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 is that I do not know how to visualize $F_{\sigma}$, hence do not know how to formulate an equivalent statement for it. 

EDIT: In this [paper][1], the authors obtained the following theorem:

> For any $\varepsilon>0$, there exists positive functions $\delta$ on $X$ such that $|f(x) - f(y)| < \varepsilon$ whenever $d_X(x,y) < \min \{ \delta(x), \delta(y) \}$ if and only if $f$ is of the first Baire class. 

So it seems that we can also have a similar definition of the above function in terms of $\varepsilon.$ However, I couldn't obtain it. 


  [1]: http://www.ams.org/journals/proc/2001-129-08/S0002-9939-00-05826-3/S0002-9939-00-05826-3.pdf