# Convergence almost surely, cryptic sentence in wikipedia article

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.

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.

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.

• @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 {}). Jun 23, 2019 at 9:09
• 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. Jun 23, 2019 at 9:10
• Between the brackets () we have a $\liminf$ on sets here. Jun 23, 2019 at 9:15
• 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? Jun 23, 2019 at 9:38
• 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 $$P\left(\liminf\left\{ \left|X_{n}-X\right|<\epsilon\right\} \right)=1\iff P\left(\limsup\left\{ \left|X_{n}-X\right|\geq\epsilon\right\} \right)=0$$ Jun 23, 2019 at 9:57