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.
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- 3$\begingroup$ 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. $\endgroup$Dave L. Renfro– Dave L. Renfro2012-05-19 10:38:20 +00:00Commented May 19, 2012 at 10:38
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2 Answers
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6 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.
- $\begingroup$ 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. $\endgroup$Rudy the Reindeer– Rudy the Reindeer2012-05-19 17:01:29 +00:00Commented May 19, 2012 at 17:01
- 1$\begingroup$ @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. $\endgroup$Brian M. Scott– Brian M. Scott2012-05-19 18:44:54 +00:00Commented May 19, 2012 at 18:44
- $\begingroup$ Ok, thanks. But I don't see how $C_k \cap C_i$ is non-empty. : / $\endgroup$Rudy the Reindeer– Rudy the Reindeer2012-05-19 18:57:20 +00:00Commented May 19, 2012 at 18:57
- $\begingroup$ @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 $C_1$. $\endgroup$Brian M. Scott– Brian M. Scott2012-05-19 19:03:54 +00:00Commented May 19, 2012 at 19:03
- $\begingroup$ 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 $C(a,b)$? : ) $\endgroup$Rudy the Reindeer– Rudy the Reindeer2012-05-19 19:09:06 +00:00Commented May 19, 2012 at 19:09
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1 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.
- 2$\begingroup$ Why is the complement of $A$ also measure dense? $\endgroup$2012-05-19 08:25:05 +00:00Commented May 19, 2012 at 8:25