I am trying to read Exploring the toolkit of Jean Bourgain, a beautiful article by Terrence Tao, that can be found here, and I have what I believe to be a simple question.
Given
$$ \int_{\mathbb{R}^2} \int_{S^1} 1_B(x) 1_B(x+t_j \omega ) d \sigma(\omega) dx ,$$
where $B \subset [-1,1]^2$, $1_B$ are indicator functions, $0<t_j \le 1$, and $d \sigma$ is the surface measure on the unit circle $S^1$, Tao states that, using the Fourier transform $\hat{f}(\xi) = \int_{\mathbb{R}^d} f(x) e^{2\pi i x \xi} dx $, the above double integral can be rewritten as
$$ \int_{\mathbb{R}^2} | \hat{1}_B(\xi) |^2 \hat{\sigma}(t_j\xi) d \xi $$
where $\hat{\sigma}(\xi) = \int_{S^1} e^{2\pi i x \xi} d \sigma( \omega) $ is the Fourier transform of the surface measure $d \sigma$. What are the steps to show this?
I think this could possibly be the solution, feedback appreciated. Note that $u(t_j\omega)=\int_{\mathbb{R}^2}\mathbf{1}_B(x)\mathbf{1}_B(x+t_j\omega)dx$ is an autocorrelation. Thus its Fourier transform over $\omega$ is $\hat{u}(\xi)=\hat{\mathbf{1}}_B(\xi/t_j)(\hat{\mathbf{1}}_B(\xi/t_j))^*/|t_j|^2=|\hat{\mathbf{1}}_B(\xi/t_j)|^2/|t_j|^2$ by convolution theorem and properties of the Fourier transform. The Fourier transform is a symmetric operator, thus for finite measure $\sigma$ on $S_1$ $$\int_{\mathbb{R}^2}\frac{1}{|t_j|^2}|\hat{\mathbf{1}}_B(\xi/t_j)|^2\hat{\sigma}(\xi)d\xi=\int_{S_1}u(t_j\omega)d\sigma(\omega)$$ set $t_j\xi'=\xi$ then $\xi'=\xi/t_j$ and $d\xi'=d\xi/|t_j|^2$.