In the proof I am following of the divergence theorem, we define an integrable function $F : U\rightarrow \mathbb{R}$, which has compact support in some open and bounded subset $U \subseteq \mathbb{R}^n$. We need to show $\int_U \partial_jF(u) \textrm{d}\lambda^n = 0$ for $1 \le j \le n$. I came up with the following solution, but somehow it doesn't make sense intuitively. My solution:
$$ \int_{\mathbb{R}^n} \partial_jF(u) \textrm{d}\lambda^n = \int_{\mathbb{R}^{n-1}} \int_{\mathbb{R}} \partial_jF(u) \textrm{d}u_j \;\textrm{d}\lambda^{n-1} $$
Furthermore $$ \int_{\mathbb{R}} \partial_jF(u) \textrm{d}u_j = \lim_{b \rightarrow\infty} \lim_{a \rightarrow\infty}(F(u_1,\dots, b,\dots,u_n)- F(u_1,\dots, a,\dots,u_n)) = 0 - 0 = 0 $$ since $U$ is bounded. Since $\text{supp} (\partial_jF) \subseteq \text{supp} (F) \subseteq U$, we have $$ \int_{\mathbb{R}^n} \partial_jF(u) \textrm{d}\lambda^n = \int_{U} \partial_jF(u) \textrm{d}\lambda^n = 0 $$
However, even after having proven it (hopefully correctly) I still feel like I should be able to come up with a counter-example. Can anybody provide some intuition?