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Lecture Notes 9: Measure and Probability Part A: Measure and Integration

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Lecture Notes 9: Measure and Probability Part A: Measure and Integration Lecture Notes 9: Measure and Probability Part A: Measure and Integration Peter Hammond, based on early notes by Andr´esCarvajal September, 2011; latest revision September 2020, typeset from Probab20A.tex; Recommended textbooks for further reading: H.L. Royden Real Analysis; R.M. Dudley Real Analysis and Probability University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 57 Outline Measures and Integrals Measurable Spaces Measure Spaces Lebesgue Integration Integrating Simple Functions Integrating Integrable Functions The Lebesgue Integral as an Antiderivative Leibnitz's Formula Revisited Products of Measure Spaces Definition The Gaussian Integral Circular Symmetry and Shell Integration University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 57 Power Sets Fix an abstract set S 6= ;. In case it is finite, its cardinality, denoted by #S, is the number of distinct elements of S. The power set of S is the family P(S) := fT j T ⊆ Sg of all subsets of S. Exercise Show that the mapping S P(S) 3 T 7! f (T ) 2 f0; 1g := fhxs is2S j 8s 2 S : xs 2 f0; 1gg defined by ( 1 if s 2 T f (T )s = 1T (s) = 0 if s 62 T is a bijection. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 57 Constructing a Bijection Proof. S Evidently the mapping T 7! f (T ) = h1T (s)is2S 2 f0; 1g defines a unique point of the Cartesian product f0; 1gS for every T ⊆ S | i.e., for every T 2 P(S). S Conversely, for each hxs is2S 2 f0; 1g , there is a unique set T (hxs is2S ) = fs 2 S j xs = 1g Now, given this hxs is2S , for each r 2 S, one has ( 1 if xr = 1 1fs2Sjxs =1g(r) = 0 if xr = 0 So f (T (hxs is2S )) = h1fs2Sjxs =1g(r)ir2S = hxr ir2S . −1 This implies that T (hxs is2S ) = f (hxs is2S ). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 57 Counting a Finite Power Set Lemma Let fAn j n 2 Kg be any finite collection of finite sets. Q Q Then its Cartesian product n2K An has n2K #An members. Applied to the Cartesian product f0; 1gS , the lemma implies that #f0; 1gS = (#f0; 1g)#S = 2#S . The existence of the bijection P(S) 3 T 7! f (T ) 2 f0; 1gS proves that when S is finite, then the cardinality of the power set is #P(S) = 2#S . This helps explain why the power set P(S) is often denoted by 2S . University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 57 Boolean Algebras, Sigma-Algebras, and Measurable Spaces The family A ⊆ P(S) is a Boolean algebra on S just in case 1. ; 2 A; 2. A 2 A implies that the complement S n A 2 A; 3. if A; B lie in A, then the union A [ B 2 A. The family Σ ⊆ P(S) is a σ-algebra just in case it is a Boolean algebra with the following stronger property: 1 whenever fAngn=1 is a countably infinite family of sets in Σ, 1 then their union [n=1An 2 Σ. The pair (S; Σ) is a measurable space just in case Σ is a σ-algebra. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 57 Exercise on Boolean Algebras and Sigma-Algebras Exercise 1. Let A be a Boolean algebra on S. Prove that if A; B 2 A, then A \ B 2 A. 2. Let Σ be a σ-algebra on S. 1 Prove that if fAngn=1 is a countably infinite family of sets 1 in Σ, then \n=1An 2 Σ. Hint 1. For part 1, use de Morgan's laws S n (A \ B) = (S n A) [ (S n B) S n (A [ B) = (S n A) \ (S n B) 2. For part 2, use the infinite extension of de Morgan's laws: 1 1 1 1 S n (\n=1An) = [n=1(S n An); S n ([n=1An) = \n=1(S n An) University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 57 Generating a Sigma-Algebra Theorem Let fΣi j i 2 I g be any indexed family of σ-algebras. \ Then the intersection Σ := \i2I Σi is also a σ-algebra. Proof left as an exercise. Let X be a space, and F ⊂ 2X any family of subsets. Since 2X is obviously a σ-algebra, there exists a non-empty set S(F) of σ-algebras that include F. Let σ(F) denote the intersection \fΣ j Σ 2 S(F)g; it is the smallest σ-algebra that includes F. Exercise Let X be any uncountably infinite set, and let F := ffxg j x 2 X g denote the family of all singleton subsets of X . Show that σ(F) consists of all subsets of X that are either countable, or co-countable (i.e., have a countable complement). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 57 Topological Spaces Given a set X , a topology T on X is a family of open subsets U ⊆ X satisfying: 1. ; 2 T and X 2 T ; 2. if U; V 2 T , then U \ V 2 T ; 3. if fUα j α 2 Ag is any (possibly uncountable) collection of open sets Uα 2 T , then the union [α2AUα 2 T . Thus, finite intersections and arbitrary unions of open sets are open. A topological space( X ; T ) is any set X together with a topology T that consists of all the open subsets of X . University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 57 Closed Sets, Closures, Interiors, and Boundaries Recall that, in a topological space (X ; T ), a set S is closed just in case its complement X n S is open. Let S be an arbitrary subset of the topological space (X ; T ). 1. The closure cl S of S is the intersection of all the closed supersets of S. 2. The interior int S of S is the union of all the open subsets of S. 3. The boundarybd S of S is cl S n int S, the complement of the interior in the closure. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 10 of 57 The Metric Topology Let (X ; d) be any metric space. The open ball of radius r centred at x is the set Br (x) := fy 2 X j d(x; y) < rg The metric topology Td of (X ; d) is the smallest topology that includes the entire family fBr (x) j x 2 X & r > 0g of all open balls in X . University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 57 Borel Sets and the Borel Sigma-Algebra Let (X ; T ) be any topological space. Its Borel σ-algebra is defined as σ(T ) | i.e., the smallest σ-algebra containing every open set of X . Each set B 2 σ(T ) is then a Borel set. Suppose the topological space is a metric space (X ; d) with its metric topology Td . Then the Borel σ-algebra is generated 0 0 by all the open balls Br (x) := fx 2 X j d(x; x ) < rg in X . For the case of the real line when X = R, its Borel σ-algebra is generated by all the open intervals of R. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 57 Outline Measures and Integrals Measurable Spaces Measure Spaces Lebesgue Integration Integrating Simple Functions Integrating Integrable Functions The Lebesgue Integral as an Antiderivative Leibnitz's Formula Revisited Products of Measure Spaces Definition The Gaussian Integral Circular Symmetry and Shell Integration University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 13 of 57 Finitely Additive Set Functions Let R¯ := R [ {−∞ + 1g = [−∞; +1] denote the extended real line which, at each end, has an endpoint added at infinity. ¯ ¯ Let R+ := R+ [ f+1g = [0; +1] be the non-negative part of R. Any family F of subsets A ⊆ X is said to be pairwise disjoint just in case A \ B = ; whenever A; B 2 F with A 6= B. Definition Let (X ; Σ) be a measurable space. ¯ A mapping µ :Σ ! R+ is said to be a set function. ¯ The set function µ :Σ ! R+ is said to be additive (or finitely additive) just in case, for any pair fA; Bg of disjoint sets in Σ, one has µ(A [ B) = µ(A) + µ(B). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 57 Implications of Finite Additivity Lemma ¯ If the set function µ :Σ ! R+ is finitely additive, then µ(;) = 0. Proof. For any non-empty A 2 Σ, the sets A and ; are disjoint. Additivity implies that µ(A) = µ(A [;) = µ(A) + µ(;), so µ(;) = 0. Exercise k For any finite collection fAngn=1 of pairwise disjoint sets in Σ, prove by induction on k that finite additivity implies k Xk µ [ An = µ(An) n=1 n=1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 15 of 57 Disjoint Does Not Imply Pairwise Disjoint Example Suppose that S = fa; b; cg where a; b; c are all different. Then the three pair subsets S−c := fa; bg; S−b := fa; cg; S−a := fb; cg obviously satisfy S−a \ S−b \ S−c = ;, so are disjoint. Yet S−a \ S−b = fcg, S−a \ S−c = fbg, and S−b \ S−c = fag are all non-empty, so the sets are not pairwise disjoint.
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