Measure Theory, MA 359 Handout 1

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1 Measure Theory, M 359 Hadout 1 Valeriy Slastikov utum, Measure theory 1.1 Geeral costructio of Lebesgue measure I this sectio we will do the geeral costructio of σ-additive complete measure by extedig iitial σ-additive measure o a semi-rig to a measure o σ-algebra geerated by this semi-rig ad the completig this measure by addig to the σ-algebra all the ull sets. This sectio provides you with the essetials of the costructio ad make some parallels with the costructio o the plae. Throughout these sectio we will deal with some collectio of sets whose elemets are subsets of some fixed abstract set X. It is ot ecessary to assume ay topology o X but for simplicity you may imagie X = R. We start with some importat defiitios: Defiitio 1.1 oempty collectio of sets S is a semi-rig if 1. Empty set S; 2. If S, B S the B S; 3. If S, 1 S the = k=1 k, where k S for all 1 k ad k are disjoit sets. If the set X S the S is called semi-algebra, the set X is called a uit of the collectio of sets S. Example 1.1 The collectio S of itervals [a, b) for all a, b R form a semi-rig sice 1. empty set = [a, a) S; 2. if S ad B S the = [a, b) ad B = [c, d). Obviously the itersectio B is either empty set or a iterval. Therefore B S; 3. if S, 1 S the = [a, b) ad 1 = [c, d), where c a ad d b. Obviously you may fid two itervals [a, c) S ad [d, b) S such that [a, b) = [a, c) [c, d) [d, b). 1

2 Note that S is ot a semi-algebra sice all those itervals are subsets of R but R ca ot be represeted as a iterval of form [a, b) ( / R, but R ca be represeted as a coutable uio of such itervals). Exercise 1.1 Show that the collectio S of itervals [a, b) for all a, b [0, 1] form a semi-algebra Exercise 1.2 Let S be the collectio of rectagles i the plae (x, y) defied by oe of the iequalities of the form a x b, a < x b, a x < b, a < x < b ad oe of the iequalities of the form Show that c y d, c < y d, c y < d, c < y < d. if a, b, c, d are arbitrary umbers i R the S is a semi-rig; if a, b, c, d are arbitrary umbers i [0, 1] the S is a semi-algebra. Defiitio 1.2 oempty collectio of sets R is a rig if 1. Empty set R; 2. If R, B R the B R, B R, ad \ B R. If the set X R the R is called a algebra. Exercise 1.3 We call a set R 2 elemetary if it ca be writte, i at least oe way, as a fiite uio of disjoit rectagles from exercise 1.2. Show that the collectio of elemetary sets R form a rig. Defiitio 1.3 set fuctio µ() defied o a collectio of sets S µ is a measure if 1. Its domai of defiitio S µ is a semi-rig; 2

3 2. µ() 0 for all S µ ; 3. If 1, 2 S µ are disjoit sets ad S µ = 1 2 the µ() = µ( 1 ) + µ( 2 ) Note that µ( ) = 0 sice µ( ) = µ( ) + µ( ) Example 1.2 If we defie a set fuctio m o a semi-rig of rectagles S from exercise 1.2 like: 1. m( ) = 0; 2. If R S is a oempty rectagle (closed, ope or half ope) defied by the umbers a, b, c, d the m(r) = (b a) (d c) It is easy to check that m is a measure o S. The followig lemmas are goig to be used i the otes. The proof is ot difficult ad is left as a exercise. Lemma 1.1 The itersectio R = α R α of a arbitrary umber of rigs is a rig. Lemma 1.2 If S is a arbitrary oempty collectio of sets there exists precisely oe rig R(S) cotaiig S ad cotaied i every rig R cotaiig S. This rig R(S) is called the miimal rig over collectio S (or the rig geerated by S). Lemma 1.3 If S is a semi-rig the R(S) coicides with the collectio of sets that admit a fiite partitio = k=1 k, k S, i j =. Defiitio 1.4 measure µ is a extesio of measure m if domai of defiitio S µ of measure µ cotais domai of defiitio S m of measure m (S m S µ ) ad µ() = m() for all S m. Exercise 1.4 Take X = [0, 1] [0, 1] ad cosider the collectio S of all rectagles from exercise 1.2 that are subsets of X. Defie a measure m o S like i example 1.2. Show that it is possible to exted this measure m to a measure m o the collectio R of elemetary sets from exercise 1.3 that are subsets of X. 3

4 Note that i the above exercise the domai of defiitio of m is actually a miimal algebra R cotaiig semi-algebra S domai of defiitio of m, we write it as R m = R(S m ) (you ca also say that R m is geerated by S m ). d therefore m is actually a extesio of m from semi-algebra to a algebra. This ca be geeralized ito the followig theorem. Theorem 1.5 Every measure m() whose domai of defiitio S m is a semi-rig has uique extesio µ() whose domai of defiitio R µ is a rig geerated by S m, i.e. R µ = R(S m ). Proof For every set R µ there exists a partitio We defie = i=1b i, where B i S m (1) µ() = m(b i ) (2) i=1 The value of µ() is idepedet of the partitio (1). To see this we assume that there are two partitios of : = i=1 B i = k j=1 Q j, where B i, Q j S m ad B i B j =, Q i Q j = for i j. Sice B i Q j S m ad by additivity property of a measure m we have m(b i ) = i=1 i=1 j=1 k m(b i Q j ) = k j=1 i=1 m(b i Q j ) = k m(q j ) So µ() is well defied. Obviously µ is oegative ad additive. This takes care of existece part. Let us show its uiqueess. Suppose there are two measures µ ad µ that are extesios of m. For ay R(S m ) we have = i=1 B i, where B i S m. By defiitio of extesio µ(b i ) = µ(b i ) = m(b i ), so usig additivity of a measure: µ() = µ(b i ) = i=1 So µ ad µ coicide. Theorem is proved. µ(b i ) = µ() Note that we proved ot oly existece of a extesio but also its uiqueess. It allows us to claim that measure m defied o the collectio of elemetary sets is the oly possible extesio of the measure m defied o the collectio of all rectagles. 4 i=1 j=1

5 Defiitio 1.6 measure µ is called semiadditive (or coutably subadditive) if for ay, 1, 2,... S µ such that =1 µ() µ( ) =1 Defiitio 1.7 measure µ is called σ-additive if for ay, 1, 2,... S µ such that = =1 ad { } are disjoit µ() = µ( ) =1 Theorem 1.8 If a measure µ defied o a rig R µ is semiadditive the it is σ-additive. Proof Let, R µ for all ad = =1. For ay N N we have N =1, µ is a measure ad hece it is additive. Therefore Lettig N we obtai µ( N =1 ) = N µ( ) µ() =1 µ( ) µ() =1 O the other had, by semiadditivity we have The theorem is proved. µ( ) µ() =1 Note that here we required the domai of defiitio of µ to be a rig sice otherwise it is ot clear if µ( N =1 ) is defied. Exercise 1.5 Prove that measure m from exercise 1.4, defied o the collectio R m of elemetary sets is σ-additive. Exercise 1.6 Prove that measure m from example 1.2, defied o the collectio S m of rectagles is semiadditive. Is it σ-additive? 5

6 Theorem 1.9 If a measure m defied o a semi-rig S m is σ-additive the its extesio µ to a miimal rig R(S m ) is σ-additive. Proof ssume that R(S m ) ad B R(S m ) for = 1, 2,... are such that = =1 B ad B are disjoit. The there exist disjoit sets i S m, ad disjoit sets B j S m such that = i i ad B = j B j, where the uios are fiite. Let C ji = B j i. Obviously C ji are disjoit ad i = j C ji ad B j = i C ji. By complete additivity of m o S m we have m( i ) = m(c ji ) ad m(b j ) = j i m(c ji ). By defiitio of µ o R(S m ) we have µ() = i m( i ) ad µ(b ) = j m(b j ) Sice sums i i ad j are fiite ad the series i coverges, it is easy to see from the above equalities that µ() = m(c ji ) = m(c ji ) = µ(b ). i j j i Theorem is proved. Defiitio 1.10 oempty collectio of sets P is a σ-rig if 1. Empty set R; 2. If P, = 1, 2,... the =1 P, =1 P; 3. If, B P the \ B P. If the set X P the P is called a σ-algebra. Lemma 1.4 The itersectio P = α P α of a arbitrary umber of σ-rigs is a σ-rig 6

7 Lemma 1.5 If S is a arbitrary oempty collectio of sets there exists precisely oe σ-rig P(S) cotaiig S ad cotaied i every σ-rig P cotaiig S. This σ-rig P(S) is called the miimal σ-rig over collectio S (or the σ-rig geerated by S). Defiitio 1.11 measure µ is called fiite if for every S µ µ() <. measure µ is called σ-fiite if for every S µ there exists a sequece of sets { } S µ such that ad µ( ) <. Theorem 1.12 Every σ-additive σ-fiite measure m() whose domai of defiitio R m is a rig has uique extesio µ() whose domai of defiitio P(R m ) is a miimal σ-rig geerated by R m ad µ is σ-additive ad σ-fiite. Proof We prove this theorem for a fiite measure defied o a algebra. The existece follows from the the theorems i sectio 1.2 ad uiqueess is left as a exercise. Remark Theorems 1.5, 1.9, 1.12 tell us that if we have a σ-additive σ-fiite measure m o a semi-rig S m, there is uique extesio µ of this measure to the miimal σ-rig P(S m ) ad moreover this extesio µ is σ-additive ad σ-fiite. Therefore oe ca always start defiig the σ-additive σ-fiite measure directly o a σ-rig, ot o a semi-rig. You ca fid this approach i may textbooks. Remark Theorem 1.12 tells you that oe ca exted measure m defied o the rig R m of elemetary sets i R 2 to the uique σ-additive measure µ defied o a miimal σ-rig P(R m ). It is easy to see that P(R m ) coicides with σ-algebra of all ope sets i R 2 (or Borel algebra). Thus, startig from a measure m o rectagles oe ca costruct uique σ-additive measure µ o Borel algebra. Theorem 1.13 If is a mootoe decreasig sequece of sets i σ-rig P, µ is a σ-additive measure o P, = ad µ( 1 ) < the µ() = lim µ( ) Proof It is eough to cosider the case = sice the geeral case reduces to this o replacig by \. Now 1 = ( 1 \ 2 ) ( 2 \ 3 )..., = ( \ +1 ) ( +1 \ +2 )... 7

8 ad by σ-additivity of µ we have µ( 1 ) = µ( \ +1 ), µ( k ) = =1 µ( \ +1 ). Sice the µ( 1 ) < we have µ( k ) 0 as k. Theorem is proved. Defiitio 1.14 Let µ be a σ-additive measure defied o Borel σ-algebra of all ope sets i R. The µ is called Borel measure. Defiitio 1.15 measure µ defied o a semi-rig S µ is called complete if S µ, B ad µ() = 0 imply that B S µ (ad the µ(b) = 0). Theorem 1.16 If µ is a σ-additive measure o a σ-rig P the the class M of all sets X of the form =k = B N, (3) where B P ad N E P such that µ(e) = 0, is a σ-rig ad the set fuctio µ defied by µ() = µ(b), for all M ad B P related like i (3) is a complete σ-additive measure o M. The measure µ is called completio of µ. Proof Let us show that M is a σ-rig. 1. Obviously M, sice ca be represeted as =. 2. If i M for i = 1, 2,... the i = B i N i, where B i P ad N i E i P such that µ(e i ) = 0. We wat to check if = i i M: by defiitio = i B i i N i. Sice P is a sigma-rig B = i B i P, N = i N i i E i P ad µ( i E i ) i µ(e i) = 0. Therefore M. Now we wat to check if = i i M: by defiitio = i B i i N i. Sice P is a σ-rig B = i B i P, N = i N i E 1 P ad µ(e 1 ) = 0. Therefore M. 8

9 3. Let 1, 2 M the 1 \ 2 = (B 1 \ 2 ) (N 1 \ 2 ) = ((B 1 \B 2 )\N 2 ) (N 1 \ 2 ). Obviously B = (B 1 \B 2 ) P ad B\N 2 = (B\E 2 ) (E 2 (B\N 2 )), sice N 2 E 2. We obtai 1 \ 2 = (B\E 2 ) ((E 2 (B\N 2 )) (N 1 \ 2 )), where B\E 2 P ad (E 2 (B\N 2 )) (N 1 \ 2 ) E 2 E 1 with µ(e 1 E 2 ) = 0. So 1 \ 2 M. We showed that M is a σ-rig. It is easy to prove that if P is a σ-algebra the M is a σ-algebra. Now we have to check that the set fuctio µ is well defied, i.e. we have to show that if 1 N 1 = 2 N 2 the µ( 1 N 1 ) = µ( 2 N 2 ). By defiitio µ( 1 N 1 ) = µ( 1 ) ad µ( 2 N 2 ) = µ( 2 ), so we have to prove that µ( 1 ) = µ( 2 ). Usig the fact 1 N 1 = 2 N 2 we obtai µ( 1 ) = µ( 1 E 1 ) µ( 2 E 1 E 2 ) = µ( 2 ) µ( 2 ) = µ( 2 E 2 ) µ( 1 E 1 E 2 ) = µ( 1 ). Therefore µ is well defied. Now we have to show that µ is σ-additive measure. 1. It is obvious µ( ) = It is obvious µ() 0 for ay M. 3. If i M for i = 1, 2,..., i j = ad = i i the µ() = i µ( i ). To show this we otice that µ() = µ( i B i ) ad sice i s are disjoit the same is true for B i s (recall that i = B i N i ). Now by σ-additivity of µ we obtai µ() = µ( i B i ) = i µ(b i ) = i µ( i ). Therefore µ is a σ-additive measure. It is easy to check that it is complete: if M ad B, ad µ() = 0 the E, where E P ad µ(e) = 0. But the B = B, B E hece B M. The theorem is proved. 9

10 Defiitio 1.17 The completio of traslatio ivariat Borel measure i R is called Lebesgue measure. Remark For simplicity, we defie Borel ad Lebesgue measures i R. These defiitios may be trasferred to some topological spaces. 1.2 The extesio of a measure o a semi-algebra usig outer measure Here we are goig to itroduce the secod approach to the costructio of complete σ-additive measure. Let m be a σ-additive measure defied o a semi-algebra S m with a uit X. Defiitio 1.18 For ay set X we defie the outer measure µ () = if B m(b ), where ifimum is take over all coverigs of by coutable collectios of sets B S m. Let m be a extesio of m to a algebra R(S m ) (it exists by theorem 1.5). The we may give a equivalet defiitio of the outer measure µ. Defiitio 1.19 For ay set X we defie the outer measure µ () = if B m (B ), where ifimum is take over all coverigs of by coutable collectios of sets B R(S m ). Let µ 1 be a extesio of m to a σ-algebra P(S m ) (it exists by theorem 1.12). The we may give yet aother equivalet defiitio of the outer measure µ (we are ot goig to use this oe). Defiitio 1.20 For ay set X we defie the outer measure µ () = if B µ 1(B), where ifimum is take over all coverigs of by sets B P(S m ). It is easy to see that these three defiitios are equivalet. Let s prove some properties of µ. 10

11 Theorem 1.21 If for some coutable collectio of sets the µ () µ ( ) Proof By defiitio of µ for all ad ay ɛ > 0 there exists a coutable collectio of sets {B k } S m such that k B k ad k m(b k ) µ ( ) + ɛ 2. The k B k ad µ () m(b k ) k µ ( ) + ɛ. Takig ɛ 0 we get the result. Lemma 1.6 For ay, B X we have µ () µ (B) µ ( B). Proof Sice B ( B) ad B ( B) we have µ () µ (B) + µ ( B), µ (B) µ () + µ ( B). This implies µ () µ (B) µ ( B). The lemma is proved. It seems that µ is a very good set fuctio: we ca measure ay subset of X with it. But the bad thig about it is that µ is ot additive (i.e. µ ( B) µ () + µ (B) if B = ) ad hece it is ot a measure i the usual sese. To see this i oe particular case whe X = R we costruct Vitali set ad use this costructio to show o-additivity of µ. Example 1.3 (Vitali set) We defie the followig relatio: for x, y R we say x y if ad oly if x y Q (Q is the set of ratioal umbers). It is easy to check that 1. x x; 2. x y y x; 3. x y ad y z x z; 11

12 ad hece is equivalece relatio ad R ca be split ito disjoit equivalece classes. We defie set E as a set cotaiig exactly oe represetative from each equivalece class. Sice e ad e [e] belog to the same class we ca always choose E [0, 1]. This E is called Vitali s set. Theorem 1.22 The outer measure defied for ay R as µ () = if I i L(I ), where I R is a ope, half-ope or a closed iterval ad L(I ) is the usual legth of the iterval, is ot additive. Proof We defie a coutable set C = Q [ 1, 1] (sice C is coutable we ca say that C = {c } =1 ), the collectio of sets { } =1, where = c + E, ad = =1. Claim. 1. [0, 1] [ 1, 2]; 2. are disjoit sets. Proof Take ay x [0, 1] the there exist uique e x E ad q x Q such that x = e x + q x. (This is true by defiitio of the set E). But E [0, 1] hece q x [ 1, 1] ad therefore ay [0, 1] x q x + E for some q x C. From this it follows [0, 1]. Obviously [ 1, 2]. Let i = c i + E ad j = c j + E ad i j (c i c j ). Let s argue by cotradictio, assume i j the there exists x such that x c i + E ad x c j + E, or x = c i + y 1 = c j + y 2, where y 1, y 2 E. But the y 1 y 2 = c j c i Q ad this meas y 1 y 2. Sice E cotais exactly oe represetative from each class it follows that y 1 = y 2 which implies c i = c j. We got a cotradictio, hece i j =. Sice µ is traslatio ivariat (prove it!) we have µ (E) = µ ( ) for all. Suppose µ is additive the by semiadditivity ad additivity of µ we have σ-additivity of µ. This meas { µ () = µ if µ(e) > 0, ( ) = 0 if µ(e) = 0. =1 However [0, 1] [ 1, 2] ad therefore 1 µ () 3. cotradictio. Therefore µ is ot additive. This is a 12

13 The solutio to this problem is to restrict µ to a ice collectio of subsets where it is additive (ad therefore σ-additive). We call such subsets measurable. There are several equivalet defiitios of a measurable set. Defiitio 1.23 subset X is called measurable if for all E X, µ (E) = µ ( E) + µ ( c E). (4) Defiitio 1.24 subset X is called measurable if µ ( X) + µ ( c X) = 1. (5) Defiitio 1.25 set X is called measurable if for ay ɛ > 0 there exists a set B R(S m ) such that µ ( B) < ɛ Theorem 1.26 Defiitios 1.23, 1.24, 1.25 are equivalet. Proof Def 1.25 Def Let be measurable accordig to the defiitio Take ay ɛ > 0 the there exists a set B R(S m ) such that µ ( B) < ɛ. Sice B = c B c we have µ ( c B c ) < ɛ, where B c R(S m ). Usig lemma 1.6, for ay E X we obtai µ (E ) µ (E B) µ ((E ) (E B)) µ ( B) < ɛ, µ (E c ) µ (E B c ) µ ((E c ) (E B c )) µ ( c B c ) < ɛ. From the above iequalities we have µ (E ) + µ (E c ) µ (E B) + µ (E B c ) + 2ɛ. (6) However sice B R(S m ) the followig is true: k µ (E B) + µ (E B c ) = µ (E). Let s show this: by defiitio of µ ad sice B R(S m ) we have µ (E) = if m (B k ) = if (m (B k B) + m (B k B c )), E k B k E k B k 13 k

14 where B k R(S m ). We also have ad µ (E B) if µ (E B c ) if E B k k E B k Usig the fact if(a + b) if a + if b we obtai k m (B k B) m (B k B c ). µ (E) µ (E B) + µ (E B c ). pplyig semiadditivity of µ we have µ (E) = µ (E B) + µ (E B c ). Now usig (6) ad takig ɛ 0 we get µ (E ) + µ (E c ) µ (E). Now the result follows from semiadditivity of µ. Def 1.23 Def This oe is obvious. Def 1.24 Def Let be measurable accordig to the defiitio 1.24, i.e. µ () + µ (X\) = 1. For ay ɛ > 0 there exist sets {B } R(S m ) ad {C } R(S m ) such that B, X\ C ad m (B ) µ () + ɛ, m (C ) µ (X\) + ɛ. Sice m (B ) < (as µ () 1) there is N N such that =N+1 m (B ) < ɛ. We defie B = N =1 B R(S m ) ad wat to show that µ ( B) < 3ɛ. It is easy to check that B P Q, 14

15 where P = =N+1 B ad Q = (B C ). Obviously µ (P ) m (B ) < ɛ. =N+1 Let us estimate µ (Q). It is easy to see that ( B ) ( (C \B)) = X ad hece 1 m (B ) + m (C \B). By defiitio of B ad C we have m (B ) + m (C ) µ () + µ (X\) + 2ɛ = 1 + 2ɛ ad therefore m (C B) = m (C ) m (C \B) < 2ɛ. This implies µ (Q) < 2ɛ ad µ ( B) µ (P ) + µ (Q) < 3ɛ. This proves the result. Remark Not all sets are measurable. Vitali set, which is used to costruct a sequece of subsets of R o which µ is ot σ-additive, is a example of a omeasurable set. Defiitio 1.27 The set fuctio µ is defied o the collectio of all measurable sets M by µ() = µ () for all M. Note that we do t kow yet that µ is a measure. Let us ivestigate the properties of measurable sets ad µ. Theorem 1.28 The collectio M of all measurable sets is a algebra. Proof Let 1 ad 2 be measurable sets the for ay ɛ > 0 there exist B 1, B 2 R(S m ) such that µ ( 1 B 1 ) < ɛ 2, µ ( 2 B 2 ) < ɛ 2. 15

16 Usig the relatio ( 1 2 ) (B 1 B 2 ) ( 1 B 1 ) ( 2 B 2 ) ad the fact that B 1 B 2 R(S m ) we obtai µ (( 1 2 ) (B 1 B 2 )) µ ( 1 B 1 ) + µ ( 2 B 2 ) < ɛ, therefore 1 2 is measurable. Usig the relatio ( 1 \ 2 ) (B 1 \B 2 ) ( 1 B 1 ) ( 2 B 2 ) ad the fact that B 1 \B 2 R(S m ) we obtai 1 \ 2 is measurable. The theorem is proved. Theorem 1.29 The fuctio µ() is σ-additive o the collectio M of measurable sets Proof First we show additivity of µ o M. Let 1, 2 M ad 1 2 =. For ay ɛ > 0 there exist B 1, B 2 R(S m ) such that µ ( 1 B 1 ) < ɛ 2, µ ( 2 B 2 ) < ɛ 2. Defie = 1 2 M ad B = B 1 B 2. It is easy to show that B 1 B 2 ( 1 B 1 ) ( 2 B 2 ) ad therefore m (B 1 B 2 ) < ɛ. By lemma 1.6 we have m (B 1 ) µ ( 1 ) < ɛ 2, m (B 2 ) µ ( 2 ) < ɛ 2. Sice m is additive o R(S m ) we obtai m (B) = m (B 1 ) + m (B 2 ) m (B 1 B 2 ) µ ( 1 ) + µ ( 2 ) 2ɛ. Notig that B ( 1 B 1 ) ( 2 B 2 ) ad usig semiadditivity of µ we have µ () m (B) µ ( B) m (B) ɛ µ ( 1 ) + µ ( 2 ) 3ɛ. Sice ɛ > is arbitrary we have µ () µ ( 1 ) + µ ( 2 ). 16

17 Usig semiadditivity of µ ad the fact that 1, 2, M we obtai µ() = µ( 1 ) + µ( 2 ). ad hece µ is additive. Usig theorem 1.8 ad the fact that µ is semiadditive o M (sice o M it coicides with µ ad µ is semiadditive) we obtai the result. Now we kow that µ is a σ-additive measure. Theorem 1.30 The collectio M of all measurable sets is a σ-algebra. Proof Let i M for i = 1, 2,... ad = i=1 i. Defie = \ 1 i=1 i. It is clear that are measurable (by theorem 1.28), disjoit ad = =1. By theorem 1.29 we have: for all N N N µ( ) = µ( N =1 ) µ(). =1 Therefore the series =1 µ( ) coverges ad for ay ɛ > 0 there exists M N such that =M µ( ) < ɛ 2. The set C = M =1 M ad hece there exist B R(S m ) such that µ (C B) < ɛ 2. Sice we obtai B (C B) ( =M ) µ ( B) < ɛ ad hece is measurable. Sice M is a algebra the theorem is proved. Theorem 1.31 Measure µ is complete. Proof Let M, B ad µ() = 0, the µ (B ) µ ( ) = µ() = 0. Sice R(S m ) we obtai B M. The theorem is proved. We showed that the extesio µ of a measure m from a semi-algebra S m to the σ-algebra M S m of all measurable sets coicidig o M with the outer measure µ is complete σ-additive measure. It seems that we costructed oe complete measure µ i sectio 1.1, theorem 1.16 ad aother measure µ = µ M here. I fact these two measures coicide. 17

18 Theorem 1.32 If m is a σ-additive σ-fiite measure o a rig R ad if µ is the outer measure iduced by m the the completio of the extesio of m to the σ-algebra P(R) is idetical with restrictio of µ to the class of all µ measurable sets. Proof The proof of this theorem is left as a exercise. Problems 1. Let 1, 2,... be a icreasig sequece of subsets of X, i. e., j j+1 j N. Suppose that j is µ-measurable for all j N ad prove that µ( j=1 j) = lim j µ( j ). 2. Let be a Lebesgue measurable subset of R. Prove that, for each ε > 0, there exists a ope subset E ε of R such that E ε ad µ(e ε \ ) < ε. 3. subset of R is called a rectagle if it is a product of itervals, i.e. R R is a rectagle if there exist itervals I 1, I 2,..., I R such that R = I 1 I 2... I. Prove that every ope subset of R ca be writte as a coutable uio of ope rectagles. Deduce that the ope subsets ad closed subsets of R are all measurable. 4. i. Let be a Lebesgue measurable subset of R. Prove that, give ε > 0, there exists a closed subset F ε of R such that F ε ad µ( \ F ε ) < ε. ii. Let B be a subset of R with the property that, for each ε > 0, there exists a ope subset E ε of R such that B E ε ad µ (E ε \ B) < ε. Prove that B is Lebesgue measurable. 5. Give a sequece of subsets E 1, E 2,... of a set X we defie lim sup E j := E k ad lim if E j := E k. j=1 k=j 18 j=1 k=j

19 Note that lim sup E j is the set of poits which belog to E j for ifiitely may values of j. Suppose that E j is µ-measurable for all j N ad prove that µ(lim if E j ) lim if µ(e j ). 6. Let U be a ope subset of R. For each x U, let a x := if{a R (a, x) U}, b x := sup{b R (x, b) U}, I x := (a x, b x ). Prove that x U I x U ad that, if x, y U ad I x I y the I x = I y. Deduce that every ope subset of R is a coutable uio of disjoit ope itervals. 7. Give λ R ad R, let +λ := {x + λ x }. i. Prove that, for all R ad for all λ R, µ ( +λ ) = µ () (µ is the outer measure from the theorem 1.22). ii. Prove that if R is Lebesgue measurable ad λ R the +λ is Lebesgue measurable. 19

20 2 Measurable fuctios First we give some geeral defiitios. Defiitio 2.1 (X, M) is called a measurable space if X is some set, M is a σ-algebra o X. Defiitio 2.2 The triple (X, M, µ) is called a measure space if X is some set, M is a σ-algebra of subsets of X ad µ is a o M. Defiitio 2.3 fuctio f : X R is called µ-measurable (or just measurable) if f 1 () M for ay Borel set o R. Our mai iterest i measurable fuctios lies i the theory of Lebesgue itegratio. Therefore throughout the rest of the lecture otes X R, M is Borel algebra with all ull sets ad µ is Lebesgue measure, although the theory remais true for geeral measure spaces. Propositio 2.4 Fuctio f : X R is measurable if ad oly if for ay c R set {x X : f(x) < c} is measurable. Proof Necessity is obvious sice (, c) is Borel set ad hece measurable. Sufficiecy: It is ot difficult to show that σ-algebra created by sets (, c), where c R, coicides with Borel σ-algebra o R. If {x X : f(x) < c} is measurable for all c R the f 1 (, c) M (by defiitio of iverse image). From this it follows that P(f 1 (, c)) M ad therefore f 1 (P((, c))) M. Exercise 2.1 I the theorem we have used the fact that if is a collectio of sets the P(f 1 ()) = f 1 (P()). Prove it. Propositio 2.5 Let f : X R be some fuctio. The followig statemets are equivalet 1. {x X : f(x) < c} M for ay c R; 2. {x X : f(x) c} M for ay c R; 20

21 3. {x X : f(x) > c} M for ay c R; 4. {x X : f(x) c} M for ay c R; Proof Sice M is a σ-algebra it is easy to see that statemets 1 ad 2 are equivalet ad statemets 3 ad 4 are equivalet. Usig the facts that {x X : f(x) c} = =1 {x X : f(x) > a 1 } ad {x X : f(x) < c} = =1{x X : f(x) a 1 } We have the result. Now to fid out if the fuctio is measurable we just have to check either of poits 1 4. We also wat to kow what kid of operatio we may do with measurable fuctios that the resultig fuctio is also measurable. For istace, we wat to kow if sum, product, e.t.c of measurable fuctios is measurable. Lemma 2.1 Let f : X R be µ-measurable ad φ : R R be Borel measurable. The φ(f(x)) is µ-measurable. Proof Let g(x) = φ(f(x)) ad R be a arbitrary Borel set. The φ 1 () is Borel set sice φ is Borel measurable ad g 1 () = f 1 (φ 1 ()) is µ-measurable. Lemma is proved. Theorem 2.6 Let f : X R ad g : X R be measurable fuctios. The f + g, f g, fg, f g (if g(x) 0), max(f, g) ad mi(f, g) are measurable fuctios. Proof It is obvious that if f is measurable fuctio the so are cf ad f(x)+c for ay c R. If f ad g are measurable fuctios we show that set {x X : f(x) > g(x)} is measurable. Ideed, take {r k } k=1 - the sequece of all ratioal umbers (we ca do it sice ratioal umbers are coutable). The {x X : f(x) > g(x)} = k=1 ({x X : f(x) > r k} {x X : r k > g(x)}). Therefore we have that set {x X : f(x) > g(x) + c} is measurable. Hece we obtai f + g is a measurable fuctio. 21

22 To show that fg is measurable we use the followig idetity fg = 1 4 ((f + g)2 (f g) 2 ) Usig f + g, f g are measurable fuctios ad the fact that cotiuous fuctio of a measurable fuctio is itself measurable we coclude the proof. The rest of the proof is left as a exercise. Theorem 2.7 Let {f } =1 be a sequece of measurable fuctios. The sup f (x), if f (x), lim sup f (x) ad lim if f (x) are measurable fuctios. Proof Let g(x) = sup f (x) the for ay c R we have {x X : g(x) > c} = {x X : f (x) > c} ad hece g(x) is a measurable fuctio. Let g(x) = if f (x) the for ay c R we have {x X : g(x) < c} = {x X : f (x) < c} ad hece g(x) is a measurable fuctio. By defiitio we have lim sup f (x) = if sup f (x) k k ad lim if f (x) = sup if f (x) k k hece result follows from previous argumets.. Exercise 2.2 From this theorem it is easy to deduce that if {f } =1 is a sequece of measurable fuctios that coverges poitwise to a fuctio f(x) the f(x) is a measurable fuctio. Do it. We did ot use aythig about completeess of our measure yet. Now is the time. Defiitio 2.8 Fuctios f : X R ad g : X R are equivalet (f g) if µ({x X : g(x) f(x)}) = 0. Propositio 2.9 fuctio f : X R equivalet to some measurable fuctio g : X R is measurable itself. Proof By defiitio of equivalece sets {x X : f(x) c} ad {x X : g(x) c} may differ just by some ull set ad hece if oe is measurable the other is measurable as well. 22

23 2.1 Covergece of measurable fuctios I this sectio we defie some types of covergeces of fuctio sequeces o the space (X, µ). Defiitio 2.10 sequece of measurable fuctios {f (x)} =1, defied o (X, µ) is called coverget almost everywhere to f(x) (f (x) f(x) a.e. X) if µ({x X : lim f (x) f(x)}) = 0 Defiitio 2.11 sequece of measurable fuctios {f (x)} =1, defied o (X, µ) is called coverget i measure to f(x) (f (x) µ f(x) a.e. X) if for every δ > 0 lim µ({x X : f (x) f(x) δ}) = 0 Propositio 2.12 If a sequece of measurable fuctios {f (x)} =1 coverges almost everywhere to a fuctio f(x) the f(x) is also a measurable fuctio Proof The proof is left as a exercise. Let us first prove the theorem that relates the otio of covergece a.e. ad uiform covergece. Theorem 2.13 (Egoroff) Suppose that a sequece of measurable fuctios {f (x)} =1 coverges a.e to f(x) o X (µ(x) < ). The for every δ > 0 there exists a measurable set X δ X such that 1. µ(x δ ) > µ(x) δ; 2. the sequece f (x) coverges to f(x) uiformly o X δ. Proof Obviously f(x) is measurable. We defie X m = i {x X : f i (x) f(x) < 1 m } ad X m = =1 Xm. By defiitio of X m we see that Xm 1 X2 m... X m... By cotiuity of measure we have: for ay m ad ay δ > 0 there exists 0 (m) such that µ(x m \X m 0 (m) ) < δ 2 m. We defie X δ = m=1 Xm 0 (m). Let us show that X δ is the required set. 23

24 1. f f uiformly o X δ sice if x X δ the x X m 0 (m) for ay m ad hece f i (x) f(x) < 1 m if i 0(m). This is exactly the defiitio of uiform covergece. 2. Let us estimate µ(x\x δ ). We otice that µ(x\x m ) = 0 for ay m. Ideed, if x 0 X\X m the there exists a sequece i such that f i (x 0 ) f(x 0 ) 1 m. This meas that f i(x 0 ) does ot coverge to f(x 0 ). Sice f i (x) f(x) a.e. X we have µ(x\x m ) = 0. This implies µ(x\x m 0 (m) ) = µ(xm \X m 0 (m) ) < δ 2 m ad we obtai The theorem is proved. µ(x\x δ ) =µ(x\ m=1 Xm 0 (m) ) =µ( m=1 (X\Xm 0 (m) )) µ(x\x m 0 (m) ) δ. m=1 I the two theorems below we relate covergece a.e. ad covergece i measure. Theorem 2.14 If the sequece of measurable fuctios f (x) f(x) a.e. the f (x) µ f(x). Proof It is easy to see that f(x) is measurable. Let = {x X : lim f (x) f(x)}, obviously µ() = 0. Fix δ > 0 ad defie X k (δ) = {x X : f k (x) f(x) δ}, R (δ) = k X k (δ), ad M = =1 R (δ). Obviously R 1 (δ) R 2 (δ)... By cotiuity of the measure we have µ(r (δ)) µ(m) as. Let us show that M. Take x 0 /, for this poit we have: for ay δ > 0 there exists N such that f k (x 0 ) f(x 0 ) < δ for ay k N. Therefore x 0 / R N (δ) ad hece x 0 / M. This implies µ(r (δ)) 0 ad sice X (δ) R (δ) we obtai µ(x (δ)) 0. The theorem is proved. Theorem 2.15 If a sequece of measurable fuctios f µ f the there exists a subsequece {f k } {f } that coverges to f a.e. X. Proof Let {ɛ } be a positive sequece such that ɛ 0 ad let {η } be a positive sequece such that =1 η <. Let us build a sequece of idices 1 < 2 <... as follows: choose 1 to be such that µ{x X : f 1 (x) f(x) ɛ 1 } < η 1 ; choose 2 > 1 to be such that µ{x X : f 1 (x) f(x) ɛ 2 } < η 2 e.t.c. 24

25 We show that f k (x) f(x) a.e. X. Ideed, let R i = k=i {x X : f k (x) f(x) ɛ k }, M = i=1 R i. Obviously R 1 R 2..., usig the cotiuity of the measure we obtai µ(r i ) µ(m), but µ(r i ) k=i η k hece µ(r i ) 0, sice the series coverges. Now we have to check that f k (x) f(x) i X\M. Let x 0 X\M the there exists i 0 such that x 0 / R i0 ad hece for ay k i 0 x 0 /.{x X : f k (x) f(x) ɛ k }. But this implies f k (x) f(x) < ɛ k for ay k i 0. Sice ɛ k 0 we get that f k (x 0 ) f(x 0 ). The theorem is proved. Theorem 2.16 (Lusi) fuctio f : [a, b] R is measurable if ad oly if for ay ɛ > 0 there exists a cotiuous fuctio φ ɛ such that µ{x [a, b] : f(x) φ ɛ (x)} < ɛ Proof Let for ay ɛ > 0 there exists φ ɛ - cotiuous fuctio such that µ{x [a, b] : f(x) φ ɛ (x)} < ɛ It is easy to see that if = {x [a, b] : f(x) < c} ad B = {x [a, b] : φ ɛ (x) < c} the B {x [a, b] : f(x) φ ɛ (x)}, B {x [a, b] : f(x) φ ɛ (x)}. Therefore B {x [a, b] : f(x) φ ɛ (x)} ad µ ( B) < epsio. O the other had sice φ ɛ is cotiuous the it is measurable ad set B is measurable. Therefore there exists Borel set C such that µ (B C) < ɛ (actually equal to 0). From this it follows that µ ( C) < 2ɛ ad hece is measurable. The secod part of the proof may be doe usig Egoroff theorem. It is left as a exercise. 25

26 3 Lebesgue itegral We are goig to defie Lebesgue itegral for elemetary fuctios first. Defiitio 3.1 fuctio f : X R is called elemetary if it is measurable ad takes ot more tha a coutable umber of values. Propositio 3.2 fuctio f : X R takig ot more tha a coutable umber of values y 1, y 2,... is measurable if ad oly if all sets = {x X : f(x) = y } are measurable. Proof The ecessity follows from the fact that = f 1 (y ) ad {y } are Borel sets. The sufficiecy is clear sice for ay P(R) we have f 1 () = y, where the uio is at most coutable. Hece f 1 () M. Propositio 3.3 fuctio f : X R is measurable if ad oly if it is a limit of a uiformly coverget sequece of elemetary fuctios. Proof Let {f (x)} be a sequece of elemetary fuctios ad f f uiformly o X. The obviously f (x) f(x) a.e. ad hece f(x) is measurable by theorem Now let f(x) be a measurable fuctio. We set f (x) = m o m m+1 m = {x X : f(x) < } (m Z ad N). Obviously f (x) is elemetary ad f (x) f(x) 1 o X. Takig we get the result. Let us defie Lebesgue itegral for elemetary fuctios. Take f : X R be elemetary fuctio with values y 1, y 2,..., y,... (y i y j for i j). Let X be a measurable set. We defie f(x)dµ = y µ( ), (7) where = {x : f(x) = y }. Defiitio 3.4 elemetary fuctio f : X R is itegrable o if the series (7) is absolutely coverget. I this case (7) is called a itegral of f over. 26

27 Lemma 3.1 Let = k B k, B i B j = for i j ad o ay B k fuctio f : R takes oly oe value c k. The f(x)dµ = c k µ(b k ) (8) k ad f is itegrable o if ad oly if the series i (8) coverges absolutely. Proof Obviously f(x) is elemetary ad the we ca fid at most coutable umber of distict values y 1, y 2,..., y,... of f(x). We have = {x : f(x) = y } = ck =y B k ad therefore y µ( ) = y µ(b k ) = c k µ(b k ) c k =y Now we have to show that these two series coverge or diverge simultaeously ad this is true sice y µ( ) = y µ(b k ) = c k µ(b k ). c k =y The lemma is proved. It is easy to see that itegral of elemetary fuctio is liear fuctioal. Now we wat to exted the defiitio of Lebesgue itegral to measurable fuctios that are ot ecessarily elemetary. 3.1 Itegrable fuctios Defiitio 3.5 fuctio f : X R is itegrable o a measurable set X if there exists a sequece {f (x)} of elemetary itegrable fuctios o such that f f uiformly o. The limit I = lim f (x)dµ (9) is deoted by ad is called the itegral of f over. f(x)dµ This defiitio makes sese if the followig coditios hold: k k 27

28 1. The limit (9) exists for ay uiformly coverget sequece of elemetary itegrable fuctios. 2. For a fixed f(x) this limit is idepedet of the choice of the sequece {f (x)}. 3. If f(x) is a elemetary fuctio the this defiitio of itegrability coicides with the defiitio 3.4 Let us show that all these poits are satisfied. Notice that if {f } is a sequece of elemetary itegrable fuctios the f (x)dµ f m (x)dµ µ() sup f (x) f m (x). (10) x This iequality implies that if {f } coverges uiformly to f the f (x)dµ is a Cauchy sequece ad hece lim f (x)dµ exists. Poit 1 is proved. To show poit 2 we assume that there are two sequeces {f } ad {g } of elemetary itegrable fuctios uiformly covergig to f. Obviously we have sup x f (x) g (x) 0 as. Formula (10) implies f (x)dµ g (x)dµ 0 that proves poit2. To show poit 3 take f (x) = f(x) for all, where f is elemetary ad itegrable ad the use poit 2. Properties of the itegral Let f ad g be ay itegrable fuctios o the: 1. 1dµ = µ(); 2. for ay c R cf(x)dµ = c f(x)dµ; 3. (f(x) + g(x))dµ = f(x)dµ + g(x)dµ; 4. if f(x) 0 the f(x)dµ 0; 5. if µ() = 0 the f(x)dµ = 0; 6. if f(x) = g(x) a.e. the f(x)dµ = g(x)dµ; 7. ay bouded measurable fuctio is itegrable; 8. if h(x) is measurable fuctio o ad h(x) f(x) for some itegrable f the h(x) is itegrable; 28

29 9. for ay measurable fuctio h(x) itegrals h(x)dµ ad h(x) dµ exist or do t exist simultaeously. These properties are usually proved for the itegrals of elemetary fuctios ad the, passig to the limit, for itegrable fuctios. We prove here oly property 8.. Propositio 3.6 If a fuctio f(x) is itegrable ad measurable fuctio h(x) f(x) the h(x) is also itegrable. Proof Let f(x) ad h(x) be elemetary fuctios. The ca be writte as a uio of coutable umber of disjoit sets o each of which f(x) ad h(x) are costats: Obviously h(x) = a, f(x) = b ad a b. a µ( ) This implies h(x) is itegrable ad h(x)dµ b µ( ) = f(x)dµ. f(x)dµ. Now let f(x) be a itegrable fuctio ad h(x) f(x). We may approximate h(x) ad f(x) by sequeces of uiformly coverget elemetary fuctios {h (x)} ad {f (x)}, respectively. Sice f(x) is itegrable the f (x) ca be chose as elemetary itegrable fuctios. This implies that for ay ɛ > 0 there exists N N such that if > N the h (x) h(x) < ɛ ad f (x) f(x) < ɛ. Obviously for > N h (x) f (x) +2ɛ ad the as before h (x) are itegrable. Hece h(x) is a limit of uiformly coverget sequece of elemetary itegrable fuctios ad therefore is itegrable. Propositio is proved. Propositio 3.7 Let =, where are measurable sets ad i j = for i j, ad let f : R be a itegrable fuctio the f(x)dµ = f(x)dµ ad existece of left itegral implies existece of itegrals i the right ad absolute covergece of the series. 29

30 Proof We check the theorem for itegrable elemetary fuctios first ad the pass to the limit to get the proof for ay itegrable fuctio. Let f(x) be a elemetary itegrable fuctio takig values y 1, y 2,... Let B k = {x : f(x) = y k } ad Bk = {x : f(x) = y k } the f(x)dµ = y k µ(b k ) = y k µ(bk ) k k = y k µ(bk ) = f(x)dµ. k We ca chage summatio idices sice f is a itegrable elemetary fuctio. Now let f be ay itegrable fuctio, by defiitio 3.5 for every ɛ > 0 we may fid a elemetary itegrable fuctio g ɛ such that g ɛ (x) f(x) < ɛ o. For g ɛ we have g ɛ (x)dµ = g ɛ (x)dµ. Sice g ɛ is itegrable over each we have that f is itegrable over each ad f(x)dµ g(x)dµ ɛµ( ) = ɛµ(), f(x)dµ g(x)dµ ɛµ(). Therefore the series f(x)dµ coverges absolutely ad f(x)dµ f(x)dµ 2ɛµ(). Lettig ɛ 0 we get the result. Propositio 3.8 Let =, where are measurable sets ad i j = for i j. Let f : R be a measurable fuctio ad f(x)dµ exist for all ad the series f(x) dµ coverges. The f(x)dµ = f(x)dµ. 30

31 Proof We check the theorem for elemetary fuctios first ad the pass to the limit to get the proof for ay itegrable fuctio. Let f(x) be a elemetary fuctio takig values y 1, y 2,... Let B k = {x : f(x) = y k } ad Bk = {x : f(x) = y k } the f(x) dµ = y k µ(bk ). k Therefore f(x) dµ = y k µ(bk ) = k k y k µ(b k ). Hece f is itegrable over ad f(x)dµ = k y kµ(b k ). Now let f be ay measurable fuctio, by propositio 3.3 for every ɛ > 0 we may fid a elemetary fuctio g ɛ such that g ɛ (x) f(x) < ɛ o. For g ɛ we have g ɛ (x) dµ f(x)dµ + ɛµ( ). Therefore g ɛ (x) dµ coverges ad g ɛ (x) is itegrable. But the f(x) is itegrable too ad by previous propositio we have the result. Theorem 3.9 (Chebyshev iequality) Let f(x) 0 be itegrable fuctio o ad c > 0 be some positive costat. The µ({x : f(x) c}) 1 f(x)dµ. c Proof Take B = {x : φ(x) c} the φ(x)dµ = φ(x)dµ + φ(x)dµ B The theorem is proved. \B B φ(x)dµ cµ(b). Corollary 3.10 If f(x) dµ = 0 the f(x) = 0 a.e. o. Proof By Chebyshev iequality we have µ({x : f(x) 1 }) 31 f(x)dµ for ay.

32 This implies µ({x : f(x) 0}) The corollary is proved. =1 µ({x : f(x) 1 }) = 0. Theorem 3.11 (bsolute cotiuity of the itegral) Let f(x) be a itegrable fuctio o. The for ay ɛ > 0 there exists δ > 0 such that f(x)dµ < ɛ for all measurable E such that µ(e) < δ. E Proof Fix ɛ > 0. The theorem is obvious if f is a bouded fuctio. Let f be a arbitrary itegrable fuctio o. We defie = {x : f(x) < + 1}, B = k=0 ad C = \B. By propositio 3.7 we have Choose N such that f(x) dµ = C N f(x) dµ = =0 =N+1 f(x) dµ. f(x) dµ < ɛ 2. We ca always do it sice the series =0 f(x) dµ coverges. Now let ɛ 0 < δ < 2(N+1) ad µ(e) < δ the sice f(x) < N + 1 o B N f(x)dµ f(x) dµ = f(x) dµ + f(x) dµ ɛ. E E E B N E C N Theorem is proved. Usig the properties of the itegral proved i this sectio we may show that for ay itegrable fuctio f(s) 0 a set fuctio defied o a measurable subsets X F () = f(x)dµ is a σ-additive measure. 32

33 3.2 Passage to the limit uder the Lebesgue itegral Theorem 3.12 (Lebesgue Domiated Covergece) Let {f (x)} be a sequece of itegrable fuctios defied o, f (x) f(x) a.e. x, ad for ay f (x) φ(x), where φ(x) is some itegrable fuctio o. The f is itegrable ad f (x)dµ f(x)dµ Proof Sice f (x) φ(x) ad f (x) f(x) a.e. we have f(x) φ(x) a.e. ad therefore f(x) is a itegrable fuctio. Fix ay ɛ > 0, by theorem 3.11 we may fid δ > 0 such that B φ(x)dµ < ɛ 2 if B ad µ(b) < δ. For this particular δ, usig Egoroff s theorem 2.13, we may fid E δ such that µ(e δ ) < δ ad f f uiformly o \E δ. Now we have lim f (x)dµ f(x)dµ lim f (x) f(x) dµ \E δ +2 φ(x)dµ < ɛ E δ Sice ɛ was arbitrary we take ɛ 0 ad obtai the result. Theorem 3.13 (Mootoe Covergece) Let f 1 (x) f 2 (x)... be a sequece of itegrable fuctios o ad f (x)dµ C for all, where C is some costat idepedet of. The f (x) coverges a.e. o to some itegrable fuctio f(x) ad f (x)dµ f(x)dµ Proof Without loss of geerality we may assume f 1 (x) 0. We wat to prove that f (x) f(x) a.e. Sice f (x) is a mootoe icreasig sequece it is obvious that for every x f (x) f(x) but here the value of f(x) may be ifiite. So our first task is to show that f(x) is ifiite oly o some ull set. We defie R = {x : lim f (x) =, R k = {x : f (x) > k}. It is easy to see that R1 k Rk 2... ad R = k=1 =1 Rk. Usig Chebyshev iequality we obtai µ(r) k 1 f (x)dµ C for ay. k k 33

34 Now we have R =1 Rk for ay k ad therefore µ(r) µ( =1R k ) = lim µ(rk ) C k for ay k N. Takig k we obtai µ(r) = 0. This proves that mootoe sequece {f (x)} has a fiite limit f(x) a.e. o. Now we wat to show itegrability of f(x). If we show this the usig Lebesgue domiated covergece theorem 3.12 we get the result sice 0 f (x) f(x) ad f (x) f(x) a.e. To show itegrability of f we costruct a auxiliary fuctio φ(x): φ(x) = k o k {x : k 1 f(x) < k}. It is obvious that φ(x) is elemetary ad f(x) φ(x) f(x) + 1. By defiitio φ(x)dµ exists if ad oly if k=1 kµ( k) coverges. We defie B m = m k=1 k, obviously m k=1 kµ( k) = B m φ(x)dµ B m f(x)dµ+µ(). Sice 0 f(x) m o B m by theorem 3.12 we have B m f(x)dµ = lim B m f (x)dµ C.Therefore m kµ( k ) C + µ() k=1 ad takig m we see that this series coverges ad φ(x) is itegrable. Sice 0 f(x) φ(x) the theorem is proved. Theorem 3.14 (Fatou lemma) If a sequece of o-egative itegrable fuctios {f (x)} coverges a.e. o to a fuctio f(x) ad f (x)dµ C for all, where C is some costat idepedet of. The f(x) is itegrable o ad lim if f (x)dµ f(x)dµ Proof We prove this result usig Mootoe covergece theorem We defie φ (x) = if k f k (x), it is easy to see that 1. φ (x) is measurable for all ; 2. 0 φ (x) f (x) ad hece φ (x) is itegrable for all with φ (x)dµ f(x)dµ C; 34

35 3. 0 φ 1 (x) φ 2 (x)... ad φ (x) f(x) a.e. Usig Mootoe covergece theorem 3.13 we have lim φ (x)dµ = f(x)dµ ad hece lim if f (x)dµ lim if φ (x)dµ = f(x)dµ. The theorem is proved. 3.3 Product measures ad Fubii theorem Defiitio 3.15 The set of ordered pairs (x 1,..., x ), where x i X i for i = 1,.., is called a product of sets X 1,..., X is deoted by X X 1 X 2... X k=1 X k. I particular, if X 1 = X 2 =... = X the X X Defiitio 3.16 If S 1,..., S are collectio of subsets of sets X 1,..., X, respectively, the S S 1... S k=1 S k is the collectio of subsets of X = k=1 X k represetable i the form = 1..., where k S k. Theorem 3.17 If S 1,..., S are semi-rigs the S = k=1 S k is a semirig. Proof The proof of this theorem is left as a exercise. Defiitio 3.18 Let µ 1,..., µ be some measures defied o the semi-rigs S 1,..., S. The the set fuctio µ = µ 1... µ o a semi-rig S = S 1... S is defied as for S = 1... µ() = µ 1 ( 1 )µ 2 ( 2 ) µ ( ) Propositio 3.19 The set fuctio µ from defiitio 3.18 is a measure. Proof The proof of this propositio is left as a exercise. 35

36 Theorem 3.20 If the measures µ 1,..., µ are σ-additive the the measure µ = k=1 µ k is σ-additive. Proof The proof of this theorem is left as a exercise. For simplicity of the presetatio we cosider the case = 2 oly. We assume that X ad Y are some sets, µ x ad µ y are Lebesgue measures o these sets. We also itroduce µ = µ x µ y which is Lebesgue extesio of a measure m = µ x µ y o X Y. Defiitio 3.21 Let Z = X Y the x = {y Y : (x, y) }, y = {x X : (x, y) }. Theorem 3.22 Uder the above assumptios o X, Y, µ x, µ y ad µ we have µ() = µ x ( y )dµ y = µ y ( x )dµ x Proof We are goig to prove oly first equality µ() = φ (y)dµ y, Y Y where φ (y) = µ x ( y ), sice the secod oe ca be doe by the same argumets. By defiitio of µ it is Lebesgue extesio of m = µ x µ y defied o the collectio of sets S m of the form = y0 x0, where y0 is µ x -measurable ad y0 is µ y -measurable. For such sets we obviously have µ() = µ x ( y0 )µ y ( x0 ) = µ x ( y0 )dµ y = φ (y)dµ y, x0 Y where φ (y) = X { µ x ( y0 ) if y x0, 0 otherwise. Note that if you make a sectio of such at ay poit y Y, you obtai either if y / x0 or y0 if y x0. This meas the theorem is true for such rectagles. The geeralizatio of the result to a fiite disjoit uio of such sets is ot difficult. Sice those sets coicide with R(S m ) we have the theorem for this algebra. 36

37 Lemma 3.2 If is µ-measurable set, the there exists a set B such that B = B, B 1 B 2..., B = k B k, B 1 B 2..., where the sets B k R(S m ), B ad µ() = µ(b). The proof of this lemma is left as a exercise. Sice we ca prove the theorem for ay set i R(S m ) ad usig the above lemma approximate ay measurable by the special set B P(S m ). We first prove the theorem for this B: φ B (y) = lim k φ Bk (y), as φ B1 (y) φ B2 (y)... φ B (y) = lim k φ Bk (y), as φ B1 (y) φ B2 (y)... Sice we kow that Y φ B k (y)dµ y = µ(b k ), by cotiuity of µ we obtai µ(b k ) µ(b ). O the other had we have φ B (y) = lim φ B k (y) ad φ Bk (y)dµ y µ(b ) k Usig mootoe covergece theorem we have φ Bk (y)dµ y φ B (y)dµ y = µ(b ). Y Y By the same argumets Y φ B (y)dµ y Y φ B(y)dµ y = µ(b). This proves the theorem for this special set B. Now we prove the theorem for ay ull set. If µ() = 0 the by lemma µ(b) = 0 ad therefore φ B (y)dµ y = µ(b) = 0. Y But sice φ B (y) 0 a.e this implies µ x (B y ) = φ B (y) = 0 a.e. Sice y B y we have y is measurable for almost all y Y ad φ (y) = µ x ( y ) = 0, φ (y)dµ y = 0 = µ(). The theorem holds for ull sets. Sice by the above lemma ay measurable set = B\N we have the result. Y 37 Y

38 Theorem 3.23 The Lebesgue itegral of a oegative itegrable fuctio f(x) is equal to the measure µ = µ x µ y of the set { x M, = 0 y f(x). Proof The proof is left as a exercise. Theorem 3.24 (Fubii) Suppose that σ-additive ad complete measures µ x ad µ y are defied o Borel algebras with uits X ad Y, respectively; further suppose that µ = µ x µ y, ad that the fuctio f(x, y) is µ-itegrable o X Y. The ( ) ( ) f(x, y)dµ = f(x, y)dµ y dµ x = f(x, y)dµ x dµ y. X x Y y Proof We prove the theorem first for the case f(x, y) 0. Let us cosider the triple product U = X Y R, ad the product measure λ = µ x µ y µ 1 = µ µ 1, where µ 1 is 1 D Lebesgue measure. We defie a set W U as follows: By theorem 3.23 W = {(x, y, z) U : x x, y y, 0 z f(x, y)}. λ(w ) = o the other had by theorem 3.22 λ(w ) = X f(x, y)dµ, ν(w x )dµ x, where ν = µ y µ 1 ad W x = {(y, z) : (x, y, z) W }. But by theorem 3.23 ν(w x ) = f(x, y)dµ y. x 38

39 Therefore we obtai f(x, y)dµ = X ( x f(x, y)dµ y ) dµ x. The theorem is proved for f(x, y) 0. The geeral case reduces to this oe by f(x, y) = f + (x, y) f (x, y). Problems 1. If f(x) is a measurable fuctio, g(x) is a itegrable fuctio ad α, β R are such that α f(x) β a.e., the there exists γ R such that α γ β ad f(x) g(x) dµ = γ g(x) dµ. X 2. If {f (x)} is a sequece of itegrable fuctios such that f (x) dµ <, X the the series f (x) f(x) a.e., where f is itegrable ad f (x) dµ = f(x)dµ. X 3. Suppose µ = µ x µ y is a product measure o X Y. Show that if f is µ-measurable ad X ( x f(x, y) dµ y )dµ x exists the f is µ-itegrable o X Y ad Fubii s theorem holds. 4. Let f L 1 (X), g L 1 (Y ) ad h(x, y) = f(x)g(y) a.e. (x, y) Ω = X Y. Prove that h L 1 (Ω) ad h(x, y)dµ = f(x)dµ x g(y)dµ y. Ω 5. Costruct Lebesgue itegral usig simple fuctios. X 6. Show that a space of itegrable fuctios is complete with respect to metric d(f, g) = f(x) g(x) dµ.. X 39 X X Y

40 7. Compare Lebesgue ad Riema itegral. What is the mai differece i the costructio ad properties of these itegrals? 8. Let X = Y = [0, 1] ad µ = µ x µ y, where µ x = µ y is Lebesgue measure. Let f(x), g(x) be itegrable over X. If F (x) = f(x)dµ x, G(x) = g(x)dµ x [0,x] for x [0, 1], the F (x)g(x)dµ x = G(1)F (1) X X [0,x] f(x)g(x)dµ x. 40

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