Prove that $\int_X |f|^p = \int_0^\infty pt^{p-1} \mu (\{x:|f(x)| \geq t\}) dt$

Question

I am stuck in this problem. We can assume that we are working in a $\sigma$-finite measure space and we have some measurable function $f$.

Now, the first thing I do is obtain two integrals

$$\int_0^\infty pt^{p-1} \mu(\{x:|f(x)|\geq t\}) dt = \int_0^\infty \int_Xpt^{p-1} \chi_{\{x:|f(x)|\geq t\}} d\mu dt$$

Now, since we are working under the assumption that $p>0, t\geq 0$ and also since the indicator function $\chi$ is non-negative, we can use Fubini's theorem to swap the integrals, and get

$$\int_X \int_0^\infty pt^{p-1} \chi_{\{x:|f(x)|\geq t\}} dt d\mu$$ From here, I am unsure if I should somehow prove that the integrand is Riemann-integrable so I can perform a change of variables. On the other hand, I saw that, for the special case $p=1$, the following identity is used:

$$\int_0^\infty \chi_{\{x:|f(x)|\geq t\}} dt = \int_0^{|f(x)|} dt$$

Is it possible to use a change of variables $s = t^p$ to get the following integral?

$$\int_0^\infty \chi_{\{x:|f(x)|^p\geq s\}}ds=\int_0^{|f(x)|^p}ds$$

If so, how can this be made rigorous? Also, how do I prove the identity?

Answer 1

Hint: the integral upper limit is not really $\infty$. Notice that by definition of the indicator function, $\int_0^\infty pt^{p-1}\chi_{\{|f((x)|\geq t\}}dt = \int_0^{|f(x)|} pt^{p-1}dt$.

Answer 2

You are on the right track. I think the main issue here is defining the sets properly so you can see what the limits of integration have to be. Rudin does it like this:

$E=\{(x, t) \in X \times [0, \infty): f(x) > t\}$ and $E^t = \{x\in X:(x,t)\in E\}.$

Then, (it helps to draw a picture) $\mu(E^t)=\int_X\chi_E(x,t)d\mu$ so

$\int^{\infty}_0(pt^{p-1})\mu(E^t)dt\overset{\text{Fubini}}=\int_Xd\mu\int^{\infty}_0\chi_E(x,t)(pt^{p-1})dt.$

Conclude by recognizing that the inner integral is $\int^{f(x)}_0pt^{p-1}dt=(f(x))^p.$