To construct a product measure space generated by $(X, \mathcal{F}, \mu)$ and $(Y, \mathcal{G}, \lambda)$, we start with an algebra consisted of finite union of the generalized rectangles $A \times B \in \mathcal{F} \times \mathcal{G}$ and a premeasure on $\mathcal{F} \times \mathcal{G}$. Then we use the Catheodory's Extension Theorem to extend the algebra to a $\sigma$-algebra $\mathcal{F} \otimes \mathcal{G}$ and a measure on this $\sigma$-algebra.
Now if we are given a sequence of measure space $\{ (X_i, \mathcal{F}_i, \mu_i) \}_{i = 1} ^n$, how should we define a $n$-dimensional product measure space? It seems to me that there are two ways to do so:
Generalize the lemmas used to construct $\mathcal{F} \otimes \mathcal{G}$ and the product measure to $n$-dimensional. Then construct directly the measure space $(\prod_{i = 1} ^n X_i, \otimes_{i = 1} ^n \mathcal{F}_i, \pi)$.
Inductively define a product measure space $((\prod_{i = 1} ^{n - 1} X_i) \times X_n, (\otimes_{i = 1} ^{n - 1} \mathcal{F}_i) \otimes \mathcal{F}_n, \tilde{\pi}: (\otimes_{i = 1} ^{n - 1} \mathcal{F}_i) \otimes \mathcal{F}_n \to [0, +\infty])$.
Would we get the same product measure space with the two ways above?
I suspect the answer is no, as associativity of product measure spaces is only guaranteed with complete measure spaces. If my suspicion is correct, there should be counter-examples. Could someone provide a counter-example?