Measure Theory on Topological Spaces. Course: Prof. Tony Dorlas 2010 Typset: Cathal Ormond

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1 Measure Theory on Topological Spaces Course: Prof. Tony Dorlas 2010 Typset: Cathal Ormond May 22, 2011

2 Contents 1 Introduction The Riemann Integral Measurable Abstract Measure Theory Rings, Algebras and Measures σ-algebra and Measure Space Completeness Measurable Integration The Integral Integrability Convergence Theorems Functional Analysis L p Spaces Compactness Radon Measures Introduction Measurablity and Continuity Properties of Radon Measures Lebesgue Measure on R n Introduction Properties of µ L Product Measures Introduction Properties Local Compactness Riesz-Markov-Saks-Kakutani Theorem Signed Measures Vector Valued Measures Signed and Radon Measures

3 Chapter 1 Introduction 1.1 The Riemann Integral We recall that the integral b f(x dx exists, in the Riemann sense of the word, if the following two extrema a exist and are equal: { n ([ ] } S(f = sup min f(x [t i t i 1 ] P t i 1 x t i S(f = inf P { n ([ max f(x t i 1 x t i ] } [t i t i 1 ] where P = {a = t 1 < t 2 < < t n = b t 1,..., t n R, n N} is the set of all partitions of the interval [a, b]. We note the following theorem: Theorem If f : [a, b] R is continuous, then f is Riemann integrable. However, f(x = 1 x on [0, 1] has a well defined integral, even though it s not defined at 0, because: 1 0 dx x = [2 x] 1 0 = 2 Obviously, the upper sums aren t defined, but the lower sums are, and gives us the answer. We can also define the integral as: 1 [ dx 1 { } ] 1 := lim min x, n dx 0 x n 0 The problem lies in subdividing the x-axis. If we subdivide the y-axis into n intervals instead, we ll be able to integrate more functions (Henri Lebesgue s idea. Let s consider some function f : [a, b] R, and define A k to be the set of points x [a, b] for which f(x crosses the k th interval. Hence, we have the approximate integral of f as f c k l(a k where c k is the smallest value of f in the k th interval, and l is the length of a set. 2

4 1.2. MEASURABLE CHAPTER 1. INTRODUCTION 1.2 Measurable In this example, the length function is simple, as we re dealing with intervals on the real line. We ll define the upper measure of A [a, b] R as: { n } n m (A = inf I k A I k where I k are intervals on the real line. The lower measure is given by: m (A = (b a m ([a, b] \ A If m (A = m (A, then we say that A is measurable, an that the measure of A is given by m(a = m (A = m (A. 3

5 Chapter 2 Abstract Measure Theory Henceforth, let Ω be a fixed, arbitrary set. 2.1 Rings, Algebras and Measures Definition (Rings and Algebras. A collection R of subsets of Ω is called a Ring if the following hold: If A, B R, then A B R If A, B R, then A \ B R R is called an Algebra if Ω R also. 2.2 Definition: σ-algebra and Measure Space Definition (σ-algebra. An algebra A of subsets of Ω is called a σ-algebra if for all A 1, A 2, A, then A n A. We say that (Ω, A is a Measure Space if A is a σ-algebra on Ω. Definition (Measure. If (Ω, A is a measure space, a (positive Measure on (Ω, A is a mapping µ : A [0, ] which satisfies: ( µ A n = µ(a n for A 1, A 2, A mutually disjoint sets. For short hand, if A 1, A 2,... are mutually disjoint, we ll use the notation: A n = Note that =, which implies that µ( = 0. We will now use the notation that (Ω, A, µ is a measure space. Proposition Let µ be a measure on (Ω, A. Then: ( n µ A k = µ(a k where n N. A n 4

6 2.3. COMPLETENESS CHAPTER 2. ABSTRACT MEASURE THEORY If A, B A and A B, then µ(a µ(b. If A, B A, then µ(a B µ(a + µ(b If A 1, A 2, A, then ( µ A n µ(a n where the sets need not be mutually disjoint. If A 1, A 2, A with A n A n+1 for all n N and A = Proof. A n, then µ(a = lim µ(a n = sup µ(a n n Letting A n+1 = A n+2 = = and noting that µ( = 0 gives the desired result from the definition. n N B = A (B \ A, so µ(b = µ(a + µ(b \ A µ(a, as required. A B = (A \ B B, which gives µ(a B = µ(a \ B + µ(b. However, A \ B A, and so µ(a \ B µ(a. This gives µ(a B µ(a + µ(b. ( n 1 Let B n = A n \ A k and B 1 = A 1. These sets are clearly disjoint, so we have: as required. ( A k = B k µ A k = µ(b k µ(a k Let B n = A n \ A n 1 and B 1 = A 1. Again, we have This gives us: as required. µ(a = µ(b k := lim N A = A k = N B k µ(b k = lim N µ ( N B k = lim N µ(a N 2.3 Completeness Definition (Null Set. Given a measure space (Ω, A, µ, a set A A is called a Null Set or Negligible if µ(a = 0. Definition (Almost Everywhere. Given a measure space (Ω, A, µ, if a property holds for all x Ω except for all x N Ω where N is null, then we say that the property holds Almost Everywhere (a.e.. 5

7 2.4. MEASURABLE CHAPTER 2. ABSTRACT MEASURE THEORY Definition (Complete. A measure space (Ω, A, µ is called Complete if for all A B A with µ(b = 0, we have A A. Theorem If (Ω, A, µ is a measure space, then define: A = {A Ω A 1, A 2 A s.t. A 1 A A 2, µ(a 2 \ A 1 = 0} µ (A = µ(a 1 = µ(a 2 Then, (Ω, A, µ is a complete measure space, called the Lebesgue Completion of (Ω, A, µ. Proof. It is straightforward to prove that A is a σ-algebra. We ll need to show that µ is well defined, namely if we have A Ω, A 1, A 2, B 1, B 2 A, with A 1 A A 2, B 1 A B 2 and µ(a 2 \ A 1 = µ(b 2 \ B 1 = 0, then µ(a 1 = µ(b 1. If we assume this situation, we have: B 1 A A 2 µ(b 1 µ(a 2 = µ(a 1 A 1 A B 2 µ(a 1 µ(b 2 = µ(b 1 so µ(a 1 = µ(b 1 and hence µ is well defined. Now we ll show that µ is a measure. Let A n A for all n N be disjoint. Then there exists A n1, A n2 such that A n1 A n A n2 with µ(a n2\a n1 for all n N. This gives A n This gives And so Now we have A n1 }{{} disjoint A n2 }{{} not necessarily disjoint µ(a n1 = µ(a n A n2 ([ ] [ ] µ A n2 \ A n1 µ(a n2 \ A n1 = 0 ( n ( n µ A n = µ A n1 = µ(a n1 = µ (A To show that the space is complete, let A B A and µ(b = 0. This means that there exists some B 2 A such that B B 2 and µ(b 2 = 0. However, A B 2 and µ(b 2 = 0 also, so A A, as required Measurable Definition (Measurable Function. Given a measure space (Ω, A, µ, a function f : Ω R is called Measurable if for all y R, we have Note: R := R {, + } {x Ω f(x y} A Proposition Given a measure space (Ω, A, µ, a function f : Ω R is measurable if any of the following hold for all y R: 6

8 {x Ω f(x y} = n N 2.4. MEASURABLE CHAPTER 2. ABSTRACT MEASURE THEORY {x Ω f(x y} A {x Ω f(x < y} A {x Ω f(x > y} A {x Ω f(x y} A Proof. Note that for all A A, A A also, so the first and third statements are equivalent, as the two sets are complements of each other. Similarly, the second and fourth statements are equivalent. The first statement is simply the definition of measurable. We note that we may write the sets as follows: {x Ω f(x y} = { x Ω f(x y + 1 } n n N {x Ω f(x + 1n y } as required. Notation Henceforth, given a measure space (Ω, A, µ, the set of all measurable functions from Ω to R will be denoted M. Proposition Let (Ω, A, µ be a measure space. f, g M implies that f + g M f M and λ R implies that λf M If f n M for all n N, then inf f n M and sup f n M n N f, g M and f, g : Ω R imply that fg M. Proof. Let y R be fixed n N Note that if f(x < y g(x, then there exists some r Q such that f(x < r < y g(x. This gives us {x Ω f(x + g(x y} = [{x Ω f(x r} {x Ω g(x y r}] r Q the latter of which is a countable union of sets of A, as required. We ll consider three cases: 1. λ = 0: then λf = 0, where we take the convention that 0 (± = λ > 0: This gives {x Ω λf(x y} = {x Ω f(x y λ } A 3. λ < 0: This gives {x Ω λf(x y} = {x Ω f(x y λ } A as required. 7

9 2.4. MEASURABLE CHAPTER 2. ABSTRACT MEASURE THEORY We note that we can express the extrema of the functions as follows: {x Ω inf f n(x < y} = {x Ω f n (x < y} n N n N {x Ω sup f n (x y} = {x Ω f n (x y} n N n N which are both countable unions and intersections of sets in A, as required. Firstly, we note that f = sup{f, f}, so f is measurable whenever f is measurable. Also, f 2 M when f M, because: if y < 0, then { x Ω f 2 (x < y } = A if y > 0, then { x Ω f 2 (x < y } = { x Ω f(x < y } A Noting that fg = 1 4 [(f + g2 (f g 2 ] gives us the desired result. Note that in the fourth section, if we had f = g = ±, then (f g would be undefined, so we need to have f ±, g ±. Corollary If f n M for all n N with f n f a.e. (except on N, and the measure space is complete, then f M. Proof. For x / N, we have f(x = lim f n(x = lim sup n n As sup f k (x is measurable, we conclude that k n { f n (x := lim n sup f k (x k n {x Ω \ N f(x < y} A } { } = inf sup f k (x n N k n However, {x N f(x < y} N {x N f(x < y} A as the space is complete. Hence, f M, as required. 8

10 Chapter 3 Integration 3.1 The Integral Definition (Simple Functions. Given a measure space (Ω, A, µ, a function g : Ω R is called Simple if it is of the form: g = c k 1 Ak where c 1,..., c n R and A 1,..., A n A are disjoint. We define the integral of the above function with respect to µ as g dµ = c k µ(a k Let S denote the set of all simple functions of a measure space, and let S + be the subset of all positive simple functions. Lemma Suppose that g S + can be written Then, Proof. We ll assume that This gives us the following: g = m c k 1 Ak = d l 1 Bl l=1 m c k µ(a k = d l µ(b l l=1 m A k = B l = Ω. We also note that A k = l=1 m (A k B l B l = l=1 c k µ(a k = (B l A k ( m c k µ(a k B l = (k,l I A k B l l=1 c k µ(a k B l 9

11 3.1. THE INTEGRAL CHAPTER 3. INTEGRATION where I N 2 is the set {1, 2,..., n} {1, 2,... m}. If x A k B l, then we have f(x = c k = d l, so as required. c k µ(a k = (k,l I A k B l = = = (k,l I A k B l c k µ(a k B l d l µ(a k B l ( m n d l µ(a k B l l=1 m d l µ(b l Proposition Let (Ω, A, µ be a measure space. We have that: if f, g S +, then f + g S + and (f + g dµ = f dµ + g dµ if f S + and λ R with λ 0, then λf S + and λf dµ = λ f dµ. if f, g S + and f g, then f dµ g dµ. These may be proved be direct application of the previous definitions and lemma. l=1 Definition (Integral (Non-Negative, Measurable. Let (Ω, A, µ be a measure space and f : Ω [0, ] be non-negative and measurable. Then we define the Integral of f with respect to µ as: { } f fµ = sup g dµ g f g S + Lemma Suppose that f : Ω [0, ] is measurable and g n S for all n N satisfy g n f, then if g S + and g f, we have: g dµ lim g n dµ n Proof. Firstly, assume that g dµ =. Then, there exists some A k with µ(a k = and c k 0, where For all n N and ɛ > 0, define g = c i 1 Ai B n = {x Ω g n (x > g(x ɛ} We have that (B n A k A k as n, because g n f and as g f, we have f > g ɛ. This implies that µ(b n A k µ(a k =. Thus: g n dµ > g n 1 Bn A k dµ > (g ɛ1 Bn A k dµ = (c k ɛ µ(b n A k 10

12 3.2. INTEGRABILITY CHAPTER 3. INTEGRATION Thus, taking the limit of both sides, we have: = g dµ lim n g n dµ = where we take ɛ < c k, as required. Now, let us consider the case that the integral of g is finite. This means that µ(a k < for all k = 1,..., n (except if c k = 0, but we can ignore this case. Define B n as before. This gives: gn dµ > = g n 1 Bn A k dµ (g ɛ1 Bn A k dµ (c k ɛ µ(b n A k (c k ɛ µ(a k = g dµ ɛ µ(a k Letting ɛ 0 gives the desired result, as the final sum is finite. Proposition If f, g 0 are measurable, then (f + g dµ = f dµ + g dµ If λ 0 and f 0, f measurable, then (λf dµ = λ f dµ If 0 f g are measurable, then f dµ g dµ These may be proved by applying limits to the above proposition and lemma. 3.2 Integrability Definition (Integrable Functions. A function f : Ω R is called (µ-integrable (or (µ- Summable if: f is measurable f dµ < + In this case, we define f dµ = f + dµ + f dµ where: f + (x = { f(x f(x 0 0 f(x 0 A function f : Ω C is called Integrable if } f (x = { f(x f(x 0 0 f(x 0 } f is measurable, i.e. R(f and I(f are measurable f dµ < + 11

13 3.2. INTEGRABILITY CHAPTER 3. INTEGRATION and we define: f dµ = R(f dµ + i I(f dµ We ll denote by L 1 (Ω, A, µ or simply L 1 the set of all integrable functions f : Ω C. Lemma If f 1, f 2 0 are measurable and f = f 1 f 2 L 1, then f dµ = f 1 dµ f 2 dµ Proof. f = f 1 f 2 = f + f, and so f 1 + f = f + + f 2 0. This gives f 1 dµ + f dµ = f + dµ + f 2 dµ And so, rearranging, f dµ = f + dµ f dµ = f 1 dµ f 2 dµ as required. Theorem If f, g L 1, then f + g dµ = f dµ + g dµ If λ C and f L 1, then λf dµ = λ dµ If f L 1, then f L 1 and f dµ f dµ Proof. We ll consider two sub-cases: Assume that f, g L 1 R. Then we have f + g = (f + g + (f + g = (f + f + (g + g = (f + + g + (f + g The previous lemma gives us: (f + g dµ = (f + g + dµ (f + g dµ = (f + + g + dµ (f + g dµ = f + dµ + g + dµ f dµ = f dµ + g dµ g dµ If f, g L 1 C, then we have: (f + g dµ = = = R(f + g dµ + i I(f + g dµ R(f dµ + i I(f dµ + R(g dµ + i f dµ + g dµ I(g dµ 12

14 3.2. INTEGRABILITY CHAPTER 3. INTEGRATION We ll consider two subcases: Assume that f L 1 R. Then, we consider λ R: λ = 0: this implies that (λf + = λf + = 0 = (λf, so both sides are zero. λ > 0: this implies that (λf + = λf + and (λf = λf. Thus: (λf dµ = λf + dµ = λf dµ = λ f dµ λ < 0: this implies that (λf + = λf and (λf = λf +. Thus: (λf dµ = ( λf dµ ( λf + dµ = λ f dµ as required. If f L 1 and λ C, then we have: (λf dµ = R(λf dµ + i I(λf dµ = [R(λR(f I(λI(f] dµ + i [R(λI(f + I(λR(f] dµ = R(λR(f dµ I(λI(f dµ + i R(λI(f dµ + i I(λR(f dµ [ ] [ ] = R(λ R(f dµ + i I(f dµ + ii(λ R(f dµ + i I(f dµ [ ] = [R(λ + ii(λ] R(f dµ + i I(f dµ = λ f dµ We know that f L 1, by above. We l consider two cases: If f L 1 R, then f dµ = f + dµ f dµ f + dµ + f dµ = f dµ If f L 1 C, then define ( α = f dµ f dµ where we assume the integral is non-zero. Note that α = 1. This gives us: f dµ = α f dµ ( = R α f dµ ( = R αf dµ = R(αf dµ R(αf dµ αf dµ = α f dµ = f dµ 1 13

15 3.3. CONVERGENCE THEOREMS CHAPTER 3. INTEGRATION as required. Proposition Let f L 1 and suppose that g is measurable and f = g a.e. Then g L 1 and f dµ = g dµ. Proof. Define h = f g. Then, we have R(h = R(f R(g = 0 a.e. Similarly, I(h = 0 a.e. Assume that g n h, where g n S + for all n N. Then, h dµ = lim g n dµ n where g n = gn dµ = 0, as required. N c nk A nk = 0 a.e. This means that either c k = 0 or A k = 0 for all k = 1,..., N. Thus, Corollary If f, g L 1 and f g a.e., then f dµ g dµ Proof. Consider h = g f 0 and k = h 1 {f g} Lemma If f M, A A, then f 1 A is measurable, where f : Ω R. Proof. Omitted Note We ll use the notation: f 1 A dµ = A f dµ Proposition If f M, A A and f(x > 0 for all x A, then f dµ = 0 µ(a = 0 Proof. Suppose that A f dµ = 0, and define D n = {x A f(x = 1 n }. Then: 0 = f dµ f dµ 1 A D n n µ(d n µ(d n = 0 But D n A, so µ(a = 0. A Conversely, if µ(a = 0, then f1 A = 0 a.e., so the result follows from the above proposition. 3.3 Convergence Theorems Theorem (The Monotone Convergence Theorem. Suppose that (f n is a sequence of non-negative, measurable functions which converges monotonically a.e. to a function f. Then f L 1 iff ( f n dµ is bounded, and in either case: f dµ = lim f n dµ n 14

16 3.3. CONVERGENCE THEOREMS CHAPTER 3. INTEGRATION Proof. We ll use what is called the diagonal argument. For all n N, choose a sequence (g (n k S + so that g (n k Define h (n k f n as k. Note that g (j k f j for all j N. = sup j k g (n j. This sequence is increasing in both n and k. This gives us: g (n k h (n k h (k k sup f j f k j k where k n. Note that lim k h(k k = f and h (k k dµ f k dµ. This gives: lim k h (k k dµ lim k f k dµ But, h (k k is simple, so As f k f for all k, we have lim k h (k k dµ = lim k f k dµ f dµ f dµ which, on taking limits on both sides, gives the desired result. Note The idea behind the argument is to construct the following sequences: g (1 1 g (1 2 g (1 3 f 1 g (2 1 g (2 2 g (2 3 f 2 g (3 1 g (3 2 g (3 3 f f There is no guarantee that the diagonal sequence g (k k will converge to f, so we define our h (n k supremum of all the elements in a column, so that it will converge to f. Corollary If f n 0 for all n N, then f n dµ = f n dµ to be the Theorem (Dominated Convergence Theorem. Let f n L 1 for all n N, and suppose that f n f a.e. Assume that there exists some g L 1 with g 0 such that f n g a.e. for all n N. Then: f dµ = lim f n dµ n Proof. Define ψ n = sup f k f 0. Then ψ n 0 a.e. In particular, we have (ψ 1 ψ n ψ 1 a.e. By the k n MCT, we have ( ( (ψ 1 ψ n dµ ψ 1 dµ 15

17 3.3. CONVERGENCE THEOREMS CHAPTER 3. INTEGRATION If ψ 1 dµ is finite then we are (almost done. ψ 1 = sup f k g sup{ f k + f } 2g k 1 k 1 as g bounds f. Thus, ψ 1 dµ 2g dµ < +. We note that ψ n dµ 0 as n, and so fn f dµ 0. Thus f dµ f n f dµ + f n dµ Finally, f n dµ f dµ f n f dµ 0 as required. 16

18 Chapter 4 Functional Analysis 4.1 L p Spaces Definition (L p Spaces. Define the L p Space on the measure space (Ω, A, µ for all p 1 as follows: L p (Ω, A, µ = {f : Ω C f M, f p L 1 } For f L p, we define the p- norm by: [ N p (f = ] 1 f p p dµ A sequence (f n is said to converge in p-norm to f if lim N p(f n f = 0 n Theorem (Hölder s Inequality. If f L p and g L q where 1 p + 1 q = 1, then fg L1, and fg dµ N p(f N q (g Note that when p = q = 2, this is called Cauchy-Schwarz Inequality. Proof. We note that as we may multiply by constants on both sides, we can safely assume that N p (f = N q (g = 1. We must now show that fg dµ 1 The proof relies on the fact that log x is a concave function, i.e. for all a, b 0 with a > b, we have log(λa + (1 λb λ log a + (1 λ log b where λ [0, 1]. The proof of this will be omitted. If we let λ = 1 p, and hence 1 λ = 1 q, we get ( a log p + b 1 q p log a + 1 q log b = log(a 1 1 p b q which gives us that a 1 p b 1 q a p + b q 17

19 4.1. L P SPACES CHAPTER 4. FUNCTIONAL ANALYSIS Now, setting a = f p and b = g q, we have fg 1 p f p + 1 q g q, and so as required. fg dµ 1 p f p dµ + 1 q = 1 p (1 1 p + 1 q (1 1 q = 1 g q dµ Theorem (Minkowski s Inequality. If f, g L p, then f + g L p and N p (f + g N p (f + N p (g. Proof. We note that f + g 2 max{ f, g }, and so f + g 2 max{ f p, g p } 2( f p + g p This gives us the following: N p (f + g p = f + g p dµ = f + g p 1 f + g dµ f + g p 1 ( f + g dµ Triangle Inequality N q ( f + g p 1 (N p (f + N p (g Hölder However, [ N q ( f + g p 1 = f + g (p 1q ] 1 q = [ f + g p ] 1 q = Np (f + g p q This gives us N p (f + g p N p (f + g p q (Np (f + N p (g Which, on dividing across by N p (f + g p q, gives us the desired result. Definition (L p Spaces. We note that under N p, L p is a semi-normed space. Thus, we define an equivalence relation on L p as follows: for all f, g L p, f g iff f = g a.e. We define the L p Space of (Ω, A, µ to be the set of all equivalence classes of this equivalence relation. We define a norm on L p as follows: for all f L p, we have f = N p (f, where f f in L p. This makes L p a normed vector space over C. Theorem (Riesz-Fischer Theorem. Suppose that (f n is a Cauchy sequence of functions in L p, i.e. the functions satisfy N p (f m f n 0 as n, m. Then, there exists a function f L p such that N p (f n f 0 as n. Proof. We only need to fund some subsequence of (f n which converges, so we relabel our sequence so the subsequence has the property N p (f n+1 f n 2 n (We ll abuse notation now, and use f n to refer to the sub-sequence, as opposed to the original sequence. Define g n = f n+1 f n and note that f n+1 = f 1 + g k 18

20 4.2. COMPACTNESS CHAPTER 4. FUNCTIONAL ANALYSIS Consider the series S n = g k. By the triangle inequality, we have N p (S n N p (g k 2 k 2 k = 1 and so S n p dµ 1 <, and thus G p dµ <, where G = lim S n In particular, G < a.e. This n implies that S n converges and also that g k converges absolutely a.e. Now define s n = g n s = lim n s n where we define s n = 0 if the sum becomes infinite on a null set. This gives us: N p (s n s p = p g k dµ k=n+1 k=n+1 g k p dµ = (G S n p dµ 0 as n. Thus, we have: as required. f = lim f n+1 = f 1 + lim n n g k = f 1 + s Corollary L p is a Banach space. Theorem L 2 is a Hilbert Space with inner product f, g = fg dµ Warning In a topological space (X, τ, continuity and closed sets may propose some problems. For example, if f : X C for for all convergent sequences (x n, (f(x n converges, f may not be continuous. Similarly, F X may not be closed even if all sequences contained in F have a limit in F. 4.2 Compactness At this point we will recall that a subset K of a metric space is compact if every sequence in K converges to a point in K. Theorem If K is a compact subset of a metric space (X, d, then for all ɛ > 0 there exists a finite n set of points x 1,..., x n K such that K B(x k, ɛ Proof. Take x 1 K as arbitrary, and then choose: k 1 x k K \ B(x j, ɛ 19

21 4.2. COMPACTNESS CHAPTER 4. FUNCTIONAL ANALYSIS If this process ends, we are done. If not, we have an infinite sequence (x n in K. As K is compact, this must converge to a point in K, say x. However, for all k l, we know that d(x k, x l ɛ, which is a contradiction. Hence, the above sequence may never be infinite. Theorem (Heine Borel. If K is a compact subset of a metric space (X, d and (O i i I is an open cover of K in X, i.e. O i X for all i I and K i I O i, then there exists a finite set {i 1,..., i p } I such that p K O k Proof. We recall the following two facts about metric spaces: closed subsets of a compact set is compact compact sets are closed and we argue by contradiction. First, we cover K by finitely many open balls as above, say: K n 1 ( B x 1k, 1 2 Assume that there does not exist such a finite subcover of the open cover. Then at least one of the sets B(x 1k, 1 2 K cannot be covered by finitely many elements of the open cover. However, this set is also compact, as it is closed. Write x 1 instead of x 1k, and cover B(x 1, 1 2 K by finitely many open balls, say ( B x 1, 1 K 2 n 2 ( B x 2k, 1 4 Now, as before, at least one of the sets B(x 1, 1 2 B(x 2 k, 1 4 K must be non-empty and hence compact. We can continue in this way to construct a sequence (x k in K such that x k B(x k 1, 2 1 k. This, however, means that d(x k, x k k, and so this sequence is Cauchy and converges to, say x K. But x O i for some i. Thus, we have B(x, 2 k O i for large enough k. However, this means that B(x k, 2 k O i for large enough k, which is a contradiction, as required.. Definition (Compact. We sat that a subset K of a topological space (T, τ is Compact if every open cover of K has a finite subcover. Definition (Hausdorff. We say that a topological space (X, τ is a Hausdorff Space if for all distinct points x, y X, there exists U, V τ such that x U, y V and U V =. This is also called the separation axiom T 2. Proposition A closed subset of a compact set in a topological space is compact. Proof. Suppose that F K is closed, where K is compact in (X, τ, and let F i I O i for an open cover of F. Thus, ( K O i F c i I 20

22 4.2. COMPACTNESS CHAPTER 4. FUNCTIONAL ANALYSIS Which means that n And so F O i, as required. p=1 ( n K O i F c p=1 Proposition If (X, τ X and (Y, τ Y are topological spaces, K X is compact and f : X Y is continuous, then f(k is also compact. Proof. Consider an open cover of f(k, say f(k i I O i. Set V i = f 1 (O i. Then we know that V i is open, and that K i I V i. Thus, there exists i 1,..., i n I such that as required. K n V ik f(k Definition (Closure. Given a subset A of a topological space (X, τ, its Closure, denoted A is defined as the smallest closed set containing A. Lemma A = {x X U τ with x U, U A } Proof. Let B denote the above set, and let x B c. Then there exists some open U with x U and U A =. This means that B c is open, and so B is closed. We know that A B, so A B. Suppose that A C with C closed. Then, for all x C there exists some U open such that x U and U C =, which means that U = C c. Thus, we have U A =, and so x B c also. We conclude that C c B c, or B C, as required. Proposition If (X, τ is Hausdorff and K X is compact, then K is closed. Proof. Suppose that K K and z K \ K. Then, for all x K, x z so there exists open sets U x, V x such that x U x and z V x with U x V x =. This gives: K = {x} x K As K is compact, there exists x 1,..., x n K such that x K U x n O ik Define This is open and z V, but V U xk implies that z K, as required. n K n V = U xk V xk n = for all k. Thus, V U xk =, and so V K =, which 21

23 4.2. COMPACTNESS CHAPTER 4. FUNCTIONAL ANALYSIS Definition (Finite Intersection Property. A collection F of subsets of a topological space (X, τ is said to have the Finite Intersection Property (FIP if the intersection of any finite sub-collection has a non-empty intersection. Proposition A subset K X is compact iff any collection of closed subsets of K with the FIP has a non-empty intersection. Proof. This is a mere restatement of the definition, taking complements. Suppose that K is compact and F is such a sub-collection. Suppose that F =. Then, F c is open and F F K F F F c K n F c k This gives us as required. n n F k = K \ Fk c K \ K = 22

24 Chapter 5 Radon Measures 5.1 Introduction Definition (Borel σ-algebra. Let (X, τ be a topological space. The Borel σ-algebra, denoted B(X is the smallest σ-algebra containing all the open sets. Definition (Radon Measure. Let (X, τ be a Hausdorff space. A Radon Measure µ on X is a measure on B(X with the following additional properties: Local Finiteness: for all x X, there exists an open set x U with µ(u < Inner Regularity: for all A B(X, we have µ(a = sup µ(k K A K compact Proposition If µ is a Radon measure, K X is compact, then µ(k <. Proof. For each x K, there exists some O x open such that x O x and µ(o x <. As K have K n O xk, and so ( n µ(k µ O xk µ(o xk < x K O k, we Proposition Let µ be a Radon measure on X and K K be compact. For all ɛ > 0, there exists some open set O such that K O and µ(o \ K < ɛ. Proof. By above, there exists some open set U such that µ(u <. This gives us, µ(u \ K <. Given some ɛ > 0, there exists (by inner regularity a compact set H U \ K such that µ(h > µ(u \ K ɛ. Let O = U \ H, which is open as H is closed. This gives us that H U \ K, or that K U \ H = O. Hence: O \ K = U \ (H K = (U \ K \ H and so as required. µ(o \ K = µ(u \ K µ(h < ɛ 23

25 5.1. INTRODUCTION CHAPTER 5. RADON MEASURES Proposition Let µ be a Radon measure on (X, τ. If (O i i I is a collection of open sets with measure 0, then ( µ O i = 0 i I ( Proof. Let O = µ O i. We know that µ(o = sup µ(k for K compact. However, we know that K O i I n K O ik. This gives us and so µ(o = 0 as required. µ(k µ(o ik = 0 Corollary Given a Radon measure µ, there exists a largest open set O with µ(o = 0. Definition (Support. Given a Radon measure µ, the complement of the largest open set with measure 0 is called the Support of µ, denoted supp(µ Theorem (Bourbaki Completion. Let µ be a Radon measure on X. The Bourbaki Completion of B(X, denoted B µ, is defined as follows: where B is the Lebesgue completion of B(X. B µ = {A X A K B K compact} Proposition B µ is a σ-algebra and µ can be uniquely extended to B µ as a Radon measure. In this way, (X, B µ, µ is a complete measure space. Proof. We know that B µ is a σ-algebra because so is B. We may extend µ by inner regularity, namely for all A B µ : µ(a = sup K K K compact µ(k Suppose that A = A n, where A n B µ for all n N. If any of these disjoint sets A n have infinite measure, then so does A. If the measure of A n is finite for all n N, then given ɛ > 0, choose K n A n compact with µ(k n > µ(a n > 2 n ɛ. We must now prove that µ(a = On the other hand, we have µ(a n. Note that as K k A k, we have ( µ(k k µ A k µ(a k = µ(a µ(a n (µ(k n + ɛ2 n = µ(k n + ɛ µ(a + ɛ 24

26 5.2. MEASURABLITY AND CONTINUITY CHAPTER 5. RADON MEASURES and since ɛ > 0 is arbitrary, we have µ(a n µ(a. On the other hand, if K A, then we have ( µ(k µ (K A n = µ(k A n µ(a n Finally, to show that the space (X, B µ, µ is complete, take A B µ with µ(a = 0 and let B A. Then B K A K and µ(a K = 0, so A K B as K is compact. Thus, B K B, and so B B µ. 5.2 Measurablity and Continuity Definition (Moderate. A Radon measure µ on a topological space (X, τ is called Moderate or σ-finite if there exists a sequence (O n n N of open sets with finite measure covering X. Proposition If µ is moderate, then B µ = B. Proof. For every n N, there exists a sequence (K m (n m=1 of disjoint, compact sets contained in O n such that ( µ O n \ = 0 m=1 To see that this is true, we choose some K (n 1 O n such that µ(o n \ K (n 1 < 1 2. Then, we choose some K (n 2 O n \ K (n 1 such that µ(o n \ (K (n 1 K (n 2 < 1 4. Continuing thus, we can construct the above sequence so that ( ( µ O n \ m=1 K (n m µ K (n m O n \ M m=1 K (n m 2 M for all M, as required. We note that the collection (K m ( (n m,n N is still countable. We will abuse notation n 1 here and replace O n by O n \ O k, and so we may assume that K m (n are all disjoint. Suppose that A B µ. Then, A K (n m A B, and so m, K (n m = m, (A K m (n B Thus, we have ( A O n \ ( A O n \ ( A O n = m, m, A K (n m K (n m m, O n \ B K (n m m, ( A K (n m B ( O n \ m, K (n m B Definition (µ-measurable. A function f : X C or f : X R on a topological space (X, τ is said to be µ-measurable if it is measurable with respect to the σ-algebra B µ. 25

27 5.2. MEASURABLITY AND CONTINUITY CHAPTER 5. RADON MEASURES Definition (Semi-Continuous and Directed. A function g : X [0, is called Upper Semi-Continuous (u.g.c if for all x X, g(x lim sup g(z. z x A function g : X [0, is called Lower Semi-Continuous (l.s.c if for all x X, g(x lim inf g(z. z x A collection (f i i I is called Downward Directed if for all i, j I, there exists some k I such that f k f i and f k f j. A collection (f i i I is called Upward Directed if for all i, j I, there exists some k I such that f k f i and f k f j. Proposition Let f : X [0, ] be µ-measurable. Then, we have: { } f dµ = sup 0 g f g is u.s.c g dµ where the support of g is compact. Proof. Note that the support of g is compact if there exists some compact set K with g(x = 0 for all x / K. A typical example of a u.s.c function is 1 F where F is closed. By definition, we have: f dµ = g n dµ If we have λ < f dµ, then there exists some g = a compact set K j A j such that so g is u.s.c, as required. lim g n f g n S + c k 1 Ak such that g dµ < λ, c k > 0 and g f. Choose c k µ(k k > λ. If the sets A j are disjoint, then so are the sets K j, and Corollary f dµ = sup f dµ K compact K Theorem Let (f i i I be a directed collection of l.s.c functions on X. If f = sup f i, then f is l.s.c., i I and f dµ = sup f i dµ i I Proof. We note, firstly that a function f is l.s.c iff for all c R, {x f(x > c} is open. Indeed, suppose that f is l.s.c. If f(x > c, then lim inf f(z > c and so there exists a neighbourhood U of x such that z U z x implies that f(z > c, and so z {x f(x > c}. If λ < f(x, then there ex- f(z λ. However, λ Conversely, suppose that {x f(x > c} is open for all c, and let x X. ists some neighbourhood U of x such that z U implies that λ < f(z, and so lim inf z x is arbitrary, and so f(x lim inf f(z. z x 26

28 5.2. MEASURABLITY AND CONTINUITY CHAPTER 5. RADON MEASURES Getting back to the Theorem, we note that as f = sup f i is l.s.c, we have i I { } {x f(x > c} = x sup f i (x > c = f i (x > c} i I i I{x is open. To show that f dµ = sup f i dµ i I suppose that λ < f dµ. We will show that sup f i dµ > λ. By the proof of the previous proposition, i I 5.2.5, there exists a finite collection of disjoint compact sets K 1,..., K n and constants c 1,..., c n such that c k 1 Kk f and c k µ(k k λ In particular, f(x c j for all x K j. By reducing each c j slightly, we may assume that f(x > c j for all x K j. However, x K j means that x O = i I O i, where we denote O i = {x f(x > c i }, and so n j n j K j O. Thus, we have that K j O k. Thus, if x K j, we have that x O k, i.e. there exists some r such that f ir (x > c j. We know that, as the collection (f i i I is upward directed, there exists some i 0 such that f i0 f ir for all r = 1,..., n. Thus, if x K j, we have f i0 (x > c j. We conclude from all this that for each j = 1,..., n there exists some i j such that K j {x f ij (x > c j }. Let i be such that f i (x f ij (x for all j = 1,..., n. It thus follows that K {x f ij (x > c j } for all j. Thus, we have: f i > c k 1 Kk f i dµ > c k µ(k k > λ as required. Theorem (Choquet s Theorem. Suppose that µ is a Radon measure on (X, τ. following properties: 1. If K 1, K 2 τ are compact with K 1 K 2, then µ(k 1 µ(k 2 2. If K 1, K 2 τ are compact, then µ(k 1 K 2 µ(k 1 + µ(k 2 3. If K 1, K 2 τ are compact with K 1 K 2 =, then µ(k 1 + K 2 = µ(k 1 + µ(k 2 Then is has the 4. If K τ is compact and ɛ > 0, then there exists an open set O such that K O and for all H O with H compact, we have µ(h µ(k + ɛ Conversely, if λ : κ(x [0, has the above four properties, where κ(x is the set of compact subsets of X, then there exists a unique Radon measure µ such that µ κ(x = λ. Proof. These four properties have already been proven. In the converse, the uniqueness of such a Radon measure is guaranteed by inner regularity. To construct such a Radon measure, we consider open sets and define µ(o = sup K κ(x K O µ(k We ll now consider some facts about open sets. 27

29 5.3. PROPERTIES OF RADON MEASURES CHAPTER 5. RADON MEASURES Definition (Upper and Lower Measures, Integrable. For an arbitrary set A X in a topological space (X, τ, we define an Upper Measure µ and a Lower Measure of some Radon measure µ as follows: Clearly, we must have µ = µ. µ (A = inf µ(o µ (A = sup O τ A O We say that an arbitrary set A X is Integrable if µ (A = µ (A < 5.3 Properties of Radon Measures K κ(x K A For the following lemmas, we ll assume that µ is some Radon measure. Many of these proofs will be trivial, and so omitted. Lemma If O 1 O 2 are open sets, then µ(o 1 µ(o 2 Lemma *** Lemma If O 1, O 2 are open, then µ(o 1 O 2 µ(o 1 + µ(o 2 ( Lemma If. ( O n n N are open and increasing, i.e. O n O n+1, then we have µ(o n µ O n and µ ( n N O n = lim n µ(o n ( ( Proof. Clearly, we have µ(o n µ O n. Suppose that λ µ O n. Then there exists some n N K n N O n with K compact such that λ < µ(k. Thus, there exists some n N such that K O n, and µ(k n N n N so we must have: λ < µ(k < µ(o n < lim n µ(o n Lemma If (O n n N is a sequence of open sets, then ( µ O n µ(o n n N N Proof. Let O N = O n = O N. Thus, O n. This gives us n N ( N µ(o N = µ O n N N N µ(o n µ(o n 28

30 5.3. PROPERTIES OF RADON MEASURES CHAPTER 5. RADON MEASURES and so as required. µ ( O n lim N µ(o N µ(o n Lemma If (A n n N is a sequence of subsets of X, then we have µ ( n N A n µ (A n Proof. If µ (A n = for some n N, then we are done, so assume that the upper measure of each of the subsets is finite. Then for all n N and given ɛ > 0, there exist open sets O n such that A n O n and Then, we have the following: µ (A n > µ(o n ɛ 2 n ( µ (A n > µ (O n ɛ 2 n ( = µ (O n ɛ ( µ O n ɛ n N µ ( n N A n ɛ The result thus follows from the fact that ɛ is arbitrary. Lemma If (A n n N is a sequence of disjoint subsets of (X, τ, then ( µ A n > µ (A n Proof. If µ (A n = for some n N, then we are done, so assume that the upper measure of each of the subsets is finite. Then for all n N and given ɛ > 0, there exist compact sets K n such that K n A n and Then, we have the following: µ (A n < µ(k n + ɛ 2 n N N ( µ (A n < µ(k n + ɛ 2 n ( N N ɛ = µ(k n + 2 n µ ( n N A n + ɛ 29

31 5.3. PROPERTIES OF RADON MEASURES CHAPTER 5. RADON MEASURES and so, we conclude that µ (A n µ ( n N The result thus follows from the fact that ɛ is arbitrary. A n + ɛ Lemma If K is compact, then K is integrable, and if O is open and µ(o <, then O is integrable. Lemma (Sandwich Criterion. A set A is integrable iff for all ɛ > 0 there exists a compact set K and an open set O with K A O and µ(o \ K < ɛ. Proof. There exists K compact with K A and µ(k > µ (A ɛ 2, and O open with A O and µ(o < µ (A + ɛ 2. This gives us that K A O and µ(o \ K = µ(o µ(k < µ (A + ɛ ( 2 µ (A ɛ < ɛ 2 as required. Lemma If A 1, A 2 are integrable, then A 1 A 2, A 2 A 2 and A 1 \ A 2 are integrable. Proof. Choose K 1, K 2 compact and O 2, O 2 open such that K 1 A 1 O 1 and K 2 A 2 O 2, and for all ɛ > 0, µ(o 1 \ K 1 < ɛ and µ(o 2 \ K 2 < ɛ. For A 1 A 2, choose K = K 1 K 2 and O = O 1 O 1, so µ(o \ K = µ((o 1 O 2 \ (K 1 K 2 = µ(o 1 \ (K 1 K 2 O 2 \ (K 1 K 2 µ(o 1 \ K 1 + µ(o 2 \ K 1 < 2ɛ For A 1 A 2, choose K = K 1 K 2 and O = O 1 O 1, so µ(o \ K = µ((o 1 O 2 \ (K 1 K 2 = µ((o 1 O 2 \ K 1 (O 1 O 2 \ (K 1 K 2 µ(o 1 \ K 1 + µ(o 2 \ K 1 < 2ɛ For A 1 \ A 2, choose K = K 1 \ O 2 and O = O 1 \ K 2, so as required. Lemma Define: Then A is a σ-algebra. µ(o \ K = µ((o 1 \ O 2 \ (K 1 \ K 2 = µ((o 1 K c 2 (K 1 O c 2 c = µ((o 1 K c 2 (K c 1 O 2 = µ((o 1 O 2 \ K 2 O 1 \ (K 1 K 2 < 2ɛ A = {A X A K is integrable for all K κ(x} If A A, then A c A also, as A c K = (X \ A K = K \ (A K, and both K and A K are integrable. To finish this proof, we apply the following lemma. 30

32 5.3. PROPERTIES OF RADON MEASURES CHAPTER 5. RADON MEASURES Lemma If A A, then µ (A = µ (A < Proof. For all ɛ, choose some O open such that A O, µ (A µ(o ɛ. We know that there exists some K O such that µ(k > µ(o ɛ. Consider the following: as required. µ (A = µ ((A K (A \ K µ (A K + µ (O \ K = µ (A K + µ(o \ K < µ (A + ɛ Proof.[Proof of Lemma ] Let (A n n N be a sequence of disjoint elements of A such that To show that A K is integrable, we need to show that µ (A K = µ (A K. We note that A n = A. ( µ A n µ (A n ( A n K = (A n K Thus, we have ( N 1 Define B N = A N \ A n A. **** Lemma B(X A µ (A K µ (A K µ (A n K Proof. If F is closed, then F K is compact for all K compact, and is hence integrable, so F A, and so F c A, as F c is open. Lemma If µ = µ A, then µ is a measure. Proof. We want to show that if (A n n N with A n A disjoint, then we have ( µ A n = µ (A n ( Assume that (A n n N are such disjoint sets, and that λ < µ A n. Then we have some compact set K A n such that λ < µ(k. This gives: ( ( µ (A n µ A n µ A n µ (A n = µ (A n 31

33 5.3. PROPERTIES OF RADON MEASURES CHAPTER 5. RADON MEASURES and so as required. µ (A n > µ (A n K = µ (A K ( > µ A n K ( = µ A n K = µ (K > λ Proposition A X is integrable iff A A and µ (A < Proof. Suppose that A is integrable. Then A K is integrable for all compact K and so A A. Conversely, assume that A A and µ (A <. some compact K O such that µ(o < µ(a + ɛ For all ɛ > 0 there exists some open O A and µ(o \ K < ɛ To show that A is integrable, i.e. that µ (A = µ (A we note the following: µ (A = µ (A K + A K C µ (A K + O K c µ (A K + µ (O K C µ (A + ɛ Letting ɛ 0 gives the desired result. Proposition A = B µ Proof. If A A, then A K is integrable, and so to prove that A B µ, it suffices to show that all integrable sets are in B µ. If A is integrable, then for all ɛ > 0, then there exists O open and K compact, such that K A O and µ(o \ K < ɛ. Take ɛ = 1 n. This means that we have sequences (O n of open sets and (K n of compact sets such that K n A O n µ(o n \ K n < 1 n Let A = n N O n and A = n N K n and then we have A A A and µ(a \ A µ(o n \ K n < 1 n for all n N. This gives that µ(a \ A = 0. As A, A B(X, we must have A B. Conversely, suppose that A B µ and that K X is compact. We need to prove that A K is integrable. There exists A, A B such that A A K A and µ(a \ A = 0. We may assume that K A. Thus, µ(a = µ (A µ (A K µ (A = µ(a = µ(a < And so we have µ (A K = µ (A K <, so A K is integrable, as required. 32

34 5.3. PROPERTIES OF RADON MEASURES CHAPTER 5. RADON MEASURES Proposition If µ is moderate, then for all A B µ, then µ (A = µ (A. Proof. If µ is moderate, then there exist open sets (O n n N such that µ(o n < for all n N. Define B 1 = O 1 and: n 1 B n = O n \ This gives us X = n N O n = We know that if A B µ, then A B n B µ, µ (A B n µ (B n and µ(b n <. This gives that A B n is integrable for all n N, and so ( µ (A = µ (A B n = µ A B n ( = µ A B n = µ (A B n = µ (A as required. O k B n 33

35 Chapter 6 Lebesgue Measure on R n 6.1 Introduction To define the Lebesgue measure in n dimensions, we only need to define it on compact sets in R n. Let K R n be compact, and define: m m µ L (K = inf I j : K I j where the sets I j are n-dimensional cross products of 1-dimensional intervals, so n I = J k J k R and we define I = n J k to be the length of the interval. We wish to show that the above function is actually a measure. To do this, we will recall Choquet s Theorem, and show that the four conditions are satisfied. We remark that we can assume that each I j satisfies I j < δ for some δ > 0, and that each interval is open Proof of (1 m m Suppose that K 1 K 2 are compact and that K 2 I j. Then, K 1 I j and so µ L (K 1 µ L (K Proof of (2 m m If K 1 and K 2 I (1 j I (2 j then we have K 1 K 2 m I (1 j k l=1 I (2 l 34

36 6.2. PROPERTIES OF µ L CHAPTER 6. LEBESGUE MEASURE ON R N which gives µ L (K 1 K 2 We can choose some ɛ > 0 such that m µ L (K 1 > I (1 j + ɛ 2 This gives us and letting ɛ 0 gives the result Proof of (3 m I (1 j + k l=1 and µ L (K 2 > I (2 l k l=1 µ L (K 1 K 2 µ L (K 1 + µ L (K 2 + ɛ I (2 l + ɛ 2 Let K 1 K 2 =. This means that d(k 1, K 2 > 0. Take (I j < 1 2 d(k 1, K 2 for all j. Then if K 1 + K 2 n I j, we either have I j K 1 = or I j K 2 =. Thus, as in the previous section, we have Proof of (4 Let ɛ > 0 be fixed. µ L (K 1 + K 2 = H m I j, we must have by above m I j = I (1 j + as required. We have just proved the following: Choose some open covering K µ L (H k l=1 I (2 l > µ L (K 1 + µ L (K 2 m I j such that µ L (K > m I j < µ L (K + ɛ m I j ɛ. Then, if Theorem There exists a unique Radon measure µ L on B(R n which extends the definition of the above defined measure for compact sets. The measure actually extends to B µ L = B as µ L is moderate. 6.2 Properties of µ L We want to show that for all intervals I in R n, we have µ L (I = I. Given an open covering of I by a collection of intervals (I j m, we must show that I = m I j Firstly, we can assume that I j I for all j = 1,..., m. We define a special partition of I to be a partition of disjoint intervals (J k p which is a product of partitions of the projections of I onto each axis. For a special partition, p I = J k 35

37 6.2. PROPERTIES OF µ L CHAPTER 6. LEBESGUE MEASURE ON R N Given any arbitrary covering I = special partition m I j, take the projections on the axes an consider the corresponding m m I j = J k k R j Each I j is then a union of elements J k of the special partition. p J k = I Lemma Every open subset O R n is a countable union of disjoint intervals. Proof. consider the diadic partitions of R n, i.e. partitions of the form n [ pi 2 k, (p i + 12 k for p i Z. This is countable. Let A k be the collection of such diadic intervals of order k contained in O but not contained in A k 1. Then we have O = A k, as required. Proposition For all A R n, we have { } µ L = inf I k : A I k, I k an interval Proof. Let A I k. Then we know that ( µ L(A µ L I k µ L (I k = I k Conversely, let ɛ > 0 and choose some O A with µ L (O µ L (A + ɛ. Then we have O = O k which gives Letting ɛ 0 gives the desired result. µ L (O = µ(i K = I K < µ L(A + ɛ Theorem µ L is translation invariant. Moreover, if µ is a translation invariant Radon measure on R n, then there exists a constant c 0 such that µ = cµ L Proof. It is quite straightforward to see that µ L is translation invariant, as µ([a, b = µ([a + c, b + c in R. m Suppose that τ A is a shift in an interval by A and that K I j. Let K = τ A K. Then K m τ A I j 36

38 6.2. PROPERTIES OF µ L CHAPTER 6. LEBESGUE MEASURE ON R N and we note that τ A I j = I j. To prove the essential uniqueness, suppose that µ is some other translation invariant Radon measure. Let c = µ([0, 1 n. This gives ([ p n µ 0, 1 n = µ([0, 1 n = c p due to translation invariance. By shifting again, we obtain that ( n ( n µ [a i, b i = cµ L [a i, b i for a i, b i Q and i = 1,..., n. But if O = I k, where I k is diadic, then Finally, we note that µ(a = inf µ(o, as required. O A Notation From now on, we ll use the notation and for an arbitrary measure µ, µ(o = µ(i k = c µ L (I k = cµ L (O f dµ L = f(x d n x R n X f dµ = X f(xµ(dx Theorem µ L is rotation-invariant and if U is an orthogonal transformation, µ L (UA = µ L (A. Proof. Define µ(a = µ L (UA. This is a translation invariant Radon Measure, and so µ L (UA = cµ L (A for all A B(R n. But the unit ball B 1 in R n is invariant under U, so µ L (B 1 = µ(ub 1 and hence c = 1, as required. Theorem If T is a general linear transformation of R n, then µ L (T A = det(a µ L (A. Proof. To prove this, we write T = UDV where U and V are orthogonal matrices and D is diagonal. 37

39 Chapter 7 Product Measures 7.1 Introduction Given two Radon measures µ 1 on X 1 and µ 2 on X 2, we wish to define a product measure µ 1 µ 2 on X 1 X 2. We know that the topology on X 1 X 2 is define as follows: O X 1 X 2 is open iff for all (x 1, x 2 O, there exist open sets O 1 X 1 and O 2 X 2 such that (x 1, x 2, O 1 O 2 O. Unless otherwise specified, we will assume that we are considering two measure spaces (X 1, B(X 1, µ 1 and (X 2, B(X 2, µ 2, and letting X = X 1 X 2 be the product topology. Definition (Elementary Sets. We say that an open set O in the product topology is called Elementary if there exist open sequences (O (1 i m τ 1 and (O (2 i m τ 2 such that m O = (O (1 i O (2 i Lemma The elementary sets containing a compact set K X form a fundamental set of neighbourhoods of K. Proof. Suppose that U K is open. By definition, for every x = (x 1, x 2 K, there exist open sets O x (1 and O x (2 such that x = (x 1, x 2 O x (1 O x (2 U Then, we have K as required. x K O x (1 O x (2 and so there exist x 1,..., x m such that K m O x (1 i O x (2 i U Theorem If A 1 B(X 1 and A 2 B(X 2, then A 1 A 2 B(X 1 X 2. Also, given measures µ 1 on X 1 and µ 2 on X 2, there exists a unique µ = µ 1 µ 2 on X 1 X 2 such that µ(a 1 A 2 = µ 1 (A 1 µ 2 (A 2 Proof. Firstly, we make some observations. If f : X Y is a continuous map, then we know that f 1 (B B(X for all B B(Y. Indeed, we note that if B is open, then so is f 1 (B, and it is also Borel. Define A = { B Y f 1 (B B(X } 38

40 7.1. INTRODUCTION CHAPTER 7. PRODUCT MEASURES It can be readily shown that this is a σ-algebra, as f 1 (B c = f 1 (B c B(X. If B n A for all n N, then ( f 1 B n = f 1 (B n A As A contains all open sets, B(Y A. Getting back to the theorem, take f = π 1 : X 1 X 2 X 1, f(x 1, x 2 = x 1. Then f 1 (A 1 = A 1 X 1 B(X. Similarly, X 1 A 1 B(X. Thus, A 1 A 2 = (A 1 X 2 (X 1 A 2 B(X. Suppose that O is elementary. we must define Then we know that O = j J µ(o = j J µ 1 (A (1 j µ 2 (A (2 j A (1 j A (2 j where J is a finite set. Thus, For the algebra A of subsets of the type j J A (1 j A ( 2 j, we define an additive set function µ A (1 j A (2 j = µ 1 (A (1 j µ 2 (A (2 j j J j J This is a well define function. Thus, we can now define the following, where K is compact We will now use Choquet s Theorem: This is quite straightforward µ(k = inf{ µ(o; O K, O elementary } Let K 1, K 2 be compact and choose ɛ > 0. Then there exist elementary sets O 1 K 1 and O 2 K 2 such that µ(k 1 > µ(o 1 ɛ 2 and µ(k 2 > µ(o 2 ɛ 2. We know that O 1 O 2 is elementary, and so µ(k 1 K 2 µ(o 1 O 2 µ(o 1 + µ(o 2 µ(k 1 + µ(k 2 + ɛ If K 1 and K 2 are disjoint, then there exists some U 1 K 1 and U 2 K 2 such that U 1 U 2 =. We know that there exists elementary sets O 1 K 1 and O 2 K 2 such that O 1 U 1 and O 2 U 2, and hence O 1 O 2 =. If O K = K 1 + K 2 is elementary, then µ(o µ(k + ɛ and O O 1 and O O 2 are also elementary with µ(k 1 + µ(k 2 < µ(o O 1 + µ(o O 2 < µ(o < µ(k + ɛ Hence, µ(k 1 + µ(k 2 µ(k = µ(k 1 + K 2 Choose ɛ > 0. If K is compact then there exists O K is elementary, with µ(o < µ(k + ɛ. For all H O, we have µ(h µ(o µ(k + ɛ, as required. 39

41 7.2. PROPERTIES CHAPTER 7. PRODUCT MEASURES 7.2 Properties We will now show that the product open sets O 1 O 2 form a fundamental system of neighbourhoods of K 1 K 2. Let U K 1 K 2 be open. Fix x 1 K 1. For every x 2 K 2, there exist open sets O x (1 and O x (2, where x = (x 1, x 2, such that: x O x (1 O x (2 U Since K 2 is compact, there exists finitely many elements x 1,2, x 2,2,..., x m,2 such that Define O (1 x 1 = m K 2 m O (2 x i,2 O (1 (x 1,x j,2. This gives {x 1} K 2 O (1 x 1 O (2 x 1, and so ( K 1 K 2 O x (1 1 O x (2 1 x 1 K 1 There this exist elements x 1,1, x 2,1,..., x p,1 such that ( p K 1 K 2 Finally, we set O 1 = p O (1 x i,1 and O 2 = p we have (x i,1, x 2 O (1 x i,1 O (2 x i,1. In particular x 2 O (2 x i,1 It follows that K 1 K 2 O 1 O 2 O x (1 i,1 O x (2 i,1 O (2 x i,1. We know that these two sets are open. Given x 2 K 2, p for all i = 1,..., p, and so x 2 O 2. This gives us: O x (1 i,1 O x (2 i,1 U µ(k 1 K 2 = inf{ µ(o 1 O 2 O 1 K 1 open, O 2 K 2 open } inf{ µ(o 2 O 2 K 2 open } = µ(k 1 µ(k 2 For arbitrary A 1 B(X 1 and A 2 B(X 2, if K A 1 A 2 is compact, and K 1 = π 1 (K A 1, K 2 = π 2 (K A 2 are also compact, then µ(a 1 A 2 = sup µ(k K A 1 A 2 K compact = sup Given A B(X, we define A x with x X 1 by K 1 K 1 A 1 sup K 2 A 2 compact K compact = µ 1 (A 1 µ 2 (A 2 A x = {y X 2 (x, y A} µ(k 1 K 2 40