Basic notation about sets - University College Dublinastier/math20180/chapter2.pdf · Basic notation about sets ... We can check that A \B = (Ac [Bc)c (exercise), and similarly, if

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Basic notation about sets

Definition 1A set is a collection of elements (any collection of any elements). Wesay that two sets are equal when they contain exactly the sameelements.

NotationLet A and B denote subsets of a larger set Ω. For an element x of Ωwe use the notation:

x ∈ A means x is an element of A,x 6∈ A means x is not an element of A,A ⊂ B means A is a subset of B, i.e. every x ∈ A belongs to B:

BA

Vincent Astier MATH20180 February 15, 2018 1 / 38


NotationA ∩ B denotes the intersection of A and B, i.e.x : x belongs to both A and B,

ABA ∩ B

A ∪ B denotes the union of A and B, i.e. x : x belongs to A or B,

AB

A ∪ B

(recall that in mathematics “or” is not “either or”, but an inclusive or: if xbelongs to both A and B then x belongs to A ∪ B)



NotationAc denotes the complement of A (in Ω), i.e. x ∈ Ω : x 6∈ A.

A \ B denotes A without B, i.e. x ∈ A : x 6∈ B = A ∩ Bc .

∅ denotes the empty set, i.e. the set with no elements.

We say that A and B are disjoint if they have no elements in common,i.e. if A ∩ B = ∅.

Definition 2Let An,n = 1,2, . . ., denote subsets of the set Ω.⋃

n∈NAn = x : x ∈ An for some n = union of all An,⋂

n∈NAn = x : x ∈ An for all n = intersection of all An.



Definition 3A set A is said to be countable if A contains a finite number ofelements or there exists an infinite sequence (xn)n∈N in which everyelement in A appears. This is the same as saying that we can count,one after another, the points in A:

x1, x2, x3, x4, . . . .

Example: N,Z, and Q are countable, but R is uncountable.

If A is a set, we denote by P(A) the set of all subsets of A. If|A| = n ∈ N, then |P(A)| = 2n. It can be shown that if A is countableand infinite, then P(A) is not countable.


Plan

Here is a very rough plan of what we will see in the following few weeks

σ-fields↓

Measures, measurable functions↓

Probabilities, integrals↓

Random variables

Then we will be able to look at the price of options as randomvariables, and get to the Black-Scholes formula. . .


σ-fields

Definition 4A σ-field (also called σ-algebra) on a set Ω is a collection F of subsetsof Ω which obeys the following rules:(a) Ω ∈ F ;(b) if A ∈ F , then Ac ∈ F (F is closed under taking complements);(c) if (An)n∈N is a sequence in F , then

⋃n∈N An ∈ F

(F is closed under taking countable unions).

Remark:The elements of F are subsets of Ω.By property (c), any finite union of elements of F is also anelement of F .P(Ω) is a σ-field; it is the largest possible σ-field on Ω. So is∅,Ω; it is the smallest σ-field on Ω.


σ-fields

Remark (continued):We can check that A ∩ B = (Ac ∪ Bc)c (exercise), and similarly, ifthe Ai (for i ∈ I) are subsets of Ω, then

⋂i∈I Ai = (

⋃i∈I Ac

i )c .

Therefore:

If A,B ∈ F , then A ∩ B ∈ F and A \ B = A ∩ Bc ∈ F ;

If every Ai ∈ F , then⋂

i∈I Ai ∈ F .

Example 1: Let Ω = 1,2,3,4,5,6 and F = ∅, 1,3,5, 2,4,6,Ω.Verify that F is a σ-field on Ω.


σ-fields: examples

Example 2: Let Ω = 1,2,3 and

F = ∅, 1, 2, 1,3, 2,3,Ω.

Is F a σ-field on Ω ? If not, indicate the property that is violated. If yes,verify that F is a σ-field on Ω.

Answer: 1 and 2 are elements of F but 1 ∪ 2 = 1,2 is notan element of F , so property (c) in the definition does not hold.

(Also: 1,3 ∩ 2,3 = 3 6∈ F , so F is not closed under finiteintersections.)


More on σ-fields

Definition 5If F is a σ-field on Ω, the pair (Ω,F) is called a measurable space.

Recall that if (Ω,F) is a measurable space and A,B ∈ F , then A ∪ B,A ∩ B and A \ B = A ∩ Bc are all in F .

Proposition 6Let F1 and F2 be two σ-fields on Ω. Then F1 ∩ F2 is a σ-field on Ω.Similarly, if Fi , for i ∈ I, are all σ-fields on Ω, then

⋂i∈I Fi is a σ-field on

Ω.

Proof: On the board.


More on σ-fields

Proposition 7If A is a collection of subsets of Ω, there is a unique σ-field F(A) suchthat

1 A ⊆ F(A), i.e. every set in A is in F(A);2 If G is another σ-field with A ⊆ G, then F(A) ⊆ G.

In other words: F(A) is the smallest σ-field containing all the elementsof A. We call it the σ-field generated by A.



More on σ-fields

Proposition 8If A, A1, A2 are subsets of P(Ω), then the following hold.

1 If A1 ⊆ A2, then F(A1) ⊆ F(A2).2 If A is a σ-field, then F(A) = A.3 F(F(A)) = F(A).


If A is given, how do we determine F(A)? We must start with all thesets of A together with ∅, then build out of them all possible sets usingcomplement and countable unions. It is at best cumbersome.


Partitions

Definition 9A collection of subsets (Ai)i∈I of Ω is called a partition (of Ω) if

Ai ∩ Aj = ∅ if i 6= j , (1)⋃i∈I

Ai = Ω. (2)

In other words, a partition of a set Ω is a collection of non-overlappingsubsets of Ω whose union covers the whole space. If the indexing set Iis finite, we write (An)k

n=1 and if I is countable, we write (An)∞n=1


Partitions

Proposition 10If A = (Ai)i∈I is a partition of Ω, then the σ-field F(A) consists of

the empty set,all sets of the form

⋃j∈J Aj where J is a subset of I such that J is

countable or I \ J is countable.


Corollary 11If A = (Ai)

∞i=1 is a countable partition of Ω, then the σ-field F(A)

consists of the empty set together with all possible unions of elementsof A.


Partitions

Proposition 12Let A be a collection of subsets of Ω. Then there is a partition PA of Ωsuch that F(PA) = F(A).

Proof: No proof.


Given a collection A of subsets of Ω, we can obtain PA as follows:The sets in PA form a partition of Ω, so every element of Ω willbelong to exactly one of them.Therefore it suffices to be able to determine, for every x ∈ Ω, theset in PA that contains x .Take x ∈ Ω. The set in PA that contains x is:

SA(x) = y ∈ Ω : y belongs to exactly the same sets of A as x= y ∈ Ω : for every A ∈ A, if A contains one of x , y ,

then it contains the other.

Observe that if y ∈ SA(x) then SA(y) = SA(x).


Example: Ω = 1,2,3,4,5,6,7, A = 2,3,4, 4,5,6. ThenSA(1) = 1,7 = SA(7)), SA(2) = 2,3 = SA(3), SA(4) = 4,SA(5) = 5,6 = SA(6). Therefore

PA = 1,7, 2,3, 4, 5,6,

and

F(A) = F(PA)

= all possible unions of elements of PA and the empty set= ∅, 4, 2,3, 5,6, 1,7, 2,3,4, 4,5,6, 1,4,7,2,3,5,6, 1,2,3,7, 1,5,6,7, 2,3,4,5,6,1,2,3,4,7, 1,4,5,6,7, 1,2,3,5,6,7,Ω


Exercises

Exercise 1: Let Ω = 1,2, . . . ,7 and letA = 1,2,3,7, 2,3,4,5,6. Find P(A).

Exercise 2: Let Ω = 1,2,3,4,5,6 and F be the σ−field on Ωgenerated by the collection of subsets A = 1,3, 2,3,5, 4,a,a 6= 4. Can F have exactly 30 subsets? Find a such that F has exactly32 subsets.

Exercise 3: Let Ω = 1,2, . . . ,10 and letA = 1,2,4,8, 2,3,4,6,7,8, 2,8,9,10. Find P(A).


Partitions

Definition 13Let P1 and P2 be two partitions of the same set Ω. We say that P2 isfiner than P1 if every set in P1 can be written as a union of sets fromP2.

Proposition 14Let P1 and P2 be two countable partitions of Ω with P2 finer that P1.Then F(P1) ⊂ F(P2).

Proof: No proof (we will not use this in class).


The Borel (σ-)field, Borel sets

As usual, (a,b) = x : a < x < b and [a,b] = x : a ≤ x ≤ b.

Definition 15

The σ-field generated by the open intervals in R is called the Borelfield on R and it is denoted B(R). Subsets of R which belong to B(R)are called Borel sets.

Proposition 16The Borel field is generated by the closed intervals.



Borel sets

Let A = (xn)∞n=1 be a countable subset of R. Is A a Borel set?

Yes: By the previous proposition, each xn = [xn, xn] is a Borel set, soA is a Borel set.

In particular N, Z and Q are Borel sets.


Borel sets in Rn

We make a similar definition in Rn:

Definition 17

The σ-field generated by the subsets of Rn of the form

(a1,b1)× · · · × (an,bn)

with a1 < b1, . . . ,an < bn ∈ R, is called the Borel field on Rn and it isdenoted B(Rn). Subsets of R which belong to B(Rn) are called Borelsets.


Measuring sets

The purpose of this section is to see how we can we assign a ‘size’ to sets.

Definition 18Let (Ai )i∈I be a sequence of subsets of Ω. We say that the elements of thesequence are pairwise disjoint if Ai ∩ Aj = ∅ whenever i , j ∈ I with i 6= j .

Definition 19A measure on Ω is a map µ : F → [0,∞] = R≥0 ∪ ∞ such thatF is a σ-field on Ω,µ(∅) = 0,For every sequence (An)n∈N of pairwise disjoint elements of F , wehave

µ(⋃n∈N

An) =∑n∈N

µ(An).

(With the convention that∞+ x =∞ for every x ∈ [0,∞].)


Examples of measures

Let µ0 : P(Ω)→ [0,∞], µ0(A) = 0 for every A ∈ P(A). Then µ0 isa measure on Ω, defined on P(Ω). It is not an interesting one.Let µc : P(Ω)→ [0,∞], µc(A) = |A|. Then µc is a measure on Ω,again defined on P(Ω).

In general, a measure is only defined on a σ-field F and not on P(Ω)because we may not need to have it defined on all subsets, or it mayeven be impossible (see below).

Question 20Is there a measure µ defined on subsets of R such thatµ([a,b]) = b − a, i.e. the usual length of intervals?

Answer: Yes (Lebesgue), but it cannot be defined on all subsets of R.


The Lebesgue measure

Theorem 21There is a unique measure λ : B(R)→ [0,∞] such that for everya < b ∈ R, λ

((a,b)

)= b − a. The measure λ is called the Lebesgue

measure.

We have the same result in dimension n:

Theorem 22There is a unique measure λn : B(Rn)→ [0,∞] such that for everya1 < b1, . . . ,an < bn ∈ R, λn

((a1,b1)× · · · × (an,bn)

)=∏n

i=1(bi − ai).The measure λn is called the Lebesgue measure on Rn.

There are subsets of R, and subsets of Rn that are notLebesgue-measurable! Cf. the Banach-Tarski paradox.


Measurable functions

Let X : Ω→ R be a function, and let B ⊆ R. Recall that the preimageof B by X is

X−1(B) = ω ∈ Ω : X (ω) ∈ B

Remark: Despite the notation, we do not require X to be invertible; theset X−1(B) is defined even if X is not invertible (this is the standardnotation, but it would be better if it did not look like the inverse of X ).

Definition 23Let X : Ω→ R and c be a real number. The set X−1(c) is called alevel set of X .



Proposition 24Let X : Ω −→ R be a function. The collection

FX = X−1(B) : B ∈ B(R)

is a σ-field on Ω. We call the above collection the σ-field generated byX .

Proof: See the board.

Proposition 25Let X : Ω −→ R be a function. Then FX is the σ-field on Ω generatedby the sets X−1((a,b)

)for a < b ∈ R ∪ −∞,+∞.

Proof: No details, but it follows from the fact that the σ-field B(R) isgenerated by the sets (a,b) for a < b ∈ R ∪ −∞,+∞.



Definition 26Let (Ω,F) be a measurable space. A mapping X : Ω −→ R is calledF-measurable if FX ⊂ F .

Observe that FX is the smallest σ-field F such that X is F-measurable(because it is the smallest σ-field F with the property that FX ⊆ F).



Proposition 27Let X : Ω −→ R be a function, and let F be a σ-field on Ω. Thefollowing are equivalent:

1 X is F-measurable;2 For every open interval (a,b) of R, X−1((a,b)

)is in F .

3 For every a ∈ R, X−1((a,∞))

is in F .


Remark:X is F-measurable is often expressed (a bit abusively) by sayingthat the value of X is completely determined by the informationcontained in F .(because F will contain each level set X−1(c).)



When Ω = R and F = B(R) and thus X is a real-valued function ofa real variable, we use the expression Borel measurable in placeof B(R)-measurable.

Proposition 28Every continuous function from R to R is Borel-measurable.

No proof.



Proposition 29Let (Ω,F) be a measurable space. Let f ,g : Ω→ R be F-mesurable,let α ∈ R, and let φ : R→ R be Borel-measurable. Then

φ f , αf , f + g, fg, |f |, min(f ,g), max(f ,g)

are all F-measurable. If g(x) 6= 0 for every x ∈ Ω, then f/g is alsoF-measurable.

Proof: On the board if we have enough time.



Definition 30The indicator function (or characteristic function) of a set A in Ω,1A,is defined as follows:

1A(ω) =

1 if ω ∈ A,0 if ω 6∈ A.

Definition 31A function X : Ω→ R is called simple if it only takes a finite number ofvalues.

Observe that if X =∑n

i=1 ci1Ai is a linear combination of indicatorfunctions, then X is a simple function.



The converse is true:

Proposition 32Let X be a simple function and let c1, . . . , ck be the different values thatX takes. Define Ai = X−1(ci). Then

X =k∑

i=1

ci1Ai .

This particular representation of X as linear combination of indicatorfunctions is called the canonical representation of X .



Theorem 33Let X =

∑ni=1 ci1Ai be the canonical representation of a simple

function. Then1 FX is the σ-field generated by A1, . . . ,An.2 X is F-measurable if and only if each Ai belongs to F .

Proof:1 Let G be the σ-field generated by A1, . . . ,An. We re-order the ci

to have them in increasing order: c1 < · · · < cn.FX contains the sets X−1((ci − ε, ci + ε)) = X−1(ci) = Ai , so byProposition 7 we have G ⊆ FX .Let a < b ∈ R ∪ −∞,+∞, and let i , j be such thata < ci < ci+1 < · · · < cj < b.



1 (Continued) Then

X−1((a,b)) = Ai ∪ Ai+1 ∪ · · · ∪ Aj ∈ G,

so G is a σ-field containing all the sets X−1((a,b)). Therefore (byProposition 7) it contains the σ-field generated by the setsX−1((a,b)), which is FX by Proposition 25.

2 “⇐” We want to show that FX ⊆ F . By the first part andProposition 7, it is true since each Ai belongs to F .

“⇒” Up to renumbering the c1 we can assume c1 < c2 < · · · < ck .If a,b ∈ R are such that cj−1 < a < cj < b < cj+1 then

Aj = X−1((a,b))∈ F

(see Proposition 27).



Example 1: Let F denote the σ-field on Ω = 1,2,3,4,5,6 generatedby 1,2, 1,4 and 2,3,5 and let

X = 2 · 11,2 + 3 · 11,3,5 − 2 · 13,5.

Is X F-measurable?

Example 2: Let Ω = N and X = 5 · 11,2,3 − 2 · 12,4,5 + 2 · 13,5,6.Find FX . Let F be the σ-field on Ω generated by B = 1,4, 4,3.Is X F-measurable?



Solution: X is F-measurable means FX ⊆ F , so we need to determineFX and F . To determine FX we use theorem 33, and we first have tofind the canonical representation of X . The different values that Xtakes are −2,0,2,3,5,7 and thus the canonical representation of X is

X = −2 · 14 + 2 · 16 + 3 · 12 + 5 · 11 + 7 · 13.

Therefore FX = P(4, 6, 2, 1, 3) = P(A) withA = 4, 6, 2, 1, 3. In this case the sets SA(x) are4, 6, 2, 1, 3 and 5,7,8, . . ., and FX consists of the emptyset and all possible unions of the sets SA(x).Now, F = F(B) by definition and F(B) = F(PB). In this case we havePB = 1, 3, 4, 2,5,6,7 . . ., so F consists of the empty setand all possible unions of elements of PB.Observe that FX ⊆ F is equivalent to having all the elements of theform SA(x) in F (because the elements of FX are the unions of theSA(x)). Looking at SA(2) = 2, we see that it is not in F , so X is notF-measurable.



Example 3: Let Ω = R, F = B(R) and

X = 3 · 1[0,4] − 2 · 1[1,2] + 5 · 1[3,6].

Find FX and prove that X is Borel measurable.

Example 4: Let Ω = 1,2,3,4,5,6 and

X = 2 · 11,2,3 − 4 · 14,5,6 + 13,5 , Y = 3 · 12,3,4 − 2 · 11,2,3,5,6.

Is X+ FY -measurable? Is Y− FX -measurable?



Example 5: Let Ω = −1,1 and define X : Ω→ R by

X (−1) = −1 and X (1) = 1.

Is X FX 2-measurable?

Example 6: Let (Ω,F) be a measurable space, and let X : Ω→ R.Suppose that |X | is F-measurable. Does this imply that X isF-measurable? Either prove it or give a counterexample.


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Basic notation about sets - University College Dublinastier/math20180/chapter2.pdf · Basic notation about sets ... We can check that A \B = (Ac [Bc)c (exercise), and similarly, if