Why Borel sets require closure under both unions and intersections

[Math Amateur]

TL;DR

I need help in order to fully understand the concept of the set of Borel sets ...

I am reading Sheldon Axler's book: Measure, Integration & Real Analysis ... and I am focused on Chapter 2: Measures ...

I need help in order to fully understand the set of Borel sets ... ...

The relevant text reads as follows:

My questions related to the above text are as follows:QUESTION 1

In the above text by Axler we read the following:

" ... ... However, the set of all such intersections is not the set of Borel sets (because it is not closed under countable unions). ... ..."Can someone please explain why exactly that the set of all such intersections is not the set of Borel sets ... ? Why exactly is such a set not closed under countable unions and why is this relevant?
QUESTION 2

In the above text by Axler we read the following:

" ... ... The set of all countable unions of countable intersections of open subsets of $\mathbb{R}$ is also not the set of Borel sets (because it is not closed under countable intersections). ... ... "Can someone please explain why exactly that the set of all countable unions of countable intersections of open subsets of $\mathbb{R}$ is not the set of Borel sets ... ? Why exactly is such a set not closed under countable intersections and why is this relevant?
Help with the above two questions will be much appreciated ...

Peter

 

[member 587159]

Question 1 and question 2 are dual (use complements), so I will only answer question 1.

Q1: Axler claims that the set $\mathcal{B}:=\{\bigcap \epsilon\mid \epsilon \mathrm{\ countable \ collection \ of \ opens}\}$ is not equal to all Borel sets. His argument is: showing that there exists a sequence of sets $(B_n)_n$ in $\mathcal{B}$ such that $\bigcup_n B_n \notin \mathcal{B}$. Since the Borel sets have the property that countable unions remain Borel, we deduce that $\mathcal{B}$ cannot be equal to the Borel sets.

I'll give an explicit example now for the situation I'm describing above. First, note that every singelton in $\Bbb{R}$ is in $\mathcal{B}$. To see this, note that $\{x\}= \bigcap_{n=1}^\infty(x-1/n, x+1/n)$. Hence, $\Bbb{Q}$ is a countable union of sets in $\mathcal{B}$. It suffices to show that $\Bbb{Q}$ is not in $\mathcal{B}$. This is non-trivial (and probably why Axler does not go in detail) as it uses the Baire category theorem (this theorem is probably proven later in the book). Here you can find a proof:

https://math.stackexchange.com/questions/69451/

To understand that link, we introduce a definition: a $G_\delta$ set is by definition a countable intersection of open sets, so $\mathcal{B}$ is the set of $G_\delta$-sets.

 

[Math Amateur]

Question 1 and question 2 are dual (use complements), so I will only answer question 1.

Q1: Axler claims that the set $\mathcal{B}:=\{\bigcap \epsilon\mid \epsilon \mathrm{\ countable \ collection \ of \ opens}\}$ is not equal to all Borel sets. His argument is: showing that there exists a sequence of sets $(B_n)_n$ in $\mathcal{B}$ such that $\bigcup_n B_n \notin \mathcal{B}$. Since the Borel sets have the property that countable unions remain Borel, we deduce that $\mathcal{B}$ cannot be equal to the Borel sets.

I'll give an explicit example now for the situation I'm describing above. First, note that every singelton in $\Bbb{R}$ is in $\mathcal{B}$. To see this, note that $\{x\}= \bigcap_{n=1}^\infty(x-1/n, x+1/n)$. Hence, $\Bbb{Q}$ is a countable union of sets in $\mathcal{B}$. It suffices to show that $\Bbb{Q}$ is not in $\mathcal{B}$. This is non-trivial (and probably why Axler does not go in detail) as it uses the Baire category theorem (this theorem is probably proven later in the book). Here you can find a proof:

https://math.stackexchange.com/questions/69451/

To understand that link, we introduce a definition: a $G_\delta$ set is by definition a countable intersection of open sets, so $\mathcal{B}$ is the set of $G_\delta$-sets.

Thanks for a most helpful post Math_QED ...

Still reflecting on what you have written ...

Peter