The content of First Borel-Cantelli lemma in my textbook is:
Let $(\Omega, \mathcal{F} , \mathbb{P})$ be a probability space and {$A_n$} an infinite sequence of events, $A_n \subseteq \Omega$, such that $\Sigma_{n=1}^\infty P(A_n)< \infty$. Define
$$A = \{ \omega: \text{there is an infinite sequence } n_i(\omega) \text{ such that } \omega \in A_{n_i}, i=1,2,… \}$$ Then $P(A)=0$.
I can understand the notation that $A = \bigcap\limits_{k=1}^{\infty} \bigcup\limits_{n=k}^{\infty}A_n$.
But where is the connection between this lemma and converge almost surely? For example, someone claims that:
$X_n/c_n\to 0$ a.s.
is equivalent to saying $\mathsf{P}(|X_n/c_n|>\epsilon_n\text{
i.o.})=0$ where $\{\epsilon_n\}$ is a positive sequence converging to $0$.
Is above statement true? How to see?
Best Answer
Note that for an event $A$, $\mathbb{P}(A) = 0$ if and only if $\mathbb{P}(A^c) = 1$. In your case, the event $A$ is "$|X_n/c_n|>\varepsilon_n\text{ i.o.}$", which, as you have noted, in set notation is $$ \bigcap_{k=1}^\infty\bigcup_{n=k}^\infty \{ |X_n/c_n|>\varepsilon_n \}. $$ We are asserting that $\mathbb{P}(A) = 0$. Again, this is equivalent to $\mathbb{P}(A^c) = 1$ (i.e., $A^c$ occurs almost surely). What is $A^c$ in your case?
(Hint: Note that $\omega\in \bigcap_{k=1}^\infty\bigcup_{n=k}^\infty A_n$ if and only if for all $k\in\mathbb{N}$, there is $n\geq k$ such that $\omega\in A_n$.)