I'm attempting to prove that for a closed, path-connected, volume $V$ of $\mathbb{R}^3$,
$$\iiint_{V} (\operatorname{div} F) \, \text{d}V = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$
Where $F : \mathbb{R}^3 \to \mathbb{R}^3$ is any differentiable function.
Is there anything wrong with my proof? Can I make it simpler?
Consider a box $B$ containing $V$. Partition $B$ into nonoverlapping boxes $B_1, B_2, \ldots B_m$, such that diameter of the partition is $< 1/n$. Now consider "chunks" $C_k = B_k \cap V$. Clearly,
$$\sum_{k} \left (\iint_{\partial C_k} F \cdot \mathbf{\hat n} \, \text{d}S \right) = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$ Moreover, by my generalized mean value theorem, there exists some $p_k \in C_k$ such that
$$\iint_{\partial C_k} F \cdot \mathbf{\hat n} \, \text{d}S = \operatorname{div} F|_{p_k}\cdot {\rm m}(C_k)\ .$$
Pick all of these $p_k$ and use them to form a Riemman sum
$$S_n = \sum_k \operatorname{div} (F|_{p_k}) \cdot \operatorname{m}(C_k) = \sum_{k} \left (\iint_{\partial C_k} F \cdot \mathbf{\hat n} \, \text{d}S \right) = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$
As we assumed $\operatorname{div} F$ was integrable, $$\lim_{n \to \infty} S_n = \iiint_{V} (\operatorname{div} F) \, \text{d}V$$
But $S_n$ is a constant sequence! Namely,
$$S_n = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$
Thus, it is the only possibility that
$$\iiint_{V} (\operatorname{div} F) \, \text{d}V = \iint_{\partial V} F \cdot \mathbf{\hat n} \, \text{d}S$$