# Why is the undergraph definition of Lebesgue integral so rare?

So in Pugh's Real Mathematical Analysis, the initial definition of the Lebesgue integral is as the Lebesgue measure of the undergraph of the function (where the function is nonnegative, with the usual extension to functions whose sign changes). This is extremely intuitive, and immediately justifies all the abstract work of extending our notion of volume with the concept of measure. Another definition is given in the main exposition, and yet another is given in the exercises.

i.e. the undergraph definition is:

$\int_Ef = m(Uf)$, where $Uf = \{(x, y) \mid x \in E, \ \ 0 \leq y \leq f(x)$

Now I'm working through the more advanced Real Analysis by Folland, where the integral is defined as the supremum of integrals of simple functions, whose integrals are fairly obvious. Additionally, in a graduate course, we used an equivalent definition as the Riemann integral of the measure of the superlevel sets, i.e.

$\int_E f = \int_0^\infty m(\{ x \mid f(x) > t\})dt$

Nowhere in the book does it seem to mention the undergraph definition, and googling around it's very hard to find mention of the undergraph definition at all. I realize why the simple function definition might be better to work with in developing the theory (since it often suffices to prove our theorems for simple functions that approximate much uglier ones), but I'm wondering if there's any particular reason why the undergraph definition seems so neglected, given that the main motivation we're given for the very first time we're introduced to the concept of integration is to compute the area below a curve. Perhaps the definition suffers from a serious limitation?

• It seems easier to me to extend the definition using simple functions to more general measure spaces. If $E \in \mathbb{R}$ then $Uf \subset \mathbb{R}^2$, but if $E$ is a subset of some other measure space then it seems like defining $m(Uf)$ would at least require the introduction of product measures. – Antonio Vargas Nov 2 '13 at 18:23
• Is that the extent of it? I can certainly see why having to look at the product space $(X$ x $R)$ would be rather unnatural for real-valued functions with domains outside of $R$, but product measures are pretty important to develop anyway. Still seems better than using the Riemann integral definition as that requires an entire, separate theory of integration. – Saigyouji Nov 3 '13 at 0:56

• If $f: \mathbb R \to \mathbb R$ you would need to define $m(Uf)$. Note that $Uf \subset \mathbb R^2$ whereas you define $m$ as a measure on subsets of $\mathbb R$.
• Even once you have defined the product measure (the one that can measure sets in $\mathbb R^2$, the whole thing being dreadfull) you would need to prove that $Uf$ is measurable.
Wheeden and Zygmund's book does something very clever with this. They construct Lebesgue measure on $\mathbb{R}^d$, then define the integral via the undergraph. Have a look.