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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.

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, 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.

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 exsists open neighbourhood $U \ni x$ such that for all $y \in U,$ we have $f(x) + \varepsilon > f(y).$

If $f$ satisfies one of hte 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 $\epsilon$? 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.

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.