Contradiction of the Borel sigma-field when we think the set of irrational numbers.

Question

I understand the sigma-field as not allowing uncountable union or intersection. In set of irrational numbers, it is the Borel set because it is a complement set of rational numbers. But if we think it as a union of the irrational numbers, it is a uncountable union and it can't be the Borel set. How do I think about Borel set?

Answer 1

If there is some way to build the set from countable intersections/unions and complements from open sets, it is Borel. Otherwise it's not (and showing a set to be non-Borel is much harder than witnessing it is a Borel set).

As $\Bbb P$ (the irrationals) can be written as the complement of a countable set and each countable set is a countable union of singletons which are closed (and thus the complement of an open set), you have a valid verification that $\Bbb P$ is in $\text{Bor}(\Bbb R)$. There it stops, you don't have to think about ways it is not Borel anymore.

Most sets that you can write down explicitly will be Borel in this way. Of course it suffices to build a set up from sets that you already know to be Borel and as long as you only use the $\sigma$-algebra operations you're guaranteed to still get Borel sets and only Borel sets.

If you find an invalid "proof" of a statement it doesn't mean the statement is false. That is a logical fallacy. We just need one good proof for the statement to be true.