Clarification on Borell-Cantelli lemma

Question

Consider a sequence of decreasing events $A_n\downarrow A$ such that $\sum_{n=1}^\infty P(A_n)<\infty$. Then by the Borel-Cantelli lemma $$P(\limsup_nA_n) = 0$$ or $$P(\lim_n A_n^c) = 1$$

I need to understand why this implies that there exists $n(\omega)<\infty$ a.s. such that $$P(A_n^c)=1,\quad \forall n > n(\omega).$$

This should be obvious, but I fail to see it.

Answer 1

Note that the reason why we have $$ P(\lim A_n^c) = 1$$

is because we have $$P(\liminf A_n^c) = 1$$

---

BCL1 gives us:

$$P(\limsup A_n) = 0$$

$$\to P(\liminf A_n^c) = 1$$

Let $$\omega \in \liminf A_n^c$$

Then

$$\exists m \ge 1 \ s.t. \ \forall n \ge m,$$

$$\omega \in A_n^c$$

I guess $m$ is random i.e. $m = n(\omega)$

Answer 2

Not because of Borel-Cantelli, but because of the convergence of the sum.

In your case $A_k\uparrow A$, so $P(A_1)\le P(A_2) \cdots $.

If $\delta = P(A_n)> 0$, for some $n$, then $P(A_k)\ge \delta$ for all $k\ge n$.

But that means the sum diverges... $$ P(A_1) + \cdots +P(A_k) \ge (k -n ) \delta,$$ which diverges as $k\rightarrow \infty$. Hence $P(A_n) = 0$ for all $n$.