# Open interval as a countably infinite union of half-open intervals

I am reading Jacod and Protter's Probability Essentials in preparation to learn Itō calculus. It's quite a terse book, but the presentation is straightforward. However, I am having a hard time following one of the theorems. It is stated thus:

The Borel $$\sigma$$-algebra of $$\mathbb{R}$$ is generated by intervals of the form $$(-\infty, a]$$, where $$a \in \mathbb{Q}$$.

Proof

Let $$\mathcal{C}$$ denote all open intervals. Since every open set in $$\mathbb{R}$$ is the countable union of open intervals, we have $$\sigma(\mathcal{C})$$ = the Borel $$\sigma$$-algebra of $$\mathbb{R}$$.

Let $$\mathcal{D}$$ denote all intervals of the form $$(\infty, a]$$, where $$a \in \mathbb{Q}$$. Let $$(a, b) \in \mathcal{C}$$, and let $$(a_{n})_{n \ge 1}$$ be a sequence of rationals decreasing to $$a$$ and $$(b_{n})_{n \ge 1}$$ be a sequence of rationals increasing strictly to $$b$$. Then

\begin{align}(a, b) &= \cup_{n=1}^{\infty}(a_{n}, b_{n}] \\ &= \cup_{n=1}^{\infty}((-\infty, b_{n}]\cap(-\infty, a_{n}]^{c})\end{align}

...

I get where they're going with this but I am lost on some specific things.

1. Why do we require $$a \in \mathbb{Q}$$?
2. Why do we require the sequences $$a_{n}$$ and $$b_{n}$$ to be composed of rationals?
3. Why does $$b_{n}$$ need to be “increasing strictly”, and in fact what does “increasing strictly” even mean?
4. Most importantly, why do we not have

$$(a, b] = \cup_{n=1}^{\infty}(a_{n}, b_{n}]$$

Question 4 is particularly confusing. How does $$b$$ not get included in their proof? That is, how can a countably infinite union of half-open intervals not converge to a half-open interval?

Thanks

1. Why do we require $$a \in \mathbb Q$$?

It's not really required. You can generate the Borel $$\sigma$$-algebra of $$\mathbb R$$ with intervals of the form $$(-\infty, a]$$ where $$a \in \mathbb R$$, but you can also accomplish the same thing by restricting the finite endpoints to be rational numbers.

Why do we require the sequences $$a_n$$ and $$b_n$$ to be composed of rationals?

Because we're trying to show that we can use intervals of the form $$(-\infty, a]$$ to generate Borel $$\sigma$$-algebra of $$\mathbb R$$, with $$a \in \mathbb Q$$. That's exactly what's happening in the line $$\cup_{n = 1}^{\infty} \bigl((-\infty, b_n], \cap (-\infty, a_n]^\complement\bigr)$$. (You have a typo in your version of this union.)

Why does $$b_n$$ need to be “increasing strictly”, and in fact what does “increasing strictly” even mean?

"Increasing strictly" means that $$b_n < b_{n + 1}$$ for all $$n$$ in the sequence. I'm not sure if this is necessary, but you certainly can choose the sequence $$(b_n)$$ to behave this way, and if it makes the proof easier to think through or present, then there's nothing wrong with that.

Most importantly, why do we not have

$$(a, b] = \cup_{n=1}^{\infty}(a_{n}, b_{n}]$$

Consider the union $$\bigcup_{n = 1}^{\infty} \bigl(\frac{1}{n+1}, 1 - \frac{1}{n+1}\bigr]$$. If this is equal to $$(0, 1]$$, then we must have $$\{ 1 \} \in \bigl(\frac{1}{j+1}, 1 - \frac{1}{j+1}\bigr]$$ for some $$j \in \mathbb N$$. That's how unions work. So for which $$j$$ does $$\bigl(\frac{1}{j+1}, 1 - \frac{1}{j+1}\bigr]$$ contain $$\{ 1 \}$$?

• Thanks for your response, I'm reading through it. But also, what was the typo exactly? May 30, 2022 at 16:04
• @matheus You're missing a minus sign in "$(\infty, a_n]^c$". May 30, 2022 at 16:08
• Fixed! Now on to your solution... May 30, 2022 at 16:22