Associativity of product measures

In the following I will prove that the product of three measures is associative. This can easily be generalized to a finite number of measures and possibly to countably many measures. As the OP indicates in his/her post scriptum, it all boils down to proving the associativity result for the $\sigma$-algebras involved.

Let $(X_i,\Sigma_i, \mu_i)$ be a measure space, $i = 1,\ 2,\ 3$. In the following I will use the (standard) notations $$\Sigma_i \times \Sigma_j = \{A \times B : A \in \Sigma_i,\ B \in \Sigma_j\}$$ $$\Sigma_i \otimes \Sigma_j = \sigma\big(\Sigma_i \times \Sigma_j\big),$$ where $\sigma(\cdot)$ indicates the $\sigma$-algebra generated by the argument, i.e. the smallest $\sigma$-algebra that contains the argument. We can finally state the result we want to prove

CLAIM: $\Sigma_1 \otimes \Sigma_2 \otimes \Sigma_3 := \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) = \sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3) = \sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)).$

Synopsis of the proof: $\Sigma_i \times \Sigma_j \subset \Sigma_i \otimes \Sigma_j$ is a $\pi$-system. Define the "good set" and apply the $\pi-\Lambda$-system Theorem. Then use the minimality of $\sigma(\cdot)$ several times.

PROOF: Notice that $$(\Sigma_1 \otimes \Sigma_2) \times \Sigma_3 = \{A \times B : A \in \Sigma_1\otimes\Sigma_2,\ B \in \Sigma_3\},$$ then fix $B \in \Sigma_3$ and let $$\Lambda = \{A : A \in \Sigma_1 \otimes \Sigma_2,\ A \times B \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)\}.$$

Clearly $\Sigma_1 \times \Sigma_2 \subset \Lambda$. Let's prove that $\Lambda$ is a $\Lambda$-system.

• $X_1 \times X_2 \times B \in \Sigma_1 \times \Sigma_2 \times \Sigma_3$ then $X_1 \times X_2 \in \Lambda$.
• Let $A_1,A_2 \in \Lambda$, $A_1 \subset A_2$. We need to show that $A_2 \setminus A_1 \in \Lambda$, i.e. we need to show that $(A_2 \setminus A_1) \times B \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$. This is easy since $\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$ is a $\sigma$-algebra, indeed $$(A_2 \setminus A_1) \times B = (A_2 \times B) \setminus (A_1 \times B) = (A_2 \times B) \cap (A_1 \times B)^c \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$
• Let $\{A_i\}$ be an increasing sequence of sets in $\Lambda$. We need to show that $\cup A_i \in \Lambda$. As before, this easily follows from the fact that $\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3)$ is a $\sigma$-algebra. Let's write out the details! $$\Big(\bigcup_{i=1}^{\infty}A_i\Big) \times B = \bigcup_{i=1}^{\infty}(A_i \times B) \in \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$

We can finally apply the $\pi-\Lambda$-system Theorem to conclude that $\sigma (\Sigma_1 \times \Sigma_2) \subset \Lambda$ and hence $\sigma(\Sigma_1 \times \Sigma_2) \times B \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$ Since this is true for every $B \in \Sigma_3$ we obviously get that $$(\Sigma_1 \otimes \Sigma_2) \times \Sigma_3 \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3),$$ which in turn yields $$\sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3) \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$

Notice we can apply the same reasoning to show that $$\sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)) \subset \sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3).$$

The other inclusion is a lot simpler: $\Sigma_1 \times \Sigma_2 \times \Sigma_3 \subset (\Sigma_1 \otimes \Sigma_2) \times \Sigma_3$, so that $$\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) \subset \sigma((\Sigma_1 \otimes \Sigma_2) \times \Sigma_3),$$ and similarly $$\sigma(\Sigma_1 \times \Sigma_2 \times \Sigma_3) \subset \sigma(\Sigma_1 \times (\Sigma_2 \otimes \Sigma_3)).$$

This proves the claim. $\blacksquare$

What happens to the measures was covered by David C. Ullrich in the comments section above.

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$