$A$ a proper subset of $[0,1]$ and $m(A \cap [a,b])\leq \frac{b-a}{2}$ for every interval $[a,b]$ 
Let $A \subset [0,1]$ measurable and $m(A \cap [a,b])\leq \frac{b-a}{2}$ for every interval $[a,b]$.  Show $m(A)=0$.

I have two ideas on how to approach this, but I don't think either of them have enough momentum to be pushed forward.
(1)  First I wanted to pick a countable, disjoint set of intervals $\{I_k\}$ where $l(I_k)<\frac{b-a}{n}$ and $A \subset \cup I_k$.
(2)  Another strategy I thought about was breaking $A$ into disjoint sets via $A_n := A \cap [a_n, b_n]$.  However, I think this approach may be messy.
 A: If we apply the Lebesgue differentiation theorem to the characteristic function $\chi_A$, then we have 
$$\chi_A(x)=\lim_{h\to 0} \frac{m(A\cap[x-h,x+h])}{2h}$$ almost everywhere, but by the properties of $A$, the function we are taking the limit of is always less than or equal to $\frac{1}{2}$.  Thus, $\chi_A\leq \frac{1}{2}$ almost everywhere, but since $\chi_A$ only takes the values $0$ or $1$, it must be equal to $0$ almost everywhere.  
A: Suppose $m(A)>0$. Note that $m(A)\leq 1$ since $A\subset [0,1]$. Let $\epsilon=m(A)/4$ which is a finite positive number. For this chosen $\epsilon>0$, we can choose a countable union of open interval $(a_i,b_i)$, $i\geq 1$ such that $A\subset \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$ and 
$$\tag{1}-\epsilon+\sum_{i=1}^\infty(b_i-a_i)\leq m(A)\leq \sum_{i=1}^\infty(b_i-a_i).$$
Then 
$$\tag{2}m(A)=m(A\cap\bigcup_{i=1}^\infty(a_i,b_i))
=m(\bigcup_{i=1}^\infty A\cap(a_i,b_i))\leq\sum_{i=1}^\infty m( A\cap(a_i,b_i))$$
where the first equality follows from  $A\subset \displaystyle\bigcup_{i=1}^\infty(a_i,b_i)$, and the
last inequality follows from the subadditivity of measure. On the other hand, 
$$\tag{3}\sum_{i=1}^\infty m( A\cap(a_i,b_i))\leq\sum_{i=1}^\infty m( A\cap[a_i,b_i])
\leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}$$
where the last inequality follows from assumption. 
Now combining $(1)$, $(2)$ and $(3)$, we have
$$-\epsilon+\sum_{i=1}^\infty(b_i-a_i)\leq\sum_{i=1}^\infty\frac{b_i-a_i}{2}$$
which implies that (since $\sum_{i=1}^\infty(b_i-a_i)<\infty$ by $(1)$, we can do the subtraction)
$$\sum_{i=1}^\infty\frac{b_i-a_i}{2}\leq \epsilon=\frac{m(A)}{4}$$
which contradicts $(1)$ when $m(A)>0$. 
Therefore, we must have $m(A)=0$. 
