Integral of Distribution Function

Theorem
Let $\struct {X, \Sigma, \mu}$ be a measure space and $f$ be a $\mu$-measurable function. Let $p > 0, r \geq 0$.

For $\lambda > 0$, let $E_\lambda = \{x \in X : |f(x)| > \lambda\}$, so that $m(\lambda) = \mu(E_\lambda)$ is the distribution function of $f$.

Then


 * $\displaystyle \int_0^\infty p\lambda^{p-1} \int_{E_\lambda} |f|^r \rd\mu \rd\lambda

= \int_X |f|^{p+r} \rd\mu$

and in particular


 * $\displaystyle \int_0^\infty p\lambda^{p-1} m(\lambda) \rd\lambda = \int_X |f|^p \rd\mu.$

Proof
Note that for any measurable $A\in \Sigma$, $\mu(A) = \int_A 1 d\mu$. Therefore, for $\lambda > 0$, $\mu(E_\lambda) = \int_{E_\lambda} 1 \rd\mu$, which can also be written $\mu(E_\lambda) = \int_{E_\lambda} \vert f \vert ^0 \rd\mu$. Therefore, taking $r = 0$ in the above, we obtain


 * $\displaystyle \int_0^\infty p\lambda^{p-1} m(\lambda) \rd\lambda = \int_0^\infty p\lambda^{p-1} \int_{E_\lambda} \vert f \vert^0 \rd\mu \rd\lambda = \int_X |f|^p \rd\mu.$