Consider the collection of subsets $A$ of the unit interval $[0,1]$ which are dense, meaning that for every $x\in[0,1]$, for every $\varepsilon>0$, there exists $a\in A$ such that $|x-a|<\varepsilon$. What are the Lebesgue measures of these sets?
Clearly these sets are bounded above by the unit interval itself, which is dense and has Lebesgue measure $1$. On the other hand, the set $\Bbb Q \cap [0,1]$ is dense and has Lebesgue measure null.
My question is this: for any $m\in[0,1]$, does there exist a dense subset $A\subseteq[0,1]$ with Lebesgue measure $m$?
Edit: I found out that if $A$ has measure $m$ and satisfies $|A\cap I|=m|I|$ for every interval $I\subseteq[0,1]$ (a better, stronger condition) where $|\cdot|$ denotes Lebesgue measure, then the density at a point $x\in A$ is given by
$$ d(x) = \lim_{\varepsilon\rightarrow0}
\frac{|A\cap(x-\varepsilon,x+\varepsilon)|}{|(x-\varepsilon,x+\varepsilon)|}
= \begin{cases} |A|/2 & \text{if } x=0\text{ or }1 \\
|A| & \text{if }x\in(0,1) \end{cases}$$
Lebesgue's density theorem says that if $A$ is measurable then $d(x)=1$ for almost all $x\in A$, and since we established $d(x)=|A|$ for $x\in(0,1)$, which is almost all of $[0,1]$, this implies $|A|=1$.
Best Answer
The answer is yes. For $m\in [0,1]$ consider the set $A:=[0,m]\cup(\mathbb{Q}\cap [0,1])$. This is clearly dense and has measure $m$.