Given two finite measure spaces,

whenever A ∈ℱ and B ∈S. This is naturally related to the concept of iterated integrals similar to what is used in calculus to evaluate a multiple integral. The approach is based on something called a π system, [?].
Definition 10.12.1 Let
Definition 10.12.2 Let Ω be a set and let K be a collection of subsets of Ω. Then K is called a π system if ∅,Ω ∈K and whenever A,B ∈K, it follows A ∩ B ∈K.
Obviously an example of a π system is the set of measurable rectangles because

The following is the fundamental lemma which shows these π systems are useful. This lemma is due to Dynkin.
Lemma 10.12.3 Let K be a π system of subsets of Ω, a set. Also let G be a collection of subsets of Ω which satisfies the following three properties.
Then G⊇ σ
Proof: First note that if

then ∩ℋ yields a collection of sets which also satisfies 1  3. Therefore, I will assume in the argument that G is the smallest collection satisfying 1  3. Let A ∈K and define

I want to show G_{A} satisfies 1  3 because then it must equal G since G is the smallest collection of subsets of Ω which satisfies 1  3. This will give the conclusion that for A ∈K and B ∈G, A ∩ B ∈G. This information will then be used to show that if A,B ∈G then A ∩ B ∈G. From this it will follow very easily that G is a σ algebra which will imply it contains σ
Since K is given to be a π system, K⊆G_{A}. Property 3 is obvious because if

because A ∩ B_{i} ∈G and the property 3 of G.
It remains to verify Property 2 so let B ∈G_{A}. I need to verify that B^{C} ∈G_{A}. In other words, I need to show that A ∩ B^{C} ∈G. However,

Here is why. Since B ∈G_{A}, A ∩ B ∈G and since A ∈K⊆G it follows A^{C} ∈G by assumption 2. It follows from assumption 3 the union of the disjoint sets, A^{C} and

I just proved K⊆G_{B}. The other arguments are identical to show G_{B} satisfies 1  3 and is therefore equal to G. This shows that whenever A,B ∈G it follows A ∩ B ∈G.
This implies G is a σ algebra. To show this, all that is left is to verify G is closed under countable unions because then it follows G is a σ algebra. Let

Therefore, G⊇ σ
With this lemma, it is easy to define product measure.
Let
 (10.12.54) 
where in the above, part of the requirement is for all integrals to make sense.
Then K⊆G. This is obvious.
Next I want to show that if E ∈G then E^{C} ∈G. Observe X_{EC} = 1 −X_{E} and so
Next I want to show G is closed under countable unions of disjoint sets of G. Let
From Lemma 10.12.3, G⊇ σ
 (10.12.56) 
Now here is Fubini’s theorem.
Theorem 10.12.4 Let f : X ×Y → [0,∞] be measurable with respect to the σ algebra, σ

Proof: Let
The symbol, ℱ×S denotes σ
Of course one can generalize right away to measures which are only σ finite.
Theorem 10.12.5 Let f : X ×Y → [0,∞] be measurable with respect to the σ algebra, σ

Proof: Since the measures are σ finite, there exist increasing sequences of sets,

Passing to the limit yields

whenever f is as above. In particular, you could take f = X_{E} where E ∈ℱ×S and define

Then just as in the proof of Theorem 10.12.4, the conclusion of this theorem is obtained. This proves the theorem.
It is also useful to note that all the above holds for ∏ _{i=1}^{n}X_{i} in place of X ×Y. You would simply modify the definition of G in 10.12.54 including all permutations for the iterated integrals and for K you would use sets of the form ∏ _{i=1}^{n}A_{i} where A_{i} is measurable. Everything goes through exactly as above. Thus the following is obtained.
