Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The Riemann integral is the most common integral in use and is the first integral I was taught to use. After doing some more advanced analysis it becomes clear that the Riemann integral has some serious flaws.

The most natural way to fix all the drawbacks of the Riemann integral is to develop some measure theory and construct the Lebesgue integral.

Recently, someone pointed out to me that the Daniell integral is ‘equivalent’ to the Lebesgue integral. It uses a functional analytic approach instead of a measure theoretic one. However, most courses in advanced analysis do not cover the theory of the Daniell integral and most books prefer the Lebesgue integral.

But since these two constructions are equivalent, why do people prefer the Lebesgue integral over the Daniell integral?

share|improve this question
The selling point of the Daniell integral is that it allows one to avoid abstract measure theory. I don't want my students to avoid abstract measure theory. I want them to learn abstract measure theory, so that they can learn measure-based probability later. –  user31373 Jul 27 '12 at 23:45
@Joe: When the question is about comparing two absolute integrals, expounding the virtues of non-absolute integration is kinda off-topic. –  Michael Greinecker Jul 28 '12 at 0:03
A related question of a similar nature: why do people prefer topological vector spaces over ordered vector spaces? The "Daniell approach" (where are the names of Riesz, Kakutani, Bourbaki, etc.?) is detailed in quite a few mainstream books, e.g. Royden or Pedersen. You may wish to have a look at Fremlin's Topological Riesz spaces and measure theory for a deeper look. –  t.b. Jul 28 '12 at 8:16
@t.b. I think the most important name after Daniell's is the one of Marshall Stone. His "Notes on Integration I-IV" are absolutely seminal. –  Michael Greinecker Jul 28 '12 at 14:18
@wildildildlife: What Michael said :) The space $H$ on the Wikipedia page on Daniell's integral is by hypothesis a Riesz subspace of $\mathbb R^X$ and the integral functional is what is called a positive linear and sequentially smooth functional on it. The study of Riesz spaces is plainly a part of functional analysis since one studies spaces of functions and functionals on them. For some (not entirely transparent) reasons, the order structure on the typical function spaces draws much less attention than their norms or topologies in basic FA-books. –  t.b. Jul 28 '12 at 14:38
show 11 more comments

4 Answers

The following excerpt is from Measure Theory Vol 2 by Vladimir Bogachev:

In the middle of the 20th century there was a very widespread point of view in favor of presentation of the theory of integration following Daniell’s approach, and some authors even declared the traditional presentation to be “obsolete”. Apart the above-mentioned conveniences in the consideration of measures on locally compact spaces, an advantage of such an approach for pedagogical purposes seemed to be that it “leads to the goal much faster, avoiding auxiliary constructions and subtleties of measure theory”. In Wiener, Paley [1987, p. 145], one even finds the following statement: “In an ideal course on Lebesgue integration, all theorems would be developed from the point of view of the Daniell integral”. But fashions pass, and now it is perfectly clear that the way of presentation in which the integral precedes measure can be considered as no more than equivalent to the traditional one. This is caused by a number of reasons. First of all, we note that the economy of Daniell’s scheme can be seen only in considerations of the very elementary properties of the Lebesgue integral (this may be important if perhaps in the course of the theory of representations of groups one has to explain briefly the concept of the integral), but in any advanced presentation of the theory this initial economy turns out to be imaginary. Secondly, the consideration of measure theory (and not only the integral) is indispensable for most applications (in many of which measures are the principal object), so in Daniell’s approach sooner or later one has to prove the same theorems on measures, and they do not come as simple corollaries of the theory of the integral. It appears that even if there are problems whose investigation requires no measure theory, but involves the Lebesgue integral, then it is very likely that most of them can also be managed without the latter.

share|improve this answer
add comment

People aren't going to switch to a new formalism unless they have a compelling reason to. Everyone already knows measure theory (including the professors teaching analysis courses). It is very powerful and suffices for many applications, so until someone convinces the mathematical community that an alternate theory of integration would fix their troubles, the mathematical community is not going to care. The perceived benefit of switching needs to outweigh the transition cost of learning a new formalism.

(I have heard of Mikusinski's approach to the integral, and in fact the first functional analysis course I took used this approach specifically to avoid measure theory. It was not enjoyable. The proofs of the basic theorems involve intricate manipulations of sequences of sequences and I don't remember any of them.)

share|improve this answer
I do not see why the fact that everyone knows measure theory would imply that the Daniell integral (which is based on a functional analytic construction) is not used, since functional analysis is also well-known under professors and it is also quite powerful (like measure theory) –  gifty Jul 27 '12 at 20:56
@gifty: Interesting, "under" is also the right preposition in that place in German (or perhaps you're German, too? :-) -- in English, it's "among professors". –  joriki Jul 27 '12 at 21:02
@gifty: the point is that when people integrate they will speak the language of measure theory. If a professor wants to prepare her students to talk with other mathematicians and do mathematics, measure theory is part of the language that those people will speak. To give them a different formalism is to incur costs on everyone who has to translate between the two formalisms in the name of mathematical communication. These kinds of questions are closely analogous to questions like "why isn't society organized like X rather than Y?" Maybe life would be easier if society really were organized... –  Qiaochu Yuan Jul 27 '12 at 23:57
... like X, but actually transitioning from Y to X might incur significant costs and be deeply unpleasant until the transition is complete, so there's no way it's going to be done in practice without some significant outside pressure (e.g. some global catastrophe). –  Qiaochu Yuan Jul 27 '12 at 23:58
@QiaochuYuan: The problem I have with your answer is that the Daniell integral used to be very fashionable, probably due to the Bourbaki approach to integration. –  Michael Greinecker Jul 28 '12 at 0:06
show 2 more comments

Just a partial answer... As Qiaochu remarks, the "standard viewpoint" will not change without considerable impetus. The fact that some variant of "Lebesgue" integration is not "perfect" doesn't matter: it is "good enough".

Further, I would claim that, in fact, "the integral" people mostly use is not so much formally defined by any particular set-up, but is characterized, perhaps passively, in a naive-category-theory style (or, those might be my words) by what properties are expected. That is, for many purposes, we truly don't care about the "definition" of "integral", because we know what we expect of "integrals", and we know that people have proven that there are such things that work that way under mild hypotheses...

For all its virtues (in my opinion/taste), this "characterization" approach seems harder for beginners to understand, so the "usual" mathematical education leaves people with definitions...

My own preferred "integral" is what some people call a "weak" integral, or Gelfand-Pettis integral (to give credit where credit is due), characterized by $\lambda(\int_X f(x)\,dx)=\int_X \lambda(f(x))\,dx$ for $V$-valued $f$, for all $\lambda$ in $V^*$, for topological vector space $V$, for measure space $X$. This may seem to beg the question, but wait a moment: when $f$ is continuous, compactly-supported, and $V$ is quasi-complete, locally convex, widely-documented arguments (e.g., my functional analysis notes at my web site) prove existence and uniqueness, granted exactly existence and uniqueness of integrals of continuous, compactly-supported scalar-valued functions on $X$. Thus, whatever sort of integral we care to contrive for the latter will give a Gelfand-Pettis integral.

Well, we can use Lebesgue's construction, or we can cite Riesz' theorem, that every continuous functional on $C^o_c(X)$ (Edit: whose topology is upsetting to many: a colimit of Frechet spaces. But, srsly, it's not so difficult) is given by "an integral" (somewhat as Bourbaki takes as definition).

Either way, we know what we want, after all.

An example of a contrast is the "Bochner/strong" integral, which has the appeal that it emulates Riemann's construction, and, thus, directly engages with traditional ... worries? But, after the dust settles, there is still a bit of work to do to prove that the (as-yet-unspoken) desiderata are obtained.

Further, surprisingly often, in practice, the "weak" integral's characterization proves to be all that one really wants/needs! Who knew? :)

share|improve this answer
Just for comparison, I think the real numbers themselves are more often and more easily introduced by characterisation (ordered field, upper bound property) than by definition (Dedekind cuts, equivalence classes of Cauchy sequences, ...). –  Marc van Leeuwen Jul 28 '12 at 9:24
@Marc van Leeuwen: Indeed! Exactly so! –  paul garrett Jul 29 '12 at 0:21
add comment

Daniell apparoch is simpler if done in mikisuinski fashion provided one introduces kernel of Null mappings. Measure theory measure and sigma algebra can be defined automaticaaly . Fubini theorem has simpler proof. but as it has emerged in 21 st century Henstock-Kurzweil integral is much superior to Lebesgue integral. calling it as non absolute integration is misnomer. it is indispensable for mean value theorem and Taylor's formula for vector valued( especially Banach space valued) mappings does not require countably additive measure but only finitely additive, has i addition to monotone convergence the hakes convergence theorem ( includes improper integral). the integral needs very little formalism in euclidean spaces and with formal axiomatic development can be extended to locally compact spaces, complete separable metric spaces. in fact Feynmann path inegrals receive logical treatment only in this approach. furyer abstract formalism probably creates a larger sigma algebra than yielded by Caratheodory construction! a fact overlooked

only the inertia of people in mathematics and Denjoy-perron results lifting madness of henstock-kurzweil theorists have delayed the widespraed use iof hk integral ut let centuries pass one will henstock-kurzweil as main integral.

share|improve this answer
Why is "non-absolute integral" a misnomer? Clearly, absolute integration is extremely important in areas such as probability theory. –  Michael Greinecker Sep 19 '13 at 18:10
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.