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 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?