TITLE: Associativity of product measures
QUESTION [13 upvotes]: Suppose we have measure spaces $(X_i, M_i, \mu_i), i=1, 2,3$, that are complete and $\sigma$-finite. I learned how to form a product measure from two measure spaces, but I wasn't so sure about product of more than two. So I am learning about it at the moment. I was wondering how does one show that
$$
\mu_1 \times (\mu_2 \times \mu_3) = (\mu_1 \times \mu_2) \times \mu_3.
$$
Any hint/solution is appreciated. Thank you!
PS I was also wondering about
$$
M_1 \times (M_2 \times M_3) = (M_1 \times M_2) \times M_3.
$$

REPLY [6 votes]: I have also thought about the question concerning the associativity of the $\sigma$-algebras and will present another proof, which does not make use of the $ \pi - \Lambda ~ - $ system Theorem but uses the direct image $ \sigma - $algebra: 

As mentioned let $ (X_i, M_i, \mu_i) $ for $ i = 1, 2, 3 $ be $\sigma-$finite  measure spaces. Then using the same notation as Giovanni 
$$ M_1 \otimes M_2 \otimes M_3 = \sigma \underbrace{\bigl \{ A_1 \times A_2 \times A_3 \bigl | A_i \in M_i \bigr \}}_{= M_1 \times M_2 \times M_3} $$
and
$$ M_1 \otimes ( M_2 \otimes M_3 ) = \sigma  \underbrace{\bigl \{ A \times \Omega \bigl | A \in M_1, \Omega \in M_2 \otimes M_3 \bigr \}}_{=M_1 \times (M_2 \otimes M_3)}. $$
By using the direct image  $\sigma-$algebra we will show that these $\sigma-$algebras are equal:
First Step: It is obvious that $ M_1 \times M_2 \times M_3 \subset M_1 \times (M_2 \otimes M_3) $ since every rectangle $ A_2 \times A_3 $, $ A_i \in M_i $ lies in $ M_2 \otimes M_3 $. Hence $ M_1 \otimes M_2 \otimes M_3 \subset M_1 \otimes ( M_2 \otimes M_3) $.
Second Step: We now show that also 
$ M_1 \otimes (M_2 \otimes M_3) \subset M_1 \otimes M_2 \otimes M_3  $, proving the associativity:
Consider the projection $ \pi: X_1 \times X_2 \times X_3 \to X_2 \times X_3, ~ \pi (x,y,z) = (y,z). $ Then the direct image
$$ \pi_* (M_1 \otimes M_2 \otimes M_3) := \{ \Omega \in X_2 \times X_3 | X_1 \times \Omega = \pi^{-1} (\Omega) \in M_1 \otimes M_2 \otimes M_3 \} $$
is a $ \sigma-$algebra on $ X_2 \times X_3$. (This is a general fact). 
Claim: $ M_2 \otimes M_3 \subset \pi_* (M_1 \otimes M_2 \otimes M_3) $. 
Proof of the Claim: By the minimality of the generated $\sigma-$algebra, it suffices to show that $ M_2 \times M_3 \subset   \pi_* (M_1 \otimes M_2 \otimes M_3) $. So let $ A_2 \times A_3 \in M_2 \times M_3$. But then we have
$$ X_1 \times A_2 \times A_3 \in M_1 \times M_2 \times M_3 $$
by definition, since $ X_1 \in M_1$. (End of the proof of the claim.)
So why does that help?
Again by the minimality of the generated $\sigma-$algebra, what we have to show is that for any $ A \in M_1, ~ \Omega \in M_2 \otimes M_3 $ it holds that
$$ A \times \Omega \in M_1 \otimes M_2 \otimes M_3. $$
But $ A \times \Omega = (A \times (X_2 \times X_3)) \cap (X_1 \times \Omega)$. Obviously $ A \times (X_2 \times X_3 ) \in M_1 \times M_2 \times M_3 $. By the above claim, we also have $ \Omega \in \pi_* (M_1 \otimes M_2 \otimes M_3) $ and therefore $ X_1 \times \Omega \in M_1 \otimes M_2 \otimes M_3$. But since $M_1 \otimes M_2 \otimes M_3$ is a $ \sigma-$algebra, then also the intersection $ (A \times (X_2 \times X_3)) \cap (X_1 \times \Omega) = A \times \Omega \in M_1 \otimes M_2 \otimes M_3$. $ \square$