\cap A_2} + 1_{A_1 \cup A_2}$$ which implies that $$1_{A_1} * 1_B + 1_{A_2} * 1_B = 1_{A_1 \cap A_2} * 1_B + 1_{A_1 \cup A_2} * 1_B.$$ Observe that for each $x \in G$, we have the pointwise inequalities $$1_{A_1 \cap A_2} * 1_B(x) \leq 1_{A_1} * 1_B(x), 1_{A_2} * 1_B(x) \leq 1_{A_1 \cup A_2} * 1_B(x);$$ by the concavity of the map $x \mapsto \min(x,t)$ we therefore have the pointwise bound $$\label{a12b}
 \min( 1_{A_1} * 1_B , t ) + \min( 1_{A_2} * 1_B, t) \geq \min( 1_{A_1 \cap A_2} * 1_B, t) + \min( 1_{A_1 \cup A_2} * 1_B, t).$$ Integrating over $G$ and using the inclusion-exclusion formula $\mu_G(A_1) + \mu_G(A_2) = \mu_G(A_1 \cap A_2) + \mu_G(A_1 \cup A_2)$, we obtain as desired.

Let $A$ be such that $\mu_G(A)=a$, and let ${\varepsilon}>0$ be a small quantity such that holds. Now we observe the following application of connectedness:

\[cont\] Let $A$ be a measurable subset of $G$, and let $t$ be any real number with $\mu_G(A)^2 \leq t \leq \mu_G(A)$. Then there exists $x \in G$ such that $\mu_G(A \cap (x+A)) = t$.

The function $x \mapsto 1_A * 1_{-A}(x) = \mu_G( A \cap (x+A))$, being a convolution 