Suppose $K\subset\mathbb{R}^n$ has a smooth boundary, and let $\phi_s(x)$ be bump functions converging pointwise to the indicator of $K$, i.e. $$\underset{s\rightarrow\infty}{\lim}\phi_{s}(x)=\mathbf{1}_{K}(x)$$
I have read elsewhere on this site that "As $\phi_s$ approaches the indicator of $K$ its gradient converges to a distribution that is concentrated on $\partial K$ and models the inward normal $-n$ of $K$" and so:
$$\underset{s\rightarrow\infty}{\lim}\int_{\mathbb{R}^{n}}\nabla\phi_{s}(x)\cdot xdx=\int_{\partial K}-n\cdot xdS$$
I don't know much about distributions but I'm trying to figure out why the limit above makes sense. I've tried taking specific bump functions and checking what their gradient is, but haven't really figured it out. In general there aren't many cases where a volume integral becomes a surface-area integral. The divergence theorem comes to mind here but in second thought seems unrelated. I tried using Fubini's theorem but could only do it for specific bodies and functions, so I think I'm missing the general idea behind this. Any help would be appreciated.
We can write the foloowing: take a test function $\psi$ and integrate it against $\nabla \phi_s$:
$$\int_{\Bbb R^n}\nabla\phi_s(x)\psi(x)dx = \int_{supp\,\nabla\phi_s}\nabla\phi_s(x)\psi(x)dx=\int_{supp\,\nabla\phi_s}\nabla(\phi_s(x)\psi(x))dx-\int_{supp\,\nabla\phi_s} \phi_s(x)\nabla\psi(x)dx=\int_{\partial \{supp\,\nabla\phi_s\}} \phi_s(x)\psi(x)\cdot \vec n dx-\int_{supp\,\nabla\phi_s} \phi_s(x)\nabla\psi(x)dx.$$
In the above integrals $supp\,\nabla\phi_s$ is support of $\nabla\phi_s$, $\partial \{supp\,\nabla\phi_s\}$ is its boundary and $\vec n$ is the outward (with respect to $supp\,\nabla\phi_s$) normal to that boundary.
We can say that $\int_{supp\,\nabla\phi_s} \phi_s(x)\nabla\psi(x)dx$ converges to zero, since we integrate two bounded functions on a set with vanishing measure.
The normal $\vec n$ converges to the normal to the boundary $\partial K$, and given that $\phi_s\to0$ outside $K$ we can conclude that in the first integral only the "boundary of $supp\,\nabla\phi_s$ inside $K$" part of the integral remains, and the value of $\vec n$ there corresponds to the inner normal to $\partial K$.