I am reading "Weakly Differentiable Functions" by William P. Ziemer, and I cannot prove an equality in a remark at page $28$ in the section on Lorentz spaces.
The equality involved the so called "non-increasing" rearrangement.Let $f \in L^{1}(R)$ and define$$f^{*}(t) := \inf \{ s : \alpha_{f}(s) \leq t \},$$where $\alpha_{f}(s)$ is the measure of the super-level of $\vert f \vert$ relative to $s$; i.e,$$ \alpha_{f}(s):= \vert \{ x : \vert f(x) \vert > s \} \vert.$$
Prove that$$\int_{0}^{t} f^{*}(s)\ ds = t \cdot f^{*}(t) + \int_{f^{*}(t)}^{+\infty} \alpha_{f}(s)\ ds $$
We can use the fact that the "distributions functions" of $f$ and $f^{*}$ are the same$$\alpha_{f}(s) = \alpha_{f^{*}}(s).$$
I am almost sure that this is the integration by part formula because $t*f^{*}(t)$ is a boundary term. Why $f^{*}(t)$ and $+\infty$ are the extrema of integration? The integral on the right feels like an integral from a change of variable $s \to f^{*}(s),$ and $f^{*}(t)$ is an extreme of integration because it is the smallest value for which $\alpha_{f}(s) \leq t.$I know it's very imprecise, I think I am almost there, I just need the right hint!