# 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.

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 $r$.
• Sorry i forgot to tell E is set of all irrational in $[ 0, 1 ] Jun 2, 2017 at 18:14 • This is irrelevant. The "density" in$[0,1]$has the same properties as in$\Bbb R$. Jun 2, 2017 at 18:15 There's even an uncountable dense set on$[0, 1]$with Lebesgue measure$0$: Begin with the ordinary Cantorset$C$. Create the set$C + \mathbb Q = \{\, c+q \mid c\in C, \, q\in\mathbb Q \,\}.$Since this is a countable union of sets of measure$0$, it has measure$0$, and since$C$is uncountable, so is$C + \mathbb Q.$It's dense since$\mathbb Q$is dense. Now take$A = (C+\mathbb Q) \cap E$. We then have$m(E)=1$but$m(A)=0\$.