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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?

Some related answers: link 1, link 2, link 3

Thanks

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1 Answer 1

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  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 $\{ 1 \}$?

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

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