# Clarification on a construction of a set $E \subset I$ intersecting every subinterval of $I$ with positive, non-full measure

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?

• What’s $|J|$ here and why is it needed
– SBF
Commented May 6, 2023 at 7:09
• $J$ is any interval. Commented May 6, 2023 at 7:49
• 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. Commented May 6, 2023 at 7:56
– SBF
Commented May 6, 2023 at 8:01
• I've changed the symbol for consistency. And yes, for this problem here, they denote the same thing. Commented May 6, 2023 at 8:40

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.

• We have a Latex-like typesetting system for mathematical expressions, called MathJax. Information is here: math.meta.stackexchange.com/a/10164 Commented May 13, 2023 at 1:34

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 $$E.$$

• They have a specific question, namely showing that $m(E\cap I')<m(I')$ with their notations Commented May 10, 2023 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. Commented May 10, 2023 at 17:38
• @user14111 I didn't choose them that way.
– zhw.
Commented May 10, 2023 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 $m(E\cap I_n)\lt m(I_n)$? Commented May 10, 2023 at 19:44