Problem Set. Problem Set #1. Math 5322, Fall March 4, 2002 ANSWERS

Transcription

1 Problem Set Problem Set #1 Math 5322, Fall 2001 March 4, 2002 ANSWRS

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3 All of the problems are from Chapter 3 of the text. Problem 1. [Problem 2, page 88] If ν is a signed measure, is ν-null iff ν () 0. Also, if ν and µ are signed measures, ν µ iff ν µ iff ν + µ and ν µ For the first part of the problem, let P N be a Hahn decomposition of with respect to ν. Thus, we have ν + () ν( P ) ν () ν( N) ν ν + + ν Assume that is ν-null. This means that for all measurable F, ν(f ) 0. But then P, so ν + () 0 and N, so ν () 0. Then we have ν () ν + () + ν () 0 Conversely, suppose that ν () 0. Let F be measurable. Then 0 ν + (F ) + ν (F ) ν (F ) ν () 0, so ν + (F ) 0 and ν (F ) 0. But then ν(f ) ν + (F ) ν (F ) 0. Thus, we can conclude that is ν-null. For the second part of the problem, we want to show that the following conditions are equivalent. (1) ν µ (2) ν + µ and ν µ (3) ν µ Let s first show that (1) (2). Since ν µ, we can decompose into a disjoint union of measurable sets A and B so that A is µ-null and B is ν-null. We can also find an Hahn decomposition P N with respect to ν, as above. Since B in ν-null, we have ν + (B) ν(p B) 0 and ν (B) ν(b N) 0. Thus, B is ν + -null and ν -null. Since A is still µ-null, we conclude that ν + µ and ν µ. Next, we show that (2) (3). Since ν + µ, we can find disjoint measurable sets A 1 and B 1 so that A 1 B 1, ν + (B 1 ) 0 and A 1 is µ-null. Similarly, we can write A 2 B 2 (disjoint union) where ν (B 2 ) 0 and A 2 is µ-null. We can then partition as (A 1 A 2 ) (A 1 B 2 ) (A 2 B 1 ) (B 1 B 2 ) Now, (A 1 A 2 ) is contained in the µ-null set A 1, and so is µ-null. Similarly, A 2 B 2 and A 2 B 1 are µ-null. We have 0 ν + (B 1 B 2 ) ν + (B 1 ) 0, so ν + (B 1 B 2 ) 0. Similarly, ν (B 1 B 2 ) 0, and thus ν (B 1 B 2 ) 0. We thus have a disjoint measurable decomposition A 3 B 3 where A 3 (A 1 A 2 ) (A 1 B 2 ) (A 2 B 1 ) B 3 B 1 B 2. 1

4 The set A 3 is µ-null and B 3 is ν -null. This shows that ν µ. Finally we show that (3) (1). So, suppose that ν µ. The we have a measurable decomposition A B where ν (B) 0 and A is µ-null. However, ν (B) 0 implies that B is ν-null (as we proved above), so we have ν µ. Problem 2. [Problem 3, page 88] Let ν be a signed measure on (, M). a. L 1 (ν) L 1 ( ν ). b. If f L 1 (ν), f dν f d ν. c. If M, { ν () sup } f dν f 1 According to the definition on page 88 of the text, the definition of L 1 (ν) is L 1 (ν) L 1 (ν + ) L 1 (ν ) and if f L 1 (ν), we define f dν f dν + f dν. So, consider part (a.) of the problem. If f L 1 (ν), then, by definition the integrals ( ) f dν +, f dν are finite. On the other hand, ν is defined by ν () ν + ()+ν (). Hence, the equation χ d ν χ dν + + χ dν holds by definition. Thus, we can apply the usual argument to show that ( ) g d ν g dν + + g dν holds for all measurable g : [0, ]: This equations holds for characteristic functions, hence (by linearity) for simple functions and hence (by the monotone convergence theorem) for nonnegative functions. Thus, if the integrals in ( ) are finite, the integral f d ν is finite, so f L 1 ( ν ). Conversely, if f L 1 ( ν ), the integral f d ν is finite, and so by ( ), the integrals in ( ) are finite, so f L 1 (ν). 2

5 Next consider part (b.) of the problem. If f L 1 (ν), then by definition f dν f dν + f dν f + dν + f dν + f + dν + f dν, where all of the integrals on the right are positive numbers. Hence, by the triangle inequality, f dν f + dν + + f dν + + f + dν + f dν (f + + f ) dν + + (f + + f ) dν f dν + + f dν f d ν Finally, consider part (c.) of the problem. If f 1, then for any M, f χ χ. Thus, we have f dν f d ν f χ d ν χ d ν ν (). Thus, the sup in part (c.) is ν (). To get the reverse inequality, let P N be a Hahn decomposition of with respect to ν, and recall the description of ν + and ν in terms of P and N from the last problem. Then we have ν () ν( P ) ν( N) (χ P χ N ) dν. But, χ P χ N 1 (P and N are disjoint). This shows that ν () is an element of the set we re supping over, so ν () the sup. Problem 3. [Problem 4, page 88] If ν is a signed measure and λ, µ are positive measures such that ν λ µ, then λ ν + and µ ν. Let P N be a Hahn decomposition of with respect to ν. Then for any measurable set, we have ( ) ν + () ν( P ) λ( P ) µ( P ). 3

6 If ν + (), this equation shows λ( P ), and so λ(). Thus, ν + () λ() is true in this case. If ν + () <, then both numbers on the right of ( ) must be finite, and ( ) gives ν + () + µ( P ) λ( P ). Thus, we have ν + () ν + () + µ( P ) λ( P ) λ(). The argument for the other inequality is similar. We have ν () ν( N) λ( N) + µ( N) If ν (), then we must have µ() µ( N), so µ() ν (). Otherwise, every thing must be finite and ν () ν () + λ( N) µ( N) µ(). Problem 4. [Problem 5, page 88] If ν 1 and ν 2 are signed measures that both omitted the value + or, then ν 1 + ν 2 ν 1 + ν 2. (Use xercise 4.) The hypothesis that ± is omitted assures that the signed measure ν 1 + ν 2 is defined. We have v 1 ν 1 + ν 1 and ν 2 ν 2 + ν 2. Adding these equations gives ν 1 + ν 2 (ν ν+ 2 ) (ν 1 + ν 2 ), where the expressions in parentheses are positive measures. Thus, by xercise 4, we have (ν 1 + ν 2 ) + ν ν+ 2 (ν 1 + ν 2 ) ν 1 + ν 2 and adding these inequalities gives ν 1 + ν 2 ν 1 + ν 2. Problem 5. [Problem 12, page 92] For j 1, 2, let µ j, ν j be σ-finite measures on ( j, M j ) such that ν j µ j. then ν 1 ν 2 µ 1 µ 2 and d(ν 1 ν 2 ) d(µ 1 µ 2 ) (x 1, x 2 ) dν 1 dµ 1 (x 1 ) dν 2 dµ 2 (x 2 ). We first want to show that ν 1 ν 2 µ 1 µ 2. So, suppose that M 1 M 2 and (µ 1 µ 2 )() 0. Recall that the x 1 -section of is defined by x1 { x 2 2 (x 1, x 2 ) }. 4

7 By Theorem 2.36 on page 66 of the text, the function x 1 µ 2 ( x1 ) is measurable and 0 (µ 1 µ 2 )() µ 2 ( x1 ) dµ 1 (x 1 ) 1 Thus, µ 2 ( x1 ) 0 for µ 1 -almost all x 1. In other words, there is a µ 1 -null set N 1 so that µ 2 ( x1 ) 0 if x 1 / N. If x 1 / N, we must have ν 2 ( x1 ) 0, since ν 2 µ 2. We also have ν 1 (N) 0, since ν 1 µ 1. Thus, x 1 ν 2 ( x1 ) is zero almost everywhere with respect to ν 1. Thus, we have (ν 1 ν 2 )() ν 2 ( x1 ) dν 1 (x 1 ) 0. 1 Thus, (µ 1 µ 2 )() 0 (ν 1 ν 2 )() 0, so ν 1 ν 2 µ 1 µ 2. For notational simplicity, let f d(ν 1 ν 2 ) d(µ 1 µ 2 ), f 1 dν 1 dµ 1, f 2 dν 2 dµ 2. The, we have ( ) (ν 1 ν 2 )() f(x 1, x 2 ) d(µ 1 µ 2 )(x 1, x 2 ) and f is the unique (up to equality almost everywhere) function with this property. Of course we have ν 1 (A) f 1 (x 1 ) dµ 1 (x 1 ) A ν 2 (B) f 2 (x 2 ) dµ 2 (x 2 ) B if A 1 and B 2 are measurable. On the other hand, the functions (x 1, x 2 ) f 1 (x 1 ) and (x 1, x 2 ) f 2 (x 2 ) are measurable, so we can define a measure λ on 1 2 by ( ) λ() f 1 (x 1 )f 2 (x 2 ) d(µ 1 µ 2 )(x 1, x 2 ). Suppose that A 1 1 and A 2 2. Then, we can use Tonelli s Theorem to 5

8 calculate λ(a 1 A 2 ) as follows: λ(a 1 A 2 ) f 1 (x 1 )f 2 (x 2 ) d(µ 1 µ 2 )(x 1, x 2 ) A 1 A 2 χ A1 A 2 (x 1, x 2 )f 1 (x 1 )f 2 (x 2 ) d(µ 1 µ 2 )(x 1, x 2 ) 1 2 χ A1 (x 1 )χ A2 (x 2 )f 1 (x 1 )f 2 (x 2 ) d(µ 1 µ 2 )(x 1, x 2 ) 1 2 χ A1 (x 1 )χ A2 (x 2 )f 1 (x 1 )f 2 (x 2 ) dµ 2 (x 2 ) dµ 1 (x 1 ) 1 2 χ A1 (x 1 )f 1 (x 1 ) χ A2 (x 2 )f 2 (x 2 ) dµ 2 (x 2 ) dµ 1 (x 1 ) 1 2 χ A1 (x 1 )f 1 (x 1 )ν 2 (A 2 ) dµ 1 (x 1 ) 1 ν 2 (A 2 ) χ A1 (x 1 )f 1 (x 1 ) dµ 1 (x 1 ) 1 ν 1 (A 1 )ν 2 (A 2 ). Thus we have λ(a 1 A 2 ) ν 1 (A 1 )ν 2 (A 2 ) (ν 1 ν 2 )(A 1 A 2 ). for all measurable rectangles A 1 A 2. Since our measures are σ-finite, this shows that λ ν 1 ν 2, see the remark at the bottom of page 64 in the text. Comparing ( ) and ( ) and using the uniqueness in ( ) then shows that f(x 1, x 2 ) f 1 (x 1 )f 2 (x 2 ) Problem 6. [Problem 16, page 92] Suppose that µ, ν are measures on (, M) with ν µ and let λ µ + ν. If f dν/dλ, then 0 f < 1 µ-a.e. and dν/dµ f/(1 f). It s clear that ν λ, so f dν/dλ makes sense. We want to show 0 f < 1 µ-a.e. Suppose, for a contradiction, that the is a set with µ() > 0 and f 1 on. On the on hand, we have ν() < ν() + µ() λ(). On the other hand, fχ χ, so ν() f dλ χ f dλ χ dλ λ(), a contradiction. If is measurable, we have 1 dν ν() f dλ 6 f dµ + f dν,

9 which yields or, in other words, (1 f) dν f dµ, χ (1 f) dν χ f dµ. By the usual procedure, we can extend this equation from characteristic functions to nonnegative functions. Thus, we have g(1 f) dν gf dµ for all nonnegative measurable functions g. Hence, for any measurable set, we have ( ) g(1 f) dν gf dµ for all nonnegative measurable functions g. Since 0 f < 1 µ-a.e. the function 1/(1 f) is defined and nonnegative µ-a.e. Since ν µ, 1/(1 f) is also defined and nonnegative ν-a.e. Thus, we can set g 1/(1 f) in ( ). This gives f ν() 1 f dµ, for all measurable, whence dν/dµ f/(1 f). Problem 7. [Problem 18, page 94] Prove Proposition 3.13c. In other words, suppose that ν is a complex measure on (, M). L 1 (ν) L 1 ( ν ) and if f L 1 (ν), f dν f d ν. Then I seem to have been very confused the day I tried to do this in class! Let z x + iy be a complex number, where x, y are real. We have x 2, y 2 x 2 + y 2 z 2 so taking square roots gives x, y z. One the other hand, we have z 2 x 2 + y 2 x 2 + y 2 x x y + y 2 [ x + y ] 2, and so z x + y. 7

10 Now lets begin the solution of the problem. By the definition on page 93 of the text, L 1 (ν) is defined to be L 1 (ν r ) L 1 (ν i ) and for f L 1 (ν), we define f dν f dν r + i f dν i. On the other hand, we know from xercise 3 on page 88 of the text (one of the problems in this set) that for a signed measure λ, L 1 (λ) L 1 ( λ ) and for f L 1 (λ), f dλ f d λ. Hence in our present situation, we know L 1 (ν) L 1 ( ν r ) L 1 ( ν i ). Let s recall the reasoning of part b. of Proposition We can find some finite positive measure µ such that ν r and ν i are absolutely continuous with respect to µ (µ ν r + ν i will do). Then, by the Lebesgue-Radon-Nikodym Theorem, there are real-valued measurable functions f and g such that dν r f dµ and dν i g dµ. Then, we have ( ) ν() h dµ where h f + ig. From this equation, we have d ν h dµ, by definition. Since f, g h we have ν r () f dµ h dµ ν () ν i () g dµ h dµ ν () so we certainly have ν r, ν i ν. Thus, once again, we can find real-valued functions ϕ and η such that dν r ϕ d ν and dν i η d ν. If ϕ ψ + iη, then ν() ϕ d ν. But then, comparing with ( ), we have ϕ h dµ h dµ. By the uniqueness part of the Lebesgue-Radon-Nikodym Theorem, ϕ h h, µ-a.e., and hence ν -a.e. On the other hand if Z is the set where h 0, then ν (Z) h dµ 0 dµ 0 Z so h 0, ν -a.e. This shows that ϕ 1, ν -a.e., so we can conclude that ψ, η ϕ 1, ν -a.e., and 1 ϕ ψ + η, ν -a.e. Finally, we have d ν r ψ d ν and d ν i η d ν. Z 8

11 Now, suppose that f L 1 ( ν ). Then f d ν < Since f ψ f, ν -a.e., we have f ψ d ν f d ν < but the integral on the left is f d ν r, so f L 1 ( ν r ). Similarly, f L 1 ( ν i ) Conversely, suppose that f L 1 ( ν r ) L 1 ( ν i ) Then we have f ψ d ν f d ν r < f η d ν f d ν i <. and so we have f [ ψ + η ] d ν <. However, we have 1 ψ + η, ν -a.e., so f d ν f [ ψ + η ] d ν <. and so f L 1 ( ν ). Finally, suppose that f L 1 (ν) L 1 ( ν ), we then have f dν f dν r + i f dν i by definition fψ d ν + i fη d ν f[ψ + iη] d ν fϕ d ν f ϕ d ν f d ν, since ϕ 1, ν -a.e. This completes the proof. 9

12 Problem 8. [Problem 20, page 94] If ν is a complex measure on (, M) and ν() ν (), then ν ν. Let f dν/d ν, so we have ( ) ν() f d ν for all measurable sets, and we know from Proposition 3.13 on page 94 of the text that f 1 ν -a.e. We may as well suppose that f 1 everywhere. We can write the complex function f as f g + ih, where g and h are real-valued. We can also write ν ν r + iν i for finite signed measures ν r and ν i. Thus, we have ν r () + iν i () ν() f d ν g d ν + i h d ν. Comparing real and imaginary parts, we get ν r () g d ν ν i () h d ν. By hypothesis, we have ν() ν (), so g d ν + i h d ν ν (). Since the right-hand side is real, the imaginary part of the left-hand side must be zero, and we have g d ν ν () 1 d ν, and so ( ) 0 Since g is the real part of f, we have (1 g) d ν. g g g + ih f 1 so 1 g 0. But then ( ) shows that 1 g 0 a.e., so g 1 a.e. We then have 1 f 2 g 2 + h h 2 a.e., so h 0 a.e. and hence f g 1 a.e. Putting this in ( ) shows that ν ν. 10

13 Problem 9. [Problem 20, page 94] Let ν be a complex measure on (, M). If M, define n n (A) µ 1 () sup ν( j ) n N, 1,..., n disjoint, j, (B) µ 2 () sup ν( j ) 1, 2,... disjoint, j, { } (C) µ 3 () sup f dν f 1. Then µ 1 µ 2 µ 3 ν. Although the author didn t explicitly say so, the sets j in (A) and (B) should, of course, be measurable and the functions f in (C) should be measurable. Also note that the book has a typo in (C) ( dµ instead of dν). As suggested in the book, we first prove that µ 1 µ 2 µ 3 µ 1 and then that µ 3 ν. We have µ 1 µ 2, because the sums in (A) are among the sums in (B): given a finite sequence 1,..., n of disjoint sets whose union is, just set j for j > n. Then 1, 2,... is an infinite sequence of disjoint sets whose union is and n ν( j ) ν( j ). Thus, the set of numbers in (A) is a subset of the set of numbers in (B). Since the sup over a larger set is larger, µ 1 µ 2. To see that µ 2 µ 3, let 1, 2,... be a sequence of disjoint sets whose union is. Now each ν( j ) is a complex number, so there is a complex number ω j such that ω j 1 and ω j ν( j ) ν( j ). Indeed, we can say { ν(j) ν( ω j, ν( j) j) 0 1, ν( j ) 0. Consider the function f defined by f ω j χ j. If x /, f(x) 0. If x, it is in exactly one j and then f(x) ω j 1. Thus, f 1 and the partial sums n f n ω j χ j 11

14 satisfy f n 1 by the same reasoning. We then have f n dν f dν (f n f) dν f n f d ν 0 by the dominated convergence theorem, since f n f pointwise and f n f is bounded above by the constant function 2, which is integrable with respect to the finite measure ν. We then have f dν lim f n dν n ω j χ j dν ω j ν( j ) ν( j ). since f dν is positive, we have f dν ν( j ). Thus, the set of numbers in (B) is a subset of the set in (C), so µ 2 µ 3. Now we want to show that µ 3 µ 1. First, we use Proposition 3.13 (and xercise 18) to show that µ 3 is finite. If f 1, then f d ν 1 d ν ν () <, so f L 1 ( ν ) L 1 (ν) and f dν f d ν 1 d ν ν (). Thus, ν () is an upper bound for the set of numbers in (C), so µ 3 () ν () ν () <. To show that µ 3 µ 1, let ε > 0 be given. Then we can find some f with f 1 such that µ 3 () ε < f dν, which implies ( ) µ 3 () 12 f dν + ε.

15 We approximate f by a simple function as follows. Let D { z C z 1 } be the closed unit disk in the complex plane. Since f 1, all the values of f are in D. The collection of balls { B(ε, z) z D } is an open cover of D. Since D is compact, there is a finite subcover, say { B(ε, z j ) j 1,..., m }. Define B j f 1 (B(ε, z j )), which is measurable since f is measurable. The union of the sets B j is. The sets B j won t be disjoint, but we can use the usual trick and define A 1 B 1, A j B j \ to get a disjoint collection of sets with A j B j and j A j. Define a simple function ϕ by m ϕ z j χ Aj. This function takes on the values z j, which are in D, so ϕ 1. If x it is in exactly one of the sets A j. But then f(x) B(ε, z j ), so j 1 k1 B k f(x) ϕ(x) f(x) z j < ε. Thus, f ϕ < ε. We then have f dν ϕ dν f dν ϕ dν f dν ϕ dν f ϕ d ν ε d ν ε ν (). Thus, we have f dν ϕ dν + ε ν (). 13

16 Substituting this is ( ) gives ( ) µ 3 () ϕ dν + ε + ε ν (). Now define j A j, j 1,..., m. Then the j s are a finite sequence of disjoint sets whose union is and we have Substituting this into ( ) gives ϕ dν [ m ] z j χ Aj dν m z j χ Aj dν m z j χ χ Aj dν m z j χ j dν m z j ν( j ) m z j ν( j ) m ν( j ) µ 1 (). µ 3 () µ 1 () + [1 + ν ()]ε. Since ε > 0 was arbitrary, we conclude that µ 3 () µ 1 (). We ve now shown that µ 1 µ 2 µ 3. The last step is to show that µ 3 ν. We ve already shown above that µ 3 () ν (). To get the reverse inequality, let g dν/d ν. We know that g 1 ν -a.e., and we may as well assume that g 1 everywhere (by modifying it on a set of measure zero). Then the conjugate ḡ of g satisfies ḡ 1 1 and so is one of the functions in the definition of µ 3, so ḡ dν µ 3(). 14

17 But, from the definition of g, ḡ dν ḡg d ν g 2 d ν 1 d ν ν (). Thus, ν () µ 3 (), so µ 3 ν, and we re done. 15