TITLE: Prove Borel Measurable Set A with the following property has measure 0.
QUESTION [16 upvotes]: This question is exercise 4.10 of Richard F. Bass's Real Analysis for Graduate Students, 2nd edition.
Let $\epsilon \in (0,1)$, let $m$ be Lebesgue measure, and suppose $A$ is Borel Measurable subset of $\mathbb R$. Prove that if $$ m(A\cap I)\leq (1-\epsilon)m(I)$$
for every interval $I$, then $m(A)=0$.
Collection of my thoughts 
1. Try to prove by contradiction, suppose $m(A)=a>0$ Borel Measurable Set satisfies the condition given. A property of any Borel Measurable Set on $\mathbb R$ is $A$ contains a closed set $F$ and is contained in open set $O$, and the measure of $F$ and $O$ can be arbitrarily close to $a$.
2. Obviously $A$ contains no interval, but it does not lead to contradiction, because we can find an example of $m(A)>0$ that contains no interval(fat Cantor Set(remove middle 1/4 each time) has measure $1/2$ and contains no interval).
2'. For fat Cantor Set $C$ and given $\epsilon>0$ how could we construct an interval I satisfies $m(C \cap I)>(1-\epsilon)m(I) ?$

Maybe try to write Borel measurable set as countable union/intersection of intervals.

I'm stucked here and will appreciate any help, some examples/special case will be great too.

REPLY [3 votes]: The set $A$ does not need to be assumed Borel measurable (or even Lebesgue measurable if you replace $m$ by $m^*$), and it is not necessary to use approximation by open sets. This follows almost directly from the definition of Lebesgue outer measure and countable subadditivity.
Assume $0 < \alpha < 1$ and $m^*(A \cap I) \le \alpha m(I)$ for every interval $I$.
First let $A_n = A \cap [-n,n]$ so that $m^*(A) \le 2n < \infty$. If $\{I_k\}$ is any countable family of intervals whose union contains $A_n$ then
$$ m^*(A_n) = m^* \left(A_n \cap \cup I_k\right) \le m^* \left(A \cap \cup I_k\right) \le \sum_k m^*(A \cap I_k) \le \alpha \sum_k m(I_k).$$
Now take the infimum over all such families $\{I_k\}$ to obtain
$$m^*(A_n) \le \alpha m^*(A_n)$$
which implies $m^*(A_n) = 0$. Finally since $A = \cup_n A_n$ you have
$$ m^*(A) \le \sum_n m^*(A_n) = 0.$$