Lebesgue Integral – Why the Undergraph Definition is Rare

integrationlebesgue-integralreal-analysissoft-question

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?

Best Answer

There are two problems with the undergraph definition:

  • 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.

So, why this abstract definition? Simplicity.