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: 
Your proof is correct. Although, you should focus on the presentation of your proof.

> 
> $RTP$ : { $12a+4b$ : $a,b ∈ $ Z} = { $4c$ : $c ∈ $ Z}
> 
> 
> 

How about labeling

$$A = \{ 12a+4b : a,b \in\mathbb Z\}$$
$$C=\{ 4c : c \in\mathbb Z\}$$

and then starting the proof with we would like to show that $A=C$. Therefore, we need to show that $A\subseteq C$ and $C\subseteq A$.

> 
> Solution : Let's first show that $LHS$ is a subset of our $RHS$.
> 
> 
> x ∈ { $x ∣4(3a+b)=x$ : a,b ∈ Z} → x ∈ { $x ∣4c=x$ : c ∈ Z}
> 
> 
> 

Correct, assume that $x\in A$. Then $x=12a+4b$ for some $a,b\in\mathbb Z$. To show that $A\subseteq C$, you need to show that $x\in C$. As $x=12a+4b=4(3a+b)$, you have correctly deduced that $x\in C$ when $c= 3a+b$.

> 
> Now let's show that $RHS$ is a subset of our $LHS$
> 
> 
> x ∈ { $x ∣4c=x$ : c ∈ Z} → x ∈ { $x ∣4(3a+b)=x$ : a,b ∈ Z}
> 
> 
> 

Assume that $x\in C$. Then, $x=4c$ for some $c\in\mathbb Z$. To show that $C\subseteq A$, you need to show that $x\in A$. Let $a,b\in\mathbb Z$ be such that $a=0$ and $b=c$. Then, it is clear that $x=4b$. Therefore, $x\in A$.

You can end the proof with a sentence stating that as we have shown that $A\subseteq C$ and $C\subseteq A$, we have that $A=C$.
The proof is correct. In the second case, you can take $b=c$ and $a=0$ to make it more explicit.