# Measure, integration

Páginas: 72 (17882 palabras) Publicado: 3 de diciembre de 2010
MEASURE, INTEGRATION
&

PROBABILITY

Ivan F Wilde Mathematics Department King’s College London

Chapter 1 σ-algebras and Borel functions

Preliminary discussion
• Suppose that X is a continuous random variable. Then Prob(X = s) = 0 for any s ∈ R. However, the event { X ∈ R } has probability one and is the disjoint union of the events { X = s } for s ∈ R; {X ∈ R} =
s

{ X = s }.Each event on the right hand side has probability zero, so the probabilities of the events on the right hand side do not “add up” to that of the left hand side. We wish to understand this. • Suppose that X is a random variable on a sample space Ω, and suppose that X takes values x1 , x2 , . . . . Put Ai = { ω ∈ Ω : X(ω) = xi }. Then EX =
i

xi Prob(X = xi ) xi Prob(Ai ).
i

=

Inparticular, for 1 Ai (ω) = l

1, ω ∈ Ai , 0, ω ∈ Ai /

E1 Ai = 1 Prob(1 Ai = 1) l l = Prob(Ai ). Also E(xi 1 Ai ) = xi Prob(Ai ). But we can write X = l recover EX as EX =
i

l i xi 1 Ai

and we

xi Prob(Ai ) =
i

E(xi 1 Ai ). l

Here X is a “step-function” on Ω. This formula forms the basis for the “general” expectation, i.e., that for an arbitrary random variable. 1

2

Chapter 1• One must (sometimes) ask which subsets of a sample space are deemed to be events. Can one take all subsets of the sample space to be events? The answer is sometimes yes and sometimes no. For example, in the case when the probability of an event within a bounded region of R3 is required to be proportional to the volume associated with the event, then one naturally asks whether every subset of (abounded region) of R3 actually has a volume. That this is not so is demonstrated by the Banach-Tarski theorem. (This says that a ball of unit radius in R3 can be cut up into a ﬁnite number of pieces which can then be reassembled to form a ball of radius 2. The meaning of “volume” for these pieces is not clear.) We must be precise about the concept of “event”. In the “modern” (Kolmogorov) theory ofprobability, this is formulated in terms of σ-algebras. Deﬁnition 1.1. A collection Σ of subsets of a non-empty set X is called a σ-algebra if (i) X ∈ Σ, (ii) if A ∈ Σ, then Ac = X \ A ∈ Σ, (iii) if An ∈ Σ for n = 1, 2, . . . , then
∞ n=1 An

∈ Σ.

The sets in Σ are called measurable sets, and (X, Σ) is called a “measurable space”. Remarks 1.2. 1. Since ∅ = X c , it follows that ∅ ∈ Σ. 2. Forany A1 , A2 , . . . , An ∈ Σ, put An+1 = An+2 = · · · = ∅. Then we see that A1 ∪ · · · ∪ An = ∞ Ak ∈ Σ, by (1) above, and (iii). k=1 3. Let A1 , A2 , · · · ∈ Σ. Then since ∞ n=1 An ∈ Σ.
∞ n=1 An

=

∞ c c n=1 An ,

we see that

If we take An+1 = An+2 = · · · = X, then we get A1 ∩ · · · ∩ An ∈ Σ. 4. Let A, B ∈ Σ. Then A \ B = A ∩ B c and so A \ B ∈ Σ. Proposition 1.3. Let { Σα } be anarbitrary collection of σ-algebras of X. Then α Σα is a σ-algebra. Proof. We check the requirements. (i) X ∈ Σα for all α and so X ∈
α Σα .

(ii) Suppose that A ∈ α Σα . Then A ∈ Σα for all α and so Ac ∈ Σα for all α, that is Ac ∈ α Σα .
Department of Mathematics

σ-algebras and Borel functions

3

(iii) Let A1 , A2 , . . . belong to α Σα . Then, for each α, An ∈ Σα for all n and so n An ∈Σα . Hence n An ∈ α Σα . The result follows. Let C be any collection of subsets of X. Then certainly C is contained in the σ-algebra consisting of all subsets of X. If we set Σ(C) =
F

Σ

where the intersection is over the family F of all those σ-algebras Σ which contain C, then Σ(C) is the “smallest” σ-algebra containing C. It is called the σ-algebra generated by C. Deﬁnition 1.4. Let Cdenote the collection of open subsets of R. Then Σ(C) is called the Borel σ-algebra of R, usually written B(R). The elements of B(R) are called Borel sets. Similarly, one deﬁnes B(Rn ) as the σ-algebra generated by the open subsets of Rn . Proposition 1.5. The following subsets of R belong to B(R): (i) (a, b) for any a < b; (ii) (−∞, a) for any a ∈ R; (iii) (a, ∞) for any a ∈ R; (iv) [a, b] for...

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