Associativity of product sigma algebras

[psie]

Homework Statement

If $(X_j,\mathcal M_j)$ are measurable spaces for $j=1,2,3$, then $\bigotimes_1^3\mathcal M_j=(\mathcal M_1\otimes\mathcal M_2)\otimes\mathcal M_3$. Moreover, if $\mu_j$ is a $\sigma$-finite measure on $(X_j,\mathcal M_j)$, then $\mu_1\times\mu_2\times\mu_3=(\mu_1\times\mu_2)\times\mu_3$.

Relevant Equations

Above $$\bigotimes_1^3\mathcal M_j = M_1\otimes\mathcal M_2\otimes\mathcal M_3= \sigma (\{ A_1 \times A_2 \times A_3 : A_i \in \mathcal M_i \}),$$and $$(\mathcal M_1\otimes\mathcal M_2)\otimes\mathcal M_3=\sigma (\{ \Omega \times A: \Omega \in \mathcal M_1 \otimes \mathcal M_2,A \in \mathcal M_3 \}).$$ Also, the first $\sigma$-algebra is the smallest $\sigma$-algebra such that the projections are measurable.

This is an exercise from Folland's book. Here's my attempt at showing $\mathcal M_1\otimes\mathcal M_2\otimes\mathcal M_3=(\mathcal M_1\otimes\mathcal M_2)\otimes\mathcal M_3$.

$\subset$: since every measurable rectangle $A_1 \times A_2$ belongs to $\mathcal M_1 \otimes \mathcal M_2$, the generating set for $\mathcal M_1\otimes\mathcal M_2\otimes\mathcal M_3$ is a subset of $(\mathcal M_1\otimes\mathcal M_2)\otimes\mathcal M_3$, so also the $\sigma$-algebra belongs to $(\mathcal M_1\otimes\mathcal M_2)\otimes\mathcal M_3$.

$\supset$: I'm stuck here. $\pi_{1,2}:X_1\times X_2\times X_3\to X_1\times X_2$ is measurable since $\pi_{1,2}^{-1}(A_1\times A_2) = A_1\times A_2\times X_3\in \mathcal M_1\otimes\mathcal M_2\otimes\mathcal M_3$ for every $A_1\times A_2\in \mathcal M_1\otimes\mathcal M_2$. For the same reason, so is $\pi_3:X_1\times X_2\times X_3\to X_3$ measurable. I'm a bit lost at what I need to show.

I'm also not sure how to proceed with the rest of the exercise. Appreciate any help.

 

[psie]

Ah, ok. I've had some time to think this through and have a major realization. What I've realized is that $X_1\times X_2\times X_3$ is not equal to $(X_1\times X_2) \times X_3$. The author writes in the text that we identify these spaces with one another, i.e. through the bijection $(x_1,x_2,x_3)\mapsto((x_1,x_2),x_3)$, which I will show is measurable and so is its inverse (you can safely ignore my first attempt in #1):

$\pi_{12}:X_1\times X_2\times X_3\to X_1\times X_2$ is measurable since $\pi_{12}^{-1}(A_1\times A_2) = A_1\times A_2\times X_3\in \mathcal M_1\otimes\mathcal M_2\otimes\mathcal M_3$ for every $A_1\times A_2\in \mathcal M_1\otimes\mathcal M_2$. For very similar reasons $\pi_3:X_1\times X_2\times X_3\to X_3$ is measurable. So we conclude that $(x_1,x_2,x_3)\mapsto ((x_1,x_2),x_3)$ is measurable, since its components are the projections $\pi_{12}$ and $\pi_{3}$. Its inverse is also measurable since the image of $A_1\times A_2\times A_3$ is $(A_1\times A_2)\times A_3$, a measurable set in $(\mathcal M_1\otimes\mathcal M_2)\otimes\mathcal M_3$. Thus $(x_1,x_2,x_3)\mapsto ((x_1,x_2),x_3)$ is a measurable isomorphism. Hence the $\sigma$-algebras are equal in the sense that we identify $(x_1,x_2,x_3)$ with $((x_1,x_2),x_3)$.

Questions? Comments?

 

[nuuskur]

Strictly speaking, for measurability one must check preimages of all Borel sets. It is sufficient to check measurability by checking preimages of all subsets of a generating $\pi$-system, which is what you are doing.

It is also correct that direct product is not associative in the strictest sense, but is regarded as an associative operation for the reason you mentioned - there is a canonical isomorphism between. Same is true for product sigma algebras (as well as tensor product of vector spaces/modules).

The following is like "measure theoretic induction" (a form of transfinite induction to be precise), if you will.

Suppose $\rho$ is a property of a subset $A\subseteq S$, so write $\rho(A)$ if $A$ has that property. Then take some $\mathcal C \subseteq \mathcal P(S)$ (possibly empty). Suppose we have the following.

1. $\rho (S),\rho (\emptyset)$
2. $\rho(A)$ for every $A\in\mathcal C$
3. $\rho (A) \Rightarrow \rho (S\setminus A)$ for every $A\subseteq S$.
4. $\rho (\bigcup _{k\in\mathbb N}A_k)$ whenever $A_k\subseteq S$ and $\rho (A_k)$ for every $k\in\mathbb N$.

If the above conditions hold, then every element of $\sigma (\mathcal C)$ has property $\rho$. The proof follows from minimality: the collection of all those subsets with property $\rho$ is a sigma algebra, and therefore must contain the sigma algebra generated by $\mathcal C$.

Suppose the measurable spaces are labelled $(S_i,\mathcal M_i)$ for $i=1,2,3$. Now suppose $A\in\mathcal M_1\otimes \mathcal M_2$ and $B\in\mathcal M_3$. We want to show that $A\times B\in \mathcal M_1\otimes \mathcal M_2\otimes \mathcal M_3$.

Define for any subset $X\subseteq S_1\times S_2$
<br /> \rho (X) \Leftrightarrow (\forall C\in\mathcal M_3)(X\times C\in\mathcal M_1\otimes \mathcal M_2\otimes \mathcal M_3)<br />
And now apply the above "induction". It follows that for any $A\in\mathcal M_1\otimes\mathcal M_2$ we have $\rho(A)$.

The equality of the product measures can be verified by the unique extension theorem. In case of $\sigma$-finite measures, it is enough to check they coincide on a generating $\pi$-system.

 

[psie]

Strictly speaking, for measurability one must check preimages of all Borel sets. It is sufficient to check measurability by checking preimages of all subsets of a generating $\pi$-system, which is what you are doing.

It is also correct that direct product is not associative in the strictest sense, but is regarded as an associative operation for the reason you mentioned - there is a canonical isomorphism between. Same is true for product sigma algebras (as well as tensor product of vector spaces/modules).

Thank you for your suggestion. I am really unsure how this exercise was meant to be solved, especially the first assertion. I feel like the author has neither introduced isomorphisms between measurable spaces nor the measure theoretic induction you are suggesting. I still do not quite understand what the equality between the two $\sigma$-algebras actually means. If the spaces $X_1\times X_2\times X_3$ and $(X_1\times X_2)\times X_3$ are not strictly equal, then I would guess that neither are the $\sigma$-algebras strictly equal. So I don’t know what exactly is being asked in the first assertion of the exercise, what the equality truly means.

 

[nuuskur]

Corrected a typo in #3 in the third step of the "induction" scheme.

On the one hand we have the sigma algebra on the set $S_1\times S_2\times S_3$:
<br /> \mathcal M_1 \otimes \mathcal M_2\otimes \mathcal M_3 = \sigma \left( \left\lbrace A_1\times A_2\times A_3 \mid A_i \in\mathcal M_i\right\rbrace\right)<br />
(generated by all measurable cuboids, I suppose)
On the other hand we have two sigma algebras $\mathcal M_1\otimes \mathcal M_2$ and $\mathcal M_3$. We can't predict what all of the elements of the product sigma algebra $\mathcal M_1\otimes\mathcal M_2$ look like, but we do know that $\mathcal M_1\times\mathcal M_2$ is a generating $\pi$-system.

It suffices to show
<br /> (\mathcal M_1\otimes \mathcal M_2) \times\mathcal M_3 \subseteq \mathcal M_1 \otimes \mathcal M_2\otimes \mathcal M_3.<br />
(up to accuracy of the canonical isomorphism between direct products)
Then the generated sigma algebra of the LHS must also be contained in the RHS. The converse inclusion is obvious as you already noted in #1.

For a subset $\Omega\subseteq S_1\times S_2$ declare $\rho (\Omega)$ if and only if $\Omega\times C\in\bigotimes \mathcal M_i$ for every $C\in\mathcal M_3$. Now let's apply the induction as specified in #3, where we take $\mathcal C=\mathcal M_1\times\mathcal M_2$ as the generating $\pi$-system. The induction "base" cases 1. and 2. are obvious. But some work is required for the "induction step", that is for 3. and 4. Then it follows that all elements of $\mathcal M_1\otimes\mathcal M_2$ have property $\rho$ and the required inclusion is concluded.