Finding an open dense subset of a given Lebesgue measure

Question

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 $m$?

Answer 1

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 $m$.

Answer 2

Take an enumeration $\mathcal E:=\{q_n\}_{n\in\mathbb N}$ of $\mathbb Q\cap(0,1)$ satisfying that for each for each $n$ we have that $$0\leq q_n-\frac{m}{M_\mathcal E}m2^{-n-1}<q_n+\frac{m}{M_\mathcal E}m2^{-n-1}\leq1,$$ where $${M_\mathcal E}=\mu\left(\bigcup_{n\in\mathbb N}\left(q_n-m2^{-n-1},q_n+m2^{-n-1}\right)\cap[0,1]\right).$$ Note that if $q_1=\frac12$, we have that $M_\mathcal E\geq\frac m2$. This means that we can find an enumeration $\mathcal E$ satisfying the given conditions by choosing $q_i\in\left[2^{-i},1-2^{-i}\right]$. Since $\frac1n>\frac1{2^{n}}$, it is not that hard to construct such an enumeration.

Now given this enumeration $\mathcal E:=\{q_n\}_{n\in\mathbb N}$, I set $$A:=\bigcup_{n\in\mathbb N}\left(q_n-m2^{-n-1},q_n+m2^{-n-1}\right)\cap[0,1].$$ Note that $0<\mu(A)\leq m$, and take $$B:=\bigcup_{n\in\mathbb N}\left(q_n-\frac{m}{\mu(A)}m2^{-n-1},q_n+\frac{m}{\mu(A)}m2^{-n-1}\right)\cap[0,1].$$

Answer 3

You can take the complement (in $[0,1]$) of a fat Cantor set with measure $1-m$. If $K$ is such a fat Cantor set, then:

• since $K$ has Lebesgue measure $1-m$, the Lebesgue measure of $[0,1]\setminus K$ is $m$;
• since $\mathring K=\emptyset$, $[0,1]\setminus K$ is dense;
• since $K$ is compact, $K$ is a closed subset of $[0,1]$, and therefore $[0,1]\setminus K$ is an open subset of $[0,1]$.