We have a continious and differentiable vector field $ F:\ \mathbb{R}^d\rightarrow \mathbb{R}^d $, $x \in \mathbb{R}^d $ and $ (A_n)_{n\in \mathbb{N}} $ a sequence of compact connected sets with smooth boundary of $\mathbb{R}^d$ such, that
$\underline{\text{for all } \varepsilon>0 \text{ exists } N\in \mathbb{N} \text{ with }} $
$\underline{A_n\subseteq B_{\varepsilon}(x)=\{y\in \mathbb{R}^d:\ ||x-y||<\varepsilon\} \text{ for all } n\geq N }$.
I have to show with that the identity
$ div(F(x))=\lim\limits_{n\to \infty} \frac{1}{\lambda^d(A_n)}\cdot \int\limits_{\partial A_n} \langle F,\mu_n \rangle \ dS $
and $ \mu_n $ is the outer unit normal vector field of $ A_n $.
My attemp: I tried to show this estimation:
For all $ \alpha>0 $ exists $ N_{\alpha}\in \mathbb{N} $ such that $ \Bigg|\Bigg| \frac{1}{\lambda^d(A_n)}\cdot \int\limits_{\partial A_n} \langle F,\mu_n \rangle \ dS-div(F(x)) \Bigg|\Bigg|<\alpha $ for all $ n\geq N_{\alpha} $.
But I don't know how I can estimate the inequality useful to proof the claim. Especially I don't see how to use the fact from above (underlined part).
You need to invoke Stroke's theorem (or for the classical case that $n=3$, Gauss's divergence theorem). By Stroke's theorem, the RHS is the integral of divergence over $A_{n}$. Then by continuity of div $F$, we have approximation $\mbox{div}F(x)\approx\mbox{div}F(x_{0})$ when $x$ is close to $x_{0}$. Hence $\frac{1}{\mu(A_{n})}\int_{A_{n}}|\mbox{div}F(x)-\mbox{div}F(x_{0})|\,\,dx\rightarrow0$ as the set $A_{n}$ shrinks to the singleton $\{x_{0}\}$.