Borel Cantelli Lemma Implies Almost Sure Convergence

Question

I am struggling to see the exact nitty gritty details of how to prove the following theorem:

Borel Cantelli Implies Almost Sure Convergence

I am specifically getting caught up in the step of trying to show that:

\begin{equation*} \mathbb{P}(\limsup A_n(\epsilon)) =0 \qquad\Longrightarrow \qquad \mathbb{P}(\lim X_n =X) = 1 \end{equation*}

I can intuitively see the argument using the "infinitely often" definition of the limsup of the sequence of events ...but I would like to show it using $\epsilon$'s with the formal definition of convergence in limits + the definition of: \begin{equation*} \limsup A_n(\epsilon) = \bigcap_{n=1}^\infty \bigcup_{i=1}^n A_i(\epsilon) \end{equation*}

Thanks for any help in advanced!

Answer 1

Your definition of limsup is not correct: $$\limsup_n A_n(\epsilon)=\bigcap_{n=1}^{\infty}\bigcup_{i=n}^{\infty}A_i(\epsilon).$$

By definition of limit, $\omega\in\{\lim_n X_n=X\}$ if and only if for all $m\geq1$ there exists $n\geq1$ such that for every $i\geq n$ it holds $|X_i(\omega)-X(\omega)|\leq 1/m$ if and only if $$\omega\in \bigcap_{m=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{\infty} (A_i(1/m))^c=\left(\bigcup_{m=1}^{\infty}\limsup_nA_n(1/m)\right)^c.$$

Then $$P(\lim_n X_n=X)=1-P\left(\bigcup_{m=1}^{\infty}\limsup_nA_n(1/m)\right)\geq 1-\sum_{m=1}^{\infty}P(\limsup_nA_n(1/m))=1.$$