I am interested in the following statement:



> 
> Let $F$ be a vector field in $\mathbb{R}^n$ that is $C^1$-smooth in a
>  domain $U$, continuous up to the boundary $\partial U$, and vanishing on $\partial U$. Then $\int\_U div(F)=0$.
> 
> 
> 


Naively, if $U\_n$ is a smooth domain approximating $U$ from inside and $|F|\leq 1/n$ on $\partial U\_n$, then the regular divergence theorem gives $|\int\_{\partial U\_n} F\cdot N| \leq Area(\partial U\_n)/n$ which might very well diverge. So it looks like one needs a finite perimeter condition. However, is there a known example of a construction in a case of non-finite perimeter where the integral of the divergence is not zero?



A special case of the Divergence theorem
I will show that if $U$ is open and bounded, $f \in C^1 (U) \cap C(\bar{U})$, $f=0$ on $\partial U$ and $v \in \mathbb{R}^n$, then
$$
 \int\_U \partial\_v f = 0,
$$
where $\partial\_v f$ is the directional derivative of $f$ in the direction $v$.


The proof goes as follows. First, the statement can be proved in the one-dimensional case by writing the open set $U$ as a countable collection of intervals, applying the fundamental theorem on each interval.


The higher dimensional case follows from the one-dimensional case and Fubini's theorem.


If $U$ is unbounded, the statement still holds if the directional derivative $\partial\_v f$ is absolutely integrable: $\int\_U \lvert \partial\_v f\rvert < \infty$.


In fact it can be proved under these conditions the extension of $f$ to the whole space $\mathbb{R}^n$ by $0$ is weakly differentiable.


