TITLE: Integration with respect to distribution function in Stein's book
QUESTION [5 upvotes]: I'm reading Stein's book on singular integrals and there is an identity of the $\mathrm{L}^p$-norm with the distribution function that I don't understand.
For $g$ measurable he first defines the distribution function of $|g|$ as $\lambda(\alpha)=\mu(\{x: |g(x)|>\alpha\})$. Then he claims that for $g\in\mathrm{L}^p$ we have the identity $$\int_{\mathbb{R}^d} |g(x)|^p\, \mathrm{d}x = -\int_0^\infty \alpha^p\, \mathrm{d}\lambda(\alpha).$$
I actually even don't understand what integration with respect to the distribution function means. What wonders me is the minus sign, is this measure with respect to the distribution function a signed one?

REPLY [2 votes]: You can also view this as an instance of the change-of-variable formula. The image of Lebesgue measure on $\Bbb R^d$, under the transformation $x\mapsto |g(x)|$, is the measure $\mu$ given by the formula
$$
\mu(B)=|\{x\in\Bbb R^d:|g(x)|\in B\}|,\qquad B\in\mathcal B([0,\infty)).
$$
where $|A|$ is $d$-dimensional Lebesgue meaure of $A\in\mathcal B(\Bbb R^d)$.
The change-of-variable formula now asserts that
$$
\int_{\Bbb R^d}|g(x)|^p\,dx =\int_{[0,\infty)}\alpha^p\,\mu(d\alpha).
$$
The function $\lambda$ is the complementary distribution function associated with $\mu$:
$$
\lambda(s) =\mu((s,\infty)),\qquad s>0.
$$
Thus, for $0<a<b$, 
$$
\mu((a,b]) = -[\lambda(b)-\lambda(a)].
$$
Stein's formula
$$\int_{\mathbb{R}^d} |g(x)|^p\, \mathrm{d}x = -\int_0^\infty \alpha^p\, \mathrm{d}\lambda(\alpha).$$
is thus seen as an expression of the Lebesgue integral $\int_{[0,\infty)}\alpha^p\,\mu(d\alpha)$ as the Stieltjes integral $\int_0^\infty \alpha^pd(-\lambda)(\alpha)$ of the continuous function $\alpha\mapsto\alpha^p$ with respect to the increasing function $-\lambda$.