# Borel Sigma Algebra generated by Open Intervals

So I know that the Borel $\sigma$-algebra of $\mathbb{R}$ is the $\sigma$-algebra generated by open sets. I have been able to prove that this Borel $\sigma$-algebra is also generated by the family of open intervals of the form $(a,b), a,b \in \mathbb{R}$ Now I want to show that the family of open intervals $(a,\infty)$ also generate the Borel $\sigma$-algebra. So what I have done is first show that

$(a,b) = \bigcup_{q \in \mathbb{Q}, q<b} ((a, \infty)-(q,\infty))$. This shows that every open interval is a countable union of intervals of the form $(a,\infty)$ and thus $\sigma$-algebra generated by $(a,b)$ is contained in the $\sigma$-algebra generated by $(a, \infty)$ But how would I show the other way around, i.e. that every interval $(a, \infty)$ is in the Borel $\sigma$- algebra?

(Can we just say that every interval $(a, \infty)$ is infact an open interval in itself and thus belongs to the $\sigma$-algebra generated by $(a,b)$?)

Denote $$\mathcal{B}=\sigma\left(\tau\right)=\sigma\left(\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \right)$$ (this equality was found out by you allready) where $$\tau$$ denotes the topology and $$\mathcal{B}_{1}=\sigma\left(\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \right)$$.

Then $$\left\{ \left(a,\infty\right)\mid a\in\mathbb{R}\right\} \subset\tau$$ so that $$\mathcal{B}_{1}\subseteq\mathcal{B}$$.

To prove the converse inclusion it is enough to show that $$\left\{ \left(a,b\right)\mid a,b\in\mathbb{R}\right\} \subset\mathcal{B}_{1}$$.

We have $$\left(a,c\right]=\left(a,\infty\right)-\left(c,\infty\right)\in\mathcal{B}_{1}$$ for each $$c$$, and consequently $$\left(a,b\right)=\bigcup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]\in\mathcal{B}_{1}$$.

• I don't see any contradiction: $\left(a,b\right]=\cap_{n\in\mathbb{N}}\left(a,b+\frac{1}{n}\right]\in\mathcal{B}_{1}$. Maybe, $\left(a,b\right)=\cup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]\in\mathcal{B}_{1}$. Commented Nov 20, 2018 at 15:38
• @Aidas You are right, and thank you for attending me. I repaired. I really don't understand (and remember) how I could write that. Commented Nov 20, 2018 at 16:12

Hint: $$(a,\infty) = \bigcup_{n \in \mathbb{N}_0} (a+n,a+n+2).$$

And yes: If you know that the Borel-$$\sigma$$ algebra contains all open sets, then $$(a,\infty) \in \mathcal{B}(\mathbb{R})$$ follows also from the fact that $$(a,\infty)$$ is an open set.

• Ok, so even though the identity you gave is correct I think I will just use the fact $(a, \infty)$ is an open interval, hence an open set. Just one last question, how would I show the above for $[a, \infty)$. I can express $(a,b)$ as a union of intervals of the form $[a, \infty)$ but can't show the other way around Commented Apr 18, 2015 at 7:36
• @user1314 Hint: Use $$[a,\infty) = \bigcap_{n \in \mathbb{N}} (a- \frac{1}{n},\infty)$$
– saz
Commented Apr 18, 2015 at 7:43
• By the way a small correction, a bit pedantic maybe, but in the above formua, don't you mean $n\in \mathbb{N}_{0}$, since you used that in the earlier one. Commented Apr 18, 2015 at 7:50
• @user1314 You mean the formula in my previous comment? No, I don't mean $n \in \mathbb{N}_0$ - $\frac{1}{n}$ is not well-defined for $n=0$.
– saz
Commented Apr 18, 2015 at 7:52
• @user1314 Ah, I see. Well, usually $\mathbb{N}$ denotes $\{1,2,3,\ldots\}$ and $\mathbb{N}_0 = \{0,1,2,3,\ldots\}$.
– saz
Commented Apr 18, 2015 at 7:55