It is well known that if $\Omega\subset\mathbb{R}^{N}$ open, $F\subset\Omega$ closed, such that $\mathcal{H}^{N−1}(F)=0$,where $\mathcal{H}^{N−1}$ denotes (N-1) dimensional Hausdorff measure, then $W^{1,p}(\Omega)=W^{1,p}(\Omega\backslash\Sigma)$ for $p\in (1,+\infty)$.


Question: Assume $F$ and $\Omega$ satisfy the conditions prescribed above. In this case, shall we have the following equality $W^{1,p}\_{0}(\Omega)=W^{1,p}\_{0}(\Omega\backslash\Sigma)$ ?


Thanks for any suggestion.



Removable set for Sobolev space
This is not true and a counterexample is based on the example in the comment of Math777.


Note that $W^{1,p}\_0(\Omega\setminus F)\subset W^{1,p}(\mathbb{R}^n)$, because $W^{1,p}\_0$ is the closure of $C\_0^\infty(\Omega\setminus F)\subset C\_0^\infty(\mathbb{R}^n)$.
If $p>n$, then Sobolev functions $W^{1,p}\_0(\Omega\setminus F)\subset W^{1,p}(\mathbb{R}^n)$ are continuous on $\mathbb{R}^n$. That means, every function in $W^{1,p}\_0(\Omega\setminus F)$, when extended by zero outside $\Omega\setminus F$ is continuous.


Let $\Omega=B^n(0,1)$ and let $F=\{0\}$. Then $W^{1,p}\_0(\Omega\setminus F)$ functions are continuous at $0$ and vanish at $0$, but functions in $W^{1,p}\_0(\Omega)$ need not vanish at $0$. 


