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Let $\Omega\subset \Bbb R^d$ be open. The space $BV(\Omega)$ consists in functions $u\in L^1(\Omega)$ with bounded variation, i.e. $|u|_{BV(\Omega) }<\infty$ where \begin{align}\label{eq:bounded-variation} |u|_{BV(\Omega) }:= \sup\Big\lbrace \int_{\Omega} u(x) \operatorname{div} \phi(x) d x:~\phi \in C_c^\infty(\Omega, \mathbb{R}^d),~\|\phi\|_{L^\infty(\Omega)}\leq 1\Big\rbrace. \end{align}

The vector $\nabla u =(\Lambda_1, \Lambda_2,\cdots, \Lambda_d)$ can be seen as a vector valued Radon measure on $\Omega$ such that \begin{align*} \int_\Omega u(x)\frac{\partial \varphi}{\partial x_i}(x) d x= -\int_\Omega \varphi(x)d \Lambda_i(x), \quad \text{for all }\quad \varphi\in C_c^\infty(\Omega),~~i=1,\cdots, d. \end{align*} In particular if $u\in W^{1,1}(\Omega)$ then $u\in BV(\Omega)$, $\partial_{x_i} u(x)d x= d\Lambda_i(x)$ and $ |u|_{BV(\Omega)}\asymp\|\nabla u\|_{L^1(\Omega)}.$

Let us recall that $\Omega \subset \mathbb{R}^d$ is called an $W^{1,p}$-extension (resp. $BV$-extension) domain if there exists a linear operator $E:W^{1,p}(\Omega)\to W^{1,p}(\mathbb{R}^d)$ (resp. $E: BV(\Omega)\to BV(\mathbb{R}^d)$) and a constant $C: = C(\Omega, d)$ such that \begin{align*} Eu\mid_{\Omega} &= u \quad\hbox{and} \quad \|Eu\|_{W^{1,p}(\mathbb{R}^d)}\leq C \|u\|_{W^{1,p}(\Omega)} \quad\text{for all}\quad u \in W^{1,p}(\Omega)\\ (\text{resp.}\quad Eu\mid_{\Omega}& = u \quad\hbox{and} \quad \|Eu\|_{BV(\mathbb{R}^d)}\leq C \|u\|_{BV(\Omega)} \quad\, \text{for all}\quad u \in BV(\Omega)). \end{align*}

Question:

1-If $\Omega$ is an $BV$-extension domain do we have $|EU|_{BV(\Bbb R^d)}(\partial\Omega)=0?$

2-If $\Omega$ is an $W^{1,p}$-extension domain do we have $|\nabla EU|_{W^{1,p}(\Bbb R^d)}(\partial\Omega)=0?$

Intuitively, the second question is more likely to hold if we assume in addition that the boundary $\partial \Omega$ has zero Lebesgue measure. Since $\nabla Eu$ is still a function when $Eu\in W^{1,p}(\Bbb R^d)$.

Or, does the fact that $\Omega$ is an extension domain implies that $|\partial\Omega|=0$?

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    $\begingroup$ To your last question: A Sobolev extension domain necessarily satisfies the measure density condition $|B(x,r) \cap \Omega| \geq c r^d$ from which $|\partial \Omega| = 0$ follows, see Sobolev embeddings, extensions and measure density condition by Hajlasz, Koskela, Tuominen. $\endgroup$
    – Hannes
    Commented Aug 6, 2020 at 9:01
  • $\begingroup$ @Hannes Thank you for this reference. $\endgroup$
    – Guy Fsone
    Commented Aug 6, 2020 at 10:07
  • $\begingroup$ For a domain with nice enough boundary (e.g. Lipschitz), you can always construct a $BV$-extension by simply extending with zero, which then does not fulfill your condition since possibly all of the boundary is a jump set. But that is a very specific extension out of many. So a better question might be, if on a $BV$-extension domain there is an extension for which the boundary is not part of the jump set. (Which sounds like it could be true.) $\endgroup$
    – mlk
    Commented May 17, 2021 at 7:01

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