5
$\begingroup$

Can I ask a homework question here?

Let $f$ be measurable and non-negative in $\mathbb R^d.$ Using Fubini's theorem, show that for $1 \leq p \lt \infty,$

$\lVert f\rVert^p_p = \int^{\infty}_{0}pt^{p-1}\lambda(\{x:f(x)\gt t\}) \ dt.$

  • 0
    Hint: Assume first that $f$ is a simple function and show that the equation holds. Then for your non-negative measurable $f$ choose a nondecreasing sequence of simple functions converging point-wise to $f$ and use monotone convergence theorem.2012-08-13

3 Answers 3

3

It is the so-called layer-cake representation. It can be found on some books, like Analysis by Lieb and Loss. Here is a short proof based on Fubini's theorem, page 5.

  • 0
    Layer cake representation together with symmeterization is one of strongest tools to prove isoperimetric type inequalities. Im glad someone mentioned that in here.2018-10-23
2

$f(x)^p=\int_0^{f(x)}pt^{p-1}\,\mathrm dt=\int_0^{+\infty}pt^{p-1}\,\mathbf 1_{f(x)\gt t}\,\mathrm dt$

1

If you know the definition of Lebesgue integral by the so called "archimedean integral" the exercise is just a simple change of variable. Let $X$ be a non empty set, $\mathcal A$ a $\sigma$-algebra over $X$ and $\mu$ a (positive) measure on $\mathcal A$.

If $f \colon X \to \mathbb R$ is a $\mathcal A$-measurable, positive function then $ \int_X f d\mu := \int_0^{\infty} \mu( \{f>t\} ) dt $ is a possible definition of the Lebesgue integral (note that the LHS is a Lebesgue integral while the RHS is a Riemann integral: infact, the function $t \mapsto \mu( \{f>t\} )$ is Riemann-integrable, since it is monotone).

Anyway, now consider $ I=\int_0^{\infty}pt^{p-1} \mu( \{f>t\} )dt $ By a simple change of variable ($w=t^p$) we get $dw = pt^{p-1}dt$ hence $ I = \int_0^{\infty} \mu( \{f>\sqrt[p]{w}\} ) dw = \int_0^{\infty} \mu( \{f^p>w\} ) dw =\int_Xf^p d\mu = \Vert f \Vert_p^p. $