# Are Specific Facts about the Riemann Integral Logically Required?

This question is somewhat in the spirit of this one in that I am trying to understand the most efficient path to the major integral theorems (Fubini, change of variables, etc).

My question is this: If one proves theorems regarding the Lebesgue integral is it not so that analogous theorems on the Riemann integral are immediate corollaries? So, for example, can one prove that the change-of-variables theorem holds for the Lebesgue integral on Lebesge-mesurable sets and then take as a corollary the corresponding change-of-variables theorem for the Riemann integral on, say, Jordan-measurable sets? Are there cases where this approach would not hold?

Note, I am not really interested in the argument of whether one should or should not "learn" the Riemann integral or whether a given approach is pedagogically sound; I'm really only interested in "logical efficiency", for lack of a better way to state it.

Update 12/28/2011 I am providing more background for this question in the hope that the additional details will allow someone to provide some direction. I have also added the self-learning tag to more accurately reflect the intent. I am asking this question, basically, to figure out if I really need to learn the theory of the classical Riemann integral in multiple dimensions considering the fact that I plan to study the theory of the Lebesgue integral. I know, of course the basics of the $n$-dimensional Riemann integral including the statements/application of Fubini and Change of Variables. Learning the proofs of these theorems, especially Change of Variables, is a fair amount of work.

This difficulty of this is compounded by the fact that every author seems to have their own specialized notations and techniques that differ in every imaginable way. Binary grids, dyadic cubes, generalized boxes, etc. Yes, the basic idea is that we are creating a partition and inspecting the upper/lower sums for equality as the mesh size tends to $0$, but, again, there seems to be no common approach with regards to the details. Furthermore, there seems to be no common agreement on what "volume" is. Some texts even define volume in terms of the integral over a unit cube/square/box etc.

In contrast, the theory of Measure and the Lebesgue integral seems very clean and tight and there is a much higher level of consistency among presentations. The results, as I understand it, are more widely applicable and considering the fact that I want to become proficient in this theory anyway, I'm trying to understand if I need to go through all of the gory details of the Riemann integral. I already know how to apply the Riemann theory of integration to solve concrete integration problems.

So, finally, what would I lose if I ignored most of the theory of the Riemann integral and simply learned the Lebesgue integral (which I find to be a considerably more pleasant task!)?

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Certainly you need whatever theorems are required to show that the Riemann integral agrees with the Lebesgue integral... – Qiaochu Yuan Dec 20 '11 at 22:26
When a proper Riemann integral of $f$ exists it is integrable and its integral coincides with the Riemann integral, right? – Jonas Teuwen Dec 20 '11 at 23:15
Well, sometimes the arguments used to prove theorems about the Lebesgue integral can be used verbatim to prove the same facts for the Riemann integral (if they are at all true for it). For instance, to prove that $x \mapsto \int_0^x f(t)dt$ is a continuous function (for $f$ integrable) is essentially the same whether the integral is Lebesgue or Riemann... – Mark Dec 29 '11 at 6:30
It's true that the Lebesgue integral generalizes the Riemann integral but when actually computing Lebesgue integrals, oftentimes you will use Riemann integration in your computation. This occurs a lot when using the convergence theorems of Lebesgue integration which are fundamental tools in the theory. – tomcuchta Dec 29 '11 at 19:38

## 1 Answer

One difference between the Riemann and Lebesgue integrals is this: a sequence of Riemann-integrable functions can converge to a function that is not Riemann-integrable. The theorems apply in cases where all functions involved are Riemann-integrable, but if I'm not mistaken, the proofs are more straightforward if Lebesgue's definition is used.

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