Prove Borel Measurable Set A with the following property has measure 0. 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.
 A: Hint: Think about Bass's Proposition 4.14 (1): given any $\delta > 0$, there is an open set $G$ with $A \subset G$ and $m(G - A) < \delta$.  Then recall (or prove) that every open set is a countable union of disjoint open intervals.  

Here are some more details.  An argument like yours has difficulties with the possibility $m(A) = \infty$, but we can reduce to the case $m(A) < \infty$ by intersecting $A$ with large bounded sets, as below.
Suppose $\epsilon \in (0,1)$ is such that $m(A \cap I) \le (1-\epsilon)m(I)$ for each interval $I$.  Let $k$ be a positive integer and let $A_k = A \cap [-k,k]$.  Note that $m(A_k) \le 2k < \infty$, and that for each interval $I$, we have $$m(A_k \cap I) \le m(A \cap I) \le (1-\epsilon) m(I). \tag{1}$$  
Let $\alpha > 0$ be arbitrary, and using Proposition 4.14, choose an open set $G$ with $A_k \subset G$ and
$$m(G - A_k) < \alpha \epsilon. \tag{2}$$
In particular, $m(G) = m(A_k) + m(G-A_k) < 2k + \alpha \epsilon < \infty$.
  Now $G$ can be written as a disjoint union of open intervals: $G = \bigcup_{n=1}^\infty I_n$.  Note that $m(I_n) \le m(G) < \infty$ for each $n$.  Also, $G - A_k = \bigcup_{n=1}^\infty (I_n - A_k)$ which is also a disjoint union.
Now since $I_n - A_k = I_n - (I_n \cap A_k)$, we have
$$m(I_n - A_k) = m(I_n) - m(I_n \cap A_k) \ge m(I_n) - (1-\epsilon)m(I_n) = \epsilon m(I_n) \tag{3}$$ using (1).  So by countable additivity,
$$m(G - A_k) = \sum_{n=1}^\infty m(I_n - A_k) \ge \sum_{n=1}^\infty \epsilon m(I_n) = \epsilon m(G). \tag{4}$$
Combining (1) and (3), we get $\epsilon m(G) < \alpha \epsilon$, so $m(G) < \alpha$.  In particular, since $A_k \subset G$, we have $m(A_k) < \alpha$.  But $\alpha > 0$ was arbitrary, so we must have $m(A_k) = 0$.
Moreover, $k$ was arbitrary, so $m(A \cap [-k,k]) = 0$ for every $k$.  Since $A = \bigcup_{k = 1}^\infty (A \cap [-k,k])$, by countable additivity we conclude $m(A) = 0$.
Now let's drop the assumption $m(A) < \infty$.  For any $n$ and any interval $I$, we have $$m(A \cap [-n,n] \cap I) \le m(A \cap I) \le (1-\epsilon)m(I).$$
Hence $A \cap [-n,n]$ satisfies the same condition and moreover $m(A \cap [-n,n]) \le 2n < \infty$.  So by the previous case, $m(A \cap [-n,n]) = 0$.  Now since $A = \bigcup_{n=1}^\infty (A \cap [-n,n])$, by countable additivity $m(A) = 0$.

For an explicit example with a fat Cantor set, let's consider the example given on Wikipedia, where at stage $n \ge 1$ we remove $2^{n-1}$ intervals, each of length $2^{-2n}$.  The final set $C$ has measure $1 - \sum_{n=1}^\infty 2^{n-1} \cdot 2^{-2n} = 1 - \sum_{n=1}^\infty 2^{-n-1} = \frac{1}{2}$.
Let $I_k$ be the leftmost interval that remains after stage $k$.  At stage $k$ the leftmost interval that was removed was centered at $2^{-k}$ and had length $2^{-2k}$, so the leftmost interval that remains is
$$I_k =  \left[0, 2^{-k} - \frac{1}{2} 2^{-2k}\right] = [0, 2^{-k}(1-2^{-k-1})].$$
At the next stage, we will remove from $I_k$ one interval of length $2^{-2(k+1)}$ from $I$, then two intervals of length $2^{-2(k+2)}$ and so on.  So the total length of the intervals removed from $I_k$ is
$$\sum_{n=1}^\infty 2^{n-1} 2^{-2(k+n)} = 2^{-2k} \sum_{n=1}^\infty 2^{-n-1} = 2^{-2k} \frac{1}{2} = 2^{-2k-1}.$$
Therefore, we have 
$$\frac{m(C \cap I_k)}{m(I_k)} = \frac{m(I_k) - 2^{-2k-1}}{m(I_k)} = \frac{2^{-k}(1-2^{-k-1}) - 2^{-2k-1}}{2^{-k}(1-2^{-k-1})} = \frac{(1-2^{-k-1}) - 2^{-k-1}}{(1-2^{-k-1})}$$
upon cancelling a factor of $2^{-k}$.  It is clear by inspection that $$\lim_{k \to \infty} \frac{m(C \cap I_k)}{m(I_k)} = 1,$$ so given $\epsilon > 0$ you can choose $k$ so large that $m(C \cap I_k) > (1-\epsilon) m(I_k)$.
A: Here's a less computation way of doing this. 
Take $f = \chi_A$ , the indication function of $A$. Suppose A has a positive measure. Then, by the Lebesgue Differentiation Theorem, there exists an $x \in A$ and a family of intervals $I_n$ centered around $x$ which strictly decrease to $x$ with $\frac{1}{m(I_n)} \int_{I_n} f(t) dt \rightarrow f(x) = 1$.
However, the left hand side is exactly $m(I_n \cap A) / m(I_n) \leq 1 - e < 1$ for all $n$, hence when we take our limit, it cannot be one, hence a contradiction. 
Of course, that theorem is a pretty strong result, so it's probably better to have done as the other posters did it. 
A: Follow the hints given by Kevin Carlson and Nate Eldredge I think I get it. Could anyone help me checking whether the following prove is valid?
If $A$ is a Borel Measurable Set and $m(A)>0$, for a  given $\epsilon>0$, there exist open set $G$ containing $A$ and $m(G-A)<\epsilon m(A)$ (exist by proposition given).
$G$ is countable union of disjoint intervals $I_i$, by the assumption of the question for each $I_i$ we have $m(A\cap I_i)\leq m(I_i) (1-\epsilon)$. By countable additivity we conclude
$m(A)=m(A\cap G)\leq m(G) (1-\epsilon)$. Combining(adding inequality) with $m(G-A)<\epsilon m(A)$ we have $$m(G)<m(G)(1-\epsilon)+\epsilon m(A)\leq m(G) (1-\epsilon)+\epsilon m(G)=m(G)$$
By the contradiction we conclude $m(A)=0$.
A: 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
$$ m^*(A) \le \sum_n m^*(A_n) = 0.$$
