Proving the equality $\int |f|^p = \int_0^\infty pt^{p-1} \mu(\{x: |f(x)|\geq t\})) dt$ for $\sigma$-finite measure

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For some $\sigma$-finite measure space $(X,\mathcal{A},\mu)$ and a measurable function $f: X \to \mathbb{C}$, it is true that:

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

for positive, finite $p$.

I am pretty lost with this problem, and have several questions.

We can use the definition of the measure in terms of an integral to get

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

I assume the integral on the right-hand side is a Riemann integral, but can I justify doing the change of variables $u = t^{p-1}$ to obtain the following integral?

$$\int_0^\infty \mu(\{x:|f(x)|^p\geq u\}) du$$

My problem is that I have a Lebesgue integral as the integrand of (what I assume to be) the Riemann integral.

Then, how is this last integral in terms of $du$ equal to the $L_p$ norm?

Also, is it better to prove this for $f:X\to \mathbb{R}$ and then extend it to $\mathbb{C}$?

It is not assumed that$\mu$ is the Lebesgue measure (at least there's no reference to that in the question).