18
$\begingroup$

I am getting pretty frustrated with the Lebesgue integral mainly because it seems highly impractical to calculate anything non-trivial. Whenever I look for a concrete calculation all I see are encomiums about how wonderful it is and then invariably the only concrete calculation is the Dirichlet function where unsurprisingly the measures are easy to calculate. When a run of the mill function is offered to be calculated ,I have seen one of two kinds of responses:

1) The answer involves a trick that can't be generalized

2) The answer given is " The function is Riemann integrable so just use that" i.e forget about the Lebesgue integral.

The only method I have seen that aspires to practicality is to use the monotone convergence theorem i.e get a bunch of simple functions whose limit is the function you want to integrate. Integrate them and take the limit. I have tried this for $x^2$ and I run into hard sums which are summed by guess what...help of the Riemann integral.(could be that I chose an inconvenient set of simple functions but that partly proves my point-very easy to make things complicated)

So is the Lebesgue integral mostly used in formal situations and then occasionally some highly pathological function is pulled to justify the work? Are there examples where the Lebesgue integral is of practical importance and there can be no recourse to the Riemann integral? Highly discontinuous functions are not welcome.

$\endgroup$
14
  • 4
    $\begingroup$ @MattSamuel This quote comes to mind: "Does anyone believe that the difference between the Lebesgue and Riemann integrals can have physical significance, and that whether say, an airplane would or would not fly could depend on this difference? If such were claimed, I should not care to fly in that plane." - Richard Hamming $\endgroup$
    – user169852
    Jul 27, 2016 at 0:15
  • 7
    $\begingroup$ It's llike saying you're disappointed with real numbers, because the only numbers that ever come up in practice are rational. (It's exactly like that - $L^1([0,1])$ is the completion of $C([0,1])$ in a certain metric, just as the reals are the completion of the rationals.) Every number that actually gets calculated is rational - regardless, we couldn't do much without the reals. $\endgroup$
    – David C. Ullrich
    Jul 27, 2016 at 0:34
  • 4
    $\begingroup$ Nah, we just like to torture students with the stupid thing. Looks like it's working ... $\endgroup$
    – zhw.
    Jul 27, 2016 at 0:38
  • 4
    $\begingroup$ "I do not accept the analogy with real numbers." Imagine my surprise. Your version of the proper analogy more or less shows you don't understand why the reals are important. Say we like to solve equations. Of course we can't solve them exactly, but we're happy approximating the solutions to some number of decimal places. Fine. Wouldn't it be nice to know that those equations have solutions? So that there actually is something we're approximating? $\endgroup$
    – David C. Ullrich
    Jul 27, 2016 at 3:59
  • 5
    $\begingroup$ "I get the impression from the answers that I am being naive for expecting a theory of integration to help in integrating." Starting with that impression is perfectly reasonable, if naive. But the fact that you still seem to have that expectation, after it's been explained that that's not the point, is curious. Look. Helping us integrate specific functions is not the point to the Lebesgue integral. Nobody said it was. $\endgroup$
    – David C. Ullrich
    Jul 27, 2016 at 4:03

1 Answer 1

Reset to default
7
$\begingroup$

Andrew D Lewis wrote a nice essay giving what he considers to be "The correct defence of the Lebesgue integral." Here is the Abstract:

It is well-known that the Lebesgue integral generalises the Riemann integral. However, as is also well-known but less frequently well-explained, this generalisation alone is not the reason why the Lebesgue integral is important and needs to be a part of the arsenal of any mathematician, pure or applied. Those who understand the correct reasons for the importance of the Lebesgue integral realise there are at least two crucial differences between the Riemann and Lebesgue theories. One is the difference between the Dominated Convergence Theorem in the two theories, and another is the completeness of the normed vector spaces of integrable functions. Here topological interpretations are provided for the differences in the Dominated Convergence Theorems, and explicit counterexamples are given which illustrate the deficiencies of the Riemann integral. Also illustrated are the deleterious consequences of the defects in the Riemann integral on Fourier transform theory if one restricts to Riemann integrable functions.

$\endgroup$

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.