[Math] Probability of limsup

Question

Let $A_1, A_2, A_3, \dots$ be a sequence of independent events on $\left (\Omega, \mathbb A, \mathbb P\right )$ such that $\mathbb P(A_n) < 1$ and $\mathbb P\left (\bigcup\limits_{n=1}^{\infty}A_n\right ) = 1$.

Why is it valid that: $\mathbb P\left (\bigcap\limits_{m=1}^{\infty} \bigcup\limits _{n=m}^{\infty} A_n\right ) = 1$ ?

Answer 1

Since $A_1,A_2,...$ are independent it follows that $A_1^c,A_2^c...$ are independent. Therefore $$P(\cup A_n)=1\Rightarrow$$ $$P(\cap A_n^c)=0\Rightarrow$$ $$\prod P(A_n^c)=0\Rightarrow $$ $$\prod (1-P(A_n))=0$$ Now, a theorem from Real Analysis which can be easily proved , states that:

Suppose $\{a_j\}_{j=1}^\infty\subset(0,1). $ Then $$\prod(1-a_j)=0\iff \sum a_j=\infty $$

Applying this to our case gives that $\sum P(A_n)=\infty$.$$$$ Now, the 2nd Borel-Cantelli lemma states:

If $\sum P(A_n)=\infty$ and the events $\{A_n\}$ are independent, then $$P(\limsup A_n)=1$$

And this implies our desired result:$$P(\bigcap\limits_{m=1}^\infty\bigcup\limits_{n=m}^\infty A_n)=P(\limsup A_n)=1$$