(\psi_{j, \tau} \hat \mu_1)^\vee = \eta_T (\psi_{j, \tau}^\vee * \mu_1).$$

Now $\psi_{j, \tau}^\vee$ is concentrated on a $R_j^{-1/2} \times R_j^{-1}$ rectangle centered at 0 and it decays rapidly outside that rectangle. Since $x \in E_2$, the distance from $x$ to the support of $\mu_1$ is $\gtrsim 1$. Therefore, $d^x_* M_T \mu_1(t)$ is tiny unless $t \sim 1$.

To study the case when $t \sim 1$, we expand out $M_T \mu_1$:

$$M_T \mu_1 (y) = \eta_T(y) ( \psi_{j, \tau} \hat \mu_1 )^{\vee}(y)  = \eta_T(y)  \int e^{2 \pi i \omega y} \psi_{j, \tau}(\omega) \hat \mu_1( \omega) d \omega.$$

Since $| \hat \mu_1(\omega) | \le 1$, and $\psi_{j, \tau}(\omega)$ is supported on $\tau$ and bounded by 1, it suffices to check that for each $\omega \in \tau$,

$$\label{statphase} \int_{S^1(x,t)} \eta_T(y) e^{2 \pi i \omega y} dl(y) \le {\textrm{RapDec}}(R_j).$$

We will prove this rapid decay by stationary phase. There are two slightly different cases, depending on whether $T$ intersects $S^1(x, t/2)$ or not. Let us start with the case that $T$ intersects $S^1(x, t/2)$, since this case is a little harder. After a coordinate rotation, we can assume that $\omega$ has the form $(0, \omega_2)$ with $\omega_2 \sim R_j$. Recall that a tube $T 