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) := {T | T ⊆ S} of all subsets of S. Exercise Show that the mapping

S P(S) 3 T 7→ f (T ) ∈ {0, 1} := {hxs is∈S | ∀s ∈ S : xs ∈ {0, 1}}

defined by ( 1 if s ∈ T f (T )s = 1T (s) = 0 if s 6∈ 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)is∈S ∈ {0, 1} defines a unique point of the Cartesian product {0, 1}S for every T ⊆ S — i.e., for every T ∈ P(S). S Conversely, for each hxs is∈S ∈ {0, 1} , there is a unique set

T (hxs is∈S ) = {s ∈ S | xs = 1}

Now, given this hxs is∈S , for each r ∈ S, one has ( 1 if xr = 1 1{s∈S|xs =1}(r) = 0 if xr = 0

So f (T (hxs is∈S )) = h1{s∈S|xs =1}(r)ir∈S = hxr ir∈S . −1 This implies that T (hxs is∈S ) = f (hxs is∈S ).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 57 Counting a Finite Power Set

Lemma Let {An | n ∈ K} be any finite collection of finite sets. Q Q Then its Cartesian product n∈K An has n∈K #An members. Applied to the Cartesian product {0, 1}S , the lemma implies that #{0, 1}S = (#{0, 1})#S = 2#S . The existence of the bijection P(S) 3 T 7→ f (T ) ∈ {0, 1}S 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. ∅ ∈ A; 2. A ∈ A implies that the complement S \ A ∈ A; 3. if A, B lie in A, then the union A ∪ B ∈ A. The family Σ ⊆ P(S) is a σ-algebra just in case it is a Boolean algebra with the following stronger property: ∞ whenever {An}n=1 is a countably infinite family of sets in Σ, ∞ then their union ∪n=1An ∈ 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 ∈ A, then A ∩ B ∈ A. 2. Let Σ be a σ-algebra on S. ∞ Prove that if {An}n=1 is a countably infinite family of sets ∞ in Σ, then ∩n=1An ∈ Σ.

Hint 1. For part 1, use de Morgan’s laws

S \ (A ∩ B) = (S \ A) ∪ (S \ B) S \ (A ∪ B) = (S \ A) ∩ (S \ B)

2. For part 2, use the infinite extension of de Morgan’s laws:

∞ ∞ ∞ ∞ S \ (∩n=1An) = ∪n=1(S \ An); S \ (∪n=1An) = ∩n=1(S \ An)

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 57 Generating a Sigma-Algebra Theorem Let {Σi | i ∈ I } be any indexed family of σ-algebras. ∩ Then the intersection Σ := ∩i∈I 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 ∩{Σ | Σ ∈ S(F)}; it is the smallest σ-algebra that includes F. Exercise Let X be any uncountably infinite set, and let F := {{x} | x ∈ X } 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. ∅ ∈ T and X ∈ T ; 2. if U, V ∈ T , then U ∩ V ∈ T ;

3. if {Uα | α ∈ A} is any (possibly uncountable) collection of open sets Uα ∈ T , then the union ∪α∈AUα ∈ 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 \ 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 \ 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) := {y ∈ X | d(x, y) < r}

The metric topology Td of (X , d) is the smallest topology that includes the entire family {Br (x) | x ∈ X & r > 0} 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 ∈ 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) := {x ∈ X | d(x, x ) < r} 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 ∪ {−∞ + ∞} = [−∞, +∞] denote the extended real line which, at each end, has an endpoint added at infinity. ¯ ¯ Let R+ := R+ ∪ {+∞} = [0, +∞] 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 ∈ 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 {A, B} 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 ∈ 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 {An}n=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 = {a, b, c} where a, b, c are all different. Then the three pair subsets

S−c := {a, b}, S−b := {a, c}, S−a := {b, c}

obviously satisfy S−a ∩ S−b ∩ S−c = ∅, so are disjoint.

Yet S−a ∩ S−b = {c}, S−a ∩ S−c = {b}, and S−b ∩ S−c = {a} are all non-empty, so the sets are not pairwise disjoint.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 16 of 57 Additivity for Pairwise Disjoint, but not for Disjoint Sets Exercise Let S be any finite set, with power set 2S . Show that the only additive function µ on the measurable space (S, 2S ) which satisfies µ({x}) = 1 for all x ∈ S is the counting measure defined by µ(E) = #E for all E ⊆ S. Exercise Following the previous example, let S = {a, b, c} where a, b, c are all different, and let S−c := {a, b}, S−b := {a, c}, S−a := {b, c}. Following the previous exercise, let µ be the counting measure on (S, 2S ).

Verify that, though the sets S−a, S−b, S−c are disjoint, one has

µ(S−a ∪ S−b ∪ S−c ) = µ(S) = 3

6= µ(S−a) + µ(S−b) + µ(S−c ) = 3 · 2 = 6 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 17 of 57 Measure as a Countably Additive Set Function

Definition ¯ The set function µ :Σ → R+ on a measurable space (X , Σ) is said to be σ-additive or countably additive just in case, ∞ for any countable collection {An}n=1 of pairwise disjoint sets in Σ, one has [∞  X∞ µ An = µ(An) n=1 n=1 A measure on a measurable space (X , Σ) ¯ is a countably additive set function µ :Σ → R+.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 18 of 57 Measure Space A measure space is a triple (X , Σ, µ) where 1. Σ is a σ-algebra on X ; 2. µ is a measure on the measurable space (X , Σ). Example A prominent example of a measure space is (R, B, `) where: 1. B is the Borel σ-algebra induced by the open sets of the real line R; 2. the measure `(J) of any interval J ⊂ R is its length, 2 defined whenever (a, b) ∈ R with a ≤ b by

`([a, b]) = `([a, b)) = `((a, b]) = `((a, b)) = b − a

3. ` is extended to all of B so as to satisfy countable additivity (it can be shown that this extension is unique).

Exercise Prove that if S ⊂ R is countable, then S ∈ B and `(S) = 0. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 19 of 57 Probability Measure and Probability Space

Consider a measure space (X , Σ, µ). The measure µ is a probability measure just in case µ(X ) = 1. Then (X , Σ, µ) is a probability space. Often one writes (Ω, F, P) in this case, where: 1. Ω is the sample space; 2. F is the σ-algebra (or σ-field) of measurable events; 3. for each event E ∈ F, the probability that E occurs is P(E). Then, because P is a measure satisfying P(Ω) = 1, one has 0 ≤ P(E) ≤ 1 for all E ∈ F.

Any measure space (X , Σ, µ) with 0 < µ(X ) < +∞ can be given a normalized measure defined for all E ∈ Σ by P(E) = µ(E)/µ(X ). This normalization makes (X , Σ, P) a probability space.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 20 of 57 Lebesgue Measurable Subsets of the Real Line A set N ⊂ R is null just in case there exists a Borel subset B ∈ B with `(B) = 0 such that N ⊂ B.

This is possible for some non-Borel subsets of R. Let N denote the family of null subsets of R. These null sets can be used to generate the Lebesgue σ-algebra of Lebesgue measurable sets, which is σ(B ∪ N ). The symmetric difference of any two sets S and B is defined as the set

S4B := (S \ B) ∪ (B \ S) = (S ∪ B) \ (S ∩ B)

of elements s that belong to one of the two sets, but not to both. One can show that S ∈ σ(B ∪ N ) if and only if there exists a Borel set B ∈ B such that S4B ∈ N — i.e., S differs from a Borel set only by a null set.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 21 of 57 The Lebesgue Real Line

There is a well-defined function λ : σ(B ∪ N ) → R¯+ that satisfies λ(S) := `(B) whenever S4B ∈ N . Moreover, one can prove that the function S 7→ λ(S) is countably additive. This makes λ a measure, called the Lebesgue measure.

The associated measure space (R, σ(B ∪ N ), λ) is called the Lebesgue real line.

Because λ(R) = +∞, the Lebesgue real line cannot be normalized to form a probability space.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 22 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 23 of 57 Measurable Functions and Measurable Partitions Let (X , Σ, µ) be a measure space, and (R, σ(B ∪ N ), λ) the Lebesgue real line. The function X 3 x 7→ f (x) ∈ R is measurable (with respect to the σ-algebras Σ and σ(B ∪ N )) just in case the set f −1(B) = {x ∈ X | f (x) ∈ B} is Σ-measurable for every Lebesgue measurable set B ∈ σ(B ∪ N ). Example Let X and Y be topological spaces. The function X 3 x 7→ f (x) ∈ Y is continuous just in case the set f −1(B) is open in X whenever B is open in Y . Then any continuous function f : X → Y is measurable provided that X and Y are each given their Borel σ-algebra.

The finite collection {Ek |k ∈ Nm} of m pairwise disjoint measurable sets Ek ∈ Σ m is a measurable partition of X just in case ∪k=1Ek = X . University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 24 of 57 Simple Functions

Given any set E ∈ Σ, the indicator function of E is defined by ( 1 if x ∈ E X 3 x 7→ 1E (x) := 0 if x 6∈ E

The function f : X 7→ R is simple just in case there exist a measurable partition {Ek |k ∈ Nm} of X m together with a corresponding collection (ak )k=1 Pm of m different real constants such that f (x) = k=1 ak 1Ek (x). Note that the range f (X ) := {y ∈ R | ∃x ∈ X : y = f (x)} of this step function is precisely the finite set {ak |k ∈ Nm} of m real constants. Let F(X , Σ) denote the set of all simple functions on the measurable space (X , Σ); in fact it is a real vector space.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 25 of 57 Integrable Simple Functions

Let (X , Σ, µ) be a measure space.

For as many functions f : X 7→ R as possible, R R we want to define the integral X f (x) dµ = X f (x) µ(dx) Pm The simple function f (x) = k=1 ak 1Ek (x) is µ-integrable Pm just in case one has k=1 |ak | µ(Ek ) < +∞.

In particular, this requires that ak 6= 0 =⇒ µ(Ek ) < +∞ or equivalently, that µ(Ek ) = +∞ =⇒ ak = 0.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 26 of 57 Integrating Indicator and Simple Functions

Let (X , Σ, µ) be a measure space. R Define the integral X f (x) µ(dx) so that in case f (x) is the: R 1. indicator function 1E (x), one has X 1E (x) µ(dx) = µ(E); Pm 2. µ-integrable simple function k=1 ak 1Ek (x), one has

" m # m Z X X Z  ak 1Ek (x) µ(dx) = ak 1Ek (x) µ(dx) X k=1 k=1 X m X = ak µ(Ek ) k=1 The second equality requires integration to be a linear operator on the vector space of µ-integrable simple functions on (X , Σ, µ).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 27 of 57 Measurable Functions

Given the measure space (X , Σ, µ), the function f : X → R is measurable just in case, for every Borel set B ⊂ R, its inverse image satisfies

f −1(B) := {x ∈ X | f (x) ∈ B} ∈ Σ

Note that we have defined a simple function to be measurable.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 28 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 29 of 57 Upper and Lower Bounds In this subsection, we consider the case of a finite measure satisfying µ(X ) < +∞.

In case X ⊆ R and µ is Lebesgue measure, this implies that X must be bounded — for example, X = [a, b]. In case µ is a probability measure satisfying µ(X ) = 1, it is automatically a finite measure.

Given any measure space (X , Σ, µ), let FS (X , Σ, µ) denote the set of µ-integrable simple functions. Given any function f : X → R, define the two sets

∗ ∗ ∗ F (f ; X , Σ, µ) := {f ∈ FS (X , Σ, µ) | ∀x ∈ X : f (x) ≥ f (x)} F∗(f ; X , Σ, µ) := {f∗ ∈ FS (X , Σ, µ) | ∀x ∈ X : f∗(x) ≤ f (x)}

of µ-integrable simple functions that are respectively upper or lower bounds for the function f .

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 30 of 57 Upper and Lower Simple Functions

When trying to find the integral of the red curve, a lower approximation is the sum of the four green rectangles, and an upper approximation adds the sum of the grey rectangles. Source: https://en.wikipedia.org/wiki/Darboux_integral. This also illustrates decreasing error as you add more steps.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 31 of 57 Upper and Lower Integrals

R ∗ The integral X f (x) µ(dx) of each µ-integrable simple function f ∗ ∈ F ∗(f ; X , Σ, µ), is an over-estimate of the true integral of f . R But the integral X f∗(x) µ(dx) of each µ-integrable simple function f∗ ∈ F∗(f ; X , Σ, µ), is an under-estimate of the true integral of f . Define the upper integral and lower integral of f as, respectively

∗ R ∗ I (f ) := inff ∗∈F ∗(f ;X ,Σ,µ) f (x) µ(dx) RX and I∗(f ) := supf∗∈F∗(f ;X ,Σ,µ) X f∗(x) µ(dx) These are respectively the smallest possible over-estimate and greatest possible under-estimate of the integral. Of course, in case f is itself a µ-integrable simple function, ∗ R one has I (f ) = I∗(f ) = X f (x) µ(dx).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 32 of 57 Integrability and Integration

Definition Let X 3 x 7→ f (x) ∈ R be defined on the measure space (X , Σ, µ). 1. The function f is integrably bounded just in case the mapping X 3 x 7→ |f (x)| ∈ R+ is bounded above by a µ-integrable simple function. 2. The function f is integrable just in case ∗ its upper and lower integrals I (f ) and I∗(f ) are equal. R 3. In case f is integrable, its integral X f (x) µ(dx) is defined as the common value ∗ of its upper integral I (f ) and its lower integral I∗(f ).

Theorem The function X 3 x 7→ f (x) ∈ R on (X , Σ, µ) is integrable if and only if it is both measurable and integrably bounded.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 33 of 57 Integration over an Interval or Other Measurable Set

Let X 3 x 7→ f (x) ∈ R be a function defined on the measure space (X , Σ, µ) that is measurable and integrably bounded.

Let E ∈ Σ be any measurable set, with indicator function 1E (x). Then the function X 3 x 7→ 1E (x)f (x) ∈ {f (x), 0} ⊂ R is also measurable and integrably bounded. So we can define the integral of f over E by Z Z f (x) µ(dx) := 1E (x)f (x) µ(dx) E X

In case (X , Σ, µ) is the Lebesgue real line, and E is the interval [a, b], R b R one usually writes a f (x) dx instead of [a,b] f (x) µ(dx).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 34 of 57 Upper and Lower Bounds on an Integral

Exercise Let X 3 x 7→ f (x) ∈ R be a function defined on the measure space (X , Σ, µ) that is measurable and integrably bounded.

Let E ∈ Σ be any measurable set, with indicator function 1E (x). Suppose that a ≤ f (x) ≤ b for all x ∈ E. ∗ ∗ 1. For any f ∈ F (f ; X , Σ, µ) and f∗ ∈ F∗(f ; X , Σ, µ), show that for all x ∈ E one has

∗ 1E (x)f (x) ≥ 1E (x)a and 1E (x)f∗(x) ≤ 1E (x)b R 2. Show that µ(E)a ≤ E f (x) µ(dx) ≤ µ(E)b.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 35 of 57 The Integral of a Nonnegative Function is a Measure

Exercise Prove the following: 1. If E and E 0 are subsets of X , 0 then 1E∪E 0 = 1E + 1E 0 if and only if E and E are disjoint. 2. If E and E 0 are disjoint measurable subsets of the measure space (X , Σ, µ), and X 3 x 7→ f (x) ∈ R is integrable w.r.t. µ, R R R then E∪E 0 f (x)µ(dx) = E f (x)µ(dx) + E 0 f (x)µ(dx).

3. If (En)n∈N is an infinite sequence of pairwise disjoint subsets of X , then: Pk I 1 k = 1E for each k ∈ ; ∪n=1En n=1 n N Pk P∞ I 1∪∞ E = sup 1 k = sup 1E = 1E . n=1 n k ∪n=1En k n=1 n n=1 n 4. If X 3 x 7→ f (x) ∈ R+ is integrable w.r.t. µ, R then Σ 3 E 7→ E f (x)µ(dx) ∈ R+ is a measure on (X , Σ).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 36 of 57 The Integral of a General Function is a Signed Measure

A general function X 3 x 7→ f (x) ∈ R may have negative values at some points x ∈ X R Then the mapping Σ 3 E 7→ E f (x)µ(dx) ∈ R will generally have negative values for some measurable sets E ∈ Σ. R So E 7→ E f (x)µ(dx) will generally not be a measure, whose values must be nonnegative. Instead, the mapping is a signed measure on (X , Σ), whose values are allowed to be negative.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 37 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 38 of 57 The Integral as a Function of Its Upper Limit

Let I 3 x 7→ f (x) ∈ R be any continuous function defined on a finite interval I = [A, B) ⊂ R. Then the function R 3 x 7→ 1I (x)f (x) is measurable and also integrably bounded, because its range f (I ) is a finite interval in R. Given any fixed a ∈ [A, B) and the Lebesgue measure λ on R, we can define the integral function of the upper limit b by

Z b [a, B) 3 b 7→ J(b) := f (x)λ(dx) a

Theorem (Leibnitz’s Formula for the Lebesgue Integral) Provided that [A, B) 3 x 7→ f (x) is continuous, the integral function J(b) is differentiable, with J0(b) = f (b).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 39 of 57 Proof of Leibnitz’s Formula

Proof. For fixed a, b ∈ [A, B) with a < b, for all small h > 0, ∗ define φ∗(h) and φ (h) respectively as the infimum and supremum of the set {f (x) | x ∈ (b, b + h)}. R b+h ∗ These definitions imply that hφ∗(h) ≤ b f (x) λ(dx) ≤ hφ (h). But the Newton quotient of J at b is

1 1 R b+h q(h) = h [J(b + h) − J(b)] = h b f (x) λ(dx)

∗ It follows that φ∗(h) ≤ q(h) ≤ φ (h) for all small h. But continuity of f implies that, for all small  > 0, there exists δ > 0 such that |x − b| < δ implies |f (x) − f (b)| <  ∗ and so |h| < δ implies that f (b) −  < φ∗(h) ≤ φ (h) < f (b) + . This proves that as h → 0, ∗ so φ∗(h), φ (h) and therefore q(h) all converge to f (b).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 40 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 41 of 57 Measurable Rectangles Let (X , ΣX ) and (Y , ΣY ) be two measurable spaces, with their respective σ-algebras ΣX and ΣY . The Cartesian product of X and Y is

X × Y = {(x, y) | x ∈ X y ∈ Y }

Let ΣX × ΣY = {A × B | A ∈ ΣX , B ∈ ΣY } denote the set of measurable rectangles that are the Cartesian product of two measurable sets.

Example Suppose that X = {a, b} and Y = {c, d}, X Y with ΣX = 2 and ΣY = 2 .

Then #ΣX = #ΣY = 4 and #(ΣX × ΣY ) = 10 after identifying E × ∅ = ∅ × F = ∅ for all E ⊆ X and all F ⊆ Y .

But then (X × Y ) \{a, c} = X × {d} = {b} × Y 6∈ ΣX × ΣY .

This implies that ΣX × ΣY is nota σ-algebra. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 42 of 57 The Product of Two Measurable Spaces

So we define the product σ-algebra, denoted by ΣX ⊗ ΣY , as σ(ΣX × ΣY ), the σ-algebra generated by ΣX × ΣY . It is the smallest σ-algebra that contains all measurable rectangles. And we define the product of the two measurable spaces (X , ΣX ) and (Y , ΣY ) as the measurable space (X × Y , ΣX ⊗ ΣY ).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 43 of 57 The Product of Two Measure Spaces

Let (X , ΣX , µX ) and (Y , ΣY , µY ) be two measure spaces, and (X × Y , ΣX ⊗ ΣY ) the product measurable space.

Say that µ on (X × Y , ΣX ⊗ ΣY ) is a product measure just in case it is a measure that satisfies µ(E × F ) = µX (E) × µY (F ) for all measurable rectangles E × F ∈ ΣX × ΣY . Typically there is a unique product measure with this property, which we denote by µX ⊗ µY .

Then (X × Y , ΣX ⊗ ΣY , µX ⊗ µY ) is the product of the two measure spaces.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 44 of 57 The Fubini Theorem

Theorem (Fubini) Provided that X × Y 3 (x, y) 7→ f (x, y) ∈ R is measurable w.r.t. ΣX ⊗ ΣY , its integral w.r.t. the product measure µX ⊗ µY satisfies Z f (x, y)(µX ⊗ µY )(dx × dy) X ×Y Z Z  = f (x, y)µY (dy) µX (dx) X Y Z Z  = f (x, y)µX (dx) µY (dy) Y X

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 45 of 57 Product Measure as a Double Integral

Corollary

For every E ∈ ΣX ⊗ ΣY , its product measure satisfies Z (µX ⊗ µY )(E) = 1E (x, y)(µX ⊗ µY )(dx × dy) E Z Z  = 1E (x, y)µY (dy) µX (dx) X Y Z Z  = 1E (x, y)µX (dx) µY (dy) Y X

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 46 of 57 The Lebesgue Plane

Example

Suppose the two measure spaces (X , ΣX , µX ) and (Y , ΣY , µY ) are both copies of the Lebesgue real line (R, L, λ) where: 1. L is the Lebesgue completion of the Borel σ-algebra on R; 2. λ is the Lebesgue measure which satisfies λ(I ) = b − a for any interval I ⊂ R with endpoints a and b satisfying a ≤ b. 2 Then the measure product (R, L, λ) is the Lebesgue plane 2 in the form of the measure space (R , A, α), where: 1. A = L ⊗ L is the product of the Lebesgue σ-algebra on R with itself; 2. α = λ ⊗ λ has the property that, for each E ∈ A, the measure α(E) is its area.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 47 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 48 of 57 Circular Symmetry of a Function 2 A function R 3 (x, y) 7→ z = f (x, y) ∈ R defined on the plane displays circular symmetry just in case, for each c ∈ R, 2 the level curve {(x, y) ∈ R | f (x, y) = c is either empty 2 2 2 2 or else is a circle {(x, y) ∈ R | x + y = r } of radius r, which depends on c, with centre at the origin. Such a function can be represented by a function R+ 3 r 7→ φ(r) ∈ R satisfying φ(r) = f (x, y) if and only if r = px2 + y 2. Example Three examples with circular symmetry: 1. f (x, y) = e−s(x2+y 2);

2. f (x, y) = 1E where E is the annulus ∗ 2 p 2 2 ∗ A(r∗, r ) := {(x, y) ∈ R | 0 ≤ r∗ ≤ x + y ≤ r }; Pm 3. the simple function f (x, y) = k=1 ak 1Ek m where the sets {Ek }k=1 are pairwise disjoint annuli.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 49 of 57 Annulus with Inner Radius r and Outer Radius R

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 50 of 57 Shell Integral

R b An integral like a f (x)dx is often introduced as the area below the graph of a function [a, b] 3 x 7→ f (x) ∈ R+. R b R d A corresponding idea is that the double integral a c f (x, y)dxdy is the volume below the graph of a function of two variables: [a, b] × [c, d] 3 (x, y) 7→ f (x, y) ∈ R+ ∗ For a function [r∗, r ] 3 r 7→ φ(r) ∈ R+ with circular symmetry, ∗ defined on the annulus A(r∗, r ), we try to define its integral as the volume below the circularly symmetric graph ∗ of A(r∗, r ) 3 (x, y) 7→ f (x, y) ∈ R+ where f (x, y) = φ(px2 + y 2). Pm We begin with a simple integrand φ(r) = k=1 ak 1A(rk−1,rk ). 2 2 Since the area of each annulus A(rk−1, rk ) is rk − rk−1, the volume beneath the graph of this simple function Pm 2 2 is V = k=1 ak (rk − rk−1).

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 51 of 57 Shell Integration Illustrated

Source: en.wikipedia.org/wiki/Shell_integration “A volume is approximated by a collection of hollow cylinders. As the cylinder walls get thinner the approximation gets better. The limit of this approximation is the shell integral.”

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 52 of 57 Leibnitz’s Formula for a Shell Integral

Theorem (Analogue of Leibnitz’s Formula) Given any pair b ≥ 0, provided that R+ 3 r 7→ φ(r) is continuous, the volume below the graph of A(0, b) 3 (x, y) 7→ f (x, y) ∈ R+ has an integral J(b) which is differentiable, with J0(b) = 2πb φ(b). R b It follows that J(b) = 0 2πr φ(r)dr. Example Two examples: 1. Suppose that φ(r) = c, a constant. R b b 1 2 2 Then J(b) = 0 2πr φ(r)dr = 2πc|0 2 r = πb c, which is the volume of a cylinder of radius b and height c. −r 2 R b R b −r 2 2. Suppose that φ(r) = e , so 0 2πr φ(r)dr = 2π 0 r e dr. b −r 2 −b2 Then J(b) = −π|0e = (1 − e )π. Note that J(b) → π as b → ∞.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 53 of 57 Gauss (1777–1855) on a Ten Deutsche Mark Note

Portrait with graph of the “bell curve” 2 1 − (x−µ) and the formula f (x) = √ e 2σ2 . σ 2π

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 54 of 57 The Gaussian Integral, I

R +b −x2 Define I (b) := −b e dx for each a ∈ R. Then

2 R +b −x2  R +b −y 2  [I (b)] = −b e dx −b e dy R +b R +b −y 2  −x2 = −b −b e dy e dx R +b R +b −x2 −y 2 = −b −b e e dxdy R −x2−y 2 = S(b) e dxdy

where S(b) := [−b, b]2 denotes the Cartesian product of the line interval [−b, b] with itself. 2 Thus, S(b) is the solid square subset of R that is centred at the origin and has sides of length 2b.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 55 of 57 Square with Inscribed and Circumscribed Circles

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 56 of 57 The Gaussian Integral, II

2 2 2 2 Let D(b) := {(x, y) ∈ R | x + y ≤ b } denote the disk of radius b centred at the origin. √ Note that D(b) ⊂ S(b) ⊂ D(b 2) and so

R −x2−y 2 R −x2−y 2 D(b) e dx dy ≤ S(b) e dx dy R √ −x2−y 2 ≤ D(b 2) e dx dy √ 2 R −x2−y 2 It follows that J(b) ≤ [I (b)] = S(b) e dxdy ≤ J(b 2) and so π(1 − e−b2 ) ≤ [I (b)]2 ≤ π(1 − e−2b2 ). Taking limits as b → ∞ one has π(1 − e−b2 ) → π and also π(1 − e−2b2 ) → π, implying that [I (b)]2 → π. Theorem The Gaussian integral R +∞ e−x2 dx = lim R b e−x2 dx √ −∞ b→∞ −b equals π.

University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 57 of 57