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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. –  Pedro Tamaroff Apr 18 '12 at 1:37
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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? –  t.b. Apr 18 '12 at 1:41
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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. –  t.b. Apr 18 '12 at 2:03
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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... –  t.b. Apr 18 '12 at 11:28
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If you refuse to show it, what exactly do you want someone to confirm? –  Did Apr 24 '12 at 7:31

1 Answer 1

Fubini's Theorem: Let $T=[a,b]X[c,d]$ and and suppose $f\in L(T)$. Then for almost all $x\in [a,b]$, $f(x,y)$ is in $L[c,d]$(as a function of $y$). Hence $$\int_{c}^{d}f(x,y)dy$$ is defined for almost all $x$. Moreover, $$\int_{c}^{d}f(x,y)dy$$ in $L[a,b]$ (as a function of $x$), and

$$\int\int_{T}f(x,y)dxdy=\int_{a}^{b}\left[\int_{c}^{d}f(x,y)dy\right]dx.\tag{1}$$ Similarly, $$ \int\int_{T}f(x,y)dx dy=\int_{c}^{d}\left[\int_{a}^{b}f(x,y)dx\right]dy.\tag{2}$$ That is, if $f$ is Lebesgue integrable as a function of two variables, then $\int\int_{T}f$ can be evaluated by perfoming an iterated integration. It follows that if $f\in L(T)$ then the right hand side of $(1)$ and $(2)$ are equal.

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You didn't answer my question. My question is about an integral on a general product measure spaces, not on the Lebesgue measure of the product of real intervals. –  Makoto Kato Apr 18 '12 at 3:28

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