A Probability Path.pdf

Thus we can generate the borel sets with any kind of

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Thus we can generate the Borel sets with any kind of interval: open, closed, semi- open, finite, semi-infinite, etc. Here is a sample proof for two of the equivalences. Let co= {(a, b), -00 b oo} be the open intervals and let c<J ={(a, b] , -00 b < oo} be the semi-open intervals open on the left. We will show a(CO) = a(C<l). Observe (a, b)= -1/n]. Now (a,b -1/n] E c<l C a(C<l), for all n implies (a, b- 1/n] E a(C<l). So (a, b) E a(C<l) which implies that co c a (C< 1). Now a (C<l) is a a -field containing co and hence contains the minimal a-field over co, that is, a(CO) c a(C<l). Conversely , (a, b] = (a, b + 1/n). Now (a, b + 1/n) E co C a(CO) so that b + 1/n) E a(CO) which implies (a, b] E a(CO) and hence c<J c a(C<>). This implies a(C<l) c a(C<>). From the two inclusions, we conclude as desired. Here is a sample proof of the fact that B(IR) = a(open sets in IR). We need the result from real analysis that if 0 C lR is open, 0 = lj, where I j are open, disjoint intervals. This makes it relatively easy to show that a( open sets) = a(Co). If 0 is an open set, then we can write

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18 1. Sets and Events We have Ij E co c a(CO) so that 0 = UJ= 1 Ij E a(CO) and hence any open set belongs to a(CO), which implies that a( open sets) c a(C<>) . Conversely, co is contained in the class of open sets and therefore a (CO) c a( open sets) . Remark. If IE is a metric space, it is usual to define 8(1E), the a-field on IE, to be the a-field generated by the open subsets of IE. Then 8(1E), is called the Borel a-field. Examples of metric spaces IE that are useful to consider are JR., the real numbers, JR.d, d-dimensional Euclidean space, !R. 00 , sequence space; that is, the space of all real sequences. C[O, oo), the space of continuous functions on [0, oo). 1.8 Comparing Borel Sets We have seen that the Borel subsets of JR. is the a-field generated by the intervals of JR.. A natural definition of Borel sets on (0, 1], denoted 8( (0, 1]) is to take C (0, 1] to be the subintervals of (0, 1] and to define 8((0, 1]) := a(C(O, 1]). If a Borel set A E 8(1R) has the property A C (0, 1], we would hope A E 8((0, 1]). The next result assures us this is true. Theorem 1.8.1 Let no c n. (1) If 8 is a a-field of subsets of n, then 8o := {A no : A E 8} is a a- field of subsets of no. (Notation: 8o =: 8 n no. We hope to verify 8( (0, 1]) = 8(1R) n (0, 1 ].) (2) Suppose C is a class of subsets of n and 8 = a (C). Set C n no= : Co= {A no: A e C} . Then a(Co) = a(C) n no In symbols (2) can be expressed as a(C n no)= a(C) n no
1.8 Comparing Borel Sets 19 so that specializing to the Borel sets on the real line we get B(O, 1] = B(lR) n (0, 1]. Proof. (1) We proceed in a series of steps to verify the postulates defining a a- field. (i) First, observe that no e Bo since nno = no and n E B. (ii) Next, we have that if B = A no e Bo, then no\ B =no\ A no= no(n \A) e Bo since n \A e B. (iii) Finally, if for n :::: 1 we have Bn = An no, and An e B, then ()() ()() ()() u Bn = u Anno= (U An) n no e Bo n=l n=l n=l since Un An e B.

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