$A$ with $\mu_G(A)=a$. Our task is to show that $f$ is non-negative on the interval $[t, 1-\mu_G(B)+t]$.

If $\mu_G(A) = 1-\mu_G(B)+t$, then by the previous discussion we have $1_A * 1_B(x) \geq t$ for all $x \in G$, and hence $c(A)=0$; hence $f(1-\mu_G(B)+t)=0$. At the other extreme, if $\mu_G(A) = t$, then $1_A * 1_B(x) \leq t$ for all $x \in G$, and hence from Fubini’s theorem we again have $c(A) = 0$.

Observe that if one modifies $A$ by a set of measure at most $\delta$, then $c(A)$ varies by $O(\delta)$. From this we conclude that $f$ is Lipschitz continuous. Thus, if we assume for contradiction that $f$ is not always non-negative; then there must exist a point $a$ in the interior of $[t, 1-\mu_G(B)+t]$ where $f$ attains a global negative minimum and is not locally constant in a neighbourhood of $a$. In particular, there exist arbitrarily small ${\varepsilon}$ such that $$\label{fae}
 f(a) < \frac{f(a-{\varepsilon}) + f(a+{\varepsilon})}{2}.$$

On the other hand, we observe the crucial submodularity property $$\label{submod}
 c(A_1) + c(A_2) \geq c(A_1 \cap A_2)+ c(A_1 \cup A_2)$$ for all measurable sets $A_1,A_2 \subset G$. To see this, we begin with the inclusion-exclusion identity $$1_{A_1} + 1_{A_2} = 1_{A_1 