4

For some bounded interval $I$, start with some compact $K \subset I$ of positive measure that contains no interval (i.e, ${m(K \cap J) < m(J)}$ for every interval $J$). The complement $I \setminus K$ is open, and thus can be expressed as an at most countable union of disjoint open intervals $V_n$. For each $V_n$, let $K_n \subset V_n$ be a compact subset containing no interval with $m(K_n) > m(V_n)/2$. Define $E_1 := \bigcup_{n=1}^\infty K_n \subset I \setminus K$. Define $E_2 \subset (I \setminus K) \setminus E_1$ similarly and so on.

Let $E := \bigcup_{n=1}^\infty E_n$. For any subinterval $I' \subset I$, we can show that $m(E \cap I') > 0$. Yet I'm not sure about the other half of the inequality $m(E \cap I') < m(I')$. Or could it be that the $E$ constructed in this way does not satisfy the given property?

shark
  • 1,551
  • What’s $|J|$ here and why is it needed – SBF May 06 '23 at 07:09
  • $J$ is any interval. – shark May 06 '23 at 07:49
  • For a compact nowhere dense subset $K \subset I$, $I \setminus K$is a countable union of open intervals. The hint suggest that we “fill in these open intervals and iterate.” I want to know the details of this construction. – shark May 06 '23 at 07:56
  • I’ve asked about absolute value sign specifically – SBF May 06 '23 at 08:01
  • It just denotes the length $m(J)$ of the interval. – shark May 06 '23 at 08:07
  • Why don’t you rewrite the OP in consistent symbols, using $m(\cdot)$ instead of $|\cdot|$. Currently it’s confusing since the latter may also mean just the cardinality of the set – SBF May 06 '23 at 08:14
  • It is such in Tao's textbook. – shark May 06 '23 at 08:17
  • Are you sure they denote the same thing then? Also there are still places in your post that use the second notation – SBF May 06 '23 at 08:22
  • 1
    I've changed the symbol for consistency. And yes, for this problem here, they denote the same thing. – shark May 06 '23 at 08:40
  • Strict inequities have a tendency to look strictness when taking limits. Why do you think it should still hold true? Could you replace it with a non strict inequality, making the measure difference uniform? – SBF May 06 '23 at 11:47
  • No. A set $E \subset I$ such that $0 < m(E \cap I’) < m(I’)$ for every subinterval $I’ \subset I$ is what we are aiming at here. $I \setminus K$ is a countable union of disjoint open intervals, and the hint explicitly suggests that “fill in these open intervals and iterate.” – shark May 06 '23 at 12:05
  • How is $E_2$ defined "similarly"? $(I\setminus K)\setminus E_1$ is not an open set unless I'm very much mistaken – FShrike May 10 '23 at 17:33
  • The answer to this question may be helpful: https://math.stackexchange.com/questions/961745/construct-a-borel-set-on-r-such-that-it-intersect-every-open-interval-with-non-z – bof May 10 '23 at 19:49
  • @FShrike: $I \setminus K$ is a countable union of open intervals, from each of these interval, we remove another copy of the $K$, the resulting set will be open. Denote the union of these copies by $E_1$, then $(I \setminus K) \setminus E_1$ is a countable union of open sets, which is again open. – shark May 10 '23 at 22:04
  • I hadn’t registered that each $K_n$ is contained in the open intervals of $I$, whoops – FShrike May 10 '23 at 22:07
  • @bof: I am aware of Rudin's elegant solution to this problem as provided in that answer. However, the original hint for these problem explicitly suggests that we " first work in a bounded interval, such as ${(-1,2)}$. The complement of the set ${K}$ is the union of at most countably many open intervals, Now fill in these open intervals and iterate." See Exercise $50$ of https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/#comments – shark May 10 '23 at 22:10

2 Answers2

0

I've done this one before but I can't find it right now.

The main idea: Let $I_n, n=1,2,\dots$ be the open intervals in $I=(0,1)$ with rational endpoints. We can inductively choose disjoint Cantor sets $K_n\subset I_n$ with $m(K_n)>0$ for all $n.$ The set $E=\bigcup_1^\infty K_n$ will do the job.

Let me expand on the above: Clearly we can choose a Cantor set $K_1\subset I_1$ with $m(K_1)>0.$ Next, note $I_2\setminus K_1,$ which is open and nonempty, contains an open interval of positive measure. Within that interval, we can choose a Cantor set $K_2$ with $m(K_2)>0.$ At this point we have chosen disjoint sets $K_1\subset I_1, K_2\subset I_2,$ with $m(K_1),m(K_2)>0.$

OK, a lttle bit more: The set $I_3\setminus(K_1 \cup K_2)$ is open and nonempty, so contains a Cantor set $K_3$ of positive measure disjoint from $K_1,K_2$. Etc ...This will lead to the desired set $E.$

zhw.
  • 107,943
  • They have a specific question, namely showing that $m(E\cap I')<m(I')$ with their notations – FShrike May 10 '23 at 17:32
  • If your sets $K_n$ are chosen so that $m(K_n)\ge\frac12m(I_n)$ for each $n$ then the set $E$ will have full measure. – user14111 May 10 '23 at 17:38
  • @user14111 I didn't choose them that way. – zhw. May 10 '23 at 19:19
  • How do we know you didn;'t choose them that way? All you told us is that you choose $K_n\subset I_n)$ with $m(K_n)\gt0$. You need to be more specific. How do you ensure that $m(E\cap I_n)\lt m(I_n)$? – user14111 May 10 '23 at 19:44
0

To understand whether the inequality m(E∩I')<m(I') holds, we need to consider the construction of E and its relationship with I'.

From your construction, E is the union of sets E_n each of which is a collection of compact subsets from I\K, with each compact subset having measure less than half of its respective interval V_n.

For each E_n, since the sum of the measures of the compact subsets K_n within each open interval V_n is less than the measure of V_n, the measure of E_n is also less than the measure of I\K.

Now consider an arbitrary subinterval I'⊂I. Because each E_n is a subset of I\K, and the E_n's are disjoint, it follows that m(E∩I') is less than m(I') since I' also contains parts of the set K which are not included in E.

So, for any subinterval I'⊂I, the inequality m(E∩I')<m(I') should hold, given the construction of E and the measure properties of the sets involved.

Sam Walker
  • 66
  • We have a Latex-like typesetting system for mathematical expressions, called MathJax. Information is here: https://math.meta.stackexchange.com/a/10164 – 311411 May 13 '23 at 01:34