Solutions to Tutorial 11 (Week 12)

THE UIVERSITY OF SYDEY SCHOOL OF MATHEMATICS AD STATISTICS Solutions to Tutorial 11 (Week 12) MATH3969: Measure Theory and Fourier Analysis (Advanced) Semester 2, 2017 Web Page: http://sydney.edu.au/science/maths/u/ug/sm/math3969/ Lecturer: Daniel Daners Material covered (1) The Radon-ikodym theorem (2) An application of the projection theorem (3) The Lebesgue outer measure Outcomes After completing this tutorial you should (1) to apply the Radon-ikodym theorem and associated properties of the density functions. (2) be able to apply the projection theorem in a simple situation. (3) work with the Lebesgue outer measure. Summary of essential material Let μ, ν be measures on X defined on the σ-algebra of subsets of X. We say ν has a density with respect to μ if there exists a measurable function g X [0, ] such that ν(a) = 1 A g dμ for all A. We call g the density of ν with respect to μ. We write say that ν is absolutely continuous with respect to μ if ν() = 0 whenever with μ() = 0. We write ν μ From the definition, any measure ν with a density with respect to μ is absolutely continuous with respect to μ. The Radon-ikodym Theorem characterises provides a converse. Radon-ikodym Theorem. Let μ, ν be σ-finite measures on X defined on the σ-algebra of subsets of X. Then, ν μ if and only if ν has a density with respect to μ. That density is unique (almost everywhere). Questions to complete during the tutorial 1. Let μ, ν be measures on X defined on the σ-algebra of subsets of X, and let a, b 0 be constants. For A define λ(a) = aμ(a) + bν(a). (a) Show that λ is a measure on X. Solution: We know that μ( ) = ν( ) = 0, so λ( ) = aμ( ) + bν( ) = 0. ext let A k, k be disjoint sets. Then, using the countable additivity of μ and ν we get λ(a k ) = lim Hence λ is a measure. ( ) aμ(a k ) + bν(a k ) = a lim Copyright 2017 The University of Sydney 1 μ(a k ) + b lim ν(a k ) ) ) = aμ A k + bν A k = λ A k ).

(b) Show that for every non-negative measurable function f X [0, ] we have f = a + b f dν. Then show that the above holds for f L 1 (X, μ) L 1 (X, ν). Solution: We start by a simple measurable function φ = n c k1 Ak with c k 0. By definition of the integral of a simple function φ = c k λ(a k ) = ( c k aμ(ak ) + bν(a k ) ) = a c k μ(a k )a + c k ν(a k ) = a φ dμ + b φ dν. To deal with non-negative functions f X [0, ] we use the existence of a sequence of simple functions φ n with the property that φ n φ n+1 f for all n and φ n f pointwise. From the above we have φ n = a φ n dμ + b φ n dν for all n. Applying the monotone convergence theorem to all three integrals we get f = a + b f dν. ote that there is no simple way to use the definition of the integral as the supremum over all simple functions 0 φ f as this only gives the inequality f a + b f dν. There is no simple way to prove the opposite inequality, and it is better to use the monotone convergence theorem. 2. Let μ, ν and λ be finite measures defined on the same σ-algebra of subsets of X. The Radon-ikodym theorem asserts that if ν μ then there exists a measurable function g X [0, ) such that f dν = fg dμ for all f L 1 (X, ν). The function g is often called the Radon-ikodym derivative and denoted dν dμ. We show that this is justified by proving familiar properties of derivatives. (a) If ν μ, then the function g from the Radon-ikodym theorem is essentially unique, that is, any two differ at most on a set of measure zero. Solution: Suppose that g 1, g 2 are two functions so that for all A. Hence ν(a) = g 1 dμ = g 2 dμ A g 1 g 2 dμ = 0 for all A. If h = g 1 g 2 0 on a set of positive measure, then P = {x X h(x) > 0} or = {x X h(x) < 0} has positive measure. If μ(p ) > 0, then P (g 1 g 2 ) dμ > 0 which is impossible. Hence μ(p ) = 0 and similarly μ() = 0, so g 1 = g 2 almost everywhere. 2

(b) Suppose that ν λ and μ λ. Prove that almost everywhere. d(ν + μ) = dν + dμ Solution: By the Radon-ikodym theorem there exist g 1 = dν and g 2 = dμ such that for all A. Hence ν(a) = g 1 and μ(a) = g 2 (ν + μ)(a) = ν(a) + μ(a) = (g 1 + g 2 ) for all A. By uniqueness of the Radon-ikodym derivative we have almost everywhere. d(ν + μ) (c) Suppose ν μ and μ λ. Prove the chain rule almost everywhere. = g 1 + g 2 = dν + dμ dν = dν dμ dμ Solution: By the Radon-ikodym theorem there exist functions g 1, g 2 X [0, ) such that f dν = fg 1 dμ and for all measurable functions f X [0, ). Hence in particular = fg 2 ν(a) 1 A dν = 1 A g 1 dμ = 1 A g 1 g 2 = g 1 g 2 for all A. By uniqueness almost everywhere of the Radon-ikodym derivative we have as claimed. (d) Suppose ν μ and μ ν. Prove that almost everywhere. dν dμ dμ g 1g 2 = dν [ dν dμ ] 1 dμ = dν Solution: By the Radon-ikodym theorem there exists a function g X [0, ) such that f dν = fg dμ for all measurable functions f X [0, ). If we apply the above identity to f replaced by the non-negative measurable function f g we get f g dν = for all measurable functions f X [0, ). By the uniqueness of the Radon-ikodym derivative dν dμ = g = 1 [ dμ 1 g = dν 3 ] 1

3. Every non-negative f L 1 (R ) can be identified with the measure μ f on R having density f with respect to the Lebesgue measure. Then, μ f (R ) = f(x) dx = f 1 <, so μ f is a finite measure on R. Using this notation, the Fourier transform of f can be written in the form R e 2πix t f(x)dx = e 2πix t dμ f (x). It makes sense to replace μ f by an arbitrary finite Borel measure μ on R. We define its Fourier transform by μ(t) = e 2πix t dμ(x). (a) Let μ be a finite Borel measure on R. Show that μ C(R ), and that μ μ(r ). Solution: The map t e 2πx t is continuous on R, and e 2πix t = 1 for all x, t R. As R 1 dμ = μ(r ) < we can apply the theorem on the continuity of integrals with parameters to conclude that μ C(R ). Also, μ(t) e 2πix t dμ = 1 dμ = μ(r ) for all t R. Hence μ μ(r ) as claimed. (b) Let f L 1 (R ) and let μ be a finite Borel measure on R. Show that R f(t) μ(t) dt = f(x) dμ(x). You may assume that Fubini s theorem applies to all finite Borel measures on R. Solution: By the definition of the Fourier transform, and interchanging the order of integration (using Fubini s theorem) R f(t) μ(t) dt = f(t) e 2πix t dμ(x) dt We can apply Fubini s theorem since = f(t)e 2πix t dt dμ(x) = f(x) dμ(x) R f(t)e 2πix t dt dμ(x) = 1 dμ(x) f(t) dt = μ(r ) f R R 1 <. (c) Given a finite Borel measure μ on R and a point x 0 R define the translation of the measure by x 0 by (τ x0 μ)(a) = μ(a x 0 ) for every Borel set A R. Then τ x0 μ is again a finite Borel measure on R. Show that τ x0 μ(t) = e 2πix 0 t μ(t), similar to the Fourier transform of a translated function. Solution: We have τ x0 μ(t) = e 2πix t d(τ x0 μ) = e 2πi(x+x0) t dμ = e 2πix 0 t e 2πix t dμ = e 2πix 0 t μ(t). R 4

as claimed. Here we used that for any integrable function f we have R f(x) d(τ x0 μ) = f(x x 0 ) dμ. To see this we can first look at indicator functions f = 1 A. We have by definition of τ x0 μ ( ) R 1 A (x) d(τ x0 μ) = (τ x0 μ)(a) = μ(a x 0 ) = 1 A (x x 0 ) dμ. Hence ( ) is valid for indicator functions, and hence for simple functions. Approximating nonnegative measurable functions from below by a sequence of simple functions, ( ) follows by applying the monotone convergence theorem. ow consider positive and negative parts, then real and imaginary parts. (d) Let f L (R ) and let μ be a finite Borel measure on R. We define the convolution of f with μ by (f μ)(x) = f(x y) dμ(y). Let now φ L 1 (R ) and set ψ(x) = φ( x). Show that R μ(t)φ(t)e 2πix t dt = (ψ μ)(x) Solution: First note that μφ L 1 (R ) because it is a product of a bounded and an integrable function. Using Questions (c) and then (b) we obtain R μ(t)φ(t)e 2πix t dt = τ x μ(t)φ(t) dt = φ(t) d(τ x μ)(t) = φ(t x) dμ(t) = φ( (x t)) dμ(t) = (ψ μ)(x) as claimed. *(e) How would you derive an inversion formula for the Fourier transform of a finite Borel measure μ using the formula in (d)? Solution: One would need to choose a sequence φ n so that φ n is an approximate identity. We can use the same as in the inversion formula, namely φ n (x) = e π x 2 n 2 and so ψ n (x) = n e π nx 2 is an approximate identity. The question is whether in some sense ψ n μ converges to μ. Since μ is a measure it does not make sense to consider any form of pointwise convergence. We may consider ψ n μ to be the density of a measure with respect to the Lebesgue measure. We may expect that (ψ n μ)(x) dx = 1 A A (x)(ψ n μ)(x) dx μ(a) ( ) as n for every Borel set A R of finite measure. We can do that by using Fubini s theorem to write = 1 A (x)(ψ n μ)(x) dx = 1 A (x) ψ n (x y) dμ(y) dx = 1 A (x)ψ n (x y) dx dμ(y) = (1 R A ψ n )(y) dμ(y), where we also used that ψ n (x y) = ψ n (y x). From lectures we know that 1 A ψ n 1 A in L 1 (R ), and hence a subsequence converges pointwise almost everywhere. In the case of convolution one can show that the convergence is in fact almost everywhere for the whole sequence. We also have (1 A ψ n 1 A )(x) ψ n 1 = 1, so 1 is a dominating function integrable with respect to the finite measure μ. By the dominated convergence theorem ( ) follows. Hence, we have a Fourier inversion formula for finite measures as well, similar to the one we know already. 5

Extra questions for further practice 4. Let μ be a probability measure defined on the σ-algebra of subsets of X. Suppose that A is such that μ(a) (0, 1). Let f L 2 (X) and define g = A 1 μ(a) A + A c μ(a c 1 ) A c. Let M be the subspace of L 2 (X) spanned by 1 A and 1 A c. Show that g is the orthogonal projection of f onto M. Solution: We only need to show that (f g 1 A ) = (f g 1 A c) = 0. Then (f g h) = 0 for all linear combinations of 1 A and 1 A c. By definition of the L 2 -inner product (f g 1 A ) = (f g)1 A dμ = (f g) dμ = f A 1 μ(a) A + A c μ(a c 1 ) A c dμ = A dμ = A μ(a) = 0. μ(a) μ(a) Similarly (f g 1 A c) = 0, completing the proof. 5. Suppose that (X,, μ) is a measure space such that μ(x) <. Let f X [0, ) be a measurable function and set Φ(t) = e tf(x) dμ(x). (a) Show that Φ C([0, ), R), that is, Φ is a continuous function of t [0, ). Solution: We verify the condition of the theorem on the continuity of integrals with parameters: By continuity of the exponential function, the map t e tf(x) is continuous for all x X. The integrand is a function of the form φ f with φ(s) = e ts continuous. Hence, for every fixed t R the function x e tf(x) is measurable, as shown in lectures (Lecture notes Prop. 9.8). Because f 0 we have 0 e tf(x) 1 for all x X. As μ(x) is finite, the constant function g(x) = 1 is an integrable dominating function. The theorem on the continuity of integrals with parameters now implies that Φ is continuous as a function of t [0, ). (b) Suppose that f L m (X) for some m, where m 1. Prove that d k Φ dt k (0) = ( 1)k f k k for k = 1,, m, where the derivative at zero is to be understood as a right derivative. Solution: We give a proof by induction, using the theorem on the differentiation of integrals with parameters. For k = 1 we have t e tf(x) = f(x)e tf(x) is continuous as a function of t and measurable as a function of x, as in the previous part, using also that the product of measurable functions is measurable. As f 0 we also have t e tf(x) f(x) By assumption f L 1 (X) and k 0, so we can use f(x) as a dominating function. 6

By the theorem on the differentiation of integrals with parameters we obtain dφ dt (t) = f(x)e tf(x) dμ(x) for all t 0. Setting t = 0 we obtain Φ (0) = f 1. As μ(x) < we have L m (X) L p (X) for all p [1, m]. Assume that f L m (X) and that the formula holds for some 1 k < m. As f 0 we have k+1 t k+1 e tf(x) k t k e tf(x) = ( 1)f(x) k+1 e tf(x) By assumption f L m (X) L k+1 (X), and f 0 so we can use f(x) k+1 as a dominating function. By the theorem on the differentiation of integrals with parameters we obtain d k+1 Φ (t) = ( 1) k+1 f(x) k+1 e tf(x) dμ(x) dt k for all t 0. Setting t = 0 we obtain Φ (k+1) (0) = ( 1) k+1 f k+1. as claimed. k+1 We do an induction because before we can compute the (k + 1)st derivative we need to know that the k-th derivative exists. (c) Assume that there exists ε > 0 such that Φ(t) < for all t ( ε, ε). Show that f p < for all 1 p <. Solution: By assumption, for εt = s < 0 the integral e sf(x) s k f(x) k dμ(x) = dμ < k! for all s (0, ε). As f 0 the monotone convergence theorem allows us to interchange the integral and the series, so e sf(x) s dμ(x) k f(x) k s dμ(x) = k k! k! f k k < for all s (0, ε). In particular f k < for all k. As L p (X) L q (X) whenever p > q if μ(x) is finite, we conclude that f p < for all p [1, ). Challenge questions (optional) The question below generalises the Radon-ikodym theorem and shows that one can finde a decomposition μ = μ a + μ s called the Lebesgue decomposition of a measure μ into a (maximal) absolutely continuous part μ a and a singular part μ s with respect to ν. In particular it shows the existence of a set P such that μ s (P ) = μ a (X P ) = 0. If that is the case we often say μ a μ s because this is reminiscent of orthogonal functions having support on disjoint sets. This question is inspired by the short recent proof of the Lebesgue decomposition theorem from Tamás Titkos. A Simple Proof of the Lebesgue Decomposition Theorem. In: Amer. Math. Monthly 122.8 (2015), pp. 793 794. DOI: 10.4169/amer.math.monthly.122.8.793. 6. Let be a σ-algebra of subsets of X and let μ, ν be finite measures defined on. Define (X) = { φ X R φ measurable and ν ( {x X φ(x) 0} ) = 0 } to be the vector space of measurable functions that are zero ν-almost everywhere. For A let μ a (A) = inf 1 φ dμ and μ φ (X) s (A) = μ(a) μ a (A). A 7

(a) Show that μ a (A) μ(a) for all A. Solution: We know that 0 (X) and hence for all A. μ a (A) = inf 1 φ dμ 1 0 dμ = 1 dμ = μ(a) φ (X) A (b) Let A, B be disjoint sets. The aim is to show that μ a (A B) = μ a (A) + μ a (B) by proving two inequalities. (i) Use the definition of an infimum to show that μ a (A B) μ a (A) + μ a (B). Solution: Using A B = and the definition of an infimum A B 1 φ dμ = 1 φ dμ + 1 φ dμ μ a (A) + μ a (B) (ii) for all φ (X). Hence μ a (A) + μ a (B) is a lower bound for A B 1 φ dμ for all φ (X). By definition of an infimum we conclude that μ a (A B) μ a (A) + μ a (B). Briefly explain why A B 1 (1 A φ 1 + 1 B φ 2 ) dμ = 1 φ 1 dμ + 1 φ 2 dμ for all φ 1, φ 2 (X) and hence show that μ a (A B) μ a (A) + μ a (B). Solution: Let φ 1, φ 2 (X). Since A B = we have 1 A (1 A φ 1 + 1 B φ 2 ) = 1 A φ 1 (X) and 1 B (1 A φ 1 + 1 B φ 2 ) = 1 B φ 2 (X). Hence A B 1 (1 A φ 1 + 1 B φ 2 ) dμ = 1 φ 1 dμ + 1 φ 2 dμ. (1) Also note that 1 A φ 1 + 1 B φ 2 (X). Taking an infimum on the left hand side of (1) we see that μ a (A B) 1 φ 1 dμ + 1 φ 2 dμ for all φ 1, φ 2 (X). We therefore have, for every fixed φ 2 (X), μ a (A B) 1 φ 2 dμ 1 φ 1 dμ for all φ 1 (X). Taking the infimum over all such φ 1 we obtain μ a (A B) 1 φ 2 dμ for every φ 2 (X). Rearranging we have inf 1 φ φ 1 (X) 1 dμ = μ a (A) A μ a (A B) μ a (A) 1 φ 2 dμ for all φ 2 (X). Taking an infimum over all φ 2 (X) we see that μ a (A B) μ B (A) μ a (A), which is equivalent to the required inequality. (c) Using the previous parts, show that μ a is a measure defined on the σ-algebra. To do so remember that we assume that μ(x) <. Solution: By part (a) and the definition 0 μ a ( ) μ( ) = 0, so that μ a ( ) = 0. To show the countable additivity let A k, k, be disjoint. Then also the union A k = A 1 A 2 A n 8 A k )

is a finite disjoint union. Hence, by part (b) and induction we have finite additivity and so ) ) μ a A k = μ a (A k ) + μ a A k (2) for all n. Using part (a) and the fact that μ is a finite measure we see that We therefore conclude that ) ) μ a A k μ A k = μ(a k ) μ(x) <. ) ) 0 lim μ a A k lim μ A k = lim Letting n in (2) therefore implies the countable additivity (d) Show that μ a ν. ) μ a A k = μ a (A k ). μ(a k ) = 0. Solution: If A and ν(a) = 0, then by definition 1 A (X). Hence by definition of μ a we have and so μ a (A) = 0. Therefore μ a ν. μ a (A) 1 1 A dμ = 0 (e) Let μ 1 be a measure on such that μ 1 μ and μ 1 ν. Show that μ 1 μ a, that is, μ 1 (A) μ a (A) for all A. Solution: Let A and fix φ (X). As μ 1 ν we have μ 1 ( {x X φ(x) 0} ) = 0 for all φ (X). As μ 1 μ we therefore have μ 1 (A) = 1 dμ 1 = 1 φ dμ 1 1 φ dμ for all φ (X). By definition of an infimum we obtain μ 1 (A) μ a (A) as claimed. (f) Suppose that η is a measure on such that η μ a and η μ s. Show that η = 0, that is, η(a) = 0 for all A. Solution: Clearly μ a + η μ a + μ s = μ. Assume that (X) with ν() = 0. Then by (d) we have μ a () = 0 and hence by assumption η() μ a () = 0. This implies that μ a () + η() = 0, showing that μ a + η ν. It follows from (e) that μ a + η μ a and hence η = 0. 9