Show that $m(A)=0$

Question

Let $​0< \delta <1$. Let ​$ A\subset \mathbb{R}$ be a lebesgue measurable set. Suppose that for any interval $(a, b)$ we have

$$m(A \cap(a,b) )\leq \delta \cdot m(a,b).$$

Here $m$ denotes the lebesgue measure on $\mathbb{R}$​. Show that $ m(A)=0$.

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Answer 1

Hint: It suffices to treat the case $m(A)<\infty.$ Let $\epsilon>0.$ Then there exist open intervals $I_n$ that cover $A$ such that $\sum_{n=1}^{\infty}m(I_n) < m(A) + \epsilon.$ We then have

$$m(A) = m(A\cap(\cup_{n=1}^{\infty}I_n)) \le \sum_{n=1}^{\infty} m(A\cap I_n) \le \delta \sum_{n=1}^{\infty} m( I_n).$$