# Application of Divergence theorem to a non-smooth surface

Divergence theorem that I learnt in analysis:

Let $$U \rightarrow \mathbb R^n$$ be an open, bounded, regular set and let $$f : \bar{U} \rightarrow \mathbb R^n$$ be such that $$f$$ is bounded and continuous in $$U$$ and there exist the partial derivatives of $$f$$ in $$\mathbb R^n$$ at all $$x \in U$$ and they are continuous and bounded. Then $$\int_U \operatorname{Div} f(x) dx = \int_{\partial U} f(x) \cdot \nu d\mathcal H^{n-1},$$

where $$\nu$$ is normal vector.

If I go back to calc 3, I see many instances where I "exploit" divergence theorem on non-smooth surfaces (such as polygons bounded by the first octant and plane $$\{x+y+z = 1\}$$). How can I justify divergence theorem on such non-smooth surfaces?

Let $$V$$ be the non-smooth open set where divergence theorem is applied. Should I take a sequence of regular set $$\{U_n\}_{n \in \mathbb N}$$ with $$U_n \subset V$$ such that $$\mu(V - U_n) \rightarrow 0$$ and take apply divergence theorem on $$U_n$$. I am assuming the contribution of vector field on $$(V - U_n)$$ would be small with $$\mu(V - U_n) < \epsilon$$. However, I am bogged down by the possibility that something strange can happen near the sharp corner of the boundary of $$V$$. Can anyone justify how I can ignore the "sharp" corner points?