Reminder Notes for the Course on Measures on Topological Spaces

Transcription

1 Reminder Notes for the Course on Measures on Topological Spaces T. C. Dorlas Dublin Institute for Advanced Studies School of Theoretical Physics 10 Burlington Road, Dublin 4, Ireland. December 6, Introduction Remember the definition of the Riemann integral of a function f : [a, b] R, where [a, b] is a closed bounded interval: f is Riemann integrable if s b a(f) = s b a(f), where the lower and upper sums are defined resp. by s b a(f) = with m i = inf ti 1 x<t i f(x), and s b a(f) = sup a=t 0 <t 1 < <t n =b inf a=t 0 <t 1 < <t n=b n m i (t i t i 1 ) (1.1) i=1 n M i (t i t i 1 ) (1.2) with M i = sup ti 1 x<t i f(x). A continuous function on [a, b] can easily be shown to be Riemann integrable. Note that the function f(x) = 1 x on [0, 1], however, is not Riemann integrable, though 1 dx = 2 x 1 = 2 x i=1

2 is well-defined and finite. It can be defined, for example as 1 dx 1 ( ) 1 = lim x n dx. 0 x n 0 This ad-hoc procedure prompted Henri Lebesgue to suggest an alternative definition of integral by subdividing the y-axis instead of the x-axis: One then needs a proper definition of the total length of a subset of the form A n = {x [a, b] : y n f(x) < y n+1 }. To do this, he proposed a similar procedure on the x-axis as Riemann had for the y-axis: Define the upper measure of a set A as m (A) = inf A p k=1 I k p I k, (1.3) where {I k } p k=1 is a finite collection of intervals, and p m (A) = sup A p k=1 I k I k, (1.4) where the intervals I k A are disjoint. Then a set A is called measurable if m (A) = m (A). A function f is measurable if {x [a, b] : y 1 f(x) < y 2 } is measurable for all y 1 y 2. The Lebesgue integral is then the limit as the subdivision of the y-axis becomes finer and finer and covers more and more of the axis. k=1 k=1 Figure 1. Subdividing the y-axis. The red and blue areas represent the contributions to the approximation from the intervals between 3.5 and 4, and between 1 and 1.5 resp. They equal the height (3.5 resp. 1) times the total length of the corresponding subsets A 9 and A 4. 2

3 In the following we make these ideas precise and generalise them to integrals over much more general spaces than just the real line. This is another advantage of Lebesgue s approach, apart from the fact that more functions are Lebesgue integrable than Riemann integrable. In particular, we shall introduce measures on general Hausdorff topological spaces, which are generalisations of metric spaces. But first we define integration for a general abstract measure on an arbitrary set Ω, which then has to assumed as given. 2 Abstract Measures and Integration 2.1 σ-algebras and Measures In the following Ω is a given arbitrary set. (Measure theory is also at the core of probability theory where the sample space is usually denoted Ω.) Definition 2.1 A collection R of subsets of Ω is called a ring if the following hold: 1. A, B R = A B R. 2. A, B R = A \ B R. It is called an algebra if in addition, Ω R. Definition 2.2 A collection A of subsets of Ω is called a σ-algebra or σ- field if it is an algebra, and it is closed under countable union, i.e. For all sequences (A n ) n N of elements A n A, n N A n A. The pair (Ω, A) is called a measurable space. Remark. Note that it follows that also n N A n A. Definition 2.3 A (positive) measure µ on (Ω, A) is a map µ : A [0, + ] with the properties µ( ) = 0 and if (A n ) n N is a sequence of disjoint elements A n A then µ A n = µ(a n ). n N (In the following we write n N A n for a disjoint union.) The triple (Ω, A, µ) is called a measure space. n=1 3

4 Proposition Let (Ω, A, µ) be a measure space. Then the following hold: 1. A, B A, A B = µ(a) µ(b). 2. A, B A = µ(a B) µ(a) + µ(b). 3. If A n A (n N) and A n A A then µ(a n ) µ(a). 4. A n A (n N) = µ ( n N A n) n=1 µ(a n). Definition 2.4 A set A A is called negligible or a null set if µ(a) = 0. A property is said to hold almost everywhere (a.e.) if it holds outside a null set. A measure space (Ω, A, µ) is called complete if the following holds: B A, A A, µ(a) = 0 = B A. Proposition Given a measure space (Ω, A, µ) we define A = {A Ω : A 1, A 2 A, A 1 A A 2, µ(a 2 \ A 1 ) = 0}. Then A is a σ-algebra and µ (A) = µ(a 1 )(= µ(a 2 )) defines a measure on A such that (Ω, A, µ ) is complete. Definition 2.5 (Ω, A, µ ) as defined in the above proposition is called the Lebesgue completion of (Ω, A, µ). Note. In the following we often assume without mention that the measure space is complete. Definition 2.6 A function f : Ω R or R = [, + ], is called measurable if for all y R, {x Ω : f(x) y} A. Proposition A function f : Ω R is measurable if and only if one (and hence each) of the following hold: 1. {x Ω : f(x) y} A y R; 2. {x Ω : f(x) < y} A y R; 3. {x Ω : f(x) y} A y R; 4. {x Ω : f(x) > y} A y R. Proposition If f, g : Ω R are measurable, then f + g is also measurable. 1. If λ R and f : Ω R is measurable, then λf is also measurable. 2. If f n Ω R are measurable (n N), then inf n N f n and sup n N f n are also measurable. 3. If f, g : Ω R are measurable, then f g is also measurable. Corollary 1 Suppose that (Ω, A, µ) is a complete measure space, and f n : Ω R are measurable (n N), f n f a.e., then f is also measurable. 4

5 2.2 Integration Definition 2.7 A function g : Ω R is said to be a simple function if it is of the form N g = c k 1 Ak, (2.5) k=1 where N N, c 1,..., c N R and A 1,..., A n A, and we use the notation { 1 if x A; 1 A (x) = (2.6) 0 if x / A. The set of simple functions will be denoted S and the subset of non-negative simple functions as S +. Proposition Every measurable function f : Ω [0, + ] is the (pointwise) limit of an increasing sequence of simple functions g n S +. For g S + of the form (2.5) we define the integral by g dµ = N c k µ(a k ). (2.7) Lemma 1 This definition is independent of the representation for g. It has the following properties: k=1 Proposition f, g S + = f + g S + and (f + g)dµ = f dµ + g dµ. 2. λ 0, f S + = λf S + and λf dµ = λ f dµ. 3. f, g S +, f g = f dµ g dµ. We can now define, for any measurable function f : Ω [0, + ], f dµ = g dµ. sup g S +,g f However, it is preferable to have a definition in terms of a limit. For this, we note: 5

6 Lemma 2 Let f : Ω [0, + ] be a measurable function and suppose that (g n ) n N is a sequence of simple functions g n S + such that g n f. If g S + and g f, then g dµ lim g n dµ. n Corollary 1 For any sequence g n S + with g n f, f dµ = lim g n dµ. n We now have the analogous properties as for simple functions: Proposition If f, g : Ω [0, + ] are measurable then (f + g)dµ = f dµ + g dµ. 2. If λ 0 and f : Ω [0, + ] is measurable then λf dµ = λ f dµ. 3. If f, g : Ω [0, + ] are measurable and f g then f dµ g dµ. Definition 2.8 A function f : Ω R is said to be µ-integrable or µ- summable if 1. f is measurable; and 2. f dµ < +. The integral of f is then defined by f dµ = f + dµ f dµ, (2.8) where f + = f 0; and f = ( f) 0. A function f : Ω C is said to be measurable if R(f) and I(f) are measurable. A function f : Ω C is called µ-integrable or µ-summable if 1. f is measurable; and 2. f dµ < +. In that case, we put f dµ = R(f) dµ + i I(f) dµ. (2.9) We write L 1 (Ω, A, µ) for the space of µ-summable functions f : Ω C. 6

7 Theorem f, g L 1 = f + g L 1 and (f + g)dµ = f dµ + g dµ. 2. λ C, f L 1 = λf L 1 and λf dµ = λ f dµ. 3. f L 1 = f L 1 and f dµ f dµ. More properties: Proposition Suppose that f L 1, g : Ω C is measurable, and g = f a.e. Then g L 1 and g dµ = f dµ. Corollary 2 If f, g L 1 R and f g a.e. then f dµ g dµ. Lemma 3 If f : Ω R (or C) is measurable and A A, then f1 A measurable. is Definition 2.9 For f L 1 or f : Ω [0, + ] measurable, and A A, we define f dµ := f1 A dµ. A Proposition Suppose that f L 1 and f(x) > 0 for x A where A A. Then f dµ = 0 µ(a) = Convergence A Theorem 2.2 (Monotone Convergence Theorem) Let (f n ) n N be a sequence of non-negative measurable functions, f n f. Then f is integrable if and only if the sequence ( f n dµ ) n N is bounded, and in any case, f dµ = lim f n dµ. n Corollary 1 For measurable functions f n : Ω [0, + ], n=1 f n dµ = 7 n=1 f n dµ.

8 Theorem 2.3 (Dominated Convergence Theorem) Let f n : Ω C be a sequence of measurable functions, f = lim n f n a.e., and suppose that there exists g L 1 + such that f n g. Then f is integrable and f dµ = lim f n dµ. n Notation. We denote, for p 1, L p = L p (Ω, A, µ) the set of measurable functions f : Ω C such that f p is integrable. We put ( N p = f p dµ) 1/p. Theorem 2.4 (Hölder s inequality) Suppose f L p and g L q, where 1 p + 1 q = 1. Then fg L1 and ( fg dµ ) 1/p ( f p dµ g q dµ) 1/q. (2.10) Theorem 2.5 (Minkowski s inequality) If f, g L p then f +g L p and N p (f + g) N p (f) + N p (g). Theorem 2.6 (Riesz-Fischer) If (f n ) n N is a Cauchy sequence in L p w.r.t. N p, i.e. N p (f n f m ) 0 as n, m, then there exists f L p such that N p (f n f) 0. Corollary 2 The linear space L p f p = N p (f). := L p / is a Banach space with norm Theorem 2.7 L 2 is a Hilbert space with scalar product f g = f g dµ. 8

9 3 Topological Spaces Remember that a metric space is a set X together with a map d : X X [0, + ) with the following properties: 1. d(x, x) = 0 for all x X and if d(x, y) = 0 for x, y X then y = x. 2. x, y X : d(x, y) = d(y, x). 3. x, y, z X : d(x, z) d(x, y) + d(y, z). In such spaces one defines open sets as follows: O X is open if for all x O there exists ɛ > 0 such that B(x, ɛ) O, where B(x, ɛ) = {y X : d(x, y) < ɛ}. It now turns out that most important properties of metric spaces can be expressed in terms of open sets only, without referring to the underlying map d. For example, a sequence (x n ) n N converges to x X if for all open sets U containing x, there exists n 0 N such that x n U for all n n 0. And a map f : X Y between two metric spaces X and Y is continuous (at x X ) if for every open V Y (containing f(x)) there exists an open U X (containing x) such that f(u) V. We therefore generalise the concept of metric space to: Definition 3.1 A topological space is a set X together with a collection T of subsets of X (the open sets) satisfying: 1., X T ; 2. If O 1, O 2 T then O 1 O 2 T ; 3. If (O i ) i I is an arbitrary collection of open subsets O i T then the union i I O i T. A subset of X is called closed if its complement is open (i.e. T ). Remark. Topological spaces are more general objects than metric spaces. For example, sequential convergence is in general no longer sufficient, i.e. a function is no longer necessarily continuous at x if for every convergent sequence x n x, f(x n ) f(x). Similarly, a subset A X is not necessarily closed if for all convergent sequences x n A, lim x n A. 9

10 In fact, for most purposes, general topological spaces are slightly too general. In particular, we introduce the following additional separation axiom: Definition 3.2 A topological space X is said to be a Hausdorff space if the following additional axiom holds: For all x, y X with x y, there exist open sets U, V X such that x U, y V and U V =. Now remember that a subset K of a metric space is called compact if every sequence (x n ) n N in K has a convergent sequence (x nk ) with limit in K. For compact sets the following two theorems were proved: Theorem 3.1 Suppose that K is a compact set in a metric space X. Then for all δ > 0, there exist finitely many points x 1,..., x n K such that K n j=1 B(x j, δ). Theorem 3.2 (Heine-Borel) Let K be a compact set in a metric space X. Suppose that (O i ) i I is a collection of open sets O i covering K, i.e. K i I O i. Then there exist finitely many indices i 1,..., i n I such that K n k=1 O i k. (Every open cover has a finite subcover.) We now use this property to generalise the concept of compact set to topological spaces: Definition 3.3 A subset K of a topological space X is said to be compact if every open cover of K has a finite subcover. Remark. As above, it is in general no longer sufficient if every sequence in K has a convergent subsequence. Compact sets are very useful. Here are some elementary properties: Proposition 3.1 If K is a compact subset of a topological space X, then every closed subset F K is also compact. For the next property, first a definition, and a lemma: Definition 3.4 The closure of a subset A of a topological space X is the smallest closed set A containing A. (Note that this exists, as arbitrary intersections of closed sets are closed.) 10

11 Lemma 1 The closure of a set A X is given by A = {x X : Uopen, x U = U A }. Using this, one now proves: Proposition 3.2 If X is a Hausdorff space then every compact set K X is closed. Note that the Hausdorff property is crucial here. Proposition 3.3 If X and Y are topological spaces, and f : X Y is continuous (i.e. the inverse image of an open set is open), then f(k) is compact whenever K X is compact. Definition 3.5 A collection (F i ) i I of subsets F i X is said to have the finite intersection property if every finite subcollection has a non-empty intersection. Proposition 3.4 A subset K of a topological space X is compact if and only if every collection(f i ) i I of closed subsets of K with the finite intersection property, has non-empty intersection: i I F i. Proposition 3.5 Given two disjoint compact sets K 1, K 2 X (K 1 K 2 = ), of a Hausdorff space calx, there exist open sets U 1, U 2 such that K 1 U 1, K 2 U 2, and U 1 U 2 =. 4 Radon Measures In the following, X will always be a Hausdorff space. Definition 4.1 The Borel σ-algebra B(X ) is the smallest σ-algebra containing all open sets of X. Definition 4.2 A (positive) Radon measure on X is a measure µ on B(X ) with the following additional properties: 1. µ is locally finite: For all x X there exists an open U containing x with µ(u) < + ; 11

12 2. µ is inner regular: For all Borel sets A B(X ), µ(a) = Some elementary properties: sup µ(k). K A; Kcompact Proposition 4.1 If µis a Radon measure on a Hausdorff space X then µ(k) < + for all compact K X. Proposition 4.2 If µ is a Radon measure on X and K X is compact, then, for any ɛ > 0, there exists on open set O K such that µ(o \ K) < ɛ. Proposition 4.3 If µ is a Radon measure on X then for any family (O i ) i I of open sets with µ(o i ) = 0, the union is also a null set: µ ( i I O i) = 0. Definition 4.3 The support of a Radon measure µ is the complement of the largest open set of measure zero. Definition 4.4 Given a Radon measure µ on X, the Bourbaki completion calb µ of B(X ) is defined by B µ = {A X : K X compact, A K B }. Proposition 4.4 Given a Radon measure µ on X, B µ is a σ-algebra, and there is a unique Radon-extension of µ to B µ. The resulting measure space is complete. Proposition 4.5 Suppose that µ is a Radon measure on X for which there exists a countable set of open sets O n (n N) such that µ(o n ) < + and X n=1 O n. (A Radon measure with this property is said to be moderate(d) or σ-finite.) Then B µ = B. Definition 4.5 If µ is a Radon measure on X then a function f : X C or R is called µ-measurable if it is measurable on the measure space (X, B µ, µ). (If it is measurable on (X, B(X ), µ) then it is called Borel.) Proposition 4.6 If f : X [0, + ] is µ-measurable, then f dµ = g dµ. sup 0 g f gu.s.c.;supp(g)compact 12

13 Corollary 1 f dµ = sup f dµ. Kcompact K Theorem 4.1 Suppose that {f i } i I is a family of l.s.c. functions f i : X [0, + ], which is upward directed, i.e. i, j I k I : f i f k andf j f k. Set f = sup i I f i. Then f is l.s.c. and hence µ-measurable (even Borel), and f dµ = sup f i dµ. i I Theorem 4.2 (Choquet) Let µ be a Radon measure on X. Then it has the following properties: 1. K 1, K 2 X compact, K 1 K 2 = µ(k 1 ) µ(k 2 ). 2. K 1, K 2 X compact, K 1 K 2 = = µ(k 1 +K 2 ) = µ(k 1 )+µ(k 2 ). 3. K 1, K 2 X compact, µ(k 1 K 2 ) mu(k 1 ) + µ(k 2 ). 4. K X compact and all ɛ > 0, there exists an open set O K such that for all compact H O, µ(h) < µ(k) + ɛ. Conversely, suppose that λ is a map from the compact subsets of X to [0, + ) with the above properties. Then there exists a unique Radon measure µ extending λ to the Borel sets calb(x ). The proof of this crucial theorem is lengthy and goes via number of lemmas. Clearly, we want to define µ(a) = For open sets in particular, we then have sup µ(k). K Acompact Lemma 1 If O 1 and O 2 are open, and O 1 O 2 then µ(o 1 ) µ(o 2 ). 13

14 Lemma 2 Suppose that O 1 and O 2 are open, and K O 1 O 2 is compact. Then there exist compact subsets K 1 O 1 and K 2 O 2 such that K = K 1 K 2. Lemma 3 If O 1 and O 2 are open, then µ(o 1 O 2 ) µ(o 1 ) + µ(o 2 ). Lemma 4 Suppose (O n ) n N is an increasing sequence of open sets, O n O. Then µ(o n ) µ(o). Lemma 5 If (O n ) n N is a sequence of open sets, O = n=1 O n, then µ(o) n=1 µ(o n). Next we define, for arbitrary subsets A X, an upper measure µ (A) and a lower measure µ (A) by µ (A) = inf µ(o) (4.11) O A open and Clearly, µ (A) µ (A). µ (A) = sup µ(k). (4.12) K A compact Lemma 6 If A = n N A n, then µ(a) n N µ(a n). Lemma 7 If A = n N A n then µ (A) n=1 µ(a n). Definition 4.6 A set A X is called integrable if µ (A) = µ (A) < +. Lemma 8 Every compact set K and every open set O with µ(o) < + is integrable. Lemma 9 If (A n ) n N are disjoint integrable sets, then µ ( n N A n ) = µ ( n N A n ) = µ(a n ). Lemma 10 (Sandwich Criterion) A X is integrable if and only if for every ɛ > 0, there exist a compact set K A and an open set O A such that µ(o \ K) < ɛ. n=1 14

15 Lemma 11 The integrable sets form a ring, i.e. if A 1 and A 2 are integrable, then so are A 1 A 2, A 1 A 2 and A 1 \ A 2. Definition 4.7 A X is measurable if A K is integrable for all compact sets K. Lemma 12 The collection A of measurable sets is a σ-algebra containing B(X ). Lemma 13 µ A is a measure. We have the following additional facts: Proposition 4.7 A X is integrable if and only if A A and µ (A) < +. Proposition 4.8 The σ-algebra A coincides with the Bourbaki completion B µ. Proposition 4.9 If µ is moderate, then µ (A) = µ (A) for all A B µ. 5 Lebesgue Measure For K R n compact, we define the Lebesgue measure of K by µ L (K) = inf K N i=1 I i N I i, (5.13) where {I i } N i=1 is a covering of K with finitely many (n-dimensional) intervals I i (i.e. products of 1-dimensional intervals). Noting that for any δ > 0 we may assume that I i < δ (by simply dividing up each interval into smaller ones), and also that each I i is open, it is easy to show that the 4 conditions of Choquet s theorem are satisfied. We thus conclude: Theorem 5.1 There exists a unique extension of µ L given by (5.13) for compact K R n, to the Borel σ-algebra calb(r n ), and hence to B µ L = B (since µ L is moderate). This is called (n-dimensional) Lebesgue measure. i=1 15

16 Theorem 5.2 Let µ L be n-dimensional Lebesgue measure. Then every n- dimensional interval I is a Borel set and µ L (I) = I. This is proved by showing that every covering of I by intervals can be refined to a special partition, some of whose elements may occur more than once. Lemma 1 Every open set O R n is a disjoint union of a countable number of intervals. Proof. Consider dyadic partitions of R of increasing order N: R = [ k 2, k + 1 ). N 2 N k Z Proposition 5.1 For all A R n, µ L(A) = inf A j=1 I j. j=1 QED Theorem 5.3 Lebesgue measure is translation invariant, and if µ is another translation-invariant measure on R n then there exists a constant c 0 such that µ = c µ L. Corollary 1 Lebesgue measure is invariant under orthogonal transformations. Theorem 5.4 If T : R n R n is a bijective linear map, and A B(R n ), then T (A) B(R n ) and µ L (T (A)) = det T µ L (A). 16

17 6 Product Measures and Fubini s Theorem Theorem 6.1 Let X 1 and X 2 be Hausdorff spaces and let µ 1 and µ 2 be Radon measures on calx 1 and X 2 respectively. Then there exists a unique Radon measure µ = µ 1 µ 2 on X = X 1 X 2 satisfying for all A 1 B(X 1 ) and A 2 B(X 2 ). µ(a 1 A 2 ) = µ 1 (A 1 )µ 2 (A 2 ) Proof. (outline) First note that A 1 A 2 B(X 1 X 2 ). This follows from the fact that if π : X Y is continuous and A B(Y) then π 1 (A) B(X ). Taking for π the projections π 1,2 we obtain A 1 X 2, calx 1 A 2 B(X ) and hence A 1 A 2 = A 1 X 2 X 1 A 2 B(X ). We now call an open set O X elementary if it is of the form where O (1,2) i O = m i=1 O (1) i O (2) i, X 1,2 are open. We then have Lemma 1 For any compact K X, the elementary open sets O K form a fundamental system of neighbourhoods of K, i.e. every neighbourhood of K contains an elementary open set containing K. This lemma immediately implies uniqueness of the measure µ since every elementary open set is of the form O = j J A (1) j A (2) j, where J is a finite set,i.e. a disjoint union of product sets. To prove the existence, we first define an additive set function on such finite disjoint unions of product sets, by ( ) µ A (1) j A (2) j = µ 1 (A (1) j )µ 2 (A (2) j ). j J j J This is easily seen (by taking intersections of the parts of two different subdivisions) to be well-defined. Then we put µ(k) = inf µ(o). O Kopen,elementary 17

18 This satisfies the conditions for Choquet s theorem, and it remains to show that (6.1) holds. This is done by first showing that it holds for compact sets. For compact sets it follows from Lemma 2 Given compact sets K 1 X 1 and K 2 X 2, the open sets of the form O 1 O 2 with O 1 K 1 and O 2 K 2 open, form a fundamental system of neighbourhoods of K 1 K 2. QED Remark. n-dimensional Lebesgue measure is a product of 1-dimensional Lebesgue measures on the coordinate axes. To prove Fubini s theorem about product measures, we first proved a number of lemmas: Lemma 1 Given A X = X 1 X 2 define, for x X 1, A x = {y X 2 : (x, y) A}. Then, (A c ) x = (A x ) c, and for any collection (A i ) i I of subsets A i X 1 X 2, ( ) ( ) A i (A i ) x ; A i (A i ) x. i I x = i I Moreover, if A B(X ) then A x B(X 2 ). i I x = i I Lemma 2 Let A denote the algebra of finite disjoint unions N i=1 A 1i A 2i with A 1i B(X 1 ) and A 2i B(X 2 ) (i = 1,..., N). Then µ(a) = µ 2 (A x ) µ 1 (dx). Lemma 3 Let O calx be open. Then x µ 2 (O x ) is l.s.c. and µ(o) = µ 2 (O x ) µ 1 (dx). Lemma 4 If K X is compact, then µ(k) = µ 2 (K x ) µ 1 (dx). 18

19 Lemma 5 If A X is µ-integrable, then A x is µ 2 -integrable for a.e. x X 1, x µ 2 (A x ) is µ 1 -measurable, and µ(a) = µ 2 (A x ) µ 1 (dx). Finally, we proved: Theorem 6.2 (Fubini) Assume that the measures µ 1 and µ 2 are moderate. Then so is µ = µ 1 µ 2 and the following hold: If f 0 is measurable, then f x is µ 2 -measurable for a.e. x X 1, the map x f x dµ 2 is µ 1 - measurable, and ( ) f dµ = f x dµ 2 µ 1 (dx). Moreover, if f L 1 (µ) is arbitrary, then f x L 1 (µ 2 ) for a.e. x and x fx dµ 2 is integrable, and (6.2) holds. Remark. Note that we in fact only need f to be moderate, i.e. {(x 1, x 2 ) : f(x 1, x 2 ) 0} is contained in a countable union of open sets of finite measure. 7 The Riesz-Markov-Saks-Kakutani Theorem We need a new concept in topology: Definition 7.1 A topological space is said to be locally compact if every point has a fundamental system of neighbourhoods which are compact sets. (A neighbourhood of x is a set containing an open set which contains x.) Proposition 7.1 A Hausdorff space is locally compact if (and only if) each point has a compact neighbourhood. Corollary 1 A compact Hausdorff space is locally compact. We need an important theorem due to Urysohn: (there is an analogous result for so-called normal spaces, which is proved in essentially the same way) 19

20 Theorem 7.1 (Urysohn) Let X be a locally compact Hausdorff space, O X open, and K O compact. Then there exists a continuous function h : X [0, 1] with compact support, such that 1 K h 1 O. The proof uses the following lemma: Lemma 1 Let X be a locally compact Hausdorff space, O X open, and K O compact. Then there exists an open set U such that its closure U is compact, and K U U O. Theorem 7.2 (Riesz et al.) Let X be a locally compact Hausdorff space, and denote κ(x ) the set of continuous real-valued function on X with compact support. If µ is a Radon measure on X then κ(x ) L 1 (µ), and the integral µ(f) := f dµ has the following properties: 1. Linearity: µ(f + g) = µ(f) + µ(g) f, g κ(x ) and µ(λf) = λµ(f) for all f κ(x ) and λ R; 2. Positivity: f κ(x ), f 0 = µ(f) 0. Conversely, if l is a positive linear form on κ(x ) (i.e. it has the properties 1. and 2.), then there exists a unique Radon measure µ on X such that l(f) = µ(f) for all f κ(x ). Proof. Then first half of the theorem is clear. Uniqueness follows from Urysohn s theorem above. To prove the existence, we define for compact K, λ(k) = inf l(f). f κ(x );1 K f The conditions for Choquet s theorem are again straightforward to verify, so there exists a Radon measure µ given by µ(a) = sup K Acompact λ(k) for A B(X ). It then remains to show that l(f) = f dµ. For this, one first proves that if f 0, l(f) f dµ by approximating f by a simple function. Next this inequality is extended to arbitrary f κ(x ), which implies the equlity. QED 8 Signed Measures Definition 8.1 A δ-ring R on a set Ω is a collection of subsets of Ω such that 1. A, B R = A \ B, A B R; 2. A n R(n N) = n N A n R. 20

21 Definition 8.2 Let R be a δ-ring on a set Ω, and let E be a normed vector space. An E-valued measure on (Ω, R) is a map µ : R E which is σ-additive, i.e. if A n R (n N) are disjoint, and A = n N A n R then µ(a) = n=1 µ(a n). (In particular, the latter is norm-convergent.) Proposition 8.1 An E-valued measure µ on (Ω, R) has the following properties: 1. µ( ) = µ is additive. 3. If A n R (n N) and A n A R then µ(a) = lim n µ(a n ). 4. If A n R (n N) and A n A then A R and µ(a) = lim n µ(a n ). Definition 8.3 If µ is an E-valued measure on (Ω, R), we define for A Ω, µ(a) = sup (A i ) m i=1 R: A P m i=1 A i m µ(a i ). Proposition 8.2 µ R extends to a (positive) measure on the σ-algebra i=1 τ(r) = {A Ω : A B R B R}. Definition 8.4 The measure µ = µ big τ(r) is called the variation (measure) of µ. µ is said to have bounded variation if µ := µ (Ω) < +. Proposition 8.3 If f L 1 ( µ ) then f d µ f d µ, where the former is defined by η, f d µ = f d η, µ for every continuous linear form η on E. Definition 8.5 Let X be a Hausdorff space and let R be the collection of relatively compact Borel sets of X. An E-valued measure µ on X is called an E-valued Radon measure if µ is a Radon mesure. 21

22 We now specialise to E = C, i.e. complex-valued measures. Proposition 8.4 Suppose that X is a locally compact Hausdorff space, and µ is a complex-valued Radon measure on X. Then, for all open sets O X, µ (O) = sup φ dµ. φ κ C (X ): φ 1;supp(φ) O Definition 8.6 A signed measure is a real-valued measure. For such a measure, we define the positive- and negative parts µ ± by µ ± = 1 ( µ ± µ). 2 Proposition 8.5 If µ is a signed Radon measure on a Hausdorff space X then µ ± are Radon measures. Theorem 8.1 Let X be a locally compact Hausdorff space. If µ is a signed Radon measure on X with bounded variation then f f dµ is a continuous linear form on κ(x ), more precisely, f dµ µ f. Conversely, if L : κ(x ) R is a continuous linear form then there exists a unique signed Radon measure µ on X with bounded variation such that L(f) = f dµ for all f κ(x ). Corollary 1 Let X be a locally compact Hausdorff space. If µ is a C-valued Radon measure on X with bounded variation then f f dµ is a continuous linear form on κ C (X ), more precisely, f dµ µ f. Conversely, if L : κ C (X ) C is a continuous linear form then there exists a unique C-valued Radon measure µ on X with bounded variation such that L(f) = f dµ for all f κ C (X ). 22