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.

We have, using Fubini and denoting by$\def\o{\mathbb 1}\def\R{\mathbb R}$ $\o_A$ the indicator function of a set $A \subseteq \R$ 
\begin{align*}
  \int_\R f(x)\, dx &= \int_\R \int_{[0,\infty)}\o_{[0,f(x)]}(t)\, dt\,dx\\
        &= \int_{[0,\infty)} \int_\R \o_{[0,f(x)]}(t)\, dx\, dt\\
        &= \int_{[0,\infty)} \int_\R \o_{\{f \ge t\}}(x)\, dx\, dt\\
        &= \int_{[0,\infty)} m(\{f\ge t\})\, dt.
\end{align*}
For the third line note that 
\begin{align*}
  \o_{[0,f(x)]}(t) = 1 &\iff 0 \le t \le f(x)\\
                       &\iff f(x) \ge t\\
                       &\iff x \in \{f\ge t\}\\
                       &\iff \o_{\{f\ge t\}}(x) = 1
\end{align*}
and hence $\o_{[0,f(x)]}(t) =  \o_{\{f \ge t\}}(x)$ for all $(x,t) \in \R \times [0,\infty)$.
