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.