Closure of a function 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.
 A: 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 $\inf S\leq \inf T$.
A: I am also reading his book Convex Analysis recently. 
According to my understanding, 
$$\text{epi}~(\text{cl}~f) = \text{cl}~(\text{epi}~f) \Rightarrow \{(x,\mu) \mid \mu \ge  (\text{cl}~f)(x)\} = \text{cl}~\{(x,\mu) \mid \mu \ge f(x)\},$$
Then i get the following intersection
$$\bigcap_{\mu > \alpha}\{(x,\mu) \mid \mu \ge  (\text{cl}~f)(x)\} = \bigcap_{\mu > \alpha}\text{cl}~\{(x,\mu) \mid \mu \ge f(x)\},$$
Thus
$$\{(x,\mu) \mid \alpha \ge (\text{cl}~f)(x)\} = \bigcap_{\mu > \alpha}\text{cl}~\{(x,\mu) \mid \mu \ge f(x)\},
$$
Then just need to project $(x,\mu)$ to $x$.
I am not sure if I am doing this problem correctly.
