Riemann integral and Lebesgue integral

Question

$f:R\rightarrow [0,\infty)$ 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.

Answer 1

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, $$ \int_{\mathbb R}f(x)\,{\rm d}m ~=~ \int_{\mathbb R} \left( \int_0^{+\infty}\chi_{\{f\geq t\}}(x)\,{\rm d}t \right){\rm d}m ~=~ \int_{\mathbb R} m\left( \{f\geq t\} \right){\rm d}t $$

Answer 2

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 $(x,t) \in \R \times [0,\infty)$.