TITLE: Is this actually true?
QUESTION [6 upvotes]: This exercise appeared in my real analysis test last year, and is still puzzling me since then. Ironically, even the professor doubts if the b part is actually truth (still...)
Let $A \subset \mathbb{R}$ 
a) If $\displaystyle0<m^*(A)<\infty$, then for every $\alpha\in (0,1)$, exists an open interval $I$ such that $\displaystyle \alpha m^*(I) \leq m^*(A \cap I) $ 
b) If $ \displaystyle m^*(A \cap I) \leq  \frac{m^*(I)}{2}$ for every open interval, then $m^*(A) = 0$

REPLY [2 votes]: Consider $A_n:=A\cap\subseteq [-n,n]$ for $n\in \mathbb N$. Let $\mu:=m^*(A_n)$. Then $\mu\le 2n<\infty$.
If $\varepsilon>0$, then there exists a countable collection of Intervals $I_k$, $k\in \mathbb N$, such that $A\subseteq\bigcup_k I_k$ and $\sum_k m^*(I_k) < \mu+\varepsilon$. We obtain $\mu \le \sum_k m^*(I_k\cap A_n) \le \sum_k \frac 12 m^*(I_k)=\frac 12\sum_k  m^*(I_k)<\frac{\mu+\varepsilon}2$, hence $\mu<\varepsilon$. Since $\epsilon>0$ was arbitrary, we conclude $\mu=0$, i.e. $m^*(A_n)=0$.
Finally, $A=\bigcup_n A_n$ implies $m^*(A)\le \sum_n m^*(A_n)=0$.