Suprenum of set of lower semicontinuous functions is lower semicontinuous.

Question

From Rudin, Real and Complex Analysis We are given that a real valued function is lower semicontinuous if

{$x: f(x)> \alpha$} is open for every real $\alpha$.

I'm thinking of proving the former as follows:

Take any family of lower semicontinuous functions

{$f_a$} for some indexed $a$

Then we have that { $x: f_a (x)> \alpha$ } is open for all lower semicontinuous $f_a$.

Taking the sup of this set we know that it is still open.

Hence for $g=sup{f_a}$ we have that the set {x: g(x) > $\alpha$} must also be open...

I would like to know if there is a way to formalize this sketch?

I've seen other proofs of this statement but they seem to follow from a few more facts on lower semicontinuous functions, which are not immediately presented in this book.

Answer 1

I'm a bit confused what you're asking. It seems all you need to do is show that if $\{x : f_a(x) > \alpha\}$ is open for each $\alpha$, then $\{x : \sup_a f_a(x) > \alpha\}$ is open. But this is easy. If $\sup_a f_a(x) > \alpha$, then $f_a(x) > \alpha$ for some $a$ and so by assumption, there's some neighborhood around $x$ for which $f_a > \alpha$, and then on this neighborhood, $\sup_a f_a > \alpha$, as desired.

Answer 2

$\{x:\sup_a f_a(x)>\alpha \}=\cup_a \{x:f_a(x)>\alpha \}$ and union of open sets is open.