Integration with respect to distribution function in Stein's book 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?
 A: The problem is proving the identities 
$$\int_{\mathbb R^n}|g(x)|^q\, dx=
\int_0^\infty qs^{q-1}\lambda(s)\, ds = -\int_0^\infty s^q\, d\lambda(s).$$
The first one is the "layer cake formula", as they call it in Lieb and Loss's book. The second is integration by parts with the Lebesgue-Stieltjes integral: 
$$
\int_0^\infty qs^{q-1}\lambda(s)\, ds=\int_0^\infty d(s^q)\lambda(s) = - \int_0^\infty s^q d\lambda(s).$$
The minus sign comes from integration by parts. An example can explain why it is necessary: if $g(x)=a\mathbf{1}_B(x)$ with $a>0$, then 
$$\lambda(s)=\begin{cases} |B|, & s\le a \\ 0, &s>a\end{cases}$$
so $d\lambda(s)=-|B|\delta_a(s)$. (Here $\delta_a$ is a Dirac delta centered at $a$). 

Proof of the layer cake formula: 
$$
\int_{\mathbb R^n} |g(x)|^q\, dx = \iint \mathbf{1}_{[0<t<|g(x)|^q]}\, dtdx = \int_0^\infty \lambda(t^\frac1q)\, dt = \int_0^\infty qs^{q-1}\lambda(s)\, ds.$$
A: 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 $-\lambda$.
A: It is not a signed one. Note that $\lambda(\alpha)$ is a function, and is obviously non negative. Usually if you have a, say, nonnegative function $f$ which is measurable respect a measure space $(X, \omega, \mu)$. Then $df(a)$ is the integral with respect to the measure defined by $df(A) = \int_A f(x) d\mu(x)$. It is easy to check that this is indeed a measure and that you have $\int_A g(a) df(a) = \int_A g(a) f(a) da$ whenever this make sense. One can obviously generalize that to functions $f \in L^1$ obtaining signed measures, but it is not your case. 
Regarding the inequality it is false because on the right you get a negative value. I know the result as:
$$\int |g|^p = p\int c^{p-1} d\lambda(c)$$
