# Proving that $\sigma(\tau_{\mathbb{R}}\times\sigma(\tau_{\mathbb{R}}))=\sigma(\tau_{\mathbb{R}})\otimes \sigma(\tau_{\mathbb{R}})$

$\sigma(\tau_{\mathbb{R}})$ denotes the Borel $\sigma$-algebra ($\tau_{\mathbb{R}}$ is the usual topology on $\mathbb{R}$), $\sigma(\tau_{\mathbb{R}}\times\sigma(\tau_{\mathbb{R}}))$ is the $\sigma$-algebra generated by the product $\tau_{\mathbb{R}}\times\sigma(\tau_{\mathbb{R}})$ and $\sigma(\tau_{\mathbb{R}})\otimes \sigma(\tau_{\mathbb{R}})$ is the $\sigma$-algebra generated by sets of the form $B\times C$ where $B$ and $C$ are Borel sets. I'm asked to prove that

$$\sigma(\tau_{\mathbb{R}}\times\sigma(\tau_{\mathbb{R}}))=\sigma(\tau_{\mathbb{R}})\otimes \sigma(\tau_{\mathbb{R}}).$$

I would like a hint, if possible, for the inclusion "$\supseteq$", since the converse one is easy. I don't know how to show that a set of the form "borelian$\times$borelian" must belong to $\sigma(\tau_{\mathbb{R}}\times\sigma(\tau_{\mathbb{R}}))$.

Thank you!

• Any luck with the exercise? – Stefan Hansen Jun 10 '13 at 12:29
• @Stefan Yes, but I didn't use the fact you stated, because that fact seems to me harder to prove than my question (I'm probably mistaken, sorry). What I used was that in general $\sigma(\mathcal{E}\cap B)=\sigma(\mathcal{E})\cap B$. – Weltschmerz Jun 10 '13 at 16:03

## 1 Answer

Suppose that $(X,\mathcal{E})$ and $(Y,\mathcal{F})$ are measurable spaces. Then use that if $\mathcal{C}$ and $\mathcal{D}$ are generators for the two sigma-algebras $\mathcal{E}$ and $\mathcal{F}$ respectively, i.e. $\sigma(\mathcal{C})=\mathcal{E}$ and $\sigma(\mathcal{D})=\mathcal{F}$ and furthermore that there exist sequences $(C_n)\subseteq\mathcal{C}$ and $(D_n)\subseteq\mathcal{D}$ such that $$X=\bigcup_{n=1}^\infty C_n\quad\text{and}\quad Y=\bigcup_{n=1}^\infty D_n,$$ then $$\mathcal{E}\otimes\mathcal{F}=\sigma\left(\{C\times D\mid C\in\mathcal{C},\;D\in\mathcal{D}\}\right).$$

Now, choose $\mathcal{C}$ and $\mathcal{D}$ wisely.