Measure Theory 1 Classes of Subsets 1.1 Topology Definition A Topological space (X; T ) consists of a non-empty set X to- gether with a collection T of subsets of X such that (T1) X; Á 2 T , T (T2) If A ; :::; A 2 T then n A 2 T , 1 n i=1 i S (T3) If Ai 2 T ; i 2 I for some index set I then i2I Ai 2 T . The elements of T are called the open sets. Note Every set has at least two topologies on it, namely the trivial ones T = fX; Ág and T = P(X), i.e. the collection of all subsets of X. T Lemma 1.1 If Tj; j 2 J are topologies on X then j2J Tj is a topology on X. Proof T (T1) X; Á 2 T for all j so X; Á 2 T . j j2J j T (T2) Take any finite collection of sets A ; :::; A 2 T : Then A ; :::; A T 1 n Tj2J j T 1 n 2 T for each j and so n A 2 T for each j and so n A 2 T . j i=1 i j T i=1 i j2J j (T3) Take any collection of sets A 2 T ; i 2 I. Then A 2 T ; for S i j2J j iS j all i 2 I and all j 2 J and so A 2 T for each j and thus A 2 T i2I i j i2I i j2J Tj. ¥ Corollary 1.2 There exists a topology U on R containing all intervals (a; b) and such that if T0 is any other topology containing all such intervals then U ⊆ T0. Proof There exists at least one topology on RT containing all (a; b), namely T0 = P(R), the power set of R. Set U = T ; the intersection over all topologies containing all the intervals (a; b). This is a topology by Lemma 1.1 and the minimality property is immediate from the definition. ¥ Definition We say that U is the usual topology on R. Definition Given any collection A of subsets of X, the intersection of all topologies containing A is said to be the topology generated by A. Example 1 The usual topology on R is the topology generated by the in- tervals (a; b). There are other topologies on R, such as Example 2 The co-finite topology on R is defined by A 2 co-finite if, and only if, either A = Á or jAcj < 1. (That is, the complement is finite.) For the following definition we require the idea of pre-image of a map f : X ! Y . So if A ⊆ Y then the pre-image of A is given by 1 f ¡1(A) = fx 2 X : f(x) 2 Ag: Definition Given topological spaces (X; TX ); (Y; TY ) and a map f : X ! Y we say that f is continuous with respect to TX ; TY if ¡1 U 2 TY ) f (U) 2 TX : (That is, the preimage of an open set is open.) We will require later the following result. TheoremS 1.3 (Heine-Borel) If [a; b] ⊆ R is covered by a collection of (ci; di), so [a; b] ⊆ i2I (ci; di), then there exists a finite sub-collection of the (ci; di), SN which can relabeled as 1 · i · N such that [a; b] ⊆ i=1(ci; di). Proof See Appendix. ¥ Theorem 1.4 (Lindel¨of’s Theorem) If G = fI® : ® 2 Ag is a collection of intervals (a; b) ⊆ R, possibly an uncountable collection, then there exists a countable subcollection fIi : i ¸ 1g ⊆ G such that [ [1 I® = Ii: ®2A i=1 Proof See Appendix ¥ Corollary 1.5 Let S be the set of all countable unions of intervals (a; b) ⊆ R. Then S = U, the usual topology. S1 Proof Let S 2 S so S = i=1(ai; bi) for some (ai; bi) But (ai; bi) 2 U for all S1 i, and U is a topology so i=1(ai; bi) 2 U. Thus S 2 U and so S ⊆ U. We now show that S is a topology by verifying the properties T1, T2 and T3. S (T1) R = n2N(¡n; n) 2 S and Á = (0; 0) 2 S. (T2) If S1;S2;S3; :::; Sn 2 S then each [1 Si = Iji for some Iji = (aji ; bji ): ji=1 Then 2 \n \n [1 Si = Iji i=1 i=1 j =1 à i ! à ! à ! [1 [1 [1 = Ij1 \ Ij2 \ ::: \ Ijn j1=1 j2=1 jn=1 [1 [1 [1 = ::: (Ij1 \ Ij2 \ ::: \ Ijn ) : j1=1 j2=1 jn=1 Importantly we have a finite intersection of open intervals which is therefore an open interval. (It is possible that an infinite intersection of open intervals Tn is closed.) Thus i=1 Si is a countable union of open intervals, hence it is an element of S. S (T3)S If Sk S2 S for k 2 K, perhaps an uncountable collection, then S = 1 I is a, possibly uncountable, union of intervals k2K k k2K jk=1 jk (a ; b ). But by Lindel¨of’s Theorem this can be written as a countable jk jik S union and so k2K Sk 2 S. Hence S is a topology containing the intervals (a; b). But U is the minimal topology containing these intervals. Hence U ⊆ S. Thus U = S. ¥ 1.2 Rings Definition A collection, S, of subsets of the non-empty set X is a semi-ring if (i) Á 2 S, (ii) A; B 2 S ) A \ B 2 S, SN (iii) A; B 2 S ) A n B = i=1 Ei, a finite disjoint union of Ei 2 S. Example 3 The collection, P, of all finite intervals of the form (a; b] ⊆ R form a semi-ring. This is the most important example of a semi-ring we shall study. It should be compared with the collection of all intervals of the form (a; b) that does not form a semi-ring. Verification is left to the student. Definition A non-empty collection, R, of subsets of X is a ring if (i) A; B 2 R ) A [ B 2 R, (ii) A; B 2 R ) A n B 2 R. Note (i) R 6= Á implies that there exists a set A in R and so Á = A n A 2 R. *(ii) Such a collection, with the operations 3 A + B = A M B and A:B = A \ B is a ring as defined in the second year course. (Here A M B = (AnB)[(BnA) is the symmetric difference.) Note The collection P is not a ring, for instance (0; 3] and (1; 2] 2 P but (0; 3] n (1; 2] = (0; 1] [ (2; 3] 2= P. Definition The collection, E, of all finite unions of disjoint members of P; is called the set elementary figures in R. Example The collection E is a ring. Verification. Sm Sn Let A; B 2 E, so A = i=1 Ai and B = j=1 Bj where Ai;Bj 2 P, disjoint unions. Then à ! [m c c A n B = A \ B = Ai \ B i=1 [m c = (Ai \ B ) ; i=1 Sn a disjoint union. Writing Ai = (a; b] and B = j=1(aj; bj] we see that ( ) n[¡1 c Ai \ B = (a; b] \ (¡1; a1] [ (bj; aj+1] [ (bn; +1] j=1 which, on applying the distributive law, is seen to be in E. Then, since the Ai are disjoint we see that A n B is a disjoint union of sets from E and so A n B 2 E, i.e. condition (ii) holds. Also A [ B = (A n B) [ (A \ B) [ (B n A); a disjoint union. By what we have already proved it suffices to show that A \ B 2 E. But A \ B = Sm Sn i=1 j=1(Ai\Bj) a disjoint union of sets from P and so in E. Thus condition (i) is satisfied. Hence E is a ring. ¥ Definition A ring R is a σ-ringS if it is closed under countable unions. That is, given A 2 R; n ¸ 1 then A 2 R: n n¸1 n S Note Given A 2 R a σ-ring, write A = A . It is trivial that T n n¸1 n n¸1 An ⊆ A but this observation means that we can write à ! \ \ An = A n A n An : n¸1 n¸1 4 Here à ! à ! \ \ c [ c A n An = A \ An = A \ An n¸1 n¸1 n¸1 [ [ c = (A \ An) = (A n An): n¸1 n¸1 S Since A and A 2 R for all n ¸ 1 we have AnA 2 R, and so (AnA ) 2 n ¡ T ¢ T n n¸1 n R and thus A n A n n¸1 An or n¸1 An 2 R. Hence σ-rings are closed under countable intersections. 1.3 Fields Definition A non-empty collection F of subsets of a non-empty set X is a field (or algebra) if (i) X 2 F; (ii) F is a ring. Further X is a σ-field (or σ-algebra) if (i) X 2 F, (ii) F is a σ-ring. Note A field can be defined to satisfy (i) X 2 F, (ii) If A 2 F then Ac 2 F, (iii) If A; B 2 F then A [ B 2 F. Similarly for σ-fields.