\Delta \phi^{-1}(I_{n'}) ) \to 0$$ as $n,n' \to \infty$. By the Bolzano-Weierstrass theorem, we may thus find a compact arc $I$ independent of $n$ such that $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}( I_n \Delta I) \to 0$$ as $n \to \infty$, which implies that $$\mu_G( \phi^{-1}(I_n) \Delta \phi^{-1}(I) ) \to 0$$ as $n \to \infty$. Hence by the triangle inequality, $A$ and $\phi^{-1}(I)$ must agree $\mu_G$-almost everywhere; as $A$ is compact, it cannot omit any interior point of $\phi^{-1}(I)$ (as this would also exclude a set of positive $\mu_G$ measure from $A$, and hence $A$ must therefore consist of the union of $\phi^{-1}(I)$ and a $\mu_G$-null set $E$. Similarly, there is a compact arc $J$ such that $B$ consists of the union of $\phi^{-1}(J)$ and a $\mu_G$-null set $F$. Thus $A+B$ contains $\phi^{-1}(I+J)$, which has measure $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}(I+J) = \mu_{{\mathbb{R}}/{\mathbb{Z}}}(I)+\mu_{{\mathbb{R}}/{\mathbb{Z}}}(J) = \mu_G(\phi^{-1}(I)) + \mu_G(\phi^{-1}(J)) = \mu_G(A) + \mu_G(B);$$ since holds, we conclude that $A+B$ is in fact equal to the union of $\phi^{-1}(I+J)$ and a $\mu_G$-null set. Thus for every $a \in A$, the set $a + \phi^{-1}(J)$ lies in the union of $\phi^{-1}(I+J)$ and a $\mu_G$-null set, which forces $a$ to lie in $\phi^{-1}(I)$; thus $A = \phi^{-1}(I)$, and similarly $B = \phi^{-1}(J)$, giving the 