decompose ${\mu_{1, good}}* \hat \sigma_r$ as

$${\mu_{1, good}}* \hat \sigma_r = \sum_{T \textrm{ good}} M_T \mu_1 * \hat \sigma_r.$$

The right-hand side is essentially the wave packet decomposition of ${\mu_{1, good}}* \hat \sigma_r$. This means that $M_T \mu_1 * \hat \sigma_r |_{B^2(1)}$ is essentially supported in $T$ and its Fourier transform is essentially supported in an arc of $S^1_r$ in the direction of $T$. Since the tubes $T$ are all good, each tube $T$ has a small $\mu_2$ measure, and we will take advantage of this to bound the inner integral $\int_{E_2} | {\mu_{1, good}}* \hat \sigma_r |^2 d \mu_2(x)$.

Since $T$ has a small $\mu_2$ measure, we can immediately get a good estimate for $\int_{E_2} | M_T \mu_1 * \hat \sigma_r|^2 d \mu_2$. But to bound $\int_{E_2}| {\mu_{1, good}}* \hat \sigma_r|^2 d \mu_2$, we need to know how the wave packets $M_T \mu_1 * \hat \sigma_r$ interact with each other. Is it possible that these wave packets have a lot of positive interference on the set $E_2$? We will use decoupling theory to control such positive interference. We will discuss this further in Section \[sec:RS\].

Proof of Proposition \[mainest1\] {#sec:mainest1}
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We will study the pushforward measures $d^x_*(\mu_1)$ and 