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The usual Fubini's theorem (see the Wikipedia article for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of Fubini's theorem without those assumptions. The idea of my proof is to use a fact that if a function is integrable, the support of the function must be $\sigma$-finite. Am I mistaken?

You may wonder what the use of removing $\sigma$-finiteness is. For example, Bourbaki developed integration theory on locally compact spaces. They didn't assume $\sigma$-compactness on those spaces. So those spaces are not necessarily $\sigma$-finite on their measures. If you want to interpret their theory in the usual measure theory framework, you need to abandon the $\sigma$-finiteness condition in most cases.

Theorem Let $(X, Ψ)$ and $(Y, Φ)$ be two measurable spaces. That is, $Ψ$ and $Φ$ are sigma algebras on X and Y respectively, and let $μ$ and $ν$ be measures on these spaces. Denote by $Ψ×Φ$ the sigma algebra on the Cartesian product $X×Y$ generated by subsets of the form $A×B$, where $A ∈ Ψ$ and $B ∈ Φ$.

A product measure $μ×ν$ is any measure on the measurable space $(X×Y, Ψ×Φ)$ satisfying the property $(μ×ν)(A×B) = μ(A)ν(B)$ for all $A ∈ Ψ$, $B ∈ Φ$.

Let f be an integrable function on $X×Y$, then its integral can be calculated by iterated integrals.

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    I guess it'd be better if you provided a link or a copy of your proof.2012-04-18
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    You can have a look at the section *Handouts* on this page: http://www.math.mcgill.ca/jakobson/courses/math564.html2012-04-18
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    Here's a test case for your theorem: what do you get if you integrate the characteristic function of the diagonal of $[0,1] \times [0,1]$ when you take the product measure of Lebesgue measure on the first factor and counting measure on the second factor?2012-04-18
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    Peter, the idea of my proof depends on a fact that if a function is integrable on a product measure space, its support must be $\sigma$-finite.2012-04-18
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    There are versions of Fubini's theorem that go through without assuming completeness (I think Royden gives a version) and also without $\sigma$-finiteness on one of the factors, however, *some* assumptions must be made on the other factor and the resulting theorems are always asymmetric in nature. One of the problems that arise is what my question was supposed to show: there is a distinction between sets that are locally null (i.e. the intersection with every measurable rectangle of finite measure is a null set) and sets that *are* null sets with respect to the product measure.2012-04-18
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    To t.b.. The measure of the diagonal in your case is not finite. So it's not a test case for my version of Fubini's theorem.2012-04-18
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    @Makoto: please state the result *precisely*. It is impossible to give a counterexample to a vague assertion! What exactly is the $\sigma$-algebra where your measure is defined? *What* exactly is the measure you consider? How do you get it? It may be the case that your assertion is true and it may be the case that your assertion is wrong. This depends very much on the details. Also, I gave my "counterexample" before you even indicated which result you intended...2012-04-18
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    @t.b.: I'm talking about the usual product measure. Please see the Wikipedia article for the definition of the product measure.2012-04-18
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    @MakotoKato Your link directs to "http:// the wikipedia's article/" and not an actual article. Please post a careful statement of what you have proven.2012-04-18
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    @Chris It was added by a moderator, not by me.2012-04-18
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    @t.b. I said the *usual* Fubini theorem in the first place. Are there many variants in the *usual* Fubini theorem? Anyway, if you look at the Wikipedia article on Fubini thorem, you can find the precise description of the *usual* theorem.2012-04-19
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    Dear Makoto, From what I can tell, there is not a unique product measure in the non-$\sigma$-finite case. Can I ask: what is *the usual product measure*? Regards,2012-04-24
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    @Matt Please look at the definition of the usual product measure at the top of the Wikipedia article on product measure.2012-04-24
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    @MakotoKato, can you give a reference to the definition of the product measure in the non-sigma-finite case other than wikipedia? In the article, the measure is defined to be "the unique measure on the [product measure space]", but further down in the article is: "The uniqueness of product measure is guaranteed only in the case that both [spaces] are sigma-finite".2012-04-24
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    @BR It needs not to be unique, since, as I wrote, the support of an integrable function must be $\sigma$-finite.2012-04-24
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    @MakotoKato, my point is that there is a certain amount of sense that needs to be made of the statement of Fubini's theorem when you remove $\sigma$-finiteness, since you can no longer talk about "the" product space. There are plenty of true things you can say here, but you should edit your question with the precise proposition you think you are proving. I think you will find that you aren't proving anything stronger than the "usual" Fubini's theorem.2012-04-24
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    @BR My point is that one needs not assume $\sigma$-finiteness on measures in Fubini's theorem. And I want someone to confirm this. Anyway I'll edit my question.2012-04-24
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    Where is the (alleged) proof?2012-04-24
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    @Didier It's a bit long and I'm not used to Latex.2012-04-24
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    If you refuse to show it, **what** exactly do you *want someone to confirm*?2012-04-24
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    @Didier I just want the assertion to be confirmed.2012-04-24
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    The statement is not precise enough since you still don't say *what* measure you are considering: The property that $(\mu \times \nu)(A \times B) = \mu(A) \nu(B)$ simply doesn't determine a unique measure, not even on the $\sigma$-algebra generated by the measurable rectangles. The complete locally determined product measure assigns the value zero to the diagonal in the example I showed in my first comment. Thus its characteristic function *is* integrable, while only one iterated integral gives the right answer. See [here](http://math.stackexchange.com/q/70888/5363) for more and a reference.2012-04-24
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    On a different note, what is the purpose of editing this question *after* you [crossposted your question to MO](http://mathoverflow.net/questions/94486/) and accepted an answer there?2012-04-24
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    @t.b. I crossposted because no one gave me answer for several days. I editted this question because of the Matt's comment. So the answer in MO is not correct?2012-04-24
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    Pietro's answer on MO is correct for the measure obtained by performing Carathéodory's construction for the outer measure on $P(X \times Y)$ obtained by $$m^\ast(E) = \inf{\left\{\mu(A_n) \nu(B_n) \,:\,E \subset \bigcup_{n=1}^\infty A_n \times B_n,\,A_n \in \Psi,\,B_n \in \Phi\right\}},$$ which is probably what you intend. For this measure neither completeness nor $\sigma$-finiteness are needed for Fubini's theorem (for functions measurable with respect to the $\sigma$-algebra generated by the rectangles). However, this result is only of limited use since Tonelli breaks down for that measure.2012-04-24
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    @t.b. That's what I meant initially. Thanks for clearing my question.2012-04-24
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    @t.b. The wikipedia article on Fubini's theorem says if $X$ and $Y$ have complete measures, Fubini's theorem holds on any product measure in the sense I stated in my question above. Is that correct?2012-04-24
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    @MakotoKato: Please mention when you [crosspost](http://mathoverflow.net/questions/94486/fubinis-theorem-without-completeness-or-sigma-finiteness-conditions) on MO.2012-04-24
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    @Michael Okay, I'll do it.2012-04-24
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    As for the downgrade to my question, though a product measure may *not* be unique, I think the meaning of the theorem is clear. I thought the theorem was correct regardress whichever a product measure one took. In any case, the Wikipedia article states a similar theorem just like mine except they assume completeness on measures on factor spaces. By the way, they 2 upgraded to the same question(as before the edit) in MO.2012-04-24
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    I think this is an interesting question and contemplating on it is a good exercise on measure theory for students including me even if the measure may not be unique and my assertion is false.2012-04-24
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    @MakotoKato: Please do not use `$...$` to make your text italicized; that is not the right formatting on this site (nor is it correct in LaTeX, either). The correct way to make text italicized on this site is `*...*`. See [this page](http://meta.stackoverflow.com/editing-help) for more explanations about formatting here.2012-04-24
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    @Zev Thanks for the info.2012-04-24

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