Combining Fubini and Tonelli's in one single Assumption I am referring to the statements on Wikipedia, there it is said that Fubini's Theorem states that if $f : X\times Y \to \mathbb R$ is integrable, then
$$
 \int_X \left( \int_Y f(x,y) dy\right) dx = 
 \int_Y \left( \int_X f(x,y) dx\right) dy =
 \int_{X\times Y} f(x,y) d(x,y)
$$
and for Tonelli's Theorem instead of supposing integrability, we just assume $f$ to be non-negative $f \ge 0$, and then the above equalities hold. Then later it is written that Fubini and Tonelli could be combined, meaning that if one of the following conditions
\begin{align*}
 \int_X \left( \int_Y |f(x,y)| dy\right) dx & < \infty \\ 
 \int_Y \left( \int_X |f(x,y)| dx\right) dy & < \infty \\
 \int_{X\times Y} |f(x,y)| d(x,y) & < \infty
\end{align*}
hold, then the above formula for interated integrals is valid.
I do not understand in what sense this entails Tonelli's condition of non-negativity? For example take $f(x,y) = \max\{0, x\}, X = Y = \mathbb R$, then Tonelli's Theorem applies, and all integrals have value $\infty$, but none of the above mentioned integrals is finite, so I think this condition does not incorporate Tonelli's Theorem??
 A: I still don't understand the question. Yes, if $f\ge0$ then all three integrals could be infinite. I don't understand why you feel that's a problem; Tonelli does not say the three integrals are finite, just that they are equal. 
Not quite understanding what the issue is, I'll try to explain the facts. NOTE we're assuming that $f$ is jointly measurable in all of this. And of course the notation below is very informal:
Fubini(F): If $f\in L^1(X\times Y)$ then $\int_X\int_Y f=\int_Y\int_X f=\int_{X\times Y}f$.
Tonelli(T): If $f\ge0$ then $\int_X\int_Y f=\int_Y\int_X f=\int_{X\times Y}f$.
(Note again this includes the case $\infty=\infty=\infty$.)
Fubini-Tonelli(FT): If one of the three integrals $\int_X\int_Y|f|$, $\int_Y\int_X|f|$, $\int_{X\times Y}|f|$ is finite then $f\in L^1(X\times Y)$ and $\int_X\int_Yf=\int_Y\int_Xf=\int_{X\times Y}f$.
F+T implies FT: Suppose one of the three integrals of $|f|$ is finite. Since $|f|\ge0$, Tonelli implies they are all finite. In particular $\int_{X\times Y}|f|$ is finite, which says exactly that $f\in L^1(X\times Y)$. Hence Fubini's theorem shows that the three integrals for $f$ are equal.
FT implies F: Suppose $f\in L^1(X\times Y)$. This says that $\int_{X\times Y}|f|$ is finite. So FT says that the three integrals for $f$ are equal.
FT implies T: Suppose $f\ge 0$. So $|f|=f$. We need to show all three integrals are equal. Case I: One of the three integrals is finite. Then FT says the three integrals are equal. Case II: All three integrals are infinite. Then they are equal.
