Can functions with "big" discontinuities be in $H^1$?

Question

How can I prove that the function:

$$u:\Omega\to\mathbb{R},\ u(x)=\begin{cases} 0, x\in\omega \\[3mm] v(x), x\in\Omega\setminus\omega\end{cases}$$ is not in $H^1(\Omega)$, knowing that $v\geq 1$ is any function from $L^{\infty}(\Omega)$, where $\Omega$ is an open, nonempty,connected and bounded subset of $\mathbb{R}^N,N\geq 2$ and $\omega\subset\Omega$ is any measurable set with $\omega$ and $\Omega\setminus\omega$ both having strict positive measure.

It is a kind of characteristic function, which I know that doesn' belong to $H^1(\Omega)$. But how to prove it rigurously? I feel it is ok because there is a gap between the branches of $u$ and then and a $H^1$ function is absolutely continuous on lines and somehow this will prove it, but I don't know how exactly...

Crossposted: https://math.stackexchange.com/questions/4838545/a-basic-question-about-h1-omega-sobolev-space? (no answer)

Answer 1

The function $u$ is not in $H^1$ (but you need $\Omega$ to be connected). Assume it is, then $u\wedge 1 =\chi_{\Omega \setminus \omega} \in H^1(\Omega)$ and its gradient is zero a.e. In fact, the gradient of an $H^1$ function is zero a.e. on every set where it is constant, so it is zero both on $\omega$ and $\Omega \setminus \omega$. But then $\chi_{\Omega \setminus \omega}$ would be a constant a.e. and this is not true since both $\omega$ and $\Omega \setminus \omega$ have positive measure.