Let $g(x)= \sup_n f_n(x)$. Prove that $g^{-1}((a,\infty]) = \bigcup_{n=1}^\infty f_n^{-1}((a,\infty])$ I am doubtful of my solution. So, I wanted to make sure.
Let $g,f_n: X \rightarrow [-\infty, \infty], \forall n \in \mathbb{N}$ s.t. $g(x)= \sup_n f_n(x) \; \forall x \in X$. Prove that
$$
g^{-1}((a,\infty]) = \bigcup_{n=1}^\infty f_n^{-1}((a,\infty])
$$
where $a \in \mathbb{R}$.
Attempt: Assume $x_0 \in g^{-1}((a,\infty])$. Then,
$$
 g(x_0) \in (a,\infty] \\[5pt]
\Rightarrow \sup_{n}f_n(x_0) \in (a,\infty] \\[5pt]
$$
So, there must be some member in the sequence $(f_n(x_0))_{n=1}^{\infty} $ which belongs to $(a,\infty]$. Hence, $x_0 \in \bigcup_{n=1}^\infty f_n^{-1}((a,\infty])$.
If $x_0 \in \bigcup_{n=1}^\infty f_n^{-1}((a,\infty])$. Then, There are natural numbers $n_1,n_2,\dots$, for which $x_0 \in f_{n_k}^{-1}((a,\infty])\;  (k=1,2,\dots)$. This implies that $f_{n_k}(x_0) \in  (a,\infty] \Rightarrow \sup_{n}f_n(x_0) \in (a,\infty]$.
Thus, $x_0 \in g^{-1}((a,\infty])$
 A: Looks fine, but maybe be more specific in the following two steps: 


*

*So, there must be some member in the sequence $(f_n(x_0))_{n=1}^{\infty}$ which belongs to $(a,\infty]$. Hence,  $x_0 \in \cup_{n=1}^{\infty} f^{−1}_n ((a,\infty])$.


Use here that the lower end of the interval $(a, \infty]$ is open (Note that for a closed interval the claim were false)


*

*This implies that $f_{n_k}(x_o) \in(a, \infty] \Rightarrow \sup_n f_n (x_0) \in (a, \infty]$
Use the defining property of the supremum.
Also as another answer mentioned, we aren't guaranteed to have a sequence of $f_{n_k}$ with $x_0 \in f^{−1}_n ((a,\infty])$, but only at least one. That's sufficient though.
A: Your proof isn't quite right.  If 
$$x_0 \in \bigcup_{n=1}^\infty f_n^{-1}((a, \infty]),$$
then all you know is that $\exists n~x_0 \in f_n^{-1} ((a, \infty])$.  You don't know that there is a sequence of such indices.  But that's good enough, because it means $f_n(x_0) \gt a$, so $g(x_0)=\sup_n f_n(x_0) \geq f_n(x_0) \gt a$ and $x_0 \in g^{-1}((a, \infty])$.
A: Stick to the definitions:
Taking $x_0 \in g^{-1}[(a, \infty]]$ we know that $g(x_0) = \sup \{f_n(x_0): n \in \Bbb N\} > a$. This means that $a$ cannot be an upper bound for the set $\{f_n(x_0): n \in \Bbb N\}$ (as the sup in question is the smallest upper bound for that set by definition). This means there is some $m \in \Bbb N$ such that $f_m(x_0) > a$, or $x_0 \in (f_m)^{-1}(a, \infty]]$ and so $x_0$ is in the union $\cup_n f_n^{-1}[(a,\infty]]$.
If $x_0$ is in that union, there is some $m$ such that $x_0 \in f_m^{-1}[(a,\infty]$ and for that $m$, $f_m(x_0) > a$ which implies that $g(x_0) = \sup \{f_n(x_0): n \in \Bbb N\} \ge f_m(x_0) > a$ (here we only use that sup is an upper bound) and thus $x_0 \in g^{-1}[(a, \infty]]$, showing the reverse inclusion.
