By Claude Dellacherie, Paul-André Meyer
Read Online or Download Probabilités et Potentiel, vol.A , chap. I à IV PDF
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Extra info for Probabilités et Potentiel, vol.A , chap. I à IV
2). Letting Step 2. e. S = → 0 proves OM3 . ∀S ∈ . 3) implies the uniqueness of the extension[ ] once we know that ¯ is well-defined, that is, independent of the particular representation of sets in ∪ . L. 3) does not depend on the representation of ∪ -sets. The family ∪ is clearly stable under finite disjoint unions. If S T ∈ find (notation as before) · S ∩ T = S1 ∪ · TN = · ∪ Sj ∩ Tk ∈ j k=1 ∪, · S \ T = S1 ∪ M ∪ ∈ also · SM \ T1 ∪ · ∪ M N = · we MN · SM ∩ T1 ∪ · ∪ and, since by S3 Sj \ Tk ∈ ∪ · TN ∪ N Sj ∩ Tkc = · j=1 k=1 Sj \ Tk ∈ j=1 k=1 ∈ ∈ · -stability of where we used the ∩- and ∪ ∪.
Closed balls with centre 0 and radius k. L. Schilling Since n c = U c U ∈ n = n (and n c = n ) we have n , hence n ⊂ n and the converse inclusion is similar. n = n c ⊂ n is generated by many different systems of sets. For The Borel -algebra our purposes the most interesting generators are the families of open rectangles o = on = o n = a1 b1 × · · · × an bn aj bj ∈ and (from the right) half-open rectangles = n = n = a1 b1 × · · · × an bn aj bj ∈ We use the convention that aj bj = aj bj = ∅ if bj aj and, of course, that a1 b1 × · · · × ∅ × · · · × an bn = ∅.
Let us now assume that (i)–(iii) hold for the set-function → 0 . In order to see that is a measure, we have to check M2 . 3) j∈ k∈ Clearly Ak ↑ A, and using repeatedly property (ii) we get Bk . From (iii) we conclude Ak = B1 + · · · + k A = lim k→ Ak = lim k→ ∗ ∗ Bj = Bj j∈ j=1 ∗ Finally assume that A < for all A ∈ and that (i), (ii) and (iii ) or (iii ) hold. We will show that under the finiteness assumption (iii )⇒(iii )⇒(iii); the assertion follows then from the considerations of the first part of the proof.