6
$\begingroup$

Recently I have learned that at some math department the introductory course to Lebesgue integration not obligatory. Thus in another course on introduction to Hilbert spaces the $L^2(0,1)$ space is defined as the abstract completion of the space of continuous functions with respect to the norm defined by the Riemann integral of the square of a function.

I did not know that such an approach is possible. This motivated the question in the title. Particularly I would like to address it to algebraists and differential geometers, in other words whether the Riemann integral is sufficient for their professional purposes and the Lebesgue integral may be omitted from their background.

Improve this question
$\endgroup$
13
  • 3
    $\begingroup$ Lebesgue integral is not necessary as long as you only deal with continuous functions, at least that;s what I'm tempted to believe. $\endgroup$
    – Wojowu
    Commented Dec 23, 2016 at 14:05
  • 2
    $\begingroup$ If you take the completion of the continuous functions w. r. t. the $L^2$ norm, it is not clear that the resulting space can still be identified with a function space. $\endgroup$
    – Dirk
    Commented Dec 23, 2016 at 16:46
  • 4
    $\begingroup$ Even people who are not analysts may need to use analysis. For example, in number theory one may need to work with infinite (Euler) products or integrate complex-valued functions on real/complex/$p$-adic/adelic topological groups with respect to Haar measure and invoke the dominated convergence theorem, so the powerful theorems about integration based on Lebesgue's ideas are definitely needed. To integrate $p$-adic valued functions on $p$-adic groups, a basic approach is through $p$-adic valued measures, which are inspired by classical measure theory. $\endgroup$
    – KConrad
    Commented Dec 23, 2016 at 17:02
  • 7
    $\begingroup$ To play devil's advocate for a minute: I learned a fair bit of (soft) functional analysis without needing to use the "proper" construction of $L^2[0,1]$, and instead regarding it as "the" completion of $C[0,1]$ with respect to the natural inner product. We should also distinguish between needing to know the statement of the Dominated Convergence Theorem, and needing to know how one constructs Lebesgue measure using e.g. Caratheodory extension $\endgroup$
    – Yemon Choi
    Commented Dec 23, 2016 at 17:42
  • 6
    $\begingroup$ This question seems kind of pointless to me. You're going to get some people saying "I am an algebraist and never use this", some others saying "here's this problem in algebra that uses it", someone replying "but that's not really algebra", etc, etc. And if you're trying to get a sense of what fraction of algebraists use it, that's a poll, and polls are off topic. $\endgroup$
    – Nate Eldredge
    Commented Dec 23, 2016 at 19:40

0

Reset to default

Browse other questions tagged .