Lebesgue Integral: Integrating the Distribution function

Question

Is there an easy way to see that if $f$ is nonnegative, then

$$\int_E f\,dx=\int_0^\infty |\{x\in E: f(x)>t\}|\,dt$$? (where $|\cdot|$ denotes Lebesgue measure)

My book proves a more general version $\int_E f^p=p\int_0^\infty \alpha^{p-1}\omega(\alpha)\,d\alpha$, where my question is just the special case $p=1$, but the proof is rather long (using Riemann-Stieltjes integral).

Hence, I am curious if there is a shorter proof for my question, or even just an intuitive way to see it non-rigorously.

Thanks for any help.

Answer 1

$$\int_0^\infty |\{x\in E: f(x)>t\}|~dt$$

Let's think about what this integral means. It will be handy to draw a picture of the graph of $y = f(x)$ as you follow along. Let's pretend $f$ is a nicely regular function over some interval $[a,b]$ for the sake of illustration.

$|\{x\in E: f(x)>t\}|$ is the superlevel set of $f$, and you can find it by drawing the horizontal line at $y=t$, then projecting the line segments that get cut off by the graph of $f$ down to the $x$-axis. Then $|\{x\in E: f(x)>t\}|dt$ is a union of infinitesimal rectangles, which lie flat horizontally over the $x$-axis. Integrating over $t\in[0,\infty]$ essentially means summing up the areas of these infinitesimal rectangles.

The intuitive picture is much like Riemann integration, except instead of cutting up the graph of $f$ vertically into infinitesimal slices, you are instead cutting it up horizontally. Where Riemann integration is analogous to a loaf of sliced bread, the Lebesgue integral presented in this form is analogous to a layered cake.

Answer 2

Using just Fubini: \begin{align} \int_E f(x)dx&=\int_E\left(\int_0^{\infty}\chi_{\{t<f(x)\}}(x,t)dt\right)dx\\ &=\int_0^\infty\left(\int_E \chi_{\{t<f(x)\}}(x,t)dx\right)dt\\ &=\int_0^\infty |\{x\in E:f(x)>t\}|dt. \end{align}