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).