Why is it clear that $\sigma(\Omega_{1})\times \sigma(\Omega_{2})\subseteq \sigma( \Omega)$

Question

Let $(\Omega_{1}, \tau_{1})$ and $(\Omega_{2}, \tau_{2})$ be topological spaces and $\tau:=\tau_{1}\otimes \tau_{2}$ while $\Omega:=\Omega_{1}\otimes \Omega_{1}$

I am attempting to brush up on my topology, I want to show that:

$\mathcal{B}(\Omega_{1})\times \mathcal{B}(\Omega_{2})\subseteq \mathcal{B}( \Omega)$

where $\mathcal{B}(\Omega_{i})$ is the Borel$-\sigma-$algebra of $\Omega_{i}$

My idea: Consider the generators of $\mathcal{B}(\Omega_{1})$ and $\mathcal{B}(\Omega_{2})$, namely $\tau_{1}$ and $\tau_{2}$ respectively. Let $A \in \tau_{1}\times \tau_{2}$ then by definition there exists $A_{1} \in \tau_{1}$ and $A_{2} \in \tau_{2}$ so that $A=A_{1} \times A_{2}$. But if I remember correctly then $\{ B_{1}\times B_{2}: B_{1}\in \tau_{1} , B_{2} \in \tau_{2}\}$ is a basis of the product topology $\tau$

Thus $\tau_{1}\times \tau_{2}\subseteq \tau$, so does this imply that $\sigma(\tau_{1})\times \sigma(\tau_{2})\subseteq \sigma(\tau)$? I'm unsure about the last conclusion because is $\sigma(\tau_{1}\times \tau_{2})=\sigma(\tau_{1})\times \sigma(\tau_{2})$ since all I can conclude from $\tau_{1}\times \tau_{2}\subseteq \tau$ is that $\sigma(\tau_{1}\times \tau_{2})\subseteq \sigma(\tau)$?

Answer 1

If $U_1$ and $U_2$ are open in $\Omega_1$ and $\Omega_2$ then $U_1 \times U_2 \in \mathbb B(\Omega)$. Fix $U_2$ and consider $\{A\in \mathbb B(\Omega_1): A\times U_2 \in \mathbb B(\Omega)\}$. Verify that this is a sigma algebra containing all open sets in $\Omega_1$. This proves that $A\times U_2 \in \mathbb B(\Omega)$ for all Borel sets $A $ in $\Omega_1$. Now fix a Borel set $A$ in $\Omega_1$ and consider $\{B\in \mathbb B(\Omega_2): A\times U_2 \in \mathbb B(\Omega)\}$. This is a sigma algebra on $\Omega_2$ containing all open sets. This proves that all product sets $A \times B$ where $A$ and $B$ are Borel sets in $\Omega_1$ and $\Omega_1$ respectively belong to $\mathbb B(\Omega)$. Rest of the argument should now be clear.

Answer 2

We need the following result:

Proposition 1: Let $(X,\mathcal{M}_{X})$ and $(Y,\mathcal{M}_{Y})$ be measurable spaces. Let $\mathcal{M}$ be the product $\sigma$-algebra of $\mathcal{M}_{X}$ and $\mathcal{M}_{Y}$, i.e., $\mathcal{M}=\sigma\left(\mathcal{M}_{X}\otimes\mathcal{M}_{Y}\right),$ where $\mathcal{M}_{X}\otimes\mathcal{M}_{Y}=\{A\times B\mid A\in\mathcal{M}_{X}\mbox{ and }B\in\mathcal{M}_{Y}\}.$ Let $\mathcal{C}_{X}\subseteq\mathcal{M}_{X}$ and $\mathcal{C}_{Y}\subseteq\mathcal{M}_{Y}$, with $X\in\mathcal{C}_{X}$ and $Y\in\mathcal{C}_{Y}$. Define $\mathcal{C}=\mathcal{C}_{X}\otimes\mathcal{C}_{Y}:=\{A\times B\mid A\in\mathcal{C}_{X}\mbox{ and }B\in\mathcal{C}_{Y}\}$. If $\sigma(\mathcal{C}_{X})=\mathcal{M}_{X}$ and $\sigma(\mathcal{C}_{Y})=\mathcal{M}_{Y}$, then $\sigma(\mathcal{C})=\mathcal{M}$.

Proof: Let $\pi_{X}:X\times Y\rightarrow X$ and $\pi_{Y}:X\times Y\rightarrow Y$ be the canonical projections. Recall that $\pi_{X}$ is $\mathcal{M}/\mathcal{M}_{X}$ measurable and $\pi_{Y}$ is $\mathcal{M}/\mathcal{M}_{Y}$ measurable. Moreover, from general measure theory, we have $\pi_{X}^{-1}(\sigma(\mathcal{C}_{X}))=\sigma(\pi_{X}^{-1}(\mathcal{C}_{X}))$. Observe that $\pi_{X}^{-1}(\sigma(\mathcal{C}_{X}))=\pi_{X}^{-1}(\mathcal{M}_{X})=\{A\times Y\mid A\in\mathcal{M}_{X}\}$. On the other hand, $\pi_{X}^{-1}(\mathcal{C}_{X})=\{A\times Y\mid A\in\mathcal{C}_{X}\}\subseteq\mathcal{C}$, and hence $\sigma(\pi_{X}^{-1}(\mathcal{C}_{X}))\subseteq\sigma(\mathcal{C})$. That is, $\{A\times Y\mid A\in\mathcal{M}_{X}\}\subseteq\sigma(\mathcal{C})$. Similarly, we can prove that $\{X\times B\mid B\in\mathcal{M}_{Y}\}\subseteq\sigma(\mathcal{C})$. Finally, if $A\in\mathcal{M}_{X}$ and $B=\mathcal{M}_{Y}$, then $A\times B=(A\times Y)\cap(X\times B)\in\sigma(\mathcal{C})$. This shows that $\mathcal{M}_{X}\otimes\mathcal{M}_{Y}\subseteq\sigma(\mathcal{C})$ and hence $\mathcal{M}\subseteq\sigma(\mathcal{C})$. The reversed inclusion $\mathcal{M}\supseteq\sigma(\mathcal{C})$ is trivial.

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We go back to your question. Let $X=\Omega_{1}$, $Y=\Omega_{2}$, $\mathcal{C}_{X}=\tau_{1}$, $\mathcal{C}_{Y}=\tau_{2}$, $\mathcal{\mathcal{M}}_{X}=\mathcal{B}(\Omega_{1})$, $\mathcal{M}_{Y}=\mathcal{B}(\Omega_{2})$. Then the product $\sigma$-algebra is $\mathcal{M}=\mathcal{\mathcal{B}}(\Omega_{1})\times\mathcal{B}(\Omega_{2})$, using your notation. Note that all the conditions in Proposition 1 are satisfied. Therefore $\mathcal{M}=\sigma(\mathcal{C}),$ where $\mathcal{C}=\{A\times B\mid A\in\tau_{1}\mbox{ and }B\in\tau_{2}\}$. Obviously, $\mathcal{C}\subseteq\tau$, where $\tau$ is the product topology induced by $\tau_{1}$ and $\tau_{2}$. Therefore, $\mathcal{M}=\sigma(\mathcal{C})\subseteq\sigma(\tau)=\mathcal{B}(\Omega)$.

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Remark: In general, unless $\tau_{1}$ and $\tau_{2}$ are second countable, the reversed inclusion does not hold.