Given $m\in[0,1]$, can we find a dense subset of $[0,1]$ whose Lebesgue measure is exactly $m$?

Question

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 $m$?

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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 $|A|=1$.

Answer 1

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 $m$.

Answer 2

Sure. Take $$ [0,m]\cup(\mathbb{Q}\cap [0,1])=[0,m]\overset{\cdot}{\cup}(\mathbb{Q}\cap(m,1]). $$

Answer 3

Well, yes. Just take any set $B$ of Lebesgue measure $m$ (e.g. $B=[0,m]$) and consider

$$A := B \cup (\mathbb{Q} \cap [0,1]).$$

The set has Lebesgue measure $m$ and it is dense in $[0,1]$.

It's a bit more tricky to construct an open dense set $A$ with small Lebesgue measure $m$. Here, one approach is to consider an enumeration $(q_n)_{n \in \mathbb{N}}$ of $\mathbb{Q} \cap [0,1]$ and $$A := \bigcup_{n \in \mathbb{N}} (q_n-\epsilon 2^{-n},q_n+\epsilon 2^{-n})$$ for fixed $\epsilon>0$. The set $A$ is open and has Lebesgue measure $\leq \epsilon$.

Remark: Note that there does not exist an open dense set with Lebesgue measure zero. In this sense, the best we can achieve is to have an open dense set of arbitrarily small Lebesgue measure, as above.

Answer 4

I am sort of new to topology so I post this as a tentative answer to see if you find it correct. Its a variation of saz's answer, in order to have an open dense subset of $[0,1]$ with measure exactly $\varepsilon$.

Let $q_1,q_2,\dots$ be, as usual, an enumeration of the rationals.

Let $A_0 = \emptyset$. For $n\geq 1$ construct an open interval $E_n$ such that $q_n \in E_n$ and $$m(E_n-A_{n-1})=\frac{\varepsilon}{2^n}.$$ Then $A_n = A_{n-1}\cup E_n$ has measure $$m(A_n) = \varepsilon\sum_{k=1}^n \frac1{2^k}$$ and $$A = A_1\cup A_2\cup A_3\dots$$ has measure $\varepsilon$.