B. Extension of an Outer Measure to a Probability Measure

To use the outer measure as a probability measure for more general sets that those in F0, we have to extend the algebra F0 to a a-algebra F of events for which the outer measure is a probability measure. In this appendix it will be shown how Fcan be constructed via the following lemmas.

Lemma 1.B.1: For any sequence Bn of disjoint sets in ^, P*(U=j Bn) <

£Г=1 p "(Bn).

Proof: Given an arbitrary є > 0 it follows from (1.21) that there exists a countable sequence of sets An, j in F0 such that Bn c An, j and P *(Bn) >

ZjU P(An, j) – є2 n; hence,


£ P*(Bn) > £ £ P(An, j) – £ £ 2-n = £ £ P(An, j) – є. n=1 n=1 j=1 n=1 n=1 j=1


Moreover, UO=1 Bn c UO=1 UO=1 An, j, where the latter is a countable union of sets in Ж0; hence, it follows from (1.21) that



и BA <££ P(An, j). (1.29)

n=1 n=1 j=1

If we combine (1.28) and (1.29), it follows that for arbitrary є > 0,

£p*(Bn) > PM u bA – Є. (1.30)

n=1 n=1

Letting є I 0, the lemma follows now from (1.30). Q. E.D.

Thus, for the outer measure to be a probability measure, we have to impose conditions on the collection Ж of subsets of U such that for any sequence Bj of disjoint sets in Ж, P *(Uf=1 Bj) > Yj=1 P * (Bj). The latter is satisfied if we choose Ж as follows:

Lemma 1.B.2: Let Ж be a collection ofsubsets B of U such thatfor any subset A of U:

P*(A) = P*(A П B) + P*(A П B). (1.31)

Then for all countable sequences of disjoint sets Aj є Ж, P *(Uj= 1 Aj) = ГГ=1 P *( Aj).

Proof: Let A = Uj=1 Aj, B = A1. Then A П B = A П A1 = A1 and A П B = Uj=2Aj are disjoint; hence,

P*( Uj=1 Aj) = P*(A) = P*(A П B) + P*(A П B)

= P*(A1) + P*( Uj=2 Aj). (1.32)

If we repeat (1.32) for P *(Uj=kAj) with B = Ak, k = 2,…, n, it follows by induction that


P *(U=1 Aj) = £; P *( Aj) + P *(U°=n+1 Aj) j =1


>J2 P*(Aj) for all n > 1;


hence, P*(Uj=1 Aj) > 1 P*(Aj). Q. E.D.

Note that condition (1.31) automatically holds if B є &0: Choose an arbitrary set A and an arbitrary small number є > 0. Then there ex­ists a covering A c U“=1 Aj, where Aj є &0, such that P(Aj) < P*(A) + є. Moreover, because A П B c Ц>=1 Aj П B, where Aj П B є &0, and A П B c U=1 Aj П B, where Aj П B є &0, we have P*(A П B) < Yj= P (Aj П B) and P * (A П B) <Yj=1 P (Aj П B); hence, P *(A П B) + P*(A П B) < P*(A) + є. Because є is arbitrary, it follows now that P*(A) >

P*(A П B) + P*(A П B).

I will show now that

Lemma 1.B.3: The collection & in Lemma 1.B.2 is a a-algebra of subsets of ^ containing the algebra &0.

Proof: First, it follows trivially from (1.31) that B є & implies B є &.Now, let Bj є &. It remains to show that U j= Bj є &, which I will do in two steps. First, I will show that & is an algebra, and then I will use Theorem 1.4 to show that & is also a a-algebra.

(a) Proof that & is an algebra: We have to show that B1, B2 є & implies that B1 U B2 є &. We have

P * (A П BO = P *( A П B1 П B2) + P *( A П B1 П B2),

and because

A П (B1 U B2) = (A П B1) U (A П B2 П B1), we have

P* (A П (B1 U B2)) < P *(A П B1) + P *(A П B2 П Bf.


P*(A П (B1 U B2)) + P*(A П B1 П B2) < P*(A П B1)

+ P*(A П B2 П B1) + P*(A П B2 П Bo = P*(A П B1) + P*(A П Bo = P*(A). (1.33)

Because ~(B1 U B2) = B1 П B2 and P*(A) < P*(A П (B1 U B2)) + P*(A П (~(B1 U B2)), it follows now from (1.33) that P*(A) = P *(A П (B1 U B2)) + P *(A П (~(B1 U B2)). Thus, B1, B2 є & implies that B1 U B2 є &; hence, & is an algebra (containing the algebra &0).

(b) Proof that & is a a-algebra: Because we have established that & is an algebra, it follows from Theorem 1.4 that, in proving that & is also a a-algebra, it suffices to verify that U j= Bj є & for disjoint

sets Bj є &. For such sets we have A П (Un=1 Bj) П Bn = A П Bn and A П (Un=1 Bj) П Bn = A П (Un-1 Bj); hence,

P* I A П ( U B

image044 Подпись: (134)

= P * ^A П ^ ^ Bjj П Bnj + P* ^A П ^ ^ Bj ) П Bn = P*(A П Bn) + P* fA П Л-! Bj

Подпись: P*(A П B) < P* A П image047 Подпись: (1.35)

Next, let B = UO 1 Bj. Then B = П° 1Bj С Пп=1Bj = ~(Un=1 Bj); hence,

It follows now from (1.34) and (1.35) that for all n > 1,

Подпись: U Bj j=1 P*(A) = P* |^A П ^ U B^j + P* |^A П


> ^ P* (A П Bj) + P*(A П B);



P*(A) > ^ P*(A П Bj) + P*(A П B) > P*(A П B) + P*(A П B),



where the last inequality is due to



U (A П Bj)J < ^ P*(A П Bj).

Because we always have P*(A) < P*(A П B) + P*(A П B) (compare Lemma 1.B.1), it follows from (1.36) that, for countable unions B = U^1 Bj of disjoint

sets Bj є

P*(A) = P*(A П B) + P*(A П B);

hence, B є &. Consequently, & is a a-algebra and the outer measure P * is a probability measure on {^, &}. Q. E.D.

Lemma 1.B.4: The a-algebra & in Lemma 1.B.3 can be chosen such that P* is unique: any probability measure P* on {^, &} that coincides with P on &0 is equal to the outer measure P *.

The proof of Lemma 1 .B.4 is too difficult and too long (see Billingsley 1986, Theorems 3.2-3.3) and is therefore omitted.

If we combine Lemmas 1.B.2-1.B.4, Theorem 1.9 follows.

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