How to show that $\int u d \mu = \int_0^{\infty} \mu (u^{-1}((t,\infty))) d t$ for $u \colon X \to \mathbb{R}^+$ measurable? On a measure space $(X, \mathscr{A}, \mu)$ we have a measurable funtion $u \colon X \to \mathbb{R}^+$, and want to show that
$$\int u d \mu = \int_0^{\infty} \mu (u^{-1}(t,\infty)) dt.$$
I've been given the hint to try for simple functions first, and someone said to try MCT, but I simply can't get it to go anywhere still:
What I've tried:
We know that $\mu (u^{-1}(t,\infty)) = \int 1_{u^{-1}(t,\infty)} d\mu$ is decreasing and
\begin{align*}
\lim_{n \to \infty} \mu (u^{-1}(-n, \infty))
&= \lim_{n \to \infty} \int 1_{u^{-1}(-n,\infty)} d\mu\\
&= \int 1_X d\mu\\
&= \mu(X).
\end{align*}
\begin{align*}
\lim_{n \to \infty} \mu (u^{-1}(n, \infty))
&= \lim_{n \to \infty} \int 1_{u^{-1}(n,\infty)} d\mu\\
&= \int 1_\varnothing d\mu\\
&= \mu(\varnothing)= 0.
\end{align*}
If $u= \sum_{n=1}^m c_n 1_{A_n}, \quad c_i \geq 0$, we have
$$ \int u d\mu = \sum_{n=1}^m c_n \mu(A_n). $$
I don't see how to connect the two.

\begin{align*}
\int_0^\infty u(x)\mu(\mathrm d  x)&=\int_0^\infty \int_0^{u(x)}\,\mathrm d t\,\mu(\mathrm d x).\\
&=\int_0^\infty \int_0^{\infty }\boldsymbol 1_{\{u(x)>t>0\}}\,\mathrm d t\,\mu(\mathrm d x)\\
&\underset{\text{Fubini}}{=}\int_0^\infty \int_0^\infty \boldsymbol 1_{\{u(x)>t\}}\,\mu(\mathrm d x)\,\mathrm d t\\
&=\int_0^\infty \mu\{u(x)>t\}\,\mathrm d t.
\end{align*}
