UBC Theses and Dissertations

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UBC Theses and Dissertations

Abelian von Neumann algebras Kerr, Charles R. 1966-08-30

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ABELIAN VON NEUMANN ALGEBRAS hy Charles R. K e r r B.A., Washington State U n i v e r s i t y , 1962 A THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF ARTS i n the Department of Mathematics We accept t h i s t h e s i s as conforming t o the requ i r e d standard THE UNIVERSITY OF BRITISH COLUMBIA September 1965 I n p r e s e n t i n g t h i s t h e s i s i n p a r t i a l f u l f i l m e n t of the requirements f o r an advanced degree at the U n i v e r s i t y of B r i t i s h Columbia, I agree that the L i b r a r y shall."make i t . f r e ely, a v a i l a b l e f o r reference and study. I f u r t h e r agree that per m i s s i o n f o r extensive copying of t h i s t h e s i s f o r s c h o l a r l y purposes may be granted by the Head of my Department or by h i s r e p r e s e n t a t i v e s o I t is.understood that copying, or p u b l i  c a t i o n of t h i s t h e s i s f o r f i n a n c i a l gain s h a l l not be allowed without my w r i t t e n permission,. Department o The U n i v e r s i t y of B r i t i s h Columbia,. Vancouver 8, Canada * Date This t h e s i s c a r r i e s out some of c l a s s i c a l i n t e g r a t i o n theory i n the context of an operator a l g e b r a . The s t a r t i n g p o i n t i s measure on the p r o j e c t i o n s of an a b e l i a n von Neumann algeb r a . This y i e l d s an i n t e g r a l on the s e l f - a d j o i n t operators whose s p e c t r a l p r o j e c t i o n s l i e i n the alg e b r a . For t h i s i n t e g r a l a Radon-Nikodym theorem, as w e l l as the usual convergence the or ems ,C'I;S proved . The methods and r e s u l t s of t h i s t h e s i s generalize, to non-commutative von Neumann Algebras [2, 3j> 53- (1) J . Dixmier Les Algebres d'Operateurs dans l'Espace H i l b e r t i e n . P a r i s , 1957. (2) H.A. Dye The Radon-Nikodym theorem f o r f i n i t e r i n g s of operators, Trans. Amer. Math. S o c , 72, 1952, 243-230. (3) ' F . J . Murray and J . von Neumann, On Rings of Operators, Ann. Math. 57, ' 1936, 116-229. (4) F . RIesz and B . v. Sz.-Nagy, F u n c t i o n a l A n a l y s i s , New York, 1955. (5) I.E. Segal A non-commutative extension of a b s t r a c t i n t e g r a t i o n , Ann. of Math. (2) 57, 1953, '401-457. i i i TABLE OP CONTENTS page I n t r o d u c t i o n 1 Examples a von Neumann Algebra . . . . . 6 Weak Compactness of the Uniformly- Closed B a l l 15 S e l f - A d j o i n t Operators i n a von Neumann Algebra . ... . . . . . 21 Unbounded Operators . . . 4 9 Measure Theorems 6 9 Measure on Simple Functions . . 72 Measure on Bounded S e l f - A d j o i n t Operators . . . . . 8 6 Measure on Unbounded Operators . . . . . 9 6 Absolute C o n t i n u i t y . . . . . . . . . 102 1 INTRODUCTION Let H "be a complex H i l b e r t space. Let L(H) be the set of bounded l i n e a r transformations on H. L(H) i s an algebra over the complex f i e l d . I t has an i n v o l u t i o n T —»T*, where T* i s the unique L(H)-member which s a t i s f i e s (Tx, y) = (x> T*y) f o r a l l x, y i n H. I n what f o l l o w s L(H) w i l l be t o p o l o g i z e d three ways.. (a) The weak topology on L(H): I n t h i s topology each neighborhood of A e L(H) must co n t a i n the i n t e r s e c t i o n of a f i n i t e f a m i l y of set s of form N(A,x,y,e) = [n : TeL(H), |(Ax,y)-(Tx,y)| < ej , where x,y e -H and € > 0. (b) The strong topology on L(H): Each neighborhood of A s L(H) contains the i n t e r s e c t i o n of a f i n i t e f a m i l y of sets of form N(A,x,€) = £T : TeL(H), ||Tx-Ax|| < e j , where x € H and e > 0. (c) The uniform topology on L(H): This i s the m e t r i c topology induced by the operator norm. These three t o p o l o g i e s are comparable: Weak <= Strong c Uniform, so t h a t i f S c L(H) "5 (weak) "5" (strong) => ( u n i f o r m ) . Under each o f these t o p o l o g i e s L(H) i s a t o p o l o g i c a l a l g e b r a , t h a t i s , the o p e r a t i o n s (X,Y) - » (X+Y) (a,X) — • aX X — ^ AX X —> XA are continuous f o r A, X, Y i n L(H) and s c a l a r a. !• D e f i n i t i o n . I f R c L(H) then R', the commutant of R, i s the set o f a l l o p e r a t o r s T I n L(H) such t h a t ST = TS S*T = TS* f o r a l l S e R. F o r any subset R c L ( H ) , R* i s always an a l g e b r a : I f A and B are i n R>, then f o r S e R (A+B)S = AS + BS = SA + SB = S(A+B) (AB)S = A(SB) = S(AB). S i m i l a r l y , (A+B) and AB commute w i t h S*. Thus (A+B) and are a l s o i n R'. Note t h a t R 1 always c o n t a i n s the i d e n t i t y o p e r a t o r , and t h a t T i s i n R' i f and o n l y i f T* i s i n R'. Furthermore, R* i s c l o s e d i n L(H) under any of the above t o p o l o g i e s . To prove t h i s note t h a t f o r S e R, £S i s merely the set on which the continuous f u n c t i o n s f ( T ) ss' ST - TS g(T) = S*T - TS* so t h a t R' i s a c l o s e d s e t . 2. D e f i n i t i o n . A von Neumann a l g e b r a i s a s e t R c L(H) such t h a t R = (R')' - R". Such a s e t i s , i n view of p r e v i o u s remarks, a s t r o n g l y c l o s e d symmetric subalgebra o f L(H) which c o n t a i n s the i d e n t i t y o p e r a t o r . On the oth e r hand, a l l such a l g e b r a s a re von Neumann a l g e b r a s . T h i s f o l l o w s from the "von Neumann d e n s i t y theorem" which says t h a t R" s= rT (weak) = H (strong) f o r every symmetric subalgebra R w i t h i d e n t i t y . J>. Theorem. I f R i s any von Neumann a l g e b r a and T i s any bounded s e l f - a d j o i n t o p e r a t o r w i t h r e s o l u t i o n are equal t o zero, and hence t h a t £s two c l o s e d s e t s . F i n a l l y I s the i n t e r s e c t i o n o f R» =D (SeR) Is]' , then T € R i f and onl y i f E_ € R f o r a l l a. Proof. I f T € R then £ Ti £ R> {T}« 2 R', so t h a t f T J " c R" = R. Since E & e E € R f o r a l l a. On the other hand, a dE a, — 00 where the right-hand side i s a-uniform l i m i t of a g e n e r a l i z e d sequence of l i n e a r combinations of E . Now R c TT(uniform) <= E(weak) = R" = R, so that R i s un i f o r m l y c l o s e d . Thus jjs ^  C R i m p l i e s TeR. This proves theorem J>. 4. D e f i n i t i o n . I f R i s any von Neumann algebra R U i s the set of u n i t a r y operators i n R. 5. Theorem• I f R i s a von Neumann algeb r a , then the bounded operator T belongs t o R i f and only i f T commutes w i t h every u n i t a r y operator i n R 1, i n other words R = ( ( R « ) U ) ' (a) Proof. Since ( R ' ) U c R ' 5 ( fR 'O U ) ' 3 R" = R (b) On the o t h e r hand, every o p e r a t o r I n the u n i f o r m l y c l o s e d symmetric a l g e b r a G may be w r i t t e n as a l i n e a r combination o f u n i t a r y o p e r a t o r s i n G. To see t h i s note f i r s t t h a t i f A € G, then E1 = 1/2 (A+A*) H 2 = i / 2 (A*-A) are s e l f - a d j o i n t o p e r a t o r s i n G and A = H 1 + i H 2 Next i f H i s s e l f - a d j o i n t I n G and ||H||<1, then U « V l - H 2 + i H i s i n G ( s i n c e G i s u n i f o r m l y closed) and i s u n i t a r y . Thus i f an o p e r a t o r T commutes wit h every u n i t a r y o p e r a t o r i n G , i t commutes wit h a l l o t h e r members of G, t h a t i s G' and i n p a r t i c u l a r C(R'-)V c ( R ' V = (c) Combining (b) and (c) y i e l d s ( a ) . 6 An example of a von Neumann alge b r a . Let (X, n) be a a - f l n l t e measure space. L r o(X, u) i s the set of measurable f u n c t i o n s which are bounded almost every where i n X. L 2 ( X , |a) i s a H i l b e r t space. Given t e L there i s a corresponding l i n e a r t r a n s - formation T on L 2 : (Tg)'(x) = t ( x ) g ( x ) f o r g e Lg. Let M be the f a m i l y of l i n e a r transformations so induced CO (1) M i s a f a m i l y of bounded l i n e a r transformations of Lg onto i t s e l f . To prove this, let g e L 2 . J | T g j 2 d|i = f Itg| 2 dU X . X < iitiif / i g i 2 dM X (where ||t|| = i n f k : \x x : | t ( x ) | > k = 0 ) = . J i t I I ? I l g j l 2 < - Thus Tg e Lg and l|Tg|| < | t „ g , so t h a t T € L ( L 2 ) . 7 This l a s t i n e q u a l i t y i m p l i e s llTll < l l t l ^ To show tha t e q u a l i t y o b t a i n s , l e t € > 0 be given, then f x : | t ( x ) | > l l t H , - € j has p o s i t i v e measure, and since u i s c r - f i n i t e , t h i s set has a subset. S of f i n i t e p o s i t i v e measure. Let g be the character i s t i c f u n c t i o n of the set S. Then g € Lg. Moreover l|Tg||2 = f | t g l 2 d u  | t g | = C | t g | 2 du + f | t g | 2 du | t ( x ) | 2 du > ( l l t l l ^ - e ) 2 MS s = ( l l t l l ^ - e ) 2 l l g l l 2 . Thus f o r a l l e > 0, l|T|| > | | t | | w - € . Hence ||T|| _> Htj| . Combining t h i s w i t h the previous i n e q u a l i t y y i e l d s T = t 00 T h i s means t h a t M and L are i s o m e t r i c copies of one another. 8 ( i i ) M i s an a b e l i a n a l g e b r a . To prove t h i s l e t S and T be M-operators corresponding r e s p e c t i v e l y t o s and t i n L . CO ([S+T]g)(x) = (Sg)(x) + (Tg)(x) o s(x)g(x) + t ( x ) g ( x ) - (s(x) + t ( x ) ) g(x) Thus S+T corresponds t o s+t i n L^, so S+T e M. I f c i s any complex number, then ( [ c S ] g ) ( x ) = c(Sg)(x) - c's(x') g ( x ) , whence. cS corresponds t o c s(x) € L^, so cS € M. Next, (STg) (x) = (S[Tg]) (x) - = s(x) [Tg] (x) = s(x) t ( x ) g ( x ) . Since s t e L , ST e M. From t h i s i t i s obvious t h a t ST = TS. Thus M i s an a b e l i a n algebra: M c M' . ( i i i ) M i s a symmetric algebra. To c a l c u l a t e T* note f i r s t t h a t i t i s by d e f i n i t i o n the unique L ( L 2 ) operator such t h a t (Tf,g) = ( f , T*g) f o r a l l f , g e Lg, t h a t i s / t f g dy = / f(T*g)du or 7 f [ t g - (T*g)] du = 0 f o r a l l f 3 g e L 2 . This means t h a t f(x) [t(x) " i W - (7*g) (x)] = 0 almost everywhere f o r a i l f,g. e L 2 , hence t(x) tr^ T - (W) (x) = 0 almost everywhere f o r a l l g e Lg. Thus tha t (x) = t(x) -gTxT (a) and T*g = Tg i n L 0 . Thus T* corresponds t o T e L . T h i s shows T* e x p l i c i t l y and proves t h a t M i s a symmetric subalgebra of L ( L 2 ) . 10 ( i v ) M = M»- Since M i s a b e l i a n , M c M». To get the r e v e r s e i n c l u  s i o n l e t A € M*. A commutes wit h a l l the p r o j e c t i o n s i n M. These p r o j e c t i o n s are induced by the c h a r a c t e r i s t i c f u n c t i o n s o f measurable subsets o f X. Thus AE = AE f o r a l l p r o j e c t i o n s E e M i m p l i e s t h a t (AEf) (x) = (EAf) (x) = e ( x ) ( A f ) ( x ) , , where f e L 2 and E corresponds t o the c h a r a c t e r i s t i c f u n c t i o n e € L . I f now b o t h e and f are c h a r a c t e r i s t i c f u n c t i o n s o f s e t s o f f i n i t e measure then e , f e L 2 n L M and ( A e f ) ( x ) = e ( x ) ( A f ) ( x ) = f ( x ) ( A e ) ( x ) Since (X'^i1.) i s c r - f i n i t e , X = (J (ietu) E ± where the E^ are d i s j o i n t and |U E i < co Thus, f o r f as above, ( A f ) ( x ) =Y ( i e u )) e i ( x ) 11 (where e^ i s the c h a r a c t e r i s t i c f u n c t i o n o f E^) = £ (Ae.f) (x) = ^ f ( x ) (Ae ±) (x) = f ( x ) I (Ae ±) (x) L e t a(x) = J(ieu)) ( A e ^ ^ x ) Then (Af) (x) = a(x) f ( x ) , (*) whenever f i s the c h a r a c t e r i s t i c f u n c t i o n of a s e t o f f i n i t e measure. T h i s a ( x ) , "being the p o i n t - w i s e l i m i t o f a sequence o f measurable f u n c t i o n s , i s i t s e l f measurable. Indeed, the f u n c t i o n a(x) i s i n L (X ,uJ, f o r l e t S = [x • |a(x)| > ||A||j. Then S has a subset S» o f f i n i t e measure. I f nS' i s p o s i t i v e , and cp i s the c h a r a c t e r i s t i c f u n c t i o n of S*. then cp e L p and That i s / I M > ||A|| l l c p l l . 12 This i s a contradiction, hence US' = 0 and |aS = 0, Thus a(x) s L o and H a l i ^ < \\A\\. The l i n e a r i t y of the operator A Immediately extends (*) to the e a s e where f i s a summable simple function. Since (X,p) i s a - f i n i t e , the set of summable simple functions i s dense i n Lg. Hence given f e L 2 and e > 0, there e x i s t s a summable simple function s(x) such that 2||A|| Then llAf - af I) < ||Af - as || + ||as - af|| •p ||Af - As || + ||as - af|| (since (As)(x) = a(x)s(x)) < llA|| l|f-s|| + l l a l l ^ ||f-s|| < 2||A|| ||f-a|| . Thus ||Af. - af || < €. But € Is a r b i t r a r y , hence (*) must hold almost everyhwere f o r every f e L 0 . Since a e L , t h i s means t h a t A € M. Thus' . M = M1 . : Now, because M = M', M Is maximal abelian, i . e . M i s properly contained i n no symmetric subalgebra which i s abelian. 15 I f M C N, where N i s a b e l i a n , then N c N» and M' = N 1, o r combining M = M' D N ' S N S M , s& t h a t M = N. More t o the p o i n t i s the observation t h a t M' = M i m p l i e s t h a t M" = H» so th a t M = M» = M", that i s , M i s a von Neumann algeb r a . 6. D e f i n i t i o n . L e t R be a von Neumann al g e b r a . Then R I s the set of p r o j e c t i o n s i n R. I f R i s any von Neumann; alg e b r a , then a measure can be defined on R P. 14 p 7 . D e f i n i t i o n , m i s a measure on R i f m i s an P extended r e a l f u n c t i o n on R such t h a t ( i ) m E _> 0 f o r a l l E e R P ( i i ) I f ^E^J i s any f a m i l y of mutually orthogonal P R -members, m l\ E i ^ m % ' (Theorem 27 w i l l show t h a t ^ E ± = sup [E±] e R P ) . 8. D e f i n i t i o n . A measure m i s f i n i t e i f m I < » P P and s e m i - f i n i t e i f f o r every non-zero E e R there i s P e R such t h a t 0 < F < E and m F < <». P 9. D e f i n i t i o n . I f m and n are two measures on R , then n i s a b s o l u t e l y continuous w i t h respect t o m i f nE = 0 f o r a l l E e R P such t h a t m E = 0. P I n what f o l l o w s a s e m i - f i n i t e measure m on R - R i s a b e l i a n - i s extended t o an i n t e g r a l on the set of a i l s e l f - a d j o i n t operators (not n e c e s s a r i l y bounded) whose s p e c t r a l P r e s o l u t i o n s l i e i n R . T h i s i n t e g r a t i o n theory i s developed f a r enough t o prove a Radon-Nikodym theorem f o r s e m i - f i n i t e measures, 15 WEAK COMPACTNESS OP UNIFORMLY CLOSED BALL 10. Theorem. The uniformly closed b a l l B = [A : AeL(H) and ||A|| _< l j i s compact i n the weak topology of L(H). Proof begins with a lemma. Lemma. In i t s weak topology L(H) i s homeomorphic to a subset of the product space P = nTC(x,y) : € H x HI > where C, \ i s a copy of the complex plane. Proof. The homeomorphism i s the evaluation map e ' e t L(H) — * f 5 p ( X , y ) e W = (Ax>y)> where P/ » i s the projection of onto C/ v and A e L(H). This map i s one-to-one: I f A and B i n L(H) are such that e(A) = e(B), then ( A x , y ) = P ( x , y ) e ( A ) = p ( x , y ) e ( B ) = (Bx ,y) f o r a l l x and y i n H, that i s A = B. Both e and e" 1 are continuous: Let 5 be a f i n i t e subset of H x H, l e t € > 0 , l e t A e L(H), l e t S(c,c) be the sphere of radius e about the complex number c, and l e t 16 Z x = [T : TeL(H) and |(Ax,y) - (Tx,y) | < E f o r a l l (x,y) e>J ?-2 =n»x'y)«) P ( i , y ) [ S ( P ( X j y ) e ( A ) , « ) ] . . The s e t s o f form Z1(A,JF,e) form a neighbourhood b a s i s f o r A I n the weak topology o f L ( H ) . The s e t s o f the form e [L(H)] n Z 2(A,5,€) form a neighbourhood b a s i s f o r e(A) i n the product topology o f •ff^ r e l a t i v i z e d t o e[L(H) ]. Now, s i n c e p ( X j y ) e ( A ) « (Ax,y), e[Z 1} = [ e ( T ) : TeL(H) and | P ( X j y ) e ( A ) - P ( x ^ y ) e ( T ) | < € f o r a l l (x,y) e 3^ ' m e [ L ( H ) ] n £w : w e fp and l p(x , ,)«W'- p(.,y)<") ' l < e f o r a l l (x,y) € ^ \ = e [ L ( H ) ] n Z 2. That i s e [ Z 1 ] = e[ L ( H ) ] n Z 2, ' i • so t h a t e and e " 1 are continuous. T h i s proves the lemma. 17 T h i s lemma i m p l i e s t h a t 8 i s compact i f and o n l y i f e[iB] i s compact. Por x arid y i n H l e t ? ( x,y) = £ 2 ' z i s a complex number • M < . N I hWJ . (x,y) i s a c l o s e d and bounded subset o f the complex p l a n e , and i s thus compact f o r every x and y . Hence by the Tihohov theorem the product space $ = T T f ( f (x,y) : (x,y) e H x E] i s a l s o compact. Now e[is] c z j £ , s i n c e f o r A € © I P M e(A) | = |(Ax,y)| < ||Ax|| HarII < l|A|! !|x|| ||y|| < l|x|| ||y|| Thus the theorem w i l l be proven i f i t i s shown t h a t e[fi] i s a c l o s e d set i n the compact space $ . F i r s t l e t x, y, z be i n H and l e t a,b be complex numbers. L e t X 1 ( x , y , z ) = {w:we ^ , P ( x + y ^ z ) ( w ) = P ( X j 2 ) ( w ) + P ( y j Z ) ( w ) j X 2 ( x , y , z ) = {w:wc#, P ( x , y z ) ( w ) - P ( x , y ) 0 0 + P ( x , z ) ( - ) j 18 X 5(x,y,a,b) = [ w : w e $ , P ( a X j b x ) ( w ) = aTJ P ( x 5 y ) ( w ) e / The p r o j e c t i o n P^ u v j o f ^ onto ^ (X>Y) i s continuous. Hence X 1 ( x , y > z ) , b e i n g the set on which two continuous f u n c t i o n s agree, i s c l o s e d f o r a l l x, y, z. S i m i l a r l y Xg and X^ are c l o s e d i n 3> . Note next t h a t f o r any A i n 'ft, P ( x + y , z ) e ( A ) - (A(x+y;,z) = (Ax,z) + (Ay,2) P ( x , z ) e ( A ) + P ( y , z ) e ( A ) > so t h a t e(A) e X^(x,y,z) f o r a l l x,y,z. S i m i l a r l y e(A) i s in X 2 ( x , y , z ) . A l s o P ( a x , b y ) e ( A ) = ( A a x ^ y ) = ab~ (Ax,y) = a ^ P ( x , y ) e ( A ) ' so t h a t e(A) e X^ (x,y,a,b) f o r every x,y and a,b. Thus e[ft] c J where J= C ] { \ i x , y , z j n / ° ] { x 2 : x, y , z j n f l f x ^ : x, y , a , b j 1 9 which i s a c l o s e d s e t . "Actually e [ © ] exhaust s , f o r i f cp e then P ( x + y , z ) ^ ) = P(x,z ) ( < P ) + P ( y , z ) ( ^ P ( x , y + z ) ( c P ) . = P ( x , y ) ( c P ) + P ( x , z ) M P ( a x , h y ) ( ^ ) = ^ P ( x , y ) ^ ) ^ t h a t i s , cp determines a b i l i n e a r f u n c t i o n a l f ( x , y ) = P ( x , y ) («p) on H. Moreover s i n c e P, „ \ ( c p ) e (f 7 (x,y) . Hence f i s a bounded b i l i n e a r f u n c t i o n a l on H and ||f|| < 1 . T h i s means t h a t by the R i e s z r e p r e s e n t a t i o n theorem f o r bounded b i l i n e a r f u n c t i o n a l s there e x i s t s an o p e r a t o r F.. e'L(H) such t h a t f ( x , y ) = (Fx,y) and 11*11 = | | f t l < l . But t h i s i m p l i e s t h a t P ( X ; y ) M - * ( x , y ) = <Fx,y) = P ( x > y ) e(P) 20 for a l l x,y, so that cp = e(F) and cp € e['(j]. Thus c e[u] . and so e[8] = *f. As stated before, the compactness of B now follows from the f a c t that e[©] i s a closed subset of the compact space ^ . This proves theorem 10. 21 BOUNDED SELF-ADJOINT OPERATORS IN A VON NEUMANN ALGEBRA 11. D e f i n i t i o n . I f R i s a von Neumann algebra, R i s the set of s e l f - a d j o i n t operators i n R. With t h i s d e f i n i t i o n theorem 3 rephrases as "T e R S i f P // and only i f the spectral resolution of T i s i n R . Any family of bounded s e l f - a d j o i n t operators i s p a r t i a l l y ordered by a r e l a t i o n _< defined as follows, S _< T i f and only i f (Sx,x) _< (Tx,x) for a l l x i n H. In most cases t h i s _< does not furnish a l i n e a r ordering; f o r example, the operators E and I-E, fo r a projection E, are not comparable. 12. D e f i n i t i o n . I f 3 c L(H) i s a family of s e l f - a d j o i n t operators, then whenever i t exists , 1 sup ^ i s the smallest s e l f - a d j o i n t operator which majorizes every U-member. I f R i s a von Neumann algebra then sup 3 , R S whenever i t existSj i s the smallest R -operator majorizing every member of 3 c R . The o p e r a t o r s i n f 3. , i n f 3 R are a n a l o g o u s l y d e f i n e d . I f S and- T are i n R S then S-T i s i n R S a l s o . I f F i s the s p e c t r a l r e s o l u t i o n o f S-T, then by theorem 3 a P F_ e R . I f R i s a b e l i a n , then S,T commute w i t h S-T, and, consequently, F commutes w i t h b o t h S and T. T h i s p e r m i t s the f o l l o w i n g d e f i n i t i o n 13. D e f i n i t i o n . I f S,T e R S f o r a b e l i a n R, l e t F be the r e s o l u t i o n o f S-T. Then (S UT) = T F 0 + S(I-F 0) (S D T) = T ( I - F Q ) + SF0 O b v i o u s l y (SUT) and (SnT) are i n R. Furthermore (S U T ) * = F Q * T * + (I-F 0)*S* = F Q T + (I-F 0)S = T F Q + S(I-F Q) = (S U T) S so t h a t (S u T) e R . I n the same way (S D T) € R 14. Theorem. Under the assumptions o f d e f i n i t i o n 13, (S U T) = sup fs,T? . R Proof. Since R i s a b e l i a n , [ F 0 j < 3 R < £ R . The f o l l o w i n g proof shows th a t (Stff) i s the smallest operator i n {FQI' which majorizes both S and T. For any x e H, (S U T x,x) - (T x,x) = (TF Qx,x) + (S(I-PQ)x,x) i - (TF Gx,x) - ( T ( I - F Q ) x , x ) - ' ( (S-T)(I-P 0 )x ,x) > 0 by d e f i n i t i o n of F . Thus (S U T) > T. S i m i l a r l y (S U T) > s- Now l e t A e £ F 0 1' loe 9X1 0 P e r a ' t o r such t h a t A > S A > T . Then A = F 0 A F Q + ( I - P 0 ) A P Q •+ F Q A C I - F Q ) + ( I -F 0 ) A ( I - P 0 ) = FQ AFQ + ( I -FQ) A ( I - F Q ) 2k (since A e S i n c e ( S y T ) = TFQ + S(I-F 0) - p o T P o + (X-F(J si^o)' for any x € H (Ax,x) - (SuTx,x) = ( P Q A F ^ X ) + ( I - F 0 ) A ( I - P Q ) x , x ) - ( P Q T P 0 x , x ) + ((I-F 0)S(I-F 0)x,x) = ((A-T)F 0x,F 0x) + ((A-S)(I-F 0)x, (I-F Q)x) . This l a s t l i n e i s non-negative, since A majorizes S and T. This proves theorem 14, In the same way, 15. Corollary ( S O T ) = i n f £S,T} 16. Theorem. I f R i s any von Neumann algebra (not necessarily abelian) and 3 c R i s directed upward and bounded above by the s e l f - a d j o i n t operator S Q, then sup 3 exists and g belongs to R . Proof. For a l l F e gr l e t W(F) be the weak closure of the set V(F) = £T : T e 3 and T _> FJ Choose F Q e 3 and l e t t h e n 3 Q = V(FQ) = £T : T e 3 and T _> F Q| Observe t h a t i f £F^ : i e n j i s any f i n i t e ^ - s u b s e t , (~) [V(F1) 2 i e n j ^ 0 . To p r o v e t h i s n o t e t h a t s i n c e # i s d i r e c t e d upward t h e r e e x i s t s T-^  € 5 such t h a t T l 1 Fo T l - F l • F o r k < n, t h e r e e x i s t s T k + 1 e 2* such t h a t Tk+1 - Fk+1 T k f l - T k O b v i o u s l y T n >_ F^ f o r a l l 1 € n, so T n e f ] [ V ( F ± ) . : 1 e n | A f o r t i o r i D { w ( F ± ) : i e n j jL 0 f o r a l l f i n i t e f a m i l i e s ^ F i : i e n j £ ^ Q T h i s means t h a t ¥(F) : F €' 3>Q 26 has the f i n i t e i n t e r s e c t i o n property i n B = £ r : T € L(H), ||T|r<max [W^l, \\SQ\ Since 8 i s weakly compact D {w(F) : F € 3?0 ^ = 0 I f A i s i n t h i s i n t e r s e c t i o n , then since A i s i n the g weak closure of V(F) c 3 c R ( i ) A i s s e l f - a d j o i n t ( i i ) A > F f o r a l l F € 3 Q and hence A _> F f o r a l l F € 3>Q ( i i i ) A € R A c t u a l l y A i s the only operator i n the i n t e r s e c t i o n and i s moreover the supremum of 3. For i f T i s any s e l f - a d j o i n t operator which majorizes every 3-member, then i n p a r t i c u l a r , . T > S f o r a l l S e ¥(F) f o r a l l F e 3. Hence T > A. This proves theorem 16. 17. C o r o l l a r y . I f R i s any von Neumann al g e b r a , and s 3 c R i s d i r e c t e d downward and bounded below by the s e l f - a d j o i n t operator S , then ( i n f 3) e x i s t s and belong t o R . 2 7 P r o o f . The set - 3 ? = £-F : F e 3 J i s d i r e c t e d upward and bounded above by - S Q . Hence 1 7 f o l l o w s from 1 6 . 1 8 . D e f i n i t i o n . F o r sequences o f s e l f - a d j o i n t o p e r a t o r s , S f w i l l mean t h a t S n S n + 1 and S | S (strong) w i l l mean t h a t s n t and S = l i m (strong) S n. The e x p r e s s i o n s S T S(weak) S -^'S (uniform) are d e f i n e d a n a l o g o u s l y . 1 9 . Theorem. For monotone sequences, s t r o n g and weak convergence are e q u i v a l e n t : i f f T I i s a monotone sequence, 28 t h e n T = l im(weak) T n i f and o n l y i f T = l i m ( s t r o n g ) T R P r o o f . S i n c e t h e s t r o n g t o p o l o g y i n c l u d e s t h e weak t o p o l o g y T = l i m ( s t r o n g ) T n i m p l i e s T = l i m ( w e a k ) T n F o r t h e c o n v e r s e , suppose T n t T (weak). S i n c e T G < T n < T> l T n i l < K > where K = max ^||T0!|, Now f o r any x i n H (T-T n)x||* = | ( ( T - T n ) x , ( T - T n ) x ) | 2 < ( ( T - T n ) 2 x , ( T - T n ) x ) ( ( T - T n ) x , x ) ( s i n c e T-T ._> 0 ) . ((T-Tn)3x,x)f(T-Tn)x3x) < ||(T-Tn)5x|| ||x|| ((T-Tn)x,x) < !!T-TN||5 ||xl|2 ((T-Tn)x,x) < ( 2 K ) 3 ||x||2 ((T-Tn)x,x) ( s i n c e ||T-Tnl| < ||T|| + HTJ < 2K) Since T = l i r a (weak) T , there e x i s t s N(e,x) such t h a t n J> N i m p l i e s A and hence t h a t ||(T-Tn)x|| < e. Thus T n f T (strong) and 19 i s proven. Remark. I n view of 19 S„ t S T I T n n w i l l be w r i t t e n w i t h the understanding t h a t the convergence i s 30 b o t h weak and s t r o n g . 20. Theorem. I f R i s a b e l i a n w i t h S , S, T i n R S, t h e n i f Snt S , t h e n a l s o \ ( S n U T) f (S U T) ( s n n T) t (s n T) P r o o f depends on two lemmas: 21. Lemma. I n any R w i t h S , S e R S, i f S n f S t h e n S = sup f s j . P r o o f . f s n ^ i s d i r e c t e d upward and bounded above. Hence C = sup I Sj S e x i s t s and b e l o n g s t o R by 16. Hence S C . I f e q u a l i t y does n o t h o l d h e r e t h e n t h e r e must e x i s t x € H such t h a t (Sx,x) > ( C x , x ) , and, since s n t s, there e x i s t s N such that n J> H implies (Sx,x) > ( S n x , x ) > (Cx,x), contradicting the f a c t that C _> S n f o r a l l n. Hence S = C = sup £ s n j This proves 2 1 . 2 2 . Lemma. In any R, i f £ s n ^ c R^ i s bounded above and SNT , then (by 1 6 ) S = sup fs nJ exists and belongs to R . Moreover, S f S n Proof. Observe that f S n } c B = {A : A € L(H), HAH < max f ||SQ!|, ||s|ljj\ Since J J i s weakly compact, [&nJ n a s a subsequence ^ " s n ^ ^ such that f o r some C e s . Since £ $ n ^ k ^ c R S, C € R S also. 32 Now f o r any x.e H, ( S n ( k ) x ' x ) ? ( C x' x) ' Hence a l s o ( S n x , x ) t (Cx,x) Since a l l the operators i n v o l v e d are s e l f - a d j o i n t s n T > c . By 21 t h i s i m p l i e s t h a t G = sup ' f s n } = S Thus s n f S and 22 i s proven. Now t o complete the proof of 20. Since (S U T) > S > S Q and (S u T) > T, obv i o u s l y (S u T) > ( S n u T) (a) Furthermore, ( S n + 1 u T) > S n + 1 > S n 33 and ( S B + 1 U T ) > T , so that ( sn+l U T ) 2. ( sn u T ) • That i s ( S n U T) t 0>) By (a) and (h) and 16, there exists C e R S such that C = sup f ( S n U T)J Now (S U T) > ( S n UT) implies that (S U T) > C (c) By 21, the f a c t that s n t s implies that S - sup { S n J , hence that C > S, . since c > ( s n U T) > S, n f o r a l l n. Moreover, C _> ( S n U T) > T, so th a t hy 14, since C e R 9 e > ( s u T) Combining (c) and (d) y i e l d s C « (S U T), t h a t i s , To prove the second a s s e r t i o n of 20 note t h a t (S U T) = sup Thus by 22 (Sn U T) f ( S U T) ( s n n T) < s n < S, n+1 and (s n n T) < T, so t h a t by 15, since (Sn f| T) € R S, ( s n n T) < (S, n+1 0 T) . Thus ( s n fl T) t . That t h i s sequence tends weakly to ( S Cl T) follows i from the formula (S+T) = S F Q + S(I-P 0) + T F Q + T(I-P 0) = S F Q + T(I-P Q> + T F Q + S(I-P 0) = ( S n T) + ( S U T ) . This proves 20. SIMPLE FUNCTIONS IN R S. SP 23. D e f i n i t i o n . R i s the set of a l l hounded operators of the form P S = •£ a i E i , 1=1 where p I s a p o s i t i v e i n t e g e r , a^ i s r e a l , and ^E^J i s a P f a m i l y of R -members such t h a t P 1=1 ( t h i s i m p l i e s t h a t the E^ are mutually or t h o g o n a l ) . 24. Theorem. I f T € R then there i s a sequence { T n ] S r S P s u c h t h a t T n f T (uniform). Proof. Choose d > 0 and l e t P be any f i n i t e p o i n t set p a r t i t i o n i n g the i n t e r v a l [-d-||T|1, ||T||]. P : -d-||T|| = a Q < < ... < a ^ < a p = ||T|| P I f E Q i s the s p e c t r a l r e s o l u t i o n of T, then E„ e R a a f o r a l l r e a l a, and a, , (E - E ) < T(E - E a ) < a, ( E o - E o ) l - x a.,, a i_ ]_ - a ± a ± _ 1 1 a ±. a ± _ 1 / o r , l e t t i n g E. « E - E 1 a ± a ± _ 1 a i - l E i - T E i - a i E i so that P L(P) « £ a ^ E ± i = l P < : £ T E ± = T 1=1 P < r ai % ^ iT^p)' t h a t i s , L(P) < T < U(P) . How P U(P) - L ( P ) • - Y ( a i - a i _ 1 ) E i i = l <max {a 1 -a 1 - ] J * (max P) I , so t h a t 0 < T - L(P) < (max P) and hence, T - L(P)H < max P 38 Thus i f a sequence £l> nJ of p a r t i t i o n s of [-d-||T||, ||T||] i s chosen so tha t l i m (max P ) = 0 , then l i m 1|T - L ( P n ) | | = 0. Hence T = l i m (uniform) L ( P ) . Now ^ L ( P n ) J c R S P, and i t s convergence t o T may be made monotone by r e q u i r i n g t h a t P n £ P n + 1 > f o r , l e t P ' m f a 1 } u P, where a' / P, a j _ i < a' < a^, say. Let P L ( P - J ) - £ (1 * 0 a 1 _ 1 E ± . 1=1 Then L(P) = L ( P - j ) + a j _ L - L(P-J) + a J _ 1 ( E a j " - E a f ) . < L ( P - j ) + a' ( E a - E f t l ) + a j - l ( Ea- " \ = L ( P » ) , t h a t i s , L(P) < L(P') < T. This proves 24. POSITIVE AND NEGATIVE PARTS OF AN RS-OPERATOR R e c a l l t h a t I f T i s any bounded s e l f - a d j o i n t operator w i t h s p e c t r a l r e s o l u t i o n E . then cl T = T + - T~, where T + = T ( I - E 0 ) T" = -TE 0 are p o s i t i v e operators which commute w i t h T and w i t h each o t h e r j and f o r which T +T" = 0. The f o l l o w i n g theorem w i l l be important t o the extension P S of a measure from R t o R s- < • • . 25. Theorem. I f T i s any s e l f - a d j o i n t operator and T = A-B, where A and B are p o s i t i v e operators t h a t commute w i t h T, then A = T + + P-. B = T" + P, 1 where P i s a p o s i t i v e operator. Proof. Since A - B = T + - T" i m p l i e s : A - T + = B - T", i t remains only t o show tha t P = A - T + i s p o s i t i v e . I f E i s the r e s o l u t i o n of T, then f o r any x e H ( P x , x ) ( A - T + x , x ) = (A-T +x,F Qx) + (A-T+XjI-Ex). = ((A-T +) E 0 x , E Q x ) + ( ( A - T + ) ( I - E 0 ) x , ( I - E Q ) x ) ( s i n c e E Q commutes w i t h A-T +) = ( A E Q x , E 0 x ) •+ (.(A-T +)(I-E 0)x, (JME 0)x) ( s i n c e T + E Q = T ( I - E 0 ^ 0 =0) = ( A E o x , E 0 x ) + ( ( A - T ) ( I - E 0 ) x , ( I - E Q ) x ) ( s i n c e T + = T ( I - E Q ) ) = ( A E 0 x , E G x ) + ((A- [A-B| ) ( I - E 0 ) x , ( I - E Q ) x ) ( s i n c e T = A-B) » ( A E Q x , E 0 x ) + ( B ( I - E 0 ) x , ( I - E 0 ) x ) . T h i s l a s t l i n e i s non-negative since A and B are p o s i t i v e . 4:2 Thus P _> 0 and 25 i s proven. Important Remark. I f T e R S, then so a l s o are T +, T~. I f A , B are i n R S, so i s P. 43 SUPREMA AND INFIMA IN R P Let R be any von Neumann algebra. I f attention i s r e s t r i c t e d P P to subsets of R ••, then 16 can be improved: subsets of R need not be directed to have suprema and infima. Moreover, the supre- mum (or infimum) of d e f i n i t i o n 12 turns out to be the usual supre- mum (Infimum) of a c o l l e c t i o n of projections. 2 6 . Theorem. I f R i s any von Neumann algebra and P = p r D £rng E : E e ffj, P then P e R and P = i n f g Proof. To show that P = i n f g , suppose that S i s a s e l f - a d j o i n t operator f o r which S _< E f o r a l l E . Assume, moreover, that there e x i s t s x e H with ||x|| = 1 such that (Sx,x) > (Px,x). 44 Now (Sx,x) < 1 and (Px,x) = 1 or (Px,x) = 0 . Thus i t must he t h a t 1 _> (Sx,x) > (Px,x) = 0 But (Px,x) = 0 i m p l i e s that x £ rng E f o r some E Q e , and hence th a t 0 = (E Qx,x) >_ (Sx,x) > 0. This i s a c o n t r a d i c t i o n . Hence (Sx,x) < (Px,x) f o r a l l x e H, t h a t i s S _< P. Since P _< E f o r a l l E e ^  , t h i s means tha t P = i n f g To show that P e R, l e t U he a u n i t a r y operator i n R' 4 5 Then UE = EU U*EU = E UEU* = E (a) for a l l E e R P. I f F i s any projection such that F _< E f o r a l l E ' ( ^ , then f o r a l l x e H, (U*FUx,x) = (FUx,Ux) < (EUx,Ux) = (U*EUx,x) = (Ex,x) , by (a), so that U*FU _< E f o r a l l E e ^  . S i m i l a r l y , UFU* < E fo r a l l E e ^  . Thus, i n p a r t i c u l a r , UPU* < E U*PU < E for a l l E e . Now P m i n f £ p , thus UPU* _< P (b) U*PU < P (c) Now by (c) (U*PUx,x) < (Px,x) f o r a l l x £ H, hence f o r x = U*y, where y e H : (U*PU(U*y), (U*y)) < (PU*y,U*y), (UU*PUU*y,y) < (UPU*y,y) , whence (Py,y) < (UPU*y,y) f o r a l l y € H. Thus P _< UPU* (d) Combining (b) and (d) y i e l d s P = UPU* or PU = UP. Since U i s otherwise a r b i t r a r y , t h i s means tha t P commutes w i t h every u n i t a r y operator i n R'. By theorem 5, P e R. This proves 26. 27. Theorem. I f R i s any von Neumann algebra and cfs*- Q = p r [ ( J [ rng E : E e£j ] 47 (that i s , Q i s the p r o j e c t i o n onto the closed subspace generated by t h i s set u n i o n ) , then Q e R and Q = s u p ^ Proof. Q = I - pr [ ( J £rng E : E egf ]X = I - p r D ^ ( r n g E ) - 1 : E € g>$ = I - v^D ^ r n g (I-E) : E e^3J- = I - i n f £l - E : E e g j by 26. Since P = i n f £l-E : E e^ j e E P , P Q € R a l s o . I f now S _> E f o r a l l E € ^ , then I - S < I-E f o r a l l E hence by 26, I - S _< P and S = I - (I-S) > I-P = Q . so t h a t Q = sup ' T h i s proves 27. 48 Remark. Now that 27 i s proven, d e f i n i t i o n 7 of a measure P on R i s complete. 4 9 BT - OPERATORS 28. D e f i n i t i o n . Let R W be the set of a l l s e l f - a d j o i n t operators (including unbounded ones) whose spectral resolution l i e s i n R P. In l i g h t of theorem J> R S = L(H) n R N. Proof of the following theorem may be found i n Functional  Analysis by,Rei s z and Sz.-Nagy (Ungar, New York, 1955) , page 314. Von Neumann's Theorem. Given (a) a sequence : iewj of projections such that and (b) a sequence £A^ : ieuu^ of bounded s e l f - a d j o i n t operators such that A i E ± = E ^ A ^ for a l l i , then there e x i s t s a unique s e l f - a d j o i n t operator A (which may not be bounded) such that AE± = E iAE i = E i A i E i = A i E i f o r a l l 1. Moreover D(A) = £x : x € H, r l l A ^ x l l 2 < » J 50 and f o r x € D(A) AX .'ss ^  A i E i x * 29. Theorem. Let R he an abelian von Neumann algebra. I f the hypotheses of von Neumann's theorem are strengthened: [E±j c R P , [ A j C R S , N then A e R . Hence i f A i s any s e l f - a d j o i n t operator f o r which E ±A c AE ± e R S f o r a l l i e uu, where N then A € R . Proof. The proof i s taken up mainly with deriving the spectral resolution of A from those of the A^. Let P ( i ) be the spectral resolution of A. f o r each I. a v ' 1 Since R i s abelian, A A . E i A i f o r a l l i , and F a ( i ) E i = E i F a ( i ) f o r a l l a and i . Let F a = sup { P a ( i ) % : i e u ^ , which i s a proje c t i o n of R hy 27. Since the E^ are mutually orthogonal, P a " £ 'at 1) E i whence obviously P a <: i f a _< b. To show that P Q y i e l d s a resolution of the i d e n t i t y , i t must be shown how the properties ( i ) lim (strong) F b = P ( i i ) l i m (strong) F = I ( i i i ) l i m (strong) P Q a - o a 0 are i n h e r i t e d from the properties of the F (i)'. To prove ( i ) , l e t n f o r x e H. Then lim = x. Let e > 0 be given. Then f o r a < b, = Y (i€u>) ||(F a(i) - F j i ) ) E ^ l l 2 = £(ien) | | ( F a ( i ) - F b ( i ) ) E ^ J For each i c n there i s an a i > a such that i f a j _ > 1 3 >. a then | | ( P a ( i ) - F f c ( i ) ) E l X n l | < ^ Hence i f a _< h < min £a^ : i e n j , then ^ ( i e n ) | l ( P a ( i ) - P b ( i ) ) E i X n | | 2 < e 2 t h a t i s , ||F x - P. x_ || < €. " a n D n" Consider now the f o l l o w i n g : | ||Fax - Fbx|| - I IP^ - F bxJ| | < ||Fb - F b | | Ux-xJ Thus, since 0 < F b - F a < I, t h i s y i e l d s | ||Fax - Fbx|| - ||F a X n - Fbxn|| | < ||x - xn|| Now f o r n > N , IIx - xn|| < e/2, so t h a t . 0 i H(V Fh) x!l < €/2 + lUVV3^ f o r ' n _> N . But there i s a l s o an a(n) such t h a t a _< b < a(n) i m p l i e s t h a t tKVPa>xJ < €/2' Thus i f a < b < a ( n ) , ||(P b-F a)x|| < e /2 + c/2 = e. Since € i s a r b i t r a r y , t h i s proves ( I ) . P r o p e r t i e s ( I i ) and ( I I I ) f o l l o w i n the same manner. Thus . £ Pa : - 00 < a < + °°J i s a r e s o l u t i o n of the i d e n t i t y . I n order t o show t h a t F Q i s the s p e c t r a l r e s o l u t i o n of ct the operator A i t remains t o show t h a t ( i v ) F A c AF. f o r a l l a ct — ct (v) AF < a F and A ( I - F j > a ( I - F j on D(A). 54 To prove ( i v ) note that from the uniqueness guarantee i n von Neumann's theorem, i f • x € D(A), then Ax = Y (ietu) A^E^x Hence P Ax = P V E.A-E.x a a L* i i 1 i x = Y± \ M E i A i E ' - VaW A i E i X (sinee E ± F a ( i ) = P a ( i ) E ±) £ W a ^ V ( s i n c e A ± P a ( i ) = P & ( i ) A ±) = AP x. a This proves ( i v ) . To prove the f i r s t p a r t of ( v ) , l e t x e D(A). Then (AP ax,x) = ^ (Va(i) V EiX) < ^ a(Pa(i) E.x, E ±x) = a (F ax,x) This proves the f i r s t a s s e r t i o n ; proof of the second i s analogous. Now the s e l f - a d j o i n t operator A has a r e s o l u t i o n F '. cl 55 I n v i r t u e of ( i v ) F P. 1 = F,'F a b b a f o r a l l r e a l a and b. Hence are p r o j e c t i o n s . To show tha t these are zero, suppose th a t x e D(A) n rng F a (I-Fa») Then since x e rng F , a ((A-aI)x,x) < 0 by ( v i ) . That i s / (b-a)d (F b'x,x) < 0, J -co This means tha t (F-^'x,x) i s constant f o r b > a. Thus since F^ 1 i s r i g h t continuous (F b'x,x) = (F a'x,x) f o r b > a. But x e rng (I-F„') a l s o , so that a (Fb»x,x) = 0 f o r b '< a. Hence 56 (P b'x,x) = 0 f o r a l l b. Thus x=0 and S i m i l a r l y Hence F Q < F ' and F ' < F . so that F '= F ' f o r a l l a. ct "™" ct St —~ cl ci That i s , F a i s the spectral resolution of A. This proves that A € R N. The second part of theorem 29 follows from the f i r s t : Since E±A c AE ± e R S N the f i r s t part of 29 says that there i s an operator A ' e R such that EjA 1 c A'E ± = AE ± By the uniqueness guarantee i n von Neumann's theorem, A=A'. Hence A € RW. This completes the proof of 2 9 . 30. Theorem. I f T i s a l i n e a r transformation with": domain and range i n H such that E.T cz TE. s R S l — x ( T E i ) * = T E i f o r a l l ieuu , where R i s an a b e l i a n yon Neumann algebra and : ieuuj? c R P, ^ E± = I , N then there e x i s t s a unique T' e R such that T' => T. Proof. B y von Neumann's theorem there e x i s t s unique T' which i s s e l f - a d j o i n t and s a t i s f i e s E.T' c T'E. = TE. 1 — I i f o r a l l i . D(T') = £x : x. e H, ^ HTEix||2 < > J and i f x e D(T») T'x = T E j [ x B y 29, T' e R N. To show th a t T' extends T, l e t x e D(T). Then Tx = C)^2^) T x = I! E.Tx = ^ TE ±x (a) (sin c e E^ T- c TE^) Now since TE^ = E j LTE i , the terms of the sum (a) are mutually orthogonal. Thus from (a) that i s , x € D(T'). Now (a) a l s o i m p l i e s that Tx = T'x. Therefore T C T' * T 1 i s the only R N - operator which extends T, f o r i f T" e. R N and T c T», then TE i c. T ^ f o r a l l 1. Since TE j L € R S, i t i s defined on a l l of H. Hence a c t u a l l y , TE ± = T"E i, and moreover TE. = T"E. = T'E. i l But t h i s , together w i t h the uniqueness guarantee i n von Neumann' theorem, i m p l i e s t h a t T" = T' Th i s proves theorem 30. J I . Lemma. I f £ E i : ^ U } { F j : are f a m i l i e s of p r o j e c t i o n s such t h a t 59 and then a l s o , E i F j = F j E i > y T.j E i F j = I * Proof. Given xeH and €>0 there must e x i s t a f i n i t e subset (si'^cj/^ such t h a t |y-x| < e / 2 , where There i s a l s o a f i n i t e subset jff's^ such t h a t |y - Y ( J € ^ ) F j y l < e/2 Thus |x - £ ( i e ^ j e / ) E ^ x l = |x - £(J«UJ') P dy| < |x-y| + |y - J (je$') P j y | < e /2 + e /2 = e Hence ^ F_. = I -j and the lemma i s proven. 60 32. Lemma. Let E . F „ be two resolutions of the i d e n t i t y ——•—— a a P i n R , where R i s abelian. I f G Q = E F , p then G i s also a resolution of the i d e n t i t y i n R . a P Proof. Since R i s abelian, G e R .. I f a C b, G G, = G, G = E . F , E F a b b a b b a a = E E v . F F, a b a b = E F a a = G a > so that G a < V Moreover, i f again a _< b, ! ! G b x " G a x I 1 - ^ V b - EaFa>xH < " ( E a P a " Ea Pb) xH + N ( E a P b " E b P b > x « < II(Pa - F b)x|| + ||(Ea - E b)x||, f o r any x e H. Thus G i n h e r i t s the r i g h t continuity of E a a and F . a The properties lim (strong)G = 0 a ^ - o o a 61 and l i m ( s t r o n g ) G a = I a-* + 0 0 can be demonstrated i n s i m i l a r or simpler f a s h i o n s . This proves 32. 33- "Theorem. I f R i s a b e l i a n and A,B € R N, then there N . are unique R -operators S, T, N such that S 2 A + B T A - B N ZD AB and N D BA. Proof. L e t A and B have s p e c t r a l r e s o l u t i o n s E and F„ r e s p e c t i v e l y . Let a E ( i ) . . E. - E ± _ 1 } F ( j ) . F 6 - P . ^ where i and j are i n t e g e r s (negative t o o ! ) . By lemma 31, since R i s a b e l i a n , ^ E ( i ) F ( j ) = 1 . Now f o r any i and j E ( i ) F ( j ) (A-B) = E ( i ) F ( j ) A - E ( i ) F ( j ) B = F ( j ) E ( i ) A - E ( i ) F ( j ) B (s i n c e R i s abe l i a n ) c F ( j ) ( A E ( i ) ) - E ( i ) ( B F ( j ) ) ( s i n c e E ( i ) and F ( j ) are s p e c t r a l p r o j e c t i o n s ) 62 = ( A E ( i ) ) F ( j ) - (BF(J)) E ( i ) (a) ( s i n c e AE(1) and B F ( j ) e R S, which i s abelian) = AE(i) F ( j ) - B E ( i ) F ( j ) - (A - B) E ( i ) F ( j ) . c Since l i n e (a) i s an R -operator, E ( i ) F ( j ) (A-B) c (A-B) E ( I ) F ( j ) e R S. Since the f a m i l y of E ( i ) F ( j ) i s countable, and R i s a b e l i a n , theorem 30 a p p l i e s , so that there e x i s t s a unique T e R^ such t h a t T => A-B. The a s s e r t i o n about (A+B) f o l l o w s immediately. The case of AB and BA i s s i m i l a r , but i t r e q u i r e s some a d d i t i o n a l remarks: E ( i ) F ( j ) AB - F ( j ) E ( i ) AB c F ( j ) ( A E ( i ) ) B ( s i n c e E & i s the s p e c t r a l r e s o l u t i o n of A) = ( A E ( i ) ) F ( j ) B ( s i n c e A E ( i ) and F ( j ) are i n R, which i s abelian) c ( A E ( i ) ) ( B F ( j ) ) (b) ( s i n c e F ( j ) i s a s p e c t r a l p r o j e c t i o n of B) = A ( B F ( j ) ) E ( i ) ( s i n c e both B F ( j ) and E ( i ) are i n R, which i s ab e l i a n ) = AB E ( i ) F ( j ) . S Line (b) i s a product of R operators, hence E.(i) F ( j ) A B c (AE(i)) (BF(j)) » ABE(i) F ( j ) e RS. By symmetry E( i ) F ( j ) BA c (BF(j)) (AE(i)) = BAE(i) F(J) € R S N Thus by 30 there e x i s t s unique N,N' i n R such that N ^ AB N' 2 BA. However, since R i s abelian ABE(i) F(J) = (AE(i)) (BF(j)) = ( B F(J)) (AE(i)) U BAE(i) F(J) i n R S. Hence N =.N'» . This completes the proof of 33. Because the S, T, N of theorem 33 are unique, the following d e f i n i t i o n s are possible. 34. D e f i n i t i o n . In 33 S = A + B T = A - B N = A o B. Note that 33 implies' that f o r A,B € RW, A ° B = B o A and that A +* B = B I A and 64 A - A = 0 . Moreover, i f C e R W a l s o , then (A+B)C = AC + BC and AC + BC c (A°C) + (B.C) c (A.C) + (B.C) while on the other hand (A+B)C c (A+ B)C c (A+ B)«C. Since these extensions are unique (A+B)oC = (A«C) + (BoC). Thus R i s a commutative algebra under the operations + and o 35. C o r o l l a r y . I f A-B >_ 0 on D(A-B), then A-B _> 0 a l s o . Proof. C a r r y i n g on w i t h the n o t a t i o n of 33, by 30 D(A-B) = £x : xeH, ||(A-B)E(i) F ( j ) x | | 2 < *>} and f o r x e D(A-B) (A-B) x = ^ (A-B) E ( i ) F ( j ) x, whence i t ' s obvious that A-B > 0 i m p l i e s A - B > 0 . 36. C o r o l l a r y . I f A e R and B e R , then 65 A 0 B m B o A = AB, and so BA c AB € R W  Proof. L e t N = A»B = BoA . Then s Since B e R , D(BA) = 'D(A) so t h a t BA i s densely defined and the operator (BA)* e x i s t s , Hence N = N* c (BA)* = A*B* = AB c N whence N = AB This proves 36. 66 POSITIVE AND NEGATIVE" PART So OF? AN"R--OPERATOR Let N be an operator i n R^ w i t h s p e c t r a l r e s o l u t i o n Grt. Then a Thus Let N = N [ ( I - G 0 ) + G Q ] 2 N ( I - G Q ) 4- N G Q 2 ( I - G 0 ) N 4- GQN = [ ( I - G 0 ) + G Q ] N = N N = N ( I - G 0 ) - ( - N G G ) N + = N ( I - G Q ) N " = - N G Q . + ^ N Then both N , N are p o s i t i v e , and by 26, both are i n R . Hence as w e l l as N m N + - N" N = N + - N" 37- Theorem. I f N = A - B, where A,B are p o s i t i v e i n R N, then there e x i s t s a unique p o s i t i v e operator P i n R N such th a t , A = P + N + B = P + N" Proof. I f A.B have the r e s p e c t i v e r e s o l u t i o n s E ,F , 67 E ( i ) ==' E j [ - E ^ i e ou and F ( j ) = - J e. w, then ^ E ( i ) P ( j ) = I ; -J s i n c e A . and B are p o s i t i v e . Note f i r s t t h a t (A-B) E ( i ) F ( j ) = (A-B) E ( i ) F ( j ) = A E ( i ) F ( j ) - B E ( i ) F ( j ) , S where the l a t t e r i s i n R . A l s o (A-B) E ( i ) F ( j ) = (N+-N") E ( i ) F ( j ) = N + E ( i ) F ( j ) - N " E ( i ) F ( j ) . Thus N + E ( i ) F ( j ) - N~E(i) F ( j ) e R S. T h i s means t h a t the s e l f - a d j o i n t o p e r a t o r s N + E ( i ) P ( J ) r N~E(i) P ( j ) c are d e f i n e d everywhere. Hence they are bounded and belong t o R . Thus A E ( i ) P ( j ) - B E ( i ) F ( j ) = N + E ( i ) F ( j ) - N " E ( i ) P ( j ) , S where a l l terms are i n R . A E ( i ) F ( j ) - N + E ( i ) P ( j ) = B E ( i ) F ( j ) - N ~ E ( i ) F ( j ) (A-N +) E ( i ) F( j ) ss - (B-N~) E ( i ) F( j ) . 68 Thus A N + = B - N~ Now l e t P B - N Then f o r any PE ( i ) F(J) (B-N") E ( i ) F ( j ) = (B-(-NG n)) E ( i ) P ( j ) (since G i s the s p e c t r a l r e s o l u t i o n of Nj and N~=-NGn) Thus i f x e rng E ( i ) F ( j ) (then x € rng P, rng A, rng B ) , (Px,x) = (B(I-G Q)x,x) + (AG Qx,x) a '((I-G 0) Bx,x) + (G QAx,x) (hy 36 G QA c AG Q, etc.) = ( ( I - G Q ) B x , ( I - G Q ) x ) + (G 0Ax,G Qx) = ( B ( I - G 0 ) x , ( I - G 0 ) x ) + (AG Qx, G 0 x ) . This l a s t i s non-negative. Hence P = A - N + = B - N" > 0 = [B(I-G 0) + AG Q] E ( i ) F ( j ) and A = N4" + P B = N" + P as r e q u i r e d . This proves 37. 69 MEASURE THEOREMS 38. Theorem. [A c o v e r i n g theorem f o r s e m i - f i n i t e measures]. L e t R he any von Neumann a l g e b r a . L e t m be a s e m i - f i n i t e measure on R P ( d e f i n i t i o n s 7 and 8), and l e t E he a non-zero p r o j e c t i o n i n R P. Then there e x i s t s a f a m i l y ^ E j / £ R P such t h a t ( i ) the E^ are mutually o r t h o g o n a l ( i i ) zaE± < 0= f o r a l l E ± ( i i i ) ^ . E ± = E Pr o o f . L e t X he the set o f a l l subsets I t R ' such t h a t ( i ) ©-members are mutu a l l y o r t h o g o n a l ( i i ) i f F 6 B , then 0 < P _< E and mP < » X i s p a r t i a l l y ordered by set i n c l u s i o n , and every l i n e a r l y ordered X-subset has an upper bound i n X (namely, i t s u n i o n ) . Hence by Zorn's Lemma, X has a maximal element B. P P P Since jB c R , sup jB e R and E - sup B e R , by 27. Now E - sup B = 0, I P f o r o t herwise, s i n c e m i s s e m i - f i n i t e , t h e r e i s P € R such t h a t 0 < P.< E - sup JB and mP < <=°. That i s , the X-member © U p r o p e r l y c o n t a i n s JB. But t h i s c o n t r a d i c t s the maximal!ty o f B. Thus TO sup'j| = E and jB f u r n i s h e s a f a m i l y o f the r e q u i r e d s o r t . QED. 39. Theorem. I f R i s any von Neumann a l g e b r a and P E . E e R , where then mE n f mE Pr o o f . By lemma 21, E = sup [ E J . Now. { En+1 " E n ; n € } p I s a f a m i l y o f "mutually o r t h o g o n a l R -members. Hence E = sup { E ^ = E Q + sup I E n + 1 - E n • n e*j - E 0 + \ < V l - E n ) : ^ 0 + \ ra(En+l " E n ) m E 0 + \ ( m E n + l " ^ n ) 1 mE Q + l i m (mE n - mE Q) so t h a t mE = l i m mE . n F i n a l l y , since m i s monotone, mE n f mE . 40. C o r o l l a r y . I f E n E and mEQ < then, mE., 4. mE n Proof. I f E n 4- E, then ( E0 " En) ' ^ 0 " E ) so t h a t m(E 0 - E n) * m(E Q hy 39. 72 EXTENSION OF m TO R S P qp 4. D e f i n i t i o n . I f S i s a p o s i t i v e R - o p e r a t o r , P s - r ^ E I i = l t h e n d e f i n e „ P mS -= Y a j _ m E j _ i = l ( w h ich i s n o n - n e g a t i v e and f i n i t e o r i n f i n i t e a c c o r d i n g as 2 (&±^0) ..mE± i s f i n i t e o r i n f i n i t e ) . S i s summable i f mS < ». SP I f S i s an a r b i t r a r y R - o p e r a t o r , t h e n S m S + - S", . where S ,S a r e p o s i t i v e R - o p e r a t o r s . S i s i n t e g r a b l e i f a t l e a s t one o f S +,S" i s summable. I f S i s i n t e g r a b l e , d e f i n e mS = mS + - mS" ( t h i s I s w e l l d e f i n e d s i n c e one o f mS+,mS" i s f i n i t e ) . S i s summable i f |mS| < », t h a t i s , i f b o t h S +,S~ a r e summable. qq gp L e t R be t h e s e t o f a l l summable R - o p e r a t o r s . gp Remarks. An R - o p e r a t o r i s summable i f and o n l y i f i t s s u p p o r t has f i n i t e measure. 73 I f S € R S P I s i n t e g r a b l e , P mS = a^ mEi . i = l Henceforth R i s assumed t o he an a b e l i a n von Neumann algeb r a . 42. Theorem. I f S € R S P, then i t s i n t e g r a b i l i t y i s SP Independent of i t s r e p r e s e n t a t i o n i n R SP I f S i s i n t e g r a b l e i n R , then i t s i n t e g r a l i s SP independent o f i t s r e p r e s e n t a t i o n i n R . Proof. I t i s enough to prove the second a s s e r t i o n f o r SP p o s i t i v e R -operators. qp Let S,T e R such t h a t S = T >_ 0, where p S = T a i E i T = £ b j F j . Note f i r s t t h a t since S = T, so th a t mS, mT are f i n i t e or i n f i n i t e simultaneously. This proves qq the a s s e r t i o n f o r the case t h a t S £ R I f mS, mT are f i n i t e mS - mT = a i m E i - \\ 1 5 J " ^ j 74 £ j a i " V j - £ j *J m E i P ( s i n c e V E i ~ Z^j = 1 3 1 1 ( 1 R i s abelian) = ; (a. - b .) mE.F. i j " " I f mE iFj > 0, then E^Fj 0, so that a± = b ^. Thus i f mEj-Pj ^  ®> s o a l s o - l ) j = 0, and hence mS - mT = 0 This proves 4 2 . 4 3 . Theorem. I f S,T e R , then S+T € R S S aS e R S S f o r a l l r e a l a. Moreover m(S+T) = mS+mT maS = amS. The second a s s e r t i o n about (S+T) holds I f S,T are p o s i t i v e and one of mS,mT i s i n f i n i t e . Proof. The a s s e r t i o n s about aS are obvious, the others t e d i o u s . I f S and T are represented as i n 4 2 , then S+T = V ( a i + b j ) Xfy. since ° ^ F. = I . 3 P Note th a t E^Fj e R , since R i s a b e l i a n . Now (a^+bj) ^ 0 only i f &^ £ 0 or b j ^ 0. I f ^ 0, then mE^ < » and mE. F < mE. < °=. i 3 ~ I S i m i l a r l y b„ ^ 0 i m p l i e s t h a t mE. P. < ». Therefore ^ ( a j + b ^ O ) mEj^Pj <.» and S+T e R S S. By d e f i n i t i o n m(S+T) = Y ( a ^ b j ) mE^F^ = £ ai ^  n E i F j + E, bd ^  -Va •- I a i m E i (I ty + £, * J m F j E l ) - \ a I ^ i + 2j b J m F J = mS + mT T h i s proves 43. SS " s 44. Theorem. R i s closed under the R l a t t i c e o perations (R i s a b e l i a n ! ) . 76 qq Proof. For S,T i n R r e c a l l that (Sl/T) = TF Q + S ( I - F Q ) (SHT) = T ( I - F 0 ) + SF Q , where F Q i s the s p e c t r a l r e s o l u t i o n of S-T. Obviously a l l of a gq the terms on the r i g h t are i n R , hence by kj>, (SUT) and (SOT) SS belong to R . QED. qp 45. Theorem. I f S and T are i n t e g r a b l e i n R then S _< T Implies mS < mT Proof. I f 0 j : S _< T, then (using the re p r e s e n t a t i o n s i n 42) Thus mS = 00 i m p l i e s t h a t mT = «>. Now I f mS < 00 then as before mT - mS = ) (b .-a. )mE.F.. L±3 3 1 1 0 i f mE^Fj ^ 0, i t must be that E ± F j £ 0, and b j - a i >^  0, since S _< T. Thus mT - mS _> 0 and so mS _< mT. SP I f S and T are a r b i t r a r y R -operators which are i n t e g r a b l e , S < T 77 implies t h a i s+ ^ T+ T" <_ S", whence from above so that mS+ < mT+ mT" < mS~, mS = mS+ -•mS" < mT+ - mT~ = mT, This proves 45. 46. Theorem. I f j f T n ^ c R S S and then T n ^ 0 mT i 0 n SP Remark. This theorem i s not true i n R . Por l e t T n = i I n = 1, 2, ... Then T n l 0 (uniform), hence also T n ^ 0 . But mTn = » f o r a l l n i f ml = co. Proof. Let M(n) T n = I b i n F i n ' 1=1 Por any r e a l number c, l e t E ( n , c ) " - I ( b i Q > c) P i n 78 E(n,c) i s thus the projection onto the subspace where T n > c. Note that E(n,c) 2 E(n,c+e) for e > 0. Consequently E(n ,0) > E(n,c) and mE(n, c) < oo . Moreover, since T n ^ , E(n,c) 2 E(n+l,c). Let C(n) = max {Vj_n : f i x e d nj . Then C(n) J, since T n i • The pi^ce de resistance of t h i s proof i s the f a c t that f o r c > 0 l i m n m E(n,c) = 0 . To prove t h i s , suppose c o n t r a r i l y that lim m E(n,c) > 0. Since mE(n,c) < oo and E(n,c) ^  , 4o applies and lim mE(n,c) = m(lim[strong] E(n,c)) Let E = i n f [E(n,c) : fixed c j . Then E € R P by 2 6 , and by 2 2 E = lim (strong) E(n,c). Therefore mE = m lim (strong) E(n,c) = lim m E(n,c) > 0, so that E ^ 0. Thus there i s non-zero x i n rng E. Now x e rng E(n,c) f o r a l l n € uo. Thus (T nx,x) > c||xj|2 > 0 f o r a l l n, contradicting; the -fact ;that T R f 0. Hence lim m E(n,c) = 0 f o r c > 0. To complete the proof, for a l l non-negative integers E(0 50) _> E(n,c) f o r c > 0. T n = E(0 30) T n = E(n,c) T n + (E(0,0) - E(n,c)) T n < C(n) E(n,c) + c(E(0 50) — E ( n , c ) ) < G(0) E(n,c) + cE(0,0) . [*] Now, given e > 0, choose c so that 0 < c < — — , 2mE(0j0) then choose N so that n > N implies 80 mE(n,c) < e/2C(0). Then from [*], n > N implies that i s , m T n 1 € / / 2 + G / / 2 = e» lim mTn = 0, By 4 5 , t h i s convergence i s monotone. QED. In the next section, m w i l l he extended to R . The following theorem hints how t h i s w i l l he done. 4 7 . Theorem. I f R i s abelian, S e R S P, £s n j c R S P, s n - 0 a n d then s n f S, mSn mS. Proof; Since for a l l n, by 4 5 , and 0 < s n < S 0 < mS„ < mS„,, < mS — n — n+i — lim mSn _< mS Thus i f lim mS^ = °=, mS = *> . n 3 For the converse of t h i s , l e t E(n,c) and E(c) be the respective projections on which S n and S are greater than 81 c >_ 0. Note that E(n,c) E(n+l,c) < E(c) (a) I f c < max S, the largest of the non-zero c o e f f i c i e n t s of S (assume S^O!) and x e rng E(c) ||x|| = 1, then (Sx,x) > c and since s n rr .s, there exists an integer Q(c,x) such that n > Q implies (Sx,x) > (S nx,x) > c so that x e rng E(n,c) f o r n _> Q, that i s , using (a) E(n,c) f E ( c ) . By lemma 39 mE(n,c) T.mE(c) (b) where c < max S. Now i f c _> 0 S n > E(n,c) S n > c E(n,c). Thus mSn 2 c mE(n,'c) (c) f o r a l l c > 0. 82 I f 0 < c < min S (d) then E(c) = E(0) and so, i f mS = « and . mE(c) =' a , subject to (d), then by (b) lim mE(n, c) = ooj and by (c) lim m S = oo. n Thus mS = oo i f and only i f lim mSn = oo, and the theorem i s proven i n t h i s case. The above also shows that mS < ce l i m m S n < oo are simultaneously true. In t h i s case a l l the operators involved are i n R^°. Since also s n t S, s - s n ^ o and by lemma 46 m(S-Sn) ^ 0 . But m ( S - S n ) = mS - mSn 83 by 43. Hence mS n f mS. op 48. Lemma. I f S i s a p o s i t i v e R - o p e r a t o r f o r which mS = + « , then there e x i s t s £S f c R such t h a t S n i S> S n f ' m S n t » Proof. L e t n 3 - I a i E i 1 = 1 Then mS = » i f and only i f mE^ = » f o r some i such t h a t By 38 a ± ^ 0 E. , J ( j e J ) F j , where the are mutually orthogonal and are of f i n i t e measure Choose a nest of f i n i t e J-subsets P(n): P(n) c P(n+1) c J f o r n € ai } such t h a t Y (jeP(n)) m P j >-S . a i Then the sequence [_^nj > S n " a i I ( J e P ( n ) ) F J > has the re q u i r e d p r o p e r t i e s . T h i s proves 48. 8k k9. Independence Theorem. If £s j and £ T n | are SP sequences i n R such that S t S T f T, n n * then T < S implies lim mT < lim mS„ . n — n Proof. By 20 (Tm n s j t (T n s) , v m n< n v m ' and since then so that S > T > T n , T H S = T m m ( T m n sn) f T. m and by 4? m( Tm n Sn) ?n m T n • Thus for arbitrary m e co and Df?'< mTm, there exists R(m,<37 ) such that i f n > N Dtf < M(Tm n sn) . But (T n s j < sn , v m n' — n 3 so 3?7< m(Tm 0 S n) < mSn 85 f o r n > If. Thus Off < l i m mSn f o r a l l Off <. mTm, hence mT < l i m mS„ . m — n Since m i s a r b i t r a r y l i m mT < l i m mS . m — n T h i s proves 4 9 . Remark. I f , as above, s n f S T T. s then l i m mS„ = l i m mT . n n ) 86 EXTENDING m TO R S CJ 50. D e f i n i t i o n . L e t T be a p o s i t i v e R -operator, By 24 there I s a sequence -^T^J c R S P such that Define n mT a' l i m mT . (Theorem 49 shows t h a t mT does not depend on the choice of sequence £T j . Theorem 47 shows t h a t t h i s i s a c o n s i s t e n t extension of d e f i n i t i o n s already made.) T i s summable i f mT < <» . 51. Lemma. I f S i s a p o s i t i v e R -operator f o r which mS 3 - +<*>, then there e x i s t s [sn] « R such that n — n' 3 n T n J — R such th a t n 3 and by d e f i n i t i o n mTn t mS == oo . E i t h e r ( c R and the theorem i s proven, or there e x i s t s T N i R S S. Then T n £ R S S f o r a l l n _> N. By 48 there e x i s t s a sequence y£Lv c R such t h a t \ < T N < S, S n f , m S n t ». Thi s proves 51. s 52« C o r o l l a r y . I f T i s a p o s i t i v e R operator, then there e x i s t s £"Tnj- £ r S S s u c n that 0 < T < T T t — n — n l i m mTn = mT, g 55• Theorem. I f S and T are R operators f o r which 0 < T < S, then mT _< mS. Hence i f S i s summable, so i s T. Proof. By 24 there e x i s t ^"sn^ and ^*T nj , sequences i n -R j such that S t S T f T n n Theorem 53 thus f o l l o w s immediately from 49. 54. Theorem. I f T i s a p o s i t i v e R -operator and G m f H s H € R S, 0 _< H _< TJ B = (n : H G R S P, 0 ^ H < T j then X = [K : H e R S S, 0 _< H < T j mT = sup £mH : H e G] = sup £mH : H e B J - sup £mH : H €• , 88 Proof. By 52, x ^ 0. Obviously X c . 8 c G. Thus by 53 mT 2 sup £mH : H e GJ 2 sup £mH : H € B j . 2 sup I mH : H e X j 2 l i m mTn = mT, c 7 SS where c R i s the sequence whose existence i s guaranteed by 52. T h i s proves 54. s The extension of m to R -operators of a r b i t r a r y s i g n i s f o r m a l l y the same as the analogous extension i n d e f i n i t i o n 41. Without f u r t h e r ado the terms " i n t e g r a b l e " and "summable" S w i l l be a p p l i e d t o R -operators. The f o l l o w i n g i s a lemma toward a proof t h a t m i s l i n e a r S on the summable R -operators. g 55. Lemma. Let A and B be p o s i t i v e R -operators. (a) m(A+B) = mA + mB (b) I f one of A,B i s summable then (A-B) I s i n t e g r a b l e and m(A-B) = mA - mB. Hence i f A and B are summable, so are (A+B) and (A-B) . 89 Proof. Since A,B are p o s i t i v e , there exist"sequences SP l k i i > f B n ? i n R such t h a t A n r A B n T B and mA = l i m mA„ mB = l i m mB . n n Now (Aj, + B n) t (A + B ) , and since m(A+B) i s independent of the sequence (A n+B n), the a s s e r t i o n (a) now f o l l o w s e a s i l y : m(A+B) = l i m m(A n + B n) = l i m (mAn + mBn) (by 45) = I i i + l i m mBn = mA 4- mB, so t h a t (a) i s proven. To prove ( h ) , note t h a t by 25 A = ( A - B ) + 4- P B =a (A-B) ~ 4- P f o r P > Oj whence A _> ( A - B ) + > 0 B > (A-B)" > 0 A > P > 0. B > P VO, so t h a t by 53, since one of A,B i s summable, so i s P arid so i s one of ( A - B ) + , (A-B)". Thus (A-B) i s i n t e g r a b l e . 90 Now mA - mB a m[(A-B)++P] -m[(A-B)~+P] •a' m(A-B)++mP - m(A-B)"-mP (by part (a)) = m(A-B)+ - m(A-B)~ = m(A-B). This proves 5 5 . 5 6 . Theorem. I f S and T are summable operators, then (S+T) i s summable and m(S+T) = mS + mT Proof. I f S and T are summable, then so are S+, S", T +, T" . Since these are postive S + + T + S" + T~ are summable by 55 (a). Thus (S+T) can be expressed as the difference of two positive summable operators: S+T = (S + - S") + (T + - T~) a (S + + T +) - (S~ + T")'. By 55 (*>), S+ T i s thus also summable and m(S+T) a m(S++ T +) - m(S" + T") 91 = mS + + mT+ - (mS" + mT") (by 54 (a)') = (mS + - mS") + (mT+ - mT") = mS + mT. Th i s proves 56. g 57. Theorem. I f S i s an i n t e g r a b l e R -operator and a i s any r e a l number, then aS i s a l s o i n t e g r a b l e and m(aS) = a mS •58. Theorem. I f S and T are summable, and S >^  T / then mS _> mT Proof. The standard argument. Since S-T _> 0, m(S-T) j> 0, hence mS s> m(S-T) + mT _> mT. 59. Lebesgue's Monotone Convergence Theorem. I f { T ^ i s a sequence of p o s i t i v e summable R S-operators such t h a t then T n t T, mT s= l i m mT . n Proof. For each n there e x i s t s £S^j c R S S such t h a t and n m n lir» mS ™ = mT < *>. n n Le t Um - S U P c^q* : * < m j > SS where the sup i s taken i n the sense, o f 1 J . Then U e R n 44. Moreover n ' f o r V l - S U P f S q m + 1 : * < m + 1 J >_ sup £sqm : q _< m+lj ( s i n c e > S - .... 2 sup i^Sq 1 1 1 : q <mj . Now = U . m S„ m <; Um < T (*) q — m — m v- 7 f o r q_< m. Hence (strong l i m i t s ) so t h a t lirri S„ m = T < l i m U . < l i m T = T m n n — m — m T < l i m U < T n — m — foi * a l l n, or 93 so t h a t T = l i m T < l i m U_ < T n — m — T = l i m U m and hence U t T. Since n", € R S S m m mT = l i m mUR (**) Now hy (*) m S„ m < mU < mT q — m — m f o r q < m. Hence and whence m q q — m — n mT < l i m mU < l i m mT q — n — n mT = l i m mU = l i m mT n n hy (**). Thus mT = l i m mT . n T h i s proves 59 • The f o l l o w i n g theorem i s an analog of the d e f i n i t i o n f ( x ) dx = l i m / f ( x ) dx f o r the Riemann i n t e g r a l . 60. Theorem. I f T i s any i n t e g r a b l e R -operator and P G_ i s any r e s o l u t i o n of the i n d e n t i t y i n R , then 94 mT = l i m (a =>) mT G SI Proof. F i r s t observe t h a t i f T > 0, then the l i m i t must e x i st and mT > l i m (a -* °°) mT G Q 2 Hm mT G n I f T > 0 there e x i s t s ^T R^ c R such th a t 0 < T < T T- f T — n — n mTn t mT i o p Now T G e R f o r a l l n. Moreover n n n n Now f o r any x € H !! ( T n G n - T)x|| < l l ( T n G n - TG n)x!l + ||(TGn - T)x|l < l l f f j l II ( T n - T)x|| + llTll |](I •- G n)x| = I I ( T n - T)x|||.+ ||T|| ||(I - G n)x|j, whence since T G T T, n n 3 T„ t T G t l . n n Since mT i s independent of the sequence mT nG n T mT, 95 , Wow t h e f a c t t h a t mT G_ < mT G„ n n — n and (*)• prover. t h e theorem f o r T _> 0. q F o r a r b i t r a r y i n t e g r a b l e T i n R , mT = mT + - mT" - l i m ( a - ^ o o ) mT +G Q - l i m (a->•<») mT~G cl cl = l i m ( a - v c o ) (mT +G - mT'Gj cL St = l i m (a^») (m(T + - T") G j Cb \ = l i m (a->») mT G cl T h i s p r o v e s 60. 96 MEASURE AND POSITIVE RW-OPERATORS I f N i s a p o s i t i v e R N-operator w i t h r e s o l u t i o n .E , S then NE„ e R f o r a l l a. Moreover a NE„ < NE. a — h f o r a < b, so t h a t mNE„ i s an i n c r e a s i n g f u n c t i o n of a. This j u s t i f i e s the f o l l o w i n g . 61. D e f i n i t i o n . I f N i s p o s i t i v e i n R W, mN = l i r a (a-*®) mNE N i s "summable i f mN < <*>. 62. Theorem. I f N i s a p o s i t i v e R -operator and G r • a P i s any r e s o l u t i o n of the i d e n t i t y I n R , then mN = l i m (a-»») mN G . cl Proof. By 36, NGQ e R N f o r a l l a. The proof must — — a show tha t l l m ( a - > c o ) mNG . a e x i s t s and equals mN. Let E„ be the r e s o l u t i o n of N and l e t a [H c(a) : — < c < »J be the r e s o l u t i o n d f : the R^-operator NG &. Then NG a H c(a) e R S f o r a i l a and c, and for fix e d a, mNG H (a) i s an increasing function of c, mNG = lim(c^») mNG H fa) ct ct C so that mNG > m NG„ H (a) (a) <3» — ct C . f o r a l l c. Furthermore, f o r a l l h<; and c, NG H„(a) E, e R S a c v ' b and NG a H c(a) > NG a H j a ) E ^ so that mNGa H c(a) > mNGa H c(a) B^ .' Combining t h i s with (a) mNGa > mNGa H c(a) E f e (b) for a l l b and c. Now NG H fa) E, = NE, G H„(a), a c x / b b a c v / ' NEfe G a e R S, and by 60 mNE^ G a = lim (c->») mNE^ G a H c(a) Hence by (b) mNGa > mNE, G (c) ci D ct f o r a l l b. To obtain a further inequality on mNG . note that ct m a V * ' E b < ^ b in R f o r a l l b and c. Again by 60, since NG & H c(a) e R for a l l c, raNGa H c(a) = lim(b-»») mNGa H^a) E^ _< lim(b-^») mNE^ = mN f o r a l l c. Thus mNGQ = l l r a ( c ^ o o ) mNG H„ (a) < mN (d) f o r a l l a. I t remains to apply 60 once more: mN =» lim(b ^ e o ) mNE^ = l i m ( b l i m ' a - ^ o o ) mNE^ G & . This means that given 377 < mN, there exist a',b» such that mN > mNE^, G &, > ^7/ (e) Por a > a' mNE, . G„ > mNE, , G o, , b 1 a — b 1 a' ' . and by (c) mNG„ > mNE, . G . a — D 1 a Thus f o r a > a', (e) becomes mN _> mNG >_ m W E h « G a - " ^ b ' G a ' y ^ * Since 077 Is a r b i t r a r y , t h i s Implies that lim (a-* co) mNG ct e x i s t s , and mN = lim (a-*<») mNG_ . ct This proves 62. 99 ^3. Theorem. I f A and B are p o s i t i v e R -operators m(A + B) = mA + mB Proof. Let E a , F & he the r e s p e c t i v e s p e c t r a l r e s o l u t i o n s of A and B. Then G a = E F a a a P i s a r e s o l u t i o n of the i d e n t i t y i n R (hy 32). Now A + B 3 A + B (A '+ B) G- r> (A + B) G = AG 0 + BG . a "— a a a a Since the right-hand side i s i n R , i t i s defined everywhere i n H f o r a l l a, hence so t h a t (A + B) G a AG^ + BG a i n R S and m(A + B) G = mAG + mBG . L e t t i n g a tend to 00 and a p p l y i n g 62 y i e l d s m(A + B) = mA + mB . This proves 63. 100 MEASURE AID RN-0PERATORS OP ARBITRARY SIGN D e f i n i t i o n . I f N c R N and one of N +, N" i s summable, then H i s i n t e g r a b l e and mN « mN4" - mST ® i s summable i f |mN| < » ^ 5 . Theorem. I f A and B are p o s i t i v e R^-operators, one of which i s summable, then (A - B) i s i n t e g r a b l e and m(A - B) = mA - mB Proof. Let N'a A - B, then by' JJ there i s a p o s i t i v e N Rx -operator P such that By 6 3 , A = N + + P B = N" + P . mA = mN+ + mP mB a ml" + mP. And since N +, N", p are p o s i t i v e raA _> m P >. 0 mB > mP > 0 mA > mN+ > 0 mB >m~ ,> 0, so t h a t , since one of A,B I s summable, P and one. of N+,N" I s summable. Thus N i s i n t e g r a b l e and mN a mN+ - mN" = mlf1" + mP - mN~ - mP (sinc e mP i s f i n i t e ) = ra(N+ + P) - m(N~ + P) (by 63) =a mA - mB . Thi s proves 65. 102 ABSOLUTE CONTINUITY Let now R be an a b e l i a n von Neumann al g e b r a . L e t m and n be s e r a i - f i n i t e measures ( d e f i n i t i o n s 7 and 8) such that n i s a b s o l u t e l y continuous w i t h respect to m ( d e f i n i t i o n 9 ) . The remainder of t h i s opus w i l l be taken up w i t h p r o v i n g a Radon-Nikodym theorem f o r t h i s s i t u a t i o n . Approximately: N there e x i s t s an operator N I n R such that nT = m(N « T) f o r a l l T In R N. 66. D e f i n i t i o n . Por any r e a l number 67. D e f i n i t i o n . L e t E e G(a) . E i s a-good i f P e G(a) whenever 0 < P < E and P e R P. 68. Lemma. I f E i s a non-zero G(a)-member, then there P e x i s t s F € R such t h a t 0 < P < E and F i s a-good. Proof. Let X be the set of a l l f a m i l i e s IB er R' ,P such th a t ( i ) B-members are mutually orthogonal ( i i ) i f K e ft, then 0 < K < E and a i K < nK. X i s p a r t i a l l y ordered by set i n c l u s i o n . Obviously every nest i n X has an upper bound i n X. Thus X Contains P a maximal element ft c R . Now a m sup JB= a m ^ (Keft) K P ( s i n c e j|-members are mutually orthogonal and sup" B e R by 27) < = n sup ft that i s , a m sup B < n sup ,B Now sup ft < E, since E e G(a), t h e r e f o r e , sup B < E. The non-zero R -member F = E - sup ft 104 has the d e s i r e d p r o p e r t i e s . For i f KcRj P and 0 < K <: F a mK < nK, then £ X and c o n t r a d i c t i n g the maximality of B. T h i s proves 68. The set G(a) i s non-empty f o r a l l r e a l a ( I t contains at l e a s t the zero p r o j e c t i o n ) . Thus the f o l l o w i n g d e f i n t i o n i s j u s t i f i e d . 69. D e f i n i t i o n . (& c R i s a maximal fam i l y of mutually orthogonal a-good p r o j e c t i o n s . 70. Lemma. sup e G(a) 71. i s a-good. 105 P Proof. Let P be an R -member such t h a t 0 < F < s u p ^ 0 . T/O show F s G(a), amP = a m (P s u p ^ 0 ) » a m (F Y ( E ^ ( ? ) E ) = a m Y ( E ^ f ) F E (where FE i s a p r o j e c t i o n ^ since R i s a b e l i a n ) = Y ( E €C?) am FE 2 Y (Ee<f)n ^ ( s i n c e PE _< E and E i s a-good) = n Y ( E E < ? ) F E = n (F s u p ^ ) =» nF Thus sup (@ i s a-good. 72. D e f i n i t i o n . E a = sup^° . Since G(a) i s never empty, E e x i s t s f o r a l l r e a l a. By 27, 70 and 71, E Q e R P, E e G(a), and E i s a-good. CU CL Cv P 73. Lemma. I f F € R and 0 < F < I - E o — a then F £ G(a'), t h a t i s , amP < nF„ 106 Proof. Suppose on the contrary t h a t . F € G(a). By 68, P there I s F 1 e R such that F' i s a-good and 0 < F T < F. Now 0 < F» < I - E o — a so that F» i s orthogonal t o every E i n ^ 2 * . T h i s means tha t i s a f a m i l y of mutually orthogonal a-good elements which p r o p e r l y contains the maximal f a m i l y (@ , a c o n t r a d i c t i o n . Thus F' / G(a) and 73 i s proven. 74. Summary of the p r o p e r t i e s of the p r o j e c t i o n E . 1 1 CL For every r e a l number .a there e x i s t s a p r o j e c t i o n E Q CL such th a t (I) E a e R P ( i i ) I f F e R P and 0 < F < E . *~~ —*~ CL amF 2 n F ( i i i ) I f F e R P and 0 < F < I-E - amF < nF P ( i v ) E i s the unique such R - p r o j e c t i o n . CL Proof of ( i v ) : L e t E€R P s a t i s f y ( i i ) and ( i i i ) Since R i s a b e l i a n , P = E ( I - E a ) Q = E a ( I - E ) 107 P are p r o j e c t i o n s i n R . I f E / E . at l e a s t one of P,Q i s cl non-zero. I f P ^ 0, then since P _<'E, ( I i ) i m p l i e s t h a t amP 2 nP. But since P < I-E . ( i i i ) i m p l i e s t h a t amP < nP A s i m i t a r c o n t r a d i c t i o n a r i s e s i f Q £ 0. T h i s proves ( i v ) . The next few lemmas w i l l show tha t [ \ 1 - < a < - j P i s almost a r e s o l u t i o n of the i d e n t i t y i n R . 75. Lemma. E i s a monotone i n c r e a s i n g f u n c t i o n of a: 1 • " CL I f a < h, then E o < E. . Proof. Since R i s a h e l i a n W = V a ' so i t .remains t o show E.E^ . E.. a D a Now P =E_ - E J ^ - E j l - E ^ > 0 a a I f • P > 0, then 0 < P < E • — a and 0 < p < x~\> 108 so that hy (11) and ( i l l ) of 74 araP _> nP bmP < nP or bmP < nP _< amP, which i s a c o n t r a d i c t i o n since a _< h. Thus P = 0 and 75 i s proven. 76. Lemma. l i m (strong) (a-* +«) E =1 Proof. L e t E = sup [E : a r e a l ] then a l s o by 75, E =» sup |Ep : pcuoj Again by 75 , so t h a t by 22, E = l i m (strong) EL. ir and s i m i l a r l y E = l i m (strong) V^p) where £ a(p)J 1 8 a n v r e a l sequence such t h a t a ( p ) " t » . Thus E =• l i m (strong) (a-^eo) E Cb The second p a r t of the proof i s to show th a t E = I . I f I-E £ 0, then, since n i s a s e m i - f i n i t e measure, 109 P there e x i s t s P € R such t h a t 0 < P < I-E nP < eo. Now 0 < P < I-E < I-E„ — — a f o r a l l r e a l a. Thus hy ( i i i ) of 74 a mP < nP f o r a l l r e a l a. Thus mP => 0 nP > 0 . But n i s a b s o l u t e l y continuous w i t h r e s p e c t t o m, so t h a t mP =* 0 i m p l i e s nP =» 0 . T h i s means t h a t nP = 0 nP > 0 . Thus the assumption I-E ^ 0 has l e a d t o a c o n t r a d i c t i o n . T h i s proves 76. 77. Lemma. E_ i s s t r o n g l y r i g h t continuous: 1 1 cl l i m (strong) (b.-*a+) E. = E P r o o f . T h i s p r o o f i s e x a c t l y l i k e t h a t o f 76. L e t E =5 I n f [ \ : b > a j € RP.' Then, s i n c e E^ i s a monotone f u n c t i o n o f b, E > i n f f E a ( p ) : peu)| where £a(p)£ i s any r e a l sequence such t h a t a ( p ) ^ a . By 22 110 E a ( p ) ^ E f o r a l l such sequences, hence E = l i m (strong) (b-> a+) E^ Obviously E > E . — a I f then P = E-E f t J 0, 0 < P < I - E o — a and 0 < p < \ f o r b > a. Hence by 7 4 amP < nP and bmP 2. nP f o r b > a, so th a t amP 2 nP a l s o . T h i s i s a c o n t r a d i c t i o n . Hence P = E-E o = 0 and 77 i s proven. I f ^ E a : - oo < a < eo j i s a r e s o l u t i o n of the i d e n t i t y then l i m (strong) ( a - * - c o ) E = 0 . ct As t h i n g s now stand, t h i s i s not t r u e : l e t E e G(a) f o r a < 0. Then amE > nE I l l or 0 amE > nE. _> 0 whence amE = nE = 0 , or, s i n c e a < 0, mE == nE =* 0. Thus hy 77 l i m (strong) (a-> -®) E = P where F i s a p r o j e c t i o n of m-measure zero. The troublesome p o s s i b i l i t y that Fj^b can be e l i m i n a t e d by d e f i n i n g E = 0 f o r a < 0. This preserves the r i g h t cont- . " 1 liri" 111 ct i n u i t y and monotonicity of Ea. So now E i s a r e s o l u t i o n of ct 1 11 ct the i d e n t i t y . N 78. D e f i n i t i o n . N i s the unique p o s i t i v e R -operator which has E as a s p e c t r a l r e s o l u t i o n , a 79. Theorem. I f E € R P and mE < » 3 then nE = r a d(mE E &) Jo a, Proof. Since mEE„ i s a f i n i t e monotone f u n c t i o n of •— a s(Q) = / a d(mEE a) JQ e x i s t s and i s f i n i t e f o r a l l i n t e g e r s Q > 0. Moreover, since 112 CO oo s(Q) 1S a d(mEE ) a '0 ^0 e x i s t s , and, given such that 0 there exists a Q sueh that CD _ 0 P a r t i t i o n the i n t e r v a l [0,Q] 0 = a Q < a^ < ... < a p =» Q. Let E ( i ) = E - E a i a i - l and form the corresponding upper and lower sums: P U = Y a i m E E ( l ) i = l P L = Y, a i - l m E E ( i ) i = l Since s(Q) = sup [lj = i n f £uj , the p a r t i t i o n can he made so that oo 077 < L _< S(Q) <u < y 113 Now a ^ mEE(i) < nEE(i) _< a,± mEE(I) Thus D?y< L = a i _ i m E E ( i ) < Y1 nEE(i) = n E ( E Q - E p) - n E E Q (sin c e nE Q = 0) _< ^ a ± mEE(i) CO That i s , given Q/J'< j there e x i s t s Q such that CO 07/ < nEE Q. < Since hy 39 so t h a t E E Q f E, n E E Q t nE, 00 nE = / a d(nEE a) T h i s proves 79. 114 To get on w i t h the true business, r e c a l l that N i s the p o s i t i v e R -operator w i t h s p e c t r a l r e s o l u t i o n E o ( d e f i n t i o n 78), Let E(a,b) = E b - E & f o r a < b. Then N(a,b) = NE(a,b) e R S N and f o r any T e R T(a,b) = T • E(a,b) = TE(a,b) e R N by 36. 80. Lemma• I f T i s a p o s i t i v e R^-operator, then f o r a < b amT(a,b) _< mT o H(a i )b) _< bmT(a,b) . p Proof. Let T have the r e s o l u t i o n F i n R , then b — c 3 • T = TF„ c c S i s a p o s i t i v e R -operator f o r a l l c. Hence T 0 N(a,b) = T N(a,b) = N(a,b)T € R S. I f now x € H and c i s f i x e d ( T c - N ( a j b ) X j X ) = ( 1/2 1/2 H ( a } X ) X ) 1/? S q (T ' e x i s t s i n R , since T i s p o s t i v e i n R ) ( T c l / 2 H ( a > b ) x, T / ^ x ) (N(a,b) T c 1 / 2 x , T ^ x ) Now Thus i n R S. a(E(a,b) T ^ x , TV^) < (N(a,h) T c l / 2 x , T c l / 2 x ) < b(E(a,b). T_ l / 2x, T l / 2 x ) aT c(a,h) < T c . N(a,b) < b' T c ( a , b ) I f a < b < 0, then a l l the above terms are zero, I f a _< 0 < b, then 0 < T • N(a,b) < b T„(a,b) I n R S. Hence by 53 mTc o N(a,b) < mbT c(a,b) = bmT c(a 5b) I f 0 < a < b 0 < a T„(a,b) < T 0 N(a 5b) < T„(a,b) S i n R and again by 5~}3 maT c(a,b) _< mTc o N(a,b) < mbT c(a,b) Hence, y -. • • ;-_ ,•  ,- ,. . \vi,Lcy.\:-\r ••- amT (a,b) < mT • N(a,b) <. bmT (a,b). holds f o r a l l c and a < b. I n p a r t i c u l a r t h i s i n e q u a l i t y holds I n the l i m i t as c-*•+=>. Thus by 62, since T = TP , amT (a,b) _< mT o N(a,b) _< bmT(a,b) . This proves 80. 8l . Lemma. I f T i s a p o s i t i v e R^-operator, then f o r a < b amT(a,b) _< nT(a,b) < bmT(a,b) Proof. Let T he as i n the preceeding lemma. F i x the number c. Wow there e x i s t s a sequence -^Sp^ G R S P such that S p > ° Sp * T c(a,b) and by d e f i n i t i o n nSn.f mT (a,b) nS„ f nT (a,b) Let q 'P l i = l S- - I a i G i be an a r b i t r a r y member of t h i s sequence. Then S p .« S p E(a,to) and bmSp == b m E(a,b) ^  G ± . =» b m aj[_ G i E(a,b) •a b a i raGi E(a,b) — a i n G i E ( a > b ) - n s p > a £\ a^ mG^  E(a,b) a m S P 1 1 7 so that amS„ < nS„ < bmS^ . P P — P Passing to the l i m i t i n p, a m T (a,b) < n T„(a,b) < h mT fa,b) f o r a l l c and a < b. Employing 62 again and l e t t i n g e->+» a m T(a,b) < n T(a,b) _< h m T(a,b) f o r a < b. This proves 8 l . n N 82. Lemma. I f T i s a p o s i t i v e R operator, then mT «. N(a,b) = nT(a,b) f o r a < b. Proof. Let P denote a p a r t i t i o n of the i n t e r v a l [a,b] P : a => a. <a, < o l <a. p - 1 = b Por iep l e t E ± =»E(a i, a i + 1 ) N, = W(a. Prom 80 and 8l f o r a l l i e p . 118 L(P) = V a, mTEj T.ep Examining upper and lower sums, i± TEi ^ m T * U± •» m T • N(a,b) ^ n T E ± = n T(a,b) < \ a i + i m T E i = u<p)- Now i f m T(a,b) < «, then U(P) - L(P) = ( a ± + 1 - a ±) m TE± _< (max P) J\ m T E 1 (where max P =3 max " a i : ^ e p J ) = (max P) m T(a,b) Thus (max P) mT(a,b) > U(P) ~ L(P) > jmT • N(a,b) - nT(a,b)| I f mT(a,b) = 0, the lemma i s proven. I f mT(a,b) > 0, then, given e > 0, there ex i s t s a p a r t i t i o n P such that max P < e/mT(a,b) so that |mT • N(a,b) - nT(a,b)| < € 119 f o r a r b i t r a r y e > 0. Hence mT o N(a,b) = nT(a,b) In the case that mT(a,b) = +», the assertion i s obvious from 80 and 8 l . This proves 82. 83. Theorem. I f T i s a p o s i t i v e R^-operator mT. • N = nT Proof. Prom 82, mT • N(0,a) = nT(0,a) or m(T o N)E n ,= nT • E„ cl cl By 62, l e t t i n g a->+ «>, 83 i s proven. 84. Theorem. I f T e R K i s n-integrable, then T • N i s m- integrable and mT • "N = nT Proof. I f T i s n-integrable, then nT = nT + - nT" where at l e a s t one of nT + = mT+ 0 ET nT" = mT" 0 N (by 83) Is f i n i t e . , 120 How (T • H) s= ( T + - T") • N = ( T + • N ) - (T" > N) where one of ( T + • N), '(T" o H) i s m-summable. Therefore, hy 65 , T o N i s m-integrable and ; m(T •• N) =a m(T + • N ) - m(T" N) = nT + - nT" = nT This proves 84. Theorem 84 i s the promised Radon-Nikodym theorem. 

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