P[limsup] of independent events

[student12s]

Homework Statement

Let (A_n) be a sequence of independent events such that Pr(A_n)<1 for all n. Show that P[limsup A_n]=1 if and only if P(\cup_{n=1}^{+\infty} A_n)=1

Homework Equations

The Attempt at a Solution

Suppose P[\limsup A_n]=1. Define $B_k=\cup_{i=k}^{+\infty} A_i so that B_n\downarrow \limsup_{n\to+\infty} A_n. Thus \lim_{n\to+\infty} P(B_n)=P(\limsup_{n\to+\infty} A_n)=1. So, for all \epsilon>0, there is M so that 1\geq P(\cup_{i=1}^{+\infty} A_i)\geq P(\cup_{i=M}^{+\infty}A_i)\geq 1-\epsilon$, so P(\cup_{i=1}^{+\infty} A_i)=1.
I am having trouble with the other direction. Any hints?

 

[mighty2000]

Thank you for your post. I am a scientist and I would be happy to help you with this problem.

To prove the other direction, we can use the definition of the limit supremum of a sequence of events. Recall that the limit supremum, denoted as limsup A_n, is defined as the set of outcomes that occur infinitely often in the sequence of events A_n.

First, assume that P(\cup_{n=1}^{+\infty} A_n)=1. This means that the probability of the union of all the events A_n occurring is equal to 1. This implies that for any \epsilon>0, there exists an M such that P(\cup_{n=M}^{+\infty} A_n) \geq 1-\epsilon.

Now, we can define a new sequence of events B_n = \cup_{i=n}^{+\infty} A_i. Notice that this sequence is decreasing, as B_{n+1} \subset B_n for all n. Also, since P(\cup_{n=1}^{+\infty} A_n)=1, we have that P(B_n) \geq 1-\epsilon for all n.

Next, we can use the continuity of probability to show that P(\limsup A_n) = P(\cap_{n=1}^{+\infty} B_n) = \lim_{n\to+\infty} P(B_n) = 1. This completes the proof in this direction.

I hope this helps. Let me know if you have any further questions or if you would like me to explain any part of the solution in more detail.