Why is the canonical product $\sigma$-algebra the right $\sigma$-algebra on the product space?

Question

Let $I$ be a finite, countably infinite or uncountable index set and let $(\Omega_i,\mathcal{A}_i)_{i \in I}$ be measurable spaces. Then we can define the sigma algebra

$$ \mathcal{A} := \bigotimes_{i \in I} \mathcal{A}_i $$

on the product space $\Omega := \times_{i \in I} \Omega_i$ as the smallest sigma algebra, such that all projections $$ \pi_i\rightarrow \Omega : \Omega_i, \quad \omega \mapsto \omega_i, $$ are $\mathcal{A}$-$\mathcal{A}_i$-measurable.

This is the canonical way of defining a $\sigma$-algebra on a product space and is used (mostly without further explanation) in lots of textbooks about advanced probability.

My problem is that I am lacking intuitive motivation for this defintion. Is this the only reasonable way to define a $\sigma$-algebra on the product space? Why? If not, why do whe choose this definition over all other possible methods for constructing $\sigma$-algebras on $\Omega$? What are the essential advantages of this definition? In short: why is this the right $\sigma$-algebra on $\Omega$? Why could it not be different?

Kind regards and thanks for any help!

Answer 1

Let $(\Omega',\mathcal A')$ be a measurable space and let $f_i:\Omega'\to\Omega_i$ for $i\in I$ denote a collection of measurable maps.

Then especially if $\Omega$ is equipped with $\mathcal A=\bigotimes_{i \in I} \mathcal{A}_i$ a unique measurable $f:\Omega'\to\Omega$ such the $f_i=\pi_i\circ f$ for every $i\in I$.

Conversely every measurable map $f:\Omega'\to\Omega$ induces a family $\{f_i=\pi\circ f:\Omega'\to\Omega_i\mid i\in I\}$ of measurable maps, so the correspondence is one-to-one.

This offers the nice possibility to identify families $\{f_i\mid i\in I\}$ of measurable maps that have common domain with just one measurable map $f$.

That is for a good deal the underlying motivation for the construction of products.

Answer 2

The main reason, I think, is that the construction you describe satisfies the axioms for the categorical product in the category of measurable spaces. This, of course, raises the follow-up question: why are categorical products good? I can think of a few reasons:

• With other $\sigma$-algebras on the product it could be quite hard to check that a map into a product is measurable; with the categorical product you just need to check that all compositions with your map and the projection maps are measurable.
• Categorical products are compatible with a lot of other constructions like (co)limits and adjunctions that come up naturally in measure theory, and you get that compatibility for free.
• Most other product structures in mathematics, like the product topology or the product group structure, are also categorical products. With your definition you can prove quite easily that the Borel $\sigma$-algebra on the product of topological spaces agrees with the product of the Borel $\sigma$-algebras on the factors, for instance.

It's also worth noting that your definition gives the "right" answer in the case where the index set is finite: the product $\sigma$-algebra is the smallest $\sigma$-algebra for which products of measurable sets are measurable. You could try to just naively extend this definition to get a $\sigma$-algebra on the product of arbitrarily many measurable spaces, but experience in other settings (e.g. the product topology) suggests that it's better to generalize along universal properties than along set-theoretic constructions.

Answer 3

This sigma algebra is chosen because it always exists. For any collection of events you may consider the intersection of all sigma-algebras containing these events, and it is the smallest sigma algebra containing these events. It always works.

But for Lebesgue sigma algebra in $R$ and in $R\times R$, it is not actually that convenient, because some nice properties of the Lebesgue measure are lost when taking the product.

From Exercise 1.7.19(v) in "An introduction to measure theory" by Terence Tao:

If $\mathcal F$ is the Lebesgue sigma-algebra on $R$, then the product sigma-algebra is not the Lebesgue sigma-algebra on $R\times R$. In particular, the product sigma-algebra of two Lebesgue sigma-algebra is not complete, so the product measure here misses out the nice property that the Lebesgue measure is complete.