Question: Regarding the converse stated in Exercise 6, if the events A n ,n 1, are not independent, then P(lim sup n A n )

Regarding the converse stated in Exercise 6, if the events An,n ≥ 1, are not independent, then P(lim supn→∞ An) = 0 need not imply that 2n=1 P(A,) <∞. Give one or two concrete examples to demonstrate this assertion. Take (Ω, A, P) = ((0, 1), B(0,1), λ),λ being the Lebesgue measure. Then

(a) Take

2n=1 P(A,)

and show that

so that

Then, by Exercise 5, P(lim supn→∞, An) = 0, where An = (|Xn| ≥ l/k) for any arbitrary but fixed k = 1, 2,... Also, show that

(b) Take

and show that

so that

Again, P(lim sup n→∞ An) = = , as in (a). Also show that

P(An)

2n=1 P(A,)

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