Product measures, Toelli s ad Fubii s theorems For use i MAT3400/4400, autum 2014 Nadia S. Larse Versio of 13 October 2014. 1. Costructio of the product measure The purpose of these otes is to preset the costructio of the product of two measure spaces ad to state ad prove the mai theorems, Toelli s ad Fubii s theorems. These otes expad upo the material o product measure i [2]. The overall goal is to be able to itegrate fuctios of two or several variables. For this we eed to make sese of a multiple itegral i such a way that we ca compute its value o appropriate fuctios by chagig the order of itegratio whe we view the fuctio i a sigle variable, the others beig kept fixed. Give two measure spaces, A 1, µ 1 ad, A 2, µ 2, the goal is to costruct a σ-algebra A 1 A 2 of subsets of ad a measure ω o A 1 A 2 i such a way that the above metioed chage of order of itegratio is possible whe we cosider A 1 A 2 -measurable fuctios oegative or real-valued. For motivatio, cosider the case of Lebesgue measure o R, ad let I, J be itervals. The the area of the rectagle I J is 1 areai J = lilj = λiλj. We ca thik of area as a appropriate choice for a measure ω assiged to a rectagle I J. Next we formalise this idea for two geeral measure spaces. We defie the collectio C of measurable rectagles i A 1 A 2 to be 2 C = {A B A A 1, B A 2 }. Lemma 1.1. The collectio C of measurable rectagles is closed uder fiite itersectios. Proof. Exercise, or see Propositio 6.1 i [1]. Motivated by 1 we would like to have a a σ-algebra A 1 A 2 of subsets of such that C A 1 A 2, ad we would like to have a measure ω o A 1 A 2 which satisfies 3 ωa B = µ 1 Aµ 2 B for ay measurable rectagle A B. It is immediate to defie A 1 A 2 by 4 A 1 A 2 = σc, the σ-algebra geerated by the measurable rectagles this is the smallest σ-algebra o that cotais C. The hard work lies i provig that there exists a measure ω satisfyig 3: we call this the existece part of the product measure costructio. It turs out that uder some coditios, the measure ω satisfyig 3 is the oly oe with this property: we call this the uiqueess part of the product measure costructio. 1
2 The uiqueess part also requires some work, ad sice some of the ideas here will be useful for the existece proof, we begi with it. 2. Uiqueess of a measure defied o a collectio of sets The mai geeral result that will be used i the uiqueess part is the followig theorem. Theorem 2.1. see [2, Theorem 7.5] Suppose that X, A, µ is a measure space with the followig properties: 1 A = σe for a collectio E of subsets of X that is closed uder fiite itersectios, ad 2 there is a sequece {X } of elemets i E such that X X +1 for all 1, X = X, ad µx < for all. The µ is uiquely determied by its values o elemets i E. More precisely, if µ is a measure o A such that µa = µa for all A E, the µ = µ. To prove this theorem we will itroduce Dyki systems. For a subset A of a set X we let A c deote the complemet X \ A. Defiitio 2.2. Let X be a set. A collectio D of subsets of X is a Dyki system if the followig are satisfied: D1 X D, D2 A D A c D, ad D3 if {A } is a pairwise disjoit family i D, the A D. I a Dyki system, if A B ad B D, the B \ A = B c A c D. Give a set X ad a collectio E of subsets of X, recall that σe deotes the σ- algebra geerated by E. Sice there exists at least oe Dyki system that cotais E, amely the collectio PX, the by takig the itersectio of all Dyki systems which cotai E we obtai the Dyki system geerated by E: we deote this by DE ad poit out that this is the smallest Dyki system that cotais E. Lemma 2.3. Give a set X, if D is a Dyki system over X which is closed uder fiite itersectios or, equivaletly, uder fiite uios, the D is a σ-algebra. Proof. Let {A } be a arbitrary sequece i D. We make this ito a disjoit sequece, thus we let B 1 = A 1, B 2 = A 2 \ A 1 = A 2 A c 1, ad for all m > 2 m m B m = A m \ A k = A m A c k. The {B m } m is a disjoit uio i D, so m B m = A D. k=1 Lemma 2.4. Dyki Let X be a set ad E a collectio of subsets of X with the property that A, B E A B E. The DE = σe. k=1
Proof. Sice ay σ-algebra is also a Dyki system, ad sice E σe, we have that DE σe. To prove the coverse iclusio it suffices to show that DE is a σ- algebra. For simplicity of otatio, write D istead of DE. By Lemma 2.3, it is eough to show that D is closed uder itersectios. For A D we let D A = {B D A B D}. We claim that D A is a Dyki system that cotais E. Sice A X = A is i D, we have X D A. Let B D A. The A B c = A \ B A, which is i D by earlier observatio. So B c D A. Now let {B m } m be a pairwise disjoit sequece i D A. The A m B m = m A B m, which is i D because {A B m } m is a pairwise disjoit sequece i D. This shows that m B m D A. All i all, D A is a Dyki system. If the set A we start with has the property that A E, the for every B E we have A B E by the hypothesis. Thus A B D, so B D A. Sice B was arbitrary, we have E D A each time we start with A i E. Sice D is the smallest Dyki system o X that cotais E, we have D D A, so D = D A whe A E. It is importat to realise that this holds for ay A E. Next we let A D. We will show that E D A. Take B E. The by what we have just show, A D B. Thus A B D, which also meas that B D A. Sice B was arbitrary, we get E D A for A D, thus D = D A for all A D. This meas that D is closed uder itersectios. Proof of Theorem 2.1. We defie D = {A A µa = µa}. We claim that D is a Dyki system. By the assumptios o µ ad {X } we have µx = lim µx = lim µx = µx, so X D, which is D1. To cotiue, we cosider first the case that µx <. Let A D. Sice µ is mootoe ad µa µx <, we have µa c = µx µa = µx µa = µa c. This shows D2. To prove D3, let {A } be a pairwise disjoit sequece i D. Set B m = m k=1 A k. The B m form a icreasig sequece i A whose uio is A ad therefore belogs to A. By properties of measures ad the assumptio that A D for all 1, we have µ A = µ B = lim µb = lim µa k = lim µa k = lim k=1 k=1 µb = µ B = µ This shows that A D, which is D3. Thus D is a Dyki system. The assumptios imply that E D. The Lemma 2.4 gives that so D = A, which meas that µ = µ. σe = DE D A = σe, A. 3
4 I the geeral case we have X = X, X E ad µx < for all 1. By the first case, we have The for each A A we have µa X = µa X for all 1 ad all A A. µa = µa X = lim µa X = lim µa X = µa X = µa, ad the theorem is proved. Measures which satisfy property ii i Theorem 2.1 have a special ame, which we ow itroduce. Defiitio 2.5. A measure space X, A, µ is σ-fiite if there exists a sequece {X } 1 of elemets i A such that X X +1 for 1, X = X ad µx < for all. 3. Existece of the product measure of two measures Let, A 1, µ 1 ad, A 2, µ 2 be two measure spaces, ad let A 1 A 2 be the product σ-algebra o : this is the smallest σ-algebra o that cotais the collectio C of measurable rectagles. If A A 1 A 2 ad x 1, x 2, we cosider the followig sets A x 2 = {x 1 x 1, x 2 A}, A x 1 = {x 2 x 1, x 2 A}. Thus, if A = A 1 A 2 C, the A x 2 = A 1 if x 2 A 2 ad A x 2 = if x 2 / A 2. Similarly for A x 1. We have the followig result. Lemma 3.1. For all A A 1 A 2, x 1 ad x 2 we have A x 2 A 1 ad A x 1 A 2. Theorem 3.2. Assume that, A 1, µ 1 ad, A 2, µ 2 are two fiite measure spaces, meaig µ k X k < for k = 1, 2. Let A A 1 A 2. The, with f = χ A, f : [0, ], the followig hold: 1 The fuctio [0, ] with x 1 fx 1, x 2 is A 1 -measurable for each x 2 i. 2 The fuctio [0, ] with x 2 fx 1, x 2 is A 2 -measurable for each x 1 i. 3 The fuctio g A : [0, ] with x 1 fx 1, x 2 dµ 2 x 2 is A 1 -measurable. 4 The fuctio h A : [0, ] with x 2 fx 1, x 2 dµ 1 x 1 is A 2 -measurable. Moreover, the followig holds: 5 fx 1, x 2 dµ 2 x 2 dµ 1 x 1 = fx 1, x 2 dµ 1 x 1 dµ 2 x 2.
Proof. Let D deote the collectio of measurable sets A i A 1 A 2 such that the statemets 1-5 hold. We claim that D cotais all measurable rectagles from C. To see this, let A = A 1 A 2 C. The the fuctio i 1 is either χ A1 if x 2 A 2 or 0 if x 2 \ A 2, so is A 1 -measurable. A similar argumet applies for 2. The fuctio g A i 3 is µ 2 A 2 χ A1, so is A 1 -measurable, ad similarly for h A. Fially, ote that both sides i 5 equal µ 1 A 1 µ 2 A 2. Our claim is therefore proved. Sice sets i C geerate A 1 A 2 ad C is closed uder fiite itersectios accordig to Lemma 1.1, it will follow from Lemma 2.4 that the theorem is valid if we show that D is a Dyki system. By defiitio, D. Let A D. Sice χ A c = 1 χ A, it is easy to verify that 1-2 hold for f = χ A c. Usig that µ 2 is fiite, the fuctio i 3 is g A cx 1 = µ 2 A 2 µ 2 A x 1, ad so is A 1 -measurable. A similar argumet shows that the fuctio i 4 is measurable. To verify that 5 holds for f = χ A c, ote that sice µ 1, µ 2 are fiite measures we have g A cx 1 dµ 1 x 1 = µ 2 µ 2 A x 1 dµ 1 x 1 = µ 1 µ 2 χ A x 1, x 2 dµ 2 x 2 dµ 1 x 1 X 2 = µ 1 µ 2 χ A x 1, x 2 dµ 1 x 1 dµ 2 x 2 X 1 = h A cx 2 dµ 2 x 2, as eeded. It remais to prove that D is closed uder disjoit coutable uios. For this, we prove first that D is closed uder fiite disjoit uios. Let A i D, i = 1,..., such that A i A j = wheever i j. The also A i x 2 A j x 2 = wheever i j, for all x 2. Note that i A i x 2 = i A i x 2, for each x 2. By additivity of µ 1, it the follows that the fuctio i 4 correspodig to i=1a i is x 2 i=1 h A x 2, ad so is A i 2 -measurable. Sice the itegral ad fiite sums of measurable fuctios iterchage order, 5 follows from h i A i dµ 2 = h Ai dµ 2 i=1 = g Ai dµ 1 by 5 for each A i i=1 = g i A i dµ 1. Now we show that D is closed uder icreasig uios. Let C i D for i 1 such that C i C i+1 for i 1. Note the that C i x 2 C i+1 x 2 for each x 2. Sice µ 1 is a 5
6 fiite measure, h Ci+1 x 2 h Ci x 2 = µ 1 C i+1 x 2 µ 1 C i x 2 = µ 1 C i+1 x 2 \ C i x 2 0. Thus {h Ci } i is a odecreasig sequece of A 2 -measurable fuctios. The poitwise limit of {h Ci } i is also A 2 -measurable, ad by the mootoe covergece theorem we have lim h Ci dµ 2 = lim h Ci dµ 2. i X i 2 Now ote that cotiuity of µ 1 implies that lim i h Ci x 2 = lim i µ 1 C i x 2 = µ 1 i C i x 2. Similarly, by applyig the mootoe covergece theorem to g Ci ad by usig assumptio 5 for all C i we get that g Ci dµ 1 = lim g Ci dµ 1 = lim h Ci dµ 1 = h Ci dµ 2, X i 1 X i 1 which shows that i C i D, as claimed. I all, we have showed that D is a Dyki system. By Lemma 2.4, we have DC = σc. Sice σc = A 1 A 2, it follows that A 1 A 2 D, so the theorem is proved. Theorem 3.3. Let, A 1, µ 1 ad, A 2, µ 2 be σ-fiite measure spaces. The statemets 1-5 of Theorem 3.2 hold. Proof. By the hypothesis, there are sets X i, Σ i for 1 ad i = 1, 2 such that {X i, } is icreasig, µ i X i, < for all i = 1, 2 ad 1, ad X i = X i, for i = 1, 2. I order to reduce the problem to the fiite case we ote that the assigmets µ 1, B = µ 1 B, ad µ 2, C = µ 2 C, for B A 1, C A 2, 1, give fiite measures o A 1 ad A 2, respectively. Note for later use that h dµ 1, = hχ X1, dµ 1 for ay oegative measurable fuctio first check for h a simple fuctio, the approximate by oegative simple fuctios to get the claim i geeral. Similarly for µ 2, ad all. Applyig Theorem 3.2 to µ 1, ad µ 2, for all 1 gives 6 g A x 1 dµ 1, x 1 = h A x 2 dµ 2, x 2 for all A A 1 A 2. By the observatio i the previous paragraph, we have g A χ X1, dµ 1 = h A χ X2, dµ 2 for all 1. Now we take the limit as ad use that {, } icreases to ad {, } icreases to. Therefore the mootoe covergece theorem, applied i both the right- ad left-had side, shows that 5 holds i this case. The other statemets ca be verified i a similar way.
Theorem 3.4. Let, A 1, µ 1 ad, A 2, µ 2 be two σ-fiite measure spaces. The there exists a uique measure o, A 1 A 2, deoted µ 1 µ 2, such that 7 µ 1 µ 2 A 1 A 2 = µ 1 A 1 µ 2 A 2, A 1 A 1, A 2 A 2. Furthermore, for each A A 1 A 2 we have 8 µ 1 µ 2 A = χ A x 1, x 2 dµ 2 x 2 dµ 1 x 1 X 2 9 = χ A x 1, x 2 dµ 1 x 1 dµ 2 x 2. Proof. We defie µ 1 µ 2 A by the value of ay of the double itegrals above. These quatities are equal by Theorem 3.35. Clearly 7 holds ad µ 1 µ 2 = 0. It remais to show that µ 1 µ 2 is coutably additive. Let {A } be a disjoit sequece i A 1 A 2 ad let A = A. Note that for ay x 1 we have A x 1 = A x 1, 7 where the secod uio is disjoit i. Sice µ 2 is coutably additive, we have g A x 1 = µ 2 A x 1 = µ 2 A x 1 = g A x 1 for each x 1. Hece, by a corollary to the mootoe covergece theorem, it follows that µ 1 µ 2 A = g A dµ 1 = g A dµ 1 = µ 1 µ 2 A, showig that µ 1 µ 2 is a measure. To show uiqueess of µ 1 2 amog all measures o A 1 A 2 which satisfy 7, ote that this will follow from Theorem 2.1 because A 1 A 2 is geerated by C if we show that µ 1 µ 2 is σ-fiite. Sice µ i for i = 1, 2 are σ-fiite, there are sequeces X i, A i for 1 ad i = 1, 2 such that {X i, } icreases to X i, ad µ i X i, < for all i = 1, 2 ad 1. The {,, } icreases to. Moreover, so ideed µ 1 µ 2 is σ-fiite. µ 1 µ 2,, = µ 1, µ 2, <, Note: a alterative proof of the existece of the product measure is give by Theorem 6.4 i [1] ad of the uiqueess by Theorem 6.5 i [1]. Our proof of existece ad uiqueess ca be see as icorporatig Lemmas 6.8 ad 6.9 i [1]. 4. Toelli s ad Fubii s theorems Let X i, A i, µ i be σ-fiite measure spaces for i = 1, 2 ad let µ 1 µ 2 be the product measure o A 1 A 2 give by Theorem 3.4. Give a fuctio f : R, defie its sectios at x 1 ad x 2 by f x2 x 1 = fx 1, x 2 ad f x1 x 2 = fx 1, x 2.
8 Propositio 4.1. If f : R is A 1 A 2 -measurable, the f x2 : R is A 1 -measurable for every x 2. Similarly, f x1 : R is A 2 -measurable for every x 1. Proof. Sice f is measurable, for ay r R the set A = {x 1, x 2 fx 1, x 2 r, ]} is i A 1 A 2. But {x 1 f x2 x 1 r, ]} = {x 1 x 1, x 2 A}, i other words fx 1 2 r, ] = A x 2, which lies i A 1 by lemma. Sice r R is arbitrary, the claim for f x2 follows. The proof i the case of f x1 is similar. Note: the coclusio of the propositio is also valid for a complex-valued measurable fuctio f o. The proof is similar to the above or see Propositio 6.10 i [1]. Theorem 4.2. Toelli s theorem Let X i, A i, µ i be σ-fiite measure spaces for i = 1, 2 ad let µ 1 µ 2 be the product measure o A 1 A 2. Suppose that f : [0, ] is A 1 A 2 -measurable. The the fuctios x 1 f x1 dµ 2 ad x 2 f x2 dµ 1 are measurable, ad 10 f dµ 1 µ 2 = fx 1, x 2 dµ 1 x 1 dµ 2 x 2 X 1 11 = fx 1, x 2 dµ 2 x 2 dµ 1 x 1. Proof. The case whe f = χ A for A Σ 1 Σ 2 was doe i Theorem 3.3. By liearity the result exteds to simple fuctios o. Next let {s } be a sequece of simple fuctios o such that s 0, s s +1 for all 1, ad s f poitwise. Thus we have for every 1 that 12 s dµ 1 µ 2 = s x 1, x 2 dµ 1 x 1 dµ 2 x 2 X 1 13 = s x 1, x 2 dµ 2 x 2 dµ 1 x 1. For every fixed x 2 the sequece {s x2 } of fuctios R is icreasig ad coverget to f x2 at every poit x 1. By the mootoe covergece theorem, f x2 dµ 1 = lim s x2 dµ 1. X 1 Sice the theorem is true for simple fuctios, the fuctio F x 2 := s x2 dµ 1 is A 2 -measurable for all 1. Moreover, F F +1 for all 1 by properties of the itegral of oegative fuctios. It follows that lim F is measurable, which meas precisely that x 2 f x2 dµ 1 is A 2 -measurable. Aother applicatio of the mootoe covergece theorem gives lim F x 2 dµ 2 x 2 = lim F x 2 dµ 2 x 2, X 2
which rewrites as lim s x2 dµ 1 dµ2 x 2 = f x2 x 1 dµ 1 x 1 dµ 2 x 2. By 12, the term i the left-had side is lim s dµ 1 µ 2, which coverges to the itegral f dµ 1 µ 2 agai by the mootoe covergece theorem. Thus we obtai 10. The other equality is proved i a similar way. Theorem 4.3. Let X i, A i, µ i be σ-fiite measure spaces for i = 1, 2 ad let µ 1 µ 2 be the product measure o A 1 A 2. Suppose that f : R or to C is itegrable. The the followig assertios are true: 1 The fuctio f x2 is i L 1, A 1, µ 1 for µ 2 -a.e. x 2. Likewise, f x1 is i L 1, A 2, µ 2 for µ 1 - a.e. x 1. 2 The fuctios x 1 f x1 dµ 2 ad x 2 f x2 dµ 1 are defied µ 1 -a.e. o ad µ 2 -a.e. o, respectively. Further, they belog to L 1, A 1, µ 1 ad L 1, A 2, µ 2, respectively, ad fx 1, x 2 dµ 1 x 1 14 dµ 2 x 2 = f dµ 1 µ 2 X 2 15 = fx 1, x 2 dµ 2 x 2 dµ 1 x 1. Proof. See the proof of Theorem 6.7 i [1]. Refereces [1] J.N. McDoald ad N.A. Weiss, A course i Real Aalysis, 2d editio, Academic Press, Amsterdam, 2013. [2] G. Teschl, Topics i real ad fuctioal aalysis, http://www.mat.uivie.ac.at/ gerald/ftp/bookfa/idex.html. 9