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: 
In response to my request that you use more prose, you said, "we were asked to prove this statement clearly." Proofs can be written very clearly in prose; and they can be written inscrutably in "symbolic" form. I decided the best way to instruct you is by example.

**Proposition** The sets $ A = \{ 12a + 4b | a,b \in \mathbb{Z} \} $ and $ B = \{ 4c | c \in \mathbb{Z} \} $ are equal. 

**Proof** To prove this proposition, we need to show that $ A \subseteq B $ and $ B \subseteq A $. 

To show that $ A \subseteq B $, select any $ x \in A $. By definition, we can find $ a $ and $ b $ in $ \mathbb{Z} $ such that $ x = 12a + 4b $. Letting $ c = 3a+b $ we have $ x = 4c $ by the distributive property, whence $ x \in B $. 

To show that $ B \subseteq A $, select any $ x \in B $. By definition there exists $ c \in \mathbb{Z} $ such that $ x = 4c $. Let $ a = 0 $ and $ b = c $, and it is trivial to see that $ x = 4b $. Thus, $ x \in A $, completing the proof. $ \Box $

I hope I have clarified what I meant in my comments above.
The proof is correct. In the second case, you can take $b=c$ and $a=0$ to make it more explicit.