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: 
You want to actually find $a$ and $b$ such that $3a+b=c$, not state that they exist without proof. You are not choosing $c$ for the second part, as $c$ is already given.
The proof is correct. In the second case, you can take $b=c$ and $a=0$ to make it more explicit.