Theorem2.49 p78, Folland's Real Analysis: There is a unique Borel measure $\sigma= \sigma_{n-1}$ on $S^{n-1}$ such that $m_*= \rho \times \sigma$. If $f$ is Borel measurable on $\mathbb{R}^n$, and $f \ge 0 $ or $f \in L^1(m)$, then $$ \int_{\mathbb{R}^n} f(x) \,dx = \int_0^\infty \int_{S^{n-1}} f(rx') r^{n-1} d \, \sigma (x') dr. $$ where for $x \in \mathbb{R}^n \setminus \{0 \}$, $$ \Phi(x) : x \mapsto (|x|,\frac{x}{|x|}) \in (0,\infty) \times S^{n-1} $$ Define a push forward Borel measure $m_*$ on the measure space (with Borel $\sigma$-algebra) $\Big( (0,\infty) \times S^{n-1}, B_{(0,\infty) \times S^{n-1})} \Big)$ induced by the Borel measurable map $\Phi$. That is, $m_*(E) = m(\Phi^{-1}(E))$. We define $\rho(E):= \int _Er^{n-1} \, dr$ on $(0,\infty)$.
Here $S^{n-1}$ is the unit sphere in $\mathbb{R}^n$. $dx$ refers to Borel measure on $B_{\mathbb{R}^n}$ (i.e. the restriciton of Lebesgue measure of $\mathbb{R}^n$).
Question: How does one extend this theorem for $f$ Lebesgue measurable? The text says to consider the completion of $\sigma$.
Just the way you get two dimensional Lebesgue measure from one dimensional Lebesgue measure: take the product of $\sigma $ with $\rho$ and then complete it. Note that the product of two complete measures almost never complete so it is not enough to complete $\sigma $ first and then take the product.