How do we prove that $\sup_{n\geq 1}f_{n}$ is a measurable function when each term $f_{n}$ is measurable?

Question

Proposition

For each $n\in\mathbb{N}$, let $f_{n}:(\Omega,\mathcal{F})\to(\overline{\mathbb{R}},\mathcal{B}(\overline{\mathbb{R}}))$ be a $\langle\mathcal{F},\mathcal{B}(\overline{\mathbb{R}})\rangle$-measurable function. Then $\sup_{n\in\mathbb{N}}f_{n}$ is $\langle\mathcal{F},\mathcal{B}(\overline{\mathbb{R}}\rangle$-mesurable.

Proof

Let $g = \sup_{n\geq 1}f_{n}$. To show that $g$ is $\langle\mathcal{F},\mathcal{B}(\overline{\mathbb{R}}\rangle$-measurable, it is enough to show that $\{\omega\in\Omega : g(\omega)\leq r\}\in\mathcal{F}$ for all $r\in\mathbb{R}$.

Now, for any $r\in\mathbb{R}$, \begin{align*} \{\omega:g(\omega)\leq r\} & = \bigcap_{n=1}^{\infty}\{\omega:f_{n}(\omega)\leq r\}\\\\ & = \bigcap_{n=1}^{\infty}f^{-1}_{n}((-\infty,r)])\in\mathcal{F} \end{align*} since $f^{-1}_{n}((-\infty,r])\in\mathcal{F}$ for all $n\geq 1$, by the measurability of $f_{n}$.

My concerns

I do not know how to interpret the symbol $\sup_{n\in\mathbb{N}}f_{n}$.

As far as I have understood, for each $\omega\in\Omega$, $g(\omega) = \sup_{n\geq 1}f_{n}(\omega)$.

That is to say, for each $\omega\in\Omega$, $\sup_{n\geq 1}f_{n}(\omega)$ is the least upper bound of the sequence $f_{n}(\omega)$.

Is it correct to think so?

If it is not the case, please let me know.

Moreover, is there a more detailed way to write the proof? I've tried the following one.

Since $g(\omega)\geq f_{n}(\omega)$ for every natural $n$, one has that \begin{align*} x\in\{\omega:g(\omega)\leq r\} \Rightarrow g(x)\leq r & \Rightarrow (\forall n\in\mathbb{N})(f_{n}(x)\leq g(x) \leq r)\\\\ & \Rightarrow (\forall n\in\mathbb{N})(f_{n}(x)\leq r)\\\\ & \Rightarrow x\in\bigcap_{n=1}^{\infty}\{\omega:f_{n}(\omega)\leq r\} \end{align*}

Conversely, if $f_{n}(\omega)\leq r$ for every $n\in\mathbb{N}$, taking the sup one obtains that $g(\omega) = \sup f_{n}(\omega)\leq \sup r = r$.

This means that \begin{align*} x\in\bigcap_{n=1}^{\infty}\{\omega:f_{n}(\omega)\leq r\} \Rightarrow x\in\{\omega: g(\omega)\leq r\} \end{align*} Hence we conclude that both sets are equal.

Answer 1

You are on the right track. Just need to use the properties of the supremum to get

$$\{\sup_nf_n>a\}=\bigcup_n\{f_n>a\}$$

where $\{h>a\}:=\{\omega\in \Omega: h(\omega)>a\}$. If each $f_n$ is measurable, then each set $\{f_n>a\}$ is measurable and so the union of all of them.

Recall that a real valued function $g$ is (Borel) measurable iff $\{g>a\}$ is measurable.