t + o(1).$$

The pullback map $\phi^*: g \mapsto g \circ \phi$ is an isometry from $L^2({\mathbb{R}}/{\mathbb{Z}}, \mu_{{\mathbb{R}}/{\mathbb{Z}}})$ to $L^2(G, \mu_G)$. Taking adjoints, we obtain a pushforward map $\phi_*: L^2( G, \mu_G ) \mapsto L^2({\mathbb{R}}/{\mathbb{Z}}, \mu_{{\mathbb{R}}/{\mathbb{Z}}})$ such that $$\int_{{\mathbb{R}}/{\mathbb{Z}}} \phi_*(f)(\alpha) g(\alpha)\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}}(\alpha) = \int_G f(x) g(\phi(x))\ d\mu_G(x)$$ for all $f \in L^2(G,\mu_G)$ and $g \in L^2({\mathbb{R}}/{\mathbb{Z}}, \mu_{{\mathbb{R}}/{\mathbb{Z}}})$. It is easy to see that the map $\phi_*$ is monotone with $\phi_*(1)=1$ (up to almost everywhere equivalence). If we write $f_A \coloneqq \phi_* 1_A$ for the pushforward of $1_A$, then $f_A$ takes values in $[0,1]$ (after modifying on a set of measure zero if necessary), and we have $$\label{rza}
 \int_{{\mathbb{R}}/{\mathbb{Z}}} f_A\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}} = \int_G 1_A\ d\mu_G = \mu_G(A).$$ Also, since $1_A * 1_B = 1_A * 1_{\phi^{-1}(J)}$ is bounded by $o(1)$ outside of $C = \phi^{-1}(E)$, we see that $f_A * 1_J$ is bounded almost everywhere by $o(1)$ outside of $E$, thus $$\label{rze}
 \int_{({\mathbb{R}}/{\mathbb{Z}}) \backslash E} f_A * 1_{[0,t] \text{ mod } {\mathbb{Z}}}\ d\mu_{{\mathbb{R}}/{\mathbb{Z}}} = o(1).$$ Let $\lambda > 0$ be any fixed parameter, and let $F_\lambda \subset {\mathbb{R}}/{\mathbb{Z}}$ denote the set $F_\lambda \coloneqq \{ f_A \geq \lambda \}$, then we have $$\int_{({\mathbb{R}}/{\mathbb{Z}}) \backslash E} 1_{F_\lambda} * 1_{[0,t] \text{ mod } {\mathbb{Z}}}\ 