Show that $\{x \in X: \sup_{i} f_i(x) > r \}= \bigcup_{i} \{x \in X : f_i(x) >r \}.$ 
Show that $\{x \in X: \sup_{i} f_i(x) > r \}= \bigcup_{i} \{x \in X : f_i(x) >r \}.$

For $x \in \bigcup_{i} \{x \in X : f_i(x) >r \} \implies$ $f_i(x)>r$ for some $i$. Now by the definition of the supremum $\sup_i f_i(x) \ge f_i(x) > r$ so $x \in \{x \in X: \sup_{i} f_i(x) > r \} $.
The other side is hard. If $x \in \{x \in X: \sup_{i} f_i(x) > r \} $. Now this says that $\sup_i f_i(x) > r$. What I need know is that $\sup_i f_i(x) \ge f_i(x)$ for all $i$, but not neccessarily that $f_i(x) > r$. I don’t remember very well the properties of the supremum, but wasn’t it that I could somehow ”squeeze” something in between $\sup_i f_i(x) > r$?
 A: By the definition of supreme, if $t\in \{x \in X: \sup_{i} f_i(x) > r \}$ then for all $\epsilon>0$ there is a $j_{\epsilon}$ such that
$$
\sup_{i} f_{i}(t) \geq f_{j_{\epsilon}}(t)>\sup_{i}f_{i}(t)-\epsilon.
$$
If $\epsilon=\frac{1}{2}\left( \sup_{i} f_i(x) - r\right)>0$ then we have
\begin{align}
f_{j_{\epsilon}}(t)>& 
\sup_{i}f_{i}(t)-\epsilon
\\
=& 
\sup_{i}f_{i}(t)-\frac{1}{2}\left( \sup_{i} f_i(x) - r\right)
\\
=& 
\frac{1}{2}\sup_{i}f_{i}(t)+\frac{1}{2}r
\\
>& 
\frac{1}{2}r+\frac{1}{2}r
\\
=& 
r
\end{align}
Conclusion $f_{j_{\epsilon}}(t)>r$.
Then $t\in \{x \in X: f_{j_{\epsilon}}(x) > r \}$ and
$$
t\in \bigcup_{i} \{x \in X: f_i(x) > r \}
$$
A: Here, I assume $f_i$ maps $X$ either to $\Bbb{R}$ or $\Bbb{R}\cup \{\infty\}$ (and that in the former case $f_i(x)$ is bounded above for each $x$) so the neccessary suprema all exist. By definition,
$\sup_i f_i(x)$ is the least $s$ such that $s \ge f_n(x)$ for all $n$. Hence, if $x \in \{x \in X : \sup_i f_i(x) > r\}$, there is some $n$ such that $f_n(x) > r$ (otherwise $r \ge f_n(x)$ for all $n$, implying that $\sup_i f_i(x) \le r$). But then, for that $n$:
$$
x \in \{x \in X : f_n(x) > r\} \subseteq \bigcup_i\{x \in X : f_i(x) > r\}
$$
