Q:

Dense subset of irrational have same measure

If A is subset of E. Where E is fix subset of irrational number. Suppose A is dense in E . Can we say that both A and E have same Lebesgue measure ? ( Assume both set measurable). I guess that it is true but i don't know how to proceed.

A:

No. The set of irrational algebraic numbers is countable, so it has Lebesgue measure $0$.
This set is dense in $\Bbb R$. It is easy to prove (and a bit tedious, if you ask me) that in every open interval $(a,b)$ there is some irrational $\sqrt r$ for rational $r$.

