# Finding an open dense subset of a given Lebesgue measure

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 $$m$$?

## 3 Answers

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 $$m$$.

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 $$0\leq q_n-\frac{m}{M_\mathcal E}m2^{-n-1} where $${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 $$B:=\bigcup_{n\in\mathbb N}\left(q_n-\frac{m}{\mu(A)}m2^{-n-1},q_n+\frac{m}{\mu(A)}m2^{-n-1}\right)\cap[0,1].$$

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 $$[0,1]$$.