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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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    @Zev Thanks for the info.2012-04-24

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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.2012-04-18