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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 $|A|=1$.

4 Answers4

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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 $m$.

Jonas Lenz
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    Great! However, this set is quite "lop-sided". I am wondering if there is a more evenly distributed example. –  Aug 09 '19 at 04:59
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    Then I refer to the answer of saz. Take an evenly distributed set of Lebesgue measure $m$ and "add" $\mathbb{Q}\cap [0,1]$. – Jonas Lenz Aug 09 '19 at 05:07
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    @M.Nestor Depends on what you mean by evenly distributed. It would be cool if, given a constant $0 < c < 1$, it were possible to find a measurable set $A$ such that $m(A \cap I) = c m(I)$ for every interval $I \subseteq [0,1]$. This turns out to be impossible –  Aug 09 '19 at 05:16
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    @Bungo After reading that post, it seems that what I am after is precisely Lebesgue's density theorem. Thanks! –  Aug 09 '19 at 05:17
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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.

saz
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  • For the 2nd example, the set must be less than $\epsilon$ since there are some overlaps between those intervals, right? – efhvcjnfdbgefg Aug 09 '19 at 05:14
  • @efhvcjnfdbgefg Yes, that's correct. There are plenty of overlaps (... exactly because the rationals are dense). – saz Aug 09 '19 at 05:15
  • In fact $\varepsilon$ must always be an overestimate, since any interval $(q_n-2^{-n}\varepsilon,q_n+2^{-n}\varepsilon)$ must contain another $q_m$, guaranteeing overlap at least $\min{\varepsilon2^{-m},\varepsilon2^{-n}}$. –  Aug 09 '19 at 05:23
  • My above comment doesn't change the validity, since the overestimate varies continuously, you can always find $\varepsilon$ such that the true measure is precisely $m$. –  Aug 13 '19 at 17:47
  • could you have a look at my construction and tell me if it is fine? Thanks a lot! (see answer below) – dfnu Aug 03 '20 at 20:27
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Sure. Take $$ [0,m]\cup(\mathbb{Q}\cap [0,1])=[0,m]\overset{\cdot}{\cup}(\mathbb{Q}\cap(m,1]). $$

Nick Peterson
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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 $\varepsilon$.

dfnu
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