# $L^p$-norm of a non-negative measurable function

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 $1 \leq p \lt \infty,$

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

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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. – T. Eskin Aug 13 '12 at 12:18

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.

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

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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 $$I = \int_0^{\infty} \mu( \{f>\sqrt[p]{w}\} ) dw = \int_0^{\infty} \mu( \{f^p>w\} ) dw =\int_Xf^p d\mu = \Vert f \Vert_p^p.$$

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