Let $(X, {\cal E}, \mu)$ be a measure space and let $B,C \in {\cal E}$. Define $\mu_B:{\cal E} \to [0,\infty]$ by $\mu_B (E) = \mu(E\cap B)$. (I have already proved that this does indeed define a new measure).
Find a necessary and sufficient condition on sets $B,C \in \mathcal{E}$ such that $\mu_B + \mu_C \leq \mu$.
Any ideas?
Claim. For $\mu_{\mathrm{B}} + \mu_{\mathrm{C}} \leq \mu$ it is a necessary and sufficient condition that $\mu(\mathrm{B} \cap \mathrm{C}) = 0.$
Proof. If $\mu(\mathrm{B} \cap \mathrm{C}) = 0$ then $\mu_{\mathrm{B}} = \mu_{\mathrm{B} \setminus \mathrm{C}},$ and similarly interchanging the rôles of $\mathrm{B}$ and $\mathrm{C}.$ Hence, $\mu_{\mathrm{B}} + \mu_{\mathrm{C}} = \mu_{(\mathrm{B} \setminus \mathrm{C}) \cup (\mathrm{C} \setminus \mathrm{B})} \leq \mu_{\mathrm{X}} = \mu;$ the condition is therefore sufficient. It is also necessary for we can evaluate the inequality $\mu_{\mathrm{B}} + \mu_{\mathrm{C}} \leq \mu$ in $\mathrm{B} \cap \mathrm{C}$ to reach $0 \leq 2 \mu(\mathrm{B} \cap \mathrm{C}) \leq \mu(\mathrm{B} \cap \mathrm{C}),$ hence $\mu(\mathrm{B} \cap \mathrm{C}) = 0.$ Q.E.D.