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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: What is an example $f:E \rightarrow \mathbb{R}$ such that $f_{\xi}$ is not defined for some ordinal $\xi$?

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 $\xi=1.$

UPDATE: According to the author, there is a gap in the proof in section $3$, Lemma $3$, case $3$ (successor case). If someone can fill in the gap, I would feel appreciated.

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  • $\begingroup$ Do you know examples where $\hat f$ is not defined? $\endgroup$ – Tommi Brander Apr 13 '17 at 6:56
  • $\begingroup$ For bounded function $f $, I do not know any. $\endgroup$ – Idonknow Apr 13 '17 at 7:10
  • $\begingroup$ You write, in the question, that the authors write that $\hat f$ might not be defined. Do they give examples or do they mean that it might not be defined for unbounded functions? $\endgroup$ – Tommi Brander Apr 13 '17 at 7:17
  • $\begingroup$ They do not give any example. They define the upper regularization for bounded function but yet, they say the upper regularization might not be defined. $\endgroup$ – Idonknow Apr 13 '17 at 11:26
  • $\begingroup$ "Question: When will the function $f_{\xi}$ not [be] defined ?" -- What kind of answer do you expect ? Do you want a classification of functions $f$ such that that $f_{\xi}$ isn't defined for some ordinal $\xi$ or do you want to see an explicit example of such a function $f$ ? $\endgroup$ – Todd Leason Apr 13 '17 at 14:26

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