In what sense is Lebesgue integral the "most general"?

Question

[ UPDATE: This question is apparently really easy to misinterpret. I already know about things like the Henstock-Kurzweil integral, etc. I'm asking if the Lebesgue integral (i.e. the general measure-theoretic integral) can be precisely characterized as "the most general integral that can be defined naturally on an arbitrary measurable space." ]

Apologies as I don't have that much of a background in real analysis. These questions may be stupid from the point of view of someone who knows this stuff; if so, just say so.

My layman's understanding of various integrals is that the Lebesgue integral is the "most general" integral you can define when your domain is an arbitrary measure space, but that if your domain has some extra structure, like for instance if it's $\mathbb{R}^n$, then you can define integrals for a wider class of functions than the measurable ones, e.g. the Henstock-Kurzweil or Khintchine integrals.

[ EDIT: To clarify, by the "Lebesgue integral" I mean the general measure-theoretic integral defined for arbitrary Borel-measurable functions on arbitrary measure spaces, rather than just the special case defined when the domain is equipped with the Lebesgue measure. ]

My question: is there a theorem saying that no sufficiently natural integral defined on arbitrary measure spaces can (1) satisfy the usual conditions and integral should, (2) agree with the Lebesgue measure for measurable functions, and (3) integrate at least one non-measurable function? Or, conversely, is this false? Of course this hinges on the correct definition of "sufficiently natural," but I assume it's not too hard to render that statement into abstract nonsense.

Such an integral would, I guess, have to somehow "detect" sigma algebras looking like those of $\mathbb{R}^n$ and somehow act on this.

[ EDIT 2: To clarify, by an "integral" I mean a function that takes as input $((\Omega, \Sigma, \mu), f)$, where $(\Omega, \Sigma, \mu)$ is any measure space and $f : (\Omega, \Sigma) \to (\mathbb{R}, \mathcal{B})$ is a measurable function, and outputs a number, subject to the obvious conditions. ]

UPDATE: The following silly example would satisfy all of my criteria except naturality:

Let $(\Omega, \Sigma, \mu)$ be a measure space, let $\mathcal{B}$ denote the Borel measure on $\mathbb{R}$, and let $f : (\Omega, \Sigma) \to (\mathbb{R}, \mathcal{B})$ be a Borel-measurable function. Define

$$\int_\Omega f \, d \mu := \begin{cases} \text{the Khintchine integral} & \text{if } (\Omega, \Sigma, \mu) = (\mathbb{R}, \mathcal{L}, \mu_\text{Lebesgue});\\ \text{the Lebesgue integral} & \text{otherwise.} \end{cases}$$

Answer 1

There are other notions of integral that possess the basic properties you'd expect of an integral but which can be more general than the Lebesgue integral. For instance, the Henstock-Kurzweil (aka Perron or Luzin integral) is a notion of integral that extends the Riemann integral and has different properties than the Lebesgue integral.

Different theories of integration should be compared on the basis of the properties you mention as well as the resulting convergence theorems (i.e., exchangeability of limits and integration), properties of absolute convergence, and the the versions of the fundamental theorem of calculus they permit. Some integrals are designed to have particularly nice properties with respect to, e.g., convergence theorems (i.e., Lebesgue integration), while others are designed to have particularly strong fundamental theorems of calculus (i.e., Henstock-Kurzweil integration). A particularly nice book that studies several integration theories and compares them is this book.

Answer 2

To answer my own question, the answer seems to be "yes," with details given in the following MathOverflow thread, particularly in the answer of G. Rodrigues:

https://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view

Answer 3

The Lebesgue measure is the only complete translation invariant measure up to scalar multiples. This is something you would want for integrals ($I_{[0,1]}$ should have the same integral as $I_{[3,4]}$). If you take completeness and translation invariance as "sufficiently natural" (and I think it is completely natural to do so) then that would be your theorem.

Answer 4

henstock-kurzweil integgration can be done on locally compact hausdroff spacesv and in much easier way on complete separable metrizable spaces ( look at my thesis). regrading general measure space if it is a probability space ( bounded measure) then the lebesgue integral of a real valued function can be recovered from performing gauge integration on range with respect to probabiliy distribution function as shown in the book theory of random variation Patrick muldowney published by john wiley actually feynmann [path integrals and weiner integration the henstock kurzweil integration is integrating a very much wider class of functions due to cancellations in Riemann sums.

for mappings with values in a banach space there is a a large class of nonmeasurable but absolutely henstock kurzweil integrable mappings even when domain is real line.

on a locally compact space there isa possibility that henstock-kurzweil integration in some cases generates a larger sigma algebra than caratheodry construction. lebesgue stiltjes integra can be done much easily by using henstock-kurzweil method. anil pedgaonakr [email protected]