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$\{\omega\in\Omega : \limsup_{n\to\infty}\lvert Y_n(\omega)-Y(\omega)\rvert=0\}$ =$\liminf_{n\to\infty}\{{\omega\in\Omega:\lvert Y_n(\omega)-Y(\omega)\rvert\lt \epsilon\}}\forall\epsilon>0$.

How do we turn the LHS which is a lim sup on functions to the RHS which is a lim inf on sets.

johnson
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1 Answers1

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Remember $\limsup_{n\to\infty} \lvert x_n-x\rvert=0$ for $x_n,x\in\mathbb{R}$ means for all $\epsilon>0$, $\lvert x_n-x\rvert<\epsilon$ eventually.

So \begin{align*} &\{\omega\in\Omega:\limsup_{n\to\infty}\lvert Y_n(\omega)-Y(\omega)\rvert=0\}\\ &=\{\omega\in\Omega:(\forall\epsilon>0)(\lvert Y_n(\omega)-Y(\omega)\rvert<\epsilon\text{ eventually})\}\\ &=\bigcap_{\epsilon>0}\{\omega\in\Omega:\lvert Y_n(\omega)-Y(\omega)\rvert<\epsilon\text{ eventually}\}\\ &=\bigcap_{\epsilon>0}\liminf_{n\to\infty}\{\omega\in\Omega:\lvert Y_n(\omega)-Y(\omega)\rvert<\epsilon\}. \end{align*}

user10354138
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  • Is the set then equal to $\bigcap_{\epsilon>0}\bigcup_{n\ge 1}\bigcap_{k\ge n}{\omega\in\Omega:\lvert Y_n(\omega)-Y(\omega)\rvert<\epsilon}$ – johnson Jun 23 '19 at 15:32
  • yes, or equivalently, since $\mathbb{R}$ is Archimedean, $\bigcap_{m>0}\bigcup_{n\geq 1}\bigcap_{k\geq n}{\omega\in\Omega:\lvert Y_n(\omega)-Y(\omega)<\frac1m}$. – user10354138 Jun 23 '19 at 15:42
  • Could you have a look at my question here. https://math.stackexchange.com/questions/3271679/almost-sure-convergence-characterisations – johnson Jun 23 '19 at 15:45