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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?

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    I'd say a better option for replacement of the Lebesgue integral would be the generalized Riemann integral: https://en.wikipedia.org/wiki/Generalized_Riemann_integral2012-07-27
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    @Joe: how does the Henstock-Kurzweil integral generalize to an arbitrary measure space?2012-07-27
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    @QiaochuYuan It does not. I am only referring to $\mathbb{R}^n$.2012-07-27
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    I am no expert, but I wonder why you call the Daniell integral 'functional analytic'. Like the Lebesgue integral, it is about (say real-valued) functions on a *set*, not necessarily a topological space. I do think the Daniell approach is most often used by people who work in functional analysis, and some of those may restrict themselves to functions on a space (I believe Bourbaki is an example) Perhaps we need to define *Daniell integral* precisely; it seems several related things are called like that. But I am quite interested in an expert answer about these matters!2012-07-27
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    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.2012-07-27
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    @Joe: okay, but that doesn't seem relevant to the question. The OP's question as I read it is about integration in general, and the Henstock-Kurzweil integral isn't relevant to that setting.2012-07-28
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    @Joe: When the question is about comparing two absolute integrals, expounding the virtues of non-absolute integration is kinda off-topic.2012-07-28
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    @Leonid I completely and totally agree. Measure theory is a much more flexible and general tool for general mathematics then the functional-analytic approach.2012-07-28
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    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](http://books.google.com/books?id=3njnPQAACAAJ)* for a deeper look.2012-07-28
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    @t.b.: since you added the tag *functional-analysis*, could you explain to me what is so 'functional analytic' about the Daniell-approach (see my previous comment, in which I btw mentioned Bourbaki :) )2012-07-28
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    @wildildildlife Constructing the integral by extending a functional sounds very functional analytic to me.2012-07-28
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    @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.2012-07-28
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    @wildildildlife: What Michael said :) The space $H$ on the Wikipedia page on Daniell's integral is by hypothesis a [Riesz subspace](http://en.wikipedia.org/wiki/Riesz_space) 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.2012-07-28
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    @Michael: thanks for the pointer.2012-07-28
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    @wildildildlife: To bring the topology into play (sorry, I missed that part of your question). Then it is natural to require that your integral be defined on some subspace of the space of continuous functions. To get a decent measure you need to strengthen the sequential smoothness condition to smoothness. In the locally compact case you then get a Radon measure. Conversely every smooth functional allows you to introduce a topology on $X$ such that the corresponding measure is quasi-Radon. See ch. 7 of the book I mentioned above or 436 in vol 4I of Fremlin's measure theory (available online).2012-07-29
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    @t.b.: thanks for your answer. To me it seems the study of Riesz spaces has so many technical details. [This book](http://www.ams.org/bookstore-getitem/item=SURV-105) is on my shelf, but I haven't yet had the motivation to read it.2012-07-29

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