Distribution of a dense subset in an interval

Question

This is a very general question. Given a bounded interval $I$, such as $[0,1]$, and an open dense subset $D \subset I$ such that $m(D \cap I) < 1/2 \cdot m(I)$(say), can we say anything interesting about $m(D \cap I')$ in terms of $m(I')$ for any subinterval $I' \subset I$? in particular a similar upper bound for $m(D \cap I')$ in terms of the length $m(I')$.

Answer 1

No and here is a counterexample. Let $I=[0,1]$ and $I_n=[\frac{1}{2^{n-1}},\frac{1}{2^n}]$ for $n\in\mathbb{N}\setminus\{0\}$. Our strategy will be to define open dense subsets $D_n\subset I_n$ such that $m(D_n)=\frac{1}{2^{n+1}}$. In this case, $m(D)=m(\bigcup_n D_n)=\frac{1}{2}m(I)$ (we can easily adjust to get an inequality at the end). We can then cut each $I_n$ into two pieces: the top piece $I^1_n=[\frac{1}{2^{n-1}},\frac{3}{2^{n+1}}]$ and the bottom piece $I^2_n=(\frac{3}{2^{n+1}},\frac{1}{2^n}]$. You can check that this splits each interval exactly in half, since we cut at its midpoint. When defining each $D_n$, we can choose a proportion of $D_n$ to split between $I^1_n$ and $I^2_n$, i.e. some $r$ such that $m(D_n\cap I^1_n)/m(I^1_n)=r$ and $m(D_n\cap I^2_n)/m(I^2_n)=1-r$. Think of this as the percentage of $D_n$ inside $I^1_n$ and the percentage of $D_n$ inside $I^2_n$; since $m(I^1_n)=m(I^2_n)=m(D_n)$, we can have at most $100$% of $D_n$ in the top half $I^1_n$ and $0$% in the second half $I^2_n$, or any other such split where each halves add to the total $100$% of the measure of $D_n$.

Because there are countably many intervals $I_n$, we can index the rational numbers in the set $[0,1]\cap\mathbb{Q}$ as $r_1,r_2,...$ and associate the each $r_n$ to the system $I^1_n$, $I^2_n$ and $D_n$ as is done above with "$r$". In this case, we get for any rational proportion $r\in[0,1]\cap\mathbb{Q}$, there is some interval $I_n$ with $m(D_n\cap I^1_n)/m(I^1_n)=r$. Letting $D$ be the open dense subset $\bigcup_n D_n\subset I$, this completes the counterexample.

To ensure that $m(D)=m(\bigcup_n D_n)<\frac{1}{2}m(I)$, we can choose the interval $D_n$ associated to the ratio $r=1$ (i.e. with $m(D_n\cap I^1_n)/m(I^1_n)=1$ and $m(D_n\cap I^2_n)/m(I^2_n)=0$) and shrink both $I^1_n$ and $D_n$, while expanding $I^2_n$ so that we still have i) $I_n=I^1_n\sqcup I^2_n$, ii) $m(D_n\cap I^1_n)/m(I^1_n)=1$, and iii) $m(D_n\cap I^2_n)/m(I^2_n)=0$. In this case $m(D_n)<\frac{1}{2^{n+1}}$, and consequently total measure has the property $m(D)=m(\bigcup_n D_n)<\frac{1}{2}m(I)$.

I hope I was as clear as possible with my reasoning and explanation, but drawing out this solution will help greatly if there's any confusion about the proof or its motivation!