Rudin, Definition of lower/upper semicontinous, understanding the consequences. In Rudin (real and complex analysis) page 37 there's the following definition

Let $f$ be a real (or extended-real) function on a topological space.
  If $$ \left\{x \;:\; f(x) > \alpha \right\} $$ is open for every real
  $\alpha$, $f$ is said to be lower semicontinuous. If $$ \left\{x \;:\;
> f(x) < \alpha \right\} $$ is open for every real $\alpha$, $f$ is said
  to be upper semicontinuous.

Right after this definition is stated that the following property is an almost immediate consequence of such definition

The supremum of any collection of lower semicontinous functions is
  lower semicontinuous. The infimum of any collection of upper
  semicontinuous functions is upper semicontinuous.

Why is this true?
Let's take a family $\left\{f_j(x)\right\}_{j \in J}$ of lower semicontinuous functions, let
$$
f(x) = \sup_{j} f_j(x)
$$
let also $\alpha$ be a real number, why is the set
$$
\left\{x \;:\; f(x) > \alpha \right\}
$$
open?
 A: You can consider the set $\{x : f_j(x) > \alpha\}$ to be the pre-image of an open interval: $f_j^{-1}[(\alpha,\infty)]$, which is open by semicontinuity, and the union of open sets is open. Furthermore, we have
$$\bigcup_{j\in J} f^{-1}_j [(\alpha,\infty)] = f^{-1}[(\alpha,\infty)].$$
To show this, first let $x \in f^{-1}[(\alpha,\infty)]$. We know $f(x) > \alpha$, but we need to show that $\alpha < f_{j_0}(x) \le f(x)$ for some $j_0 \in J$. Since $f(x) = \sup_j f_j(x)$, the $f_j(x)$ must approach $f(x)$ to within any $\varepsilon > 0$. In particular, take $\varepsilon < f(x) - \alpha$, then for some $j_0 \in J$,
$$f(x) - f_{j_0}(x) < \varepsilon < f(x) - \alpha$$
which implies $\alpha < f_{j_0}(x)$ as required after subracting $f(x)$ on both ends and reversing signs.
For the reverse, take $x \in \bigcup_j f_j^{-1}[(\alpha,\infty)]$. Then $x$ is in any particular $j_0 \in J$, which means $f_{j_0}(x) > \alpha$. Now since $f(x) = \sup_j f_j(x)$, we have  $f(x) \ge f_{j_0}(x) > \alpha$. Thus, $x \in f^{-1}[(\alpha,\infty)]$.
Therefore, $\{x : f(x) > \alpha\}$ is open since it is a union of open sets.
