limsup of events, and events of limsup of a sequence of random variables

Question

Given a sequence of nonnegative random variables $\{Z_{n}\}$, for each positive integer $k$, is it true that $$ \left\{ \limsup_{n\to\infty}Z_{n} > \frac{1}{k}\right\} =\limsup _{n\to\infty}\left\{ Z_{n}>\frac{1}{k}\right\} $$

wiki: Let $Z_{n}=\left\vert Y_{n}-Y\right\vert $, since $\limsup_{n\to \infty}Z_{n}\geq0$, the sets $\{\limsup_{n\to\infty}Z_{n}=0\}$ and $\{\limsup_{n\to\infty}Z_{n}>0\}=\left\{ \limsup_{n\to\infty }Z_{n}\neq0\right\} $ form a partition of the sample space. We have $$ \left\{ \limsup_{n\to\infty}Z_{n}>0\right\} =\bigcup_{k=1}^{\infty }\left\{ \limsup_{n\to\infty}Z_{n}>\frac{1}{k}\right\} =\lim_{k\to\infty}\left\{ \limsup_{n\to\infty}Z_{n}>\frac {1}{k}\right\} $$ However, wiki gives $\lim_{k\to\infty}\limsup_{n\to\infty}\left\{ Z_{n}>\frac{1}{k}\right\} $.

Answer 1

We will show that $$ \mathsf{E}=\left\{ \limsup_{n\rightarrow\infty}Z_{n}>\frac{1}{k}\right\} \subset\limsup_{n\rightarrow\infty}\left\{ Z_{n}>\frac{1}{k}\right\} =\mathsf{F} $$

First, $w\in\mathsf{E}\implies w\in\mathsf{F}$. Let $T_{n}(w)=\sup_{m\geq n}Z_{m}(w)$ and $$ V(w)=\limsup_{n\rightarrow\infty}Z_{n}(w)=\lim_{n\rightarrow\infty}\sup_{m\geq n}Z_{m}(w)=\lim_{n\rightarrow\infty}T_{n}(w) $$ If $w\in\mathsf{E}$, then $\lim_{n\rightarrow\infty}T_{n}(w)=V(w)>\frac{1}{k} $. Since $T_{n}\geq T_{n+1}$, we have $T_{n}(w)>\frac{1}{k}$ for all $n$. We find that $w\in\bigcup_{m=n}^{\infty}\left\{ Z_{m}>\frac{1}{k}\right\} $ for all $n$. Otherwise, if $w\notin\bigcup_{m=n}^{\infty}\left\{ Z_{m}>\frac {1}{k}\right\} $ for some $n$, there must be $Z_{m}\leq\frac{1}{k}$ for all $m\geq n$, which results in a contradiction $\frac{1}{k}<T_{n}=\sup_{m\geq n}Z_{m}\leq\frac{1}{k}$. Thus, $w\in\mathsf{F}=\lim_{n\rightarrow\infty }\bigcup_{m=n}^{\infty}\left\{ Z_{m}>\frac{1}{k}\right\} $.

Second, there is $w\in\mathsf{F}$ but $w\notin\mathsf{E}$. For example, let $X$ be a random variable with $0<X\leq1$, and $Z_{n}=\frac{1}{k}+\frac{1}{n} X$, then $Z_{n}>\frac{1}{k}$ for all $n$, and $\limsup_{n\rightarrow\infty }Z_{n}=\frac{1}{k}$. Thus, if $w\in X^{-1}((0,1])$, $w\in\mathsf{F} =\limsup_{n\rightarrow\infty}\left\{ Z_{n}>\frac{1}{k}\right\} $ but $w\notin\mathsf{E}=\left\{ \frac{1}{k}>\frac{1}{k}\right\} =\emptyset$.

Answer 2

With "$>$", the inclusion only goes one way. A set of the form $\{\limsup_{n\to\infty} Z_n > a\}$ for $a > 0$ is contained in, but not necessarily equal to $\limsup_{n\to\infty}\{Z_n>a\}$. To see this, if $\omega\in\{\limsup_{n\to\infty}Z_n>a\}$, then $Z_n(\omega)>a$ for infinitely many $n$, which is precisely what it means for $\omega\in \limsup_{n\to\infty}\{Z_n>a\}$.

The reverse membership does not hold, since if $Z_n > a$ for infinitely many $n$, all we may conclude is that $\limsup Z_n \ge a$. The counterexample in the other answer shows we may have strict containment.