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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 $(a,b)$?)

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2 Answers 2

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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 $\left(a,b\right)=\bigcup_{n\in\mathbb{N}}\left(a,b-\frac{1}{n}\right]\in\mathcal{B}_{1}$.

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

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  • $\begingroup$ 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 $\endgroup$
    – user1314
    Commented Apr 18, 2015 at 7:36
  • $\begingroup$ @user1314 Hint: Use $$[a,\infty) = \bigcap_{n \in \mathbb{N}} (a- \frac{1}{n},\infty)$$ $\endgroup$
    – saz
    Commented Apr 18, 2015 at 7:43
  • $\begingroup$ 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. $\endgroup$
    – user1314
    Commented Apr 18, 2015 at 7:50
  • $\begingroup$ @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 $n=0$. $\endgroup$
    – saz
    Commented Apr 18, 2015 at 7:52
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    $\begingroup$ @user1314 Ah, I see. Well, usually $\mathbb{N}$ denotes $\{1,2,3,\ldots\}$ and $\mathbb{N}_0 = \{0,1,2,3,\ldots\}$. $\endgroup$
    – saz
    Commented Apr 18, 2015 at 7:55

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