how to prove this equality $||f||_{L^p}^{p}=p\int_0^{+\infty} \lambda^{p -1}\mu(E^f_\lambda) d\lambda$

Question

Let $(X,B(X),\mu)$ be a measure space, suppose there is a function f that is measurable

Define the distribution function ${\mu(E_\lambda^f): {\mathbb R}^+ \rightarrow [0,+\infty]}$

How to prove this equality :$$||f||_{L^p}^{p}=p\int_0^{+\infty} \lambda^{p-1} \mu(E^f_\lambda) d\lambda$$ ?

My attempt:

We know that :$||f||_{L^p}^{p}=\int_X |f| d\mu$

By the Chebyshev inequality: $\lambda\mu(E_\lambda^f)<\int_X |f|dx$

And then? I don't know how to solve this exercise

Any help is appreciated

Answer 1

The set $\{f > 0\}$ is a $\sigma$-finite subset of $X$ so we will be able to invoke the Tonelli theorem. Since $$ |f|^p = \int_0^{|f|} p\lambda^{p-1} \, d\lambda $$ you have $$ \int_X |f|^p \, d\mu = \int_{\{f > 0\}} \int_0^{|f(x)|} p\lambda^{p-1} \, d\lambda d\mu(x).$$ This can be evaluated as $$\int_{\{f > 0\}} \int_0^\infty \chi_{\{0 \le t \le |f(x)|\}} p\lambda^{p-1} \, d\lambda d\mu(x) = \int_0^\infty p\lambda^{p-1} \int_{\{f > 0\}} \chi_{\{0 \le t \le |f(x)|\}}\, d\mu(x) d\lambda$$ where $$\int_{\{f > 0\}} \chi_{\{0 \le t \le |f(x)|\}}\, d\mu(x) = \mu(\{x : |f(x)| \ge t\})$$ for all $t > 0$.