I know there's already a question with a title very similar to this, unfortunately as I understand the OP skips over the part of the proof that is not clear to me.
Let $I$ be a set and $f_\alpha$, $\alpha \in I$ be a collection of lower semicontinuous functions. Show that $g=\sup\limits_{\alpha\in I}\,f_\alpha$ is lower semicontinuous.
My attempt: let $S_f(t)=\{x\in\mathbb{R}^n: f(x)>t\}$ for $t\in\mathbb{R}$. It is easy to see that $f$ is lower semicontinuous if and only if $S_f(t)$ is an open set, now define $$S(t)=\bigcup_\limits{\alpha\in I}S_{f_\alpha}(t).$$ Obviously $S(t)$ is open. I should now show that $S_{g}(t)=S(t)$, which ends the proof.
Now, obviously $S(t)\subset S_g(t)$ because if $f_\alpha(x)>t$ for some $\alpha\in I$ then $g(x)>t$ because $g(x)>f_\alpha(x)$ for all $\alpha\in I$, but I fail to see why $S_g(t)\subset S(t)$, I think I should use the definition of supremum but I don't see how. I'm thinking I should prove it for a particular $t\in \mathbb{R}$ because that seems very untrue for any $t$.
Thank you.
Best Answer
Fix any $x \in \Bbb{R}^n$. By definition of $\sup$ you have $$\sup_{\alpha \in I} f_{\alpha} (x) > t \ \ \Longleftrightarrow \ \ \exists\alpha \in I : f_{\alpha} (x) > t$$
Hence $$x \in S_g(t) \ \ \Longleftrightarrow \ \ g(x) > t \ \ \Longleftrightarrow \ \ \exists\alpha \in I : x \in S_{f_{\alpha}} (t) \ \ \Longleftrightarrow \ \ x \in \bigcup_{\alpha \in I} S_{f_{\alpha}} (t)$$