Green for some references. The author also thanks John Griesmer and the anonymous referees for further corrections and suggestions.

Proof of Theorem \[ruzsa-thm\] {#ruz}
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We now prove Theorem \[ruzsa-thm\]. By inner regularity of Haar measure and a limiting argument we may assume $A,B$ are compact. In the case $$\mu_G(A) + \mu_G(B) - t \geq 1,$$ we see that the set $A \cap (x-B) = \{ y \in A: x-y \in B \}$ has measure at least $\mu_G(A)+\mu_G(B)-1 \geq t$ for every $x \in G$, and hence $1_A * 1_B(x) \geq t$ for all $x \in G$, giving the claim in this case. Thus we may assume that $\mu_G(A)+\mu_G(B)-t < 1$. We may also assume that $G$ is non-trivial, which (by the connectedness of $G$) implies that there exist measurable subsets of $G$ of arbitrary measure between $0$ and $1$.

Fix $G$, let $B$ be a compact subset of $G$, and let $0 \leq t \leq \mu_G(B)$ be a real number. For any compact $A \subset G$, define the quantity $$c(A) \coloneqq \int_G \min(1_A * 1_B, t)\ d\mu_G - t (\mu_G(A) + \mu_G(B) -t),$$ and then for every $a \in [0,1]$, let $f(a)$ denote the infimum of $c(A)$ over all 