Following this answer by Christophe, I'm trying to show that $\nu$ defined by $$\nu(A) := \int_{\Bbb R^2 \times \Bbb R} \int_{S^1} \mathbf{1}_{A}(|x|w,z) \,d\sigma(w)\, d\mu(x,z)$$ is a Borel measure on $\Bbb R^2 \times \Bbb R$. It is enough to check $\sigma$-additivity, i.e., if $\{A_i\}$ is a countable collection of disjoint sets in the Borel $\sigma$-algebra of $\Bbb R^2 \times \Bbb R$, then $$\nu\left(\bigcup_{i=1}^\infty A_i \right) = \sum_{i=1}^\infty \nu(A_i) $$
My work: Suppose $A = \cup_i A_i$ where $\{A_i\}$ are disjoint sets in the Borel $\sigma$-algebra of $\Bbb R^2 \times \Bbb R$. We have \begin{align*} \nu(A) &= \int_{\Bbb R^2 \times \Bbb R} \Big(\int_{S^1} \mathbf{1}_{\cup_i A_i}(|x|w,z)\, d\sigma(w) \Big)\, d\mu(x,z)\\ &= \int_{\Bbb R^2 \times \Bbb R} \Big(\int_{S^1} \int_{\Bbb N} \mathbf{1}_{A_i}(|x|w,z)\,d\lambda(i)\, d\sigma(w) \Big)\, d\mu(x,z)\\ &= \int_{\Bbb R^2 \times \Bbb R} \Big(\int_{\Bbb N} \int_{S^1} \mathbf{1}_{A_i}(|x|w,z)\, d\sigma(w)\, d\lambda(i) \Big)\, d\mu(x,z)\\ &= \int_{\Bbb N} \Big(\int_{\Bbb R^2 \times \Bbb R} \int_{S^1} \mathbf{1}_{A_i}(|x|w,z)\, d\sigma(w) \, d\mu(x,z) \Big)\, d\lambda(i)\\ &= \sum_{i=1}^\infty \nu(A_i) \end{align*} where $\lambda$ is the counting measure on $\Bbb N$. Since all the involved spaces are $\sigma$-finite, the product measure on their cartesian product is uniquely determined (and so the above integrals make sense.)
I have used Tonelli's theorem for non-negative measurable functions as stated here for the above computation. It remains to check that the function $h: (\Bbb R^2 \times \Bbb R) \times S^1 \times \Bbb N \to \Bbb C$ defined by $$h((x,z),w, i) = \mathbf{1}_{A_i}(|x|w,z)$$ is Borel measurable on $(\Bbb R^2 \times \Bbb R) \times S^1 \times \Bbb N$. Could someone kindly help me complete the proof? Thank you!
The function $T : (x,z,w) \mapsto (|x|w,z)$ from $\mathbb{R}^2 \times \mathbb{R} \times \mathbb{S}_1$ to $\mathbb{R}^2 \times \mathbb{R}$ is continuous hence Borel-measurable. The function $h$ takes values in $\{0,1\}$, and $$h^{-1}\{1\} = \bigcup_{i \in \mathbb{N}}T^{-1}(A_i) \times \{i\}$$ is a Borel subset. Hence $h$ is Borel-measurable.