For 0 < p < infinity, the layer-cake formula computes the integral of |f|^p from the distribution function

Fix $a\ge 0$. If $a=0$, both integrals below are zero. If $a>0$, apply [L2] on $[\epsilon ,a]$ and let $\epsilon \downarrow 0$. Since $p>0$, ${\epsilon }^p\to 0$; monotone convergence for the nonnegative integrand gives $a^p={\int }_0^{a}p{t}^{p-1}dt={\int }_0^{\mathrm{\infty }}p{t}^{p-1}{\mathbf{1}}_{\{t<a\}}dt\mathrm{.}$ This argument also covers $0<p<1$, when $t^{p-1}$ is unbounded at zero.

Let $s={\sum }_{j=1}^m{a}_j{\mathbf{1}}_{E_j}$ be a nonnegative simple function, with the sets $E_j$ pairwise disjoint and the coefficients $a_j\ge 0$. Then ${\int }_X{s}^{p}d\mu ={\sum }_{j=1}^m{a}_j^p\mu (E_j)\mathrm{.}$ Using step 1.1 for each coefficient and exchanging the resulting finite sum with the real integral gives ${\int }_X{s}^{p}d\mu =p{\int }_0^{\mathrm{\infty }}{t}^{p-1}\mu (\{s>t\})dt\mathrm{.}$

By [L3], choose simple functions $s_n\uparrow \mathrm{\mid }f\mathrm{\mid }$. Then $s_n^p\uparrow \mathrm{\mid }f{\mathrm{\mid }}^p$, and for each $t\ge 0$ one has ${\mathbf{1}}_{\{s_n>t\}}\uparrow {\mathbf{1}}_{\{\mathrm{\mid }f\mathrm{\mid }>t\}}$, hence $\mu (\{s_n>t\})\uparrow \mu (\{\mathrm{\mid }f\mathrm{\mid }>t\})=A_f(t)$ by [L1]. Applying [L4] first on $X$ and then on $(0,\mathrm{\infty })$ to the identities from step 2.1 yields ${\int }_X\mathrm{\mid }f{\mathrm{\mid }}^{p}d\mu =p{\int }_0^{\mathrm{\infty }}{t}^{p-1}\mu (\{\mathrm{\mid }f\mathrm{\mid }>t\})dt=p{\int }_0^{\mathrm{\infty }}{t}^{p-1}{A}_f(t)dt\mathrm{.}$