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

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

$${\int}_{\mathrm{\Omega}}f\phantom{\rule{thinmathspace}{0ex}}d\mu :=\{\begin{array}{ll}\text{the Khintchine integral}& \text{if}(\mathrm{\Omega},\mathrm{\Sigma},\mu )=(\mathbb{R},\mathcal{L},{\mu}_{\text{Lebesgue}});\\ \text{the Lebesgue integral}& \text{otherwise.}\end{array}$$