INTEGRATION THEORY AND FUNCTIONAL ANALYSIS MM-501

INTEGRATION THEORY AND FUNCTIONAL ANALYSIS M.A./M.Sc. Mathematics (Fial) MM-50 Directorate of Distace Educatio Maharshi Dayaad Uiversity ROHTAK 4 00

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Cotets 3 UNIT I: Siged Measure 5 UNIT II: Normal Liear Spaces 4 UNIT III: Secod Cojugate Spaces 04 UNIT IV: Compact Operatios o Normal Spaces 38 UNIT V: Orthoormal Sets 6

4 M.A./M.Sc. Mathematics (Fial) INTEGRATION THEORY AND FUNCTIONAL ANALYSIS MM-50 Max. Marks : 00 Time : 3 Hours Note: Questio paper will cosist of three sectios. Sectio I cosistig of oe questio with te parts coverig whole of the syllabus of marks each shall be compulsory. From Sectio II, 0 questios to be set selectig two questios from each uit. The cadidate will be required to attempt ay seve questios each of five marks. Sectio III, five questios to be set, oe from each uit. The cadidate will be required to attempt ay three questios each of fiftee marks. Uit I Siged measure, Hah decompositio theorem, Jorda decompositio theorem, Mutually sigular measure, Rado- Nikodym theorem. Lebesgue decompositio, Lebesgue-Stieltjes itegral, Product measures, Fubii s theorem. Baire sets, Baire measure, Cotiuous fuctios with compact support, Regularity of measures o locally compact support, Riesz-Markoff theorem. Uit II Normed liear spaces, Metric o ormed liear spaces, Holder s ad Mikowski s iequality, Completeess of quotiet spaces of ormed liear spaces. Completeess of l p, L p, R, C ad C [a, b]. Bouded liear trasformatio. Equivalet formulatio of cotiuity. Spaces of bouded liear trasformatios, Cotiuous liear fuctioal, Cojugate spaces, Hah-Baach extesio theorem (Real ad Complex form), Riesz Represetatio theorem for bouded liear fuctioals o L p ad C[a,b]. Uit III Secod cojugate spaces, Reflexive spaces, Uiform boudedess priciple ad its cosequeces, Ope mappig theorem ad its applicatio, projectios, Closed Graph theorem, Equivalet orms, weak ad strog covergece, their equivalece i fiite dimesioal spaces. Uit IV Compact operatios ad its relatio with cotiuous operator. Compactess of liear trasformatio o a fiite dimesioal space, properties of compact operators, Compactess of the limit of the sequece of compact operators. The closed rage theorem. Ier product spaces, Hilbert spaces, Schwarz s iequality, Hilbert space as ormed liear space, Covex sets i Hilbert spaces. Projectio theorem. Uit V Orthoormal sets, Bessell s iequality, Parseval s idetity, Cojugate of Hilbert space, Riesz represetatio theorem i Hilbert spaces. Adjoit of a opertor o a Hilbert space, Reflexivity of Hilbert space, Self-adjoit operator, Positive operator, Normal ad uitary operators, Projectios o Hilbert space, Spectral theorem o fiite dimesioal spaces, Lax-Milgiam theorem.

SIGNED MEASURE 5 Uit-I Siged Measure Siged Measure We defie a measure as a o-egative set fuctio, we will ow allow measure to take both positive ad egative values. Suppose that µ ad µ are two measures, o the same measurable space (X, B). If we defie a ew measure µ 3 o (X, B) by settig µ 3 (E) = C µ (E) + C µ (E) C, C 0. The it is clear that µ 3 is a measure, thus two measures ca be added. This ca be exteded to ay fiite sum. Aother way of costructig ew measures is to multiply a give measure by a arbitrary o-egative costat. Combiig these two methods, we see that if {µ, µ,, µ }. is a fiite set of measures ad {α, α,, α } is a fiite set of o egative real umbers. The the set f µ defied for every set E i X by µ E = α i µ i E is a measure. Now what happes if we try to defie a measure by ve = µ E µ E The first thig may occur is that v is ot always o-egative ad this leads to the cosideratio of siged measure which we shall defie ow. Also we get more difficulty from the fact that v is ot defied whe µ E = µ E =. For this reaso, we should have either µ E or µ E fiite with these cosideratio i mid, we make the followig defiitio Defiitio :- Let (X, B) be a measurable space. A exteded real valued fuctio, v : B R defied o the σ algebra B is called a siged measure if it satisfies the followig coditios.

6 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS () v assumes at most oe of the values + ad () v(φ) = 0 (3) For ay sequece {E i } of disjoit measurable sets. v Υ E i = i = v E i the equality here meas that the series o the R.H.S. coverges absolutely if v Υ Ei is fiite ad that it properly diverges otherwise i.e. defiitely diverges to + or. Thus a measure is a special case of siged measure but a siged measure is ot i geeral a measure. Defiitio :- Let (X, B) be a measurable space ad let A be a subspace of X. We say that A is a positive set w.r. to siged measure v if A is measurable ad for every measurable subset E of A we have ve 0. Every measurable subset of positive set is agai positive ad if we take the restrictio of v to a positive set, we obtai a measure. Similarly a set B is called egative if it is measurable ad every measurable subset E of it has a o-positive v measure i.e. ve 0. A set which is both positive ad egative with respect to v is called a ull set. Thus a measurable set is a ull set iff every measurable subset of it has v measure zero. Remark :- Every ull set have measure zero but a set of measure zero may be a uio of two sets whose measures are ot zero but are egatives of each other. Similarly a positive set is ot to be cofused with a set which merely has positive measure. Lemma :- The uio of a coutable collectio of positive sets is positive. Proof :- Let A = A be the uio of a sequece <A > of positive sets. Let E be a measurable subset of A. Sice A are measurable, A is measurable ad A c are measurable. Set E = E C A C A A C The E is a measurable subset of A ad ve 0. Sice the sets E s are disjoit ad. E = E. Therefore we have

SIGNED MEASURE 7 v(e) = v(e ) 0. = Thus we have proved that A = A is a measurable set ad for every measurable subset E of A we have v(e) 0. Hece A is a positive set. Lemma :- If E ad F are measurable sets ad v is a siged 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 is ifiite the so is v(f). If they are both ifiite, [sice v assumes at most oe of the values + ad.) They are equal ad agai ifiite. Thus oly oe possibility remais that both terms are fiite ad this proves that every measurable subset of a set of fiite siged measure has fiite siged measure. Theorem :- Let E be a measurable set such that 0 < ve <. The there is a positive set A cotaied i E with va > 0. Proof :- If E is a positive set the we take A = E ad thus va = ve > 0 which proves the theorem. We cosider the case whe E is ot positive, the it cotais sets of egative measure. Let be the smallest positive iteger such that there is a measurable set E E with ve < Now E = (E E ) E ad E E ad E are disjoit Therefore ve = v(e E ) + v(e ) v(e E ) = ve ve () Sice ve is fiite (give). It follows that v(e E ) ad ve are fiite. Moreover ve > 0 ad ve is egative, it follows from () that v(e E ) > 0. Thus 0 < v (E E ) <. If E E is positive, we ca take

8 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS A = E E. Hece the result Suppose that E E is ot positive. The it cotais set of egative measure. Let be the smallest positive iteger such that there is a measurable set E E E with ve < Now E = E Υ E i (E E ) ad E Υ E i ad (E E ) are disjoit. Therefore ve = v E Υ E i + v[e E ] v E Υ E i = v(e) v[e E ] = v(e) [ve + ve ] = ve ve ve Sice ve ad ve are egative, it follows that v E Υ E i 0. If E Υ Ei is positive, we ca take A = E Υ E i ad the Theorem is established. If it is ot so, the it cotais sets of egative measures, let 3 be the smallest iteger such that there is a measurable set E 3 E Υ E i with ve 3 <. Proceedig by iductio, let k be the smallest iteger for 3 which there is a measurable set E k E ve k < k k Υ Ei ad

SIGNED MEASURE 9 If we set A = E Υ E k= (3) The as before k E = A Υ E k= Sice this is a disjoit uio, we have k VE = va + v Υ E k= = va + ve k k < va k= k. Sice ve is fiite, the series o the R.H.S. coverges absolutely. Thus coverges ad we have k. Sice ve k 0 ad ve > 0, we must have va > 0. It remais to show that A is positive set. Let >0 be give. It is clear from (3) that A is the differece of two measurable sets ad therefore A is measurable. k Let >0 be give. Sice choose k so large that ( k ) < coverges, this implies that k, we may k Sice A E Υ k E j j= A ca cotai o measurable sets with measure less tha k which is greater tha. Thus A cotais o measurable sets of measure less tha. Sice is a arbitrary positive umber, it follows that A ca cotai o sets of egative measure ad so must be a positive set. Defiitio :- Hah Decompositio

0 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS A decompositio of X ito two disjoit sets A ad B such that A is positive with respect the siged measure v ad B is egative with respect the siged measure v is called a Hah Decompositio for the siged measure v. Hah Decompositio Theorem Statemet :- Let v be a siged measure o the measurable space (X, B). The there is a positive set A ad a egative set B such that X = A B ad A B = φ. Proof :- Let v be a siged measure defied o the measurable space (X, B). By defiitio v assumes at most oe of the values + ad. Therefore w.l.o.g. we may assume that + is the ifiite value omitted by v. Let λ be the sup of va over all sets A which are positive with respect to v. Sice the empty set is positive, λ 0. Let {A i } be a sequece of positive sets such that λ = lim va i i ad set A = Υ A i Sice coutable uio of positive sets is positive. Therefore λ va. () But A A i A ad so v(a A i ) 0. Sice A = (A A i ) A i va = v(a A i ) + v(a i ) v(a i ) Hece va λ () Thus we have from () ad () va = λ. which implies that va = λ ad λ <. Let B = A C ad let E be a positive subset of B. The E ad A are disjoit ad E A is a positive set. Hece λ v (E A) = ve + va

SIGNED MEASURE = ve + λ. ve = 0 [where 0 λ < ] Thus B cotais o positive subset of positive measure ad hece o subset of positive measure by the previous Lemma. Cosequetly B is a egative set ad A B = φ. Remark :- The above theorem states the existece of a Hah decompositio for each siged measure. Ufortuately, a Hah-decompositio eed ot be uique. Ifact, it is uique except for ull sets. For if X = A B ad X = A B are two Hah decompositios 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 this, 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 Sigular Measures Defiitio :- v + (E) = v(e A) ad v (E) = v(e B) are called respectively positive ad egative variatios of v. The measure v defied by v (E) = v + E + v E

INTEGRATION THEORY AND FUNCTIONAL ANALYSIS is called the absolute value or total variatio of v. Defiitio :- Two measures v ad v o a measurable space (X, B) are said to be mutually sigular if there are disjoit measurable sets A ad B with X = A B such that v (A) = v (B) = 0 Thus the measures v + ad v defied above are mutually sigular sice v + (B) = v(b A) = v (φ) = 0 ad v (A) = v(a B) = v(φ) = 0 Jorda Decompositio Defiitio :- Let v be a siged measure defied o a measurable space (X, B). Let v + ad v be two mutually sigular measures o (X, B) such that v = v + v. The this decompositio of v is called the Jorda Decompositio of v. Sice v assumes at most oe of values + ad, either v + ad v must be fiite. If they are bot fiite, we call v, a fiite siged measure. A set E is positive for v if v E = 0. It is a ull set if v (E) = 0. Defiitio :- A measure v is said to be absolutely cotiuous with respect to measure µ if va = 0 for each set A for which µa = 0. We use the symbol v < < µ whe v is absolutely cotiuous with respect to µ. I the case of siged measures µ are v, we say that v µ if v < < µ ad v µ if v µ Defiitio :- Let µ be a measure ad f, a o-egative measurable fuctio o X. For E i B, set ve = f dµ E The v is a set fuctio defied o B. Also v is coutably additive ad hece a measure ad the measure v will be fiite if ad oly if f is itegrable sice the itegral over a set of µ-measure zero is zero. Jorda Decompositio Theorem Propositio :- Let v be a siged measure o a measurable space (X, B). The there are two mutually sigular measures v + ad v o (X, B) such that v = v + v. Moreover, there is oly oe such pair of mutually sigular measure.

SIGNED MEASURE 3 Proof :- Sice by defiitio 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 sigular sice where X = A B. v + B = v(a B) = v(φ) = 0 v A = v(b A) = v(φ) = 0. Sice each such pair determies a Hah decompositio ad also we have. Hah-decompositio is uique except for ull sets. Thus there is oly oe such pair of mutually sigular measures. Also v takes at most oe of the values + ad implies that at least oe of the set fuctios v + ad v is always fiite. Rado Nikodym Theorem Let (X, B, µ) be a σ-fiite measure space ad let v be a measure defied o B which is absolutely cotiuous w.r.t µ. The there is a o-egative measurable fuctio f such that for each set E i B, we have ve = f dµ, E E B. The fuctio f is uique i the sese that if g is ay measurable fuctio with this property, the g = f a.e. i X w.r.t µ. Proof :- We first assume that µ is fiite. The v αµ is a siged measure for each ratioal α. Let (A α, B α ) be a Hah-Decompositio for v αµ ad take A 0 = X ad B 0 = φ. Now sice X = A β B β, C B = A β B α B β B α ad is egative B α B β A β ad is positive. Now B α B β = B α C B = B α A β

4 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Thus (v αµ) (B α B β ) 0 (v βµ) (B α B β ) 0 If β > α µ(b α B β ) = 0, therefore there is a measurable fuctio f such that for each ratioal umber α, we have f α a.e. o A α ad f α a. e o B α. Sice B 0 = φ, we may take f to be o-egative. Let E be a arbitrary set i B ad set Bk+ Bk E k = E ~ N N E = E ~ B k N The E = E [ E k ] Ad this is a disjoit uio. Hece ve = ve + v[ E k ] = v E + k=0 ve k Sice E k = Bk N + Bk N E Thus E k = E Bk N + Bk N C = E Bk N + A k N sice B k C = A k Hece E k ad so B k+ N A k, we have N k k f + o E k from the above existece of f. N N k µ E N k E k f dµ k + µ E k () N

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

6 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS O E, we have f = a.e. ve k µ Ek f dµ ve k + µ Ek. (6) N Ek N If µ E > 0, we must have v E =. Sice (γ αµ) E is positive for each α. If µ E = 0, we must have ve = 0 sice v < < µ. I either case, we ca write ve = E f dµ (7) Thus from (6) ad (7), we have ve E N E f d ve + N µ E Sice µ E is fiite ad N arbitrary, we must have ve = E f dµ. To show that the theorem is proved for σ-fiite measure µ, decompose X ito coutable uio of X i s of fiite measure. Applyig the same argumet for each X i, we get the required fuctio f. To show the secod part, let g be ay measurable fuctio satisfyig ve = E g du, E B For each N defie A = x X,f (x) g(x) B ad B = x X, g(x) f (x) B Sice f(x) g(x) x A A (f g)d µ A

SIGNED MEASURE 7 By liearity, we have f d g d A A µ A v A v A µ A 0 µ A µ A 0 Sice µ A ca ot be egative, we have µ A = 0. Similarly we ca show that µ B = 0 If we take C = {x X, f(x) g(x)} = {A B } But µ A = 0 = µ B µc = µ A + µ B = 0 µc = 0 Hece f = g a.e w.r.t. measure µ. Remark :- The fuctio f give by above theorem is called Rado-Nikodym dv derivative of v with respect to µ. It is deoted by du Lebesgue Decompositio Theorem Let (X, B, µ) be a σ-fiite measure space ad v a σ-fiite measure defied o B. The we ca fid a measure v 0 which is sigular w.r.to µ ad a measure v which is absolutely cotiuous with respect to µ such that v = v 0 + v where the measures v 0 ad v are uique. Proof :- Sice µ ad v are σ-fiite measures, so is the measure λ = µ +v. Sice both µ ad v are absolutely cotiuous with respect to λ. The Rado-

8 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Nikodym theorem asserts the existece of o-egative measurable fuctios f ad g such that for each x E B µ E = E f dλ, v E = E g dλ. Let A = {x ; f(x) >0} ad B = {x ; f(x) = 0}. The X is the disjoit uio of A ad B while µ B = 0. If we defie v 0 by v 0 E = v(e B) The v 0 A = v(a B) = v(φ) = 0 Sice A ad B are disjoit. ad so v 0 µ. Let v (E) = v(e A) = g dλ E A Thus v = v 0 + v 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) It remais to show that v < < µ. Let E be a set of µ-measure zero. The 0 = µ E = E f dλ. ad therefore f = 0 a.e. w.r.t. λ o E. Sice f > 0 o A E, we must have λ(a E) = 0. Hece v(a E) = 0 ad so v E = v(a E) = 0 v < < µ Thus v is absolutely cotiuous w.r.t. µ

SIGNED MEASURE 9 Now to prove the uiqueess of v 0 ad v, let v 0 ad v be measures such that v = v 0 + v which has the same properties as that of measures v 0 ad v. The v. v = v 0 + v ad v = v 0 + v are two Lebesgue decompositio of Thus v 0 v 0 = v v. Takig the uio 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 µ(e 0 ) = 0 But v v is absolutely cotiuous w.r.t. µ ad therefore zero o E 0 sice µ 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 sice v v is zero o E 0 Thus v 0 E = v 0 E ad v E = v E for all measurable sets E which proves the uiqueess of v 0 ad v. Remark :- The idetity v = v 0 + v provided by the precedig theorem (where v 0 is sigular w.r.t. µ ad v is absolutely cotiuous with respect to µ) is called the Lebesgue Decompositio of v with respect to µ. Lebesgue-Stieltjes Itegral Let X be the set of real umbers ad B the class of Borel sets. A measure µ defied o B ad fiite for bouded sets is called a Baire measure (o the real lie) to each fiite Baire measure, we associate a fuctio F by settig. F(x) = µ(, x] The fuctio F is called the cumulative distributio fuctio of µ ad is real valued ad mootoe icreasig we have µ(a, b] = F(b) F(a)

0 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Sice (a, b] is the itersectio of the sets a,b + ad so µ (a, b] lim µa,b + F(b) F(a) = lim F b + F(a) F(b) = lim Fb + = F(b +) Thus a cumulative distributio fuctio cotiuous o the right. Similarly µ(b) = = lim µ b, b lim F (b) F b = F(b) F(b ) Hece F is cotiuous at b iff the set {b} cosistig of b aloe has measure zero. Sice φ = (, ] µ φ = lim µ(, ] 0 = lim [F( )] lim F() = 0 lim F(x) = 0. x Sice F is mootoic.

SIGNED MEASURE Thus we have proved that if µ is fiite Baire measure o the real lie, the its Commulative Distributio fuctio F is a mootoe icreasig bouded fuctio which is cotiuous o the right. ad lim F(x) = 0 x Defiitio :- If φ is a o-egative Borel measurable fuctio ad F is a mootoe icreasig fuctio which is cotiuous o the right. We defie Lebesgue-Stieltjes Itegral of φ with respect to F as φ df = φ dµ. where µ is the Baire measure havig F as it cumulative distributio fuctio. If φ is both positive ad egative, we say that it is itegrable w.r.t F if it is itegrable w.r.t. µ. Defiitio :- If F is ay mootoe icreasig fuctio the F* is a mootoe icreasig fuctio defied by F*(x) = lim F(y) y x+ which is cotiuous o the right ad equal to F where ever F is cotiuous o the right. Also (F*)* = F* ad if F ad G are mootoe icreasig fuctios wherever they both are cotiuous, the F* = G*. Thus there is a uique fuctio F* which is mootoe icreasig cotiuous o the right ad agrees with F wherever F is cotiuous o the right. The we defie L-Stieltjes itegral of φ w.r.t. F by φ df = φ df*. Propositio :- Let F be a mootoe icreasig fuctio cotiuous o the right. If (a, b] Υ (a i, b i ]. The F(b) F(a) [F(b i ) F(a i )] Proof :- Write i = (a i, b i ) ad select itervals as follows. Let a k, say b k b. Let k be such that b etc. By the iductio, this sequece comes to ed whe b km k k > b. Reuwhereig the itervals, we have chose U, U,, U m where a i+ < b i < b i+, i =,,, m

INTEGRATION THEORY AND FUNCTIONAL ANALYSIS F(b) F(a) F(b m ) F(a ) m [F(b i ) F(a i )] [F(b i ) F(a i )] Product Measures Let (X, S, u) ad (Y,, v) be two fixed measure spaces. The product semirig S of subsets of X Y is defied by S = {A B; A S ad B } The above collectio S is ideed a semirig of subsets of X Y. Now defie the set fuctio u v : S [0, ] by for each A B S. u v (A B) = u(a). µ(b) This set fuctio is a measure o the product semirig S, called the product measure of µ ad v. (proof give below) Theorem :- The set fuctio u v : S [0, ] defied by u v (A B) = u(a). v(b) for each A B S is a measure Proof :- Clearly u v(φ) = 0. For the subadditivity of u v, let A B S ad (A B ) be a sequece of mutually disjoit sets of S such that A B = Υ = A B. It must be established that u(a). v(b) = = u(a ). v(b ) ( ) Obviously ( ) holds if either A or B has measure zero. Thus we ca assume that u(a) 0 ad v(b) 0.

SIGNED MEASURE 3 Sice χ A B = = A B, we see that χ A (x). χ B (y) = = A (x). B (y) holds for all x ad y. Now fix y B. Sice (y) equals oe or zero, it follows that B χ A (x) = i k A (x), where k = {i N, y B i } i observe that the collectio {A i ; i k}must be disjoit ad thus u(a) = i k u(a i ) holds. Therefore ( ) u(a). χ B (y) = u(a ). (y) = B holds for all y Y. Sice a term with µ(a ) = 0 does ot alter the sum i ( ) or ( ), we ca assume that u(a ) 0 for all. Now if both A ad B have fiite measures, the itegratig term by term, we see that ( ) holds. O the other had if either A or B has ifiite measure, the = u(a ). v(b ) = must hold. Ideed if the last sum is fiite, the u(a)χ B (y) defies a itegrable fuctio which is impossible. Thus i this case ( ) holds with both sides ifiite. Hece the result. The ext few results will uveil the basic properties 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 fiite measure, the (u v)* (A B) = u* v*(a B) = u*(a). v*(b) Proof :- Clearly S Λ Λ holds. Now let {A B } be a sequece of S such that A B Υ = measure o the semirig Λ u u v (A B ). Sice by the last theorem, u* v* is a Λ v, it 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, if >0 is give, choose two sequeces {A } S ad {B } with A Υ = A, B Υ = B such that = u(a ) < u*(a) + ad = v(b ) < v*(b) +. But the A B Υ Υ = 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 ). m= v(b m ) for all >0, that is Therefore < [u*(a) + ]. [v*(b) + ] (u v)* (A B) u*(a). v*(b) = u* v*(a B) (u v)* (A B) = u* v* (A B) holds as required. Theorem :- If A is a u-measurable subset of X ad B, a v-measurable subset of Y, the A B is a u v measurable subset of X Y. Proof :- Let C D S with

SIGNED MEASURE 5 u v(c D) = u(c). v(d) <. To establish u v measurability of A B, it is 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 iequality is obvious. 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 with every member of the above uio havig fiite measure. Now the subadditivity of (u v)* combied with the last theorem gives (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 required. Remark :- I geeral, it is ot true that the measure u* v* is the oly extesio of u v from S to a measure o Λ Λ. However if both (X, S, u) ad (Y,, v) are σ-fiite measure spaces, the (X Y, S, u v) is likewise a σ-fiite measure space, ad therefore u* v* is the oly extesio of u v to a measure o Λ u Λ v. Moreover sice Λ u Λ v Λ u v ad the fact that (u v)* is a measure o Λ, it follows i this case that (u v)* = u* v* holds o Λ u Λ v. u v Defiitio :- If A is a subset of X Y, ad x X, the the x-sectio of A is defied by A x = {y Y ; (x, y) A} u v

6 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Clearly A x is a subset of Y. Similarly if y Y, the the y-sectio of A is defied by A y = {x X; (x, y) A} Clearly A y is a subset of Y. Remark :- The followig theorem shows that the relatio 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 with (u v)* (E) <. The for u-almost all x, the set E x is a v-measurable subset of Y, ad the fuctio X v*(e x ) defies a itegrable fuctio over X such that () (u v)* (E) = X v*(e x ) d u(x). Similarly, for v-almost all y, the set E y is a u-measurable subset of X ad the fuctio y u* (E y ) defies a itegrable fuctio over such that () (u v)* (E) = Y u*(e y ) dv(y) Proof :- Due to symmetry of () ad (), it is eough to establish the first formula. The proof goes by steps. Step I :- Assume E = A B S. Clearly E x = B if x A ad E x = φ if x A. Thus E x is a v-measurable subset of Y for each x X ad (3) v(e x ) = v(b) χ A (x) holds for all x X. Sice (u v)* (E) = (u v) (A B) = u(a). v(b) <, two possibilities arise : (a) Both A ad B have fiite measure. I this case (3) shows that x v*(e x ) is a itegrable fuctio (actually, it is a step fuctio). Such that (b) X v*(e x ) d u(x) = v(b) χ A du = u(a). v(b) = (u v)* (E). (c) Either A or B has ifiite measure.

SIGNED MEASURE 7 I this 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 ) defies the zero fuctio ad therefore X v*(e x ) du(x) = 0 = (u v)* (E) Step II :- Assume that E is a σ-set of S. Choose a disjoit sequece {E }of S such that E = Υ = E. I view of E x = Υ = (E ) x ad he precedig step, it follows that E x is a measurable subset of Y for each x X. Now defie f(x) = v*(e x ) ad f (x) = defies a itegrable fuctio ad v((e i ) x ) for each x X ad all. By step I, each f f du = X v ((E i ) x ) d u(x) = u v (E i ) (u v)*(e) < Sice {(E ) x } is a disjoit sequece of, we have v*(e x ) = = v((e ) x ) ad so f (x) f(x)., holds for each x X. Thus by Levi s theorem Assume that a sequece {f } of itegrable fuctios satisfies f f + a.e. for all ad lim f du <. The there exists a itegrable fuctio f such that f f a.e. ad hece f du f du holds f defies a itegrable fuctio ad X v*(e x ) dux = f du = lim f du = u v (E i ) = (u v)* (E) Step III :- Assume that E is a coutable itersectio of σ-sets of fiite measure. Choose a sequece {E } of σ-sets such that E = Ι = E, For each, let g (x) = 0 if (u v)* (E )< ad E + E for all.

8 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS v*((e ) x ) = ad g (x) = v* ((E ) x ) if v*((e ) x ) <. By step II, each g is a itegrable fuctio over X such that g du = (u v)* (E ) holds. I view of E x = Ι = (E ) x, it follows that E x is a v- measurable set for each x X. Also sice v*((e ) x )< holds for u-almost all x, it follows that g (x) = v*((e ) x ) v*(e x ) holds for u-almost all x. Thus x v* (E x ) defies a itegrable fuctio ad X v*(e x ) du(x) = lim g du = lim (u v)* (E ) = (u v)* (E) Step IV :- Assume that (u v)* (E) = 0, thus there exists a measurable set G, which is a coutable itersectio of σ-sets of fiite 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 view of E x G x for all x, we must have v*(e x ) = 0 for u-almost all x. Therefore E x is v-measurable for u- almost all x ad x v*(e x ) defies the zero fuctio. Thus X v*(e x ) du(x) = 0 = (u v)* (E). Step V :- The geeral case. Choose a u v measurable set F that is a coutable itersectio of σ-sets all of fiite measure such that E F ad (u v)* (F) = (u v)* (E). Set G = F ~ E. The G is a ull set ad thus by step IV, v*(g x ) = 0 holds for u-almost all x. Therefore E x is v-measurable ad v*(e x ) = v*(f x ) holds for u-almost all x. By step III x v* (F x ) defies a itegrable fuctio ad so x v*(e x ) defies a itegrable fuctio 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 is ow complete. Defiitio :- Let f : X Y R be a fuctio. The the iterated itegral f du dv is said to exist if f y is a itegrable fuctio over X for v-almost all y ad

SIGNED MEASURE 9 the fuctio g(y) = f y du = X Y. f(x, y) du(x) defies a itegrable fuctio over If E is a u v measurable subset of X Y with (u v)* (E) <, the both iterated itegrals χ E du dv ad χ E dv du exist ad χ E du dv = χ E dv du = χ E d(u v) = (u v)* (E) Sice every u v step fuctio is a liear combiatio of characteristic fuctios of u v measurable sets of fiite measure, it follows that if φ is a u v step fuctio, the both iterated itegrals φ du dv ad φ dv du exist ad moreover φ du dv = φ dv du = φ d(u v) The above idetities regardig iterated itegrals are special cases of a more geeral result kow as Fubii s theorem. Fubii s Theorem Let f : X Y R be u v itegrable fuctio. The both iterated itegrals exist ad holds. fd (u v) = fdu dv = f dv du Proof :- Without loss of geerality, we ca assume that f(x, y) 0 holds for all x. Choose a sequece {φ } of step fuctios such that Thus 0 φ (x, y) f(x, y) holds for all x ad y. () X [ φ (x, y) dv (y)] du(x) = φ (u v) fd(u v) < Y Now by the last theorem, for each, the fuctio g (x) = (φ ) x dv = Y φ (x, y) dv(y) defies a itegrable fuctio over X ad clearly g (x) holds for u-almost all x. But the by Levi s Theorem Assume that a sequece {f } of itegrable

30 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS fuctios satisfies f f + a.e. for all ad lim f du <. The there exists a itegrable fuctio f such that f f a.e., there exists a u-itegrable fuctio g : X R such that g (x) g(x) u a.e. holds, that is there exists a u- ull subset A of X such that (φ ) x dv g(x)< holds for all x A. Sice (φ ) x f x holds for each x, it follows that f x is v-itegrable for all x A ad holds for all x A. g (x) = (φ ) x dv = Y φ (x, y) dv(y) Y f x dv Now () implies that the fuctio x Y such that f x dv defies a itegrable fuctio f d(u v) = X Y f x dv du = f dv du Similarly, f d(u v) = fdu. dv ad the proof of the theorem is complete. Remark :- The existece of the iterated itegrals is by o meas eough to esure that the fuctio is itegrable over the product space. As a example of this sort, cosider X = Y = [0, ] u = v = λ (the Lebesgue measure) ad f(x, y) = (x (x y + y ) ) if (x, y) (0, 0) ad f(0, 0) = 0. The f du dv = 4 ad f dv du = 4 Fubii s theorem shows of course that f is ot itegrable over [0, ] [0, ] There is a coverse to Fubii s theorem however accordig to which the existece of oe of the iterated itegrals is sufficiet for the itegrality of the fuctio over the product space. This result is kow as Toell s Theorem ad this result is frequetly used i applicatios. Measure ad Topology We are ofte cocered with measures o a set X which is also a topological space ad it is atural to cosider coditios o the measure so that it is coected with the topological structure. There seem to be two classes of topological spaces for which it is possible to carry out a reasoable theory. Oe

SIGNED MEASURE 3 is the class of locally compact Hausdorff spaces ad other is the class of complete metric spaces. The preset chapter develops the theory for the class of locally compact Hausdorff spaces. Baire Sets ad Borel Sets Let X be a locally compact Hausdorff space. Let C c (X) be the family cosistig of all cotiuous real-valued fuctios that vaish outside a compact subset of X. If f is a real valued fuctio, the support of f is the closure of the set {x ; f(x) 0}. Thus C c (X) is the class of all cotiuous real valued fuctios o X with compact support. The class of Baire sets is defied to be the smallest σ-algebra B of subsets of X such that each fuctio i C c (X) is measurable with respect to B. Thus B is the σ-algebra geerated by the sets {x; f(x) α} with f C c (X). If α > 0, these sets are compact G δ s. Thus each compact G δ is a Baire set. Cosequetly B is the σ-algebra geerated by the compact G γ s If X is ay topological space, the smallest σ-algebra cotaiig the closed sets is called the class of Borel sets. Thus if X is locally compact, every Baire set is a Borel set. The coverse is true whe X is a locally compact separable metric space, but there are compact spaces where the class of Borel sets is larger tha the class of Baire sets. Baire Measure Let X be a locally compact Hausdorff space. By a Baire measure o X, we mea a measure defied for all Baire sets ad fiite for each compact Baire set. By a Borel measure, we mea a measure defied o the σ-algebra of Borel sets or completio of such a measure. Defiitio :- A set E i a locally compact Hausdorff space is said to be (topologically) bouded if E is cotaied i some compact set i.e. E is a compact. A set E is said to be σ-bouded if it is the uio of a coutable collectio of bouded sets. From ow owards, X will be a locally compact Hausdorff space. Now we state a umber of Lemmas that are useful i dealig with Baire ad Borel sets. Lemma :- Let K be a compact set, O a ope set with K O. The K H O where U is a σ-compact ope set ad H is a compact G δ. Lemma :- Every σ-compact ope set is the uio of a coutable collectio of compact G δ s ad hece a Baire set.

3 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Lemma 3 :- Every bouded set is cotaied i a compact G δ. Every σ- bouded set E is cotaied i a σ-compact ope set O. If E is bouded, we may take O to be compact. Lemma 4 :- Let R be a rig of sets ad let R = {E; E R}. The either R = R or else R R = 0. I the latter case R R is the smallest algebra cotaiig R. If R is a σ-rig, the R R is a σ-algebra. Lemma 5 :- If E is a Baire set, the E or E is σ-bouded. Both are σ-bouded if ad oly if X is σ-compact. Lemma 6 :- The class of σ-bouded Baire sets is the smallest σ-rig cotaiig the compact G δ s. Lemma 7 :- Each σ-bouded Baire set is the uio of a coutable disjoit uio of bouded Baire sets. Remark :- The followig Propositio gives useful meas of applyig theorems about Baire ad Borel sets i compact spaces to bouded Baire ad Borel sets i locally compact spaces. Propositio :- Let F be a closed subset of X. The F is a locally compact Hausdorff space ad the Baire sets of F are those sets of the form B F, where B is a Baire set i X. Thus if F is a closed Baire set, the Baire subsets of F are just those Baire subsets of X which are cotaied i F. The Borel sets of F are those Borel sets of X which are cotaied i F. Proof :- Let R = { E ; E = B F; B Ba(X)} where Ba(X) is the class of Baire sets. The R is a σ-algebra which icludes all compact G δ s cotaied i F. Thus Ba(F) R ad each Baire set of F is of the form B F. Let B = {E X ; E F Ba(F)}. The B is a σ-algebra. Let K be a compact G δ i X. The K F is a closed subset of K ad hece compact. Sice K is a G δ i x, K F is a G δ i F. Thus K F is a compact G δ of F ad so is i Ba(F). Cosequetly Ba(X) B ad so each Baire set of X iterests F i a Baire set of F. If F is a closed Baire subset of X, the B F is a Baire subset of X wheever B is. Thus each Baire subset of F is of this form. O the other had for each Baire subset B of X with B F we have B = B F ad so B is a Baire subset of F. Cotiuous Fuctios with Compact Support Let X be a locally compact topological space. If φ : X R ad S = {x X; φ(x) 0}. The the closure K of S is called the support of φ. Suppose that φ has support K where K is a compact subset of X. The φ vaishes outside S. Coversely if φ vaishes outside some compact set C ad S C as C is closed,

SIGNED MEASURE 33 the closure K of S is cotaied i C, ow K is a closed subset of the compact set C ad as such K is compact. The φ has compact support. Theorem :- Let X be a locally compact Hausdorff space, A ad B o-empty disjoit subsets of X, A closed ad B compact. The there is a cotiuous fuctio ψ : X [0, ] C of compact support such that ψ(x) = 0 for all x i A ad ψ(x) = for all x i B. First we give some Lemma. Lemma :- Let X be a Hausdorff space, K a compact subset ad p K c. The there exists disjoit ope subsets G, H such that p G ad K H. Proof :- To ay poit x of K, there exist disjoit ope sets A x, B x such that p A x, x B x. From the coverig {B x } of K, there is a fiite subcoverig Bx, Bx,, Bx ad the sets G = Ι A, H = satisfy the required coditios. Υ B xi x i Lemma :- Let X be a locally compact Hausdorff space, K a compact subset, U a ope subset ad K U. The there exists a ope subset V with compact closure V such that K V V U. Proof :- Let G be the ope set with compact closure G. If U = X, we simply take V = G. I geeral G is too large, the ope set G U is compact as its closure is a subset of G but its closure may still cotai poits outside U. We assume that the complemet F of U is ot empty. To ay poit p of F, there are disjoit ope sets G p, H p such that p G p, K H p. As F G is compact, there are poits p, p,, p i F such that G,G,..., G cover F G. We ow verify at oce that the ope set V = G H... H satisfy the coditios of the Lemma. p p Proof of the theorem :- Let U be the complemet of A. Accordig to Lemma, there is a ope set V / with compact closure such that B V V U, ad the there are ope sets, V 3 4 4 p p V with compact closure such that B V3 V3 V V V V 4 4 4 4 U p

34 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Cotiuig i this way we obtai ope sets V r for each dyadic ratioal umber r = p/ m i (0, ) such that ad B V r V U Vr V s for r > s. () r Now we must costruct a cotiuous fuctio ψ : X [0, ]. To this ed, defie for each r = p/ m i (0, ). ad the ψ = ψ r (x) = r if x V r. sup. r r = 0 otherwise It follows at oce that 0 ψ, that ψ = 0 o A ad ψ = o B. It follows that ψ r ad ψ are lower semicotiuous. To prove that ψ is cotiuous, we itroduce the upper semicotiuous fuctio θ r ad θ defied by θ r (x) = if x V, ad θ = If r. = r otherwise It is sufficiet to show that ψ = θ. r r We ca oly have ψ r (x) > θ s (x) if r > s x V r ad x V. But this is impossible by (), whece ψ r θ s for all r, s ad so ψ θ. O the other had, suppose that ψ(x) < θ(x) the there are dyadic ratioals r, s i (0, ) such that ψ(x) < r < s < θ(x). As ψ(x) < r, we have x V r ad as θ(x) > s we have x V which agai cotradicts (). Thus ψ θ, combiig these iequalities gives ψ = θ ad establishes the cotiuity of ψ. Hece the result. Regularity of Measure Let µ be a measure defied o a σ-algebra M of subsets of X where X is a locally compact Hausdorff space. ad suppose that M cotais the Baire sets. A set E M is said to be outer regular for µ(or µ is outer regular for E) if It is said to be ier regular if µe = If {µo : E O, ope, O M} s s

SIGNED MEASURE 35 µe = sup {µk : K E, K compact, K M} The set E is said to be regular for µ if it is both ier ad outer regular for µ. We say that the measure µ is ier regular (outer regular, regular) if it is ier regular (outer regular, regular) for each set E M. Lebesgue measure is a regular measure. For compact spaces X, there is complete symmetry betwee ier regularity ad outer regularity. A measurable set E is outer regular if ad oly if its complemet is ier regular. A fiite measure o X is ier regular if ad oly if it is outer regular, ad hece regular. Whe X is compact, every Baire measure is regular. Remark :- Whe X is o loger compact, we lose this symmetry because the complemet of a ope set eed ot be compact. Propositio :- Let µ be a fiite measure defied o a σ-algebra M which cotais all the Baire sets of a locally compact space X. If µ is ier regular, it is regular. Proof :- Let E M, the µ E = sup {µk; K E, K M ad K compact}. But for each such a K, we have K ope ad E K. Hece Thus µe = µx µ E = If{µX µk} = If µ K If {µo; E O} µe = If {µo : E O; O ope ad O M}. Theorem : - Let µ be a Baire measure o a locally compact space X ad E a σ- bouded Baire set i X. The for > 0, (i) There is a σ - compact ope set O with E O ad µ (O ~ E) <. (ii) µ E = sup {µk ; K E, K a compact G δ }. Proof :- Let R be the class of sets E that satisfy (i) ad (ii) for each > 0. Suppose E = UE, where E R. The for each, there is a σ-compact ope set O with E O ad µ (O ~ E ) <. The O = U O is agai a σ-compact ope set with ad so µ (O ~ E) U (O ~ E ) µ(o ~ E) µ(o ~ E ) <

36 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Thus E satisfies (i) If for some,, we have µ E =, the there are compact G δ s of arbitrary large fiite measure cotaied i E E. Hece (ii) holds for E. If u E < for each, there is a K E, K, a compact G δ ad The µ(e ~ K ) < µ E = sup µ N Υ N = E sup µ N Υ N = K + Thus E satisfies (ii). If E is a compact G δ, the there is a cotiuous real valued fuctio φ with compact support such that 0 φ ad E = {x ; φ(x) = }. Let O = {x ; φ(x) > /}. The O is a σ-compact ope set with O compact. Sice µo <, we have µe = If µ O. Thus each compact G δ satisfies (i) ad it trivially satisfies (ii) Let X be compact. The E satisfies (i) if ad oly if E satisfies (ii) ad so the collectio R of sets satisfyig (i) ad (ii) is a σ-algebra cotaiig the compact G δ s. Thus R cotais all Baire sets ad the prop. holds whe X is compact. For a arbitrary locally compact space X ad bouded Baire set E, we ca take H to be a compact G δ ad U to be a σ-compact ope subsets of X such that E U H. The E is a Baire subset of H ad so µ (W E) < Sice W ad U are σ-compact, so is O = W U. Thus O is a σ-compact ope set with E O W. ad O ~ E W ~E µ(o ~ E) <. Thus E satisfies (i). Therefore all bouded Baire sets are i R. Sice R is closed uder coutable uios ad each σ-bouded Baire set is a coutable uio of bouded Baire sets, we see that every σ-bouded Baire set belogs to R. Remark :- If we had defied the class of Baire sets to be the smallest σ-rig cotaiig the compact G δ s ad take a Baire measure to be defied o this σ- rig, the the above theorem takes the form

SIGNED MEASURE 37 Every Baire measure is regular. If X is σ-compact, the σ-rig ad the σ- algebra geerated by the compact G δ s are the same. Hece we have the followig corollary. Corollary :- If X is σ-compact, the every Baire measure o X is regular. Quasi-Measure :- A measure µ defied o σ-algebra M which cotais the Baire sets is said to be quasi-regular if it is outer regular ad each ope set O M is ier regular for µ. A Baire measure o a space which is ot σ-compact eed ot be regular but we ca require it to be ier regular or quasi-regular without chagig its values o the σ-bouded Baire sets. Propositio :- Let µ be a measure defied o a σ-algebra M cotaiig the Baire sets. Assume either that µ is quasi-regular or that µ is ier regular. The for each E M with µe<, there is a Baire set B with µ(e B) = O Proof :- We cosider oly the quasi-regular case. Let E be a measurable set of fiite measure. Sice µ is outer regular, we ca fid a sequece <O > of ope sets with ad O O + E µ O < µ E + Sice µ is quasi-regular, there is a compact set K m O m with µk m > µo m m ad we may take K m to be a G δ set by Lemma. Now Set µk m > µo m m µe m H m = Υ j=m > µo m K j The H m is a Baire set, H m O m O for m. Also H m H m+, ad µ H m µk m > µo m. Let B = H m. The B is a Baire set, B O ad Thus µb = lim µ H m µb µo,

38 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS Sice B O ad E O, we have ad so This is true for ay ad so B E (O ~B) (O ~ E) µ(b E) µ(o ~ B) + µ(o ~ E) µ(b E) = 0. < + = + Propositio :- Let be a o-egative exteded real valued fuctio defied o the class of ope subsets of X ad satisfyig (i) O < if O compact. (ii) O O if O O. (iii) (O O ) = O + O if O O = φ. (iv) (UO i ) O i (v) (O) == sup {µ U; U O, U compact} The the set fuctio µ* defied by µ*e = If { O; E O} is a topologically regular outer measure. Proof :- The mootoicity ad coutable subadditivity of µ* follow directly from (ii) ad (iv) ad the defiitio of µ*. Also µ*o = O for O ope ad so coditio (ii) of the defiitio of regularity follows from hypothesis (iii) of the propositio ad the coditio (i) from the defiitio of µ*. Sice O < for O compact, we have µ* E < for each bouded set E. Riesz-Markov Theorem Let X be a locally compact Hausdorff space. By C c (X), we deote as usual, the space of cotiuous real valued fuctios with compact support. A real valued liear fuctioal I o C c (X) is said to be positive if I(f) 0 wheever f 0. The purpose of the followig theorem is to prove that every positive liear fuctioal o C c (X) is represeted by itegratio with respect to a suitable Borel (or Baire) measure. I particular we have the followig theorem : Statemet of Riesz-Markov Theorem Let X be a locally compact Hausdorff space ad I a positive liear fuctioal o C c (X). The there is a Borel measure µ o X such that I(f) = f du

SIGNED MEASURE 39 For each f C c (X). The measure µ may be take to be quasi-regular or to be ier regular. I each of these cases it is the uique. Proof :- For each ope set O defie O by O = sup {I(f); f C c (X), O f, sup f O} The is a exteded real valued fuctio defied o all ope sets ad is readily see to be mootoe, fiite o bouded sets ad to satisfy the regularity (v) of the above Propositio. To see that is coutably subadditive o ope sets, let O = UO i ad let f be ay fuctio i C c (X) with O f ad sup f O. Thus there are o-egative fuctios φ, φ,, φ i C c (X) with sup φ i O i ad φ i =. o sup f. The f = φ i f, o φ i f ad sup (φ i f) O i. Thus If = I(φ i f) O i O i Takig the sup over all such f gives O O i ad is coutably subadditive. If O = O O with O O = φ ad f i C c (X), 0 f i ad sup f i O i, the the fuctio f = f + f has sup f O ad 0 f. Thus I f + If O. Sice f ad f ca be chose arbitrarily, subject to 0 f i ad sup f i O i, we have whece O + O O, O + O = O Thus satisfies the hypothesis of the above propositio so exteds to a quasi-regular Borel measure.

40 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS We ext proceed to show that If = f du for each f C c (X). Sice f is the differece of two o-egative fuctios i C c (X), it is sufficiet to cosider f 0. By liearity we may also take f. Choose a bouded ope set O with sup f O. Set O k = {x; f(x) > k } ad O o = O. The O + = φ ad O k + Ok. Defie φ k = f (x) 0 i k+ k + i Ok Ok+ i O O k The f = k= φ k We also have sup φ k for k. Also for k. Hece O k O k ad φ K = o O k+. Thus O k+ Iφ k O k O k+ φ k d O k Cosequetly µo k= (Iφ k φ k ) O 0 + O sice is arbitrary, If fdu O If = f d. Thus there is a ier regular Borel measure µ which agrees with o the σ-bouded Borel sets. Sice oly the values of µ o σ-bouded Baire sets eter ito f du, we have If = f du. The uicity of ad µ is obvious.

NORMED LINEAR SPACES 4 Uit-II Normed Liear Spaces First of all we itroduce some sort of distace measurig device to vector spaces ad ultimately itroduce limitig otios. I other words, our aim is to study a class of spaces which are edowed with both a topological ad algebraic structure. This combiatio of topological ad algebraic structures opes up the possibility of studyig liear trasformatios of oe such space ito aother. First of all we give some basic cocepts ad defiitios. Defiitio. A vector space or liear vector space X is a additive Abelia group (whose elemets are called vectors) with the property that ay scalar α ad ay vector x ca be combied by a operatio called scalar multiplicatio to yield a vector αx i such a way that (i) (ii) (iii) (iv) α(x +y) = αx + αy (α+β)x = αx + βx, (αβ)x = α(βx).x = x x, y X ad α, β are scalars. The two primary operatios i a liear space additio ad scalar multiplicatio are called the liear operatios. The zero elemet of a liear space is usually referred to as the origi. A liear space is called a real liear space or a complex liear space accordig as the scalars are real umbers or complex umbers. Defiitio. A isomorphism f betwee liear spaces (over the same scalar field) is a bijective liear map that is f is bijective ad f(αx + βy) = αf(x) + β f(y) Two liear spaces are called isomorphic (or liearly isomorphic) if ad oly if there exists a isomorphism betwee them. Defiitio 3. A semi-orm o a liear space X is a fuctio ρ : X R satisfyig (i) ρ(x) 0 x X.

4 INTEGRATION THEORY AND FUNCTIONAL ANALYSIS (ii) (iii) ρ(αx) = α ρ(x) for all x X ad α(scalar) ρ(x +y) ρ(x) + ρ(y) for all x, y X. Property (i) is called absolute homogeeity of ρ ad property (ii) is called subadditivity of ρ. Thus a semi-orm is o-egative real, subadditive, absolutely homogeeous fuctio of the liear space e.g. ρ(x) = x is a semiorm o the liear space C of complex umbers. Similarly if f : X C is a liear map, the ρ(x) = f(x) is a semi-orm o X. Thus a semi-ormed liear space is a ordered pair (X, ρ) where ρ is a semiorm o X. Defiitio 4. A orm o a liear space X is a fuctio : X R satisfyig (i) (ii) (iii) x 0 ad x = 0 if ad oly if x = 0 for x X αx = α. x x+y x + y we observe that a semi-orm becomes a orm if it satisfies oe additioal coditio i.e. x = 0 iff x = 0 Further, x is called orm of x. The o-egative real umber x is cosidered as the legth of the vector x. A ormed liear space is a ordered pair (X, ) where is a orm o X. Metric o Normed liear Spaces Defiitio 5. Let X be a arbitrary set. It is called a metric space if there exists a fuctio d : X X R (called distace or metric fuctio) satisfyig (i) d(x, y) 0 (ii) d(x, y) = 0 if ad oly if x = y. (iii) d(x, y) = d(y, x) (iv) d(x, z) d(x, y) +d(y, z) [Triagle iequality] for ay x, y, z X (X, d) is called a metric space.

NORMED LINEAR SPACES 43 Let N be a ormed liear space. We itroduce a metric i N defied by d(x, y) = x y This metric (distace fuctio) satisfies all axioms of the defiitio of orm. Hece a ormed liear space N is a metric space with respect to the metric d defied above. But every metric space eed ot be a ormed liear space sice i every metric space there eed ot be a vector space structure defied e.g. the vector space X 0 with the discrete metric defied by d(x, y) = 0 if if x = y x y is ot a ormed liear space. Remark :- I the defiitio of orm x = 0 x = 0 is equivalet to the coditio x 0 if x 0 Also the fact that x > 0 is implied by the secod ad third coditio of orm 0 = 0. = 0. = 0 ad 0 = x x x + x = x x 0 x 0. Remark :- As i the case of real lie, the cotiuity of a fuctio ca be give i terms of covergece of certai sequece. We ca alteratively defie cotiuity i terms of covergece of sequece i ormed liear space also. Defiitio 6. Let (E,. E ) ad (F,. F ) be two ormed liear spaces respectively. We say that f is cotiuous at x 0 E if give >0, δ > 0 wheever x x 0 E < δ f(x) f(x 0 ) F < Sice every ormed liear space is a metric space, this defiitio of cotiuity is same i it as the defiitio of cotiuity i metric space. Thus f is cotiuous at x 0 E iff wheever x x 0 i E