Hello,

I wish to show the following:

The Borel-sets of the real line $\displaystyle \mathbb{R}$ are also generated by the system $\displaystyle \left\{(-\infty,a):a\in \mathbb{Q}\right\}$, i.e. that $\displaystyle \mathcal{B}=\mathcal{B(\mathbb{R})}=\sigma(\left\{ (-\infty,a):a\in \mathbb{Q}\right\})$.

This is what I have done:

1)

I will use the following information:

(a) the Borel $\displaystyle \sigma$-algebra is also generated by the system $\displaystyle \left\{[a,b):a,b\in \mathbb{Q}\right\}$

(b) if $\displaystyle \mathcal{F} \subset \mathcal{G} \subset \mathcal{A}$ then $\displaystyle \sigma(\mathcal{F}) \subset \sigma(\mathcal{G}) \subset \sigma(\mathcal{A})$

2)

Notice that $\displaystyle (\infty,a) \in \mathcal{O}$, where $\displaystyle \mathcal{O}$ is the system of open sets.

Notice also that $\displaystyle \mathcal{O} \subseteq \mathcal{B}=\mathcal{B(\mathbb{R})}$.

3)

Using (b) we have that $\displaystyle \mathcal{O} \subseteq \mathcal{B(\mathbb{R})} \Leftrightarrow \sigma(\mathcal{O}) \subseteq \sigma(\mathcal{B(\mathbb{R})})=\mathcal{B(\mathbb {R})} $

Since $\displaystyle (\infty,a)$ is only an element of $\displaystyle \mathcal{O}$ we have that $\displaystyle \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\}) \subseteq \sigma(\mathcal{O})$.

Therefore we have that $\displaystyle \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\}) \subseteq \mathcal{B(\mathbb{R})}$

4)

Using (a) we have for $\displaystyle a,b \in \mathbb{Q}$ and $\displaystyle a<b$ that:

$\displaystyle \mathcal{B}=\left\{[a,b):a,b\in \mathbb{Q}\right\}= \left\{ (-\infty,b) \setminus (-\infty,a):a,b\in \mathbb{Q}\right\} = \left\{(-\infty,b) \cap (-\infty,a)^{c}:a,b\in \mathbb{Q}\right\} $

By the properties of a $\displaystyle \sigma$-algebra we have that $\displaystyle (-\infty,b) \cap (-\infty,a)^{c} \in \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$.

We therefore have that $\displaystyle \mathcal{B}= \sigma(\left\{[a,b):a,b\in \mathbb{Q}\right\}) \subseteq \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$

5)

Since we have shown that $\displaystyle \mathcal{B} \subseteq \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$ and $\displaystyle \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\}) \subseteq \mathcal{B}$ we conclude that $\displaystyle \mathcal{B}= \sigma(\left\{(-\infty,a):a\in \mathbb{Q}\right\})$

Could someone comment on the above proof? Is it fine? Is there something missing?

Thanks.