I am trying to find an open dense subset of $[0,1]$ with Lebesgue measure exactly $m$. I know that I can look at an enumeration $\{q_n\}_{n\in \mathbb{N}}$ of rationals in $[0,1]$ and look at this set:
$$A=\cup_n \left( q_n-m2^{-n},q_n+m2^{-n}\right)$$
which is an open dense subset of measure atmost $m$ but may I know how to get a set of measure exactly $m$?
Best Answer
As you suggested, define for $t \in [0,1]$ $$A:=A(t) := \bigcup_{n \in \mathbb{N}} (q_n-t2^{-n},q_n+t 2^{-n}).$$ If we set $$f(t) := \lambda(A(t)), \qquad t \in [0,1],$$ then $f$ is continuous and satisfies $f(0)=0$, $f(1)\geq1$. By the intermediate value theorem, there exists for any $m \in [0,1]$ some $t \in [0,1]$ with $\lambda(A(t))=f(t)=m$; the set $A(t)$ is open, dense and has Lebesgue measure $m$.