How can one prove (or where can I find a proof) that if $u \in W^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^N$, then $u$ cannot have a $(N-1)$-manifold of discontinuity points? 



Sobolev functions on $\mathbb{R}^N$ cannot be discontinuous on a $(N-1)$-dimensional submanifold
Any function in $W^{1,p}$, $p>N$, has a continuous representative by the Sobolev embedding theorem so there is no issue here. However: 



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> **Proposition.** There is a function $f\in W^{1,p}$, $p\leq N$, that is essentially discontinuous everywhere. In fact you can find a function such that the essential supremum on every open set is $+\infty$ and the essential infimum on every open set is $-\infty$. 
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See Example 2.26 in [2].
For a similar construction see also: 


[A function in $W^{1,p}(\Omega)$ for $1 < p < n$ which is not differentiable a.e](https://mathoverflow.net/q/38264/121665).


Thus the answer to the question the way it is stated is **no**, the function can be essentially discontinuous everywhere.


There is however, a different point of view which shows that, in fact, a Sobolev function behaves nicely when restricted to an $(N-1)$-dimensional manifold and I will present two different approaches to it.


**Approach 1.**
According to Theorem 2 p. 164 in [1] (I am referring to the first edition)
any function $f\in W^{1,p}$ has a representative that is absolutely continuous on almost all lines. Here by a representative I mean a Borel function defined everywhere and equal to $f$ almost everywhere.


If $M\subset\Omega$ is an $(N-1)$-dimensional manifold, then almost all lines pass through almost all points on $M$ so we can define restriction of $f$ to $M$ by looking at values of $f$ at the points where the lines intersect with $M$. Such a restriction is called a trace. Therefore $f$ may be discontinuous on $M$, but still it behaves nicely on $M$.


**Approach 2.**
Any Function in $W^{1,p}$, $p>N$, has a continuous representative by the Sobolev embedding theorem so there is no issue here.If $f\in W^{1,N}$, then $f\in W^{1,p}\_{\rm loc}$ for ant $1\leq p<N$ so it suffices to discuss the case $1\leq p<N$ only.


The following result is Theorem 1 on p. 160 in [1].



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> **Theorem 1.** There is a representative of $f\in W^{1,p}(\Omega)$, $1\leq p<N$, $\Omega\subset\mathbb{R}^N$ that is $p$-quasicontinuous.
>  That means for any $\epsilon>0$, there is an open set $V\subset\Omega$ with $\operatorname{Cap}\_p(V)<\epsilon$
>  such that $f|\_{\Omega\setminus V}$ is continuous.
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Here $\operatorname{Cap}\_p$ stands for the $p$-capacity.


Capacity is a certain outer measure. While I will not recall its definition I will explain how it is related to the Hausdorff measure. The next result is Theorem 4 p. 156 and and Theorem 3 p. 193 in [1].



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> **Theorem 2.** If $1\leq p<N$ and $\operatorname{Cap}\_p(A)=0$, then $\mathcal{H}^s(A)=0$ for all $s>n-p$. Moreover if $A$ is compact, then $\operatorname{Cap}\_1(A)=0$ if and only if $\mathcal{H}^{N-1}(A)=0$.
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Now Theorem 1 says that away of a set of arbitrarily small capacity, $f$ is continuous so the exceptional set has capacity zero and Theorem 2 says that this exceptional set has vanishing $(N-1)$-dimensional measure. Therefore is we consider an $(N-1)$-dimensional manifold $M$ in $\Omega$ this exceptional set has measure zero. 


**[1] L. C. Evans, R. F. Gariepy,** *Measure theory and fine properties of functions.* Studies in Advanced Mathematics. CRC Press, Boca Raton, FL, 1992.


**[2] P. Hajłasz,** [Non-linear elliptic partial differential equations](http://www.pitt.edu/~hajlasz/Notatki/Corlona%20lectures.pdf) .


