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MEASURE THEORY LECTURE NOTES JANUSZ ADAMUS XI. Complex Measures Definition 1. Let M be a σ-algebra on a set X.A complex measure on M is a complex-valued function µ : M! C satisfying [ X (1) µ( En) = µ(En) ; n n for every denumerable collection fEngn of pairwise disjoint elements of M. Remark 2. 1. It follows from the definition that µ(?) = 0. Indeed, we have ? = ? t ? and hence µ(?) = 2µ(?), which in light of µ(?) 2 C implies µ(?) = 0. 2. Since the left side of (1) is a complex number, it follows that the series on the right side of (1) is always convergent. Moreover, the sum of the series is independent of the ordering of the sets En, and hence the series is absulutely convergent. Example 3. The following is a model example of a complex measure. By the Radon-Nikodym Theorem below, all complex measures arise in this way modulo a set of measure zero. Let (X; M; µ) be a σ-finite measure space, and let h : X ! C be an integrable function. (Recall that a complex-valued function f = u + iv, where u and v are real-valued, is called integrable when both u and v are integrable as real-valued functions. In this case, one defines R f := R u + i R v.) Then, the function ν : M! C defined as Z (2) ν(A) := h dµ A is a complex measure. The proof is an elementary exercise. Definition 4. Let µ : M! C be a complex measure on a σ-algebra M. The function jµj : M! R defined as 1 1 X [ jµj(E) := supf jµ(En)j : fEngn ⊂ M;E = En;Ej \ Ek = ? for j 6= kg n=1 n=1 is called the total variation measure of µ. Remark 5. Observe that we always have jµj(A) ≥ jµ(A)j, for every A 2 M. Indeed, the family fA; ?; ?;:::g forms a measurable partition of A, and hence 1 X jµj(A) ≥ jµ(A)j + jµ(?)j = jµ(A)j : n=2 1 2 JANUSZ ADAMUS Exercise 6. For a signed measure µ on a σ-algebra M, one defines the total variation measure of µ as jµj := µ+ + µ−, where µ+ and µ− are the unique positive measures from the Jordan Decomposition Theorem for µ. Note that a real-valued (i.e., finite) signed measure µ may be regarded as a complex measure. Prove that if µ is a real-valued signed measure, then the two definitions of jµj coincide. Theorem 7. The total variation jµj of a complex measure µ on a σ-algebra M is a positive measure on M. 1 Proof. Let fEngn=1 ⊂ M be an arbitrary collection of pairwise disjoint measurable S sets. Set E := n En. We want to show that 1 X jµj(E) = jµj(En) : n=1 To this end, for every n 2 Z+, choose an arbitrary real number tn satisfying jµj(En) > tn. Then, by definition of jµj, there exists for every n a measurable 1 P1 partition fAnkgk=1 of En such that k=1 jµ(Ank)j ≥ tn. Since the (countable) 1 family fAnkgn;k=1 forms a partition of E, we get that 1 1 1 ! 1 X X X X jµj(E) ≥ jµ(Ank)j = jµ(Ank)j ≥ tn : n;k=1 n=1 k=1 n=1 (Note that above we used the absolute convergence of the series in question to rearrange its terms without altering the sum.) 1 Taking supremum over all sequences (tn)n=1 as above, we get 1 X jµj(E) ≥ jµj(En) : n=1 1 For the proof of the opposite inequality, let fAkgk=1 ⊂ M be another arbitrary 1 partition of E. Then, for every k, fAk \ Engn=1 is a measurable partition of Ak, 1 and for every n, fAk \ Engk=1 is a measurable partition of En. Thus, again by absolute convergence, we get 1 1 1 1 1 ! X X X X X jµ(Ak)j = µ(Ak \ En) ≤ jµ(Ak \ En)j k=1 k=1 n=1 k=1 n=1 1 1 ! 1 X X X = jµ(Ak \ En)j ≤ jµj(En) : n=1 k=1 n=1 Taking supremum over all countable measurable partitions fAkgk of E, we get P1 jµj(E) ≤ n=1 jµj(En). This proves countable additivity of jµj. The equality jµj(?) = 0 follows from µ(?) = 0 and the fact that the only countable measurable partition of ? consists of copies of ?. Perhaps even more interestingly, the total variation of any complex measure is a finite positive measure, as the following theorem shows. Theorem 8. If µ is a complex measure on a measurable space (X; M), then the total variation measure jµj satisfies jµj(X) < +1. We shall first establish an auxiliary lemma, which is a somewhat surprising complex analysis result of independent interest. MATH4122B/9022B - INTRODUCTION TO MEASURE THEORY 3 Lemma 9. Given any collection fz1; : : : ; zN g of (not necessarily pairwise distinct) N complex numbers, there exists an index subset S ⊂ f1;:::;Ng such that N X 1 X z ≥ · jz j : n π n n2S n=1 iαn Proof. For 1 ≤ n ≤ N, let αn 2 (−π; π] be such that zn = jznje . For # 2 [−π; π], let S(#) := fn 2 f1;:::;Ng : cos(αn − #) > 0g : Then, 0 1 X −i# X X −i# X i(αn−#) j znj = je j · j znj = j e znj ≥ Re @ jznje A n2S(#) n2S(#) n2S(#) n2S(#) N X X + = jznj cos(αn − #) = jznj cos (αn − #) ; n2S(#) n=1 where, as usual, cos+ = maxfcos; 0g. Now, choose #0 2 [−π; π] so as to maximize the latter sum, and set S := S(#0). By the Mean Value Theorem for Riemann integral, we have N N ! X 1 Z π X jz j cos+(α − # ) ≥ jz j cos+(α − #) d# ; n n 0 2π n n n=1 −π n=1 and hence N ! X 1 Z π X j z j ≥ jz j cos+(α − #) d# n 2π n n n2S −π n=1 N N N X 1 Z π X 1 1 X = jz j · cos+(α − #) d# = (jz j · ) = · jz j ; n 2π n n π π n n=1 −π n=1 n=1 where the penultimate equality follows from the fact that cos+ is periodic with R π + period 2π, and hence −π cos (αn − #) d# = 2 independently of the choice of αn. Proof of Theorem 8. Suppose first that, for some E 2 M, we have jµj(E) = +1. Set t := π(1 + jµ(E)j). (Of course, t < +1, since µ is a complex measure.) Since jµj(E) = +1, then by definition of jµj there exist a measurable partition 1 fEngn=1 of E and a positive integer N such that N X jµ(En)j > t : n=1 Applying Lemma 9 with zn := µ(En), we conclude that there is a measurable set A ⊂ E (namely, the union of some of the E1;:::;EN ) such that t jµ(A)j > ≥ 1 : π Moreover, for B := E n A, we also have t jµ(B)j = jµ(E n A)j = jµ(E) − µ(A)j ≥ jµ(A)j − jµ(E)j > − jµ(E)j = 1 : π 4 JANUSZ ADAMUS We thus have E = A t B, with jµ(A)j > 1 and jµ(B)j > 1. By additivity of jµj, at least one of the A; B must be of infinite jµj-measure. 1 Now, if jµj(X) = +1, we construct recursively an infinite sequence (An)n=1 ⊂ M of pairwise disjoint sets with jµ(An)j > 1 for all n, as follows: By the first part of the proof, we can partition X into A1 and B1 such that jµ(A1)j > 1 and jµj(B1) = +1. Having defined A1;:::;Ak and B1;:::;Bk, partition Bk into Ak+1 and Bk+1 such that jµ(Ak+1)j > 1 and jµj(Bk+1) = +1. S1 P1 Then, by additivity of µ, we have µ( n=1 An) = n=1 µ(An). In particular, the P1 series n=1 µ(An) is absolutely convergent. Hence, by the Basic Divergence Test, lim jµ(An)j = 0, contradicting the choice of the An. n!1 Exercise 10. (a) Show that if µ, ν are complex measures on a σ-algebra M, then so are µ + ν and c · µ for any c 2 C (where (µ + ν)(E) = µ(E) + ν(E) and (cµ)(E) = c · µ(E) for E 2 M). Therefore the set of all complex measures on M forms a complex vector space. (b) Prove that the function kµk := jµj(X) defines a norm on that vector space. Absolute Continuity. Definition 11. Let µ be a positive measure on a σ-algebra M, and let λ be an arbitrary measure on M. We say that λ is absolutely continuous with respect to µ, and write λ µ, when λ(E) = 0 for all E 2 M with µ(E) = 0. For the following proposition, recall that a (positive, signed, or complex) measure µ is said to be concentrated on a measurable set A, when µ(E) = 0 for every measurable E ⊂ Ac. Proposition 12. Suppose λ, λ1; λ2 are arbitrary measures on a σ-algebra M, and µ is a positive measure on M. Then: (i) If λ is concentrated on A 2 M, then so is jλj. (ii) If λ1?λ2, then jλ1j?jλ2j. (iii) If λ1?µ and λ2?µ, then (λ1 + λ2)?µ. (iv) If λ1 µ and λ2 µ, then (λ1 + λ2) µ.
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