Real Analysis, 2nd Edition, G.B.Folland Signed Measures and Differentiation

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1 Real Analysis, 2nd dition, G.B.Folland Chapter 3 Signed Measures and Differentiation Yung-Hsiang Huang 3. Signed Measures. Proof. The first part is proved by using addivitiy and consider F j = j j, 0 =. For the second part, say j, note µ( ) = µ() + µ( 2 ) + µ( 2 3 ) + = µ() + µ( ) µ( 2 ) + µ( 2 ) µ( 3 ) + = µ() + µ( ) lim k µ( k ). The second equality is by finiteness. 2. Proof. Let M with ν () = 0 = ν + ()+ν (). Since ν + () and ν () are nonnegative, ν + () = ν () = 0. So ν() = ν + () ν () = 0. Conversely, if is ν-null, then ν + () = ν( P ) = 0 = ν( N) = ν () since P and N are contained in. So ν () = ν + () + ν () = 0. The second assertion is to proved the followings are equivalent, ()ν µ, (2)ν + µ and ν µ, (3) ν µ : (3) () and () (2) are by the definition and the first part of this exercise. To prove (2) (3), we note that there are measurable sets A, B, A 2, B 2 such that A B = = A 2 B 2, A B = = A 2 B 2, and µ(a ) = 0 = ν + (B ) = µ(a 2 ) = ν (B 2 ). Let = (A A 2 ) (A B 2 ) (B A 2 ) and F = B B 2, then F =, F =, and by definition, µ() = 0 = ν + (F ) = ν (F ). Hence ν (F ) = 0, and therefore ν µ. 3. Proof. (a)(b) and (c) ν () sup{ f dν : f } are trivial. To prove that they are equal, let = P N be the Hahn decomposition for ν. We note that ν () = ν + () + ν () = ν( P ) ν( N) = χ P χ N dν. Since P and N are disjoint, χ P χ N. Hence the proof is completed. Last Modified: 206/08/5. Department of Math., National Taiwan University. mail: d042200@ntu.edu.tw

2 4. (The minimality of Jordan decomposition of ν.) Proof. Let = P N be the Hahn decomposition for ν. For any measurable set, we have ν + () = ν( P ) = λ( P ) µ( P ). If ν + () =, then λ( P ) = and hence λ() =. So ν + () λ(). If ν + () <, then λ( P ), µ( P ) <. This implies ν + () ν + () + µ( P ) = λ( P ) λ(). The second assertion ν µ is proved in a similar way. 5. Proof. Since ν = ν + ν and ν 2 = ν 2 + ν2 (), ν + ν 2 = (ν + + ν 2 + ) (ν + ν2 ). By ex 4, (ν + + ν 2 + ) (ν + ν 2 ) + and (ν + ν2 ) (ν + ν 2 ). The proof is completed by adding these two inequalities. 6. Proof. The Hahn decomposition is P N := {f 0} {f < 0}. ν ± () = f ± dµ. 7. Proof. Let = P N be the Hahn decomposition. (a)given F, ν(f ) = ν(f P ) + ν(f N) ν(f P ) ν( P ) which yields the inequality. The equality is true by taking F = P directly. The second case is similar. (b)given, n partition, note that ν( ) + ν( n ) ν ( ) + + ν ( n ) = ν () = ν(f P ) + ν(f N) (the proof of this inequality is similar to exercise 3). 3.2 The Lebesgue-Radon-Nikodym Theorem 8. Proof. ν µ ν µ is proved by exercise 2. Using this equivalence, we know ν µ ν + µ and ν µ. The converse is by definition. 9. Proof. For first assertion, we know from the assumption there is a sequence of {( j, F j } such that for each j, = j F j, ν j (F j ) = 0 and µ( j ) = 0. Take = j j and F = j F j, then µ() = 0 and j ν j(f ) = 0. The second assertion is much easier. 0. Nothing to comment.. This problem is easy, we omit the proof here and remark that the converse of (b) is also true (try to apply goroff s Theorem). This is known as Vitali Convergence Theorem stated in xercise 6.5, page 87. Necessity of the hypothesis and further discussions are given in Rudin [3, xercise 6.0-]. Also see the additional exercises to Chapter Proof. 2

3 3. Proof. (σ-finiteness condition can NOT be omitted in Radon-Nikodym Theorem.) (a) Suppose there is an extended µ-integrable function f such that dm = fdµ. Since for each x [0, ], 0 = m({x}) = f dµ = f(x). This leads to a contradiction since {x} = m([0, ]) = f dµ = 0. (b) Suppose that µ has a Lebesgue decomposition λ + ρ with respect to m, with λ m and ρ m. Then for all x [0, ], ρ({x}) = 0 and hence λ({x}) =. Since there exists disjoint measurable sets A, B partition [0, ] with A is λ-null, and m(b) = 0, we see for any x A, 0 = λ({x}) =, so A =. But then 0 = m(b) = m([0, ]) =, a contradiciton! 4. Proof. 5. Note that some textbooks referred to call this measure almost decomposable since (iii) is required to be true for µ() < only, and they call a measure decomposable if (iii) is true for any measurable set. A subtle difference is marked in the remark after xercise [0,] Proof. (a) Since = n= n for some n M and µ( n ) <. Take F be the collection of F k = k \ k j= j which possess the desired property (i)-(iv). (b) 6. According to Prof. Folland s errata sheet, we assume µ, ν are σ-finite. Proof. 7. (xistence of conditional expectation of f on N.) Remark. I think we need to assume ν = µ N is σ-finite on (, N ) to make every hypothesis in Radon-Nikodym Theorem satisfied. A counterexample is µ = Lebesgue measure on real line and N = the σ-algebra of countable or co-countable sets. Proof. Let (, M, µ) be a σ- finite measure space, N be a sub-algebra of M, ν = µ N and f L (µ). We define λ() := f dµ to be a signed measure on (, N ) (by considering f + dµ, f dµ) Note that given A N with ν(a) = 0, then µ(a) = 0, and hence λ ν. By Radon-Nikodym Theorem, there exists extended ν-integrable g such that dλ = g dν and any two such functions are equal ν- a.e.. Since one of g + dν, g dν is finite and g dν = f dµ <, 3

4 we know g dν is finite. Hence g L (ν). The uniqueness assertion is easy to proved (cf: Proposition 2.6.) Remark 2. The following alternative approach to this problem is taken from Williams [5, chapter 9]. A little difference here is that we work on the σ-finite measure space (, M, µ) instead of finite (probability) measure space. Remark 3. Williams [5, p.85] mentions the following theorem to explain that conditional expectation (and the martingale theory) is crucial in filtering and control- of space-ships, of industrial process, or whatever. Theorem 4. (Conditional expectation as least-squares-best predictor) Let f L (µ) L 2 (µ), N is a sub σ-algebra of M and ν = µ N. Then the conditional expectation (f N ) is the minimizer of mean square error to f, (f g) 2 := (f g)2 dµ in the space of L (ν) L 2 (ν). We need the following lemma : Lemma 5. If f, g is N - measurable and g, fg <, then (fg N ) = f(g N ). Proof. It s easy to see the right-hand side is N - measurable. Given N, if f = χ B with B N, then χ B (g N ) dν = (g N ) dν = g dν = χ B g dν = χ B g dµ. B B By linearity and monotone convergence theorem, this can be extended to any f, g 0, and then the desired result is true by splitting them into the positive and negative parts. Proof of Theorem 4. (Taken from Durrett [2, p.229]) For any h L (ν) L 2 (ν), (hf N ) = h(f N ), (by Cauchy-Schwarz, fh <.) Integrate both sides with respect to µ, we see h(f N ) dµ = And hence we see (hf N ) dµ = ( ) h (f N ) f dµ = 0 hf dµ Then given g L (ν) L 2 (ν), take h = (f N ) g L (ν). Note h L 2 (ν) is equivalent to (f N ) L 2 (ν), and the latter is proved by the Jensen s inequality. 4

5 Therefore, (f g) 2 dµ = = which shows the desired result. ( ) 2 f (f N ) + (f N ) g dµ = ( ) 2 f (f N ) dµ h 2 dµ ( f (f N ) + h) 2 dµ Proof of Jensen s inequality. The proof is taken from Chung, [, p38-39]. For any x and y: ϕ(x) ϕ(y) ϕ (y)(x y) where ϕ is the right-hand derivative of ϕ. Hence ϕ(f) ϕ((f N ) ϕ ((f N ))(f (f N )) The right member may not be integrable; but let Λ = {w : (f N )(w) A} for A > 0. Replace f by fχ Λ in the above, Observe that????????????????????????????????? 3.3 Complex Measures 8. Proof. From xercise 3, we know that L (ν) = L (ν r ) L (ν i ) = L ( ν r ) L ( ν i ). By Radon- Nikodym Theorem, for µ := ν r + ν i, there are real-valued µ-integrable functions f, g such that dν r = fdµ and dν i = gdµ. Then we have ν() = f + ig dµ =: h dµ. From this equation, we have d ν = h dµ. Since f, g h, we have ν r () ν (), ν i () ν () By Radon-Nikodym Theorem again, there exist real-valued µ-integrable functions φ, ϕ such that dν r = φd ν and dν i = ϕd ν. Therefore h dµ = ν() = φ + iϕ d ν = [φ + iϕ] h dµ. By the uniqueness part of Radon-Nikodym Theorem, [φ + iϕ] h = h, µ-a.e., and hence ν -a.e. Let Z be the set where h = 0, then it has ν measure zero since ν (Z) = h dµ = 0 5 Z

6 This shows that φ + iϕ =, ν -a.e. Now suppose that f L ( ν ). Since d ν r = φ d ν and d ν i = ϕ d ν, we have f d ν r = f φ d ν f d ν < and f d ν i = f ϕ d ν f d ν <. So f L ( ν r ) L ( ν i ). Conversely, suppose f L ( ν r ) L ( ν i ), then we have f d ν = f φ + iϕ d ν f [ φ + ϕ ] d ν = So f L ( ν ). f d ν r + f d ν i <. Finally, supppose that f L (ν) = L ( ν ), we then have f dν = f dν r + i f dν i = f[φ + iϕ] d ν f φ + iϕ d ν = f d ν. 9. If ν, µ are complex measures and λ is a positive measure, then ν λ ν λ and ν µ ν µ. Proof. The first ( ) is proved by Radon-Nikodym Theorem. Both ( ) are proved by definition and the fact ν () ν(). To prove the second ( ), we note there exist positive measure ρ = µ r + µ i and σ = ν r + ν i, and some functions f L (ρ) and g L (σ) such that dµ = fdρ, d µ = f dρ, dν = gdσ, and d ν = g dσ. Since ν µ, for each pair a, b {r, i} there exists disjoint measurable sets A ab, B ab such that = A ab B ab, A ab is ν a -null and B ab is µ a -null. It follows that A := (A rr A ri ) (A ir A ii ) is both ν r -null and ν i -null. In particular, for each n N the subsets {x A : Re(f(x)) > n } and {x A : Re(f(x)) < n } of A has ν r and ν i - measure zero, which implies f = 0 on A and hence A is ν -null. Moreover B := A c = (B rr B ri ) (B ir B ii ) is both µ r -null and µ i -null, and a similar argument implies B is µ -null. 20. Proof. Given M, since ν () + ν ( ) = ν () = ν() = ν( ) + ν(), we have ν () ν() = ν( ) ν ( ). 6

7 Since the left-hand side have nonnegative real part and the right-hand have nonpositive one, Re(ν()) = ν () ν() = Re(ν()) 2 + Im(ν()) 2. So Im(ν()) = 0 and therefore ν() = Re(ν()) = ν (). Since is arbitrary, ν = ν. 2. Proof. We are going to show µ µ 2 µ 3 µ and then µ 3 = ν. The first inequality is trivial. For the second one, since there exists w j C, w j = such that ν( j ) = w j ν( j ). Consider the function f = j w jχ j. Since { j } is mutually disjoint, f. Take f n = n w jχ j. We then have, by dominate convergence theorem, f n f dν f n f d ν 0, We then have f dν = lim f n dν = lim n n n w j ν( j ) = ν( j ). Then ν( j) = f dν µ 3(). Since { j } is arbitrary, µ 2 µ 3. For the third inequality, given ɛ > 0, then we can find some f with f such that µ 3 () < f dν + ɛ We approximate f by a simple function as follows. Let D C be the closed unit disc. The compactness of D implies that there are finite many z j D such that B ɛ (z j ) covers D. Define B j = f (B ɛ (z j )), which is measurable since f is. The union of B j is. Let A = B, A j = B j \ j i= B i be the disjoint sets with A j B j and A j =. Define the simple function φ = m z jχ Aj, then φ and f(x) φ(x) < ɛ for all x by the construction. Then f dν φ dν f φ dν < ν (). µ 3 () < f dν + ɛ < φ dν + ɛ + ɛ ν () Now we define F j = A j, then F j is a finite partition of and φ dν = z j χ Aj dν = z j ν(f j ) ν(f j ) µ (). j j j Since ɛ is arbitrary, we have µ 3 () µ (). We have already show µ 3 ν in the above argument. To get the reverse inequality, let g = dν/d ν. We know g = ν -a.e., and hence µ 3 () ḡ dν = ḡg d ν = ν(). 7

8 3.4 Differentiation on uclidean Space 22. Proof. Since M := f > 0, there exists R > 0 such that B R f > M/2. For x > R, the (0) ball B 2 x (x) B R (0) and hence Hf(x) f f > 2 n x n B 2 x (x) 2 n x n B R (0) For every small α > 0, there is an inclusion {x : R x < (c/α) /n } = {x : x R, and M 2 n+ x n. C > α} {x : Hf(x) > α} x n Thus, m({x : Hf(x) > α}) m({x : R x < (C/α) /n }) = w n (C R n α)/α > w n C/2α, provided C R n α > C/2, that is, α < C/2R n. Remark 6. maximal inequality 23. Proof. Hf H f is trivial. H f 2 n Hf is proved by the fact x B r (y) B 2r (x). 24. Obvious. 25. Proof. (a) Apply Lebesgue Differentiation Theorem to χ and χ c (b)the first example can be found by considering as sector of angle 2πα and x is the origin, the second example is = [2 n, 2 n + 2 n ], x = 0. Fixed N N, we compute m(b 2 N (0)) ) m(b 2 N (0)) does NOT exist. = /4 and m(b 2 N +2 N (0) ) m(b 2 N +2 N (0)) = /3. So the limit 26. Proof. Given compact set K, ν(k), λ(k) < (ν + λ)(k) <. Let = A B be the singular decomposition of λ and ν. Given Borel set R n, there exists open sets U n such that (λ + ν)(u n ) (λ + ν)() 0. Note that (λ + ν)(u n ) = λ(u n A) + ν(u n B) and (λ+ν)() = λ( A)+ν( B). Since ν(u n B) ν( B) 0 and λ(u n A) λ( A) 0, λ(u n A)+ν(U n B) λ( A) ν( B) λ(u n A) λ( A) = λ(u n ) λ() 0. Since the left-hand side tends to 0 as n, so does λ(u n ) λ(). Similarly, ν(u n ) ν(). 3.5 Functions of Bounded Variation 27. Proof. 28. Proof. 8

9 29. Proof. 30. Proof. Let {r n } be the set of all rational numbers. Define f : R R by x {j:r j <x}. It s 2 j easy to see it s increasing and discontinuous at any r n since f(z) f(r n ) 2 n for all z > r n. Given x Q c and ɛ > 0, there is N N such that N < ɛ. 2 j Since x is irrational, let δ = 2 min{ x r j : j =, N } > 0. Then for any y B δ (x), f(x) f(y) N 2 j < ɛ. Therefore, f is continuous at any irrational number. 3. We omit (a) since it is standard. (b) is included in the following problem taken from Stein- Shakarchi [4, xercise 2.]. If a, b > 0, let Proof. 32. Proof. 33. Proof. By Theorem 3.23, we know 0 F exists a.e. We may instead F by the function, still call it F, which equal to F (x) if x < b and equal to F(b) if x b. Consider f k (x) = {F (x + h) F (x)}/h where h = /k, then f k f a.e. and Fatou s lemma implies b a F (x) dx lim inf k = lim inf h 0 + b a ( h f k (x) dx = lim inf b+h b h 0 + F (x) dx h b F (x + h) F (x) dx h ) F (x) dx F (b) F (a). a a+h In fact, we have proved that b a F (x) dx F (b ) F (a+). 34. Proof. 35. Proof. 36. Proof. 37. Proof. 38. Proof. 39. Proof. 9 a

10 40. Proof. 4. Proof. 42. Proof. References [] Kai Lai Chung. A course in probability theory. Academic press, 3 edition, 200. [2] Rick Durrett. Probability: theory and examples. Cambridge university press, 4 edition, 200. [3] Walter Rudin. Real and complex analysis. Tata McGraw-Hill ducation, 3 edition, 987. [4] lias M Stein and Rami Shakarchi. Real analysis: measure theory, integration, and Hilbert spaces. Princeton University Press, [5] David Williams. Probability with martingales. Cambridge university press, 99. 0