Is it possible to prove Fubini’s Theorem without Dynkin’s Theorem or the Monotone Class Theorem? Fubini’s Theorem for Lebesgue integrals states that if $X$ and $Y$ are Sigma-finite measure spaces then the integral of a (well-behaved) function $f(x,y)$ with respect to the product measure on $X\times Y$ is equal to the iterated integral of $f$ with respect to the measure on $X$ and the measure on $Y$. The standard way to prove Fubini’s theorem is to prove it first for characteristic functions, then for simple functions, then for non-negative measurable functions, etc.  
The hard part is proving it for characteristic functions.  You first prove it for characteristic functions of measurable rectangles, i.e. Cartesian products of measurable sets in $X$ and measurable sets in $Y$. Then you have to somehow use that to prove it for characteristic functions of all measurable sets in the product measure space.  This is usually done using one of two theorems:


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*Dynkin’s pi-lambda theorem, which states that if a pi-system of sets is contained in a lambda-system of sets, then the sigma algebra generated by the pi-system is also contained in the lambda system.

*Halmos’ monotone class theorem, which states that if an algebra of sets is contained in a monotone class of sets, then the sigma algebra generated by the algebra is also contained in the monotone class.


Both these theorems apply because the collection of measurable rectangles is both a pi-system and an algebra, and the collection of sets whose characteristic functions satisfy Fubini’s theorem is both a lambda-system anda monotone class.
My question is, is it possible to prove Fubini’s theorem without using either of these results?  I assume that there would be some way, considering that Fubini proved his theorem long before Eugene Dynkin and Paul Halmos were even born.
 A: Real Analysis by Stein and Shakarchi has a proof that avoids the monotone class theorem and Dykin's $\pi$-$\lambda$ theorem; I would not be surprised if this proof is Fubini's original proof. See Chapter 2 Section 3 for the proof for Lebesgue measure. See Chapter 6 Section 3 for the proof for general measures. The proof in the general case is indeed a generalization of the proof in the Lebesgue case. Introduction to Real Analysis by Christopher Heil gives essentially the same proof as Stein and Shakarchi, in the case of Lebesgue measure. 
The proof is long, so I will only give a sketch here. By the usual linearity and convergence arguments, it suffices to prove the theorem for indicator (i.e., characteristic) functions of sets $E$. The idea is to first prove the theorem for measurable rectangles $E$, then for sets $E$ in $\mathcal{A}$ (where $\mathcal{A}$ is the algebra of finite disjoint unions of measurable rectangles), then for sets $E$ in $\mathcal{A}_{\sigma}$ (countable unions of sets in $\mathcal{A}$), then for sets $E$ in $\mathcal{A}_{\sigma \delta}$ (countable intersections of sets in $\mathcal{A}_{\sigma}$). Finally, to prove the theorem for arbitrary measurable sets, we use the following proposition whose proof is implicit in the Caratheodory construction. 
Proposition 1.6 (Chapter 6, Section 1). Let $\mu_{*}$ be the outer measure generated by a  premeasure $\mu_0$ on an algebra $\mathcal{A}$ on a set $X$. For any set $E \subseteq X$ and any $\epsilon>0$, there is an $E_1 \in \mathcal{A}_{\sigma}$ and an $E_2 \in \mathcal{A}_{\sigma \delta}$ such that $E \subseteq E_2 \subseteq E_1$ and 
$$\mu_{*}(E_1) - \epsilon \leq \mu_{*}(E) = \mu_{*}(E_2).$$ 
Note: I have not carefully verified the correctness of Stein and Shakarchi's proof of Fubini's theorem.  
