Bohr sets such that $$\mu_G( A \Delta \phi^{-1}_n(I_n) ), \mu_G( B \Delta \phi^{-1}_n(J_n) )  = o(1),$$ where in this section we use $o(1)$ to denote a quantity that goes to zero as $n \to \infty$. In particular, the arcs $I_n,J_n$ in the circle ${\mathbb{R}}/{\mathbb{Z}}$ have measure $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}(I_n) = \mu_G(A) + o(1), \mu_{{\mathbb{R}}/{\mathbb{Z}}}(J_n) = \mu_G(B) + o(1).$$ Taking Fourier coefficients, we see that $$\begin{aligned}
 \left| \int_G 1_A(x) e^{2\pi i \phi_n(x)}\ d\mu_G(x) \right| &= \left|\int_{I_n} e^{2\pi i \alpha}\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}}(\alpha)\right| + o(1) \\
&= \frac{1}{\pi} \sin(\pi \mu_{{\mathbb{R}}/{\mathbb{Z}}}(I_n)) + o(1) \\
&= \frac{1}{\pi} \sin(\pi \mu_G(A)) + o(1).\end{aligned}$$ On the other hand, from Plancherel’s theorem we have $$\sum_{\phi \in \hat G}  \left| \int_G 1_A(x) e^{2\pi i \phi(x)}\ d\mu_G(x) \right|^2 = \mu_G(A)$$ where the Pontryagin dual group $\hat G$ consists of all continuous homomorphisms $\phi$ from $G$ to ${\mathbb{R}}/{\mathbb{Z}}$. Thus, for $n$ large enough, there are only boundedly many possible choices for $\phi_n$, and by passing to a subsequence if necessary we may assume that $\phi_n = \phi$ does not depend on $n$. For $n,n' \to \infty$, we now have $$\mu_G( A \Delta \phi^{-1}(I_n) ), \mu_G( A \Delta \phi^{-1}(I_{n'}) ) \to 0,$$ and hence by the triangle inequality $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}( I_n \Delta I_{n'} ) = \mu_G( \phi^{-1}(I_n) 