Show $m(A \cap I) \leq (1-\epsilon)m(I)$ for every $I$ implies $m(A) = 0$ The problem I stuck was following. 

Suppose $m$ is Lebesgue measure and $A$ is a Borel measurable subset of $\mathbb{R}$. Prove that if 
  $$ 
m(A\cap I) \leq (1-\epsilon)m(I)
$$
  for every open interval $I$, then $m(A) = 0$.

I asked about this problem to the TA, and they just suggested me to use open interval with outer measure. 
The outer measure I am using is 
$$
m^\star (E) = \inf  \{ \sum_{i=1}^{\infty} l(A_{i}) : E \subset \cup_{i=1}^{\infty}A_{i} \}
$$ 
Thanks in advance.
 A: Here is a simple argument which does not require regularity or even measurability and is probably what the TA had in mind: assume that $m^*(A \cap I) \le (1-\epsilon) l(I)$ for any open interval $I$.
If $m^*(A) < \infty$ cover $A$ with a sequence $\{I_k\}$ of open intervals. The monotonicity of the outer measure and the stated assumption imply $$m^*(A) = m^*\left( A \cap \bigcup_k I_k \right) \le \sum_k m^*(A \cap I_k) \le (1-\epsilon) \sum_k l(I_k).$$
Now take the infimum over all coverings. What does that tell you about $m^*(A)$?
If $m^*(A) = \infty$ let $A_m = A \cap (-m,m)$ and use the fact that $$m^*(A_m \cap I) = m^*(A \cap (I \cap (-m,m)) \le (1-\epsilon) l(I \cap (-m,m)) \le (1-\epsilon) l(I).$$
A: Hint: If $A$ is measurable and $m(A)>0$ then there is an open $B \supset A$ with $m(B) < (1+\epsilon) m(A)$.
A: What about this? 
Let $ A \subset \cup_{i=1}^{\infty}I_{i}$ and let all $\{ I_{i}\}$ be disjoint open interval. It is possible to cover $A$ with disjoint open interval because open set can be expressed as the union of open interval. 
Then for each $I_{i}$ and fixed $\epsilon \in (0,1)$, 
$$
m(A \cap I_{i}) \leq (1-\epsilon)m(I_{i}) 
$$
by given assumption. 
Then $$
\sum_{i=1}^{\infty} m(A \cap I_{i}) \leq \sum_{i=1}^{\infty}(1-\epsilon)m(I_{i})
$$
and
$$ \sum_{i=1}^{\infty}m(A \cap I_{i}) = m(A) $$
Then 
$$
m(A) \;\;\leq \;\; \sum_{i=1}^{\infty}(1-\epsilon)m(I_{i}) \;\; \leq \;\; (1-\epsilon)\sum_{i=1}^{\infty}m(I_{i}) \;\; \leq \;\;(1-\epsilon)\sum_{i=1}^{\infty}m(\cup_{i=1}^{\infty}I_{i})
$$
Then 
$$
(1-\epsilon)\sum_{i=1}^{\infty}m(\cup_{i=1}^{\infty}I_{i}) \;\;\leq \;\;(1-\epsilon)l(\cup_{i=1}^{\infty}I_{i}) \;\;\leq \;\;(1-\epsilon)\left[m(A) + \epsilon^\star \right]
$$
for arbitary $\epsilon^\star$
Thus, as 
$$
m(A) \;\; \leq \;\;(1-\epsilon)\left[m(A) + \epsilon^\star \right]  
$$ 
for arbitary $\epsilon^\star$, the only $m(A)$ satisfying the inequality is $m(A) = 0$
