TITLE: An unbounded subset of the reals containing no intervals with the following property.
QUESTION [0 upvotes]: I'm trying to find an unbounded subset of the reals, $A$, with the property that:
$m(A\cap I)<0.9m(I) $ for all intervals, $I$. And $m(A)\neq0$. Where $m(X)$ denotes the outer measure of $X\subset\mathbb{R}$.
Thus far, I have concluded that if such a set does exist, clearly it must be uncountable, it must contain no intervals otherwise you cannot satisfy "$m(A\cap I)<0.5m(I) $", but I can't think of much else.

REPLY [4 votes]: Let $0\lt\alpha\lt1$, and suppose $A$ is a subset of $\mathbb R$ such that $m(A\cap I)\lt\alpha\ m(I)$ for every interval I, where $m$ is Lebesgue outer measure on $\mathbb R$.
Consider a fixed interval $I$. Choose an open set $B$ such that $A\cap I\subseteq B$ and $m(B)\lt\alpha\ m(I)$. Write $B=\bigcup_{n=1}^{\infty}I_n$ where the $I_n$'s are pairwise disjoint open intervals, so that $m(B)=\sum_{n=1}^{\infty}m(I_n)$. Then $m(A\cap I)\le\sum_{n=1}^{\infty}m(A\cap I_n)\lt\sum_{n=1}^{\infty}\alpha\ m(I_n)=\alpha\ m(B)\lt\alpha^2\ m(I)$.
Repeating this argument, we have $m(A\cap I)\lt\alpha^km(I)$ for every interval $I$ and every natural $k$, whence $m(A)=0$.
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