I need a proof for this theorem and have absolutely no idea how to do it.
Let $ U \subseteq \mathbb{R}^n $ be an open set, $ F : U \to \mathbb{R}^n $ a $C^1$ vector field. Let $ A_k \subseteq U, k \in \mathbb{N}$ be a series of compact, non-empty subsets with smooth border, which converges to $ x \in U $.
Then: $$ {\rm div}\, F(x) = \lim_{k \to \infty} \frac{1}{{\rm volume}(A_k)} \int_{\partial A_k} \langle F, \nu \rangle\, {\rm d}S$$
where $ \nu $ is a normal field at $ \partial A_k $.
Thanks in advance.
Let $ B(x,r) \subset U $ be the metric ball centred at $ x $ with radius $ r $.
The Lebesgue-Besicovith theorem asserts that
$$ \lim_{r \rightarrow 0} \frac{1}{vol(B(x,r)} \int_{B(x,r)} f(y)\,dy = f(x) \; \; \; (1) $$
for every $ f \in L^{1}_{loc}(U) $.
The Divergence theorem asserts that if $ F $ is a $ C^1 $ vector field on $ U $ it holds that:
$$ \int_{B(x,r)}div F \, dy = \int_{\partial B(x,r)}\nu \cdot F \, dS \; \; \; (2) $$
Then from (1) and (2) the proof of your statement readily follows in the case $ A_k = B(0,1/k) $. For general $ A_k $ you should approximate these sets exteriorly and interiorly by balls.