Irrational numbers and Borel Sets

Is the set of all irrational numbers in [0; 1] a Borel set? If yes, what is its Lebesgue measure? I have been trying to answer this for a long time now. I know that the set of rational numbers is infinitely countable but I am having trouble with the proof of this question. I am not necessarily looking for an answer just confirmation that I am doing the correct thing.

I have already tried the following: The set of rationals is countable between 0 and 1 and is therefore a Borel Set. Since it is countable its lebesgue measure is 0

Solution 1:

The set of all rational numbers in $[0,1]$ is countable and hence a Borel set. Therefore, also its complement is a Borel set.

The Lebesgue measure of $[0,1]$ is $1$, the lebesgue measure of all rational numbers in $[0,1]$ is $0$ since it is countable. Therefore,...