[proofplan]
We decompose $\mathbb{R}^n \setminus \{0\}$ using the coarea formula applied to the radial map $\rho(x) = |x|$, whose level sets are the spheres $\partial B(0,r)$. The coarea formula expresses the integral of $g$ as an integral over level sets weighted by the Jacobian of $\rho$. We verify that $|\nabla \rho| = 1$ on $\mathbb{R}^n \setminus \{0\}$, apply the coarea formula, and then convert the integral over $\partial B(0,r)$ to an integral over the unit sphere $\partial B(0,1)$ using the Hausdorff measure scaling law.
[/proofplan]
[step:Define the radial map and verify $|\nabla \rho| = 1$ on $\mathbb{R}^n \setminus \{0\}$]
Define the radial distance map
\begin{align*}
\rho: \mathbb{R}^n \setminus \{0\} &\to (0, \infty) \\
x &\mapsto |x| = \left(\sum_{i=1}^n x_i^2\right)^{1/2}.
\end{align*}
This map is $C^\infty$ on $\mathbb{R}^n \setminus \{0\}$ and Lipschitz with constant $1$ on all of $\mathbb{R}^n$ (since $\big| |x| - |y| \big| \leq |x - y|$ by the reverse triangle inequality). Computing the gradient:
\begin{align*}
\partial_{x_i} \rho(x) = \frac{x_i}{|x|}, \qquad \nabla \rho(x) = \frac{x}{|x|}.
\end{align*}
Therefore $|\nabla \rho(x)| = \frac{|x|}{|x|} = 1$ for all $x \in \mathbb{R}^n \setminus \{0\}$.
The level sets of $\rho$ are the spheres of radius $r > 0$:
\begin{align*}
\rho^{-1}(r) = \{x \in \mathbb{R}^n : |x| = r\} = \partial B(0,r).
\end{align*}
[/step]
[step:Apply the coarea formula to decompose the integral over level sets of $\rho$]
The coarea formula for a Lipschitz function $u: \mathbb{R}^n \to \mathbb{R}$ states: for any Borel measurable $h: \mathbb{R}^n \to [0, \infty]$,
\begin{align*}
\int_{\mathbb{R}^n} h(x) |\nabla u(x)| \, d\mathcal{L}^n(x) = \int_{-\infty}^\infty \left(\int_{u^{-1}(t)} h(x) \, d\mathcal{H}^{n-1}(x)\right) d\mathcal{L}^1(t).
\end{align*}
We apply this with $u = \rho$ and $h = g$. Since $\mathcal{L}^n(\{0\}) = 0$, the origin does not affect the left-hand side. Since $|\nabla \rho(x)| = 1$ on $\mathbb{R}^n \setminus \{0\}$, the coarea formula gives
\begin{align*}
\int_{\mathbb{R}^n} g(x) \, d\mathcal{L}^n(x) = \int_0^\infty \left(\int_{\partial B(0,r)} g(x) \, d\mathcal{H}^{n-1}(x)\right) d\mathcal{L}^1(r),
\end{align*}
where the outer integral runs over $(0, \infty)$ because $\rho^{-1}(t) = \varnothing$ for $t \leq 0$.
[/step]
[step:Convert the integral over $\partial B(0,r)$ to the unit sphere via the scaling law]
For each fixed $r > 0$, consider the dilation
\begin{align*}
\delta_r: \partial B(0,1) &\to \partial B(0,r) \\
\theta &\mapsto r\theta.
\end{align*}
This map is bi-Lipschitz with constants $c = C = r$, so by the [Bi-Lipschitz Maps Preserve Hausdorff Dimension](/theorems/3061) (applied as the Lipschitz bound to both $\delta_r$ and $\delta_{1/r}$), $\mathcal{H}^{n-1}$ transforms under dilation by $r$ as
\begin{align*}
\mathcal{H}^{n-1}(r \cdot E) = r^{n-1} \mathcal{H}^{n-1}(E)
\end{align*}
for every Borel set $E \subset \partial B(0,1)$. This scaling law applied to the change of variables $x = r\theta$ gives
\begin{align*}
\int_{\partial B(0,r)} g(x) \, d\mathcal{H}^{n-1}(x) = \int_{\partial B(0,1)} g(r\theta) \, r^{n-1} \, d\mathcal{H}^{n-1}(\theta).
\end{align*}
Substituting into the coarea decomposition from the previous step:
\begin{align*}
\int_{\mathbb{R}^n} g(x) \, d\mathcal{L}^n(x) = \int_0^\infty \left(\int_{\partial B(0,1)} g(r\theta) \, d\mathcal{H}^{n-1}(\theta)\right) r^{n-1} \, d\mathcal{L}^1(r).
\end{align*}
[/step]
