I would like to write this inequality:
$$\sup\left\{ \int_{A \cup (\mathbb{R}^d \setminus F)} \operatorname{div} T \space \rm d m^d : T \in C_0^1(\mathbb{R}^d,\mathbb{R}^d), |T(x)| \le 1 \right\} \le \\ \le \sup\left\{ \int_A \operatorname{div} T \space \rm d m^d : T \in C_0^1(\mathbb{R}^d,\mathbb{R}^d), |T(x)| \le 1 \right\} +\\ +\sup \left\{ \int_{\mathbb{R}^d \setminus F} \operatorname{div} T \space dm^d : T \in C_0^1(\mathbb{R}^d,\mathbb{R}^d),|T(x)| \le 1 \right\}$$
Where $\operatorname{div} T$ is the divergence of the field $T$ and $A,F \subset \mathbb{R}^d$. The problem is when the divergence is negative. But, in my opinion, when I consider the supremum, if $T$ is one candidate then also $-T$ is, and so I can consider field that have positive divergence, for the regularity of that field in which I take the supremum.
Following the comments by Alex M: write $$A= \sup\left\{ \int_A \operatorname{div} T \space \rm d m^d : T \in C_0^1(\mathbb{R}^d,\mathbb{R}^d), |T(x)| \le 1 \right\}$$ and $$B = \sup\left\{ \int_{\mathbb{R}^d \setminus F} \operatorname{div} T \space dm^d : T \in C_0^1(\mathbb{R}^d,\mathbb{R}^d),|T(x)| \le 1 \right\}$$ to shorten notation. Note that $A\ge 0$ and $B\ge 0$ because $T\equiv 0$ is one of eligible fields.
For every field $T \in C_0^1(\mathbb{R}^d,\mathbb{R}^d)$ such that $|T(x)| \le 1$ we have $$ \int_A \operatorname{div} T\, dm^d\le A\quad\text{and}\quad \int_{\mathbb{R}^d \setminus F} \operatorname{div} T \, dm^d \le B $$ hence $$ \int_{A\cup (\mathbb{R}^d \setminus F)} \operatorname{div} T\, dm^d\le A+ B $$ Since $A+B$ is an upper bound for such integrals, it follows that their supremum (the least upper bound) is $\le A+B$.