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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 $(a,b), a<b$ be an open interval in $\mathbb{R}$ and take $c\in\mathbb{R}, a<c<b$, 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.

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    $\begingroup$ It's perfect. At which parts do you have doubts? $\endgroup$
    – Berci
    Apr 22, 2021 at 23:15
  • $\begingroup$ @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). $\endgroup$
    – lorenzo
    Apr 22, 2021 at 23:33

1 Answer 1

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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 $c\leq a_n<c+\frac{1}{n}$ and $d-\frac{1}{n}<b_n\leq d.$
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 $\mathbb{R}$.

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