# Generalize upper semicontinuous regularization using Borel Hierachy

Let $X$ be a metric space. Suppose a real-valued function $f:X\rightarrow \mathbb{R}$ is upper semicontinuous class $2$ if for all $c \in \mathbb{R},$ its preimage $f^{-1}(-\infty,c)$ is $F_{\sigma}$.

In this post, I asked whether we can approximate upper semicontinuous class $2$ by class of upper semicontinuous functions. The answer is negative.

Question: Suppose $f$ is upper semicontinuous class $2$. Does there exist a sequence $(f_n)_{n \in \mathbb{N}}$ of upper semicontinuous class $2$ such that $f(x) = \inf_{n \in \mathbb{N}}f_n(x)$ for all $x \in X?$ In other words, does there exists a strictly decreasing sequence $(f_n)_{n \in \mathbb{N}}$ such that it decreases pointwise to $f$?

UPDATE: As mentioned in comment, one can simply take $f_n(x) = f(x) + \frac{1}{n}$ for all $n \in \mathbb{N},$ and $f$ still remains upper semicontinuous class $2$.

Denote the class of upper semicontinuous class 2 as $SB(2).$

Question: For any bounded function $f$ on $X$, define regularization of $f$ as $\hat{f}=\inf\{ g: g \in SB(2),g \geq f \}$. Is $\hat{f} \in SB(2)?$

For any real number $c$, one has $(\hat{f})^{-1}(-\infty,c) = \cup_{g \in SB(2)}g^{-1}(-\infty,c)$. If $SB(2)$ is uncountable (which I believe it is), then one cannot conclude that $\hat{f} \in SB(2).$ However, this does not disprove that $f \in SB(2).$

• Can you provide some context? – Liviu Nicolaescu Apr 16 '17 at 10:59
• You haven't stated in the question that you want the $f_n$ also to be upper semi-continuous class 2. – Joel David Hamkins Apr 16 '17 at 11:05
• But now I wonder if that really is what you mean. Nothing seems to prevents me taking $f_n=f$. Do you mean that $f_n$ should be upper semicontinuous, or what? – Joel David Hamkins Apr 16 '17 at 11:18
• If we assume that $f_n$ is upper semicontinuous, then the answer is negative, as stated in the link in the post. If $f_n = f$, then $(f_n)$ is not strictly decreasing to $f$. – Idonknow Apr 16 '17 at 11:21
• Ah, I missed that detail. But can't you just take $f_n(x)=f(x)+\frac 1n$? This shifts the whole graph up by a constant, makes it strictly decreasing, and doesn't seem to change the upper semi-continuity class 2 part. – Joel David Hamkins Apr 16 '17 at 11:29