$d^x_*({\mu_{1, good}})$. Recall by definition that

$$\int \psi(t) d^x_*(\mu) = \int \psi(|x-y|) d \mu(y).$$

In particular, if $\psi$ is the characteristic function of the interval $t_0 \le t \le t_0 + \Delta t$, then we see that

$$\int_{t_0}^{t_0 + \Delta t} d^x_*(\mu) = \int_{t_0 \le |x-y| \le t_0 + \Delta t} d \mu.$$

To evaluate $d^x_*(\mu)$ at $t$, we can take the limit as $\Delta t \rightarrow 0$. If we think of $\mu$ as $\mu(y) dy$, then we get

$$d^x_*(\mu) (t) = \int_{S^1(x,t)} \mu(y) dl(y),$$

where $dl(y)$ denotes the arc length measure on the circle $S^1(x,t)$.

To control $\| d^x_*(\mu_1) - d^x_*({\mu_{1, good}}) \|_{L^1}$ we will start by studying $ d^x_*(M_T \mu_1)$ for different $T$. For a tube $T \in {\mathbb{T}}$, let $2T$ denote the concentric tube of twice the radius. If $x \notin 2T$, we show that $d^x_*( M_T \mu_1)$ is negligible.

\[xnotinT\] If $T \in {\mathbb{T}}_{j, \tau}$, and $x \in E_2$, and $x \notin 2T$, then

$$\| d^x_* (M_T \mu_1) \|_{L^1} \lesssim {\textrm{RapDec}}(R_j).$$

We will prove the stronger estimate that for every $t$:

$$d^x_* (M_T \mu_1) (t)  \lesssim {\textrm{RapDec}}(R_j).$$

Recall that

$$\label{dxMT} d^x_*  (M_T \mu_1)(t) = \int_{S^1(x,t)} M_T \mu_1(y) dl(y).$$

We also recall that

$$M_T \mu_1 = \eta_T 