I am working through Evan's Partial Differential Equations, and at one point (Page 20), he says, "In view of the Gauss-Green Theorem [divergence theorem], we have $$ \int_{V} \text{div } \mathbf{F} dx = \int_{\delta V} \mathbf{F} \cdot \mathbf{\nu} dS = 0 $$ and so $$ \text{div }\mathbf{F} = 0 \text{ in } U, $$ since $V$ was arbitrary."
Perhaps I am missing something, but why are we able to make this conclusion for $U$? My initial thoughts are that since $V$ is arbitrary, that means every integral of divergence of $F$ is 0, for every possible surface $V$ (location, shape, size, etc can all vary). The only way this can be true is if div $F = 0$ identically. I'm not sure if this is correct or how to put this in a more rigorous way though.
Edit: Here is the first part of this section in the textbook:
Laplace's equation comes up in a wide variety of physical contexts. In a typical interpretation $u$ denotes the density of some quantity (e.g. a chemical concentration) in equilibrium. Then if $V$ is any smooth subregion within $U$, the net flux of $u$ through $\delta V$ is zero: $$ \int_{\delta V} \mathbf{F} \cdot \mathbf{\nu} dS = 0, $$ $\mathbf{F}$ denoting the flux density and $\nu$ the unit outer normal field. In view of the Gauss-Green Theorem, we have...