Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm going over old comprehensive exams and part of one question is giving me a bit of trouble. It asked for an example of a subset of the real numbers such that the set and its complement were measure dense (with respect to the Lebesgue measure). A set is measure dense if its intersection with any open interval has positive measure. Any help would be greatly appreciated.

share|improve this question
3  
You can find a detailed discussion of such sets in this Math StackExchange question from 13 August 2011: Construction of a Borel set with positive but not full measure in each interval. –  Dave L. Renfro May 19 '12 at 10:38
add comment

2 Answers

This article explicitly constructs a Cantor set $C\subseteq[0,1]$ of measure $1/2$. The same construction can clearly be used to build a Cantor set $C(a,b)\subseteq[a,b]$ of measure $(b-a)/2$ for any $a,b\in\Bbb R$ with $a<b$.

Let $D_0=C_0=C(0,1)$. Let $\{(a_0(n),b_0(n)):n\in\Bbb N\}$ be an enumeration of the maximal open subintervals of $[0,1]\setminus D_0$, let $$C_1=\bigcup_{n\in\Bbb N}C\Big(a_0(n),b_0(n)\Big)\;,$$ and let $D_1=D_0\cup C_1$.

In general, given $C_k$ and $D_k$ for some $k\in\Bbb N$, let $\{(a_k(n),b_k(n):n\in\Bbb N\}$ be an enumeration of the maximal open subintervals of $[0,1]\setminus D_k$, let $$C_{k+1}=\bigcup_{n\in\Bbb N}C\Big(a_k(n),b_k(n)\Big)\;,$$ and let $D_{k+1}=D_k\cup C_{k+1}$.

Now let $A=\bigcup_{k\in\Bbb N}C_{2k}$ and $B=\bigcup_{k\in\Bbb N}C_{2k+1}$. $A$ and $B$ are both measure dense in $[0,1]$, since any non-empty open set in $[0,1]$ contains intervals $\big(a_{2k}(n),b_{2k}(n)\big)$ and $\big(a_{2k+1}(m),b_{2k+1}(m)\big)$ for some $k,m,n\in\Bbb N$. Moreover, $|A\cap B|$ is countable and therefore a null set so for every open $U\subseteq[0,1]$ we have $$m(U\setminus A)\ge m\big(U\cap (B\setminus A)\big)=m(U\cap B)>0\;,$$ and hence $[0,1]\setminus A$ is measure dense in $[0,1]$ as well.

Now just extend this to $\Bbb R$ by replacing $A$ by $\bigcup\limits_{n\in\Bbb Z}(A+n)$, the union of its integer translates.

share|improve this answer
    
Why did you define $D_{k+1} = C_k \cup C_{k+1}$? If you define it to be $D_{k+1} = D_k \cup C_{k+1}$ then all the $C_k$s are disjoint and $A \cap B = \varnothing$, I think. Wouldn't that be easier? Because I don't yet see why $|A \cap B|$ is countable. –  Matt N. May 19 '12 at 17:01
1  
@Matt: Typo: it was supposed to be $D_k$. No, $A\cap B\ne\varnothing$: they intersect at the points $a_k(n)$ and $b_k(n)$. But there are only countably many such points. –  Brian M. Scott May 19 '12 at 18:44
    
Ok, thanks. But I don't see how $C_k \cap C_i$ is non-empty. : / –  Matt N. May 19 '12 at 18:57
    
@Matt: To keep it simple, look at $C_0$ and $C_1$. $C_0$ has $(3/8,5/8)$ as its central ‘hole’, so $3/8$ and $5/8$ will be $a_0(n)$ and $b_0(n)$ for some $n$. They already belong to $C_0$, just as $1/3$ and $2/3$ belong to the middle-thirds Cantor set, but they also belong to $C(a_0(n),b_0(n))$: they’re its outermost endpoints, analogous to $0$ and $1$ of the middle-thirds Cantor set. Thus, $C_0$ has two points in common with each of the $C(a_0(n),b_0(n))$ and hence $\omega$ points in common with $C_1$. –  Brian M. Scott May 19 '12 at 19:03
    
Oh. I read $C(a,b)$ as the interval $(a,b)$ with holes removed to get a fat Cantor set. : ( I'm not making any sense, since it was clear to me that $C(0,1)$ is $[0,1]$ with holes removed. But why on earth don't you write $C[a,b]$ rather than $C(a,b)$? : ) –  Matt N. May 19 '12 at 19:09
show 1 more comment

Note that one can construct a "fat Cantor set" as a subset of some interval with any measure strictly between the measure of the interval. Let $A_0$ be a fat cantor set on $[0,1]$. There are countably many intervals deleted during the construction, now fill those in with smaller fat cantor sets of appropriate size. One obtains $A_1$. Continue recursively. At each stage, one filled in fat cantor sets on those intervals. Let $A = \bigcup A_n$. Now copy translate $A \subset [0,1]$ all over $\mathbb{R}$.

Now $0 < m(I \cap A) < m(I)$. $A$ is measure dense.

share|improve this answer
1  
Why is the complement of $A$ also measure dense? –  Asaf Karagila May 19 '12 at 8:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.