I am trying to show that if the limit of $Z_n$ converges to $Z$, where $Z_n$ is defined to be a general sequence of events, that the $\limsup Z_n$ and $\liminf Z_n$ must equal to $Z$ as well. My approach is the following, first define as follows:
\begin{align*} \limsup_{n\in\mathbb{N}} Z_n &= \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} Z_j\right)\\\ \liminf_{n\in\mathbb{N}} Z_n &= \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} Z_j\right). \end{align*}
Then show that $\liminf Z_n \subset \limsup Z_n$, which is trivial and holds under any case. Then, I would like to show that $\limsup Z_n \subset \liminf Z_n$. However, I am having trouble here as I am not sure how to use the fact that the limit converges. From the definition, the convergence means that for each $\omega \in \Omega$, there will exist an $N$ such that it is either the case that $\omega \in Z_n$ for $N\geq n$ or $\omega \notin Z_n$ for $N\geq n$. It seems to me that the right path would involve using the fact that the $Z_n$ eventually reach $Z$ and use that to change something in the limsup and liminf definitions. Would anyone have any idea if this approach is in the right direction?
Hint. In order to show that $\limsup Z_n\subseteq \liminf Z_n$ you assume that $\omega\in\limsup Z_n$ for some $\omega$ and seek to show that this same $\omega$ is also in $\liminf Z_n$.
Because you're also assuming that your definition of convergence is true about the $Z_n$s, you also know that there is an $N$ such that $\omega$ is either in all $Z_n$ for $n\ge N$ or in none of them (note that the inequalities in the definition you're quoting are the wrong way around). Which of these can be the case if $\omega\in\limsup Z_n$?