I've been trying to solve the following question from Bass:
Fix $\epsilon \in (0,1).$ If a Lebesgue measurable set $A$ is such that for every bounded interval $I$, $m(A \cap I) \leq (1-\epsilon) m(I)$, prove that $m(A)=0$
My attempt : I tried choosing a collection of intervals $\{C_i\}$ such that $\sum m(C_i) \leq \frac{1}{1-\epsilon} m(A)$. After that I am unable to make a fruitful move. Can anyone please provide a hint for this ?
Assume that $m(A) > 0$, since $A$ is Lebesgue measureable it's measure has to be equal to the outer measure, that is:
$m(A) = \operatorname{inf} \{\sum m(I_k) : I_k \text{ is a sequence of open intervals with } A\subseteq \bigcup I_k\}$
Let's choce a such sequence $I_k$ (with $\sum m(I_k) \ge m(A)$) such that ${1\over1-\epsilon/2}m(A) \ge \sum m(I_k)$ or $m(A) \ge (1-\epsilon/2) \sum m(I_k)$ (which is possible since $m(A) \gt 0$).
Since $A \subseteq \bigcup I_k$ we also have $A \subseteq \bigcup A \cap I_k$, so we have
$m(A) \le m(\bigcup A \cap I_k) \le \sum m(A \cap I_k) \le \sum (1-\epsilon)m(I_k) = (1-\epsilon)\sum m(I_k)$
Now $(1-\epsilon)\sum m(I_k) \ge m(A) \ge (1-\epsilon/2)\sum m(I_k)$, but since $\sum m(I_k) \ge m(A) \gt 0$ we would have $1-\epsilon \ge 1-\epsilon/2$ which can't be.