Problem: Let $\mathcal{S}=\{(-b,b): b\geq 0\}$. Is $\sigma(\mathcal{S})=\mathfrak{B}o(\mathbb{R})$?
Notation: $\sigma(\mathcal{S})$ is the sigma algebra generated by $\mathcal{S}$ and $\mathfrak{B}o(\mathbb{R})$ is the Borel sigma algebra.
My attempt: I think it's true. My approach to this problem is as follows:
By definition in theory of sets, we need to prove two parts:
- $\sigma(\mathcal{S})\subseteq \mathfrak{B}o(\mathbb{R})$.
- $\mathfrak{B}o(\mathbb{R})\subseteq \sigma(\mathcal{S})$.
Now, for to prove $\boxed{1}$, let's define $$\mathcal{O}(\mathbb{R})=\text{collection of all open sets of $\mathbb{R}$}.$$
and since that by definition $\boxed{\mathfrak{B}o(\mathbb{R})=\sigma(\mathcal{O}(\mathbb{R}))}$, we can see that $$\mathcal{S}\subseteq \mathcal{O}(\mathbb{R}) \implies \sigma(\mathcal{S}) \subseteq \sigma(\mathcal{O}(\mathbb{R})) \quad \text{and since that $\sigma(\mathcal{O}(\mathbb{R}))\subseteq \mathfrak{B}o(\mathbb{R})$ we have} \quad \sigma(\mathcal{S})\subseteq \sigma(\mathcal{O}(\mathbb{R}))\subseteq \mathfrak{B}o(\mathbb{R}).$$
Finally, for to prove $\boxed{2}$, we need to remember that and every open set is the countable and disjoint union of open intervals, so we obtain $\mathcal{O}(\mathbb{R})\subseteq \sigma(\mathcal{S})$, so $$\mathfrak{B}o(\mathbb{R})\subseteq \sigma(\mathcal{S}).$$
Is it correct? any suggestion?
Thanks!
Best Answer
It is true as you showed that $\sigma(\mathcal S)$ is contained in the Borel $\sigma$-algebra. It is true that an open set can be written as a countable disjoint union of open intervals. But you have also to prove that any open interval belongs to $\sigma(\mathcal S)$, which is actually not the case.
Let $$ \mathcal B:=\{A\in\mathcal B(\mathbb R), A=-A\}, $$ where $-A:=\{-a,a\in A\}$. It is possible to show that $\mathcal B$ is a $\sigma$-algebra containing $\mathcal S$ and that it is the smallest $\sigma$-algebra having this feature.