If $m^*(A\cap(a,b))\leq\frac{b-a}{2}$. Show that $m(A)=0$. Let $A\subset\mathbb{R}$ such that $m^*(A\cap(a,b))\leq\frac{b-a}{2}$ for any $a,b\in\mathbb{R}$. Show that $A$ is measurable and $m(A)=0$.
Here $m^*(A)$ is the Lebesgue outer measure of $A$.
In this question I'm trying to prove that $m(A)=0$ (So it will also imply the measurability of $A$).
So for that, what I was trying to do is, to write $A$ as a union of small intervals so that if I can get something like:
$$m^*(A)\leq m^*(A\bigcup_{k}(a_k,b_k))\leq\sum\limits_k m^*(A\cap(a_k,b_k))<\epsilon$$
But I really cannot see properly how to connect the dots.
Appreciate your help
 A: Suppose that $m^*(A)>0.$ Then, there is a closed interval $I$ such that $m^*(A\cap I)=r>0.$ Without loss of generality, $I=[0,1]$. Now, since $r\le 1/2$, so by definition of the outer measure, $A$ must be contained in a union of intervals of length $strictly\  less$ than $1$. Using translation invariance of the Lebesgue measure, we may assume $A\cap I\subseteq [0,\delta_1]: \delta_1<1$. It follows that  $r\le \delta_1/2<1/2$ so (since $r$ is strictly less than $1/2$), there is a $0<\delta_2<1/2$ such that $A\cap I\subseteq [0,\delta_2]$. It follows that  $r\le \delta_2/2<\delta_1/2^2$. Continuing, we may inductively define a sequence $(\delta_n)$ such that $A\cap I\subseteq [0,\delta_n]$ and $r<\delta_1/2^n$, which of course, is a contradiction, as soon as $n$ is big enough.
A: By the definition for outer measure, there exists an open set such that $m^*(O)-m^*(A)< \epsilon$ for any $\epsilon > 0$ where $A \subset O$.
Let $O$ be the union of disjoint intervals $(a_k,b_k)$ where $k \in N$.
Now, $$A= A \cap O= A\cap (\cup (a_k,b_k))=\cup (A \cap (a_k,b_k))$$.
So, $$m^*(A)=m^*(\cup (A \cap (a_k,b_k))) \leq m^*(A\cup (a_k,b_k)) \leq \Sigma(b_k-a_k)/2=m^*(O)/2$$.
Therefore, $m^*(O)/2 \leq m^*(O)-m^*(A) < \epsilon$ which reduces to $m^*(O) \leq 2\epsilon$.
