How can I prove that this integrals are equal? [duplicate]

The quick way is to use Fubini:

$$ \begin{align} & \int_{(0,\infty)} \mu(\{f>t\}) \, dt=\int^\infty_0 \int _X \chi _{(\{f>t\})} \, d\mu \, dt = \int_X \int^\infty_0 \chi _{(\{f>t\})} \, dt \, d\mu \\[10pt] = {} & \int_X \int_0^{f(x)} 1 \, dt \, d\mu = \int_X f(x) \, d\mu \end{align} $$

but it's kind of cheating I think because Fubini is non-trivial.

But here's one very basic way:

Define $E_{nk} = \left\{x \in X\mid f(x) > \frac{k}{2^n}\right\}$

Then set $f_n = \frac{1}{2^n} \sum_{k=1}^{4^n} \chi _{E_{nk}}$ so that

$$\tag1\int_X f_n \, d\mu = \frac{1}{2^n} \sum_{k=1}^{4^n} \mu E_{nk}$$

Now, $f_n$ increases to $f$ so the MCT says $$\tag2\int_X f\,d\mu = \lim_{n\to\infty} \int_X f_n\,d\mu.$$

As $\left \{ x\in X\mid f(x)>t \right \} = \bigcup_{n \in \mathbb{N}} \left \{ x \in X\mid f_n(x)>t \right \}$

we have $\mu\left \{ x\in X\mid f(x)>t \right \} = \lim_{n\to\infty} \mu\left \{ x \in X\mid f_n(x)>t \right \}$

so we may apply MCT again:

$$\tag3\int_0^\infty \mu \left \{ x\in X\mid f(x)>t \right \} \, dt=\lim_{n\to \infty} \int_0^\infty \mu \left \{ x\in X\mid f_n(x)>t \right \} \, dt$$

Observe now that $\mu E_{nk}=\mu\left\{x \in X\mid f(x) \gt \frac{k}{2^n} \right\} = \mu \left \{ x\in X\mid f_n(x)>t\right \}$ whenever $1\leq k \leq 4^n$ and $\frac{k-1}{2^n} < t\leq \frac{k}{2^n}$ and $0$ if $t>4^n$.

But then this means that $$\tag4\int_0^\infty \mu \left \{ x\in X\mid f_n(x) > t\right \} \, dt = \frac{1}{2^n} \sum_{k=1}^{4^n} \mu E_{nk}$$

$(1)$ and $(4)$ give $\int_X f_n \, d\mu = \int_0^\infty \mu \left \{ x\in X\mid f_{n}(x)>t\right \} \, dt$

and now taking limits and invoking $(2)$ and $(3)$ give the result.