# Working with the product of sigma algebras

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 :)

• aren't all elements of $\mathcal{E}$ in $\sigma(\mathcal{E})$ by definition? What am I missing? – Robson Dec 2 '18 at 4:32

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.