that B M. This proves that M is closed uder disjoit coutable uios ad it is a easy exercise 2
to show that this etails that M is closed uder arbitrary coutable uios, i.e. M is a σ-algebra. Moreover, takig E = B gives µ (B) = µ (A j ) provig that the restrictio of µ to M is a measure. It remais to prove completeess. Assume that N M, µ (N) = 0 ad A N. The µ (A) = 0 by the defiitio of outer measure. Therefore µ (E) µ (E A) + µ (E A c ) = µ (E A c ) µ (E) provig that A M. Next assume that µ 0 is a coutably additive set fuctio o the algebra A. Defie µ : P(X) [0, ] by µ (E) = if{ µ 0 (A j ) : A j A, A j E}. () Claim II. µ is a outer measure. Proof. It is trivial that µ ( ) = 0 ad E F µ (E) µ (F ). It remais to prove coutable subadditivity. Fix ɛ > 0. If E j P(X), j =, 2,..., the for each j oe ca fid A j (k) A, k =, 2,... so that k A j(k) E j ad k µ 0(A j (k)) µ (E j ) + ɛ2 j. Sice j,k A j(k) j E j, we get µ ( j E j ) j,k µ 0 (A j (k)) j µ (E j ) + ɛ ad sice ɛ was arbitrary, µ ( j E j ) j µ (E j ) as desired. For the fial two claims, it is assumed that µ is defied by () ad M is the σ-algebra of µ -measurable sets. Claim III. µ (E) = µ 0 (A) for all E A. 3
Proof. If E A, take E = A ad E 2 = E 3 =... = i the defiitio of µ to see that µ (E) µ 0 (E). Provig the reverse iequality amouts to showig that µ 0 (A) j µ 0(A j ) wheever A j A ad j A j E. Let B = E (A \ A j. The the B s are disjoit ad B = E. By the coutable additivity of µ 0, it follows that µ 0 (E) µ 0 (B ) µ 0 (A ). Claim IV. A M. Proof. Pick A A ad arbitrary E X ad ɛ > 0. By the defiitio of µ, there exist B j A such that j B j E ad j µ 0(B j ) < µ (E) + ɛ. We get, by the additivity of µ 0 o A, µ (E) + ɛ > j µ 0 (B j A) + j µ 0 (B j A c ) µ (E A) + µ(e A c ) where the last equality follows from the defiitio of µ. Take together, these four claims prove Carathéodory s Theorem. 3.8 The Lebesgue measure ad Lebesgue-Stieltjes measures Up to ow, we have ot see ay cocrete examples of o-trivial measures. Whe X is a coutable space, X = {x, x 2,...}, the it is easy to costruct such measures. Take e.g. M = P(X), let {w(x } = be ay collectio of oegative umbers ad let µ be defied by µ(a) = x A w(x). We have also see that for X = (0, ] ad M = P(X), o sesible legth measure exists. We are ow equipped with the tools eeded to costruct a proper legth measure o R. Sice it is ot possible to do this for all subsets, we have to settle for a smaller σ- algebra. Clearly sets of the form costructed i Sectio 2 via the axiom of choice, are uatural to expect to be able to measure i terms of legth. O the other had, ay sesible legth measure must be able to measure the legth of a iterval. If we could also measure the legth of ay set that ca be costructed from a coutable umber of set operatios o itervals, the it is difficult eough to come up with a example of a set which would ot have a legth (such as the set A i 4
Sectio 2) ad eve harder to motivate why oe would eve wish to give such a set a legth if doig so causes problems. This poit of view is what we are goig to adopt. Now recall that the Borel σ-algebra is the σ-algebra geerated by all itervals ad hece, by virtue of beig a σ-algebra, cotais all sets we wish to assig a legth to. Hece the aim is to costruct a legth measure o B(R). It turs out to be slightly more comfortable to restrict to (0, ] ad B(0, ]. Havig doe so, we obviously also have legth measures o (, + ] for all Z by traslatio ad ca exted to the whole real lie by lettig, for E B(R), defiig the legth of E be the sum of the legths of E (, + ], Z. Let X = (0, ] ad let A be the algebra cosistig of fiite disjoit uios of itervals of the type (a, b], 0 a b. Hece ay A A ca be writte as (a j, b j ] for some Z + ad the (a j, b j ] s disjoit. Defie µ 0 : A [0, ] by µ 0 ( (a j, b j ]) = (b j a j ). Clearly the legth of ay set i A must be give by µ 0 (A), so we would like to exted µ to a measure o B(0, ] = σ(a). By Carathéodory s Extesio Theorem, there is a uique such extesio, provided that µ 0 is a coutably additive set fuctio o A. It is trivial that µ 0 ( ) = 0 ad that µ 0 is additive, but coutable additivity is ot so clear. It must be proved that µ 0 ( A ) = µ 0(A ) wheever A, A 2 are disjoit sets i A ad A. Sice µ 0 is fiitely additive, we may assume without loss of geerality that the A s ad A cosist of a sigle iterval: A = (a, b ] ad A = (a, b]. O oe had, by fiite additivity, µ 0 (A) = µ 0 (A \ A j ) + µ 0 ( ( ) A j ) µ 0 A j = µ 0 (A j ) for every, so lettig gives µ 0 (A) µ 0 (A j ). Now we focus o the reverse iequality. Fix ɛ > 0. The sets (a ɛ2, b ) form a ope cover of the set compact set [a, b ɛ] ad ca hece be reduced to a fiite subcover (a ɛ2, b ), =,..., N. Let c = a ɛ2 ad assume 5
without loss of geerality that c c 2... c N (otherwise just reorder). We may also assume without loss of geerality that b b 2... b N, otherwise discard those itervals that are cotaied i oe of the others; this caot icrease N (b j a j ). The, sice b j a j+ for all i =,..., N, b ɛ a b N c Hece N (b j c j ) µ 0 (A) = b a N (b j a j + ɛ2 j ) µ 0 (A j ) + 2ɛ. (b j a j ) + ɛ. This establishes that µ 0 is coutably additive. Hece µ 0 exteds to a uique legth measure µ o B(0, ]. This measure is kow as the Lebesgue measure ad the otatio we will use for it is m. Lookig back o the proof of Carathéodory s Extesio Theorem, we fid that for sets E B(0, ] that are ot i A, m(e) is explicitly expressed i terms of µ 0 by µ (E) = if{ µ 0 (A ) : A A, A E} (2) ad m the restrictio of the outer measure µ to B(0, ]. Moreover, we recall that µ 0 actually exteds to a complete measure o the σ-algebra M of µ -measurable sets. This σ-algebra cotais A ad hece B(0, ], but othig says that it could ot be larger. Ideed, it turs out that M equals the completio of B(0, ] with respect to m ad that this σ-algebra is strictly larger tha the Borel σ-algebra. The larger σ-algebra M is called the Lebesgue σ-algebra, deoted L(0, ]. Sice this extesio comes at o extra cost, it will be assumed throughout that the Lebesgue measure is the complete measure defied o L(0, ], uless otherwise stated. The costructio of the Lebesgue measure ca easily be geeralized i the followig way. Let F : R R be a o-decreasig right-cotiuous fuctio. Redefie the µ 0 above by µ 0,F ( A j ) = (F (b j ) F (a j )). A aalogous argumet shows that µ 0 is coutably additive o A ad hece exteds to a uique measure µ F o B(R). For sets E B(R) \ A, (2) becomes µ F (E) = if{ µ 0,F (A ) : A A, A E} (3) 6
ad µ F the restrictio of µ F to B(R). As for the Lebesgue measure, the σ-algebra M F of µ F -measurable sets is strictly larger tha B(R) ad the restrictio of µ F to M F coicides with the completio of µ F. I aalogy with the Lebesque measure, we will heceforth take the otatio µ F to deote this completio uless otherwise stated. The measure µ F thus costructed is called the Lebesgue-Stieltjes measure associated to F. From (3) it follows (exercise!) that a Lebesgue-Stieltjes measure satisfies the followig regularity properties, called outer regularity ad ier regularity respectively. Propositio 3.5 For all E M F, µ F (E) = if{µ F (U) : U ope, U E} = sup{µ F (K) : K compact, K E}. Aother property i the same vei is the followig. Propositio 3.6 For all E M F ad ɛ > 0, there exists a set A, which is a fiite uio of ope itervals, such that 3.9 The Cator Set µ F (A E) < ɛ. For ay x R, we have m({x}) = 0, so for ay coutable subset E R, m(e) = 0. Does the reverse implicatio also hold? I.e. are coutable sets the oly oes to have Lebesgue measure 0? The aswer is o. The most well-kow example is the Cator set. It is costructed the followig way. Let for =, 2,..., D = 3 j=0 ((3j + )3, (3j + 2)3 ). Let C = [0, ] \ D ad recursively C = C \ D. Let C = C. The set C is the Cator set. I words, the process is the followig. Start with the closed uit iterval with the ope mid third removed; this is C. From the two closed itervals that make up C, remove from each of them the ope mid third to get C 2. Now C 2 is the uio of four closed itervals. Remove from each of these the ope mid third to get C 3, 7
etc. The Cator set is the limitig set of this process. Clearly m(c ) = (2/3), so by the cotiuity of measures m(c) = 0. O the other had, C has the same cardiality as (0, ]. To see this, write each umber x [0, ] by its triary expasio: x = a (x)3 = where a (x) {0,, 2}. The expasio is uique for all x except those that are of the type x = j3, j Z +, for which oe ca either choose a expasio edig with a ifiite sequece of 0 s or oe edig with a ifiite sequece of 2 s. I such cases, we pick the latter expasio. The C = {x {0, } : a (x) {0, 2} for each }. Hece, by mappig each 2 to, we see that C is i a --correspodece with the set of all biary expasios b 2, i.e. with (0, ]. 4 Measurable fuctios / radom variables Let (X, M, µ) be a measure space ad let (Y, N ) be a measurable space. Defiitio 4. A fuctio f : X Y is said to be (M, N )-measurable if f (A) M for all A N. So f is (M, N )-measurable if {x X : f(x) A} is M-measurable wheever A is N -measurable. I words, this could be phrased as that f is measurable if statemets that make sese i terms of the values of f also make sese i terms of the values of x. See the probabilistic iterpretatio of this i the example below. Whe oe of the σ-algebras is uderstood, we may speak of f as simply M- measurable or N -measurable ad if M ad N are both uderstood, we may speak of f as simply measurable. If (X, M, µ) is a probability space, a (M, N )- measurable fuctio is usually called a (Y -valued) radom variable. Example. Let (X, M, P) be a probability space ad suppose Y = (R, B(R)). Let ξ : X R be a radom variable. This meas that ξ is a (M, B(R))-measurable fuctio, i.e. ξ (B) = {x X : ξ(x) B} M 8
wheever B B(R). Hece P(ξ (B)) = P(ξ B) is defied for all Borel sets B. I.e. measurability meas that it makes sese to speak of the probability that ξ belogs to B for ay give Borel set B. Clearly the compositio of two measurable fuctios is measurable. More specifically, if (Z, O) is a third measurable space, f : X Y is (M, N )- measurable ad g : Y Z is (N, O)-measurable, the, sice (g f) (A) = f (g (A)), g f is (M, O)-measurable. The followig result is a idispesable tool for provig that a give fuctio is measurable. Theorem 4.2 Let E be a class of subsets of Y ad assume that N = σ(e). The f : X Y is measurable if ad oly if f (A) M for all A E. Proof. The oly if directio is trivial. Let F = {A N : f (A) M}. Sice F E, it suffices to show that F is a σ-algebra. The key is the to recall that f commutes as a operator with the basic set operatios, i.e. f (A c ) = f (A) c ad ( ) f A α = ( ) f (A α ), f A α = f (A α ) α α α for all A ad A α ad α ragig over arbitrary idex sets. Hece X = f (Y ) ad X M (sice M is a σ-algebra), so Y F, A F f (A) M f (A) c M f (A c ) M A c F, A F, =, 2,... f (A ) M f (A ) M f ( A ) M A F. α Corollary 4.3 If X ad Y are topological spaces ad M ad N are the Borel σ-algebras, the ay cotiuous fuctio is measurable. Proof. Let f be cotiuous ad let T be the topology (i.e. the family of ope sets) of Y. By the defiitio of cotiuity, f (U) is ope for all U T ad hece measurable by the defiitio of the Borel σ-algebra o X. Sice B(Y ) = σ(t ) a applicatio of Theorem 4.2 with E = T gives the result. 9
Corollary 4.4 A map f : X R is (Borel)-measurable i either of the followig cases f [, a] M for all a R, f [, a) M for all a R, f [a, ] M for all a R, f (a, ] M for all a R. Sice either of the four classes geerate B(R), the proofs follow o mimickig the proof of Corollary 4.3. Of course aalogous statemets are valid if R is replaced with R, R + or R +. Example. Let X be the sample space of a radom experimet. The ξ : X R is a radom variable iff {ξ a} is a evet for all a R. This is sometimes take as the defiitio of a radom variable i courses which wat to preset the ecessary fudametals without ivolvig uecessary measure-theoretic detail. Theorem 4.5 Let f, g : X R be measurable ad λ R a costat. The f + g, λf ad fg are all measurable fuctios. The same is true for /f provided that f(x) > 0 for all x X. Proof. We do f + g ad leave the other cases as exercises. By Corollary 4.4 it suffices to show that {x : f(x) + g(x) < a} M for all a R. However {x : f(x) + g(x) < a} = ( ) {x : f(x) < q} {x : g(x) < a q} M q Q sice Q is coutable ad f ad g are measurable. Theorem 4.6 Assume that f, f 2,... are measurable. The sup f, if f, lim sup f ad lim if f are measurable. Moreover, the set {x : lim f (x), exists} is measurable ad if lim f (x) exists for all x, the lim f (x) is a measurable fuctio. 20
Proof. That sup f is measurable follows from the observatio that {x : sup f (x) a} = {x : f (x) a}, a coutable uio of measurable sets. Sice costat fuctios are trivially measurable, we get that if f = 0 sup ( f ) is measurable. Sice lim sup f = if m sup m f ad lim if f = sup m if m f, these are the also measurable. If lim f (x) exists for all x, the lim f = lim if f = lim sup f ad is hece measurable. Fially {x : lim f (x) exists} = {x : lim sup (x) lim if(x) = 0} is measurable by Theorem 4.5 (sice {0} B(R)). Example. Costructio of a uiform radom variable. Let (X, M, P) = ([0, ], B, m) ad ξ(x) = x, x X. The ξ is cotiuous ad hece a radom variable ad P(ξ a) = m{x : ξ(x) x} = m{x : x a} = m[0, a] = a. Example. Costructio of a radom variable with give distributio. Assume that F : R R is o-decreasig ad right cotiuous with lim F (x) = 0, lim F (x) =. x x We wat to costruct a radom variable ξ so that P(ξ a) = F (a). Recall the Lebesque-Stieltjes measure µ F. The coditios o F imply that µ F is a probability measure, so let (X, M, P) = (R, B, µ F ) ad ξ(x) = x, x R. The P(ξ a) = µ F (, a] = F (a). A alterative costructio is the followig, which is most coveietly described i the case whe F is cotiuous ad strictly icreasig. The F exists, so we ca take (X, M, P) = ([0, ], B, m) ad ξ(x) = F (x) ad get P(ξ a) = m{x : F (x) a} = m[0, F (a)] = F (a). I the geeral case, oe ca replace F with the geeralized iverse, which maps all poits i [F (x ), F (x+)] to x ad poits y [0, ] for which F ({y}) is a 2
iterval, which must have the form [c, d) or [c, d] sice F is right cotiuous, to c. Example. Costructio of a sequece of uiform radom variables. Agai take (X, M, P) = ([0, ], B, m). Represet each x [0, ] with its biary expasio x = a (x)2. Each a (x) is a {0, }-valued measurable fuctio of x, sice a ({}) is a uio of 2 itervals (of legth 2 ). Let { ij } j=, i =, 2,... be disjoit sequeces ad let ξ i (x) = a ij 2 j. j= The ξ is measurable for each i by Theorems 4.5 ad 4.6 (why do we eed them both?) ad clearly P(ξ i a) = a as i the first of the previous examples. Example. Costructio of a sequece of fair coi flips. With the same settig as i the previous example, let simply ξ i (x) = a i (x). We ed this sectio with a few otes o completeess. Suppose that g is M-measurable ad that f = g a.e. If µ is complete, the this implies that f is measurable. However if µ is ot complete, the this may ot be the case. O the other had, by the costructio of the completio µ of µ, it is clear that f is M- measurable. Similarly, if µ is complete, f, f 2,... measurable ad f f a.e., the f is measurable. (These facts make up Propositio 2. i Follad.) Vice versa, if f is M-measurable, the there exists a M-measurable fuctio such that f = g µ-a.e. (This last fact is Propositio 2.2 i Follad.) 4. Product-σ-algebras ad complex measurable fuctios Let (Y, N ) be a measurable space ad f : X Y. The the σ-algebra o X geerated by f is give by σ(f) := σ{f (A) : A N }. I other words, σ(f) is the smallest σ-algebra o X that makes f measurable. (I fact {f (A) : A N } is a σ-algebra (prove this!), so σ(f) equals this set.) 22
More geerally, if F is a family of fuctios from X to Y, the σ(f) := σ{f (A) : f F, A N }. Now let (X, M ) ad (X 2, M 2 ) be two measurable spaces. The projectio maps π ad π 2 are give by i =, 2. π i : X X 2 X i, π i (x, x 2 ) = x i Defiitio 4.7 The product σ-algebra of M ad M 2 is give by More geerally M M 2 := σ(π, π 2 ) = σ{e E 2 : E i M i, i =, 2}. M = σ{π : =, 2,...} = σ{ E : E M } ad for a geeral idex set I M α = σ{π α : α I} α I = σ{ α I E α : E α Mα ad E α = X α for all but coutably may α}. Make sure that you uderstad the equalities i the defiitios. Propositio 4.8 Let (X, M) ad (Y α, N α ), α I, be measurable spaces. A map h = (f α ) α I : X α I Y α is (M, α I N α)-measurable if ad oly if each f α is (M, N α )-measurable. Proof. Sice f α = π α h, a compositio of two measurable maps, the oly if directio holds. O the other had, if all f α are measurable, the for ay α ad A N α, h (πα (A)) = (π α h) (A) = fα (A) M. Sice α N a is geerated by π α, α I, the if directio ow follows from Theorem 4.2. 23
Propositio 4.9 B(R 2 ) = B(R) B(R). Proof. Let A = {(a, b ) (a 2, b 2 ) : a, b, a 2, b 2 Q}. Sice ay ope set i R 2 ca be writte as a coutable uio of sets i A, we have B(R 2 ) = σ(a). By defiitio B(R) B(R) cotais A ad hece B(R) B(R) B(R 2 ). O the other had, B(R) B(R) is geerated by π (A), A B(R), i =, 2. We have π (A) = A R, so it suffices to show that A R B(R 2 ) for every A B(R). (The similar statemet for π 2 is of course aalogous.) Sice A R is ope i R 2 wheever A is ope i R, this holds for all ope A. Hece, the family {A B(R) : A R B(R 2 )} cotais all ope sets, so if we ca show that it is also a σ-algebra, we are doe. This, however, is obvious. Two immediate corollaries follow. Corollary 4.0 B(C) = B(R) B(R). Corollary 4. A fuctio f : X C is (M, B(C))-measurable if ad oly if Rf ad If are both measurable. 4.2 Idepedet radom variables I the ext sectios (X, M, P) will be a probability space. Defiitio 4.2 Let I be a arbitrary set ad let E α, α I, be subclasses of M. We say that {E α } α I is idepedet if P( j J E j ) = j J P(E j ) for all fiite J I ad all E j E j, j J. The family of radom variables {ξ α } α I said to be idepedet if {σ(ξ α )} α I is idepedet. The family of evets {E α } α I, is said to be idepedet if {χ Eα } α I is idepedet. The give defiitio is completely geeral i terms of the idex set I. Although havig I ucoutable ca be useful sometimes, e.g. whe defiig Gaussia white oise, it will ot be so here, so i the sequel I will be either fiite or coutably ifiite. 24
Lemma 4.3 Assume that I, J M are two π-systems ad let N = σ(i) ad O = σ(j ). The {N, O} is idepedet if ad oly if {I, J } is idepedet. Proof. The oly if directio is trivial. The if directio will be proved by a two-step procedure. First fix arbitrary I I ad defie two measures o O by, for each B O, settig µ (B) = P(I B) µ 2 (B) = P(I)P(B) By hypothesis µ ad µ 2 agree o J ad µ (X) = µ 2 (X) <, so by the Uiqueess Theorem for measures, µ = µ 2. Next fix arbitrary B O ad defie two measures o N by settig, for each A N, µ 3 (A) = P(A B) µ 4 (A) = P(A)P(B). By what we just proved, µ 3 ad µ 4 agree o I. They are also fiite ad agree o X, ad are hece equal. This proves idepedece. Clearly Lemma 4.3 exteds to all fiite collectios of π-systems ad their geerated σ-algebras. Sice idepedece of a ifiite family of σ-algebras is equivalet to idepedece of fiite subfamilies, Lemma 4.3 also exteds to: Corollary 4.4 Let I, I 2,... M be π-systems. If {I, I 2,...} is idepedet, the also {σ(i ), σ(i 2 ),...} is idepedet. The followig two examples are importat. First observe the followig useful fact. Let f : X (Y, N ) ad suppose that E P(Y ) geerates N. The {f (E) : E E} geerates σ(f); this is so sice {E Y : f (E) σ(f)} is a σ-algebra, by the commutativity of iverse images ad basic set operatios. Example. Let ξ ad η be two radom variables. The {ξ (, a] : a R} ad {η (, b] : b R} are π-systems ad geerate σ(ξ) ad σ(η) respectively. Hece by Lemma 4.3 {ξ, η} is idepedet iff P(ξ (, a] η (, b]) = P(ξ (, a])p(η (, b]) for all a, b, i.e. if P(ξ a, η b) = P(ξ a)p(η b) for all a, b R. More geerally, by Corollary 4.4, {ξ, ξ 2,...} is idepedet iff P(ξ i a,..., ξ i a ) = P(ξ ik a k ) 25 k=
for all =, 2,..., all i <... < i ad all a,..., a R. For trivial reasos, {f(ξ), g(η)} are idepedet wheever ξ ad η are idepedet. (Check that you uderstad why!). Aalogously, if {{ξ, ξ 2,...}, {η, η 2,...}} is a idepedet pair of families of radom variables (i.e. a idepedet pair of R -valued radom variables; there is othig i the above defiitios that prevets us from cosiderig radom variables takig o values i a arbitrary space), the f(ξ, ξ 2,...) ad g(η, η 2,...) are idepedet. It is ituitively clear that if {ξ, ξ 2,...} is idepedet, the, if we extract two disjoit subfamilies, these two should make a idepedet pair of R -valued radom variables. The ext example shows that this is ideed the case. Example. Let ξ, ξ 2,... be idepedet radom variables ad let I ad J be two disjoit idex sets (i.e. I, J N ad I J = ). The {{ξ i a,..., {ξ i a } : =, 2,..., i <... < i, a,..., a R} is a π-system that geerates σ(ξ i : i I) ad the aalogous π-system geerates σ(ξ j : j J). By the previous example, the two π-systems are idepedet. Hece the collectios (ξ i : i I) ad (ξ j : j J) are idepedet, by Corollary 4.4. To relax our laguage a bit, let us take the statemet ξ, ξ 2,... are idepedet to mea that the family {ξ, ξ 2,...} is idepedet. Note that it is actually importat to spell this out, sice aother iterpretatio of the statemet could have bee that the radom variables are all pairwise idepedet. This, however, is a much weaker statemet. Cosider for example the three {0, }-valued radom variables ξ, ξ 2, ξ 3 give by P(ξ = 0, ξ 2 = 0, ξ 3 = ) = P(ξ = 0, ξ 2 =, ξ 3 = 0) = P(ξ =, ξ 2 = 0, ξ 3 = 0) = P(ξ =, ξ 2 =, ξ 3 = ) = /4, which are pairwise idepedet, but clearly ot idepedet sice ay of them is the xor sum of the other two. Hece, i the sequel, sayig that a set of radom variables are idepedet meas somethig stroger tha sayig that the same radom variables are pairwise idepedet. Theorem 4.5 (Borel-Catelli s Secod Lemma) Let E, E 2,... be a sequece of idepedet evets. If P(E ) =, the P(lim sup E ) =. Proof. Note that (lim sup E ) c = ( m m E ) c = m m E c 26
so by the cotiuity of measures, it suffices to show that P( m Ec ) = 0 for all m. This i tur follows from the followig computatios P( r E) c = lim P( E c r ) = lim r m = m ( P (E )) m r P(E) c m m e P(E) = e m P(E) = 0 Example. Let ξ, ξ 2,... be idepedet radom variables with expoetial() distributio, i.e. P(ξ > x) = e x, x 0. The ( ξ ) P log > a = e a log = a. Hece P(ξ > a log ) is fiite for a > ad ifiite for a. By the Borel-Catelli Lemmas, this etails that if a, the almost surely ξ > a log for ifiitely may, if a >, the almost surely ξ > a log for oly fiitely may. 4.3 Kolmogorov s 0--law Let ξ, ξ 2,... be idepedet radom variables. For each, let T = σ(ξ +, ξ +2,...) ad T = T. The σ-algebra T is called the tail-σ-algebra (w.r.t. ξ, ξ 2,...). A set E T is called a tail evet ad a radom variable which is T -measurable is called a tail fuctio of the ξ s. 27
A tail evet does ot, for ay, deped o the first of the ξ k s, so at a first glace it may seem that T should be trivial. This, however, would be the wrog impressio, sice T actually cotais a lot of iterestig evets. E.g. the evet {x X : lim ξ (x) exists} is a tail evet ad η = lim sup ( ) ξ k is a tail fuctio; they are T -measurable of every ad hece T -measurable. Kolmogorov s 0--law states that the probability for a tail evet must be either 0 or ad that ay tail fuctio must be a costat a.s. Theorem 4.6 (Kolmogorov s 0--law) Let ξ, ξ 2,... be idepedet radom variables. (i) If E T, the P(E) {0, }, (ii) If η is T -measurable, the there exists a costat c R such that η = c a.s. Proof. (i) Let F = σ(ξ,..., ξ ), =, 2,.... By the above example, F ad T are idepedet. Sice T T, F ad T are idepedet for every. Hece F ad T are idepedet. Sice F is a π-system, it follows that σ( F ) ad T are idepedet. However T σ{ξ, ξ 2,...) = σ( F ), so T is idepedet of itself. This meas that for each E T, P(E) = P(E E) = P(E) 2 which etails that P(E) is either 0 or. (ii) For all a R, P(η a) {0, } by (i). Let c = if{a : P(η a) = }. The ( P(η c) = P {x : η(x) c + ) } = ad ( P(η < c) = P {x : η(x) c ) } = 0. Example. (Mokey typig Shakespeare) Suppose that a mokey is typig uiform radom keys o a laptop. There are, say, N keys o the laptop. The collected works of Shakespeare (to be abbr. CWS) 28
comprises, say, M symbols. Let E be the evet that the mokey happes to type CWS evetually. Will E occur? If we let F be the evet that the mokey types CWS ifiitely may times, the by Kolmogorov s 0--law, P(F ) is 0 or. Let F be the evet that the mokey types CWS with the M + th to ( + )M th symols it types. The P(F ) = /N m, so P(F ) = ad hece P(lim sup F ) = by Borel-Catelli. Hece P(E) P(F ) P(lim sup F ) =. So the aswer is yes, the mokey will evetually type CWS (but, of course, very much provided that it has a ifiite life ad ca be persuaded to sped a ifiite amout of time at the laptop). Note that they key i the example was really Borel-Catelli s Secod Lemma ad that the iformatio provided by Kolmogorov s 0--law was oly that P(F ) {0, }. I the ext example, the 0--law plays a more vital role. Example. (Percolatio) Cosider the two-dimesioal iteger lattice, i.e. the graph obtaied by placig a vertex at each iteger poit (, k) i the Euclidea plae ad placig a edge betwee (, k) ad (m, j) if either = m ad k j = or k = j ad m =. Now remove edges at radom by lettig each edge be kept (or ope ) with probability p ad removed (or closed) with probability p, idepedetly of other edges. The resultig radom graph will of course a.s. fall ito (ifiitely may) coected compoets. However, will there be a ifiitely large coected compoet? Let E be the evet that a ifiite coected compoet exists. Let ξ i be the status, i.e. kept or removed, of edge umber i; here assume that edges are umbered accordig to their distace from the origi ad arbitrarily amog those edges that are equally far away. Now observe that E is a tail evet. This is so sice the presece or absece of ifiite compoets caot be chaged by chagig the status of the first edges o matter the value of. (For a outcome where ifiite compoets exist, chagig a fiite umber of edges ca chage the umber of such compoets, but ever chage presece/absece.) Hece, by Kolmogorov s 0--law, P(E) is 0 or. Determiig for what p we have P(E) = 0 ad for what p we have P(E) = Percolatio theory has its origis i the study of water flow through porous materials. The edges the represet microscopic chaels which may or may ot be ope for water flow. 29
is a differet story. This is of course a geeral fact about applicatios of Kolmogorov s 0--law; it tells us that a tail evet has probability 0 or, but ever tells which it is. However, kowig that P(E) is 0 or is still very helpful sice if we ca also show that P(E) > 0, the it follows immediately that P(E) =. I the percolatio settig of this example, cosider the probability that o vertex i the 2 2-box cetered at the origi, is part of a ifiite path of kept edges. It ca be show that this probability is bouded by (3( p)). (This is doe by boudig the umber of ways that the box ca be cut off from ifiity.) This is less tha for large eough if p > 2/3. Hece P(E) = for p > 2/3. O the other had, by similar coutig, it is easy to see that P(E) = 0 for p < /3. I fact, the critical probability for whe P(E) switches from 0 to is p = /2. This a cetral ad highly o-trivial fact of percolatio theory. (Whe p = /2, the P(E) = 0.) 5 Itegratio of oegative fuctios Defiig the Lebesgue itegral is a stepwise procedure. It starts with oegative simple fuctios. Defiitio 5. A fuctio φ : (X, M, µ) C is said to be simple if it is of the form φ(x) = z j χ Ej (x) for some, where z j C ad {E,..., E } is a partitio of X such that E j M for all j. Let L + (X, M, µ) deote the set of all M-measurable fuctios f : X [0, ]. Depedig o the level of risk for cofusio, we ofte use shorthad otatios such as L + (X), L + (M) or simply L +. Defiitio 5.2 Let φ = a jχ Ej, a j R + be simple. The the itegral of φ with respect to µ is give by X φ(x)dµ(x) := a j µ(e j ). 30
Example. Let (X, M, µ) = ([0, ], L, m) ad φ = χ Q [0,]. Sice Q [0, ] is coutable, it is measurable, so φ is a simple fuctio ad φdm = 0. Compare this with what happes if we try to calculate the Riema itegral of this fuctio. Sice the Riema itegral is defied i terms of approximatios of φ from above ad from below by simple fuctios that are costat itervals, we fid that the Riema itegral of φ is ot defied. Thus, there are fuctios defied o a iterval of the real lie which the Lebesgue itegral ca hadle, but which the Riema itegral caot. Later, we will also see that ay Riema itegrable fuctio o a iterval is Lebesgue itegrable ad that for such fuctios, the two methods give the same result. Alterative ad/or shorthad otatios for the itegral are φ(x)µ(dx), φdµ X ad φ. The represetatio of a simple fuctio as a fiite liear combiatio of characteristic fuctios is of course ot uique, but it is easy to see that differet represetatios give the same result, so the itegral is well-defied. For A M, write φdµ := φχ A dµ. A This is well-defied sice φχ A = a jχ A Ej + 0 χ Ac is simple. A few basic facts follow. Propositio 5.3 Let c R + ad φ = a jχ Ej, ψ = m b jχ Fj L + be simple fuctios. The (a) cφ = c φ, (b) (φ + ψ) = φ + ψ, (c) φ ψ φ ψ, (d) The map A φ, A M, is a measure. A Proof. Part (a) is trivial. For part (b) observe that φ + ψ = (a i + b j )χ Ei F j. i j Hece (φ + ψ) = i = i X (a i + b j )µ(e i F j ) = a i µ(e i F j ) + j i j j a i µ(e i ) + b j µ(f j ) = φ + ψ. j b j µ(e i F j ) i 3
For part (c) use the represetatios φ = i j a iχe i F j ad ψ = i j b jχe i F j. O each E i F j we have a i b j, so the result follows immediately from the defiitio. To prove part (d), it must be show that if A, A 2,... are disjoit sets i M, the k A k φ = k A k φ. We have k= A k φ = = ( a j µ E j ( k a j µ(e j A k ) = j= k= j= ) A k ) = j= k= k= a j µ(e j A k ) A k φ where the secod equality is coutable additivity of µ. The ext step is to defie itegrals of arbitrary fuctios i L + by approximatig with simple fuctios. The followig approximatio result tells us that it makes sese to do so. Theorem 5.4 (a) Let f L +. There are simple fuctios φ L + such that φ (x) f(x) for every x X. (b) Let f : X C be measurable. The there are simple fuctios φ such that φ φ 2... f ad φ f poitwise. Proof. I (a), let A j = {x : f(x) [j2, (j+)2 )}, j = 0,..., 2 ad let A 2 = {x : f(x) }. Sice f is measurable, all these sets are measurable, so lettig 2 φ (x) = j2 χ Aj (x) 0 gives φ s of the desired form. For (b), apply the proof of (a) to all four parts of f; see below for defiitios. I the light of Theorem 5.4, we make the followig defiitio. Defiitio 5.5 Let f L +. The { } f(x)dµ(x) := sup φ(x)dµ(x) : 0 φ f, φ simple. X X 32
For A M, A fdµ := fχ A dµ. It is obvious that if c R +, the cf = c f ad if f g, f, g L +, the f g. The ext result is oe of the key results i itegratio theory. Theorem 5.6 (The Mootoe Covergece Theorem) Assume that f, f L + ad f f poitwise. The f dµ fdµ. Proof. Sice {f } is icreasig, { f } is icreasig ad hece lim f exists (but may be equal to ). Sice f f for all, lim f f. Now pick a arbitrary simple fuctio φ L + such that φ f ad a arbitrary a (0, ). Sice f f poitwise, the sets A := {x : f (x) > aφ(x)} are icreasig i ad A = X. Sice the map A φ is a measure, it follows A from the cotiuity of measures that A φ φ. Therefore lim f a lim if φ = a φ. A Sice a was arbitrary, lettig a etails that lim f φ. The result ow follows from the defiitio of f. The first cosequece of the MCT is that the itegral is additive. Ideed, it is i fact coutably additive: Theorem 5.7 Let f L +. The ( f )dµ = f dµ. Proof. First cosider fiite additivity. By Theorem 5.4, there are simple oegative fuctios φ ad ψ such that φ f ad ψ f 2 poitwise. By the MCT ad Propositio 5.3, (f + f 2 ) = lim (φ + ψ ) = lim φ + lim ψ = f + f 2. Now fiite additivity follows by iductio. Sice N f f as N, aother applicatio of the MCT ow shows that N N ( f ) = lim ( f ) = lim f = f. N N 33
Corollary 5.8 Let f L +. The the map A fdµ, A M, is a measure. A The hypothesis i the MCT is that f f poitwise. This ca be relaxed a bit; it suffices to have f f a.e. To see this, first observe that if f = 0, the we ca fid simple φ L + with φ f poitwise ad φ = 0. However, sice φ is simple, this trivially meas that φ = 0 a.e. Now if x is a poit such that f(x) > 0, the φ (x) > 0 for all sufficietly large. Hece ( ) µ{x : f(x) > 0} µ {x : φ (x) > 0} = 0. I summary Propositio 5.9 Let f L +. The fdµ = 0 if ad oly if f = 0 a.e. Suppose ow that f f a.e. ad let E = {x : f (x) f(x)}. The f χ E fχ E poitwise so by the MCT, f χ E fχ E. Sice f fχ E L + ad f fχ E = 0 a.e., Propositio 5.9 implies that fχ E = f. From the same argumet, f χ E = f. Puttig these facts together gives f f. (This result is Corollary 2.7 i Follad.) The MCT states that if f L + ad f f a.e., the f f, but what about whe f f without beig icreasig i? Does this also imply f f? The aswer is o, as the followig example shows. Let (X, M, µ) = ([0, ], L, m) ad f (x) = χ [0,/] (x). The f 0 a.e. (but ot poitwise, sice f (0) ), but f = for every. Hece some further assumptio is eeded to guaratee that f f a.e. etails that f f. Such a coditio will be give i the Domiated Covergece Theorem below. Before that, we will exted the itegral from oegative real fuctios to geeral complex fuctios. First however, we fiish the preset sectio with the importat Fatou s Lemma ad a ote o σ-fiiteess. Theorem 5.0 (Fatou s Lemma) If f L +, =, 2,..., the (lim if f )dµ lim if f dµ. 34