let $ D_1 = \left[0,1\right]\cap \mathbb{Q}$ and $ D_2 = \left[0,1\right]\cap \mathbb{Q}^C$, show that $m( D_1)=0$ and $m(D_2)=1$

Question

let $ D_1 = \left[0,1\right]\cap \mathbb{Q}$ and $ D_2 = \left[0,1\right]\cap \mathbb{Q}^C$, show that $m( D_1)=0$ and $m(D_2)=1$

$m(D1​)=0$

The set $D_1$​ is the set of all rational numbers in the interval $[0,1]$. Rational numbers are dense on the interval $[0,1]$, which means that there are infinitely many rational numbers on any non-empty subinterval of $[0,1]$. Therefore, $D_1$​ is also dense in the interval $[0,1]$.

The Lebesgue measure of a dense set on an interval is zero. This is because the Lebesgue measure of an interval is its length, and the length of a dense set on an interval is zero.

Therefore, $m(D1​)=0$.

$m(D_2​)=1$

The set $D_2$​ is the set of all irrational numbers in the interval $[0,1]$. Irrational numbers are a complementary set of rational numbers in the interval $[0,1]$, which means that $D_2​\cup D_1=[0,1]$ and $D_2​ \cap D_1​=∅$.

Since $m(D_1​)=0$ and $D_2​\cup D_1​=[0,1]$, then $m(D_2​)=m([0,1])−m(D_1​)=1−0=1$.

Therefore, $m(D_2​)=1$

Is this proof correct?

Are there any errors?

Answer 1

Your first proof ($D_1$) is incorrect in a fundamental way : to show that $D_1$ has measure zero, you shouldn't be able to argue by saying that $D_1$ is big (here, dense in $[0, 1]$). Sets with measure zero are meant to be small in a certain way, so you want to explain why $D_1$ is small. Here, it has something to do with its cardinality.

Your second proof ($D_2$) is perfectly correct though, once you have effectively proved that $m(D_1) = 0$.