MATS113 ADVANCED MEASURE THEORY SPRING 2016

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1 MATS113 ADVANCED MEASURE THEORY SPRING 2016 Foreword These are the lecture notes for the course Advanced Measure Theory given at the University of Jyväskylä in the Spring of The lecture notes can by found (with a possible delay) from the course website (or in Koppa) In the course we study basics about Borel and Souslin sets, measurable choices and partitions, conditional measures, disintegration of measures and ergodic theorems. Other possible topics include Egoroff and Lusin theorems, extensions of measures, weak convergence and Prokhorov theorem. Grading for the course: 3 sets of exercises whose solutions are handed in to the lecturer. Each exercise set has maximum points at least 5. The final exam has maximum 30 points. Both, the exercises and the exam are optional. The grade is given by the generous 1 formula ( ( exam points 12 + ) ) 3 min(5, i:th set points) max min, 5, 0. 3 Literature: Bogachev, Measure Theory mostly volume 2 Fremlin, Measure Theory a five volume book that is available (at least in.tex) at Holopainen online notes (in Finnish) Version: May 8, The lecturer reserves the right to diminish generosity by a difficult final exam. 1

2 2 ADVANCED MEASURE THEORY Contents 1. Measures and Algebras 3 2. Borel, Baire and Souslin sets Borel and Baire sets Souslin sets 9 3. Measurable selections Borel, Baire and Radon measures Extensions, products and the Kolmogorov theorem Weak convergence Conditional measures Ergodicity 45 References 47 Solutions to Exercises 48

3 ADVANCED MEASURE THEORY 3 1. Measures and Algebras Definition 1.1. A class A of subsets of a set is called an algebra if (i) A (ii) A, B A A \ B A. Remark 1.2. For an algebra A it holds A and A, B A A B, A B A. Definition 1.3. An algebra A of sets is called a σ-algebra if for any {A n } A one has A n A. Definition 1.4. A pair (, A) consisting of a set and a σ-algebra of its subsets is called a measurable space. Definition 1.5. Given a family F of subset of the smallest σ-algebra of containing F is denoted by σ(f ) and is called the σ-algebra generated by F. Proposition 1.6. For any family F of subset of there exists a unique σ-algebra generated by F. Proof. Set I := A σ-algebra F A Notice that 2 := {A : A } is a σ-algebra and F 2. By definition F I. Let us show that I is a σ-algebra. Since for all algebras A of we have A, we also have I. Similarly, if {A n } I and A is a σ-algebra of containing F, we have {A n} A and thus A 1 \ A 2, A n A. Consequently, A 1 \ A 2, A n I. Thus I is a σ-algebra. To see the minimality of I suppose that there exists a σ-algebra B of containing F. Then by definition I B. To see the uniqueness, suppose B is a minimal σ-algebra of containing F. Then F B I I, B and by minimality I = B since I B is also a σ-algebra by the above. We can also define an algebra generated by a family of subsets F of similarly as the σ-algebra generated by F. Theorem 1.7. Let E be a class of subsets of a space containing the empty set. Let F be the smallest class of sets in with the following properties: (i) If A E, then A, \ A F. (ii) If {A i } i N F is disjointed, then i N A i F. (iii) If {A i } i N F, then i N A i F. Then F = σ(e). Proof. Let A = {A F : \A F}. We will show that A is closed with respect to countable unions of pairwise disjoint sets and closed with respect to arbitrary countable intersections. Since A is closed under complementation, the claim will then follow. If {A i } i N A is disjointed, then i N A i F by (ii) and \ i N A i = i N ( \ A i) F by (iii) and the fact that \ A i F by the definition of A. Thus i N A i A. If {A i } i N A is an arbitrary collection, then i N A i F by (iii). For the complement we write \ A i = ( \ A i ) = ( ) i 1 ( \ A i ) ( A k ). i N i N i N A. k=1

4 4 ADVANCED MEASURE THEORY This is a countable union of pairwise disjoint sets ( \ A i ) ( i 1 k=1 A k) each of which belongs to F by (iii) and the assumption \ A i F. Thus i N A i A and we are done. Definition 1.8. Let A be a σ-algebra on. A function µ: A [0, ] that satisfies (i) µ( ) = 0 (ii) µ( A i) = µ(a i) for all pairwise disjoint sets A i A (countable additivity) is called a measure (with values in [0, ]). The measure µ is called σ-finite, if = for some i A with µ( i ) < for all i N. Definition 1.9. A function µ: 2 [0, ] that satisfies (i) µ( ) = 0 (ii) µ(a) µ(b) for all A B (monotonicity) (iii) µ( A i) µ(a i) for all A i (countable subadditivity) is called a (Carathéodory) outer measure. Definition Let µ be an outer measure on. A set A is called µ-measurable if µ(e) = µ(e A) + µ(e \ A) for all E. Theorem Let µ be an outer measure on. Then the collection of µ-measurable subsets of form a σ-algebra. Proof. Let us denote the µ-measurable subsets of by A. Since for all E it holds µ(e) = µ(e) + µ( ) we have A. Let A, B A. Then for any E we get µ(e) = µ(e \ B) + µ(e B) = µ((e \ B) A) + µ((e \ B) \ A) + µ(e B) = µ(e (A \ B)) + µ((e \ (A \ B)) \ B) + µ((e \ (A \ B)) B) = µ(e (A \ B)) + µ(e \ (A \ B)) and thus A \ B A. Suppose now A n A for n N. By the fact that A is an algebra, we already know that B N := N A n A for all N N. Now let E. By subadditivity of µ we have µ(e) µ(e \ n N A n ) + µ(e n N A n ). So it remains to prove the converse inequality. For all N N we have µ(e) = µ(e \ B N ) + µ(e B N ) µ(e \ n N A n ) + µ(e B N ) and thus it remains to show that lim µ(e B N) µ(e A n ) (1.1) N n N Write C N := E (B N \ N 1 B n). By countable subadditivity, µ(e A n ) = µ( C N ) µ(c N ) = lim µ(e B N), N n N N N N N

5 ADVANCED MEASURE THEORY 5 where the last equality follows from the µ-measurability of B N : µ(c 1 ) + µ(c 2 ) = µ(e B 1 ) + µ(e (B 2 \ B 1 )) = µ((e B 2 ) B 1 ) + µ((e B 2 ) \ B 1 ) = µ(e B 2 ) and by induction N µ(c n) = µ(e B N ). Thus (1.1) holds. Theorem Let A be a σ-algebra on and µ a measure on it. Then µ : 2 [0, ] defined as µ (B) = inf{µ(a) : B A A} is an outer measure on. Proof. First of all µ ( ) = µ( ) = 0. Let B 1 B 2. Then {A A : B 1 A} {A A : B 2 A} and thus µ (B 1 ) µ (B 2 ). Finally, let B n for n N. For all n N take B n A n A. Since A is a σ-algebra, also n N A n A. Now from countable additivity of µ we get Thus infimizing over {A n } n N we get µ( A n ) µ(a n ). n N n N µ ( n N B n ) n N µ (B n ). Outer measures can be defined in a similar fashion also from less information. Let A 2 any collection and µ: A [0, ] any function. Define for all B { } µ (B) := inf A n, I N (1.2) n I µ(a n ) : A n A, B n I with the interpretation that the infimum over an empty set is +. Definition An outer measure µ on is called regular if for every A there exists a µ-measurable set B such that A B and µ(a) = µ(b). Theorem Let µ be a finite regular outer measure on. Then A is µ-measurable if and only if µ() = µ(a) + µ( \ A). Proof. The necessity of the condition is obvious. Let us show the sufficiency. Assume that for A we have µ() = µ(a) + µ( \ A) and let E. The inequality µ(e) µ(e A)+µ(E\A) follows from subadditivity, so we only need to show the converse inequality. By the regularity of µ there exists a µ-measurable set B E such that µ(e) = µ(b). Now µ() = µ(a) + µ( \ A) = µ(a B) + µ(a \ B) + µ(( \ A) B) + µ(( \ A) \ B) = µ(b A) + µ(b \ A) + µ(( \ B) A) + µ(( \ B) \ A) µ(b) + µ( \ B) = µ().

6 6 ADVANCED MEASURE THEORY In other words, the inequality in the previous chain is an equality. Thus µ(e) = µ(b) = µ(b A) + µ(b \ A) + µ(( \ B) A) + µ(( \ B) \ A) µ( \ B) µ(b A) + µ(b \ A) µ(e A) + µ(e \ A) Borel and Baire sets. 2. Borel, Baire and Souslin sets Definition 2.1. Let be a topological space. The Borel σ-algebra B() is the σ-algebra generated by all the open subsets of. Definition 2.2. Let and Y be topological spaces. A map f : Y is called Borel (measurable) if f 1 (B) B() for all B B(Y ). More generally, if (, Ω), (Y, A) are measurable spaces, a function f : Y is called measurable (with respect to A and A ) if f 1 (B) A for all B A. Lemma 2.3. Continuous maps between topological spaces are Borel measurable. Proof. Let f : Y be a continuous map between topological spaces and Y. Let us define D := {B B(Y ) : f 1 (B) B()}. By continuity of f the class D contains all open subsets of Y. Therefore, in order to show that D = B(Y ) and thus the claim, it suffices ot check that D is a σ-algebra. As an open set Y D. Suppose A, B D. Then f 1 (A \ B) = f 1 (A) \ f 1 (B) B() and thus A \ B D. Therefore D is an algebra. Suppose then that A i D for all i N. Then ( ) f 1 A i = f 1 (A i ) B() i i and thus i A i D. Therefore D is a σ-algebra. Proposition 2.4. Let (, A) be a measurable space and Y a metric space. Then any pointwise limit of measurable maps f i : (, A) (Y, B(Y )) is measurable. Proof. Let us show the measurability via limits of real-valued functions. Let f be the pointwise limit of f i and let C Y be a closed set. To prove the measurability of f it suffices to show that f 1 (C) A. Consider the function ϕ: Y R: x d(x, C). Then C = ϕ 1 (0). Since f i := ϕ f i converges pointwise to ϕ f it now suffices to prove the claim of the proposition in the case Y = R. This can be reduced to showing the following lemma on supremum of measurable functions since and inf i f i = sup i f i. lim i f i = lim sup f i = inf i sup f i n N i n Lemma 2.5. Let (, A) be a measurable space. Then any pointwise supremum of measurable functions f i : (, A) (R, B(R)) is measurable.

7 ADVANCED MEASURE THEORY 7 Proof. Let f := sup f i. We will show that for any a R it holds f 1 ((a, )) A. This follows if we can show that f 1 ((a, )) = fi 1 ((a, )). Since f i f for all i N, we have fi 1 ((a, )) f 1 ((a, )) and thus fi 1 ((a, )) f 1 ((a, )). Suppose then that x / x / f 1 ((a, )). f i 1 ((a, )) then f i (x) a for all i N and thus f(x) a and Remark 2.6. Notice that in Proposition 2.4 the metric space Y could be replaced by a perfectly normal topological space Y, i.e. one for which any closed set can be realized as the zero set of a continuous function ϕ: Y R. As will be seen from the following example, the assumption cannot be weakened to completely regular topological space Y, i.e. for which for every x Y and closed C Y not containing x there exists a continuous function f : Y [0, 1] with f(c) = {0} and f(x) = 1. Let us recall a few definitions from topology. Definition 2.7. Let t, t T be a collection of topological spaces. The product topology on the product space := t T t := {(x t ) t T : x t t } is the topology where unions of sets of the form are the open sets. {(x t ) : x ti U i, i = 1,..., n, n N, U i open in ti } Definition 2.8. Let be a set, Y a topological space and F = {f : Y }. The topology of pointwise convergence in F is defined as the product topology of Y via the identification (y x ) x f : x y x. Lemma 2.9. A sequence f n F converges to f F in the pointwise convergence topology if and only if f n (x) f(x) for all x. Proof. Suppose first that f n f in the poinwise convergence topology. Let x and let U Y be an open neighbourhood of f(x). Then by the assumption there exists some n 0 N such that f n {g : Y : g(x) U} for all n n 0. In other words, f n (x) U for all n n 0. By the arbitrariness of U and of x we have f n (x) f(x) for all x. For the other direction assume f n (x) f(x) for all x. Let x i for i = 1,..., k and U i Y be open neightbourhoods of f(x i ). By assumption there exists n 0 N such that f n (x i ) U i for all n n 0 and i = 1,..., k. Therefore f n {g : Y : g(x i ) U i, i = 1,..., k} for all n n 0. Since sets of this type form the basis 2 of the topology in pointwise convergence, we are done. 2 Recall that a basis of a topology is a collection of open sets such that any open set can be written as the union of sets from this collection.

8 8 ADVANCED MEASURE THEORY Example Consider the mappings f i : [0, 1] F ([0, 1]) := {g : [0, 1] [0, 1]} defined as f i (x)(y) = max{1 i x y, 0} and the space [0, 1] equipped with its usual topology and F ([0, 1]) with the pointwise convergence topology. The mappings f i converge pointwise to the mapping f(x) = χ {x}. Now each mapping f i is continuous and as such Borel measurable. However, the limit map f is not Borel: take a non-borel C [0, 1] and consider the open set U C := {x F ([0, 1]) : x(y) > 0}. Now f 1 (U C ) = C. y C Definition Let be a topological space. The Baire σ-algebra Ba() is the σ-algebra generated by the class {f 1 ((0, )) : f : R continuous}. By definition the Baire σ-algebra Ba() is the smallest σ-algebra with respect to which every continuous function f : (R, B(R)) is measurable. Since by Lemma 2.3 this measurability holds also with the Borel σ-algebra, we have Ba() B(). In perfectly normal topological spaces (like metric spaces), Ba() = B() since by definition any closed set can be realized as the zero set of a continuos function. By looking at the proof of Proposition 2.4 we immediately (exercise) notice that with Borel replaced by Baire, we may drop the extra assumption on the target Y. Proposition Let (, A) be a measurable space and Y a topological space. Then any pointwise limit of measurable maps f i : (, A) (Y, Ba(Y )) is measurable. Definition Let ( t, A t ) t T be measurable spaces. The product σ-algebra t T A t on the product space t T t is the σ-algebra generated by sets of the form {(x t ) : x ti B i, i = 1,..., n, n N, B i A ti }. Proposition Let and Y be separable metric spaces. Then B( Y ) = B() B(Y ). Proof. Let us first show that every open set U in Y belongs to B() B(Y ). Since is separable, there exists a countable basis {U n } n of the topology of. Then U is the union of sets of the form U n W α, where W α is open in Y. Now we can write U = n (U n V n ) B() B(Y ), where V n is the union of all open W α such that U n W α U. Therefore B( Y ) B() B(Y ). For the other direction take U open and notice that U Y B( Y ) since it is an open set. Since B( Y ) is a σ-algebra, this shows that for any A B() we have A Y B( Y ). Similarly A B( Y ) for all A B(Y ). Therefore for any A B() and B B(Y ) we have A B = (A Y ) ( B) B( Y ). Since such sets generate the σ-algebra B() B(Y ), we are done. Notice that in Proposition 2.14 we only used the assumption (in one of the inclusions) that one of the spaces has a countable basis. For the conlcusion an assumption to this direction is necessary as is shown by the next proposition.

9 ADVANCED MEASURE THEORY 9 Proposition Let be a Hausdorff space with cardinality greater than that of the continuum. Then B( ) B() B(). Proof. Let E denote the class of sets in E such that E and ( ) \ E are representable by continuum unions of sets of the form A B, A, B. Let us show that E is a σ-algebra. Clearly E. Let A n E, n N. Then A n = α I A n,α B n,α and ( ) \ A n = α I A n,α B n,α where the index set I in the unions is the continuum. Now A n,α B n,α and ( ) \ n N n N A n = n N A n = ( ) \ A n = n N n N α I A n,α B n,α = α I (α k ) I n N A n,α n B n,α n, where the sets n N A n,α n B n,α n are still rectangles and I has the same cardinality as the continuum. Therefore n A n E. Now let A, B E. Then A \ B = ( ) \ ((( ) \ A) B) E. Therefore E is a σ-algebra. Moreover, since B() B() is generated by sets of the form A B and ( ) \ (A B) = (( \ A) ) (A ( \ B)), we have B() B() E. Now we observe that the diagonal := {(x, x) : x } does not belong to E although it is closed in. Proposition Let and Y be Hausdorff spaces such that Y := {(y, y) : y Y } B(Y ) B(Y ). Then the graph of any Borel function f : Y is in B() B(Y ). Proof. The map F : Y Y Y : (x, y) (f(x), y) is measurable with respect to the algebras B() B(Y ) and B(Y ) B(Y ). Since by assumption Y B(Y ) B(Y ), we get 2.2. Souslin sets. graph(f) = F 1 ( Y ) B() B(Y ). Definition A subset of a Hausdorff space is called a Souslin set if it is the image of a Polish space under a continuous mapping. We denote by S() the class of Souslin sets in. A Hausdorff space is called a Souslin space if it is a Souslin set. Souslin sets are also called Analytic sets. Recall that a topological space is Hausdorff if every two points have separate neighbourhoods. A Polish space is a topological space that is homeomorphic to a complete separable metric space (thus Souslin sets are exactly images of separable metric spaces under continuous mappings). A model Polish space is the Baire space N equipped with the product topology. Proposition N is a Polish space.

10 10 ADVANCED MEASURE THEORY Proof. Define a distance on N by setting d((a i ), (b i )) = 2 inf{i : a i b i } for all (a i ), (b i ) N. This distance induces the product topology on N. The set {(a i ) N : a i = 42 for all i n, n N} (2.1) is countable and dense in (N, d). In order to see completeness of (N, d) take a Cauchy sequence ((a j i ) ) j=1 in (N, d). For all n N there exists j n such that d((a j i ), (ak i )) < 2 n for all j, k j n. Consequently a j n = a k n for all j, k j n. Thus we can define (a i ) N by setting a i = a j i i. Now d((aj i ), (a i)) 0 as j, and we have shown that (N, d) is complete. Proposition N is homeomorphic to the irrational numbers in (0, 1) (equipped with the usual topology from R). Proof. Let us construct a homeomorphism h from Z to P := (0, 1) \ Q. We first enumerate the rationals in (0, 1): {q n } n N = (0, 1) Q. Next we define inductively collections of nontrivial open intervals I s1,s 2,s 3,...,s n (0, 1) with s i Z and n N that have the following properties: (1) s 1,...,s I n Z s 1,...,s n P for all n N. (2) The right endpoint of I s1,...,s n is the left endpoint of I s1,...,s n+1 for all n N and s n Z. (3) The length of I s1,...,s n is less than 1 n for all n N, s 1,..., s n Z. (4) I s1,...,s n+1 I s1,...,s n for all n N, s 1,..., s n+1 Z. (5) q n is the endpoint of some interval I s1,...,s n for all n N. With these intervals we define h by setting for each (s n ) n Z I s1,...,s n =: {h((s n ) n )}. n N First of all, this intersection is always nonempty due to the fact that I s1,...,s n+1 I s1,...,s n for all (s n ) n Z, n N. It is a singleton since the diameters of the intervals go to zero as n. By (5) we have h((s n ) n ) / Q. Thus h: Z P. Since the intervals of level n are disjoint, the function h is injective. For all z P and n N there exist s 1,..., s n Z such that z I s1,...,s n. Thus h is surjective. The bijection h is homeomorphism since h({s 1 } {s n } Z ) = I s1,...,s n P for all n N, s 1,..., s n Z, and {s 1 } {s n } Z form a basis in Z and I s1,...,s n P in P. Proposition Every Polish space is the image of N under a continuous mapping. Proof. Let N be equipped with the distance defined in (2.1). Let ρ be a distance on such that (, ρ) is a complete separable metric space. Let {x i } i N be dense in. Let us define C i = B(x i, 1) for all i N and inductively for all n 2 and i 1,..., i n N take {x i1,...,i n,j} j N C i1,...,i n dense in C i1,...,i n and define closed nonempty sets C i1,...,i n,j = B(x i1,...,i n,j, 2 n ) C i1,...,i n for all j N. Now define {f((i j ) j N )} = n N C i 1,...,i n for all (i j ) j N N. Since C i1 C i1,...,i n C i1,...,i n,i n+1

11 ADVANCED MEASURE THEORY 11 are nested closed subsets of a complete metric space with diam(c i1,...,i n ) 0 as n the function f is well defined. By construction f(n ) =. Let (i j ) j N, (k j ) j N N with 0 < d((i j ), (k j )) < 1 2. Let m = inf{j : i j k j }. By definition of the distance d we have d((i j ), (k j )) = 2 m and by the assumption d((i j ), (k j )) < 1 2 we have m > 1. Then f((i j )), f((k j )) C i1,...,i m 1. Since C i1,...,i m 1 B(x i1,...,i m 1, 2 m+2 ) we have ρ(f((i j )), f((k j ))) 2 m+3 = 2 3 d((i j ), (k j )). Thus f is continuous. Corollary A subset of a Hausdorff space is Souslin if and only if it is the image of the Baire space under a continuous mapping. Consequently, a set is Souslin if and only if it is the image of the irrational numbers in (0, 1) under a continuous mapping. Theorem Let be a Hausdorff space and A i S(), i N. Then also A i S() and A i S(). Proof. First take sequences of complete separable metric spaces (E i, d i ) and continuous mappings f i : E i such that f i (E i ) = A i. Define a space E := i E i with the distance { min{d i (x, y), 1}, if x, y E i d(x, y) := 1, if x E i, y E j, i j. The space (E, d) is still a complete separable metric space. Define a mapping f : E by f Ei = f i. Then f is continuous and f(e) = A i. For the intersection, consider the product space Ẽ := E i. The space Ẽ is complete and separable for example with the distance d i (x i, y i ) d((x i ), (y i )) = 2 i (1 + d i (x i, y i )) (or with the distance Now the set min{d i (x i, y i ), 1}). D = {(x i ) i N Ẽ : f 1(x 1 ) = f j (x j ) for all j N} is closed in Ẽ. In order to see this take (x i) i N Ẽ \ D. By definition there exists some j N such that f 1 (x 1 ) f j (x j ). Since is a Hausdorff space there exist U 1, U j disjoint open neighbourhoods of f 1 (x 1 ) and f j (x j ). By continuity f1 1 (U 1) is an open neighbourhood of x 1 and fj 1 (U j ) an open neighbourhood of x j. Now U := U 1 E 2 E j 1 U j k=j+1 E k is an open neighbourhood of (x i ) i N and U Ẽ \ D. Thus D is closed. Define f : D as f((x i ) i N ) = f 1 (x 1 ). Now f is continuous from the complete separable metric space (D, d) to. Thus A i = f(d) is Souslin. Corollary Let be a Polish spaces. Then B() S(). Proof. Since is Polish there exists a distance d making it complete and separable. Then any closed set A is trivially Souslin via the identity map. If A is open, then it is a

12 12 ADVANCED MEASURE THEORY countable union of closed sets A = {x A : d(x, \ A) 1 n } and thus by Theorem 2.22 it is Souslin. Now consider the collection E of subsets A of such that both A and \ A are Souslin. Then E contains all open sets, is stable under taking complements by definition and is stable under countable unions and intersections by Theorem Thus E is a σ-algebra and B() E S(). The inclusion in Corollary 2.23 can be strict as we will later (soon) see. Theorem Let be a Hausdorff space and A i S(), i N be pairwise disjoint. Then there exist pairwise disjoint B i B() such that A i B i for all i N. Proof. Let us first reduce the proof to the case of two Souslin sets. Suppose that the claim holds for two sets and let A i S(), i N be pairwise disjoint. Then there exist disjoint B 1, B 2 B() such that A 1 B 1 and i=2 B 2. We continue inductively. Inside B 2 there exist B 2, B 3 B() such that A 2 B 2 and i=3 B 3, and so on. Let us now prove the claim for two disjoint sets A 1, A 2 S(). Then there exist (see the proof of Theorem 2.22) separable metric spacee E, closed sets C 1, C 2 E and a continuous map f : E such that f(c 1 ) = A 1 and f(c 2 ) = A 2 Suppose that there does not exist Borel sets separating A 1 and A 2. Writing for j = 1, 2 the sets as C j = C j,i with C j,i closed and diam(c j,i ) < 1 we observe that there exist i 1 and k 1 with f(c 1,i1 ) and f(c 2,k1 ) that cannot be separated by Borel sets. Indeed, if all such pairs could be separated by Borel sets B 1,i1,k 2 and B 2,i1,k 2, then so would C 1 and C 2 be separated by the Borel sets and B 2,j,i. j=1 B 1,i,j j=1 Now we find inductively closed sets C 1,i1,...,i n C 1,i1,...,i n 1 and C 2,k1,...,k n C 2,k1,...,k n 1 with diameters less than 2 n such that f(c 1,i1,...,i n ) and f(c 2,k1,...,k n ) cannot be separated by Borel sets. Since E is complete, there exist e 1, e 2 E such that {e 1 } = C 1,i1,...,i n and {e 2 } = C 2,k1,...,k n. Since is Hausdorff, the points f(e 1 ) and f(e 2 ) can be separated by disjoint open sets U 1 f(e 1 ) and U 2 f(e 2 ). By continuity of f there exists n N such that f(c 1,i1,...,i n ) U 1 and f(c 2,k1,...,k n ) U 2. Thus f(c 1,i1,...,i n ) and f(c 2,k1,...,k n ) are separated by the Borel sets U 1 and U 2, which is a contradiction. Corollary Let be a Hausdorff space. Suppose that A S() and \ A S(). Then A B(). Corollary 2.25 shows that the σ-algebra E in the proof of Corollary 2.23 was actually equal to B(). Theorem Let be a Polish space and A. Then the following are equivalent (i) A S() (ii) A is the projection of a closed set in N (iii) A is the projection of a Borel set in R.

13 ADVANCED MEASURE THEORY 13 Proof. Since N is a Polish space and the projection map is continuous, (ii) implies (i). Since R is Polish, by Corollary 2.23 any Borel set in R is Souslin and hence as a continuous image of a Souslin set also A is Souslin. Thus (iii) implies (i). Now let A S(). Then by Corollary 2.21 there exists a continuous map f : N such that f(n ) = A. The graph of f is a closed subset of N and its projection is A. Thus (i) implies (ii). Again assuming A S(), by Corollary 2.21 there exists a continuous map g : (0, 1) \ Q such that g((0, 1) \ Q) = A. Let x A and define g : R { g(y), if y (0, 1) \ Q g(y) := x, otherwise. Then g(r) = A and g is Borel. Now the graph of g is Borel in R by Propositions 2.14 and Thus (i) implies (iii). Theorem Let and Y be Souslin spaces and f : Y a Borel mapping. Then for all A S() and B S(Y ) also f(a) S(Y ) and f 1 (B) S(). If f is injective, then f 1 : f 1 () is Borel. Proof. By Propositions 2.14 and 2.16 we know that graph(f) B() B(Y ) (details as an exercise). The sets A Y and B are also Souslin in Y. Therefore as a projection to Y of the Souslin set (A Y ) graph(f) the set f(a) is Souslin. Similarly, as the projection to of the Souslin set ( B) graph(f) also f 1 (B) is Souslin. Assume then that f is injective. Take C B(). We know by the above that f(c) is Souslin. Since \ C B(), also f() \ f(c) = f( \ C) is Souslin. Now by Corollary 2.25 we have f(c) B(f()). Definition A bijection f : Y is called an isomorphism between measurable spaces (, A) and (Y, C) if f(a) = C and f 1 (C) = A. An isomorphism between two topological spaces equipped with their Borel σ-algebras is often called a Borel isomorphism. Theorem Let be a Souslin space. isomorphism f : S. Then there exists S S([0, 1]) and a Borel Proof. By Theorem 2.27 it suffices to construct an injective Borel map h: [0, 1]. First of all, the space is separable since it is the continuous image of a Polish space Y : the image of a countable dense set in Y is countable and dense in. Therefore there exists a countable collection {A i } i N of open sets of such that for every x, y there exist i, j N with A i A j = and x A i, y A j. Using these sets we construct the map h by setting h(x) = 3 i χ Ai (x), where χ Ai is the indicator function of A i. The map h is Borel and injective (exercise). Theorem S(N ) B(N ). Proof. The proof is based on finding a suitable Souslin set in A N N whose projection to the first N has the desired propoerty. First let {U i } i N be a countable basis of the topology in N N. We define { C := (x, (y i ) i N ) (N N ) N : x / } U yi. i N

14 14 ADVANCED MEASURE THEORY Then C is a closed set having the property that for any closed C N N there exists (y i ) i N such that C coincides with the section C (yi ), i.e. C (N N {(y i )}) = C {(y i )}. Using the set C we define B := {(x, y) N N : (x, z, y) C for some z N }. The set B is Souslin since it is the projection of a closed set in N N N. Since any Souslin set in N is a projection of some closed set in N N, that is, the projection of a section C y for some y N, any Souslin set in N is also a section B y for some y N. Finally, define A := B {(x, x) : x N } which is Souslin as the intersection (in a Polish space) of a Souslin set and a closed set (the diagonal). The projection S of A S = {x N : (x, x) B} is hence also Souslin. Let us show that N \ S is not Souslin. Suppose it is. Then there exists some y N with N \ S = B y. Assuming y S we have (y, y) B by the definition of S and hence y B y = N \ S. To the other direction, supposing y N \ S = B y we get that (y, y) B and thus y S. In both cases we arrive at a contradiction. Thus S S(N ) \ B(N ). In Corollary 2.21 we observed that Souslin sets are exactly the continuous images of N. Next we will show that Borel sets are exactly the continuous injective images from closed subsets of N. We start with a result similar to Theorem 2.26 on projections of closed sets in N. Theorem Let be a Polish space. Then every Borel set in is the injective image of some closed set in N under the natural projection to the space. Proof. Let E denote the class of Borel sets A in with the property that there exists a closed set in N whose injective projection A is. Let us use Theorem 1.7 to show that E the Borel σ-algebra. We need to show that E contains open and closed subsets of and that it is closed under countable unions of pairwise disjoint sets and closed under arbitrary countable unions. Clearly closed subsets belong to E. Let A be open. If we use (0, ) instead of N we can easily realize A as the injective projection of the closed set E := {(x, t) (0, ) : dist(x, \ A) = 1 t }. In order to replace (0, ) by N we need a continuous bijection f : D (0, ) from a closed set D N. Once we have found the map f the set {(x, n) N : (x, f(n)) E} is the required closed set projecting injectively to A. Let us construct the map f. We already know by Proposition 2.19 that (0, 1) \ Q is homeomopphic to N. Therefore there is a homeomorphism f : {1} N (0, 1) \ Q which we consider to be a (closed) subset of N. We enumerate {q n } = (0, 1) Q and map f((2, n, 1, 1,... )) = q n. The interval (0, 1) can then be trivially mapped to (0, 1 ) by a homeomorphism h. We let f = h( f) and observe that f is a continuous injective map from the closed set D = {1} N {2} N {(1, 1,... )}

15 ADVANCED MEASURE THEORY 15 with image (0, ). Suppose then that {A i } i N E is disjointed. Let C i N be a closed set whose injective projection is A i. Now we define a new closed set C = i N {i} C i N N. The projection of C to is injective and has i N A i as the image. Since N N is homeomorphic to N we have that i N A i E. Finally, let {A i } i N E and let C i N be a closed set whose injective projection is A i. Consider the set C = {(x, v 1, v 2,... ) : x, V i N, (x, v i ) C i for all i N} (N ). Then C is closed in (N ) and its injective projection is i N A i. homeomorphic to N (exercise) we are done. Since (N ) is Theorem Let be a Polish space and Y a Souslin space. Then for any B B() and f : B Y injective Borel mapping we have f(b) B(Y ). Proof. By Theorem 2.29 we may assume Y = [0, 1]. Since graph(f) B( [0, 1]), by Theorem 2.31 it is the injective projection of a closed set in [0, 1] N. By injectivity of f the projection of graph(f) to [0, 1] is also injective. Thus we may assume B to be a complete separable metric space (the closed subset of [0, 1] N ) and f to be continuous (the composition of the two projections). As in the proof of Proposition 2.20 we cover B by closed subsets E(n 1,..., n k ) of B with E( ) = B, and diam(e(n 1,..., n k )) < 2 k 2 and E(n 1,..., n k ) = E(n 1,..., n k, j) for all n i N and k N {0}. Define A( ) = B and inductively A(n 1,..., n k ) = A(n 1,..., n k 1 ) E(n 1,..., n k ) \ j=1 j<n k A(n 1,..., n k 1, j). For every k N the collection {A(n 1,..., n k )} is a (disjointed) Borel partition of B. Thus by injectivity of f the collection {f(a(n 1,..., n k ))} is a (disjointed) Souslin partition of f(b). Now by Theorem 2.24 there exist disjoint Borel sets B(n 1,..., n k ) in Y such that f(a(n 1,..., n k )) B(n 1,..., n k ) f(a(n 1,..., n k )) (2.2) (for the second inclusion consider the intersection with the Borel set f(a(n 1,..., n k ))). Using these we can define a new collection of Borel sets B(n 1,..., n k ) in Y inductively by setting B(n 1 ) = B(n 1 ) and B(n 1,..., n k ) = B(n 1,..., n k 1 ) B(n 1,..., n k ). Still (2.2) holds with B(n 1,..., n k ) in place of B(n 1,..., n k ). Let us then show that f(b) = B(n 1,..., n k ), (2.3) k=1 (n 1,...,n k ) N k from which the claim easily follows.

16 16 ADVANCED MEASURE THEORY Since f(b) = f(a(n 1,..., n k )) B(n 1,..., n k ), (n 1,...,n k ) N k (n 1,...,n k ) N k the inclusion f(b) B(n 1,..., n k ) k=1 (n 1,...,n k ) N k is obvious. To see the other inclusion, take y B(n 1,..., n k ). k=1 (n 1,...,n k ) N k By definition, for every k N there exists (n 1 (k),..., n k (k)) N k such that y B(n 1 (k),..., n k (k)). Since the sets B(n 1,..., n k ) are disjoint the element (n 1 (k),..., n k (k)) is unique and since the sets B(n 1,..., n k ) nested we have n i (k) = n i (k + 1) for all i k. Therefore y B(n 1 (1), n 2 (2),..., n k (k)) In particular k=1 Now take (n 1,...,n k ) N k B(n 1,..., n k ) y k=1 (n i ) N k=1 (n i ) N k=1 B(n 1,..., n k ) f(e(n 1,..., n k )). (n i ) N k=1 f(e(n 1,..., n k )). Then there exist for every k N a point x n1,...,n k E(n 1,..., n k ) such that f(x n1,...,n k ) y as k. Since B is complete and {x n1,...,n k } k is Cauchy is converges to a point x B. By continuity, f(x) = y. Thus (2.3) is established. Corollary A set in a Polish space is Borel if and only if it is the image of a closed subset of N under a continuous injective mapping. Proof. the if part follows from Theorem For the only if part we know by Theorem 2.31 that every Borel set can be obtained as the injective image of a closed set in N under the projection to. What remains to show is that any Polish space can be obtained as a continuous injective image of a closed set in N. (Exercise)

17 ADVANCED MEASURE THEORY 17 Exercise To be returned to the lecturer at the latest on Feb. 25th. (i) Show that Definition 2.13 gives a σ-algebra. (This is accidentally trivial.) (1p) (ii) Show that a graph of any Borel function between separable metric spaces and Y is in B() B(Y ). (1p) (iii) Show that (N ) is homeomorphic to N, when the spaces are equipped with the product topologies. (1p) (iv) Provide the details to the proof of Theorem (Why does there exists a countable collection of open sets A i with the mentioned property? (It might help to prove that is hereditarily Lindelöf.) Why is h Borel and injective?) (2p) (v) Show that any nonempty complete metric space without isolated points has a non-borel Souslin subset. (Hint: Show that such space contains a subset homeomorphic to N.) (3p) (vi) Show that there exists a Borel set B R 3 such that y B {x R3 : d(x, y) = 1} is not Borel. (Hint: Combine Theorem 2.26 (iii) and exercise (v).) (2p) (vii) Show that for any Polish space there exists a closed set C N and a continuous injective map f : C with f(c) =. (Hint: First show that there exists such closed set for = R. Then show that is homeomorphic to a closed subset of R. Finally use exercise (iii).) (3p)

18 18 ADVANCED MEASURE THEORY 3. Measurable selections Definition 3.1. Let f : Y be a map. A map g : f() with f(g(y)) = y for all y f() is called a selection (or section) of the map f. In this section we take a quick look at the problem of making measurable selections. It is clear that for a map f : Y one can always find a selection g by picking an element g(y) f 1 ({y}). In applications we would like to make the selection to be as good as possible. For example, if f is Borel, we would like g to be Borel also. However, this is not always possible. What is possible is to find a selection which is measurable with respect to the σ-algebra generated by the Souslin sets. Before proving this we prove the existence of a Borel selection in a nice situation. Before proving the selection theorems let us look at a model case where such theorems could be used. Consider the map f : C([0, 1]; R n ) R n R n : γ (γ(0), γ(1)) where the space C([0, 1]; R n ) of continuous maps from [0, 1] to R n is equipped with the distance d(γ 1, γ 2 ) = sup γ 1 (t) γ 2 (t). t [0,1] The map f is Lipschitz and there is an obvious way to find a selection for it. Namely, for each pair (x, y) R n R n we take g((x, y)) to be the constant speed geodesic from x to y, or in other words the affine map from [0, 1] to R n having the correct endpoints. In this case, the selection g will also be Lipschitz. Consider now the same question for the map f : C([0, 1]; S 1 ) S 1 S 1 : γ (γ(0), γ(1)). Again, we could take the selection to be constant speed geodesics between the given endpoints. However, in S 1 the geodesic is not unique for antipodal points. We are then forced to make a selection. In this case it is still easy to make the selection: for example, out of the two possible geodesics between antipodal points we can choose the one going in the clockwise direction. This selection is no longer continuous, but it is Borel. It is also clear that one can make a bad selection here by taking a non-borel set where we choose the clockwise direction and for the rest the anticlockwise direction. Consider then the question for a (complete separable) geodesic metric space. Again we could choose to select among geodesics connecting the given endpoints. However, now there could be infinitely many geodesics joining the given endpoints. How do we make a consistent selection? Is it even possible? It turns out that if is compact, we can make a Borel selection and in the noncompact case we can at least make a Souslin selection. Notice that in the case of the sphere we made a consistent selection by saying that clockwise direction is always preferred over the anticlockwise. Copying this idea, in the general situation we would then like to order all the possible choices in a way that respects the measurability. In the proofs of the following two theorems we use this idea; first by mapping to [0, 1] and using its natural order, and in the second proof by putting the lexicographic order on N. Theorem 3.2. Let be a compact metric space, Y a Hausdorff space and f : Y a continuous map. Then there exists a Borel set B such that f(b) = f(), f is injective on B and f 1 : f() B is Borel. Proof. Since is second countable and compact, so is f(). Therefore as a second countable compact Hausdorff space, f() is metrizable. Let us reduce the case of general compact to the case [0, 1]. For this we need a continuous map ϕ: K with K [0, 1] compact and ϕ(k) =. Such continuous map can be constructed similarly as the map in Proposition

19 ADVANCED MEASURE THEORY from K a 1 3-Cantor set using the fact that is compact. (Exercise) Now, supposing the claim holds for domains in [0, 1], we find a Borel set B 0 [0, 1] such that f ϕ: B 0 Y is injective and f(ϕ(b 0 )) = f(ϕ(k)) = f(). Then ϕ is injective on B 0 and hence by Theorem 2.32 also B = ϕ(b 0 ) is Borel. By injectivity of f ϕ also f is injective on B. Let us then prove the claim in the case [0, 1]. Set for every y f() g(y) = inf{x : f(x) = y}. This way we obtain a function g : f() [0, 1]. The function g is Borel, since by continuity of f {y : g(y) c} = f([0, c]) is closed for all c R. Clearly f(g(y)) = y for all y f() and thus g is injective, B = g(f(y )) is Borel and f(b) = f(). Theorem 3.3. Let and Y be Souslin spaces and let f : Y be a Borel mapping such that f() = Y. Then there exists a mapping g : Y such that f(g(y)) = y for all y Y and g is measurable between (Y, σ(s(y ))) and (, B()). Moreover, g(y ) σ(s()). Proof. Let us first prove the claim for a continuous f. Since is Souslin, it is the image of N under a continuous map p. Therefore we may assume = N. As in Theorem 3.2 we would like to select the minimal element from f 1 ({y}). In order to do this we need to order N. A natural order is the lexicographic order: (n i ) < (k i ) if there exists m N such that n m < k m and n i = k i for all i {1,..., m 1}. Let us now define for all y Y the point g(y) N to be the minimal element in f 1 ({y}) in the lexicographic order. Such minimal element exists since f 1 ({y}) is nonempty and closed by the continuity of f. By construction f(g(y)) = y. Let us show that for every B B(N ) we have g 1 (B) σ(s(y )). Consider first a set B of the form: for a fixed b N, B = {n N : n b}. On one hand, if g(y) B, then y f(b). On the other hand, if y = f(z) for some z B, then g(y) z and thus y g 1 (B). Therefore g 1 (B) = f(b) S(Y ). It remains to observe that the sets B of the above type generate the σ-algebra B(N ). Let us then consider the general case with Borel f. As was observed in the proof of Theorem 2.27, graph(f) B() B(Y ), hence in particular the graph of f is Souslin. Since the projection π Y of graph(f) to Y is continuous, by the proof of the theorem in the continuous case we know that there exists a measurable map ϕ: (Y, σ(s(y ))) (graph(f), B(graph(f)) with π Y (ϕ(y)) = y for all y Y. Let π : graph(f) be the natural projection and let g = π ϕ. Now f(g(y)) = f(π (ϕ(y))) = π Y (ϕ(y)) = y for all y Y. By the measurability of ϕ and continuity of π the map g : (Y, σ(s(y ))) (, B()) is measurable. Finally, let us show that g(y ) σ(s()). Define h(x) = g(f(x)) for all x. Then g(y ) = {x : h(x) = x} = {x : f n (x) = f n (h(x))}, where f n : R form a collection of Borel functions with the property that for any x 1, x 2 there exists n N such that f n (x 1 ) f n (x 2 ). (The existence of such functions

20 20 ADVANCED MEASURE THEORY follows for example from the existence of sets A i in the proof of Theorem 2.29.) Since the difference of two measurable functions is measurable (exercise), it suffices to show that f n h = f n g f : (, σ(s)) (R, B(R)) is measurable. We already know that f : (, B()) (R, B(R)), g : (Y, σ(s(y ))) (, B()) have the right measurability. From the measurability of f : (, B()) (Y, B(Y )) it follows by Theorem 2.27 that f : (, σ(s())) (Y, σ(s(y ))) is measurable. Let us next give part of the proof of the existence of a Borel function on the unit interval [0, 1] for which there does not exist a Borel selection. Theorem 3.4. There exist two disjoint sets A and B in N such that their complements are Souslin 3 and A and B cannot be separated by Borel sets. Proof. We will skip the proof. Theorem 3.5. There exists a continuous surjective map F : N {0, 1} N such that there is no Souslin set S with F (S) = N and F restricted to S injective. In particular, there is no Borel selection for F. Proof. By Theorem 3.4 there exist two disjoint sets C 0 and C 1 in N with Souslin complements A 0 and A 1 such that there is no Borel set separating C 0 and C 1. Since A 0 and A 1 are Souslin, there exist continuous surjections F 0 : N A 0 and F 1 : N A 1. Let F : N {0, 1} N be defined by F (v, i) = F i (v). This way we obtain a continuous surjection, since N = A 0 A 1. Assume now that there exists Souslin S such that F (S) = N and F restricted to S is injective. Define S i = {v N : (v, i) S} for i {0, 1}. Now the sets B i = F i (S i ) for i {0, 1} are Souslin and disjoint by the injectivity of F on S. Since B 0 B 1 = N, by Corollary 2.25 B 0 and B 1 are Borel. But now B i A i and thus C i B 1 i contradicting the fact that C 0 and C 1 can be separated by Borel sets. Since by Theorem 2.32 any injective image of N is Borel, there does not exist a Borel selection. Corollary 3.6. There exists a surjective Borel function f : [0, 1] [0, 1] such that there is no Borel function g : [0, 1] [0, 1] with f(g(y)) = y for all y [0, 1]. 4. Borel, Baire and Radon measures For simplicity, in this section we will assume measures to be finite. Many of the statements are easily modified for more general measures, for example those giving finite measure for compact sets. Definition 4.1. Let be a topological space. A finite measure on the Borel σ-algebra B() is called a Borel measure on. Similarly, a finite measure on the Baire σ-algebra B() is called a Baire measure on. Definition 4.2. A Borel measure µ on a topological space is called a Radon measure if for every A B() and ε > 0 there exists a compact set K ɛ A such that µ(a \ K ɛ ) < ε. We will later see that on complete separable metric spaces all Borel measures are Radon, but will also see an example of a non-radon Borel measure on a compact topological space. Before this let us give a few more definitions. 3 In such case, A and B are called co-souslin, or coanalytic

21 ADVANCED MEASURE THEORY 21 Definition 4.3. Let be a topological space, A a σ-algebra on. A measure µ on A is called tight (on A) if for every ε > 0 there exists a compact set K ε such that µ(b) < ε for all B A with B K ε =. Notice that for a Borel measure µ on a Hausdorff space tightness can be defined equivalently by just requiring that for every ε > 0 there exists a compact set K ε such that µ(\k ε ) < ε. Clearly any Radon measure is tight. Definition 4.4. Let be a topological space, A a σ-algebra on. A measure µ on A is called regular if for every B A and every ε > 0 there exists a closed set F ε B with B \ F ε A and µ(a \ F ε ) < ε. Clearly any Radon measures is also regular. For the converse direction we have: Lemma 4.5. Let µ be a Borel measure on a Hausdorff space. Suppose that µ is tight and regular. Then µ is Radon. Proof. Let A B() and ε > 0. By tightness of µ there exists a compact set K ε such that µ( \ K ε) < ε. By regularity of µ there exists a closed set F ε A with µ(a \ F ε ) < ε. Now define K ε := K ε F ε A. As the intersection of a closed set with a compact set K ε is compact. Since (A \ K ε ) (( \ K ε) (A \ F ε )), we have µ(a \ K ε ) µ( \ K ε) + µ(a \ F ε ) < 2ε, and thus the measure µ is Radon. Theorem 4.6. Let (, d) be a metric space. Then every Borel measure µ on is regular. Proof. Let us define A := {B B() : for every ε > 0 there exist a closed F ε B and an open U ε B such that µ(u ε \ F ε ) < ε}. Our aim is to show that A = B(). Let B be closed. Then for any ε > 0 we can take F ε = B. For the approximation by open sets from the outside, consider open sets O n := B(B, 1 n ). Then µ(o n) µ(b) since B = n N O n. By construction A is closed under taking complements. Thus it only remains to prove that A is closed under countable unions. Let B j A for j N. Let ε > 0. Then for every j N there exist closed F j B j and open U j B j with µ(u j \ F j ) < ε2 j. Then Since there exists k N such that µ U j \ F j µ (U j \ F j ) < ε2 j = ε. j N j N j N n µ j=1 F j µ F j as n, j N µ U j \ j N k j=1 F j < ε.

22 22 ADVANCED MEASURE THEORY Finally we observe that k j=1 F j is a closed subset of j N B j which is a subset of the open set j N U j. The assumption of (, d) being a metric space is not sufficient for tightness of Borel measures on. This is seen from the next example. Example 4.7. Let λ be the Lebesgue measure on [0, 1] and [0, 1] be such that and inf{λ(a) : A B([0, 1])} = 1 sup{λ(a) : A B([0, 1])} = 0. (The existence of such set is left as an exercise.) Equip the space with the restriction of the Euclidean distance on [0, 1] and consider the measure µ on B() defined by µ(b) = inf{λ(a) : B A B([0, 1])}. (The fact that µ is a measure on B() is left as an exercise.) By Theorem 4.6 as a Borel measure on a metric space µ is regular. However, since any compact K is also compact in [0, 1], we have µ(k) = 0 and since µ() = 1 the measure µ is not tight. The metric space in the previous example was separable. If we further add the assumption that (, d) is complete, we get tightness of Borel measures. Theorem 4.8. Let (, d) be a complete separable metric space. Then every Borel measure µ on is Radon. Proof. By Lemma 4.5 and Theorem 4.6 we only need to show that µ is tight. Let ε > 0. Since is separable, there exists a countable dense set {x i } i N in. By countable additivity, for any n N there exists N n N such that µ( \ W n ) < ε2 n, where W n = N n B(x i, 2 n ). Let W = n N W n. Then µ( \ W ) n N µ( \ W n ) n N ε2 n = ε and W is compact by the completeness of (, d) and the fact that by construction W is totally bounded. Example 4.9 (Dieudonné measure). Let be the set of all ordinals not exceeding the first uncountable ordinal ω 1. Then is an uncountable well-ordered set with the maximal element ω 1. Equip with the order topology whose base consists of all sets of the form {x : x < α}, {x : α < x < β} and {x : x > α}, where α, β ω 1. Notice that {x : x α} is at most countably infinite for α ω 1. Let 0 = \ {ω 1 } and denote by F 0 the class of uncountable closed subsets of 0 (in the topology induced by ). Let us then define a measure µ by setting { 1, if there exists A B with A F 0, µ(b) = 0, otherwise. We need to show that µ is a measure on B(). To this end we define a larger σ-algebra E that contains B() and show that µ is a measure on E. We define E = {A : exists B F 0 such that B A or B \ A}.

23 ADVANCED MEASURE THEORY 23 Clearly E and E is closed under taking complements. Next we show that E is closed under countable intersections. Let A n E for n N. Suppose there exists n 0 N such that \ A n0 contains an element of F 0. Then so does A = n N A n. Suppose then that for every n N there exists B n F 0 such that B n A n. We claim that B = n N B n F 0. Suppose this is not the case. That means that B is countable and thus there exists α < ω 1 such that B [0, α]. By induction we find a strictly increasing sequence of ordinals α j (α, ω 1 ) such that (α j ) j N contains infinitely many elements from every B n. Then α j has a limit α B for which α > α. This contradicts the choice of α. Therefore E is closed under countable intersections and thus a σ-algebra. For a countable closed set A there exists α < ω 1 such that A [0, α] and thus an element of F 0 is contained in its complement. For an uncountable closed set A we can simply take A 0 F 0. Thus B() E. Let us check that µ is a measure on E. Clearly µ( ) = 0. Let {A n } n N E be disjointed. Suppose that k l and A k and A l both contain an element of F 0 ; call them B k and B l. With the above shown existence of an increasing sequence (α j ) j N containing infinitely many elements from B k and B l we see that the limit α B k B l. Thus A k A l which contradicts their choice. Therefore at most one A n contains an element of F 0. If no A n contains an element of F 0, then their complements do and by the above also n N ( \ A n) implying that n N A n does not contain an element of F 0. In conclusion, µ is countably additive on E. Every point x ω 1 has a neighbourhood of measure zero, hence µ(k) = 0 for every compact set K 0. Since µ({ω 1 }) = 0, µ is not Radon. Definition Let µ be a measure on the measurable space (, A). The Lebesgue extension of (, A, µ) is the triple (, A µ, µ), where the extension of µ is defined as the outer measure µ in (1.2) and A µ is the σ-algebra of all µ -measurable sets. Definition Let be a topological space. A set A is called universally measurable if A B µ () for all Borel measures µ. Similarly, a set A is called universally Radon measurable if A B µ () for all Radon measures µ on. A function f : Y between topological spaces and Y is called universally measurable if f 1 (B) is universally measurable for all B B(Y ). Let us now give versions of the classical Egoroff and Lusin theorems. On the real analysis course the Egoroff and Lusin theorems were proven for real valued functions from a metric space. Theorem 4.12 (Egoroff s Theorem). Let (, A) be a measure space with a finite measure µ and let (Y, d) be a separable metric space. Suppose we have a sequence of measurable mappings f i : (, A) (Y, B(Y )) converging µ-a.e. to a mapping f. Then for every ε > 0 there exists a set E ε A with µ( \ E ε ) ε so that f i converges to f uniformly on E ε. Proof. The proof is essentially the same as in the more classical case. By Proposition 2.4 we may assume f is also measurable. Define A k,m := {x : d(f i (x), f(x)) < 1k } for all i m. Compared to the proof given on the real analysis course, we only need to show that A k,m A for all k, m N. To see this, notice that A k,m = Fi 1 ([0, 1/k)), where F i (x) = d(f i (x), f(x))