[Math] Closure of a function

Question

Consider a function $f: \mathbb{R}^n \rightarrow \left\{-\infty, + \infty \right\}$. The epigraph of the function is the subset of $\mathbb{R}^{n+1}$
given by $\operatorname{epi}(f) = \left\{(x,\mu): \, f(x) \le \mu \right\}$. Given a set $F$ of $\mathbb{R}^{n+1}$, one may define a function
$\psi: \mathbb{R}^n \rightarrow \left\{-\infty, + \infty \right\}$, by
$\psi(x) = \inf \left\{ \mu: \, (x,\mu) \in F \right\}$. Now, the way i understand it, in his book Convex Analysis, at page 52 Rockafellar defines the
closure of $f$, to be the function $\psi$ corresponding to the
closure of $\operatorname{epi}(f)$. Let us denote this function by
$f_{cl}$. According to my understanding
\begin{align}
f_{cl}(x) = \inf \left\{ \mu: \,x \in \bigcap_{\alpha> \mu}
cl\left\{y: \, f(y) \le \alpha \right\} \right\}, \, \, \, (*)
\end{align} where
$cl\left\{y: \, f(y) \le \alpha \right\}$ denotes the closure of the set
$\left\{y: \, f(y) \le \alpha \right\}$. However, Rockafellar says towards the bottom of page 52 that
\begin{align}
f_{cl}(x) = \inf \left\{ \mu: \,x \in
cl\left\{y: \, f(y) \le \mu \right\} \right\}, \, \, \, (**).
\end{align} How do we see that the two values given in $(*)$ and $(**)$ coincide? One idea is to try and show that $\bigcap_{\alpha> \mu}
cl\left\{y: \, f(y) \le \alpha \right\} =cl\left\{y: \, f(y) \le \mu \right\}$. It is clear that the RHS is inside the LHS, but i have trouble proving the other inclusion.

Answer 1

As you have said, $$\bigcap_{\alpha> \mu} cl\left\{y: \, f(y) \le \alpha \right\} \supseteq cl\left\{y: \, f(y) \le \mu \right\},$$ and it follows that the right-hand side of $(**)$ is greater than or equal to the right-hand side of $(*)$ (since it is the infimum of a smaller set). So it suffices to show that the right-hand side of $(**)$ is less than or equal to the right-hand side of $(*)$. That is, it suffices to show that $$\inf S\leq\inf T$$ where $$S=\left\{ \mu: \,x \in cl\left\{y: \, f(y) \le \mu \right\} \right\}$$ and $$T=\left\{ \mu: \,x \in \bigcap_{\alpha> \mu} cl\left\{y: \, f(y) \le \alpha \right\} \right\}.$$

To prove this, suppose $\mu\in T$. Then for every $\alpha>\mu$, $x\in cl\left\{y: \, f(y) \le \alpha \right\}$. It follows that every $\alpha>\mu$ is an element of the set $S$. This implies $\inf S\leq \mu$. Since $\mu\in T$ was arbitrary, this means $\inf S$ is a lower bound for the set $T$, so $\inf S\leq \inf T$.