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

Question

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?

Answer 1

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.

Answer 2

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 $E.$