Let $X$ be a metric space and denote $f:X \rightarrow \mathbb{R}.$

It is easy to show that the following two statements are equivalent:

$(1)$ For any real number $c$, we have $f^{-1}(-\infty,c)$ is open in $X$.

$(2)$ For any $x \in X$ and any $\varepsilon > 0$, there exists open neighbourhood $U \ni x$ such that for all $y \in U,$ we have $f(x) + \varepsilon > f(y).$

If $f$ satisfies one of the above statements, then $f$ is said to be upper semicontinous.

If we change $(1)$ from open to $F_{\sigma}$, do we have a similar characterization which involves $\varepsilon$? To be more precise,

If for any $c \in \mathbb{R}$, we have $f^{-1}(-\infty,c)$ is $F_{\sigma}$, then can we obtain a conclusion which is similar to $(2)?$

The problem that I encounter here is that I do not know how to vissualize $F_{\sigma}$, hence do not know how to formulate an equivalent statement for it.