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COMPLEX MEASURES Absolute Continuity and Representation Theorems

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COMPLEX MEASURES Absolute Continuity and Representation Theorems COMPLEX MEASURES Absolute Continuity and Representation Theorems Lakhdar Meziani Introduction Let (X; ; ) be a measure space and let f : X C be a complex - integrable function.F Let us consider the set function !given by: (A) = f d; A ( ) A 2 F Such set function has the followingR properties: (1) ( additivity): For any sequence (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 bounded variation: 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 bijection) : N N, the series z(n) converges to M. For real ! n or complex numbers series unconditional convergenceP is equivalent to absolute convergence 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. Total variation 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 ofj 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 real number 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 vector space on C, and is a norm on M (X; ) Moreover M F(X; ), endowed with the normk k , is a Banach space.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.
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