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 $E_\lambda = \set {x \in X: \size {\map f x} > \lambda}$, so that $\map m \lambda = \map \mu {E_\lambda}$ is the distribution function of $f$.

Then:


 * $\ds \int_0^\infty p \lambda^{p - 1} \int_{E_\lambda} \size f^r \rd \mu \rd \lambda = \int_X \size f^{p + r} \rd \mu$

and in particular:


 * $\ds \int_0^\infty p \lambda^{p - 1} \map m \lambda \rd \lambda = \int_X \size f^p \rd \mu$

Proof
We have that for any measurable $A \in \Sigma$:
 * $\map \mu A = \ds \int_A 1 \rd \mu$

Therefore, for $\lambda > 0$:
 * $\map \mu {E_\lambda} = \ds \int_{E_\lambda} 1 \rd \mu$

which can also be written:
 * $\map \mu {E_\lambda} = \ds \int_{E_\lambda} \size f^0 \rd \mu$

Therefore, taking $r = 0$ in the above, we obtain: