I am currently taking a course in measure theory and I am struggling to grasp the concept well of the product of sigma algebras from an exercise.
Suppose we have $\mathcal{A} = \sigma_X (\mathcal{E})$ and $B=\sigma_Y (\mathcal{F})$ as the smallest $\sigma$-algebras on $X$ and $Y$ generated by some $\mathcal{E} \subseteq \mathcal{P}(X)$ and $\mathcal{F} \subseteq \mathcal{P}(Y)$ respectively.
Now let $\mathcal{C} = \sigma_{X \times Y}(\{E \times F : E \in \mathcal{E}, F \in \mathcal{F}\})$, I first want to show that $\mathcal{C} \subseteq \mathcal{A} \times \mathcal{B}$.
After writing down examples of smaller sets with $\sigma$-algebras to get an idea of what was going on I feel like the right sort of approach would be to find elements that generate $\mathcal{C}$ and show that these are contained in $\mathcal{A} \times \mathcal{B}$, as then the $\sigma$-algebra would ensure that the rest of the elements of $\mathcal{C}$ are also contained in this set.
To do this I thought that $\mathcal{C}$ was generated by all elements of $\mathcal{E} \times \mathcal{F}$ if I could show any element of $\mathcal{E} \times \mathcal{F}$ is contained in $\mathcal{A} \times \mathcal{B}$ then I would be done, which I feel can be reduced to showing all elements of $\mathcal{E}$ are contained in $\sigma_X(\mathcal{E})$, though I am not sure as I do not know how to proceed from here.
When I wasn't sure I tried the next part which is that the reverse inclusion holds under the condition that $X \in \mathcal{E}$ and $Y \in \mathcal{F}$, that is that $\mathcal{A} \times \mathcal{B} = \mathcal{C}$, which I had no idea how to proceed with.
Any help or insight would be greatly appreciated thanks 🙂
Best Answer
By definition, $\mathcal{A} \otimes \mathcal{B}$ is the $\sigma$-algebra on $X \times Y$ generated by the set $\mathcal{A} \times \mathcal{B}=\{A \times B: A \in \mathcal{A}, B \in \mathcal{B}\}$, (or equivalently the smallest $\sigma$-algebra that makes the projections measurable).
As $\mathcal{E} \subseteq \sigma_X(\mathcal{E}) = \mathcal{A}$ by definition and likewise $\mathcal{F} \subseteq \sigma_Y(\mathcal{F}) = \mathcal{B}$, we have that
$$\mathcal{E} \times \mathcal{F} = \{E \times F: E \in \mathcal{E}, F \in \mathcal{F}\} \subseteq \mathcal{A} \times \mathcal{B}$$
so $$\mathcal{C} = \sigma_{X \times Y}(\mathcal{E} \times \mathcal{F}) \subseteq \sigma_{X \times Y}(\mathcal{A} \times \mathcal{B}) = \mathcal{A} \otimes \mathcal{B}$$
This uses the obvious fact (by the definitions) that if $\mathcal{G},\mathcal{G}'$ are families of subsets of any set $Z$, then $\mathcal{G} \subseteq \mathcal{G}'$ implies $\sigma_Z(\mathcal{G}) \subseteq \sigma_Z(\mathcal{G}')$ as well.
The simple example in this post shows that we indeed need some condition like $X \in \mathcal{E}$ and $Y \in \mathcal{F}$ to show the reverse inclusion $\mathcal{A} \otimes \mathcal{B} \subseteq \mathcal{C}$ as well. For this inclusion it suffices to show that $$\mathcal{A} \times \mathcal{B} \subseteq \sigma_{X \times Y}(\mathcal{E} \times \mathcal{F}) = \mathcal{C}\tag{1}$$ and this is more subtle:
Define $$\mathcal{A}' = \{A \subseteq X: (\pi_X)^{-1}[A] \in \mathcal{C}\}$$
where $\pi_X: X \times Y \to X$ is the projection.
It is easy to check that this defines a $\sigma$-algebra on $X$, by the properties of inverse images and the fact that $\mathcal{C}$ is a $\sigma$-algebra. Also, for $E \in \mathcal{E}$ (and because $Y \in \mathcal{F}$), we have that $$(\pi_X)^{-1}[E] = E \times Y \in \mathcal{E} \times \mathcal{F} \subseteq \mathcal{C}$$
so that $\mathcal{E} \subseteq \mathcal{A}'$ which means that $\sigma_X(\mathcal{E}) = \mathcal{A} \subseteq \mathcal{A}'$ as well, or equivalently:
$$\forall A \in \mathcal{A}: A \times Y \in \mathcal{C}\tag{2}$$
Using the analogous argument for $\mathcal{B}$ and $\pi_Y$ and the assumption that $X \in \mathcal{E}$ we also get:
$$\forall B \in \mathcal{B}: X \times B \in \mathcal{C}\tag{3}$$
And then note that $(2)$ together with $(3)$ imply $(1)$ by the simple fact that
$$A \times B = (X \times B) \cap (A \times Y)$$
using that $\mathcal{C}$ is closed under intersections. This concludes the proof of the reverse inclusion.