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Let $W^{k,p}$ denote the sobolev space of all $k$-times weak-differentiable $L^p$ functions such that each derivative is $L^p$ and $W_0^{k,p}(\Omega)$ be the closure of $\mathcal C^\infty_c(\Omega)$ in this space.

Given $\Omega\subset\subset\mathbb R^N$, a Caccioppoli set $E\subset \Omega$, $U\subseteq \Omega$ open and $w\in W_0^{1,p}(\Omega)$. Can we conclude in the case of $$ \int_E div(\eta w) = 0 \qquad\forall\;\eta\in\mathcal C^1_c(U;\mathbb R^N) $$ that there exists a sequence $w_n\in\mathcal C^1_c(\Omega)$ with $\mathcal H^{N-1}(U\cap\partial^*E \cap [w_n\neq 0])=0$ which converges to $w$ in $W^{1,p}(\Omega)$? (Where $[w_n\neq 0]=\Omega\setminus w_n^{-1}(0)$ and $\mathcal H^{N-1}$ is the $N-1$-dimensional Hausdorff-measure).


The assertion seems naturally to me, because for $w\in\mathcal C_c^1(\Omega)$ the equality would per definition of a caccippoli set imply $$ \int_{\partial^*E} w\eta \nu_E = 0 \qquad\forall\;\eta\in\mathcal C^1_c(U,\mathbb R^N)$$ where $\nu_E$ is the inner normal of $E$ existing almost-everywhere on the reduced boundary. This would imply that $w=0$ $\mathcal H^{N-1}$-almost everywhere on the reduced boundary, wouldn't it? Just the same is by the divergence-theorem true if $\partial E$ is lipschitz and $w$ sobolev, because in that case the trace of $w$ on $\partial E$ is well defined, which means that there exists a sequence of $\mathcal C_c^1(\Omega)$ functions vanashing on $\partial E\cap U$ approximating $w$ in $W^{1,p}(\Omega)$.

But I didn't find informations about this special case, except in the case of $w\in W_0^{1,p}(\Omega)\cap L^\infty(\Omega)$ in which this seems also to be correct -- see Divergence-Measure Fields, Sets of Finite Perimeter, and Conservation Laws, by Gui-Qiang Chen, Monica Torres in Arch. Rational Mech. Anal. 175, 2005. They do actually not need $W^{1,p}$ but it works in that case...

Are at all the statements above for $w\in\mathcal C^1_c$ or $\partial E$ lipschitz correct? What is with the case of $w$ only in $W^{1,p}$ and $E$ only caccioppoli?

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Can you clarify what you mean by Caccioppoli set? –  timur Aug 30 '12 at 14:43
    
A set $E\subseteq\mathbb R^n$ is a Caccioppoli set iff its characteristic function $\mathbb 1_E$ has bounded variation, iff there is a vector-valued radon measure $D\mathbb1_E$ with $$\int_{\mathbb R^n}\mathbb1_Ediv\varphi d\mathcal L^n=-\int_{\mathbb R^n}\varphi d(D\mathbb 1_E)$$ for all $\mathbb R^n$ valued compactly-supported $\mathcal C^1$-functions $\varphi$, where $\mathcal L^n$ is the n-dimensional Lebesgue measure. In this case $D\mathbb 1_E$ is the distributional derivation of $\mathbb 1_E$. –  Elgrimm Aug 31 '12 at 19:03
    
Oh in the notation of the question $D\mathbb 1_E=\nu_E\cdot\mathcal H^{N-1}|_{\partial^*E}$, where $\mathcal H^{N-1}|_{\partial^*E}(A):=\mathcal H^{N-1}(\partial^*E\cap A)$ –  Elgrimm Sep 1 '12 at 11:03
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