Let us say we have the indicator function $\chi_{\{|x|\leq 1\}}$ in $\mathbb{R}^2$.
How can I write out the weak derivative of this indicator function?
Is it $\delta_{|x|=1}$? Or it should be vector valued measure like $\bigg(\frac{\partial}{\partial x_1} \chi_{{|x|\leq 1}},\frac{\partial}{\partial x_2} \chi_{{|x|\leq 1}}\bigg)$, but now $\frac{\partial}{\partial x_1} \chi_{{|x|\leq 1}}$ is something that depends on the value of $x_2$.
The weak gradient of the characteristic function of a domain $\Omega$ with a smooth boundary is the vector-valued measure $\nu(x)d\sigma(x)$ where $\nu(x)$ is the outward unit normal at $x\in\partial\Omega$ and $d\sigma$ is the surface measure.
Why so? By the divergence theorem. Recall that by definition, the weak gradient satisfies $$ \int \nabla u\cdot \nabla \varphi = -\int u\operatorname{div}\varphi $$ for every compactly supported smooth vector field $\varphi$. With $u=\chi_\Omega$ and $\nabla u$ as above this is exactly the divergence theorem.