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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!

User086688
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1 Answers1

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

user39756
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  • Yes you are correct, sorry for the typo. Thank you for that - that was perfect and exactly what I was looking for!!! =) – User086688 Jan 02 '17 at 23:28
  • 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.)

     – Mathguest Dec 03 '19 at 09:25
  • (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 :)

     – Mathguest Dec 03 '19 at 09:44
  • (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 :)

     – Mathguest Dec 03 '19 at 10:09