I am working through some revision notes, and I have come across this proof in my lecture notes that the class of $\mathcal{F}$-measurable functions is closed under limit operations:
Let $\{f_n\}$ be a sequence of $\mathcal{F}$-measurable functions on
$\Omega$. Recall our definitions of $\limsup$ and $\liminf$, and apply
them here pointwise to our sequence of functions $\{f_n\}$. That is,
$(\limsup_{n\to\infty} f_n)(x) = \sup_{m\geqslant n}\{f_m(x)\}$.Now let $g_n(x)=\sup_{m\geqslant n}\{f_m(x)\}$, so that $(g_n)$ is a
pointwise increasing sequence, and $g_n\uparrow g$, where
$g=\limsup_{n\to\infty} f_n$. Similarly define $(h_n)$ to be the
pointwise decreasing sequence where $h_n(x)=\inf_{m\geqslant n}\{f_m(x)\}$,
so that $h_n\downarrow h$, where $h = \liminf_{n\to\infty} f_n$.For any $a\in\mathbb{R}$ we have $$
\{g_n \leqslant a\} =
\bigcap_{m=n}^{\infty} \{f_m \leqslant a\} $$ which implies that $g_n$ is $\mathcal{F}$-measurable.
I don't understand where the conclusion in the last line comes from though, since nowhere have I seen any properties of countable intersections concerning measurable spaces.
Am I missing something very obvious, or is there maybe another line that I could add in to make the conclusion a bit more clear?
Edit: Since I am not too sure how commonplace this notation is, for $f\colon\Omega\to\mathbb{R}$, $\{f < a\}=\{\omega\in\Omega \colon f(\omega) < a\}$
Best Answer
Here $g_{n}\left(\omega\right)=\sup\left\{ f_{m}\left(\omega\right)\mid m\geq n\right\} $ for each $\omega\in\Omega$ right?
Then $g_{n}\left(\omega\right)\leq a$ if and only if $f_{m}\left(\omega\right)\leq a$ for each $m\geq n$.
You could also say that $\omega\in\left\{ g_{n}\leq a\right\} $ if and only $\omega\in\cap_{m\geq n}\left\{ f_{m}\leq a\right\} $.
This is true for every $\omega\in \Omega$ so that means exactly that the sets $\left\{ g_{n}\leq a\right\} $ and $\cap_{m\geq n}\left\{ f_{m}\leq a\right\} $ are the same.
It shows that set $\left\{ g_{n}\leq a\right\} $ is a countable intersection of measurable sets, and this implies that the set is measurable.
addendum:
Characteristic for a $\sigma$-algebra is that it is closed under complements and countable unions. Then it must also be closed under countable intersections. This because: $$\cap_{n=1}^{\infty}A_{n}=\left(\cup_{n=1}^{\infty}A_{n}^{c}\right)^{c}$$