Proposition 15.3.2. Let $(X, \cm)$ be a measurable space, $\seq{f_n}$ be $(\cm, \cb_{\ol \real})$-measurable functions, then the following functions are $(\cm, \cb_{\ol \real})$-measurable:
$F = \sup_{n \in \natp}f_{n}$.
$f = \inf_{n \in \natp}f_{n}$.
$G = \limsup_{n \to \infty}f_{n}$.
$g = \limsup_{n \to \infty}f_{n}$.
$\limv{n}f_{n}$ (if it exists).
In addition, if the above functions are $\real$-valued, then they are $(\cm, \cb_{\real})$-measurable.
Proof. (1): Let $\alpha \in \real$, then for any $x \in X$, $F(x) > \alpha$ if and only if there exists $n \in \natp$ such that $f_{n}(x) > \alpha$. Thus
(2): Let $\alpha \in \real$, then
By Proposition 13.1.10 and Lemma 15.1.4, $F$ and $f$ are both $(\cm, \cb_{\ol \real})$-measurable.
(3): $\limsup_{n \to \infty}f_{n} = \inf_{n \in \natp}\sup_{k \ge n}f_{k}$.
(4): $\liminf_{n \to \infty}f_{n} = \sup_{n \in \natp}\inf_{k \ge n}f_{k}$.
(5): If $\limv{n}f_{n}$ exists, then it is equal to (3) and (4). In which case, it is measurable.
Finally, if the above functions are $\real$-valued, then they are $(\cm, \cb_{\real})$-measurable by Lemma 15.3.1.$\square$