I asked the following question in [maths stack exchange][1] but does not receive any response. 

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][2], 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?


  [1]: http://math.stackexchange.com/questions/2221806/when-will-the-upper-regularization-of-a-bounded-function-not-defined
  [2]: http://www.ams.org/journals/tran/1990-318-01/S0002-9947-1990-0946424-3/S0002-9947-1990-0946424-3.pdf