The usual Fubini's theorem(see the [Wikipedia article](http://en.wikipedia.org/wiki/Fubini%27s_theorem) for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of the Fubini's theorem without those assumptions. Am I mistaken?


I restate the theorem to avoid confusion: If a function is integrable on a product measure space, its integral can be calculated by iterated integrals.


The idea of my proof is to use a fact that if a function is integrable on a product measure space, the function must be zero outside a $\sigma$-finite subset of the product measure space.



Fubini's theorem without completeness or $\sigma$-finiteness conditions
You do not need $\sigma$-finiteness of the measure in Fubini theorem, although it is an hypothesis that can be assumed with no loss of generality, in that the support of an integrable function is, of course, $\sigma$-finite. 


On the opposite, Tonelli theorem deals with non-negative measurable functions, whose support may well be non-$\sigma$-finite (as in the quoted example) and $\sigma$-finiteness is really needed. 


As to the hypothesis of completeness of the factor measures spaces $(X,\mathcal A,\mu)$ and $(Y,\mathcal B, \nu)$, it is not necessary, in either theorems (but, again, it could be assumed w.l.o.g., since the two measures can be completed). 
Notice however that some care is needed in stating the theorem if you consider a function on the product space which is measurable *with respect to the $\mu\otimes\nu$ -completion of the product $\sigma$-algebra* $\mathcal {A}\otimes\mathcal B$, more generally than just "measurable wrto the product $\sigma$-algebra". This generalization occurs quite naturally dealing with a Lebesgue measurable function on ${\bf R}^n\times{\bf R}^m$ (because the completion of the product measures is the Lebesgue measure on the product space). The result of taking this more general hypothesis is that you have to stuff the statement of the theorem with a sequel of "*a.e.*", which wouldn't be necessary in the case $f\in \mathcal{L}^1 (X\times Y, \mathcal {A}\otimes\mathcal B, \mu\otimes\nu)\, .$


