d\mu_{{\mathbb{R}}/{\mathbb{Z}}} = o(1).$$ From Markov’s inequality, we conclude that for any fixed ${\varepsilon}>0$, all but $o(1)$ in measure of the set $F_\lambda +_{\varepsilon}J$ is contained in $E$, thus $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}( F_\lambda +_{\varepsilon}J ) \leq \mu_{{\mathbb{R}}/{\mathbb{Z}}}(E) + o(1) = \mu_G(A) + t + o(1).$$ On the other hand, from Corollary \[kemp-cor\] we have $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}( F_\lambda +_{\varepsilon}J ) \geq \min( \mu_{{\mathbb{R}}/{\mathbb{Z}}}(F_\lambda) + t, 1 ) - 2 \sqrt{{\varepsilon}};$$ combining the two bounds and sending ${\varepsilon}$ to zero, we conclude using that $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}(F_\lambda) \leq \mu_G(A) + o(1).$$ for any fixed $\lambda>0$. Sending $\lambda$ sufficiently slowly to zero as $n \to \infty$, we conclude on diagonalising that $$\mu_{{\mathbb{R}}/{\mathbb{Z}}}( F_\kappa ) \leq \mu_G(A) + o(1)$$ for some infinitesimal $\kappa>0$. Combining this with and the pointwise bound $f_A \leq 1_{F_\kappa} + o(1)$, we conclude that $$\mu_G(A) = \int_{{\mathbb{R}}/{\mathbb{Z}}} f_A\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}} \leq \int_{{\mathbb{R}}/{\mathbb{Z}}} 1_{F_\kappa}\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}} + o(1) \leq \mu_G(A) + o(1)$$ which implies in particular that $$\label{mfk}
\mu_{{\mathbb{R}}/{\mathbb{Z}}}(F_\kappa) = \mu_G(A) + o(1)$$ and $$\label{rzf}
 \int_{{\mathbb{R}}/{\mathbb{Z}}} |1_{F_\kappa} - f_A|\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}} = o(1).$$ Pulling back to $G$, this implies that $$\int_{G} |1_{\phi^{-1}(F_\kappa)} - 1_A|\ d\mu_G = o(1),$$ thus $A$ is asymptotically equivalent to $\phi^{-1}(F_\kappa)$. Thus to establish the proposition, it suffices to show that $F_\kappa$ is asymptotically equivalent to an 