Real ad Complex Aalysis, 3rd ditio, W.Rudi Chapter 6 Complex Measures Yug-Hsiag Huag 206/08/22. Let ν be a complex measure o (X, M ). If M, defie { } µ () = sup ν( j ) : N,, 2, disjoit, = j { } ν () = sup ν( j ) :, 2, disjoit, = j { µ 3 () = sup Show that µ = ν = µ 3. } f dν : f Remark. We exted the problem slightly, which comes from [3, xercise 3.2]. To prove the origial problem, we ca simplify the proof give here, that is, without µ 3. Proof. We are goig to show µ ν µ 3 µ. The first iequality is trivial. For the secod oe, sice there exists w j C, w j = such that ν( j ) = w j ν( j ). Cosider the fuctio f = j w jχ j. Sice { j } is mutually disjoit, f. Take f = w jχ j. We the have, by domiate covergece theorem, f We the have f dν f dν = lim f dν = lim f f d ν 0, w j ν( j ) = ν( j ). The ν( j) = f dν µ 3(). Sice { j } is arbitrary, ν µ 3. For the third iequality, give ɛ > 0, the we ca fid some f with f such that µ 3 () < f dν + ɛ Departmet of Math., Natioal Taiwa Uiversity. mail: d042200@tu.edu.tw
We approximate f by a simple fuctio as follows. Let D C be the closed uit disc. The compactess of D implies that there are fiite may z j D such that B ɛ (z j ) covers D. Defie B j = f (B ɛ (z j )) X, which is measurable sice f is. The uio of B j is X. Let A = B, A j = B j \ j i= B i be the disjoit sets with A j B j ad A j = X. Defie the simple fuctio φ = m z jχ Aj, the φ ad f(x) φ(x) < ɛ for all x by the costructio. The f dν φ dν f φ dν < ν (). µ 3 () < f dν + ɛ < φ dν + ɛ + ɛ ν () Now we defie F j = A j, the F j is a fiite partitio of ad φ dν = z j χ Aj dν = j j Sice ɛ is arbitrary, we have µ 3 () µ (). z j ν(f j ) j ν(f j ) µ (). 2. Fid two measures µ, λ that are σ-fiite positive measure o some (X, M ), but the fuctio h i the absolutely cotiuous part of the Lebesgue-Rado-Nikodym decompositio of λ with respect to µ is ot µ-itegrable. Proof. Let µ be the Lebesgue measure o ((0, ), L ). The desired λ is defied by λ() = x dµ(x). which is easy to see it s positive, σ-fiite, absolutely cotiuous with respect to µ. Remark 2. This example also provide us a reaso why we eed to assume the measure λ is fiite i Theorem 6.. 3. Proof. Ay multiple of complex regular Borel measure is still a complex regular Borel measure. Give two complex regular Borel measures µ, µ 2. Give a Borel set ad k N, the there exist ope sets O, O 2 cotaiig such that µ i (O i \ ) < 2k (i =, 2), the µ + µ 2 (O O 2 ) = µ + µ 2 () + µ + µ 2 ((O O 2 ) \ ) µ + µ 2 () + µ (O \ ) + µ 2 (O 2 \ ) µ + µ 2 () + k, which implies the outer regularity of µ + µ 2. Give ay ope set or Borel set with fiite measure, we see for each k N, the there exist ope sets K, K 2 cotaied i such that µ i ( \K i ) < 2k (i =, 2), the µ +µ 2 (K K 2 ) = 2
µ + µ 2 () µ + µ 2 ( \ (K K 2 )) µ + µ 2 (), which implies the ier regularity k of µ + µ 2. Hece µ + µ 2 is a complex regular Borel measure. Therefore M(X) is a vector space over C. For each ν M(X), by Rado-Nikodym theorem, there is a itegrable fuctio f with respect to µ = ν r + ν i such that dν = fdµ ad the d ν = f dµ. So ν (X) = ν = 0 iff f is zero µ-a.e. iff ν = 0. For each a C, aν = aν (X) = af dµ = a ν (X) = a ν. The triagle iequality is proved by the X same method with µ = µ + µ 2 := ν,r + ν,i + ν 2,r + ν 2,i. Therefore, is a orm o M(X). Give ν be a absolutely coverget series i M(X). Sice for each A Σ, ν (A) ν (A) ν (X) = ν <, we may defie ν : Σ C by ν(a) = ν (A). Clearly, ν( ) = 0. If A k is a sequece of disjoit measureable sets, sice by Toelli s theorem ν (A k ) =,k ν (A k ) = k ν ( k A k ) ν < ad therefore ν( k A k ) = ν ( k A k ) = ν (A k ) = k ν (A k ) = k ν(a k ). k Hece, ν is a complex measure. ν M(X). Give X X m covers X, m j= ( ν N ν )(X j ) = = m ν (X j ) j= =N+ m ν (X j ) = ν (X) = ν. j= =N+ =N+ =N+ By xercise, ν N = ν =N+ ν 0 as N 0. So M(X) is complete. 4. A special case was discussed i xercise 4.7 earlier. st proof. Suppose there is o C > 0, such that fg C f p. The there exist f such that f p = ad f g > 3. Let f = 2 f, the f p ad for each fg > f g 2 > ( 3 2 ) This cotradicts to fg L ad therefore the map f fg from L p R or C is bouded. if p <, the by Riesz Represetatio theorem, g = g L q a.e.. For p =, if g is ot itegrable, the it cotradicts to the hypothesis by takig f. 2d proof. This is ispired by xercise 5.0, which is a applicatio of Baach-Steihaus Theorem. I thik this method is easier tha the first proof sice we do t eed duality theorem. 3
This time, we defie a liear map Λ : L p C by Λ (f) = fg, where g = gχ { x : g(x) } (this is the place we use the σ-fiiteess of µ). Note that Λ = g q is bouded for all ad {Λ (f)} is bouded by fg for each f L p. Therefore g L q by Baach-Steihaus Theorem ad mootoe covergece theorem. There is a theorem so called the coverse Hölder s iequality: Theorem 3. Let (X, Σ, µ) be a measure space, q <. If g L q, the g q = sup{ fg dµ : f p = }. If q =, this result holds if µ is semi-fiite. This exercise seems to be the coverse of the theorem (but it s NOT) if we assume M q (g) = sup{ fg dµ : f is simple ad its support has fiite measure, f p = } <. I see this result i Follad [3, Theorem 6.4]. Compare the differeces betwee their proofs. I thik they ca NOT be replaced by each other. (Reaso: xercise 6.4 do NOT permit M q (g) <. O the other had, I thik we ca t use approximate argumet to see M q (g) < implies fg L for all f L p.) 5. Proof. No, sice L (µ) is oe dimesioal, so does its dual space. But L (µ) is two dimesioal. Remark 4. This exercise shows that the map from L (L ) defied by g φ g is ot ijective if µ is ot semifiite. This problem, however, ca be remedied by redefiig L. See Follad [3, xercise 6.23-25]. 6. Proof. For each σ-fiite set X there is a a.e.-uique g L q () such that Φ(f) = fg for all f L p () ad g q Φ. If F is σ-fiite ad F, the g F = g a.e. o, so g F q g q. Let M be the supremum of g q as rages over all σ-fiite sets, otig that M Φ. Choose a sequece { } so that g q M, ad set F =. The F is σ-fiite ad g F q g q for all, whece g F q = M. Now, if A is a σ-fiite set cotaiig F, we have g F q + g A\F q = g A q M q = g F q, ad thus g A\F = 0 ad g A = g F a.e. (Here we use the fact q <.) But if f L p, the A f = F {x : f(x) 0} is σ-fiite, so Φ(f) = fg Af = fg F. Thus we may take g = g F, ad the proof is complete. 4
7. Proof. As Hit, the hypothesis holds for ay f dµ with f is a trigoometric polyomial, ad hece with f is cotiuous fuctio by Weierstrauss theorem, ad therefore with f is ay bouded Borel fuctio by Lusi s theorem. Sice by Theorem 6.2, dµ = h d µ for some h = a.e., µ () 0 as. Sice µ () = µ () = µ ( ), we kow µ (m) 0 as m. By the same argumet, it holds with f d µ for ay f is bouded. I particular, we pick hd µ = dµ. 8. Proof. Assume that there is some k N such that for all, e it dν := e it (e ikt ) dµ = µ( + k) µ() = 0 Sice trigoometric polyomials are dese i C[0, 2π], f dν = 0 for ay cotiuous f. Give ope set U, there is a icreasig sequece of cotiuous fuctio f =, if dist(x, U) > ad f = dist(x, U) otherwise, that approximate to the charactertic fuctio χ U poitwise. By mootoe covergece theorem, we see ν(u) = 0. So ν r (U) = ν i (U) = 0, that is ν + r = ν r, ν + i = ν i for all ope sets. Sice they are outer regular (Theorem 2.7), ν + r = ν r, ν + i = ν i, that is, ν r = ν i = 0 = ν. Sice (e ikt ) 0 iff t 2πj (j N), µ = k k j=0 a jδ 2πj k a j C. for some 9. Proof. We are goig to costruct h 0 such that the coditios (i)(ii) are satisfied ad for each f C(I), fh dm f dm, more precisely, the Riema itegral of f. The desired g = ( + ) (h + ) > 0 ad satisfies coditios (i)-(iii). Let δ = (2 3 ). Cosider h = 2 2 h k (x) = liear k= hk where for x 2k δ 2 2, for 2 δ < x 2k 2 + 2 δ, 0 for 2 δ < x 2k 2. () Now observe that it s easy to check they satisfies (i)(ii). Moreover, fh dm k= sup k=,, x 2k 2 k= f( 2k 2 ) 2 2 f(x) f( 2k ( 2 ) dx + 2 2 f δ ) + f 2 2 δ sup{ f( 2k 2k ) f(x) : x 2 2 2 δ 2 } + f which teds to zero as 0 by the uiform cotiuity of f. This is what we wat to see. 0. (a) it s obvious. 5
(b) The Vitali s Covergece Theorem is easy to prove if we kow goroff s theorem, Fatou s lemma, ad the equivalece betwee the uses of f ad f i the defiitio of uiform itegrability. (c) Show that we ca ot omit the tightess coditio (iii): for every ɛ > 0 there exists X such that µ() < ad f c p < ɛ for all, eve if { f } is bouded. Proof. Cosider the Lebesgue measure o (, ) with f = χ (,+). (d) To apply Vitali s theorem i fiite measure space, sometimes we see f(x) < a.e. is automatically true, but sometimes it s ot. Give examples. Proof. (i) Cosider the Lebesgue measure o [0, ], by the uiform itegrability we kow there exist ooverlappig closed itervals I, I k whose uio is [0, ], ad for all j =, k ad N, I j f <. (It s easy to show it s equivalet to use f ad f i the defiitio of uiform itegrability.) By Fatou s Lemma, [0,] f lim if [0,] f < k. (ii) We eed to fid out a fiite measure space (X, M, µ) ad a uiformly itegrable sequece of L fuctios f with f f a.e., f(x) is ot fiite a.e. ad f f i L. O R, let M is the σ-algebra of coutable or co-coutable sets. µ() := 0 if is coutable, µ() := if is co-coutable, which is easy to check µ is a measure o M. Cosider f, which is the desired example. (e) It s easy to see Vitali s Theorem implies Lebesgue Domiated Covergece Theorem i fiite measure space. The sequece f (x) = χ x ( +, )(x) is a example i which Vitali s theorem applies although the hypothesis of Lebesgue s theorem do ot hold. (f) The sequece f = χ (0,/) χ (,) o [0,] shows the assumptio that f 0 is sometimes importat i some applicatios. Note that f (x) 0 for every x [0, ], f (x) dx = 0, but f is ot uiformly itegrable. (g) However, the followig coverse of Vitali s theorem is true: Theorem 5. If µ(x) <, f L (µ), ad lim f dµ exists for every M, the {f } is uiformly itegrable. 6
Proof. As hit, we defie ρ(a, B) = χ A χ B dµ. The (M, ρ) is a complete metric space (modulo sets of measure zero), ad f dµ is cotiuous for each, (deote this map by F.) If ɛ > 0, cosider A N = { : F () F m () < ɛ, if, m N}. Sice X = A N by hypothesis, Baire Category theorem implies that some A N has oempty iterior, that is, there exist 0 M, δ > 0, N N so that (f f N ) dµ < ɛ if ρ(, 0 ) < δ, > N. (2) If µ(a) < δ, (2) holds with B = 0 \ A ad C = 0 A i place of. Thus, (f f N ) dµ = (f f N ) dµ < 2ɛ. A C B By cosiderig {f, f N }, there exists δ > 0, such that f dµ < 3ɛ if µ(a) < δ, =, 2, 3, A Remark 6. The Duford-Pettis theorem: [2, p.466 ad 472]. Proof. By Fatou s lemma X f p < C ad hece f is fiite a.e.. Give ɛ > 0, by goroff s theorem, there is a measurable set F with C /p µ(f ) /p < ɛ, such that f f uiformly o X \ F. The there is a N(ɛ) N such that for > N(ɛ), f f f f + f + f < ɛ + 2C /p µ(f ) /p < 3ɛ. X X\F F F 2. The assertios i this problem are easy to prove, we omit the proof ad remark several thigs: First, this problem shows that the map g Φ g from L (L ) is ot surjective if the measure space is ot σ-fiite; secod, that map is also ot ijective if µ is ot semifiite, this ca be see from the example I leared from Follad [3, p.9-92]: Let X be a set of ifiite measure that cotais o subset of positive fiite measure. The for ay f L, the set {x : f(x) 0} itersects i a ull set. It follows that Φ χ = 0 although χ 0 i L. 3. Apply Hah-Baach Theorem, see Theorem 5.9, p.07. Additioal Results We preset the proof that the costat i Sec 6.3 is sharpest. (Due to Bledsoe [] which is a π simpler treatmet of Kaufma-Rickert [4]). 7
Theorem 7. Let C 0 be the supremum of the umber C for which C z z S z T where T is ay oempty fiite set of complex umbers ad S is ay subset of T. The C 0 = π. Proof. By Theorem 6.3, C 0 π. O the other had, cosider z j = exp[i(2π/)j], j =,,. For this choice of poits we have for each θ, let S θ be the set of those z j such that θ arg(z j ) π 2 Thus, C 0 π. ( lim z Sθ / z j= ) z j = lim z z Sθ = lim 2π π /2 = 2π θ+π/2 θ π/2 z z S θ e iφ dφ = π. Refereces [] WW Bledsoe. A iequality about complex umbers. The America Mathematical Mothly, 77(2):80 82, 970. [2] Haim Brezis. Fuctioal Aalysis, Sobolev Spaces ad Partial Differetial quatios. Spriger Sciece & Busiess Media, 200. [3] Gerald B Follad. Real aalysis: Moder Techiques ad Their Applicatios. Joh Wiley & Sos, 2d editio, 999. [4] Robert P Kaufma ad Neil W Rickert. A iequality cocerig measures. Bulleti of the America Mathematical Society, 72(4):672 676, 966. 8