[Ian Morris](https://mathoverflow.net/questions/242295/continuous-upper-envelope-of-upper-semicontinuous-function) quoted the following:



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> For any upper semi-continuous function $f \colon X \to [-\infty,+\infty)$ defined on a nonempty topological space $X$ there exists a nonempty set $\mathcal{F}\subset C(X,\mathbb{R})$ such that $f(x)=\inf\{g(x)\colon g \in \mathcal{F}\}$ for every $x \in X$. **If $X$ is metrisable then $\mathcal{F}$ may be taken to be countable.** 
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I am interested in the bold sentence: If $X$ is metrizable, then the class of continuous functions $\mathcal{F}$ can be countable. 


Question: If we define a function $f:X \rightarrow [\infty,+\infty)$ such that for any real number $c \in \mathbb{R},$ its pre-image $f^{-1}(-\infty,c)$ is a $F\_{\sigma}$ set, can we get a class $\mathcal{F}$ of real-valued upper semicontinuos functions on $X$ such that $f(x) = \inf \{ g(x):g \in \mathcal{F} \}$ and **$\mathcal{F}$ is countable**?



Does there exist a class of real-valued upper semicontinuos functions on $X$ such that $\mathcal{F}$ is countable?
Based on the comment by Nik Weaver, the answer to my question is negative, that is, there exists a function $f:X \rightarrow \mathbb{R}$ such that it is not an infimum of any upper-semicontinuous functions:


Define $f:X \rightarrow \mathbb{R}$ such that $f(x) = 0$ if $x \leq 0$ and $f(x) =1$ if $x>0.$ Clearly $f$ is lower semicontinuous everywhere and not continuous at $x=0.$ 


If $f$ is the infimum of a class of upper semicontinuous functions, then $f$ should be upper semicontinuous. A lower and upper semicontinuous function should be continuous, a contradiction. So $f$ is not the infimum of a class of upper semicontinuous functions.


