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).