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.
$\begingroup$
$\endgroup$
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
2
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 $r$.
-
$\begingroup$ Sorry i forgot to tell E is set of all irrational in $ [ 0, 1 ] $\endgroup$ Commented Jun 2, 2017 at 18:14
-
3$\begingroup$ This is irrelevant. The "density" in $[0,1]$ has the same properties as in $\Bbb R$. $\endgroup$– ajotatxeCommented Jun 2, 2017 at 18:15
$\begingroup$
$\endgroup$
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 $m(A)=0$.
