1. σ-algebras Definition 1. Let X be any set and let F be a collection of subsets of X. We say that F is a σ-algebra (on X), if it satisfies the following. (1) X 2 F. (2) If A 2 F, then Ac 2 F. 1 (3) If A1;A2; · · · 2 F, a countable collection, then [n=1An 2 F. In the above situation, the elements of F are called measurable sets and X (or more precisely (X; F)) is called a measurable space. Example 1. For any set X, P(X), the set of all subsets of X is a σ-algebra and so is F = f;;Xg. Theorem 1. Let X be any set and let G be any collection of subsets of X. Then there exists a σ-algebra, containing G and smallest with respect to inclusion. Proof. Let S be the collection of all σ-algebras containing G. This set is non-empty, since P(X) 2 S. Then F = \A2SA can easily be checked to have all the properties asserted in the theorem. Remark 1. In the above situation, F is called the σ-algebra generated by G. Definition 2. Let X be a topological space (for example, a metric space) and let B be the σ-algebra generated by the set of all open sets of X. The elements of B are called Borel sets and (X; B), a Borel measurable space. Definition 3. Let (X; F) be a measurable space and let Y be any topo- logical space and let f : X ! Y be any function. We say that f is measurable, if for any open set U ⊂ Y , f −1(U) 2 F. Example 2. (1) If (X; F) is a measurable space with X a topologi- cal space and B ⊂ F, then any continuous function from X to any topological space is measurable. (2) If f : X ! Y is measurable and g : Y ! Z is continuous, g ◦ f is measurable. Theorem 2. Let (X; F) be a measurable space and let f : X ! Y be a function, where Y is any set. (1) If G is defined as the collection of all subsets A of Y such that f −1(A) 2 F, then G is a σ-algebra on Y . (2) If f : X ! [−∞; 1] is any function, f is measurable if and only if f −1((a; 1]) 2 F for any a 2 [−∞; 1]. 1 2 Proof. The first part is mere checking. For the second part, notice that since (a; 1] is open in [−∞; 1], the condition is clearly necessary. To 1 check sufficiency, one notes that [−∞; b) = [n[−∞; b− n ] for any b and [−∞; c] = (c; 1]c. So, one gets that f −1([−∞; b)) 2 F too for any b and thus we get f −1((a; b)) 2 F for any a; b, since (a; b) = [−∞; b) \ (a; 1] and these generate all open sets. Theorem 3. Let fn : X ! [−∞; 1] be a sequence of measurable functions. Then, supfn; inffn; lim supfn; lim inffn are all measurable. −1 −1 Proof. Let g = supfn. Then g ((a; 1]) = [nfn ((a; 1]) and so we are done by the previous theorem. A similar proof applies to infimum. Let gn = supk≥nfk. Then all the gn's are measurable by the previous part. Then lim supfn = infgn and hence it too is measurable. Similar argument applies to lim inf. We get as a corollary the following results. Corollary 1. Let f; g : X ! [−∞; 1] be measurable. Then so are maxff; gg and minff; gg. In particular, writing f + = maxff; 0g and f − = − minff; 0g, we see that these are measurable. Remark 2. In all the above, we have dealt with [−∞; 1] which has the advantage that supremum, lim sup etc. make sense. But, notice that if a; b 2 [−∞; 1], then a + b may not make sense. On the other hand, if we worked with R, then the latter obviously makes sense, but not necessarily the former. In most cases, it should be clear which space we are working in and why. Definition 4. A function f : X ! [0; 1) is called simple, if it takes only finitely many values. Example 3. (1) If A ⊂ X, the characteristic function χA defined as χA(x) = 1 if x 2 A an zero otherwise, is a simple function. (2) More generally, if s is a simple function, let a1; : : : ; an be its −1 Pn finitely many values and let Ai = s (ai). Then s = i=1 aiχAi . Notice that [iAi = X and Ai \ Aj = ; if i 6= j (pairwise disjoint). (3) If s is a simple function as above from a measurable space X, s is measurable if and only if all the Ai's are measurable. P P (4) If s = aiχAi and t = bjχBj are simple (measurable) func- tions, so is s + t. Letting Cij = Ai \ Bj, one can see that P s + t = (ai + bj)χCij . Theorem 4. Let f : X ! [0; 1] be a measurable function. Then there exists simple measurable functions s1 ≤ s2 ≤ · · · ≤ f such that lim sn = f. 3 n −1 i−1 i Proof. For any n and an i such that 1 ≤ i ≤ n2 , let En;i = f ([ 2n ; 2n )) −1 and let Fn = f ([n; 1]). Then these are all measurable sets, pairwise Pn2n i−1 disjoint for a fixed n and cover all of X. So, sn = i=1 2n χEn;i + nχFn is a simple measurable function. Easy to check that sn ≤ sn+1 ≤ f for all n and lim sn = f. 2. Measures Definition 5. Let (X; F) be a measurable space. A (positive) measure on X is a function µ : F! [0; 1] which is countably additive. That is, if A1;A2;::: 2 F is a countable collection of pairwise disjoint sets, 1 P1 then µ([n=1An) = n=1 µ(An). Since µ(A) = 1 for all A 2 F is trivially a measure by the above, to avoid this case, we will always assume that there exists some A 2 F with µ(A) < 1 for our definition. Example 4. (1) Let X be any set and consider the function µ : P(X) ! [0; 1] given by µ(A) is the cardinality of A if A is finite and 1 otherwise. Then µ is a positive measure. This is called the counting measure. (2) Again consider (X; P(X)) and let a 2 X, fixed. Define µ(A) = 1 if a 2 A and zero otherwise. This is a positive measure. Theorem 5. Let µ be a positive measure on (X; F). (1) µ(;) = 0. n (2) If A1;A2;:::;An 2 F, pairwise disjoint, then µ([i=1Ai) = Pn i=1 µ(Ai). (Finite additivity) (3) If A ⊂ B are measurable, µ(A) ≤ µ(B). (Monotonicity) 1 (4) If A1 ⊂ A2 ⊂ · · · are measurable sets and A = [n=1An, then limn!1 µ(An) = µ(A). 1 (5) If A1 ⊃ A2 ⊃ · · · and A = \n=1An with µ(A1) < 1, then lim µ(An) = µ(A). Proof. (1) Let A be such that µ(A) < 1. Let A1 = A and An = ; for n > 1. Then these are pairwise disjoint and its union is A. So, we get µ(A) = µ(A) + P µ(;). Rest is clear, since we can cancel µ(A) < 1 from bot sides. (2) This is clear by taking Am = ; for m > n and using the first. (3) We can write B = A [ (B − A), pairwise disjoint and thus µ(B) = µ(A) + µ(B − A) ≥ µ(A). (4) Let Bn = An − An−1 for n > 1 and B1 = A1. Then fBng are 1 n pairwise disjoint and [n=1Bn = A. Also, [i=1Bi = An. So, we Pn P1 get µ(An) = i=1 µ(Bi) and µ(A) = n=1 µ(Bn). 4 1 (5) For this, let Cn = A1 − An. Then Cn ⊂ Cn+1 and [n=1Cn = A1 −A. So, from the previous part, we have lim µ(Cn) = µ(A1 − A) = µ(A1) − µ(A), since A ⊂ A1. We also have by the same reason, µ(Cn) = µ(A1) − µ(An). So, taking limits, we get, lim µ(Cn) = µ(A1) − lim(An) = µ(A1) − µ(A) and canceling µ(A1) < 1, we are done. 3. Lebesgue Integral Definition 6. Let (X; F; µ) be a measure space and let s : X ! [0; 1) P be a simple positive measurable function. Writing s = aiχA , define R P i X sdµ = aiµ(Ai), where we use the convention, 0 · µ(A) = 0 even if µ(A) = 1. If 0 ≤ f is any positive measurable function, we define R R X fdµ to be the supremum of X sdµ for simple measurable functions 0 ≤ s ≤ f. Remark 3. The above gives two possible definitions for the integral of a simple measurable function, but it is clear that the definitions coincide. We collect a few easy facts which follow from the definitions. Theorem 6. Let (X; F; µ) be a measure space. All functions men- tioned below are measurable and so are all the subsets of X. R R (1) If 0 ≤ f ≤ g, then X fdµ ≤ X gdµ. (2) If A ⊂ B in F, for 0 ≤ f on X, we have, R fdµ ≤ R fdµ. R RA B (3) If 0 ≤ f and 0 ≤ c < 1, then X cfdµ = c X fdµ. (4) If f(x) = 0 for all x 2 A, R fdµ = 0, even if µ(A) = 1. A R (5) If µ(A) = 0, for any measurable positive f, A fdµ = 0 even if f(x) = 1 for some or all x 2 A.