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 :)

  • 1
    $\begingroup$ aren't all elements of $\mathcal{E}$ in $\sigma(\mathcal{E})$ by definition? What am I missing? $\endgroup$ – 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.