An application of Borel-Cantelli Lemma?

Question

I decided to dust off my measure theory notes and try some problems. I saw that I left this homework problem blank a few years ago.

Consider a probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Let $\{A_n\}_{n=1}^{\infty}$ be a sequence of independent events defined on $(\Omega, \mathcal{F}, \mathbb{P})$ such that $\mathbb{P}(A_n)<1$ for all $n$ and $\mathbb{P}(\bigcup_{n=1}^{\infty}A_n)=1.$ Find the value of $\mathbb{P}(\bigcap_{i=1}^{\infty}\bigcup_{n=i}^{\infty}A_n).$

I see the definition of limsup in $\mathbb{P}(\bigcap_{i=1}^{\infty}\bigcup_{n=i}^{\infty}A_n)$. I had a knee jerk reaction that this is $1$. Am I missing something in all of this?

Answer 1

Yes, it's $1$.

Since $\ A_i\ $ are independent, \begin{align} 0&=\mathbb{P}\left(\left(\bigcup_{i=1}^\infty A_i\right) ^c\right)\\ &=\mathbb{P}\left(\bigcap_{i=1}^\infty A_i^c\right)\\ &=\prod_{i=1}^\infty \mathbb{P}\left(A_i^c\right)\ . \end{align} But since $\ \mathbb{P}\left(A_n^c\right)=1-\mathbb{P}\left(A_n\right)>0\ $ for all $\ n\ $, it follows that \begin{align} \mathbb{P}\left(\bigcap_{n=i}^\infty A_n^c\right)&= \prod_{n=i}^\infty \mathbb{P}\left(A_n^c\right)\\ &=\frac{\prod_{n=1}^\infty \mathbb{P}\left(A_n^c\right)}{\prod_{n=1}^{i-1} \mathbb{P}\left(A_n^c\right)}\\ &=0 \end{align} for all $\ i\ $. Therefore \begin{align} \mathbb{P}\left(\left(\bigcap_{i=1}^\infty\bigcup_{n=i}^\infty A_i\right)\right)&= 1-\mathbb{P}\left(\left(\bigcap_{i=1}^\infty\bigcup_{n=i}^\infty A_n\right)^c\right)\\ &=1-\mathbb{P}\left(\bigcup_{i=1}^\infty\bigcap_{n=i}^\infty A_n^c\right)\\ &\ge1-\sum_{i=1}^\infty \mathbb{P}\left(\bigcap_{n=i}^\infty A_n^c\right)\\ &=1\ . \end{align}

Answer 2

Indeed, the answer is $1$. One can prove by induction on $n$ that $p_n:=\mathbb P\left(\bigcup_{i=n}^\infty A_i\right)=1$. For $n=1$ this is the assumption; if $p_n=1$, then by independence $$ 1=p_{n}=\mathbb P\left(A_n\cup\bigcup_{i=n+1}^\infty A_i\right)=\mathbb P(A_n)+p_{n+1}-\mathbb P(A_n)p_{n+1} $$ hence $\left(1-p_{n+1}\right)\left(1-\mathbb P(A_n)\right)=0$ and $\mathbb P(A_n)<1$ forces $p_{n+1}=1$.