COMPLEX MEASURES and Representation Theorems

Lakhdar Meziani

Introduction Let (X; ; ) be a space and let f : X C be a complex - integrable .F Let us consider the function  !given by:

 (A) = f d; A ( ) A 2 F  Such set function has the followingR properties:

(1) ( additivity): For any (An) of pairwise disjoint sets An in F we have  An =  (An) [n n (2) (absolute  continuity):P Let A with  (A) = 0 then  (A) = 0, because 2 F f:IA = 0  a:e, we say that  is absolutely continuous with respect to . This relation will be denoted by    + (3). If f is real valued let us write f = f f then +  (A) = A f d = A f d A f d so we have  (A) = 1 (A) 2 (A), with + 1 (A) = f d and 2 (A) = f d positive and  additive. R A R R A In this chapter we consider complex valued  additive set functions R R  : C and we will show successively: (a)F !is of : more precisely there is a positive …nite measure  on such that  (E)  (E), E j j F  is called the total variationj ofj:  j j 8 2 F j j + (b) if  is real valued then it can be written as  =   + where  ;  are …nite positive measures. This is called the Jordan decomposition. (c)  has the integral form ( ) for some complex -integrable function f provided   for some  …nite positive measure : This is the Radon-Nicodym Theorem.

1. Complex Measures property

De…nition 1.1 Let (X; ) be a measurable space and  : C a complex set function. F F ! We say that  is a complex measure if for every sequence (An) of pairwise disjoint sets in we have  An =  (An) : F [n n Remark (1) Let zn = M be a convergent P series of real or complex numbers. n We say that the seriesP is unconditionally convergent to M if for any permuta- tion (i.e )  : N N, the series z(n) converges to M. For real ! n or complex numbers series unconditional convergenceP is equivalent to by Riemann series theorem.

1 (2) Let  : N N be a permutation of N then An = A(n) = A and the ! [n [n sets A(n) are pairwise disjoint so  (A) =  (An) =  A(n) , since  n n is arbitrary  this implies that the series  (APn) is unconditionallyP convergent n and then absolutely convergent. P (3) If  is a complex measure on (X; ) then one can write  = Re ()+i Im (), where it is easy to see that Re () andF Im () are real  additive set functions on (X; ). This simple observation leads to the following de…nition: F De…nition 1.2 Let (X; ) be a measurable space and  : R a real set function. F F ! We say that  is a real measure if for every sequence (An) of pairwise disjoint sets in we have  An =  (An) : F [n n Let  : N N be a permutation P of N then An = A(n) = A and the ! [n [n sets A(n) are pairwise disjoint so  (A) =  (An) =  A(n) , since  n n is arbitrary  this implies that the series  (APn) is unconditionallyP convergent n and then absolutely convergent (see RemarkP (1))

2. of a complex measure

Let  be a complex measure on (X; ) Among all positive measures  on (X;F ) satisfying  (E)  (E), E , there one andF only one called the Total variation jof  andj  given by8 the2 following F theorem: Theorem 2.1 If  is a complex measure on (X; ) let us de…ne the positive set function  : [0F; ] by: j j F ! 1 E ,  (E) = sup  (En) ; (En) partition of E in 2 F j j j j F  n  the supremum being takenP over all partitions (En) of E in : Then:  is a positive bounded measure on (X; ) satisfying F j j  (E)  (E), E F moreover  is thej smallestj  j j positive8 bounded2 F measure with this property. Proof. Letj jE be a set in F If (En) is a sequence of pairwise disjoint sets in we have to prove that: F  En =  (En) j j [n n j j let us put E= PEn and take a partition (Am) of E in then we have: [n F (Am En)n 1 is a partition of Am \  (Am En)m 1 is a partition of En \  so  (Am) =  (Am En)  (Am En) ; m 1 and then j j \  j \ j 8  n n

 (Am) P  (Am E n) P=  (Am En) m j j  m n j \ j n m j \ j P PP PP

2 but we have from the de…nition of   (Am En)  (En) n 1 j j m j \ j  j j 8  therefore we deduce that  (Am) P  (En) inequality valid for every m j j  n j j P P partition(Am) of E and implies  (E)  (En) by the de…nition of  (E) : j j  n j j j j P It remains to prove that  (En)  (E), to do this we use characteristic n j j  j j property of the supremum: for each n 1 let an > 0 be any P  such that an <  (En), then from the de…nition of  (En) there is a partition j j j j (Amn)m 1 of En with an <  (Amn) , but we have E = : :En = :Amn  m j j [n mn[ and an <  (Amn) .P Since (Amn)m;n 1 is a partition of E we deduce n n m j j  that P  (PAmnP )  (E) and so an <  (E), but this is true for all n m j j  j j n j j an > P0 satisfyingP an <  (En) thisP implies that  (En)  (E) : The n n j j n j j  j j proof of the boundednessP P of  is left to the reader. P j j  Theorem 2.2 If  is a complex measure on (X; ) F For any increasing or decreasing sequence (An) in we have F  limAn = lim (An) n n wherelimAnstands for An in the increasing case n [n and for An in the decreasing one. \n Proof. use the  additivity of  and the fact that  (E) < , E . j j 1 8 2 F  Theorem 2.3 Let M (X; ) be the family of all complex measures on (X; ) F F let ;  be in M (X; ) and let C, then de…ne:  + ; :; F , by the following2 recipe ( + )(E) =  (E)k +k (E), ( :)(E) = : (E) , E  =  (X) 2 F k k j j Then M (X; ) is a on C, and  is a on M (X; ) Moreover M F(X; ), endowed with the normk k , is a .F F k k Proof. see any standard book on measure theory for a classical proof.e.g.[R F ] 

3 3. Hahn-Jordan Decomposition of a Real Measure

Theorem 3.1 Jordan Decomposition of a Real Measure Let (X; ) be a measurable space and  : R a real measure (De…nition 1.2) F F ! + Let us de…ne the set functions  ;  as follows: for each E + (E) = sup  (F ): F ;F E 2 F f 2 F  g  (E) = inf  (F ): F ;F E + f 2 F  g Then  ;  are positive bounded measures on (X; ) satisfying: F + 1 1 (i)  = (  + )(ii)  = (  ) 2 j j 2 j j + (iii)  =   (Jordan Decomposition) + (iv)  =  +  Proof j j First it is enough to prove that + is a positive bounded measure on (X; ) + F because  = ( ) : From the de…nition of the total variation of  we can see that for E F ;F E::::: (F )  (F )  (F )  (E)  (X) <2 F so we8 deduce2 F that + is positive j bounded;j  j j  j j  j j 1 it remains to prove that + is  additive Let (En) be pairwise disjoint sets in ; we have to prove: + + F  En =  (En) [n n + + Since n 1  P(En) < , we have from the de…nition of  : 8  1 +  let  > 0 then for each n 1, Fn En such that  (En) <  (Fn)  9  2n + summing over n we get  (En)  <  (Fn) =  Fn n n [n   P + P + + but Fn En =  Fn  En ; therefore  (En)  <  En [n  [n ) [n  [n n [n    +  +   since  > 0 is arbitrary, we obtain  (En)  EPn after making  0: n  [n !   On the other hand let F En:::FP , then F = (En F ) and  [n 2 F [n \ +  (F ) =  (En F )  (En) because En F En n \  n \  P P + + since F En::is arbitrary in , we deduce that  En  (En)  [n F [n  n so + is a positive measure on (X; ) :   P F We have (i) = (ii) (iii) and (iv) = (ii) + (iii) + it is enough to prove (i) because (ii) follows with  = ( ) then we get (iii) with (ii) (i) and (iv) with (ii) + (iii) : 1 So let us prove (i) that is + = (  + ): 2 j j …x  > 0 then from the de…nition of the total variation  j j there is a partition (En) of E in such that ( )  (E)  <  (En) : F  j j n j j P

4 Let us put L = l :  (El) 0 and K = k :  (Ek) < 0 , we get: f  g f g  (En) =  (El)  (Ek) n j j L K nowP de…ne F = PEl and GP= Ek, so by the  additivity of  we deduce [L K[  (F ) =  (El) and  (G) =  (Ek), then  (En) =  (F )  (G) L K n j j therefore weP obtain from the inequalityP ( ) that P  (2 )  (E)  <  (En) =  (F )  (G)  j j n j j but since E = F G weP have (3 )  (E) = [(F ) +  (G) adding (2 ) + (3 ) we get:    (E) +  (E)  < 2 (F ) thej de…nitionj of +implies  (F ) + (E), because F E 1   …nally (  (E) +  (E) ) < + (E) with nothing depending on  apart ; 2 j j 1 making  0 we get (  (E) +  (E)) + (E) : ! 2 j j  This inequality cannot be strict: 1 indeed suppose that we have (  (E) +  (E)) < + (E), this would imply 2 j j 1 the existence of an F in with F E and (  (E) +  (E)) <  (F ) < + (E) F  2 j j 1 but the set function (  (E) +  (E)) is a positive measure on (X; ) ; 2 j j F 1 1 since F E we should have (  (F ) +  (F )) (  (E) +  (E))  2 j j  2 j j 1 1 since  (F )  (F ), we have  (F ) = ( (F ) +  (F )) (  (F ) +  (F ))  j j 2  2 j j  1 (  (E) +  (E)) <  (F ) which is absurd 2 j j 1 so we deduce that (  (E) +  (E)) = + (E) : 2 j j  Remark. The Jordan decomposition of a real measure  as di¤erence of two positive + bounded measures  =   is not unique, since for any positive bounded + measure  one can write  = ( + ) ( + ). However such decomposition is minimal in the sense that if  =   + with ;  positive bounded measures then   and  ; to see this use  +  the facts   and ( )  in the de…nition of  ; :   Theorem 3.2 The Hahn Decomposition Let  : R be a real measure on the measurable space (X; ). There existsF ! a partition of X in two sets A; B in such that: F F (1)  (F ) 0 for every F A and  (A) = 0 (2)  (F )  0 for every F  B and + (B) = 0 The partitionX = A B satisfying (1) ; (2) is unique in the following sense: if C;D is another partition[ of X satisfying (1) ; (2) then  (AC) = 0 and  (BD) = 0, (see [R F ] for the Proof). j j j j

5 5. Absolute Continuity of Measures Radon-Nikodym Theorem

De…nition 5.1 Let  be a complex measure on (X; ) and  a positive measure. We say that  is absolutely continuousF with respect to  if: for E satisfying  (E) = 0 =  (E) = 0 notation:2 F   ) Example 5.2  Let f L1 (), then the complex measure  on (X; ) given by: 2 F E ,  (E) = f:d ( ) 2 F  EZ is absolutely continuous with respect to , indeed suppose  (E) = 0 then .f.IE = 0  a:e and f:d = f:IE:d = 0 =  absolutely continuous ) EZ XZ with respect to :(by the property of the integral) This example is fundamental in the following sense: If  is  …nite then the complex measure in the integral form ( ) is the only one which is absolutely continuous with respect to   this is due to the Radon-Nikodym Theorem. First let us look at some properties of the absolute continuity. Theorem 5.3 Let  be a complex measure on (X; ) and  a positive measure. The following properties are equivalent:F (a)    (b) For any  > 0 there is  =  > 0 such that for A  (A) <  =  (A) <  2 F Proof. ) j j (b) = (a) if  (A)) = 0 then  (A) <   > 0, so  (A) <   > 0 that is  (A) = 0 8 j j 8 j j (a) = (b) we prove) that not (b) = not (a) suppose not (b) then there) is  > 0 such that for each n 1 1  there exists En with  (En) < and  (En)  2 F 2n j j  put E = lim supEn = : :Ek, then since  is bounded we get: n \n k[n j j 

 (E) = lim  :Ek lim  (En)  > 0: j j n j j k[n  n j j     1 On the other hand since we have  (En) < n < , we can apply n n 2 1 P P the Borel-Cantelli Lemma to the sequence En to get  (E) =  lim supEn = 0 n so not (a) is satis…ed and not (b) = not (a) is proved.   ) 

6 De…nition 5.4 (Singular Measures) Let ;  be positive measures on (X; ). We say that ;  are singular if thereF is a partition A; B of X such that  (A) =  (B) = 0 f g notation:   If ;  are complex? measures then they are singular if   j j ? j j Remark. Suppose ;  positive or complex measures and   if A; B is the partition of X de…ning the singularity? then we have ffor Eg  (E) =  (E B) and  (E) =  (E A) 2 F \ \ Proposition 5.5 Let ; 1; 2; be measures on (X; ) with  positive and 1; 2; complex then F (a) 1  and 2  = 1 + 2  ? ? ) ? (b) 1  and 2  = 1 + 2    )  (c) 1  = 1   ) j j  (d).1  and 2  = 1 2  ? ) ? (e).1  and 1  = 1 = 0 Proof  ? ) (b) and (c) are deduced from de…nition, (e) is a consequence of (d) so let us prove (a) and (d) : To see (a) let A1;B1 ; A2;B2 be two partitions of X with f g f g 1 (A1) =  (B1) = 0 and 2 (A2) =  (B2) = 0 j j j j put A = A1 A2 and B = B1 B2, then A; B is a partition of X and we have \ [ f g 1 + 2 (A) ( 1 + 2 )(A) = 0 and  (B)  (B1) +  (B2) = 0 j j  j j j j  therefore 1 + 2  this proves (a) : ? To prove (d): since 2  let A; B be a partition of X with 2 (A) =  (B) = 0 ? f g j j but we have also 1  so we will have 1 (B) = 0 by (c), …nally we get  j j 2 (A) = 1 (B) = 0 j j j j that is 1 2, then we deduce (d) : ?  Theorem 5.6 Radon-Nikodym-Lebesgue Theorem. Let ; ; be measures on (X; ) with  positive  …nite and  complex then: F (1) There is a unique pair of complex measures a; s such that a  , s  and a s  ? ? moreover  = a + s Lebesgue Decomposition (2) There is a unique function h L1 () such that 2 a (E) = h:d for every E :Radon-Nikodym Theorem 2 F EZ Proof Let us point out: 0 0 (i) uniqueness in (1) of a; s: if a; s is an other pair of complex measures 0 0 0 0 with a  , s  and  = a + s = a + s then we have 0  0? 0  a = s  ; we deduce from Proposition 5.5 (a) that s   a s s ?

7 0 and from Proposition 5.5 (b) that a a  0 0 so part (e) of the same Proposition implies a a = s s = 0, whence the uniqueness of the decomposition. 0 (ii) uniqueness in (2) of the function h: if h L1 () is an other function with 2 0 0 a (E) = h :d for every E then h:d = h :d, E , and so by 2 F 8 2 F EZ EZ EZ the property of the integral we get h = h0  a:e: We prove the theorem when ;  are positive bounded measures the proof in this case is due to John Von Neuman. (for the general case see reference [R F ]). Let ;  be positive bounded measures on(X; ) and put m =  +  then m is a positive bounded measure on (X; F) and we have F f:dm = f:d + f:d

XZ XZ XZ for any measurable positive function f on X this can be checked easily for simple positive functions by using Beppo-Levy convergence theorem one can prove it for measurable positive functions.

Observe that L1 (m) = L1 () L1 () because f :dm = f :d+ f :d \ j j j j j j XZ XZ XZ Now take f in the Hilbert space L2 (m) then we have by the Schwarz inequality 1 2 1 f:d f :d f :dm f 2 :dm : (m (X))2  j j  j j  0 j j 1 Z Z Z Z X X X X @ A consequently the linear functional f f:d is continuous on the Hilbert ! XZ Space L2 (m). Therefore there is a unique function g in L2 (m) such that :

( ) f:d = f:g:dm, by the isomorphism between L2 (m) and its strong dual  XZ XZ take f = IE in equation ( ) to get  ( )  (E) = g:dm, then since 0  (E) m (E) E    8 2 F EZ we obtain 0 g 1 m a:e, but m =  + , from which ( ) gives    ( ) f: (1 g) :d = f:g:d:    EZ EZ Now put A = 0 g < 1 , B = g = 1 and de…ne measures a; s as follows f  g f g a (E) =  (A E) and s (E) =  (E B) E \ \ 8 2 F putting E = X, f = IB in the relation ( ) gives   

8 (1 g) :d = g:d =  (B) BZ BZ since g = 1 on the set B we have (1 g) :d = 0, so  (B) = 0 BZ but s (E) = 0 for every E A, therefore s :  ? 2 n Again consider ( ) but with f = 1 + g + g + ::: + g :IE, we get:    1 gn+1 :d = 1 gn+1 :d + 1 gn+1 :d = 1 gn+1 :d EZ EZA EZB EZA since g = 1 on the set\ B.  \  \  On the other hand we have gn+1 (x) 0 x E A, since A = 0 g < 1 so 1 gn+1 (x) 1 x E A and by# Beppo-Levy8 2 \ convergencef  theoremg we get: " 8 2 \

n+1 (4 ) lim 1 g :d =  (E A) = a (E)  n \ EZA \  g on A but we have g: 1 + g + g2 + ::: + gn h = 1 g " ( on B ) since  (B) = 0, we deduce that  1

(5 ) g: 1 + g + g2 + ::: + gn :d h:d  " Z Z E  E now properties (4 ) and (5 ) jointly imply   a (E) = h:d and h L1 () 2 EZ which ends the proof of the Theorem. the function h is called the Radon- d Nikodym density of  with respect to  and is denoted by h = : d  The following theorem is a version of the preceding one in the case ;  positive  …nite measures, the proof can be found in [R F ] : Theorem 5.7 Let ; ; be positive  …nite measures on (X; ) F (1) There is a unique pair of positive measures a; s such that a  , s  and a s  ? ? moreover  = a + s Lebesgue Decomposition (2) There is a unique positive function h locally  integrable such that a (E) = h:d for every E :Radon-Nikodym Theorem 2 F EZ h locally  integrable means that there is a partition (Xn) of X in such that F h:d < n: 1 8 XZn

9 6. Applications We recall that the structures given in Theorem 5.6 are valid on (X; ) with  positive  …nite and  complex although its proof has been given forF;  posi- tive bounded. So hereafter we consider the general context of complex measures. Let us start by the relation of a complex mesure and its total variation. Proposition 6.1 Let  be a complex measure on (X; ) then F there exists a h : X C such that ! h (x) = 1 for every x X and  (E) = h:d  E j j 2 j j 8 2 F EZ where  is the total variation of  (see Theorem.2.1 for total variation of  ) j j Proof Observe that   and use the Radon-Nikodym Theorem.  j j  Proposition 6.2 Let (X; ; ) be a measure space and let f be in L1 (). F Consider the complex measure  on (X; ) given by  (E) = f:d F EZ then we have  (E) = f :d E : j j j j 8 2 F EZ Proof Let us consider the set measure  (E) = f :d on (X; ) then we have: j j F EZ

 (E) = f:d f :d =  (E) =  (E)  (E) j j  j j ) j j  Z Z E E but we now that the total variation  is the smallest positive measure satisfying  ( E)  by theorem 2.1j j, so we deduce that   and therefore j j  j j j j   : Since  is bounded by the Radon-Nikodym theorem there is ' L1 () j j  2 with ' positive and  (E) = ':d: The integral form of  allows to write j j EZ  (E) = ': f :d: By Proposition 6.1 there exists a measurable function h : j j j j EZ

X C such that h (x) = 1 for every x X and  (E) = h:d  E . ! j j 2 j j 8 2 F EZ We deduce from the integration process that  (E) = h:d  = h:': f :d. j j j j EZ EZ By hypothesis  (E) = f:d so we get h:': f :d = f:d; E and j j 8 2 F EZ EZ EZ

10 then h:': f = f  a:e: But h (x) = 1 for every x X so ' = 1  a:e j j j j 2 because ' 0, …nally  (E) = ':d =  (E) = f :d:  j j j j  EZ EZ Proposition 6.3 Let  be a  …nite measure on (X; ) and let us denote by F  the family of all complex measures absolutely continuous with respect to  A Then  is a closed subspace of the Banach space M (X; ). Moreover A F there is a linear isometry from L1 () onto : A Proof Recall the Banach space M (X; ) of all complex measures on (X; ) with the norm  =  (X) de…nedF in Theorem 2.3. F k k j j It is easy to check that  is a subspace of M (X; ). We prove that it is closed: A F let (n) be a sequence in  converging to  A that is lim n  = lim n  (X) = 0: This implies that (n (E)) n k k n j j converges to  (E) even uniformly with respect to E. If we have  (E) = 0 then n (E) = 0 n, so  (E) = 0 that is   8 2 A this shows that  is closed. A Now we de…ne the linear isometry from L1 () onto  as follows: A for f L1 () put (f) = , where  is the complex measure on (X; ) given by 2 F  (E) = f:d. Then it is clear that is linear, moreover it is invertible,

EZ indeed if   then   and since  is  …nite there is a unique f L1 () such that2 A  2  (E) = f:d = (f), by the Radon-Nikodym Theorem. On the other hand

EZ is an isometry since we have (f) =  =  (X) = f :d = f L () : k k k k j j j j k k 1 XZ

References 1. Royden-Fitzpatrick. Real Analysis Fourth Edition Prentice Hall. 2. W. Rudin Real and Complex Analysis Mc GRAW-HILL Third Edition .

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