I am confused about which, if any, of these integrals is the most general, assuming we allow "improper" versions of each (where applicable).

The reason I am confused is by this passage in Wikipedia on Riemann-Stieltjes integrals:

An important generalization is the Lebesgue–Stieltjes integral which generalizes the Riemann–Stieltjes integral in a way analogous to how the Lebesgue integral generalizes the Riemann integral. If improper Riemann–Stieltjes integrals are allowed, the Lebesgue integral is not strictly more general than the Riemann–Stieltjes integral.

I am in no way well-versed in measure theory, but I was under the impression that the Lebesgue integral is considered the most general form of the above integrals. However, the above seems to call this into question.

Also, I recently had the chance to peruse Abbott's "Understanding Analysis", where he introduces the "Generalized Riemann Integral"...a version I have never heard of before. A really nice exposition of it is given by Bartle, where he argues that we should stop relying on the Lebesgue integral as the "most general" and instead switch to this new type of extended Riemann integral (the key element of which is the use of a non-constant gauge over the partitions in the Riemann sum).

I was introduced to the Riemann-Stieltjes integral while studying from Ross' "Elementary Analysis", where he devotes a considerable amount of real-estate to the, quite useful, generalization of the Riemann integral. In fact, it seems that for the vast majority of probability and statistics applications, the Riemann-Stieltjes integral is sufficient for calculations, without the need for extensive measure-theoretic constructions.

It turns out that someone has gone the extra step to generalize the Riemann-Stieltjes integral using a similar construction to the Generalized Riemann integral.

Based on Bartle's discussion of the Genearlized Riemann integral, it would appear that this Generalized Riemann-Steiltjes integral can encompass more integrand and domains than any other. However, this is just my impression on reading this.

I would appreciate if someone could explain which of the integrals I've discussed is, in fact, the most general in the following, hopefully precise, sense:

Let $I_i(f,g)$ be an integral of type $i$, with integrand $f$ and integrator $g$ (e.g., a normal Riemann integral has integrator $g(x)=x$). Finally, let the notation $f \in I_i$ mean that there there exists an integrator such that $f$ is integrable under $I_i$.

I will define $I_i$ as being more general than $I_j$ ($I_j \prec I_i$) iff:

$$f \in I_i\;\forall f\in I_j\;\textrm{ and}\; \exists h \in I_i: h \notin I_j$$

I would consider $I_j$ equivalent to $I_i$ iff:

$$ I_j \nprec I_i \;\textrm{ and } I_i \nprec I_j$$


So, given the above definition for ordering by generality, what is the relationship between the improper Riemann-Stieltjes, Generalized Riemann-Stieltjes, and the most general form of the Lebesgue-Stieltjes integral?

  • $\begingroup$ Remember, of course, that the Lebesgue integral works in a much more general setting than the Riemann integral: any measure space. $\endgroup$
    – Chappers
    Oct 4, 2015 at 15:17
  • $\begingroup$ @Chappers yea, I guess if we're talking about arbitary measure spaces, then I guess you can't get away from using a measure as the integrator.....does that negate the statement in the wikipedia article? $\endgroup$
    – user237392
    Oct 4, 2015 at 15:22
  • $\begingroup$ @Chappers or...is it the case that the generalized Riemann-stieltjes integral and Lebesgue-Stieltjes are equivalent under my definition? $\endgroup$
    – user237392
    Oct 4, 2015 at 15:23
  • $\begingroup$ This sort of discussion of the generality of integrals normally refers solely to their use on $\mathbb{R}$: I suppose what I'm trying to say more concretely is that because the Stieltjes definitions require some sort of path be defined, you can't make an equivalent of even, say, a surface integral, as far as I'm aware. $\endgroup$
    – Chappers
    Oct 4, 2015 at 15:27
  • $\begingroup$ So Riemann-Stieltjes are solely one-dimensional? $\endgroup$
    – user237392
    Oct 4, 2015 at 15:35

1 Answer 1


No, the Lebesgue integral is not more general than the improper Riemann one, it just has some very nice properties that make it convenient to work with. Remember that, once you define the concept of Lebesgue integrability, an important theorem says that $f$ is Lebesgue integrable if and only if $|f|$ is so. Consider now the function $\Bbb e^{\Bbb i x^2}$: its modulus is $1$, which is clearly not integrable on $\Bbb R$; nevertheless, its improper Riemann integral exists as $\lim \limits _{R \to \infty} \int \limits _{-R} ^R \Bbb e^{\Bbb i x^2} \Bbb d x = \sqrt {\pi \Bbb i}$, so you may still assign a value to it.

As you can see, there are moments when the "humbler" improper Riemann integral is capable of producing better results than the Lebesgue one. Let us see why and when. When mathematicians use the Lebesgue integral, they usually do so in order to use the already established (and very powerful) theory of Lebesgue spaces, which are Banach spaces. Being Banach spaces, we usually use various inequalities regarding their norms; nevertheless, most of our approaches rely on the following starting point: $| \int f | \le \int |f|$ (or something similar). If you think of this, and of the example in the above paragraph, you will see that there is a class of functions (those that are not absolutely integrable) for which these techniques will not produce useful results.

(If you think further about this issue, Lebesgue integrability is like absolute convergence for series: it works well and produces powerful results for many series, but what do you do with the following example: $\sum \limits _n (-1)^n \dfrac {\Bbb e ^{\Bbb i n x}} n$? Absolute integrability will betray you in this case, you'll have to resort to the Abel-Dirichlet test that will confirm the convergence of the above.)

To summarize, for absolutely $p$-integrable functions the Lebesgue integral works best, no need to use anything else; for non-absolutely $p$-integrable functions, you'll have to try alternative approaches, such as improper Riemann integrability.

Of course, the above makes sense on spaces of infinite measure. If your space is a compact subset of $\Bbb R^n$ with the usual measure and $f$ is a Riemann integrable function, then it will also be Lebesgue integrable and its two integrals will coincide, which is very nice if you think about it (the advantage of Riemann integrability on compact spaces being that we know how to explicitly compute things, essentially repeatedly simplifying our problem until we may use the Leibniz-Newton theorem).

  • 1
    $\begingroup$ "If your space is a compact subset of $R^n$ with the usual measure, then Lebesgue integrability and Riemann integrability coincide" – Really? What about the characteristic function of the rational numbers which is zero almost everywhere? Perhaps I am completely misunderstanding something, it is many years ago that I learned this stuff :) $\endgroup$
    – Martin R
    Oct 7, 2015 at 14:06
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    $\begingroup$ @MartinR: Your example is not Riemann integrable. My statement says that if a function is integrable in both senses, then its Riemann and Lebesgue integrals coincide. $\endgroup$
    – Alex M.
    Oct 7, 2015 at 14:18
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    $\begingroup$ You said "... then Lebesgue integrability and Riemann integrability coincide" which can easily be understood as "f is Lebesgue integrable if and only if it is Riemann integrable". $\endgroup$
    – Martin R
    Oct 7, 2015 at 14:21
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    $\begingroup$ @Bey: No, there are also functions that the Riemann integral fails to integrate, but that the Lebesgue one has no problem with, such as the function that is $1$ on $\Bbb Q$ and $0$ outside of it (this one makes the Riemann integral choke, while its Lebesgue integral is easily seen to be $0$). To summarize, the two types of integrals coincide on a fairly large class of functions, and each one is able to integrate some functions that the other one can't. Therefore, none is more powerful (or more general) than the other, both are important (or else we wouldn't speak about one of them anymore). $\endgroup$
    – Alex M.
    Oct 7, 2015 at 16:09
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    $\begingroup$ @Bey: To complicate things, on locally-compact topological groups there is a third integral, called the Haar integral, that does not necessarily coincide with the other two discussed. As you can see, there are many types of integrals and no general theory of integration. (For instance, there are the Wiener and the Feynman integrals on spaces of curves, that deserve a separate discussion, that do things the ones mentioned previously cannot do.) $\endgroup$
    – Alex M.
    Oct 7, 2015 at 16:12

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