Let $A\subset\mathbb{R}$ a measurable and bounded set. Show that exists for each $0<\alpha<1$ an interval $I$ such that $m(A\cap I)/m(I)>\alpha$.
Solution 1:
Here is another approach (I assume $m$ is Lebesgue measure): if, to the contrary, $m(A \cap I) \le \alpha m(I)$ for every interval $I$ you would have $$m(A) \le m \left( A \cap \cup I_k \right) \le \sum_k m(A \cap I_k) \le \alpha \sum_k m(I_k)$$ whenever $\{I_k\}$ is a covering of $A$ by intervals. What is the definition of $m(A)$?