Exercise 1. Show that the Radon-Nikodym theorem for a finite measure implies the theorem for a σ-finite measure.

Transcription

1 Real Variables, Fall 2014 Problem set 8 Solution suggestions xercise 1. Show that the Radon-Nikodym theorem for a finite measure implies the theorem for a σ-finite measure. nswer: ssume that the Radon-Nikodym theorem holds for finite measures and let ν µ where (X, B, µ) is a σ-finite measure space. Since µ is σ-finite we find a sequence (B n ) B of disjoint measurable sets so that X = B n and µ(b n ) < for all n N. For each n N apply Radon-Nikodym theorem on the set B n to find a measurable function f n : B n [0, ] so that ν() = f n dµ for all B n with B. Let f n : X [0, ] be such that f n = f n in B n and f n 0 otherwise, and define f = f nχ Bn. Note that this sum converges for all x X because the sets B n are disjoint, and it is clearly measurable and non-negative since each f n is. Now for any B we have ( ) ν() = ν ( B n ) = = f n dµ = B n ν( B n ) = f n χ Bn dµ = B n f n dµ f dµ. So f is the desired Radon-Nikodym derivative. The function f is also almost everywhere unique since each f n was, and a countable union of sets of measure zero has measure zero. xercise 2. xtend the Radon-Nikodym Theorem to the case of signed measures. nswer: We first prove the Radon-Nikodym theorem for a σ-finite measure space (X, B, µ) and ν µ where ν is a signed measure. Firstly, we can write ν = ν + ν as the Jordan decomposition of ν. Now ν + µ and ν µ, so we can apply exercise 1 to find Radon-Nikodym derivatives f + and f of ν + µ and ν µ respectively. Since ν is a finite measure, the function f is ν integrable and thus ν almost everywhere finite. 1

2 We can thus choose a representative of the Radon-Nikodym derivative that is everywhere finite. So we can define f = f + f. For this f we have by construction that for all B, ν() = ν + () ν () = f + dµ f dµ = f + f dµ = f dµ. So f is a Radon-Nikodym derivative of ν with respect to µ. To show that it is almost everywhere unique, let g be another such candidate that satisfies ν() = g dµ for all B. Let U, F be the sets given by the Jordan decomposition of ν = ν + ν so that U F = X and ν + is supported on U and ν is supported on F. Now for any U with B we have g dµ = ν() = ν + () = f + dµ. Since this holds for all U with B, then g = f + for µ-almost every point of U. Similarly one sees that g = f for µ-almost every point of F. So g = f + f = f for µ-almost everywhere in X. So we have shown the Radon-Nikodym theorem in the case where µ is a measure and ν a signed measure. We then assume that µ is also a signed measure. Take the Jordan decomposition of µ to µ = µ + µ and note that ν µ implies that ν µ + and ν µ. Use then the earlier part of this exercise to find the Radon-Nikodym derivatives h + and h of ν µ + and ν µ respectively. Let, D be the sets X = D where µ + and µ are supported. In particular we can choose h + to vanish in D and h to vanish in. Then for all B we have ν() = ν(( ) ( D)) = ν( ) + ν( D) = h + dµ + + h dµ = h + dµ + D = h + + h dµ = h + + h dµ. ( ) ( D) D h dµ So h + + h is the desired Radon-Nikodym derivative. Uniqueness argument is analogous to the one before. 2

3 xercise 3. lternate proof of Radon-Nikodym theorem: Use the fact that if F is a bounded linear functional on a Hilbert space H, then there exists a unique g H so that F (f) = (f, g) for all f H. Fill in the following sketch of the proof: (a) Let µ and ν be finite measure on a measure space (X, B) and let λ = µ + ν. Define F (f) = f dµ. Then F is a bounded linear functional on L 2 (λ). nswer: We have F (af + bg) = af + bg dµ = a f dµ + b g dµ = af (f) + bf (g), so F is linear. It is bounded because for all f L 2 (λ) we have F (f) = f dµ f dµ f dµ + f dν = f d(µ + ν) = f dλ Holder λ(x) 1 2 f L 2 (λ). This shows that F λ(x) 1 2 so F is bounded because λ(x) <. (b) The function g L 2 (λ) so that F (f) = (f, g) = fg dλ has the property that 0 g 1 and µ() = g dλ, ν() = (1 g) dλ. nswer: Let g L 2 (λ) be the function in the statement. Define = {g > 1} and B = {g < 0}. Now µ() = χ dµ = F (χ ) = (χ, g) = χ g dλ = g dλ λ() µ(). Since g > 1 in then g dλ > λ() unless λ() = 0. But the first option implies a contradiction with the above so we must have λ() = 0. So g 1 for λ-almost every point. We also have by similar reasoning as above that µ(b) = g dλ 0 3 B

4 as g < 0 in B. Since µ is a measure then we must have λ(b) = 0. So 0 g 1 for λ almost every point. The above reasoning also yields for any set B we have µ() = g dλ, and thus also ν() = λ() µ() = χ dλ gχ dλ = (1 g)χ dλ = (1 g) dλ. (c) If ν µ, then λ µ and g = 0 only on a set of µ-measure zero, and hence on a set of λ-measure zero. In this case, λ() = g 1 dµ. nswer: ssume that ν µ and fix a set B so that µ() = 0. Then ν() = 0, so λ() = µ() + ν() = 0. Thus λ µ. Denote F = {g = 0}. Since µ(f ) = g dλ = 0, F then λ µ implies that λ(f ) = 0. In other words, g = 0 only a set of λ-measure zero. By exercise 6 of problem set 7 we have g 1 dµ = g 1 g dλ = dλ = λ() for all B. (d) If ν µ then (1 g)g 1 is integrable with respect to µ, and ν() = (1 g)g 1 dµ 4

5 nswer: By exercise 6 of problem set 7 and part (b) of this exercise we have (1 g)g 1 dµ = (1 g)g 1 g dλ = (1 g) dλ = ν(). Since ν() < then the integrability follows as well. xercise 4. Give an example to show that the hypothesis in the Radon-Nikodym theorem that µ is σ-finite can not be omitted. nswer: Let (X, B) = ([0, 1], Bor([0, 1])) and let µ be the counting measure on B and ν the Lebesgue measure on [0, 1]. In other words, µ() is the number of elements on if is finite and µ() = otherwise. Now µ() = 0 implies that =, which gives ν() = 0. Hence ν µ. Note that all sets of finite µ-measure must have finitely many elements, and countable unions of finite sets are countable. So [0, 1] can not be the union of countably many sets of finite µ-measure because [0, 1] is uncountable. Hence µ is not σ-finite. ssume towards contradiction that Radon-Nikodym theorem would hold. Then we find a non-negative measurable function f so that ν() = f dµ for all B. Fix x X. Then for = {x} we have 0 = ν() = f dµ = f(x). Since this is true for all x [0, 1] then f 0. Thus ν() = f dµ = 0 for all B so ν 0, which is a contradiction. Hence the Radon- Nikodym theorem fails to hold. 5

6 xercise 5. Show by example that the Riesz representation theorem is not necessarily true for p = 1. nswer: We look again at the counting measure on [0, 1]. Let B = { [0, 1] : or c is countable}. This collection is a σ-algebra. Let µ be the counting measure on B, i.e. µ() equals the number of elements in if is finite and µ() = otherwise. Note that µ is not σ-finite because sets of finite measure must have finitely many elements and a countable union of finite sets is countable. But [0, 1] is uncountable. We then fix f L 1 (µ). Now f dµ = sup ϕ dµ where ϕ f is a non-negative simple function. Since ϕ dµ < for any such simple function, then ϕ must take each of its non-zero values only on a finite set. So the support of ϕ is finite. Since we can take a sequence of such simple functions (ϕ n ) so that ϕ n f pointwise, then f must have a countable support. So we can write f L 1 (µ) = f(x) as µ is the counting measure. Define then T : L 1 (µ) R by T (f) = xf(x). This is a linear functional that is bounded because T (f) = xf(x) x f(x) f(x) = f L 1 (µ) for all f L 1 (µ), so T 1. ssume towards contradiction that the Riesz representation theorem would hold. In other words, we would find g L (µ) so that T (f) = fg dµ for all f L 1 (µ). Let x [0, 1]. Then for χ x L 1 (µ) we have x = xχ x = T (χ x ) = χ x g dµ = g(x). In other words, g(x) = x for all x [0, 1]. But g 1 [0, 1] = [0, 1] 2 2 and this set is uncountable and its complement is also uncountable, so g 1 [0, 1 ] / B. So g is not measurable and thus we get a contradiction. 2 6

7 There is another counter example in ric M. Vestrup s book The Theory of Measures and Integration, page 413. Here is a sketch of it. Take X = [0, 1] [0, 1] and let B be the Borel subsets of X and m the Lebesgue measure on R restricted to the Borel subsets of [0, 1]. We define ν and µ on B by setting ν() = m({y [0, 1] : (x, y) }) and µ() = ν() + y [0,1] m({y [0, 1] : (x, y) }) for all B, and the uncountable sum is defined as the supremum over all finite partial sums. Since m is a measure on B one can verify that both µ and ν are measures on B and by construction ν µ. We then define T : L 1 (X, B, µ) R by T (f) = f dν. This definition makes sense because L 1 (X, B, µ) L 1 (X, B, ν) by construction. lso if f L 1 (µ) = 1 then f L 1 (ν) 1, so T 1. Hence T is bounded. Now assume towards contradiction that the Riesz Representation theorem would hold. So we find g L (X, B, µ) so that T (f) = fg dµ for all f L 1 (X, B, µ). Then (without a proof) a point x 0 [0, 1] is chosen so that 1 g(x 0 0, y) dm(y) = 0. For the set V = {x 0 } [0, 1] and f = χ V one can show that ν(v ) = 1 and f L 1 (µ). But 1 = ν(v ) = f dν = T (f) = fg dµ = g dµ g(x 0, y) dm(y) = 0, which is a contradiction. V [0,1] 7