# Borel Cantelli Lemma Implies Almost Sure Convergence

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!

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 $$P(\lim_n X_n=X)=1-P\left(\bigcup_{m=1}^{\infty}\limsup_nA_n(1/m)\right)\geq 1-\sum_{m=1}^{\infty}P(\limsup_nA_n(1/m))=1.$$
• Sorry but I'm stuck here: "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 $$\omega\in \bigcap_{m=1}^{\infty}\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{\infty} (A_i(1/m))^c$$" From my understanding, the $n$ you mentoned in "there exists" part, depends on both $\omega, m$. So calling this $n(\omega,m)$, we get: $\omega\in\{\lim_n X_n=X\}$ if and only if $$\omega\in \bigcap_{m=1}^{\infty}\bigcap_{i=n(\omega,m)}^{\infty} (A_i(1/m))^c$$. (contd.) Dec 3, 2019 at 9:25
• (contd. from previous paragraph) Denoting $\bigcup_{n=1}^{\infty}\bigcap_{i=n}^{\infty} (A_i(1/m))^c$ by $S_{1m}$, and $\bigcap_{i=n(\omega,m)}^{\infty} (A_i(1/m))^c$ by $S_{2m; \omega}$, we see that: $S_{1m} \supset S_{2m; \omega}$, and hence following your argument in probabilitis computation: $$P(\lim_n X_n=X)= P(\bigcap _{m=1}^{\infty} S_{2m; \omega}) = 1 - P(\bigcup _{m=1}^{\infty} S_{2m; \omega})^{C} < 1 - P(\bigcup _{m=1}^{\infty} S_{1m})^{C}= 1$$. Hence I can't see why the argument you gave is going through...thanks for correcting :) Dec 3, 2019 at 9:44
• (contd. from last paragraph) $P(\lim_n X_n=X)$ = $P(\bigcup_{\{\omega: \lim_n X_n(\omega)= X(\omega)\}} \bigcap _{m=1}^{\infty} S_{2m; \omega}) =$ $1 - P(\bigcup _{m=1}^{\infty} S_{2m; \omega})^{C})$ $\leq 1 - P(\bigcup _{m=1}^{\infty} S_{1m})^{C}= 1$. Hence I can't see why the argument you gave is going through...thanks for correcting :) Dec 3, 2019 at 10:09