# $\mathcal{S}$, smallest $\sigma$-algebra on $\mathbb{R}$ containing $\{(r,s]:r,s\in\mathbb{Q}\}$, is the collection of Borel subsets of $\mathbb{R}$

I have proved the following statement and I would like to know if my proof is correct, thank you:

"$$\mathcal{S}$$, smallest $$\sigma$$-algebra on $$\mathbb{R}$$ containing $$\{(r,s]:r,s\in\mathbb{Q}\}$$, is the collection of Borel subsets of $$\mathbb{R}$$"

DEF. (Borel set): The smallest $$\sigma$$-algebra on $$\mathbb{R}$$ containing all open subsets of $$\mathbb{R}$$, $$\mathcal{B}$$, is called the collection of Borel subsets of $$\mathbb{R}$$. An element of this $$\sigma$$-algebra is called a Borel set.

LEMMA (1): Let $$x\in\mathbb{R}$$: then there exists both a decreasing sequence and an increasing sequence of rational numbers converging to $$x$$.

LEMMA (2): Every open subset of $$\mathbb{R}$$ can be written as a countable union of disjoint open intervals.

My proof:

Let $$(a,b), a be an open interval in $$\mathbb{R}$$ and take $$c\in\mathbb{R}, a, arbitrary: then by Lemma (1) there exist a decreasing sequence of rational numbers $$(a_i)$$ such that $$a_i\overset{i\to +\infty}{\to} a$$ and an increasing sequence of rational numbers $$(c_i)$$ such that $$c_i\overset{i\to +\infty}{\to} c$$ so $$(a_i, c_i]\in \{(r,s]:r,s\in\mathbb{Q}\}\subset\mathcal{S}$$ and $$(a,c]=\bigcup_{i=1}^{\infty} (a_i, c_i]\in S$$ since $$\mathcal{S}$$, being a $$\sigma$$-algebra, is closed under countable unions. We thus have that every half-open interval $$(a,c]\in\mathbb{R}$$ belongs to $$\mathcal{S}$$ so it must also be that $$(a,b)=\bigcup_{n=1}^{\infty}(a,b-\frac{1}{n}]\in S$$ (i.e. every open interval in $$\mathbb{R}$$ also belongs to $$\mathcal{S}$$) hence, by LEMMA (2), every open set in $$\mathbb{R}$$. So, since every open set belongs to $$\mathcal{S}$$, a $$\sigma$$-algebra, and $$\mathcal{B}$$ is the smallest $$\sigma$$-algebra containing them by definition, we have that $$\mathcal{B}\subset\mathcal{S}$$. Since each half-open interval with rational endpoints $$(r,s]$$ belongs to $$\mathcal{B}$$ ($$(r,s)\in\mathcal{B}$$ because it is open, $$\{s\}\in\mathcal{B}$$ because closed sets are Borel sets so their union also belongs to $$\mathcal{B}$$) and $$\mathcal{S}$$ is the smallest $$\sigma$$-algebra containing them by hypothesis we also have that $$\mathcal{S}\subset\mathcal{B}$$ thus $$\mathcal{S}=\mathcal{B}$$, as desired.

• It's perfect. At which parts do you have doubts? Apr 22, 2021 at 23:15
• @Berci Thank you for your interest in my question; I don't really have doubts I just want other people to evaluate it so I know there's nothing missing in my proof (something I think I have got a proof but someone later points out to me there is a case I have not considered or something like that). Apr 22, 2021 at 23:33

I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
This exercise is Exercise 3 on p.38 in Exercises 2B in this book.

My solution is here:

Let $$a,b\in\mathbb{Q}.$$
$$(a,b)=\bigcup_{n=1}^\infty (a,b-\frac{1}{n}]\in\mathcal{S}.$$
Let $$c,d\in\mathbb{R}$$.
For any $$n\in\{1,2,\dots\}$$, there exists $$a_n,b_n\in\mathbb{Q}$$ such that $$c\leq a_n and $$d-\frac{1}{n}
Then, $$(c,d)=\bigcup_{n=1}^\infty (a_n,b_n).$$
Since $$(a_n,b_n)\in\mathcal{S}$$ for each $$n\in\{1,2,\dots\}$$, $$(c,d)\in\mathcal{S}$$.
A subset of $$\mathbb{R}$$ is open if and only if it is the union of a disjoint sequence of open intervals by 0.59 on p.30 in "Supplement for Measure, Integration & Real Analysis" by Sheldon Axler.
So, $$G\in\mathcal{S}$$ for any open subset $$G$$.
So, the collection of Borel subsets of $$\mathbb{R}$$ is a subset of $$\mathcal{S}.$$

Let $$r,s\in\mathbb{Q}.$$
Then, $$(r,s]=\bigcap_{n=1}^\infty (r,s+\frac{1}{n}).$$
Since each $$(r,s+\frac{1}{n})$$ is a Borel set, $$(r,s]$$ is also a Borel set by 2.25(c) on p.27 in "Measure, Integration & Real Analysis" by Sheldon Axler.
So, $$\mathcal{S}$$ is a subset of the collection of Borel subsets of $$\mathbb{R}$$.