Lower semi continuous envelope is lower semi continuous

Question

Let X be a topological space and $F:X \rightarrow \overline{\mathbb{R}}$. The lower semi continuous envelope of $F$ is defined by $sc^-F(u)=\sup\{\phi(u)\;|\; \phi :X \rightarrow \overline{\mathbb{R}} \; \mathrm{is \; l.s.c. \; and} \; \phi \leq F\}$. Show that $sc^-F$ is lower semi continuous.

Here is my proof, but it seems a bit too easy, so it would be great if somebody could check if it is correct.

Let $u_k \rightarrow u$ in $X$. The set $M:=\{\phi(u)\;|\; \phi :X \rightarrow \overline{\mathbb{R}} \; \mathrm{is \; l.s.c. \; and} \; \phi \leq F\}$ is a family of l.s.c. functions, because every $\phi \in M$ is l.s.c. Then for every $\phi \in M$ we obtain: $\phi(u) \leq \liminf \; \phi(u_k) \\ \leq \liminf \; \sup \{\phi(u_k)\;|\; \phi :X \rightarrow \overline{\mathbb{R}} \; \mathrm{is \; l.s.c. \; and} \; \phi \leq F\} = \liminf \; sc^-F(u_k).$

Because this is true for every $\phi \in M$ it is also true for the supremum of all $\phi$ in $M$ (at this point I'm not really sure if it is correct) an we get

$\sup\{\phi(u)\;|\; \phi :X \rightarrow \overline{\mathbb{R}} \; \mathrm{is \; l.s.c. \; and} \; \phi \leq F\} = sc^-F(u) \leq \liminf \; sc^-F(u_k)$.

So the envelope of $F$ is l.s.c.

Answer 1

It is correct but the notation can be improved:

Then for every $\phi \in M$ we obtain: $\phi(u) \leq \liminf \; \phi(u_k) \\ \leq \liminf \; \sup \{\phi(u_k)\;|\; \phi :X \rightarrow \overline{\mathbb{R}} \; \mathrm{is \; l.s.c. \; and} \; \phi \leq F\} = \liminf \; sc^-F(u_k).$

Here I strongly suggest to choose different variables for $\phi$ inside the set (or outside). For example:

$ \liminf \phi(u_k) \leq \liminf \; \sup \{\psi(u_k)\;|\; \psi :X \rightarrow \overline{\mathbb{R}} \; \mathrm{is \; l.s.c. \; and} \; \psi \leq F\} = \liminf \; sc^-F(u_k).$