While trying to prove the following statement:
$U\subseteq\mathbb{R}^n$ is an open set, $p\in U$ and $F:U\rightarrow\mathbb{R}^n$ a smooth vector field, Prove that $$\textrm{div}(F)(p)=\underset{r\to0}{\textrm{lim}}\frac{1}{\textrm{Vol}(B_{r}(p))}\intop_{\partial B_{r}(p)}\langle F,\hat{n}\rangle d\sigma$$
I was told that I could use some form of MVT for integrals of the form: For $f$ a scalar function which is continuous on $B_{r}(p)$ There exists $q\in B_{r}(p)$ such that the following holds $$f(c)=\frac{1}{\textrm{Vol}(B_{r}(p))}\intop_{B_{r}(p)}f(x) dx$$
I have been trying to find proof for this and couldn't, I tried proving it myself and am not too sure where to even start. Is this even correct?
EDIT Assuming this is true, using Divergence theorem I'm able to prove the statement but the question remains. $$\underset{r\to0}{\textrm{lim}}\frac{1}{\textrm{Vol}(B_{r}(p))}\intop_{\partial B_{r}(p)}\langle F,\hat{n}\rangle d\sigma = \underset{r\to0}{\textrm{lim}}\frac{1}{\textrm{Vol}(B_{r}(p))}\intop_{B_{r}(p)}\mathrm{div}\left(F\right)dx = \underset{r\to0}{\textrm{lim}}\frac{1}{\textrm{Vol}(B_{r}(p))}\intop_{B_{r}(p)}\sum_{i=1}^{n}\frac{\partial F_{i}}{\partial x_{i}}\left(x\right)dx = \sum_{i=1}^{n}\underset{r\to0}{\textrm{lim}}\frac{\intop_{B_{r}(p)}\frac{\partial F_{i}}{\partial x_{i}}\left(x\right)dx}{\textrm{Vol}(B_{r}(p))} = \sum_{i=1}^{n}\underset{r\to0}{\textrm{lim}}\frac{\partial F_{i}}{\partial x_{i}}\left(c\left(i,r\right)\right) = \sum_{i=1}^{n}\frac{\partial F_{i}}{\partial x_{i}}\left(\underset{r\to0}{\textrm{lim}}c\left(i,r\right)\right)=\sum_{i=1}^{n}\frac{\partial F_{i}}{\partial x_{i}}\left(p\right)=\textrm{div}(F)(p)$$
Suppose not, i.e. there does not exist any $c\in B_r(p)$ such that $f(c)=f_{avg}$, where I denote $f_{avg}$ as the average integral of $f$ over $B_r(p)$.
We must have that either $f>f_{avg}$ or $f<f_{avg}$ on the whole of $B_r(p)$. This is due to assuming there are $x,y\in B_r(p)$ with $f(x)<f_{avg}<f(y)$, then restricting $f$ onto the line segment connecting $x,y$ contradicts intermediate value theorem.
Of course $f>f_{avg}$ or the reverse case are impossible.