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This article says https://en.wikipedia.org/wiki/Set-theoretic_limit#Almost_sure_convergence.

The event that a sequence of random variables $Y_1, Y_2, \dots$ converges to another random variable $Y$ is formally expressed as $\{{\limsup _{n\to \infty }|Y_{n}-Y|=0\}}$. It would be a mistake, however, to write this simply as a limsup of events. That is, this is not the event $ \limsup _{n\to \infty }\{|Y_{n}-Y|=0\}$ !

I was wondering if the second expression is simply incorrect notation because it did not have a pair of parentheses around the whole expression. Both expressions mean the same thing to me.

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Remember the event $$ \{\limsup_{n\to\infty}\lvert Y_n-Y\rvert=0\} $$ is $$ \{\omega\in\Omega : \limsup_{n\to\infty}\lvert Y_n(\omega)-Y(\omega)\rvert=0\}. $$ But the limsup of events $$ \limsup_{n\to\infty}\{\lvert Y_n-Y\rvert=0\} $$ is $$ \{\omega\in\Omega : \lvert Y_n(\omega)-Y(\omega)\rvert=0 \text{ infinitely often}\} $$ which is not the same. For example, for some $\omega\in\Omega$ we could have $Y_n(\omega)-Y(\omega)=n^{-1}$ and this $\omega$ would be in the first, but clearly $\lvert Y_n(\omega)-Y(\omega)\rvert$ is never zero so is not in the second.

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The first expression concerns a $\limsup$ of functions.

The second expression concerns a $\limsup$ of sets (which again is a set).


$$\omega\in\{\limsup_{n\to\infty}|Y_n-Y|=0\}\iff\limsup_{n\to\infty}|Y_n(\omega)-Y(\omega)|=0\iff\lim_{n\to\infty}Y_n(\omega)=Y(\omega)\tag1$$ and:$$\omega\in\limsup\{|Y_n-Y|=0\}\iff\{n\in\mathbb N|Y_n(\omega)-Y(\omega)|=0\}\text{ is infinite}\tag2$$

Note that $(1)$ does not imply $(2)$ and also $(2)$ does not imply $(1)$.

If e.g. $Y_n(\omega)=\frac1n$ and $Y(\omega)=0$ then $(1)$ is true and $(2)$ is not true.

If e.g. $Y_n(\omega)=0=Y(\omega)$ for $n$ odd, and $Y_n(\omega)=1$ for $n$ even then $(2)$ is true and $(1)$ is not true.

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    $\begingroup$ @johnson Yes $\{\limsup_{n\to\infty}|Y_n-Y|=0\}$ is a set all right. I am not denying that but only state that in this expression we deal with a limsup of functions (between the brackets {}). $\endgroup$
    – drhab
    Jun 23, 2019 at 9:09
  • $\begingroup$ I thought that the first expression is also a lim inf of sets. What about this expression from en.wikipedia.org/wiki/Convergence_of_random_variables $\operatorname {Pr} {\Big (}\liminf _{n\to \infty }{\big \{}\omega \in \Omega :|X_{n}(\omega )-X(\omega )|<\varepsilon {\big \}}{\Big )}=1\quad {\text{for all}}\quad \varepsilon >0. $ Is this a lim inf of functions or a lim inf of sets. $\endgroup$
    – johnson
    Jun 23, 2019 at 9:10
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    $\begingroup$ Between the brackets () we have a $\liminf$ on sets here. $\endgroup$
    – drhab
    Jun 23, 2019 at 9:15
  • $\begingroup$ But isn't the complement of that equal to this $\lim_{n\to\infty}\mathbb{P}[\omega:\sup_{k>n}|X_k(\omega) - X(\omega)|>\epsilon] = 0$ for some $\epsilon>0$ which is equal to the first expression. So a lim inf on sets becomes a lim sup on functions? $\endgroup$
    – johnson
    Jun 23, 2019 at 9:38
  • $\begingroup$ The complement of set $\liminf\left\{ \left|X_{n}-X\right|<\epsilon\right\} $ is set $\limsup\left\{ \left|X_{n}-X\right|\geq\epsilon\right\} $ so that $$P\left(\liminf\left\{ \left|X_{n}-X\right|<\epsilon\right\} \right)=1\iff P\left(\limsup\left\{ \left|X_{n}-X\right|\geq\epsilon\right\} \right)=0$$ $\endgroup$
    – drhab
    Jun 23, 2019 at 9:57

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