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In the development of complex analysis you use Riemann integration and not Lebesgue integration to define line integrals. My questions are:

Are the theories developed the same? (i.e. does it not matter which integral you use in the development? Since all the functions usually involved are analytic or meromorphic can you use things such as analytic is equivalent to having a power series representation and uniform convergence within the radius of convergence to somehow show that the choice doesn't matter. I feel as if the function involved is analytic and has a finite radius of convergence this should be the case but I'm not so sure about what would happen if the function was meromorphic and/or has an infinite radius of convergence (Am I on the right track?))

If the theories developed are the same, does it become significantly easier to develop the theory with Riemann integration rather than Lebesgue integration.

If they are not the same, what are examples to show that show they are different?

UserB1234
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2 Answers2

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It does not matter much, since most of the things one integrates are a-priori continuous and compactly supported. For that matter, one could easily abstract the properties of "integrals" one needs, without specifying a construction of such integrals.

Unsurprisingly, at the time Cauchy developed the basic ideas, there was scarcely any formalization of any notion of "integral", but "everyone knew how they behaved (in nice circumstances)". By later in the 19th century, the formalization of integrals as Riemann integrals more-than-sufficed for basic complex analysis, although occasionally things like Lebesgue Dominated Convergence or Monotone Convergence would make things simpler to explain.

In terms of textbooks and coursework or logical development of any sort, it is obviously simpler to develop basic complex analysis early, as soon as one has any reasonable notion of integral, rather than waiting for a more sophisticated notion (such as Lebesgue's), because the issues addressed in the more sophisticated scenarios mostly are irrelevant to basic complex analysis.

Also, until relatively recently, complex analysis was often studied by engineering and physics students who had most definitely not encountered Lebesgue integration, but who had seen some version of Riemann's construction. So, again, since Riemann's construction more than suffices, there was no reason to add "burdens" for this population.

paul garrett
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  • Thank you. You said it didn't matter much. Can you think of an example where it does matter? – UserB1234 Dec 23 '13 at 13:59
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    @DanulG, as mentioned in Daniel Fischer's answer, in fact the strict notion of Lebesgue integral sometimes fails while the Riemann integral is ok! (Rather than, as one might imagine, the opposite.) On another hand, as in Gerald Edgar's remark, apparently Fatou prove "his" lemma to address some technical issues in complex analysis. Things start to become trickier (as Daniel Fischer notes) when looking at boundary values of holomorphic functions, e.g., at the edge of disk of convergence, maybe merely distributions, not classical functions. But then it's not about measure, but more general. – paul garrett Dec 23 '13 at 14:20
  • It matters when you are looking at the behavior of an analytic function on the boundary of its domain, for example. For example, if f is bounded and holomorphic on the open unit disk, then what can you say about the radial limits of f? Boundary behaviors require full use of Lebesgue integration with respect to measures. – Disintegrating By Parts Dec 23 '13 at 18:03
  • @T.A.E., I think measure theory is inadequate to talk about boundary values. Distributions, and/or hyperfunctions (almost by definition) are necessary. To my mind, this actually does illustrate the point that, to address issues of "modern analysis", having Lebesgue's integration theory is just a step. Need Sobolev and Schwartz, too. – paul garrett Dec 23 '13 at 18:11
  • Hardy spaces, and log-summable classes do include a lot of examples, and those are handled well by Lebesgue integration. – Disintegrating By Parts Dec 23 '13 at 20:55
  • @T.A.E., yes, certainly, but I merely wanted to make the point that it is very easy to write down things whose boundary values exist as distributions, but not as measures: on the unit disk, $f(z)=\sum_{n\ge 0} c_n,z^n$ where $c_n$ are of polynomial growth, and do definitely grow without bound. As to why one might care or not, that's a different story. – paul garrett Dec 23 '13 at 21:01
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Mostly it doesn't matter. Typically, the paths of integration are piecewise continuously differentiable, and the integrated functions are continuous, so for these cases both theories are equivalent. The Riemann integral is simpler, and since not everybody hearing complex analysis may be familiar with the Lebesgue integral, using the Riemann integral may have pedagogical advantages.

There are cases where the two theories differ, though. For example when considering integrals like

$$\int_{-\infty}^\infty \frac{\sin x}{x}\,dx,$$

in the Riemann theory this exists as an ordinary improper integral, since

$$\lim_{\substack{R\to\infty\\S\to -\infty}} \int_S^R \frac{\sin x}{x}\,dx$$

exists when the bounds approach their limits independently. In the Lebesgue theory, the integral does not exist, one has to interpret it as a principal value integral or explicitly as a limit of integrals over bounded intervals.

That is not a big problem, however.

On the other side, if one considers Cauchy integrals of boundary values of holomorphic functions (Hardy spaces for example), one needs the Lebesgue integral (or some other integration theory that makes them well-defined), since the boundary values in general are not regular enough to be Riemann-integrable, but they are Lebesgue-integrable for interesting classes of functions.

Daniel Fischer
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    In fact, Fatou proved his famous lemma for Lebesgue integration because he needed it in a paper on complex analysis. – GEdgar Dec 23 '13 at 14:15
  • Interesting. Do you know what for in particular? – Daniel Fischer Dec 23 '13 at 14:17
  • I understand Fatou's Lemma was a lemma for Fatou's Theorem: under mild conditions, a holomorphic function in the unit disk has a radial limit at almost every boundary point of the disk. – GEdgar Dec 23 '13 at 14:54
  • See https://hsm.stackexchange.com/questions/12759/history-of-fatous-lemma for a discussion of the context in which Fatou proved his famous lemma and the paper where it appeared. – KCd Jan 29 '21 at 08:01
  • Thanks for the reference, @KCd. Hadn't seen that. – Daniel Fischer Jan 29 '21 at 13:20