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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:

  1. 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.
  2. 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.

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  • $\begingroup$ Good question. I have the same one. It might be worth asking on Mathoverflow. $\endgroup$ – JasonJones Apr 26 at 1:54

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