Riemann integral and Lebesgue integral $f:R\rightarrow [0,\infty)$ is a Lebesgue-integrable function. Show that
$$
\int_R f \ d m=\int_0^\infty m(\{f\geq t\})\ dt
$$
where $m$ is Lebesgue measure.
I know the question may be a little dump.

First note that
$$
\int_0^{f(x)}1\,{\rm d}t
~=~
f(x)
$$
since $f(x)$ is constant with respect to variable $t$, so
$$
\int_{\mathbb R}f(x)\,{\rm d}m
~=~
\int_{\mathbb R}
\left(
  \int_0^{f(x)}1\,{\rm d}t
\right){\rm d}m
$$
Now let $\chi_A(x)$ be the characteristic function of the set $A$, i.e.
$$
\chi_A(x) =
\begin{cases}
1 & \text{if }x\in A \\
0 & \text{if }x\notin A
\end{cases}
$$
so that the previous expression turns out to be equal to
$$
\int_{\mathbb R}f(x)\,{\rm d}m
~=~
\int_{\mathbb R}
\left(
  \int_0^{+\infty}\chi_{[0,f(x)]}(t)\,{\rm d}t
\right){\rm d}m
$$
Now the crucial observation is that since $f(x)\geq 0$ then
$$
\chi_{[0,f(x)]}(t)
~=~
\chi_{\{f\geq t\}}(x)
\quad \forall t\geq 0
$$
Now, going back to the integral,
$$
\int_{\mathbb R}f(x)\,{\rm d}m
~=~
\int_{\mathbb R}
\left(
  \int_0^{+\infty}\chi_{\{f\geq t\}}(x)\,{\rm d}t
\right){\rm d}m
~=~
\int_{\mathbb R}
m\left(
 \{f\geq t\}
\right){\rm d}t
$$
