FUNDAMENTALS OF REAL ANALYSIS by. V.1. Product measures

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1 FUNDAMENTALS OF REAL ANALSIS by Doğa Çömez V. PRODUCT MEASURE SPACES V.1. Product measures Let (, A, µ) ad (, B, ν) be two measure spaces. I this sectio we will costruct a product measure µ ν o that coicides with µ ad ν whe restricted to ad, respectively. Defiitio. If A ad B, the set A B is called a rectagle; if A A ad B B, the A B is called a measurable rectagle. Fact.1 Let R = {A B : A A, B B}, the collectio of of all measurable rectagles. The: a) R is a coverig class for subsets of. b) If U, V R, the U V R. c) If U R, the U c is a fiite disjoit uio of sets i R. d) If U, V R, the U V is a fiite disjoit uio of sets i R. Proof. Exercise. Remark. (b)-(d) of Fact.1 says that the class S of all fiite disjoit uio of sets i R forms a algebra. Defiitio. The σ-algebra geerated by S is called the product σ-algebra (o ) ad is deoted by A B. Remark. By defiitio, A B is the smallest σ-algebra cotaiig S. O A B oe ca defie umerous measures; however, our aim is to defie (costruct) a measure which is compatible with µ ad ν. For, we will follow the procedure that eabled us to costruct the Lebesgue measure o R (with appropriate modificatios, of course). Give the measure spaces (, A, µ) ad (, B, ν), defie a set fuctio λ : R R # by λ(a B) = µ(a)ν(b), A A, B B. The λ( ) = 0, ad is compatible with µ ad ν i the sese that λ(a ) = µ(a) ad λ( B) = ν(b). Fact.2 Let R ad λ be as above. The, a) λ is mootoe o R. b) If {A i B i } i=1 is a coutable disjoit collectio i R whose uio is also i R, say A B = i A i B i, the µ(a)ν(b) = i=1 µ(a i)ν(b i ). Proof. (a) Obvious. (b) Fix x A. The, for each y B, the poit (x, y) belogs exactly oe A i B i by disjoitess; the same also holds if x ad y are iterchaged. Thus, for ay x ad y, χ A (x)χ B (y) = χ A B (x, y) = i 1 χ Ai B i (x, y) = i χ Ai (x)χ Bi (y).

2 2 Now, itegratig w.r.t. x (i.e., µ) ad usig MCT, we obtai µ(a)χ B (y) = i µ(a i)χ Bi (y), ad itegratig this w.r.t. y (i.e., ν), it follows that µ(a)ν(b) = i=1 µ(a i)ν(b i ). It follows from Fact.2 that the set fuctio λ is also a coutably additive set fuctio o S. Therefore, λ geerates a outer measure (µ ν) o subsets of. Let G be the σ-algebra of (µ ν) -measurable subsets of of ad let µ ν = (µ ν) G. The, µ ν is a measure o G. Notice that, sice A B is the smallest σ-algebra cotaiig S, we see that A B G; ad hece, µ ν is also a measure o A B. Defiitio. The measure µ ν is called the product measure of µ ad ν, ad A B is called the product σ-algebra of A ad B. The triple (, A B, µ ν) is called the product measure space of (, A, µ) ad (, B, ν). Remarks. 1. The triple (, G, µ ν) is also a measure space; ideed, it is a complete measure space (see Sectio II. 5) (?). The differece betwee G ad A B is the sets of µ ν-measure zero. 2. If (, A, µ) ad (, B, ν) are fiite (σ-fiite) measure spaces, the so is (, A B, µ ν). 3. If (, A, µ) ad (, B, ν) are σ-fiite, the (µ ν) is the uique measure o A B such that (µ ν)(a B) = µ(a)ν(b) for all A B R. If (, A, µ) ad (, B, ν) are ot σ-fiite, the this assertio may ot hold. 4. A typical elemet of A B eed ot be a measurable rectagle; ad uio of elemets of G eed ot be a measurable rectagle, either. Termiology. If = = R, A = B = F ad µ = ν = m, the µ ν = m m is called the two-dimesioal Lebesgue measure. Exercise. Take two of your favorite measure spaces (they ca be the same) ad costruct measures o a σ-algebra of subsets of their product (ot ecessarily compatible with the compoet measures). V.2. Double ad Iterated Itegrals Now, havig product measure spaces defied, we will proceed to defie itegratio o product measure spaces. For, we will cotiue followig the process of defiig Lebesgue measurable sets ad costructio of Lebesgue itegral i Chapter.III. I what follows (, A, µ) ad (, B, ν) are (fiite or σ-fiite) measure spaces, the so is G, A B, ad µ ν are as defied above. Defiitio. A fuctio f : R # is called µ ν-measurable (or A B-measurable) iff f 1 (O) A B for every ope set O R. Similarly, f : R # is called G-measurable iff f 1 (O) G for every ope set O R. Now, as i Chapter.III, begiig with simple fuctios, we ca defie µ ν-itegral of ay A B-measurable (or G-measurable) fuctio. Defie f : R + as µ ν-itegrable if fd(µ ν) <, ad f : R as µ ν-itegrable if both f + ad f are µ ν-itegrable. Now, we will explore the ramificatios of the property (µ ν)(a B) = µ(a)ν(b), for all A B R, of the measure µ ν i the itegral of ay fuctio f : R #. Let E, x ad y. The we defie

3 3 E x = {y : (x, y) E} the x-sectio of E, ad E y = {x : (x, y) E} the y-sectio of E. Similarly, if f : R # ad x, y, the we defie f x (y) := f(x, y) (x is fixed) the x-sectio of f, ad f y (x) := f(x, y) (y is fixed) the y-sectio of f. Observe that, if f : E R, E, the f x (y) is a fuctio o E x ad f y (x) is a fuctio o E y. Remark. For ay measurable rectagle E = A B, x ad y, { B if x A E x = (A B) x = otherwise, { A if y B E y = (A B) y = otherwise. Exercise. For {E } A B, show that ( E ) x = (E ) x ad ( E ) x = (E ) x. The same also hold for the y-sectios of E s. Furthermore, for E A B, (E c ) x = (E x ) c, the same also holds for y-sectio of E. Fact.3 For E A B, we have (i) E x B x, ad (ii) E y A y. Proof. It is eough to prove (i). Let D = {E A B : E x B x }. By the Remark above, D cotais all measurable rectagles. If {E } D, the ( E ) x = (E ) x B x. Thus, E D. Also, if E D, the (E c ) x = (E x ) c B x ; hece, E c D. Sice both ad are i D, we obtai that D is a σ-algebra. It follows that A B D, which implies (i). Remark. The same method of proof of Fact.3 shows that for ay E G, E x B, x, ad E y A, y. Corollary. If f : R is A B -measurable, the f x is B-measurable for all x ad f y is A-measurable for all y. Proof. By Fact.3, we have (f x ) 1 (O) = (f 1 (O)) x ad (f y ) 1 (O) = (f 1 (O)) y for ay O R ope. Recall that R is the collectio of all measurable rectagles i, S is the algebra geerated by R ad A B is the σ-algebra geerated by S (hece, by R). Notatio. By R σ we will deote the collectio of coutable uios of sets i R : ad by R σδ we will deote the collectio of coutable itersectios of sets i R σ. Fact.4 Let E A B (G). The there exists F R σδ such that E F ad (µ ν)(e) = (µ ν)(f ). Proof. Exercise. [Hit: See the proof of Theorem.2 i Chapter.II.] Fact.5 Let E A B (G) such that (µ ν)(e) <. The (i) the fuctio g(x) = ν(e x ) is µ-measurable ad gdµ = ν(e x )dµ = (µ ν)(e), ad (ii) the fuctio h(y) = ν(e y ) is ν-measurable ad hdν = µ(e y )dν = (µ ν)(e).

4 4 Proof. We ll prove (i) oly, (ii) is left to the reader. By Fact.3 (i), E x B; hece, it is ν- measurable ad g(x) = ν(e x ) is well-defied. If E = A B, A A, B B, the the assertio is trivial. Now, assume E R σ such that E = E, E R, is a disjoit uio. So E + (A B ). Fix x, the E x = (A B ) x = B, ad by coutable additivity of ]u, we have ν(e x ) = ν(b ) ν((a B ) x ). Let g (x) = ν((a B ) x ) ad g(x) = g (x). So, g(x) = ν(e x ). Sice each g is µ- measurable, g is µ-measurable. Also, g (x) 0 for all 1; hece, if h k (x) = k i=1 g i(x), the h g. Therefore, by MCT, gdµ = lim k k 1 g k dµ = lim k which proves the assertio for sets i R σ. k (µ ν)(a k B k ) = (µ ν)(e), 1 If E R σδ, the there is {E } R σ such that E ad E = E. By hypothesis, (µ ν)(e 1 ) <. Let g (x) = ν((e ) x ), the g 1 dµ = (µ ν)(e 1 ) <, which implies that g 1 < µ-a.e. x. Hece, for each x with g 1 (x) <, the family {E ) x } is decreasig, each has fiite ν-measure, ad E x = (E ) x. Thus, by cotiuity of the measure ν, g(x) = ν(e x ) = lim ν((e ) x ) = lim g (x), ad hece, g g µ-a.e. Sice each g is µ-measurable, so is g. From 0 g g 1 for all 1, applyig DCT, ad from the fact that g dµ = (µ ν)(e ), we see that gdµ = lim g dµ = lim[(µ ν)(e )] = (µ ν)(e), provig the assertio for sets i R σδ. The assertio for arbitrary sets i A B (G) follows from Fact.4. Remark. By Fact.5, if E A B, the the fuctios x ν(e x ) ad y µ(e y ) are measurable ad ν(e x )dµ = (µ ν)(e) = µ(e y )dν. Corollary. Let (, A, µ) ad (, B, ν) be σ-fiite measure spaces. If if E A B, the the fuctios x ν(e x ) ad y µ(e y ) are µ-measurable ad ν-measurable, respectively, ad ν(e x )dµ = (µ ν)(e) = µ(e y )dν. Proof. Case (, A, µ) ad (, B, ν) are fiite measure spaces follow from Fact.5. For the geeral case, let = i ( i i ), where { i i } i is a icreasig sequece of rectagles of fiite measure. The apply Fact.5 to E ( i i ). For a µ ν-measurable fuctio f : R, the itegral of f w.r.t. µ ν is defied as usual, i.e., f(x, y)d(µ ν). We call this itegral as the double itegral of f. Sice the the fuctios f x (y) ad f y (x) are ν-measurable ad µ-measurable fuctios, respectively, we ca also cosider the itegrals g(x) = f x(y)dν ad h(y) = f y (x)dµ, if they are defied. I

5 tur, if g ad h are µ- ad ν-measurable, respectively, the we ca thik about defiig their itegrals w.r.t. µ ad ν, respectively, as g(x) = [ f x (y)dν]dµ := f(x, y)dνdµ h(y) = [ f y (x)dµ]dν := f(x, y)dµdν. These are called the iterated itegrals of f. Now, observe that, if φ = χ E is the characteristic fuctio of a set E A B with fiite measure, the by Fact.5, (i) for µ-a.e. x, the fuctio φ x (y) = φ(x,.) is ν-itegrable o ad the fuctio φ y (x) = φ(., y) is µ-itegrable o, respectively, ad (ii) φd(µ ν) = ( φ x (y)dν)dµ = ( φ y (x)dµ)dν. Cosequetly, the same also follows for simple fuctios φ : R vaishig outside sets of fiite µ ν-measure. Theorem.1 (Fubii s Theorem) Let Let (, A, µ) ad (, B, ν) be σ-fiite measure spaces ad let f : R be a µ ν-itegrable fuctio. The a) for µ-a.e. x, f x (y) := f(x,.) is ν-itegrable, b) for ν-a.e. y, f y (x) := f(., y) is µ-itegrable, ad c) fd(µ ν) = ( f x(y)dν)dµ = ( f y (x)dµ)dν. Proof. WLOG we ca assume that f 0. The there exists a mootoe icreasig sequece {φ } of simple fuctios such that φ f µ ν-a.e., where each φ vaish outside a set of fiite µ ν-measure. The, by the observatio above, for each 1, φ d(µ ν) = [ (φ ) x (y)dν]dµ = [ (φ ) y (x)dµ]dν. 5 Now, by MCT, ( ) ad ( ) fd(µ ν) = lim fd(µ ν) = lim φ d(µ ν) = lim φ d(µ ν) = lim [ (φ ) x (y)dν]dµ [ (φ ) y (x)dµ]dν. Fix x. The {(φ ) x } is a mootoe icreasig sequece of simple ν-measurable fuctios such that (φ ) x f x ν-a.e. Hece, f x is ν-measurable, ad by MCT, we have f xdν = lim (φ ) x dν. Next, let h(x) = f xdν ad h (x) = (φ ) x dν. The, by the observatio above, each h is µ-itegrable. Sice h h µ-a.e., by MCT, we obtai that hdµ = lim h dµ = lim [ φ dν]dµ. Sice hdµ = [ f xdν]dµ, this combied with (*) gives that fd(µ ν) = [ f x dν]dµ. Similarly, ext by fixig y ad repeatig the above for {(φ ) y x}, we also obtai that fd(µ ν) = [ f y dµ]dν, provig the theorem.

6 6 Remark. Fubii s Theorem requires f to be µ ν-itegrable, which is ot easy to show. Also, recall that, the usual itegratio process goes through defiig itegrals of measurable fuctios, ad the defie itegrable fuctios. Toelli s Theorem closes this gap. Theorem.2 (Toelli s Theorem) Let (, A, µ) ad (, B, ν) be σ-fiite measure spaces ad let f : R be a o-egative A B-measurable fuctio. The a) for µ-a.e. x, f x (y) := f(x,.) is ν-measurable ad the fuctio f xdν is µ- measurable, b) for ν-a.e. y, f y (x) := f(., y) is µ-measurable ad the fuctio f y dµ is ν- measurable, ad c) fd(µ ν) = ( f x(y)dν)dµ = ( f y (x)dµ)dν. Proof. Sice ad are σ-fiite, so is. Now, pick a sequece {f } of simple fuctios that vaish outside a set of fiite µ ν-measure such that f f poitwise. Hece, each f is µ ν-itegrable. If g (x) = (f ) x dν, the g f x ν-a.e. Thus, [ (f ) x dν]dµ = fd(µ ν). Now, the rest of the proof is the same as that of Fubii. Exercise. Let = = N, A = B = P(N), µ = ν = c, where c is the coutig measure. Restate Fubii s ad Toelli s Theorems i this settig. Remarks. 1. The itegrals i Toelli s Theorem ca be ; however, if ay oe of the itegrals i (c) is fiite, so are the other two. 2. Typical applicatio goes as follows: first, oe applies Toelli s Theorem ad show that fd(µ ν) <, ad the uses Fubii to calculate the iterated itegrals. 3. σ-fiiteess i both Fubii ad Toelli s Theorems is essetial as the followig example shows. Example. Cosider ([0, 1], F, m) ad ([0, 1], P([0, 1]), c), where c is the coutig measure. Observe that ([0, 1], P([0, 1]), c) is ot a σ-fiite measure space. Let E = {(x, y) : x = y}. The [ χ E (x, y)dc]dm = 1 0 = [ χ E (x, y)dm]dc. (Also χ E(x, y)d(m c) = 0.) 4. As the example below shows, o-egativity i Toelli s Theorem is also essetial. Example. Let = = N, A = B = P(N), µ = ν = c, where c is the coutig measure. Notice that, o (N, P(N, c), if f : N R, the f is measurable, ad f is itegrable if ad oly if f() < (ad fdc = N f()). Now, cosider a fuctio g : N N R defied by g(, ) = 1 1 g( + 1, ) = 1 1 g(m, ) = 0 for m, m + 1. Hece, g is bouded ad c c-measurable. So [ g(m, )dµ]dν = (1 1) + (1 1) + = 0, N N [ g(m, )dν]dµ = 1 + ( 1 + 1) + ( 1 + 1) + = 1. N N

7 This does ot cotradict with Toelli s Theorem, or Fubii s Theorem, sice g is ot oegative. Also observe that g d(µ ν) = g + d(µ ν) + g d(µ ν), N N N N but both the itegrals o the RHS are ifiite. Exercises. 1. Let (, A, µ) ad (, B, ν) be σ-fiite measure spaces ad let f : R, g : R be A- ad B-measurable fuctios, respectively. The the fuctioh(x, y) = f(x)g(y) is a itegrable fuctio o ad hd(µ ν) = ( fdµ)( gdν). 2. Let = = R, A = B = F, ad µ = ν = m. Let f : R 2 R be defied by { x 2 y 2 if (x, y) (0, 0) f(x) = (x 2 +y 2 ) 2 0 if (x, y) = (0, 0). Show that f is m m-measurable ad ( f dm(x))dm(y) N N ( f dm(y))dm(x). Remark. I geeral the product space (, A B, µ ν) is ot complete, eve if both (, A, µ) ad (, B, ν) are complete. For istace, let = = R, A = B = F, µ = ν = m. If B P(R) \ F ad A F, the A B / A B by Fact.3. Sice A B A R, we have (µ ν)(a B) = (m m)(a B) is ot defied, whereas, µ ν)(a ) = (m m)(a R) = 0. However, we have the followig: Propositio. Let (, A, µ) be a (σ-fiite) measure space, (, B, ν) be a complete a (σ-fiite) measure space, ad E G be such that (µ ν)(e) = 0. The for µ-a.e. x, the set E x is ν-measurable ad ν(e x ) = 0. [Hece (µ ν)(e) = ν(e x)dµ.] Proof. Sice (µ ν)(e) <, there exists A R σδ such that E A ad (µ ν)(a) = (µ ν)(e) = 0. By Fact.4, for all x, A x is ν-measurable ad 0 = (µ ν)(a) = ν(a x)dµ. Thus ν(a x ) = 0 for µ-a.e. x. Sice E x A x for all x, ad sice (, B, ν) is complete, for µ-a.e. x, the set E x is ν-measurable ad ν(e x ) = 0. Questio. Let = = R, A = B = F, µ = ν = m, ad E = {a} [0, 1], for some a R. The (µ ν)(e) = 0, but ν(e a ) 0. Does this cotradict Propositio above? Although the product space (, A B, µ ν) is ot complete i geeral, the product space (, G, µ ν) is complete, by costructio. Actually the σ-algebra G is the completio of A B. Despite this fact, the relatioship betwee a G-measurable fuctio ad its x-sectio f x ad y-sectio f y is much more complicated; hece we will ot deal with it here. Istead, we will simply state ad prove the versio of Fubii-Toelli Theorem i this settig. Theorem.3 (Fubii-Toelli Theorem for complete spaces) Let (, A, µ) ad (, B, ν) be complete σ-fiite measure spaces ad let f : R be a o-egative G-measurable fuctio. The a) for µ-a.e. x, f x (y) := f(x,.) is ν-measurable ad the fuctio f xdν is µ- measurable, 7

8 8 b) for ν-a.e. y, f y (x) := f(., y) is µ-measurable ad the fuctio f y dµ is ν- measurable, ad c) fd(µ ν) = ( f x(y)dν)dµ = ( f y (x)dµ)dν. If f : R is µ µ-itegrable (ot ecessarily o-egative), the i the statemets (a) ad (b) above measurable is replaced by itegrable ad (c) holds with fiite value. Proof. Exercise. [Hit: Follow the steps of the proof of Theorem 2 ad Theorem.2 while usig Propositio.] Remarks. 1. The costructio of product measure ca be repeated to obtai product measure space of (ideed, ifiitely may) measure spaces. The versios of Fubii ad Toelli s Theorems are also valid i their respective settigs. 2. The special case, whe = = R, A = B = F, µ = ν = m (or whe i = R, A i = F, µ i = m) is called two () dimesioal Lebesgue measure space. All the properties above, of course, are valid i this case; furthermore, like the Lebesgue measure, the -dimesioal Lebesgue measure, 2, is ivariat uder cogruet maps (i.e., uder traslatios, reflectios, ad rotatios) o R. Exercise. Cosider the fuctio f(x, y) = e xy 2e 2xy. Prove that: a) f(x, y)dm(x)dm(y) <. [1, ) b) f(x, y)dm(y)dm(x) <. [1, ) Observe that f(x, y)dm(x)dm(y) f(x, y)dm(y)dm(x). Does this cotradict to the Fubii-Toelli Theorem? [1, ] [1, ] Explai.