Basic notation about sets - University College Dublinastier/math20180/chapter2.pdf · Basic...
Transcript of Basic notation about sets - University College Dublinastier/math20180/chapter2.pdf · Basic...
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
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Basic notation about sets
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)
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Basic notation about sets
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.
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Basic notation about sets
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.
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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. . .
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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.
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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.
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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.)
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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.
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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.
Proof: On the board.
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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).
Proof: On the board.
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.
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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
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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.
Proof: On the board.
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.
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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.
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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).
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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
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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).
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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).
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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.
Proof: On the board.
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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.
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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.
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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,∞].)
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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.
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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.
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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 .
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Measurable functions
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 ∪ −∞,+∞.
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Measurable functions
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).
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Measurable functions
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 .
Proof: On the board.
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).)
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Measurable functions
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.
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Measurable functions
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.
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Measurable functions
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.
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Measurable functions
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 .
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Measurable functions
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.
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Measurable functions
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).
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Measurable functions
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?
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Measurable functions
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.
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Measurable functions
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?
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Measurable functions
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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