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Let $\Omega\subseteq\mathbb{R}^N$ be an open and bounded set with Lipschitz boundary. Consider a sequence of open sets $\omega_n\subseteq\Omega,\ n\in\mathbb{N}^*$ such that there is a Lebesgue measurable set $\omega\subseteq\Omega$ with:

$$\chi_{\omega_n}\to \chi_{\omega}\ \text{in}\ L^1(\Omega).$$

Is it true that $\omega$ is an open set? (not just measurable), knowing that the perimeters of $\omega_n$ and $\omega$ are less than a constant $c>0$ for all $n\in\mathbb{N}^*$ (which means that $\chi_{\omega_n},\chi_{\omega}\in BV(\Omega)$)

It is known that $\omega$ can be chosen (up to a set of zero measure) to be a Borel set (see Delfour, Zolesio - Shapes and Geometries, page 221, theorem 3.3). From there we also know that my question has an afirmative answer in the case in which the boundary of $\omega$ has null Lebesgue measure.

It is also known that $BV(\Omega)$ is compactly imbedded in $L^1(\Omega)$ if $\Omega$ has a Lipschitz boundary.

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    $\begingroup$ It seems unlikely that $\omega$ should be an open set. What if e.g. you take $\omega$ to be a set with empty interior, but positive measure and $\omega_n = \{ x \in \Omega \mid \mathrm{dist}(x,\omega) < \frac{1}{n} \}$? I suspect a suitable choice for $\omega$ (so not dense for example) might lead to a counterexample. $\endgroup$
    – Leo Moos
    Commented Feb 20, 2021 at 12:59
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    $\begingroup$ Actually, you may be able to get a counterexample by just considering a fat Cantor set in $\mathbf{R}$. $\endgroup$
    – Leo Moos
    Commented Feb 20, 2021 at 13:10
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    $\begingroup$ Thanks a lot! I see. $\endgroup$
    – Bogdan
    Commented Feb 20, 2021 at 13:34
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    $\begingroup$ By outer regularity, $\omega$ can be any measurable set. $\endgroup$
    – Christian Remling
    Commented Feb 20, 2021 at 15:25
  • $\begingroup$ @LeoMoos: We can take $\omega_n\supseteq\omega$ in the approximation (that's what outer signifies in "outer regularity"). $\endgroup$
    – Christian Remling
    Commented Feb 20, 2021 at 19:32

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