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There are some nice connections between properties of sets and properties of their characteristic functions.

For instance:

  1. a set $C\subset \mathbb{R}$ is closed (resp. open) IFF the characteristic function $\mathbb{1}_C$ is upper semi-continuous (resp. lower semi-continuous).

  2. A set $E\subset \mathbb{R}^d$ has finite perimeter IFF $\mathbb{1}_E$ has bounded variation (in the distributional sense).

There are also interesting results in one direction, i.e. without an equivalence.

I would be interested in similar equivalences/implications for quasicontinuity or any related notion like cliquishness. The former harks back to Baire and Volterra and is defined as follows:

a function $f:\mathbb{R}\rightarrow \mathbb{R}$ is quasicontinuous at $x\in \mathbb{R}$ if for any $\epsilon>0$ and $\delta>0$ there is some interval $(a, b)\subset B(x, \delta)$ such that $(\forall y\in (a, b))(|f(x)-f(y)|<\epsilon)$.

Note that in this definition, we need not have that $x\in (a, b)$.

I welcome any (fairly abstract) notion/equivalence, also stemming from logic/set theory.

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  • $\begingroup$ Why do you need both $\epsilon$ and $\delta$? Isn't it equivalent to take $\epsilon=\delta$? Replace each with the minimum... $\endgroup$
    – Joel David Hamkins
    Commented Nov 12, 2022 at 20:30
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    $\begingroup$ $\chi_A$ quasi-continuous is (obviously) equivalent to both $A$ and $A^c$ having dense open subsets, and I'm not sure if there is a better way of saying this. $\endgroup$
    – Christian Remling
    Commented Nov 12, 2022 at 21:48
  • $\begingroup$ Along similar lines: a set is regular open iff its characteristic function is normal lower semicontinuous (meaning that $f(x) = \operatorname{lim.sup}_{y\to x} \operatorname{lim.inf}_{z\to y} f(z)$). $\endgroup$
    – Gro-Tsen
    Commented Nov 12, 2022 at 22:10
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    $\begingroup$ @ChristianRemling Or equivalently, $\operatorname{int}\overline A\subseteq A\subseteq\overline{\operatorname{int}A}$. $\endgroup$
    – Emil Jeřábek
    Commented Nov 12, 2022 at 22:47
  • $\begingroup$ @EmilJeřábek: This is actually a nicer way of stating the condition. $\endgroup$
    – Christian Remling
    Commented Nov 12, 2022 at 23:01

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