Question

3. Let (12, F,P) be a probability space, and A1, A2, ... be an increasing sequence of events; that is, A1 CA2 C.... Using only the Kolmogorov axioms, prove that P is continuous from belour: lim P(An) = P(U=1 An). Hint: Work with a new sequence of events By := A and B := An An-1. n+00 [1]

Answer 1

Let (r, F.P) be a probability annend space, A.,A. be an increasing sequence of events, the is, A CĄC . Using only Kolmogoro anons prove that pus continuans from below:- lim P (A) = P (An) let can be a measurable space of events A probability measure is a real valued function mapking P AR satisfying: - (1) for any event & EF, P(E)>0 [Non negativity of (2) prob. measure P(2)=1 (3) for any countably infinite sequence of events (Eilisl, that are mutually exclusive. P (UE)= P(E;) known as countable Also there is one aniom of continuity, ie. If BoB . Bn be a countable sequence such that (i Bi Biti Br- & their lim P(Bn) = 0 - 0 also. P(A)= P (A) = & P.) if Ais are pai disjoint According to question, A CAÇ ... An using BIA .. B = A ₂ 'A – – Bar = An Any + B, 2B, ). Bm As these are also Plu And = {P(An) not disjoint so o e have a result that ① and 2 are same as {A} is disjoint A-Ů ACAU (BA) no! no

Assume A. (int nts occured emplying occurre n-> let Buy be Bry i Obnoudly it is decreasing also A (A) & But I Ais are pararse disjoint Ain Bnt = 6 (i=1,2,...,n) A. (i=ntinta,...) as an event Bohas. implying occurrence of Andere, Since A's are disjoint => Air, Aitz, do not occure leading to conclusion that Bit Bit 2 will not occure i.e. Be = 0 hence lim P(B) =0 But according to our question, they are not pairwise disjoint lim P(B) & Elf no 0 P(Ac) = lim & Plai) + lim (Brt) Plu ) - lim Plan) - lim P (An) = Plon) Mence broved Pq-@ no nas n !