we have $$\mu_G(A +_{\varepsilon}B) \geq \min( \mu_G(A) + \mu_G(B), 1 ) - 2 \sqrt{{\varepsilon}}$$

One can improve the error term $2\sqrt{{\varepsilon}}$ slightly, but we will not need to do so here.

From the pointwise bound $$\min( 1_A * 1_B, \sqrt{{\varepsilon}} ) \leq {\varepsilon}+ \sqrt{{\varepsilon}} 1_{A +_{\varepsilon}B}$$ one has $$\int_G \min( 1_A * 1_B, \sqrt{{\varepsilon}} )\ d\mu_G \leq {\varepsilon}+ \sqrt{{\varepsilon}}  \mu_G( A +_{\varepsilon}B)$$ and hence by Theorem \[ruzsa-thm\], we have $$\mu_G(A +_{\varepsilon}B)  \geq \min( \mu_G(A) + \mu_G(B) - \sqrt{{\varepsilon}}, 1 )  - \sqrt{{\varepsilon}},$$ giving the claim.

Since the set $A +_{\varepsilon}B$ is contained in $A+B$, the claim follows from this corollary (in the case $\mu_G(A), \mu_G(B) > 0$) by sending ${\varepsilon}$ to $0$, noting that is trivial when $\mu_G(A)=0$ or $\mu_G(B)=0$.

There are several cases in which the estimate is sharp. Firstly, one has the trivial cases in which $A$ or $B$ is a point; there are some further examples of this type where (say) $A$ is a coset of a measure zero subgroup of $G$, and $B$ is a union of cosets of that group. Secondly, if one has $\mu_G(A) + \mu_G(B) \geq 1$, then the compact sets $A$ and $x-B$ cannot be disjoint (as this would disconnect $G$, since 