2
$\begingroup$

This wiki article opens with the two sentences

In measure theory, Lebesgue's dominated convergence theorem provides sufficient conditions under which almost everywhere convergence of a sequence of functions implies convergence in the $L_1$ norm. Its power and utility are two of the primary theoretical advantages (my boldface) of Lebesgue integration over Riemann integration.

I thought the term "theoretical advantages" was interesting. I only know enough about Lebesgue integration to be able to understand terms like "Lebesgue measurable" when they crop up, and always thought of Lebesgue integration as a way of making a wider class of functions integrable than would be for the Riemann integral alone.

Is there a generally agreed upon set of "theoretical advantages" like this for Lebesgue integration?

$\endgroup$
2
$\begingroup$

One theoretical advantage comes from the fact that Lebesgue integration makes certain function spaces complete.

For example, consider the real vector space $\mathcal{R}$ of real-valued Riemann integrable functions on $[0,1]$, with addition and scalar multiplication defined in the usual way. One can define an inner product and associated norm on $\mathcal{R}$ in the standard way via $$\langle f,g\rangle = \int_{0}^1 f(x)g(x)\,dx.$$

As it turns out, the space $\mathcal{R}$ is not complete (which is the essential obstacle for it being a Hilbert space), because there are Cauchy sequences of functions $f_1,f_2,\ldots\in\mathcal{R}$ which converge to a function that is not bounded and thus not Riemann integrable.

The Lebesgue integral gives us a way to form a completion of $\mathcal{R}$, the Lebesgue class $L^2([0,1])$. The introduction of this function space was convenient in a way similar to the way we introduced the real numbers as a completion for the rationals.

Source: Stein and Shakarchi, Volume I: Fourier analysis.

Also, Volume III by the same authors should have some nice sources.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.