the complement of $A \cup (x-B)$ would be an open null set and hence empty); hence $A+B=G$ and holds with equality. Define a *Bohr set* to be a subset of $G$ of the form $\phi^{-1}(I)$, where $\phi: G \to {\mathbb{R}}/{\mathbb{Z}}$ is a continuous surjective homomorphism and $I$ is a compact arc in ${\mathbb{R}}/{\mathbb{Z}}$ (i.e., a set of the form $I = [a,b] \text{ mod } {\mathbb{Z}}$ for some $a < b$, where $x \mapsto x \text{ mod } {\mathbb{Z}}$ is the projection from ${\mathbb{R}}$ to ${\mathbb{R}}/{\mathbb{Z}}$), and say that two Bohr sets $\phi^{-1}(I), \psi^{-1}(J)$ are *parallel* if $\phi=\psi$. If $A = \phi^{-1}(I)$ and $B = \phi^{-1}(J)$ are two parallel Bohr sets, then $A+B = \phi^{-1}(I+J)$ is also a Bohr set, and (by the uniqueness of Haar measure) the Haar measure of $A,B,A+B$ is equal to the measures of $I,J,I+J$ respectively on the unit circle. One can then easily verify that holds with equality in these cases.

The main result of this paper is an inverse theorem that asserts, roughly speaking, that the above examples are essentially the only situations in which equality can occur. More precisely, we have

\[inv-1\] Let ${\varepsilon}>0$, and suppose that $\delta>0$ is sufficiently small depending on 