Aren't functions in Sobolev spaces defined only upto a set of measure zero? I see some people on math.SE arguing all the functions of the form $f:\Omega\to\mathbb{R}$, where $\Omega$ is a convex open subset of $\mathbb{R}^N$, they say if $f\in W^{1,N+1}$, then by default $f$ is continuous inside $\Omega$. I agree $f$ cannot have jump discontinuities, but what about isolated discontinuities? especially given that functions in Sobolev spaces are defined only upto a set of measure zero? I am really confused between these contradicting definitions and facts.

  • $\begingroup$ You are correct that they are only defined up to sets of measure zero but we typically brush this fact under the rug. When we talk about a function $f$ in a Sobolev space, we are really talking about the equivalence class $[f]$ so when we say that such a function is continuous, we mean that there is a function in the equivalence class which is continuous. $\endgroup$ – User8128 Jul 17 '17 at 12:33
  • $\begingroup$ @User8128 : I have a problem which is very sensitive to this brush off. I have to define a Sobolev space but in the course of the problem, I have to differentiate between functions even if dont agree on a set of measure zero. How can I name such a space. I have to define this space as jsut functions whose Sobolev norm is finite. I had asked a recent question of mine and came under pressure due to this fact and deleted that question. Can someone please undelete that question of mine. $\endgroup$ – Rajesh D Jul 17 '17 at 12:38
  • $\begingroup$ It is Here : math.stackexchange.com/q/2361235/2987 Can you ressurect this brush off by Gerw. $\endgroup$ – Rajesh D Jul 17 '17 at 12:39

Yes, Sobolev functions are in $L^p$ so they are not functions but equivalence classes $[f]$ of functions. However, if $p>N$ then Morrey's theorem says that there is a representative in the equivalence class $[f]$ of a Sobolev function which is Holder's continuous.

  • $\begingroup$ Can you please undelete a question of mine I had just deleted, if you have the power. It is very much relevant to this fact and I had to delete it under pressure. $\endgroup$ – Rajesh D Jul 17 '17 at 12:33
  • $\begingroup$ I don't have that power, sorry! $\endgroup$ – Gio67 Jul 17 '17 at 12:35
  • $\begingroup$ It is here : math.stackexchange.com/q/2361235/2987 $\endgroup$ – Rajesh D Jul 17 '17 at 12:39
  • $\begingroup$ Can you ressurect this brush off by Gerw $\endgroup$ – Rajesh D Jul 17 '17 at 12:40
  • $\begingroup$ I saw your post. If you have a function $f\in C^1(\overline{\Omega})$, then $f$ belongs to $W^{1,N+1}(\Omega)$, so you can consider its equivalence class $[f]$ and it is true that you can modify it to make it discontinuous, but why would you want to do that? If you have a nice function, you keep it as it is, you don't change the representative to make it worse. $\endgroup$ – Gio67 Jul 17 '17 at 12:45

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