INTEGRATION THEORY AND FUNCTIONAL ANALYSIS MM-501

INTEGRATION THEORY AND FUNCTIONAL ANALYSIS M.A./M.Sc. Mathematcs (Fal) MM-50 Drectorate of Dstace Educato Maharsh Dayaad Uversty ROHTAK 4 00

Copyrght 004, Maharsh Dayaad Uversty, ROHTAK All Rghts Reserved. No part of ths publcato may be reproduced or stored a retreval system or trasmtted ay form or by ay meas; electroc, mechacal, photocopyg, recordg or otherwse, wthout the wrtte permsso of the copyrght holder. Maharsh Dayaad Uversty ROHTAK 4 00 Developed & Produced by EXCEL BOOKS PVT. LTD., A-45 Naraa, Phase, New Delh-0 08

Cotets 3 UNIT I: Sged Measure 5 UNIT II: Normal Lear Spaces 4 UNIT III: Secod Cojugate Spaces 04 UNIT IV: Compact Operatos o Normal Spaces 38 UNIT V: Orthoormal Sets 6

4 M.A./M.Sc. Mathematcs (Fal) INTEGRATION THEORY AND FUNCTIONAL ANALYSIS MM-50 Max. Marks : 00 Tme : 3 Hours Note: Questo paper wll cosst of three sectos. Secto I cosstg of oe questo wth te parts coverg whole of the syllabus of marks each shall be compulsory. From Secto II, 0 questos to be set selectg two questos from each ut. The caddate wll be requred to attempt ay seve questos each of fve marks. Secto III, fve questos to be set, oe from each ut. The caddate wll be requred to attempt ay three questos each of fftee marks. Ut I Sged measure, Hah decomposto theorem, Jorda decomposto theorem, Mutually sgular measure, Rado- Nkodym theorem. Lebesgue decomposto, Lebesgue-Steltjes tegral, Product measures, Fub s theorem. Bare sets, Bare measure, Cotuous fuctos wth compact support, Regularty of measures o locally compact support, Resz-Markoff theorem. Ut II Normed lear spaces, Metrc o ormed lear spaces, Holder s ad Mkowsk s equalty, Completeess of quotet spaces of ormed lear spaces. Completeess of l p, L p, R, C ad C [a, b]. Bouded lear trasformato. Equvalet formulato of cotuty. Spaces of bouded lear trasformatos, Cotuous lear fuctoal, Cojugate spaces, Hah-Baach exteso theorem (Real ad Complex form), Resz Represetato theorem for bouded lear fuctoals o L p ad C[a,b]. Ut III Secod cojugate spaces, Reflexve spaces, Uform boudedess prcple ad ts cosequeces, Ope mappg theorem ad ts applcato, projectos, Closed Graph theorem, Equvalet orms, weak ad strog covergece, ther equvalece fte dmesoal spaces. Ut IV Compact operatos ad ts relato wth cotuous operator. Compactess of lear trasformato o a fte dmesoal space, propertes of compact operators, Compactess of the lmt of the sequece of compact operators. The closed rage theorem. Ier product spaces, Hlbert spaces, Schwarz s equalty, Hlbert space as ormed lear space, Covex sets Hlbert spaces. Projecto theorem. Ut V Orthoormal sets, Bessell s equalty, Parseval s detty, Cojugate of Hlbert space, Resz represetato theorem Hlbert spaces. Adjot of a opertor o a Hlbert space, Reflexvty of Hlbert space, Self-adjot operator, Postve operator, Normal ad utary operators, Projectos o Hlbert space, Spectral theorem o fte dmesoal spaces, Lax-Mlgam theorem.

SIGNED MEASURE 5 Ut-I Sged Measure Sged Measure We defe a measure as a o-egatve set fucto, we wll ow allow measure to take both postve ad egatve values. Suppose that ad are two measures, o the same measurable space (X, B). If we defe a ew measure 3 o (X, B) by settg 3(E) = C (E) + C (E) C, C 0. The t s clear that 3 s a measure, thus two measures ca be added. Ths ca be exteded to ay fte sum. Aother way of costructg ew measures s to multply a gve measure by a arbtrary o-egatve costat. Combg these two methods, we see that f {,,, }. s a fte set of measures ad {,,, } s a fte set of o egatve real umbers. The the set f defed for every set E X by E = E s a measure. Now what happes f we try to defe a measure by ve = E E The frst thg may occur s that v s ot always o-egatve ad ths leads to the cosderato of sged measure whch we shall defe ow. Also we get more dffculty from the fact that v s ot defed whe E = E =. For ths reaso, we should have ether E or E fte wth these cosderato md, we make the followg defto Defto :- Let (X, B) be a measurable space. A exteded real valued fucto, v : B R defed o the algebra B s called a sged measure f t satsfes the followg codtos.

6 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS () v assumes at most oe of the values + ad () v( ) = 0 (3) For ay sequece {E } of dsjot measurable sets. v E ve the equalty here meas that the seres o the R.H.S. coverges absolutely f v E to + or. s fte ad that t properly dverges otherwse.e. deftely dverges Thus a measure s a specal case of sged measure but a sged measure s ot geeral a measure. Defto :- Let (X, B) be a measurable space ad let A be a subspace of X. We say that A s a postve set w.r. to sged measure v f A s measurable ad for every measurable subset E of A we have ve 0. Every measurable subset of postve set s aga postve ad f we take the restrcto of v to a postve set, we obta a measure. Smlarly a set B s called egatve f t s measurable ad every measurable subset E of t has a o-postve v measure.e. ve 0. A set whch s both postve ad egatve wth respect to v s called a ull set. Thus a measurable set s a ull set ff every measurable subset of t has v measure zero. Remark :- Every ull set have measure zero but a set of measure zero may be a uo of two sets whose measures are ot zero but are egatves of each other. Smlarly a postve set s ot to be cofused wth a set whch merely has postve measure. Lemma :- The uo of a coutable collecto of postve sets s postve. Proof :- Let A = A be the uo of a sequece <A > of postve sets. Let E be a measurable subset of A. Sce A are measurable, A s measurable ad A c are measurable. Set E = E C A A C C A The E s a measurable subset of A ad ve dsjot ad. E = E. 0. Sce the sets E s are Therefore we have

SIGNED MEASURE 7 v(e) = v (E ) 0. Thus we have proved that A = A s a measurable set ad for every measurable subset E of A we have v(e) 0. Hece A s a postve set. Lemma :- If E ad F are measurable sets ad v s a sged measure such that E F ad vf < The v E <. Proof :- We have vf = v(f E) + v(e) If exactly oe of the term s fte the so s v(f). If they are both fte, [sce v assumes at most oe of the values + ad.) They are equal ad aga fte. Thus oly oe possblty remas that both terms are fte ad ths proves that every measurable subset of a set of fte sged measure has fte sged measure. Theorem :- Let E be a measurable set such that 0 < ve < postve set A cotaed E wth va > 0.. The there s a Proof :- If E s a postve set the we take A = E ad thus va = ve > 0 whch proves the theorem. We cosder the case whe E s ot postve, the t cotas sets of egatve measure. Let be the smallest postve teger such that there s a measurable set E E wth ve < Now E = (E E ) E ad E E ad E are dsjot Therefore ve = v(e E ) + v(e ) v(e E ) = ve ve () Sce ve s fte (gve). It follows that v(e E ) ad ve are fte. Moreover ve > 0 ad ve s egatve, t follows from () that v(e E ) > 0. Thus 0 < v (E E ) <. If E E s postve, we ca take

8 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS A = E E. Hece the result Suppose that E E s ot postve. The t cotas set of egatve measure. Let be the smallest postve teger such that there s a measurable set E E E wth ve < Now E = E E (E E ) ad E E ad (E E ) are dsjot. Therefore ve = v E E + v[e E ] v E E = v(e) v[e E ] = v(e) [ve + ve ] = ve ve ve Sce ve ad ve are egatve, t follows that v E E 0. If E E s postve, we ca take A = E E ad the Theorem s establshed. If t s ot so, the t cotas sets of egatve measures, let 3 be the smallest teger such that there s a measurable set E 3 E E wth ve 3 <. Proceedg by ducto, let k be the smallest teger for 3 whch there s a measurable set E k E k E ad ve k < k

SIGNED MEASURE 9 If we set A = E (3) E k k The as before E = A E k k Sce ths s a dsjot uo, we have VE = va + v E k k = va + ve k < va k. k Sce ve s fte, the seres o the R.H.S. coverges absolutely. Thus coverges ad we have k. Sce ve k 0 ad ve > 0, we must have va > 0. It remas to show that A s postve set. Let >0 be gve. It s clear from (3) that A s the dfferece of two measurable sets ad therefore A s measurable. k Let >0 be gve. Sce choose k so large that coverges, ths mples that k, we may k ( k ) < Sce A E k j E j A ca cota o measurable sets wth measure less tha k whch s greater tha. Thus A cotas o measurable sets of measure less tha. Sce s a arbtrary postve umber, t follows that A ca cota o sets of egatve measure ad so must be a postve set. Defto :- Hah Decomposto

0 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS A decomposto of X to two dsjot sets A ad B such that A s postve wth respect the sged measure v ad B s egatve wth respect the sged measure v s called a Hah Decomposto for the sged measure v. Hah Decomposto Theorem Statemet :- Let v be a sged measure o the measurable space (X, B). The there s a postve set A ad a egatve set B such that X = A B ad A B =. Proof :- Let v be a sged measure defed o the measurable space (X, B). By defto v assumes at most oe of the values + ad. Therefore w.l.o.g. we may assume that + s the fte value omtted by v. Let be the sup of va over all sets A whch are postve wth respect to v. Sce the empty set s postve, 0. Let {A } be a sequece of postve sets such that = lm va ad set A = A Sce coutable uo of postve sets s postve. Therefore va. () But A A A ad so v(a A ) 0. Sce A = (A A ) A va = v(a A ) + v(a ) v(a ) Hece va () Thus we have from () ad () va =. whch mples that va = ad <. Let B = A C ad let E be a postve subset of B. The E ad A are dsjot ad E A s a postve set. Hece v (E A) = ve + va

SIGNED MEASURE = ve +. ve = 0 [where 0 < ] Thus B cotas o postve subset of postve measure ad hece o subset of postve measure by the prevous Lemma. Cosequetly B s a egatve set ad A B =. Remark :- The above theorem states the exstece of a Hah decomposto for each sged measure. Ufortuately, a Hah-decomposto eed ot be uque. Ifact, t s uque except for ull sets. For f X = A B ad X = A B are two Hah decompostos of X, the we ca show that for a measurable set E, v(e A ) = v(e A ) ad v(e B ) = v(e B ) To see ths, we observe that E (A A ) (E A ) so that v[e (A A )] 0 Moreover E (A A ) E B v[e (A A )] 0 Hece v[e (A A )] = 0 ad by symmetry v[e (A A )] = 0 Thus v(e A ) = v[e (A A )] = v[e A ] Mutually Sgular Measures Defto :- v + (E) = v(e A) ad v (E) = v(e B) are called respectvely postve ad egatve varatos of v. The measure v defed by v (E) = v + E + v E

INTEGRATION THEORY AND FUNCTIONAL ANALYSIS s called the absolute value or total varato of v. Defto :- Two measures v ad v o a measurable space (X, B) are sad to be mutually sgular f there are dsjot measurable sets A ad B wth X = A B such that v (A) = v (B) = 0 Thus the measures v + ad v defed above are mutually sgular sce v + (B) = v(b A) = v ( ) = 0 ad v (A) = v(a B) = v( ) = 0 Jorda Decomposto Defto :- Let v be a sged measure defed o a measurable space (X, B). Let v + ad v be two mutually sgular measures o (X, B) such that v = v + v. The ths decomposto of v s called the Jorda Decomposto of v. Sce v assumes at most oe of values + ad, ether v + ad v must be fte. If they are bot fte, we call v, a fte sged measure. A set E s postve for v f v E = 0. It s a ull set f v (E) = 0. Defto :- A measure v s sad to be absolutely cotuous wth respect to measure f va = 0 for each set A for whch A = 0. We use the symbol v < < whe v s absolutely cotuous wth respect to. I the case of sged measures are v, we say that v f v < < ad v f v Defto :- Let X. For E B, set ve = be a measure ad f, a o-egatve measurable fucto o f d E The v s a set fucto defed o B. Also v s coutably addtve ad hece a measure ad the measure v wll be fte f ad oly f f s tegrable sce the tegral over a set of -measure zero s zero. Jorda Decomposto Theorem Proposto :- Let v be a sged measure o a measurable space (X, B). The there are two mutually sgular measures v + ad v o (X, B) such that v = v + v. Moreover, there s oly oe such par of mutually sgular measure.

SIGNED MEASURE 3 Proof :- Sce by defto v + (E) = v(e A) v (E) = v(e B) v(e) = v(e A) + v(e B) = v + v Also v + ad v are mutually sgular sce where X = A B. v + B = v(a B) = v( ) = 0 v A = v(b A) = v( ) = 0. Sce each such par determes a Hah decomposto ad also we have. Hah-decomposto s uque except for ull sets. Thus there s oly oe such par of mutually sgular measures. Also v takes at most oe of the values + ad mples that at least oe of the set fuctos v + ad v s always fte. Rado Nkodym Theorem Let (X, B, ) be a -fte measure space ad let v be a measure defed o B whch s absolutely cotuous w.r.t. The there s a o-egatve measurable fucto f such that for each set E B, we have ve = E f d, E B. The fucto f s uque the sese that f g s ay measurable fucto wth ths property, the g = f a.e. X w.r.t. Proof :- We frst assume that s fte. The v s a sged measure for each ratoal. Let (A, B ) be a Hah-Decomposto for v ad take A 0 = X ad B 0 =. Now sce X = A B, C B β = A B B B ad s egatve B B A ad s postve. Now B B = B C B β = B A

4 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Thus (v ) (B B ) 0 (v ) (B B ) 0 If > (B B ) = 0, therefore there s a measurable fucto f such that for each ratoal umber, we have f a.e. o A ad f a. e o B. Sce B 0 =, we may take f to be o-egatve. Let E be a arbtrary set B ad set E k = E Bk N Bk ~ N E = E ~ B k N The E = E [ E k ] Ad ths s a dsjot uo. Hece ve = ve + v[ E k ] = v E + k 0 ve k Sce E k = B N Bk N k E Thus E k = E B N k Bk N C = E Bk Ak sce B C k = A k N N Hece E k B k N A k, we have N k N f k N o E k from the above exstece of f. ad so k N E k E k f d k N E k ()

SIGNED MEASURE 5 Thus we have k μe N ve k N k k E k k N E k v E k () ad ve k k N E k (3) Now from (), we have ve k k N E k ve k + N E k k N E k + N E k = k μ E k f d [from ()] N Ek Hece ve k + N E k f d. Ek or f d v E k + Ek N E k (4) Smlarly from (3), we have ve k k N E k ve k N E k k N E k N E k or ve k N E k k N E k Ek f d [from ()] Thus ve k N E k Ek f d (5) Combg (4) ad (5) we have

6 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS ve k N E k Ek f d ve k + N E k. (6) O E, we have f = a.e. If E > 0, we must have v E =. Sce ( ) E s postve for each. If E = 0, we must have ve = 0 sce v < <. I ether case, we ca wrte ve = E f d (7) Thus from (6) ad (7), we have ve μe N E f dμ ve N E Sce E s fte ad N arbtrary, we must have ve = E f d. To show that the theorem s proved for -fte measure, decompose X to coutable uo of X s of fte measure. Applyg the same argumet for each X, we get the requred fucto f. To show the secod part, let g be ay measurable fucto satsfyg ve = E g du, E B For each N defe A = x X,f (x) g(x) B ad B = x X, g(x) f (x) B Sce f(x) g(x) x A A (f g)dμ A

SIGNED MEASURE 7 By learty, we have A f dμ gdμ A A v A v A A 0 A A 0 Sce A ca ot be egatve, we have A = 0. Smlarly we ca show that B = 0 If we take C = {x X, f(x) g(x)} = {A B } But A = 0 = B C = A + B = 0 C = 0 Hece f = g a.e w.r.t. measure. Remark :- The fucto f gve by above theorem s called Rado-Nkodym dv dervatve of v wth respect to. It s deoted by du Lebesgue Decomposto Theorem Let (X, B, ) be a -fte measure space ad v a -fte measure defed o B. The we ca fd a measure v 0 whch s sgular w.r.to ad a measure v whch s absolutely cotuous wth respect to such that v = v 0 + v where the measures v 0 ad v are uque. Proof :- Sce ad v are -fte measures, so s the measure = +v. Sce both ad v are absolutely cotuous wth respect to. The Rado-

8 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Nkodym theorem asserts the exstece of o-egatve measurable fuctos f ad g such that for each x E B E = E f d, v E = E g d. Let A = {x ; f(x) >0} ad B = {x ; f(x) = 0}. The X s the dsjot uo of A ad B whle B = 0. If we defe v 0 by v 0 E = v(e B) The v 0 A = v(a B) = v( ) = 0 Sce A ad B are dsjot. ad so v 0. Let v (E) = v(e A) = E A g d v 0 E + v E = v(e B) + v(e A) = v[(e B) (E A)] = v[e (A B)] = v[e X] = v(e) Thus v = v 0 + v It remas to show that v < <. Let E be a set of -measure zero. The 0 = E = E f d. ad therefore f = 0 a.e. w.r.t. o E. Sce f > 0 o A E, we must have (A E) = 0. Hece v(a E) = 0 ad so v E = v(a E) = 0 v < < Thus v s absolutely cotuous w.r.t.

SIGNED MEASURE 9 Now to prove the uqueess of v 0 ad v, let v 0 ad v be measures such that v = v 0 + v whch has the same propertes as that of measures v 0 ad v. The v. v = v 0 + v ad v = v 0 + v are two Lebesgue decomposto of Thus v 0 v 0 = v v. Takg the uo of the support sets of v 0 ad v 0, we have a set E 0 such that (v 0 v 0 ) (E) = (v 0 v 0 ) (E E 0 ) ad (E 0 ) = 0 But v v s absolutely cotuous w.r.t. ad therefore zero o E 0 sce E 0 = 0. Thus for ay measurable set E, we have (v v )E = (v 0 v 0 )E = (v 0 v 0 ) (E E 0 ) = (v v ) (E E 0 ) = 0 sce v v s zero o E 0 Thus ad v 0 E = v 0 E v E = v E for all measurable sets E whch proves the uqueess of v 0 ad v. Remark :- The detty v = v 0 + v provded by the precedg theorem (where v 0 s sgular w.r.t. ad v s absolutely cotuous wth respect to ) s called the Lebesgue Decomposto of v wth respect to. Lebesgue-Steltjes Itegral Let X be the set of real umbers ad B the class of Borel sets. A measure defed o B ad fte for bouded sets s called a Bare measure (o the real le) to each fte Bare measure, we assocate a fucto F by settg. F(x) = (, x] The fucto F s called the cumulatve dstrbuto fucto of valued ad mootoe creasg we have ad s real (a, b] = F(b) F(a)

0 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Sce (a, b] s the tersecto of the sets a,b (a, b] lm a,b ad so F(b) F(a) = lm F b F(a) F(b) = lm F b = F(b +) Thus a cumulatve dstrbuto fucto cotuous o the rght. Smlarly (b) = lm b, b = lm F (b) F b = F(b) F(b ) Hece F s cotuous at b ff the set {b} cosstg of b aloe has measure zero. Sce = (, ] = lm (, ] 0 = lm [F( )] Sce F s mootoc. lm F() = 0 lm F(x) = 0. x

SIGNED MEASURE Thus we have proved that f s fte Bare measure o the real le, the ts Commulatve Dstrbuto fucto F s a mootoe creasg bouded fucto whch s cotuous o the rght. ad lm F(x) = 0 x Defto :- If s a o-egatve Borel measurable fucto ad F s a mootoe creasg fucto whch s cotuous o the rght. We defe Lebesgue-Steltjes Itegral of wth respect to F as df = d. where s the Bare measure havg F as t cumulatve dstrbuto fucto. If s both postve ad egatve, we say that t s tegrable w.r.t F f t s tegrable w.r.t.. Defto :- If F s ay mootoe creasg fucto the F* s a mootoe creasg fucto defed by F*(x) = y lm F(y) x whch s cotuous o the rght ad equal to F where ever F s cotuous o the rght. Also (F*)* = F* ad f F ad G are mootoe creasg fuctos wherever they both are cotuous, the F* = G*. Thus there s a uque fucto F* whch s mootoe creasg cotuous o the rght ad agrees wth F wherever F s cotuous o the rght. The we defe L-Steltjes tegral of w.r.t. F by df = df*. Proposto :- Let F be a mootoe creasg fucto cotuous o the rght. If (a, b] (a, b ]. The F(b) F(a) [F(b ) F(a )] Proof :- Wrte = (a, b ) ad select tervals as follows. Let a k, say b k b. Let k be such that bk k etc. By the ducto, ths sequece comes to ed whe > b. Reuwhereg the tervals, we have chose b km U, U,, U m where a + < b < b +, =,,, m

INTEGRATION THEORY AND FUNCTIONAL ANALYSIS F(b) F(a) F(b m ) F(a ) m [F(b ) F(a )] [F(b ) F(a )] Product Measures Let (X, S, u) ad (Y,, v) be two fxed measure spaces. The product semrg S of subsets of X Y s defed by S = {A B; A S ad B } The above collecto S s deed a semrg of subsets of X Y. Now defe the set fucto u v : S [0, ] by for each A B S. u v (A B) = u(a). (B) Ths set fucto s a measure o the product semrg S product measure of ad v. (proof gve below), called the Theorem :- The set fucto u v : S [0, ] defed by u v (A B) = u(a). v(b) for each A B S s a measure Proof :- Clearly u v( ) = 0. For the subaddtvty of u v, let A B S ad (A B ) be a sequece of mutually dsjot sets of S such that A B = A B. It must be establshed that u(a). v(b) = u(a ). v(b ) ( ) Obvously ( ) holds f ether A or B has measure zero. Thus we ca assume that u(a) 0 ad v(b) 0.

SIGNED MEASURE 3 Sce A B = χ A B, we see that A(x). B(y) = χ A (x).χ B (y) holds for all x ad y. Now fx y B. Sce (y) equals oe or zero, t follows that χ B A(x) = k χ A (x), where k = { N, y B } observe that the collecto {A ; k}must be dsjot ad thus u(a) = k u(a ) holds. Therefore ( ) u(a). B(y) = u(a ).χ (y) B holds for all y Y. Sce a term wth (A ) = 0 does ot alter the sum ( ) or ( ), we ca assume that u(a ) 0 for all. Now f both A ad B have fte measures, the tegratg term by term, we see that ( holds. O the other had f ether A or B has fte measure, the u(a ). v(b ) = must hold. Ideed f the last sum s fte, the u(a) B (y) defes a tegrable fucto whch s mpossble. Thus ths case ( ) holds wth both sdes fte. Hece the result. The ext few results wll uvel the basc propertes of the product measure u v. As usual (u v)* deotes the outer measure geerated by the measure space (X Y, S, u v) o X Y. Theorem :- If A X ad B Y are measurable sets of fte measure, the (u v)* (A B) = u* v*(a B) = u*(a). v*(b) Proof :- Clearly S u v holds. Now let {A B } be a sequece of S such that A B measure o the semrg u (A B ). Sce by the last theorem, u* v* s a v, t follows that

4 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS u* v*(a B) u* v* (A B ) = u v (A B ) ad so u* v* (A B) (u v)* (A B) O the other had, f >0 s gve, choose two sequeces {A } S ad {B } wth A A, B B such that u(a ) < u*(a) + ad v(b ) < v*(b) +. But the A B m A B m holds ad so (u v)* (A B) m u v (A B m ) = m u(a ). v(b m ) = u(a ). v(b ) < [u*(a) + ]. [v*(b) + ] m m for all >0, that s (u v)* (A B) u*(a). v*(b) = u* v*(a B) Therefore (u v)* (A B) = u* v* (A B) holds as requred. Theorem :- If A s a u-measurable subset of X ad B, a v-measurable subset of Y, the A B s a u v measurable subset of X Y. Proof :- Let C D S wth

SIGNED MEASURE 5 u v(c D) = u(c). v(d) <. To establsh u v measurablty of A B, t s eough to show that (C D) (u v)* ((C D) (A B)) + (u v)* ((C D) (A B) C ) u v If u v (C D) = 0, the the above equalty s obvous. So we ca assume u(c) < ad v(d)<. Clearly (C D) (A B) = (C A) (D B) (C D) (A B) C = [(C A C ) (D B)] [(C A) (D B c )] [(C A C ) (D B C )] hold wth every member of the above uo havg fte measure. Now the subaddtvty of (u v)* combed wth the last theorem gves (u v)* ((C D) (A B)) + (u v)* ((C D) (A B) C ) u*(c A). v*(d B) + u*(c A C ). v* (D B) + u*(c A). v*(d B C ) + u*(c A C ). v*(d B C ) B C )] = [u* (C A) + u*(c A C )]. [v*(d B) + v*(d = u(c). v(d) = u v (C D) as requred. Remark :- I geeral, t s ot true that the measure u* v* s the oly exteso of u v from S to a measure o u v. However f both (X, S, u) ad (Y,, v) are -fte measure spaces, the (X Y, S, u v) s lkewse a -fte measure space, ad therefore u* v* s the oly exteso of u v to a measure o u v. Moreover sce u v u v ad the fact that (u v)* s a measure o u v, t follows ths case that (u v)* = u* v* holds o u v. Defto :- If A s a subset of X Y, ad x X, the the x-secto of A s defed by A x = {y Y ; (x, y) A}

6 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Clearly A x s a subset of Y. Smlarly f y Y, the the y-secto of A s defed by A y = {x X; (x, y) A} Clearly A y s a subset of Y. Remark :- The followg theorem shows that the relato betwee the u v measurable subsets of X Y ad the measurable subsets of X ad Y. Theorem :- Let E be a u v measurable subset of X Y wth (u v)* (E) <. The for u-almost all x, the set E x s a v-measurable subset of Y, ad the fucto X v*(e x ) defes a tegrable fucto over X such that () (u v)* (E) = X v*(e x ) d u(x). Smlarly, for v-almost all y, the set E y s a u-measurable subset of X ad the fucto y u* (E y ) defes a tegrable fucto over such that () (u v)* (E) = Y u*(e y ) dv(y) Proof :- Due to symmetry of () ad (), t s eough to establsh the frst formula. The proof goes by steps. Step I :- Assume E = A B S. Clearly E x = B f x A ad E x = f x A. Thus E x s a v-measurable subset of Y for each x X ad (3) v(e x ) = v(b) A(x) holds for all x X. Sce (u v)* (E) = (u v) (A B) = u(a). v(b) <, two possbltes arse : (a) Both A ad B have fte measure. I ths case (3) shows that x a tegrable fucto (actually, t s a step fucto). Such that v*(e x ) s (b) X v*(e x ) d u(x) = v(b) A du = u(a). v(b) = (u v)* (E). (c) Ether A or B has fte measure.

SIGNED MEASURE 7 I ths case, the other set must have measure zero ad so (3) shows that v(e x ) = 0 for u-almost all x. Thus x v* (E x ) defes the zero fucto ad therefore X v*(e x ) du(x) = 0 = (u v)* (E) Step II :- Assume that E s a -set of S {E }of S such that E = E. I vew of E x =. Choose a dsjot sequece (E ) x ad he precedg step, t follows that E x s a measurable subset of Y for each x X. Now defe f(x) = v*(e x ) ad f (x) = v((e ) x ) for each x X ad all. By step I, each f defes a tegrable fucto ad f du = X v ((E ) x ) d u(x) = u v (E ) (u v)*(e) < Sce {(E ) x } s a dsjot sequece of, we have v*(e x ) = v((e ) x ) ad so f (x) f(x)., holds for each x X. Thus by Lev s theorem Assume that a sequece {f } of tegrable fuctos satsfes f f + a.e. for all ad lm f du <. The there exsts a tegrable fucto f such that f f a.e. ad hece f du f du holds f defes a tegrable fucto ad X v*(e x ) dux = f du = lm f du = u v (E ) = (u v)* (E) Step III :- Assume that E s a coutable tersecto of measure. Choose a sequece {E } of -sets such that E = For each, let g (x) = 0 f (u v)* (E )< ad E + E for all. E, -sets of fte

8 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS v*((e ) x ) = ad g (x) = v* ((E ) x ) f v*((e ) x ) <. By step II, each g s a tegrable fucto over X such that g du = (u v)* (E ) holds. I vew of E x = (E ), t follows that E x s a v- measurable set for each x X. Also sce v*((e ) x )< holds for u-almost all x, t follows that g (x) = v*((e ) x ) v*(e x ) holds for u-almost all x. Thus x v* (E x ) defes a tegrable fucto ad x X v*(e x ) du(x) = lm g du = lm (u v)* (E ) = (u v)* (E) Step IV :- Assume that (u v)* (E) = 0, thus there exsts a measurable set G, whch s a coutable tersecto of -sets of fte measure such that E G ad (u v)* (G) = 0. By step III, X v*(g x ) du(x) = (u v)* (G) = 0 ad so v*(g x ) = 0 holds for u-almost all x. I vew of E x G x for all x, we must have v*(e x ) = 0 for u-almost all x. Therefore E x s v-measurable for u- almost all x ad x v*(e x ) defes the zero fucto. Thus X v*(e x ) du(x) = 0 = (u v)* (E). Step V :- The geeral case. Choose a u v measurable set F that s a coutable tersecto of -sets all of fte measure such that E F ad (u v)* (F) = (u v)* (E). Set G = F ~ E. The G s a ull set ad thus by step IV, v*(g x ) = 0 holds for u-almost all x. Therefore E x s v-measurable ad v*(e x ) = v*(f x ) holds for u-almost all x. By step III x v* (F x ) defes a tegrable fucto ad so x v*(e x ) defes a tegrable fucto ad (u v)*(e) = (u v)*(f) = X v*(f x ) d u(x) = X v*(e x ) d ux. holds. The proof of the theorem s ow complete. Defto :- Let f : X Y R be a fucto. The the terated tegral f du dv s sad to exst f f y s a tegrable fucto over X for v-almost all y ad

SIGNED MEASURE 9 the fucto g(y) = f y du = Y. X f(x, y) du(x) defes a tegrable fucto over If E s a u v measurable subset of X Y wth (u v)* (E) <, the both terated tegrals E du dv ad E dv du exst ad E du dv = E dv du = E d(u v) = (u v)* (E) Sce every u v step fucto s a lear combato of characterstc fuctos of u v measurable sets of fte measure, t follows that f s a u v step fucto, the both terated tegrals du dv ad dv du exst ad moreover du dv = dv du = d(u v) The above dettes regardg terated tegrals are specal cases of a more geeral result kow as Fub s theorem. Fub s Theorem Let f : X Y ad R be u v tegrable fucto. The both terated tegrals exst fd (u v) = fdu dv = f dv du holds. Proof :- Wthout loss of geeralty, we ca assume that f(x, y) x. Choose a sequece { } of step fuctos such that 0 holds for all 0 (x, y) f(x, y) holds for all x ad y. Thus () X [ (x, y) dv (y)] du(x) = (u v) fd(u v) < Y Now by the last theorem, for each, the fucto g (x) = ( ) x dv = Y (x, y) dv(y) defes a tegrable fucto over X ad clearly g (x) holds for u-almost all x. But the by Lev s Theorem Assume that a sequece {f } of tegrable

30 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS fuctos satsfes f f + a.e. for all ad lm f du <. The there exsts a tegrable fucto f such that f f a.e., there exsts a u-tegrable fucto g : X R such that g (x) g(x) u a.e. holds, that s there exsts a u- ull subset A of X such that ( ) x dv g(x)< holds for all x A. Sce ( ) x f x holds for each x, t follows that f x s v-tegrable for all x A ad g (x) = ( ) x dv = Y (x, y) dv(y) Y f x dv holds for all x A. Now () mples that the fucto x such that Y f x dv defes a tegrable fucto f d(u v) = X Y f x dv du = f dv du Smlarly, f d(u v) = fdu. dv ad the proof of the theorem s complete. Remark :- The exstece of the terated tegrals s by o meas eough to esure that the fucto s tegrable over the product space. As a example of ths sort, cosder X = Y = [0, ] u = v = (the Lebesgue measure) ad f(x, y) = (x (x y y ) ) f (x, y) (0, 0) ad f(0, 0) = 0. The f du dv = π ad 4 f dv du = 4 π Fub s theorem shows of course that f s ot tegrable over [0, ] [0, ] There s a coverse to Fub s theorem however accordg to whch the exstece of oe of the terated tegrals s suffcet for the tegralty of the fucto over the product space. Ths result s kow as Toell s Theorem ad ths result s frequetly used applcatos. Measure ad Topology We are ofte cocered wth measures o a set X whch s also a topologcal space ad t s atural to cosder codtos o the measure so that t s coected wth the topologcal structure. There seem to be two classes of topologcal spaces for whch t s possble to carry out a reasoable theory. Oe

SIGNED MEASURE 3 s the class of locally compact Hausdorff spaces ad other s the class of complete metrc spaces. The preset chapter develops the theory for the class of locally compact Hausdorff spaces. Bare Sets ad Borel Sets Let X be a locally compact Hausdorff space. Let C c (X) be the famly cosstg of all cotuous real-valued fuctos that vash outsde a compact subset of X. If f s a real valued fucto, the support of f s the closure of the set {x ; f(x) 0}. Thus C c (X) s the class of all cotuous real valued fuctos o X wth compact support. The class of Bare sets s defed to be the smallest -algebra B of subsets of X such that each fucto C c (X) s measurable wth respect to B. Thus B s the -algebra geerated by the sets {x; f(x) } wth f C c (X). If > 0, these sets are compact G s. Thus each compact G s a Bare set. Cosequetly B s the -algebra geerated by the compact G s If X s ay topologcal space, the smallest -algebra cotag the closed sets s called the class of Borel sets. Thus f X s locally compact, every Bare set s a Borel set. The coverse s true whe X s a locally compact separable metrc space, but there are compact spaces where the class of Borel sets s larger tha the class of Bare sets. Bare Measure Let X be a locally compact Hausdorff space. By a Bare measure o X, we mea a measure defed for all Bare sets ad fte for each compact Bare set. By a Borel measure, we mea a measure defed o the -algebra of Borel sets or completo of such a measure. Defto :- A set E a locally compact Hausdorff space s sad to be (topologcally) bouded f E s cotaed some compact set.e. E s a compact. A set E s sad to be -bouded f t s the uo of a coutable collecto of bouded sets. From ow owards, X wll be a locally compact Hausdorff space. Now we state a umber of Lemmas that are useful dealg wth Bare ad Borel sets. Lemma :- Let K be a compact set, O a ope set wth K K H O where U s a -compact ope set ad H s a compact G. O. The Lemma :- Every -compact ope set s the uo of a coutable collecto of compact G s ad hece a Bare set.

3 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Lemma 3 :- Every bouded set s cotaed a compact G. Every - bouded set E s cotaed a -compact ope set O. If E s bouded, we may take O to be compact. Lemma 4 :- Let R be a rg of sets ad let R = {E; E R}. The ether R = R or else R R = 0. I the latter case R R s the smallest algebra cotag R. If R s a -rg, the R R s a -algebra. Lemma 5 :- If E s a Bare set, the E or E s -bouded. Both are -bouded f ad oly f X s -compact. Lemma 6 :- The class of -bouded Bare sets s the smallest -rg cotag the compact G s. Lemma 7 :- Each -bouded Bare set s the uo of a coutable dsjot uo of bouded Bare sets. Remark :- The followg Proposto gves useful meas of applyg theorems about Bare ad Borel sets compact spaces to bouded Bare ad Borel sets locally compact spaces. Proposto :- Let F be a closed subset of X. The F s a locally compact Hausdorff space ad the Bare sets of F are those sets of the form B F, where B s a Bare set X. Thus f F s a closed Bare set, the Bare subsets of F are just those Bare subsets of X whch are cotaed F. The Borel sets of F are those Borel sets of X whch are cotaed F. Proof :- Let R = { E ; E = B F; B Ba(X)} where Ba(X) s the class of Bare sets. The R s a -algebra whch cludes all compact G s cotaed F. Thus Ba(F) R ad each Bare set of F s of the form B F. Let B = {E X ; E F Ba(F)}. The B s a -algebra. Let K be a compact G X. The K F s a closed subset of K ad hece compact. Sce K s a G x, K F s a G F. Thus K F s a compact G of F ad so s Ba(F). Cosequetly Ba(X) B ad so each Bare set of X terests F a Bare set of F. If F s a closed Bare subset of X, the B F s a Bare subset of X wheever B s. Thus each Bare subset of F s of ths form. O the other had for each Bare subset B of X wth B F we have B = B F ad so B s a Bare subset of F. Cotuous Fuctos wth Compact Support Let X be a locally compact topologcal space. If : X R ad S = {x X; (x) 0}. The the closure K of S s called the support of. Suppose that has support K where K s a compact subset of X. The vashes outsde S. Coversely f vashes outsde some compact set C ad S C as C s closed,

SIGNED MEASURE 33 the closure K of S s cotaed C, ow K s a closed subset of the compact set C ad as such K s compact. The has compact support. Theorem :- Let X be a locally compact Hausdorff space, A ad B o-empty dsjot subsets of X, A closed ad B compact. The there s a cotuous fucto : X [0, ] C of compact support such that (x) = 0 for all x A ad (x) = for all x B. Frst we gve some Lemma. Lemma :- Let X be a Hausdorff space, K a compact subset ad p K c. The there exsts dsjot ope subsets G, H such that p G ad K H. Proof :- To ay pot x of K, there exst dsjot ope sets A x, B x such that p A x, x B x. From the coverg {B x } of K, there s a fte subcoverg Bx, Bx,, Bx ad the sets G = A x, H satsfy the requred codtos. x B Lemma :- Let X be a locally compact Hausdorff space, K a compact subset, U a ope subset ad K U. The there exsts a ope subset V wth compact closure V such that K V V U. Proof :- Let G be the ope set wth compact closure G. If U = X, we smply take V = G. I geeral G s too large, the ope set G U s compact as ts closure s a subset of G but ts closure may stll cota pots outsde U. We assume that the complemet F of U s ot empty. To ay pot p of F, there are dsjot ope sets G p, H p such that p G p, K H p. As F G s compact, there are pots p, p,, p F such that G,G,..., G cover F G. We ow verfy at oce that the ope set V = G H... H satsfy the codtos of the Lemma. p p Proof of the theorem :- Let U be the complemet of A. Accordg to Lemma, there s a ope set V / wth compact closure such that B V V U, ad the there are ope sets B V, V3 4 4 p p V wth compact closure such that 3 V3 V V V V 4 4 4 4 U p

34 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Cotug ths way we obta ope sets V r for each dyadc ratoal umber r = p/ m (0, ) such that ad B V r V r U V r V s for r > s. () Now we must costruct a cotuous fucto : X [0, ]. To ths ed, defe for each r = p/ m (0, ). ad the = sup ψr. r r(x) = r f x V r. = 0 otherwse It follows at oce that 0, that = 0 o A ad = o B. It follows that r ad are lower semcotuous. To prove that s cotuous, we troduce the upper semcotuous fucto r ad defed by r(x) = f x ad = If θr. V r, = r otherwse It s suffcet to show that =. r We ca oly have r(x) > s(x) f r > s x V r ad x by (), whece r s for all r, s ad so. V s. But ths s mpossble O the other had, suppose that (x) < (x) the there are dyadc ratoals r, s (0, ) such that (x) < r < s < (x). As (x) < r, we have x V r ad as (x) > s we have x V s whch aga cotradcts (). Thus, combg these equaltes gves = ad establshes the cotuty of. Hece the result. Regularty of Measure Let be a measure defed o a -algebra M of subsets of X where X s a locally compact Hausdorff space. ad suppose that M cotas the Bare sets. A set E M s sad to be outer regular for (or s outer regular for E) f It s sad to be er regular f E = If { O : E O, ope, O M}

SIGNED MEASURE 35 E = sup { K : K E, K compact, K M} The set E s sad to be regular for f t s both er ad outer regular for. We say that the measure s er regular (outer regular, regular) f t s er regular (outer regular, regular) for each set E M. Lebesgue measure s a regular measure. For compact spaces X, there s complete symmetry betwee er regularty ad outer regularty. A measurable set E s outer regular f ad oly f ts complemet s er regular. A fte measure o X s er regular f ad oly f t s outer regular, ad hece regular. Whe X s compact, every Bare measure s regular. Remark :- Whe X s o loger compact, we lose ths symmetry because the complemet of a ope set eed ot be compact. Proposto :- Let be a fte measure defed o a -algebra M whch cotas all the Bare sets of a locally compact space X. If s er regular, t s regular. Proof :- Let E M, the E = sup { K; K But for each such a K, we have K ope ad E Thus E = X E = If{ X K} E = If { O : E E, K M ad K compact}. = If K If { O; E O} K. Hece O; O ope ad O M}. Theorem : - Let be a Bare measure o a locally compact space X ad E a - bouded Bare set X. The for > 0, () There s a - compact ope set O wth E O ad (O ~ E) <. () E = sup { K ; K E, K a compact G }. Proof :- Let R be the class of sets E that satsfy () ad () for each > 0. Suppose E = UE, where E R. The for each, there s a -compact ope set O wth E O ad (O ~ E ) <. The O = U O s aga a -compact ope set wth ad so (O ~ E) U (O ~ E ) (O ~ E) (O ~ E ) <

36 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Thus E satsfes () If for some,, we have E =, the there are compact G s of arbtrary large fte measure cotaed E E. Hece () holds for E. If u E < for each, there s a K E, K, a compact G ad The (E ~ K ) < E = sup N N E sup N N K + Thus E satsfes (). If E s a compact G, the there s a cotuous real valued fucto wth compact support such that 0 ad E = {x ; (x) = }. Let O = {x ; (x) > /}. The O s a -compact ope set wth O compact. Sce O <, we have E = If O. Thus each compact G satsfes () ad t trvally satsfes () Let X be compact. The E satsfes () f ad oly f E satsfes () ad so the collecto R of sets satsfyg () ad () s a -algebra cotag the compact G s. Thus R cotas all Bare sets ad the prop. holds whe X s compact. For a arbtrary locally compact space X ad bouded Bare set E, we ca take H to be a compact G ad U to be a -compact ope subsets of X such that E U H. The E s a Bare subset of H ad so (W E) < Sce W ad U are -compact, so s O = W set wth E O W. ad O ~ E W ~E (O ~ E) <. Thus E satsfes (). Therefore all bouded Bare sets are R. U. Thus O s a -compact ope Sce R s closed uder coutable uos ad each -bouded Bare set s a coutable uo of bouded Bare sets, we see that every -bouded Bare set belogs to R. Remark :- If we had defed the class of Bare sets to be the smallest -rg cotag the compact G s ad take a Bare measure to be defed o ths - rg, the the above theorem takes the form

SIGNED MEASURE 37 Every Bare measure s regular. If X s -compact, the -rg ad the - algebra geerated by the compact G s are the same. Hece we have the followg corollary. Corollary :- If X s -compact, the every Bare measure o X s regular. Quas-Measure :- A measure defed o -algebra M whch cotas the Bare sets s sad to be quas-regular f t s outer regular ad each ope set O M s er regular for. A Bare measure o a space whch s ot -compact eed ot be regular but we ca requre t to be er regular or quas-regular wthout chagg ts values o the -bouded Bare sets. Proposto :- Let be a measure defed o a -algebra M cotag the Bare sets. Assume ether that s quas-regular or that s er regular. The for each E M wth E<, there s a Bare set B wth (E B) = O Proof :- We cosder oly the quas-regular case. Let E be a measurable set of fte measure. Sce s outer regular, we ca fd a sequece <O > of ope sets wth ad O O + E O < E + Sce s quas-regular, there s a compact set K m O m wth K m > O m m ad we may take K m to be a G set by Lemma. Now Set K m > O m m E m H m = j m > O m K j The H m s a Bare set, H m O m O for m. Also H m H m+, ad H m K m > O m. Let B = H m. The B s a Bare set, B O ad Thus B = lm H m B O,

38 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Sce B O ad E O, we have ad so Ths s true for ay ad so B E (O ~B) (O ~ E) (B E) (O ~ B) + (O ~ E) (B E) = 0. < + = + Proposto :- Let μ be a o-egatve exteded real valued fucto defed o the class of ope subsets of X ad satsfyg () μ O < f O compact. () μ O μ O f O O. () μ (O O ) = μ O + μ O f O O =. (v) μ (UO ) μ O (v) μ (O) == sup { U; U O, U compact} The the set fucto * defed by *E = If { μ O; E O} s a topologcally regular outer measure. Proof :- The mootocty ad coutable subaddtvty of * follow drectly from () ad (v) ad the defto of *. Also *O = μ O for O ope ad so codto () of the defto of regularty follows from hypothess () of the proposto ad the codto () from the defto of *. Sce μ O < for O compact, we have * E < for each bouded set E. Resz-Markov Theorem Let X be a locally compact Hausdorff space. By C c (X), we deote as usual, the space of cotuous real valued fuctos wth compact support. A real valued lear fuctoal I o C c (X) s sad to be postve f I(f) 0 wheever f 0. The purpose of the followg theorem s to prove that every postve lear fuctoal o C c (X) s represeted by tegrato wth respect to a sutable Borel (or Bare) measure. I partcular we have the followg theorem : Statemet of Resz-Markov Theorem Let X be a locally compact Hausdorff space ad I a postve lear fuctoal o C c (X). The there s a Borel measure o X such that I(f) = f du

SIGNED MEASURE 39 For each f C c (X). The measure may be take to be quas-regular or to be er regular. I each of these cases t s the uque. Proof :- For each ope set O defe μ O by μ O = sup {I(f); f C c (X), O f, sup f O} The μ s a exteded real valued fucto defed o all ope sets ad s readly see to be mootoe, fte o bouded sets ad to satsfy the regularty (v) of the above Proposto. To see that μ s coutably subaddtve o ope sets, let O = UO ad let f be ay fucto C c (X) wth O f ad sup f O. Thus there are o-egatve fuctos,,, C c (X) wth sup O ad =. o sup f. The f = f, o f ad sup ( f) O. Thus If = I( f) μ O μ O Takg the sup over all such f gves μ O μ O ad μ s coutably subaddtve. If O = O O wth O O = ad f C c (X), 0 f ad sup f O, the the fucto f = f + f has sup f O ad 0 f. Thus I f + If μ O. Sce f ad f ca be chose arbtrarly, subject to 0 f ad sup f O, we have whece μ O + μ O μ O, μ O + μ O = μ O Thus μ satsfes the hypothess of the above proposto so μ exteds to a quas-regular Borel measure.

40 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS We ext proceed to show that If = f du for each f C c (X). Sce f s the dfferece of two o-egatve fuctos C c (X), t s suffcet to cosder f 0. By learty we may also take f. Choose a bouded ope set O wth sup f O k = {x; f(x) > k } ad O o = O. The O + = ad O k Ok. Defe k = The f = We also have sup k for k. Also f(x) 0 k O k k k O k ad K = o O k+. Thus μ O k+ I k μ O k μ O k+ k d μ μ O k O. Set O O O k k k O k for k. Hece O k (I k k) μ O 0 + μ O Cosequetly sce s arbtrary, If fdu If = f d μ. μ O Thus there s a er regular Borel measure whch agrees wth μ o the -bouded Borel sets. Sce oly the values of o -bouded Bare sets eter to f du, we have If = f du. The ucty of μ ad s obvous.

SIGNED MEASURE 4 Measure ad Topology We are ofte cocered wth measures o a set X whch s also a topologcal space ad t s atural to cosder codtos o the measure so that t s coected wth the topologcal structure. There seem to be two classes of topologcal spaces for whch t s possble to carry out a reasoable theory. Oe s the class of locally compact Hausdorff spaces ad other s the class of complete metrc spaces. The preset chapter develops the theory for the class of locally compact Hausdorff spaces. Bare Sets ad Borel Sets Let X be a locally compact Hausdorff space. Let C c (X) be the famly cosstg of all cotuous real-valued fuctos that vash outsde a compact subset of X. If f s a real valued fucto, the support of f s the closure of the set {x ; f(x) 0}. Thus C c (X) s the class of all cotuous real valued fuctos o X wth compact support. The class of Bare sets s defed to be the smallest -algebra B of subsets of X such that each fucto C c (X) s measurable wth respect to B. Thus B s the -algebra geerated by the sets {x; f(x) } wth f C c (X). If > 0, these sets are compact G s. Thus each compact G s a Bare set. Cosequetly B s the -algebra geerated by the compact G s If X s ay topologcal space, the smallest -algebra cotag the closed sets s called the class of Borel sets. Thus f X s locally compact, every Bare set s a Borel set. The coverse s true whe X s a locally compact separable metrc space, but there are compact spaces where the class of Borel sets s larger tha the class of Bare sets. Bare Measure Let X be a locally compact Hausdorff space. By a Bare measure o X, we mea a measure defed for all Bare sets ad fte for each compact Bare set. By a Borel measure, we mea a measure defed o the - algebra of Borel sets or completo of such a measure. Defto :- A set E a locally compact Hausdorff space s sad to be (topologcally) bouded f E s cotaed some compact set.e. E s a compact. A set E s sad to be -bouded f t s the uo of a coutable collecto of bouded sets. From ow owards, X wll be a locally compact Hausdorff space. Now we state a umber of Lemmas that are useful dealg wth Bare ad Borel sets. Lemma :- Let K be a compact set, O a ope set wth K K H O O. The

4 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS where U s a -compact ope set ad H s a compact G. Lemma :- Every -compact ope set s the uo of a coutable collecto of compact G s ad hece a Bare set. Lemma 3 :- Every bouded set s cotaed a compact G. Every - bouded set E s cotaed a -compact ope set O. If E s bouded, we may take O to be compact. Lemma 4 :- Let R be a rg of sets ad let R = {E; E R}. The ether R = R or else R R = 0. I the latter case R R s the smallest algebra cotag R. If R s a -rg, the R R s a -algebra. Lemma 5 :- If E s a Bare set, the E or E s -bouded. Both are - bouded f ad oly f X s -compact. Lemma 6 :- The class of -bouded Bare sets s the smallest -rg cotag the compact G s. Lemma 7 :- Each -bouded Bare set s the uo of a coutable dsjot uo of bouded Bare sets. Remark :- The followg Proposto gves useful meas of applyg theorems about Bare ad Borel sets compact spaces to bouded Bare ad Borel sets locally compact spaces. Proposto :- Let F be a closed subset of X. The F s a locally compact Hausdorff space ad the Bare sets of F are those sets of the form B F, where B s a Bare set X. Thus f F s a closed Bare set, the Bare subsets of F are just those Bare subsets of X whch are cotaed F. The Borel sets of F are those Borel sets of X whch are cotaed F. Proof :- Let R = { E ; E = B F; B Ba(X)} where Ba(X) s the class of Bare sets. The R s a -algebra whch cludes all compact G s cotaed F. Thus Ba(F) R ad each Bare set of F s of the form B F. Let B = {E X ; E F Ba(F)}. The B s a -algebra. Let K be a compact G X. The K F s a closed subset of K ad hece compact. Sce K s a G x, K F s a G F. Thus K F s a compact G of F ad so s Ba(F). Cosequetly Ba(X) B ad so each Bare set of X terests F a Bare set of F. If F s a closed Bare subset of X, the B F s a Bare subset of X wheever B s. Thus each Bare subset of F s of ths form. O the other had for each Bare subset B of X wth B F we have B = B F ad so B s a Bare subset of F. Cotuous Fuctos wth Compact Support

SIGNED MEASURE 43 Let X be a locally compact topologcal space. If : X R ad S = {x X; (x) 0}. The the closure K of S s called the support of. Suppose that has support K where K s a compact subset of X. The vashes outsde S. Coversely f vashes outsde some compact set C ad S C as C s closed, the closure K of S s cotaed C, ow K s a closed subset of the compact set C ad as such K s compact. The has compact support. Theorem :- Let X be a locally compact Hausdorff space, A ad B oempty dsjot subsets of X, A closed ad B compact. The there s a cotuous fucto : X [0, ] C of compact support such that (x) = 0 for all x A ad (x) = for all x B. Frst we gve some Lemma. Lemma :- Let X be a Hausdorff space, K a compact subset ad p K c. The there exsts dsjot ope subsets G, H such that p G ad K H. Proof :- To ay pot x of K, there exst dsjot ope sets A x, B x such that p A x, x B x. From the coverg {B x } of K, there s a fte subcoverg Bx, Bx,, Bx ad the sets G = A x, H x satsfy the requred codtos. B Lemma :- Let X be a locally compact Hausdorff space, K a compact subset, U a ope subset ad K U. The there exsts a ope subset V wth compact closure V such that K V V U. Proof :- Let G be the ope set wth compact closure G. If U = X, we smply take V = G. I geeral G s too large, the ope set G U s compact as ts closure s a subset of G but ts closure may stll cota pots outsde U. We assume that the complemet F of U s ot empty. To ay pot p of F, there are dsjot ope sets G p, H p such that p G p, K H p. As F G s compact, there are pots p, p,, p F such that G,G,..., G cover F G. We ow verfy at oce that the ope set V = G p p p p H... H satsfy the codtos of the Lemma. p Proof of the theorem :- Let U be the complemet of A. Accordg to Lemma, there s a ope set V / wth compact closure such that B V V U,

44 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS ad the there are ope sets B V V wth compact closure such that, V3 4 4 3 V3 V V V V 4 4 4 4 Cotug ths way we obta ope sets V r for each dyadc ratoal umber r = p/ m (0, ) such that ad B V r V r U V r V s for r > s. () Now we must costruct a cotuous fucto : X [0, ]. To ths ed, defe for each r = p/ m (0, ). ad the = sup ψr. r r(x) = r f x V r. = 0 otherwse It follows at oce that 0, that = 0 o A ad = o B. It follows that r ad are lower semcotuous. To prove that s cotuous, we troduce the upper semcotuous fucto r ad defed by r(x) = f x ad = If θr. V r, = r otherwse It s suffcet to show that =. r We ca oly have r(x) > s(x) f r > s x V r ad x V s. But ths s mpossble by (), whece r s for all r, s ad so. O the other had, suppose that (x) < (x) the there are dyadc ratoals r, s (0, ) such that (x) < r < s < (x). As (x) < r, we have x V r ad as (x) > s we have x V s whch aga cotradcts (). Thus, combg these equaltes gves = ad establshes the cotuty of. Hece the result. Regularty of Measure U