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 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 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. 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 measurable function 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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