Integral of a function with compact support I am reading a book with the following statement 
Since $g$ has compact support $$\sum_{i=1}^d \int_{\mathbb{R}^d} \frac{\partial (fF_i g)}{x_i} dx = 0$$
where 


*

*$\frac{dx}{dt} = F(x)$ with $x\in \mathbb{R}^d$  

*$g: \mathbb{R}^d \rightarrow \mathbb{R}$ is continuously differentiable with compact support 

*$f: \mathbb{R}^d \rightarrow \mathbb{R}$
The compact support means that if it is zero outside of a compact set.
How can we say the sum of integral is zero?  
 A: The key of the reasoning is that, defined
$$
V(x)=\left(
\begin{matrix}
fF_1 g\\
fF_2 g\\
\vdots\\
fF_{d-1} g\\
fF_d g
\end{matrix}
\right) \implies V\text{ has compact support in }\mathbb{R}^d
$$
Since
$$
\sum_{i=1}^d \int\limits_{\mathbb{R}^d} \frac{\partial (fF_i g)}{\partial x_i} \mathrm{d}x =  \int\limits_{\mathbb{R}^d} \nabla\cdot V\, \mathrm{d}x,
$$
then, by considering a ball $B(0,r)\in\mathbb{R}^d$ of radius $r>0$, we have that
$$
\begin{split}
\sum_{i=1}^d \int\limits_{\mathbb{R}^d} \frac{\partial (fF_i g)}{\partial x_i} \mathrm{d}x & =  \int\limits_{\mathbb{R}^d} \nabla\cdot V\, \mathrm{d}x\\
&\triangleq\lim_{r\to\infty}\int\limits_{B(0,r)} \nabla\cdot V\, \mathrm{d}x\\
&\text{ by Gauss-Green}\\
&=\lim_{r\to\infty}\int\limits_{\partial B(0,r)} V\cdot\nu_x\, \mathrm{d}\sigma_x=0
\end{split}
$$
where $\nu_x$ is the normal unit vector to $\partial B(0,r)$ in the point $x\in \partial B(0,r)$, since $V$ has compact support $\iff$ $V\cdot\nu_x=0$ for all $\nu_x$ and all $r$ larger than a fixed finite value $r_0>0$.
