Suppose $E$ is a compact metric space.

A function $f :E \rightarrow \mathbb{R}$ is upper semicontinous if for all $c \in \mathbb{R}$, $f^{-1}(-\infty, c)$ is open in $E.$

For any real-valued bounded function $f$ defined on $E$, we can define the upper regularization of $f$ as $$\hat{f} = \inf\{ g: g \text{ is upper semicontinuous on }E, g \geq f \}$$

In Kechris and Louveau paper, they define a sequence of functions using upper regularization:

$f_1 = \hat{f}$. For sucessor ordinal $\xi,$ if $f_{\xi}$ is defined, then $f_{\xi+1}=\widehat{\widehat{f_{\xi}-f}+f}.$ For limit ordinal $\xi,$ if $f_{\lambda}$ is defined for all $\lambda < \xi,$ then $f_{\xi} = \widehat{\sup_{\lambda < \xi} f_{\xi}}.$

In the same paper, the authors stated that the function $\hat{f}$ may not be defined.

Question: When will the function $f_{\xi}$ not defined?

EDIT: The authors prove the following proposition (page $220$, part of Proposition $2$ in the same paper:

If $f = u - v$ where $u,v$ are upper semicontinuous functions, then for all countable $\xi$, $f_{\xi}$ is defined.

When the authors prove the above proposition, they apply transfinite induction to prove that for all countable $\xi $, $u \geq f_{\xi}$. Hence, $f_{\xi}$ is defined.

From the proof above, I 'guess' that when $f_{\xi}$ is not defined, it should be the case that $f_{\xi}(x) = \infty$ for some $x \in E.$ However, I fail to construct such an example, even at $\xi=1.$