Fubini's theorem without completeness or $\sigma$-finiteness conditions 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 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.
 A: Folland's Real Analysis: Modern Techniques and Their Applications states the Fubini-Tonelli Theorem as

Suppose that $(X, \mathcal{M},\mu)$ and $(Y, \mathcal{N},\nu)$ are $\sigma$-finite measure spaces.

a. (Tonelli) If $f \in L^+(X \times Y)$, then the functions $g(x) = \int f_x d\nu$ and $h(y) = \int f^y d\mu$ are in $L^+(X)$ and $L^+(Y)$, respectively, and $$ \int f d(\mu \times \nu) = \int \left[ \int f(x,y) d\nu(y) \right] d\mu(x) = \int \left[ \int f(x,y) d\mu(x) \right] d\nu(y).$$
b. (Fubini) If $f \in L^1(\mu \times \nu)$, then $f_x \in L^1(\nu)$ for a.e. $x \in X, f^y \in L^1(\mu)$ for a.e. $y \in Y$, the a.e.-defined functions $g(x) = \int f_x d\nu$ and $h(x) = \int f^y d\nu$ are in $L^1(\mu)$ and $L^1(\nu)$, respectively, and the above equation holds.

(Page 67)

So completeness is not necessary.  On the other hand, $\sigma$-finiteness is necessary.  This exercise follows the Fubini-Tonelli theorem in Folland.  Here $\mathcal{B}_{[0,1]}$ denotes the usual Borel $\sigma$-algebra on $[0,1]$.

46. Let $X = Y = [0,1]$, $\mathcal{M} = \mathcal{N} = \mathcal{B}_{[0,1]}$, $\mu = $ Lebesgue measure, and $\nu =$ counting measure.  If $D = \{(x,x): x\in [0,1]\}$ is the diagonal in $X \times Y$, then $\iint \chi_D  \; d\mu\; d\nu$, $\iint \chi_D \; d\nu \; d\mu$, and $\int \chi_D \; d(\mu \times \nu)$ are all unequal.
(Page 68)

A: This is a complement to the above answers. With which I agree.
I think the definition of the product measure is not unique.
This is best seen in that the (usual definition of the) product of 
the complete Lebesgue  measure on ${\bf R}$ give us not the complete Lebesgue 
measure on the product space ${\bf R}^2$.
What we want is a measure $\mu\otimes\nu$ such that 
$\mu\otimes\nu(A\times B)=\mu(A)\nu(B)$ for $A$ and $B$ measurables
and of finite measure. 
Therefore it is natural to consider here the sigma algebra $\Sigma_0$ generated 
by the products $A\times B$ with $\mu(A)$ and $\nu(B)$ finite.
Usually we consider the sigma algebra $\sigma$ generated by the the products 
$A\times B$ of measurable sets.  In the case of $\sigma$-finite measures 
the two sigma algebras coincide.
If you consider only the measure on $\Sigma_0$ (this is not what it is usually
done) you obtain a product measure that is unique in $\Sigma_0$ and both 
theorems Fubini-Tonelli and Fubini are true, without assuming anything about
the measures.
The usual example, given above by Adam Saltz, is not a counterexample because 
the diagonal is not measurable (that is, it is not in $\Sigma_0$).
With this definition of the product we get the same integrable function that
with the usual one.  This is what make the theorem of Fubini-Tonelli true
because the support of an integrable function is sigma finite. 
So I propose to define the product measure always on $\Sigma_0$. We get 
the usual definition in the $\sigma$-finite case. In other case we get 
always Fubini-Tonelli and Tonelli theorems without restrictions.
I have experimented this many times in my classes of Measure Theory (now 
dead by Bolonia reform).
A: 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 $f\in \mathcal{L}^1 (X\times Y, \mathcal {A}\otimes\mathcal B,  \mu\otimes\nu)\, .$
A: Here is another complement to the above answers:
With Fubini on non-$\sigma$-finite spaces you have to be careful about what you mean by 'product measure'.
Suppose we use the Carathéodory method to extend the premeasure from measurable rectangles to a complete measure.
Let $\mu\otimes\nu$ mean this measure restricted to $\mathcal A \times \mathcal B$. (The restricted measure may no longer be complete.)
Then Fubini holds for any function integrable with respect to $\mu \otimes \nu$.
Now consider the problem from Folland above (#46), 
except define $(\mu \times \nu) (S)$ to be the sum of the $\mu$ measures of all horizontal sections of $S$.
(Note: Folland's $\mu \times \nu$ is the same as our $\mu \otimes \nu$, 
but even so, he presents Fubini only for $\sigma$-finite measures.)
This is a product measure, since it satisfies $(\mu \times \nu)(A \times B) = \mu(A) \; \nu(B)$ for all measurable rectangles 
(defining $0 \cdot \infty = 0$).
Then $\int \chi_D \; d(\mu \times \nu) = \iint \chi_D  \; d\mu \; d\nu = 0$, but $\iint \chi_D \; d\nu \; d\mu = 1$. 
Thus, Fubini fails for this measure and integrable function.
