of measure strictly between $0$ and $1$, which would of course yield a counterexample to .

In a blog post [@blog] of the author, it was observed that one could use an argument of Ruzsa [@ruzsa] to obtain the following stronger bound (cf. Pollard’s bound [@pollard] for cyclic groups):

\[ruzsa-thm\] Let $A,B$ be measurable subsets of a compact connected abelian group $G$. Then $$\int_G \min( 1_A * 1_B, t )\ d\mu_G \geq t \min( \mu_G(A) + \mu_G(B) -t, 1 )$$ for any $0 \leq t \leq \min(\mu_G(A), \mu_G(B))$, where $$1_A * 1_B(x) \coloneqq \int_G 1_A(y) 1_B(x-y)\ d\mu_G(y)$$ is the convolution of $1_A$ and $1_B$, and $1_A$ denotes the indicator function of $A$.

For the convenience of the reader, we give the proof of this theorem in Section \[ruz\]. To see why this result implies , we observe the following corollary of Theorem \[ruzsa-thm\]. Given two measurable subsets $A,B$ of $G$ and a parameter ${\varepsilon}>0$, we define the partial sumset $A +_{\varepsilon}B$ by the formula $$A +_{\varepsilon}B := \{ x \in G: 1_A * 1_B(x) \geq {\varepsilon}\}.$$ This is a compact subset of $A+B$.

\[kemp-cor\] Let $G, A, B$ be as in Theorem \[ruzsa-thm\]. Then for any $0 < {\varepsilon}< \min(\mu_G(A),\mu_G(B))^2$, 