Measure of irrational numbers on an bounded interval $[a,b]$

Question

Since I am new to measure theory, I would like to ask the following question:

I do know that the measure of the rational numbers $\mathbb{Q}$ in a real interval $[a,b]$ with $b>a$, has a measure of zero. But how can I prove that the measure of the irrational numbers in the same $[a,b]$ is equal to $(b-a)$? (for which I am not sure).

Thank you all in advance!

Answer 1

Note that $[a,b]=([a,b] \cap \mathbb{Q}) \cup ([a,b] \cap ([a,b] \setminus \mathbb{Q} ))$ and $([a,b] \cap \mathbb{Q}) \cap ([a,b] \cap ([a,b] \setminus \mathbb{Q} ))=\emptyset$, so:

$$b-a=\lambda([a,b])=\lambda(([a,b] \cap \mathbb{Q}) \cup ([a,b] \cap ([a,b] \setminus \mathbb{Q} )))= \\= \lambda([a,b] \cap \mathbb{Q})+ \lambda([a,b] \cap ([a,b] \setminus \mathbb{Q} ))=0+ \lambda([a,b] \cap ([a,b] \setminus \mathbb{Q} ))$$

Answer 2

Measure is additive: $$ m([a,b]\cap\mathbb Q) + m([a,b]\setminus\mathbb Q) = m([a,b])=b-a. $$ If one of the two terms on the left is $0$, then you can find the other.