3.7 The Lebesgue integral

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1 3 Mesure d Itegrtio The f is simple fuctio d positive wheever f is positive (the ltter follows from the fct tht i this cse f 1 [B,k ] = for ll k, ). Moreover, f (x) f (x). Ideed, if x, the there exists 0 N such tht f (x) 2 0. The f (x) f (x) 2 for ll 0. If f 0 the f f +1. Ideed, if f (x) = k2 the f (x) [k2, (k + 1)2 ). But the either f (x) [(2k)2 (+1), (2k + 1)2 (+1) ), i which cse f +1 (x) = (2k)2 (+1) = f (x), or f (x) [(2k + 1)2 (+1), (2k + 2)2 (+1) ), i which cse f +1 (x) = (2k + 1)2 (+1) > f (x). I the cse where K = C, we fid sequeces of simple fuctios (g ) d (h ) tht coverge to Re f d Im f, respectively. The we set f := g + ih for ll N. For exmple, this is turl for stoppig times of rdom processes. There exist metrics o R resp. [0, ] such tht B(R) resp. B([0, ]) is the Borel σ-lgebr for this metric. We sometimes prefer to work with slightly more geerl otio of mesurble fuctios. It hs techicl dvtges to llow fuctios to tke the vlues or. Remrk 3.54 (Exteded rel lie). We put R := { } R { } which we edow with the σ-lgebr B(R), defied s σ(b(r) {{ }, { }}). It follows tht fuctio f : (, Σ) R is mesurble if d oly if f 1 [{ }], f 1 [{ }] Σ d f 1 [A] Σ for ll A B(R). Similrly, B([0, ]) is defied s σ(b([0, )) {{ }}). We lso remrk tht Propositio 3.53 geerlizes to this situtio. Exercise Let (, Σ) be mesurble spce d f : R be mesurble for ll N. The the fuctios lim if f, lim sup f, if N f d sup N f re mesurble. Hit: Observe tht {x : sup{ f (x) : N} > } = N{x : f (x) > }. 3.7 The Lebesgue itegrl Give mesure spce (, Σ, µ), we ow itroduce the Lebesgue itegrl f dµ for suitble complex-vlued, mesurble fuctios tht we cll itegrble. We proceed i severl steps d first defie the itegrl for (rel-vlued) mesurble fuctios f tkig vlues i [0, ]. Defiitio Let (, Σ, µ) be mesure spce. If f : [0, ] is simple fuctio, with stdrd represettio f = k1 Ak, oe defies the Lebesgue itegrl of f by f dµ := k µ(a k ) 72

2 3.7 The Lebesgue itegrl where, by covetio, 0 := 0 d + =. The followig lemm shows tht the Lebesgue itegrl for oegtive simple fuctios is positively lier d mootoe. Lemm If (, Σ, µ) is mesure spce, f, g : [0, ] re simple fuctios d λ > 0, the λ f + g dµ = λ f dµ + g dµ. Moreover, if f g, the f dµ g dµ. The covetio 0 = 0 esures tht the Lebesgue itegrl of the zero fuctio o R is zero. Note tht λ(r) =. Proof. If k1 Ak is the stdrd represettio of f, the λ k1 Ak is the stdrd represettio of λ f. Now λ f dµ = λ f dµ follows from the defiitio of the itegrl. Now let λ = 1 d m l=1 b l1 Bl be the stdrd represettio of g. If w c j1 Cj is the stdrd represettio of f + g, the C j = k +b l =c j A k B l. Thus f dµ + g dµ = = k µ(a k ) + w c j µ(c j ) = m l=1 b l µ(b l ) = f + g dµ, m l=1 ( k + b l )µ(a k B l ) where we hve used the fiite dditivity of µ d the fct tht the sets A k B l re pirwise disjoit. Now ssume tht f g. Cosider the fuctio g f, with the covetio tht = 0 d c = for ll c [0, ). The g f is oegtive, simple fuctio. Hece, g dµ = (g f ) + f dµ = g f dµ + f dµ f dµ, sice the itegrl of oegtive, simple fuctio is clerly oegtive. Lemm Let (, Σ, µ) be mesure spce d f : [0, ] be simple. The ν : Σ [0, ] defied by ν(a) = 1 A f dµ is mesure o (, Σ). Proof. As 1 f is the zero fuctio, the mp ν clerly stisfies property (M1). It remis to very (M2). So let (A k ) be sequece of disjoit sets i Σ. Defie B := A k. The ν(b ) = 1 B f dµ = 1 Ak f dµ = 1 Ak f dµ = ν(a k ). Clerly, B B +1 for ll N d A := N B = k N A k. Sice f is simple fuctio, tke f = m j1 Cj to be its stdrd represettio. 73

3 3 Mesure d Itegrtio Observe tht ν(a) = 1 A f dµ = m j 1 Cj A dµ = m j µ(c j A) = lim m j µ(c j B ) m = lim j 1 Cj B dµ = lim 1 B f dµ = lim ν(b ) = ν(a k ) I the previous clcultio we used the defiitio of the itegrl for simple fuctios, tht it is positively lier, the cotiuity of the mesure µ, d i the lst step the idetity ν(b ) = ν(a k) tht ws estblished bove. This proves (M2). Hece ν is mesure. We c ow defie the Lebesgue itegrl by pproximtio for rbitrry mesurble fuctio f : [0, ]. Defiitio Let (, Σ, µ) be mesure spce d f : [0, ] mesurble. Oe defies { } f dµ = sup g dµ : 0 g f, g simple. We sy tht f is (Lebesgue) itegrble, if f dµ <. Exmple If δ is the Dirc mesure o (, Σ) for, the for ll mesurble f : [0, ] we hve f dδ = f () d f is itegrble if d oly if f () <. Proof. First, let f be simple fuctio. The δ ( f 1 [{t}]) = 1 if f 1 [{t}], i.e. f () = t d δ [ f 1 [{t}]] = 0 else. Thus i this cse f dδ = f (). Now let f : [0, ] be mesurble fuctio. If g is simple fuctio with g f, the g dδ = g() f (). Tkig the supremum over such g, it follows tht f dµ f (). 74

4 3.7 The Lebesgue itegrl For the reverse iequlity, let g(x) := f ()1 f 1 [{ f ()}](x). Sice f is mesurble d { f ()} B(R), it follows tht f 1 [{ f ()}] Σ. So g is simple fuctio. Moreover, g f. Hece, f dδ g dδ = f (). Exercise Cosider the mesure spce (N, P(N), ζ). A mesurble fuctio f : N [0, ] is sequece ( ) N = ( f ()) N i [0, ]. Show tht f is itegrble if d oly if ( ) l 1 d i this cse, N f dζ =. =1 Theorem 3.62 (Mootoe covergece theorem). Let (, Σ, µ) be mesure spce d f : [0, ] be mesurble fuctios for ll N tht re mootoiclly icresig i. The f (x) := sup N f (x) is mesurble d f dµ = sup N f dµ. Proof. As cosequece of Exercise 3.55, f is mesurble. Oe deduces from f f tht sup N f dµ f dµ. It remis to show the coverse iequlity. To this ed, let 0 ϕ f be simple d β > 1. It suffices to show tht ϕ dµ β sup N f dµ. Isted of sup we could write lim i the sttemet of Theorem Of course this requires to cosider improper limits, thereby for exmple settig f (x) to where ( f (x)) is ubouded. The sufficiecy follows from first tkig the supremum over ll simple 0 ϕ f d the tkig the ifimum over ll β > 1. Defie B := {x : β f (x) ϕ(x)}. The B Σ, B B +1 d N B =. Moreover, we hve β f ϕ o B. Defie ν(a) := 1 A ϕ dµ for ll A Σ. The ν is mesure by Lemm Hece ϕ dµ = ν() = lim ν(b ) = lim ϕ1 B dµ sup β f dµ N = β sup f dµ. N This estblishes the coverse iequlity. Hece sup N f dµ = f dµ. Remrk Let f : [0, ] be mesurble. Let ( f ) be sequece of oegtive mesurble fuctios such tht f f poitwise. Note 75

5 3 Mesure d Itegrtio tht such sequece exists by Propositio It ow follows from Theorem 3.62 tht { } f dµ = sup ϕ dµ : 0 ϕ f simple = sup N f dµ. So while the itegrl for oegtive mesurble fuctios is defied s the supremum over very lrge clss of simple fuctios, it is ctully lredy obtied by the supremum over pproximtig sequece s i Propositio Corollry Let (, Σ, µ) be mesure spce, f, g : [0, ] mesurble d λ > 0. The λ f + g dµ = λ f dµ + g dµ. Moreover, if f g, the f dµ g dµ. Proof. Let ( f ) d (g ) be sequeces of simple fuctios such tht f f d g g. The λ f + g λ f + g. Usig mootoe covergece d Lemm 3.57, we obti λ f + g dµ = lim = lim λ λ f + g dµ f dµ + g dµ = λ f dµ + g dµ The secod ssertio is cler from the defiitio d ws lredy observed i the proof of Theorem Note tht the otio of ull set crucilly depeds o the mesure, of course. So sometimes it is ecessry to poit out this depedece by spekig bout µ-ull set. Defiitio Let (, Σ, µ) be mesure spce. A ull set is set N such tht there exists set M Σ with N M d µ(m) = 0. Now let P = P(x) be property which, depedig o x, my be true or flse. We sy tht P holds lmost everywhere or for lmost every x, if {x : P(x) is flse} is ull set. If µ is probbility mesure, we lso sy tht P holds lmost surely if it holds lmost everywhere. Remrk Note tht we do ot ssume tht ull set N is mesurble. A mesure spce i which ll ull sets re mesurble is clled complete. It is ot hrd to see tht the mesure spce (, M, µ ) i the proof of Crthéodory s theorem, Theorem 3.42, is lwys complete. We ote tht there is strightforwrd procedure to complete mesure spce by miimlly elrgig the σ-lgebr d thus the domi of the mesure to mke ll ull sets mesurble. Exmple Cosider (R, B(R), δ 0 ). The lmost every (or more precisely, δ 0 -.e.) x R is equl to 0. Cosider (R, B(R), λ). The lmost every (or more precisely, λ-.e.) x R is ot equl to 0. I fct, lmost every x R is irrtiol. Corollry Let (, Σ, µ) be mesure spce d f : [0, ] be mesurble. The f dµ = 0 if d oly if f (x) = 0 for lmost every x. 76

6 3.7 The Lebesgue itegrl Proof. If f is simple fuctio, the the ssertio is obvious. I the geerl cse, let mesurble f : [0, ] with f dµ = 0 be give. Let ( f ) be icresig sequece of simple fuctios covergig to f. Such sequece exists by Propositio It follows tht f dµ = 0 for ll N. Hece, by the cse bove, {x : f (x) = 0} is ull set, whece there exists M Σ with µ(m ) = 0 such tht f (x) = 0 for ll x M. Put M := N M. The M Σ d µ(m) =1 µ(m ) = 0. Moreover, x M implies tht f (x) = 0. Hece f = 0 lmost everywhere. If, coversely, f = 0 lmost everywhere, the there exists mesurble set M with f (x) = 0 for ll x M. If g is simple fuctio with 0 g f the g 1 [{x}] M for ll x > 0. By the defiitio of the itegrl for simple fuctios, g dµ = 0 d hece, sice g f ws rbitrry, f dµ = 0. Exercise Let (, Σ, µ) be mesure spce d f : [0, ] be mesurble. Show tht ν : Σ [0, ], defied by ν(a) := A f dµ := 1 A f dµ, defies mesure o (, Σ). Moreover, show tht if µ(a) = 0, the ν(a) = 0. I other words, ν is bsolutely cotiuous with respect to µ, which is usully deoted by writig ν µ. Remrk I the settig of Exmple 3.69 the fuctio f is clled the desity of ν with respect to µ. I probbility theory desity for distributio is commoly tke with respect to the Lebesgue mesure. Accordig to the Rdo Nikodym theorem probbility mesure P o B(R) hs desity with respect to the Lebesgue mesure if d oly if it is bsolutely cotiuous with respect to the Lebesgue mesure. Equivletly, this is the cse if d oly if its distributio fuctio F(x) := P((, x]) is bsolutely cotiuous fuctio. Not every cotiuous rel fuctio is bsolutely cotiuous, but every Lipschitz fuctio is. Theorem 3.71 (Ftou s Lemm). Let (, Σ, µ) be mesure spce, f : [0, ] be mesurble d set f (x) := lim if f (x). The f is mesurble d f dµ lim if f dµ. Proof. Note tht f (x) = sup k 1 if k f (x). Let g k (x) := if k f (x). The g k is mesurble by Exercise Clerly g k f. So it follows from mootoe covergece tht sup k N gk dµ = f dµ. O the other hd, f dµ = sup g k dµ lim if f dµ. k N 77

7 3 Mesure d Itegrtio Ideed, for every k we hve g k f d thus g k dµ f dµ. Sice this is true for ll k, we hve g k dµ if k f dµ. By tkig the supremum over k N o both sides, the bove iequlity follows. Defiitio Let (, Σ, µ) be mesure spce, f : K be mesurble. The f is clled itegrble if f dµ <. We write f L 1 (, Σ, µ). If K = R, we set f dµ := f + dµ f dµ. Note tht if f is itegrble the f + d f re both itegrble oegtive fuctios. If K = C, we set f dµ = Re f dµ + i Im f dµ. Note tht if f is itegrble, the Re f d Im f re both itegrble relvlued fuctios. If f is itegrble (rel or complex) mesurble fuctio d A Σ, we defie f dµ := f 1 A dµ Lemm Let (, Σ, µ) be mesure spce. () For ll itegrble f, we hve f dµ f dµ. A (b) If f is itegrble d λ K, the λ f is itegrble d λ f dµ = λ f dµ. (c) If f d g re itegrble, the f + g is itegrble d f + g dµ = f dµ + g dµ. Remrk Note tht (b) d (c) c be expressed by syig tht the itegrble fuctios form vector spce d the mp f f dµ is lier mp from the itegrble fuctios to K. Proof. Let us first cosider the cse where K = R. () We hve f dµ = f + dµ f dµ f + dµ + f dµ = f dµ, where we hve used Corollry 3.64 i the lst step. (b) First ote tht by Corollry 3.64 oe hs λ f dµ = λ f dµ < if f is itegrble. This proves tht λ f is itegrble wheever f is. 78

8 3.7 The Lebesgue itegrl Now, if λ > 0, the (λ f ) + = λ f + d (λ f ) = λ f. Thus, usig Corollry 3.64, λ f dµ = λ f + dµ λ f dµ = λ f + dµ λ f dµ = λ f dµ. If, o the other hd, λ < 0, the (λ f ) + = λ f d (λ f ) = λ f +. Thus, i this cse, λ f dµ = λ f dµ λ f + dµ = λ f + dµ λ f dµ = λ f dµ. (c) Sice f + g f + g, it follows tht f + g dµ f + g dµ = f dµ + g dµ < if f d g re itegrble. Moreover, by defiitio d Corollry 3.64, f dµ + g dµ = f + dµ f dµ + g + dµ = f + + g + dµ ( f + g ) dµ ( ) = ( f + g) + dµ ( f + g) dµ = f + g dµ. g dµ Here, ( ) follows from itegrtig the idetity f + + g + + ( f + g) = ( f + g) + + f + g d usig Corollry I the cse where K = C, for (b) d (c) oe uses tht f dµ = Re f dµ + i Im f dµ. We omit the esy computtios. For (), we use tht for every complex umber z oe hs z = sup t R Re(e it z). So for every t R oe hs Re (e it ) f dµ = Re e it f dµ = Re(e it f ) dµ f dµ. Tkig the supremum over t R, sttemet () follows. Theorem 3.75 (Domited covergece theorem). Let (, Σ, µ) be mesure spce d ( f ) be sequece of itegrble fuctios with the followig two properties. () f (x) := lim f (x) exists for lmost every x, sy outside the set N Σ with µ(n) = 0. 79

9 3 Mesure d Itegrtio The techiclities with f d f re required s f might ot be mesurble. Altertively, to void spekig bout f d N, oe could ssume tht mesurble f : K is give tht is poitwise lmost everywhere the limit of the f. (b) There exists itegrble fuctio g with f (x) g(x) for lmost every x d ll N. The f : K, defied by f (x) = f (x) if x N d f (x) = 0 if x N, is itegrble d lim f f dµ = 0. I prticulr, lim f dµ = f dµ. Proof. Chgig f d f o set of mesure zero, we my ssume tht () d (b) hold everywhere. By Propositio 3.30, f is mesurble. Sice f g, it follows tht f is itegrble. Now observe tht f f 2g d hece 2g f f 0. By Ftou s Lemm 3.71, 2g dµ = lim if (2g f f ) dµ lim if 2g f f dµ = 2g dµ lim sup f f dµ. So lim sup f f dµ = 0, d therefore lim f f dµ = 0. By Lemm 3.73, This proves the clim. f dµ f dµ f f dµ 0. I the lecture we discussed here how the expected vlue of rdom vribles is usully computed usig desities for the push-forwrd mesure. Recll tht µ Φ (A) = µ(φ 1 [A]) for ll A F, see Lemm Exmple Let us give exmple tht coditio (b) i Theorem 3.75 is ecessry. Cosider (R, B(R), λ). If we set f := 1 (0, 1 ), the ( f ) is sequece of simple fuctios covergig to 0 everywhere. However, R f dλ 1 0 = R 0 dλ. Exercise Cosider the situtio of Exercise 3.69, i.e. (, Σ, µ) is mesure spce, f : [0, ] is mesurble d ν(a) := A f dµ. Show tht g is itegrble with respect to ν if d oly if g f is itegrble with respect to µ d i this cse, g dν = g f dµ. We close this sectio by cosiderig the itegrtio uder push-forwrd mesure, which is of gret importce for pplictios. Theorem Let (, Σ, µ) be mesure spce, (M, F ) be mesurble spce d Φ : (, Σ) (M, F ) be mesurble. We deote the push-forwrd of µ uder Φ by µ Φ. The for mesurble f : (M, F ) (K, B(K)) we hve 80

10 3.8 O the coectio betwee the Lebesgue d Riem itegrl f Φ L 1 (, Σ, µ) if d oly if f L 1 (M, F, µ Φ ). I this cse, f Φ dµ = M f dµ Φ. Proof. First, let f = k1 Ak be oegtive, simple fuctio. The f Φ = k 1 Φ 1 [A k ] d thus, by the defiitio of the push-forwrd mesure, f Φ dµ = k µ(φ 1 [A]) = k µ Φ (A) = f dµ Φ. M It follows tht for oegtive, simple fuctios the ssertio holds true. Now let f : M [0, ] be mesurble d ( f ) be sequece of simple fuctios with f f poitwise. The, by mootoe covergece d the bove, f Φ dµ = sup f Φ dµ = sup f dµ Φ = N N M M f dµ Φ. This shows tht the ssertio holds for rbitrry mesurble positive f. Sice f Φ = f Φ, it follows tht f L 1 (M, F, µ Φ ) if y oly if f Φ L 1 (, Σ, µ). The geerl formul follows by splittig rel vlued fuctios f ito the positive fuctios f + d f d complex vlued fuctios f ito Re f d Im f. 3.8 O the coectio betwee the Lebesgue d Riem itegrl We ext compre the Lebesgue itegrl with the Riem itegrl. As is well-kow, every cotiuous fuctio f : [, b] R is Riem itegrble. We ow show tht such fuctios re lso Lebesgue itegrble d the Lebesgue itegrl grees with the Riem itegrl. Theorem If f : [, b] K is cotiuous, the f is Lebesgue itegrble fuctio o ([, b], B([, b]), λ). Moreover, [,b] b f dλ = R- f (t) dt. Proof. Let sequece of prtitios π := (t () 0,..., t() k ) with π 0 be give d let ξ = (ξ () 1,..., ξ () ) be sequece of ssocited smple k 81

11 3 Mesure d Itegrtio poits. Put f := k f (ξ () j )1 [t () j 1,t() j ). The f is simple fuctio d [,b] f dλ = S( f, π, ξ ). Moreover, () f f d the ltter is itegrble o our mesure spce d (b) f (t) f (t) for ll t [, b]. Ideed, for fixed t [, b], we hve f (t) f (t) = f (ξ () j the itervl [t () the ξ () j j 1, t() j t t () j ) f (t), where ξ () j is the smple poit i ] d j is chose such tht t lies i this itervl. But t () j 1 π 0 d hece, by the cotiuity of f, it follows tht f (ξ () j ) f (t) 0. Hece the domited covergece theorem, Theorem 3.75, pplies d shows tht f is itegrble d [,b] f dλ = lim f dλ = lim S( f, π, ξ ). [,b] Sice, o the other hd, the Riem sums coverge to R- b ssertio follows. f (t) dt, the Remrk Actully, the cotiuity ssumptio i Theorem 3.79 is ot eeded, if oe is willig to elrge the σ-lgebr. It c be proved tht if f : [, b] K is Riem itegrble the it is lmost everywhere equl to mesurble fuctio tht is Lebesgue itegrble d the Riem d the Lebesgue itegrl coicide. There is lso extesio of Theorem 3.79 to improper Riem itegrls. We recll tht if < < b d f : [, b) R is cotiuous, the f is clled improperly Riem itegrble o [, b) if the limit lim r b R- r f (t) dt exists. The limit is the clled the improper Riem itegrl of f over [, b) d deoted by R- b f (t) dt. Theorem Let < < b d f : [, b) R be cotiuous, such tht the improper Riem itegrl R- b f (t) dt exists, the f is itegrble o ([, b), B([, b)), λ), the improper Riem itegrl R- b f (t)dt exists d b f dλ = R- f (t) dt. [,b) Proof. Pick sequece (b ) (, b) with b b. By mootoe covergece d Theorem 3.79, [,b) f dλ = lim f dλ [,b ] = lim R- b f (t) dt = R- b f (t) dt <. 82

12 3.8 O the coectio betwee the Lebesgue d Riem itegrl It follows tht f is itegrble o [, b). Moreover, sice 1 [,b ) f coverges to f poitwise d 1 [,b ) f f, the domited covergece theorem yields [,b) b f dλ = lim f dλ = lim R- [,b ) where we hve used Theorem 3.79 i the secod step. b f (t) dt = R- f (t) dt, Remrk Similr results s i Theorem 3.81 lso hold for improper Riem itegrls tht re improper o the left-hd side or o both sides. Exmple I Theorem 3.81, the ssumptio tht R- [,b) f (t) dt exists is crucil d cot be omitted. A exmple is give by f : [1, ) R, defied by f (t) = si t t I this cse, by itegrtio by prts, we obti x si t R- 1 t dt = cos t t x x R- 1 1 cos t cos t t 2 dt cos 1 R- 1 t 2 dt s x. The ltter improper Riem itegrl exists sice t 2 cos t t 2 d the ltter is itegrble. It follows tht the improper Riem itegrl R- cos t 1 dt exists. t 2 O the other hd, o ech itervl [kπ, (k + 1)π), we hve f (t) si(t) ((k + 1)π) 1. It thus follows tht [1, ) 1 f dλ (k + 1)π = 1 ( 1 ) π R- k + 1 [kπ,(k+1)π) π 0 si(t) dλ(t) si(t) dt. Sice the hrmoic series diverges, it follows tht f is ot itegrble o [1, ). Remrk I wht follows, we will lso use the differetil dt i Lebesgue itegrls isted of the (formlly correct) dλ. We will thus write b f (t) dt or f (t) dt [,b] to deote the Lebesgue itegrl of f o the itervl [, b]. This is prticulrly helpful whe the fuctio f depeds o more th oe vrible. To hve this feture lso t hd for geerl mesures, we will frequetly write f (x) dµ(x) isted of f dµ to emphsize tht we re itegrtig with respect to the vrible x. 83

13 3 Mesure d Itegrtio 3.9 Itegrls depedig o prmeter Isted of [0, 1] we could use geerl metric spce here. The mi topic of this sectio is to iterchge opertios like itegrtio, differetitio d tkig limits. This is topic t the very hert of lysis. Suppose tht (, Σ, µ) is mesure spce. If we re give mp f : [0, 1] C such tht f (t, ) is itegrble for ll t [0, 1], we my defie F(t) := f (t, x) dµ(x). It is the turl to sk how F depeds o the prmeter t [0, 1]. I this short sectio, we use the domited covergece theorem to prove some results i this directio. Propositio Let (, Σ, µ) be mesure spce. Furthermore, let f : [0, 1] K be such tht the followig three properties hold. () x f (t, x) L 1 (, Σ, µ) for ll t [0, 1]. (b) t f (t, x) is cotiuous for lmost ll x. (c) There exists g L 1 (, Σ, µ) such tht f (t, x) g(x) for ll (t, x) [0, 1]. The F : [0, 1] C defied by F(t) = f (t, x) dµ(x) is cotiuous. Proof. Let t t i [0, 1]. The f (t, x) f (t, x) for lmost ll x by (b). Sice f (t, x) g(x) for ll x by ssumptio d g L 1 (), it follows from the domited covergece theorem, Theorem 3.75, tht F(t ) = f (t, x) dµ(x) f (t, x) dµ(x) = F(t). This proves the cotiuity of F. Propositio Let I be itervl i R d (, Σ, µ) be mesure spce. Furthermore, let f : I K be such tht the followig three properties hold. () x f (t, x) L 1 (, Σ, µ) for ll t I. (b) t f (t, x) is differetible for ll x. (c) There exists g L 1 (, Σ, µ) such tht t f (t, x) g(x) for ll (t, x) I. The F : I K defied by F(t) = f (t, x) dµ(x) is differetible. Moreover, f (t, x) is itegrble for ll t I d t F (t) = d dt f (t, x) dµ(x) = f (t, x) dµ(x). t Proof. Fix t I d let (t ) be sequece i I tht coverges to t. Defie h, h : C by h (x) := (t t) 1 ( f (t, x) f (t, x)) d h(x) = t f (t, x). The h is itegrble for every N s lier combitio of itegrble fuctios. Moreover, h (x) h(x) for ll x by ssumptio. By the me-vlue theorem, h (x) = t f (ξ, x) for some ξ betwee 84

14 3.10 Product mesures t d t. I prticulr, h g. Thus the domited covergece theorem shows tht h is itegrble d F(t ) F(t) t t = This fiishes the proof. h (x) dµ(x) h(x) dµ(x) = f (t, x) dµ(x). t 3.10 Product mesures I this sectio we costruct σ-lgebr d correspodig mesure o the product of two suitble mesure spces. Our motivtio is to exted the theory i order to del with iterted itegrls. The product mesure will llow us to write iterted itegrl s sigle itegrl with resprect to the product mesure. Defiitio Let ( k, Σ k ) be mesurble spce for k = 1,...,. The product (mesurble spce) of the spces ( k, Σ k ) is the mesurble spce ( k, Σ k ), where i is the Crtesi product of the sets k, i.e., the set of ll tuples (x 1,..., x ) where x k k d Σ k is geerted by the cuboids A 1 A where A k Σ k. Exercise Let (, Σ) d, for k = 1,...,, lso ( k, Σ k ) be mesure spces. Let f k : k be fuctio d defie f : k by f (x) = ( f 1 (x),..., f (x)). Show tht f is Σ/ Σ k -mesurble if d oly if f k is Σ/Σ k -mesurble for ll k = 1,...,. I the followig, let ( i, Σ i, µ i ) be σ-fiite mesure spces for i = 1, 2. We defie mesure µ 1 µ 2 o the σ-lgebr Σ 1 Σ 2 which is the product of the mesures µ 1 d µ 2 i the sese tht µ 1 µ 2 (A B) = µ 1 (A)µ 2 (B) for ll A Σ 1 d B Σ 2. Note tht s µ 1 d µ 2 re σ-fiite, by Corollry 3.38 there exists t most oe such mesure. For set Q 1 2 d x 1, y 2, we defie the cuts [Q] x d [Q] y by [Q] x := {y 2 : (x, y) Q} d [Q] y := {x 1 : (x, y) Q}. Lemm For x 1, y 2 d Q Σ 1 Σ 2 we hve [Q] x Σ 2 d [Q] y Σ 1. Observe tht the 2-dimesiol Lebesgue mesure λ 2 is the product mesure of the oe-dimesiol Lebesgue mesure with itself. Proof. We put G := {Q Σ 1 Σ 2 : [Q] x Σ 2 }. We clim tht G is σ- lgebr o 1 2. Clerly (S1) holds, sice [ 1 2 ] x = 2. (S2) d 85