Understanding $\nu$ given by $\int f\, d\nu = \int_{\Bbb R^2\times\Bbb R} \int_{S^1} f(|x|w,z) \, d\sigma(w)\, d\mu(x,z)$ I have a question about the proof of Theorem $2.1$ here. The proof is on Pg. $10$. I am trying to work out the $d = 2$ case in particular. Here is an excerpt from the paper:

Suppose for a contradiction that $\dim_F \mathcal C^2 > 1$. Then there exists $\alpha > 1$ and a Borel probability measure $\mu$ on $\mathcal C^2$ such that $$|\hat\mu(\xi)| \le |\xi|^{-s/2}, \forall \xi\in \Bbb R^3$$
By replacing $\mu$ with $f\mu$ for an appropriate bump function $f$, it may be assumed that for some $\epsilon > 0$, $$\operatorname{supp} \mu \subset \{(\xi, |\xi|) \in \Bbb R^2\times \Bbb R: \epsilon \le |\xi| \le 1/\epsilon\}$$
Let $\nu$ be the Borel probability measure on $\mathcal C^2$ defined by
$$\color{blue}{\int \varphi\, d\nu = \int_{\Bbb R^2\times\Bbb R} \int_{S^1} \varphi(|x|\omega,z) \, d\sigma(\omega)\, d\mu(x,z)}$$
for any non-negative Borel function $\varphi$, where $\sigma$ is the rotation invariant Borel probability measure on $S^1$.

Question: How is $\nu$ defined? It is not clear why the above integral makes sense, i.e., why is $\nu$ given by the above integral representation a measure? Why does the product of measures $\sigma$ and $\mu$ make sense? The introduction of an "extra" variable $\omega$ seems quite unmotivated too. It would be quite helpful if someone could clarify why the integral on the right side of the blue equation makes sense, and where it comes from! Thank you.

Notation: The authors denote the $d$-dimensional cone by $$\mathcal C^d = \{(x_1, \ldots, x_{d+1}): |(x_1, \ldots, x_d)| = |x_{d+1}|\} \subset \Bbb R^{d+1}$$
The author uses the following definition of the Fourier dimension of $A\subset\Bbb R^{d+1}$:
$$\dim_F A = \sup\{0\le s\le d+1: \exists\mu\in \mathcal P(A) \text{ s.t. } |\hat\mu(\xi)| \lesssim |\xi|^{-s/2},\, \forall \xi\in \Bbb R^{d+1}\}$$
where $\mathcal P(A)$ is the set of Borel probability measures on $\Bbb R^{d+1}$ satisfying $\mu(A) = 1$.
 A: There are many questions.
One can define $\nu$ by
$$\forall A \in \mathcal{B}(R^2 \times R), \quad \nu(A) := \int_{R^2 \times R} \Big(\int_{S_1} 1_A(|x|w,z) d\sigma(w) \Big) d\mu(x,z) \in [0,+\infty].$$
The $\sigma$-additivity of $\nu$ follows from the Fubini theorem for positive functions (view a discrete sum over al $n \in \mathbb{N}$ as an integral with regard to the counting measure on $\mathbb{N}$).
By linearity, the blue equality in your question holds for all non-negative simple functions $f$. Using Beppo Levi's Theorem, one extends it to all measurable non-negative function. By difference, it holds for all $\nu$-integrable function.
Last, what does this measure $\nu$ represent? Taking the average over $S_1$ with respect to $\sigma$ is a standard way to transform $\nu$ into a measure which is invariant by rotation on the first factor $R^2$.
Here is an alternative description of $\nu$. I denote by $R_\theta$ the rotation of $\theta$ in the vector space $\mathbb{R^2}$.
$$\forall A \in \mathcal{B}(R^2 \times R), \quad \nu(A) := \int_{R^2 \times R} \Big( \frac{1}{2\pi}\int_0^{2\pi} 1_A(R_\theta x,z) d\theta \Big) d\mu(x,z) \in [0,+\infty].$$
By Fubini theorem for non-negative functions,
\begin{eqnarray*}
\forall A \in \mathcal{B}(R^2 \times R), \quad \nu(A) 
&=&  \frac{1}{2\pi}\int_0^{2\pi}  \Big(\int_{R^2 \times R} 1_A(R_\theta x,z) d\mu(x,z) \Big) d\theta \\
&=&  \frac{1}{2\pi}\int_0^{2\pi}  \Big(\int_{R^2 \times R} 1_A(x,z) d\mu_\theta(x,z) \Big) d\theta,
\end{eqnarray*}
where $\mu_\theta$ is the image of $\mu$ by the map $(x,z) \mapsto (R_\theta x,z)$
so $\nu$ is an average of the measures $\mu_\theta$ over all $\theta \in [0,2\pi)$.
To see that this description gives the same result, Use Fubini theorem for positive functions and integrate in polar coordinates, namely set $x = (r\cos\theta,r\sin\theta)$ with $r>0$ and $\theta \in [0,2\pi)$, so $dx=rdrd\theta$.
For positive measurable functions on $\mathbb{R}^2$, $$\int_{R^2} f(x) \mathrm{d}x = \int_0^{+\infty} \Big( \int_0^{2\pi} f(r\cos\theta,r\sin\theta) \mathrm{d}\theta \Big) r\mathrm{d}r.$$
The uniform measure $\sigma$ on $S_1$ is the image of the uniform measure on $[0,2\pi)$ by the map $\theta \mapsto (\cos\theta,\sin\theta)$, so $$\frac{1}{2\pi} \int_0^{2\pi} f(r\cos\theta,r\sin\theta) \mathrm{d}\theta = \int_{S_1} f(rw) \mathrm{d}\sigma(w).$$
