Asking for a hint: proving a set is measurable 
Suppose $b < c$ and $A \subset (b,c)$. Prove that $A$ is Lebesgue
  measurable if and only if $|A| + |(b,c) \setminus A| = c- b$.

where $|A|$ denotes the outer measure of $A$.
I know how to prove that if $A$ is measurable, $|A| + |(b,c) \setminus A| = c -b$. However, I'm stuck in proving the other direction.
I've tried to use these facts: if $A$ is measurable, then there exists an open set $G \supset A$ such that $|G\setminus A| < \epsilon$ or there exists a closed set $ F \subset A$ such that $|A \setminus F| < \epsilon$, but to no avail. Can I get a hint?
 A: This is more than a hint but there are many details for you to fill in.
Recall the definition of the inner measure, $m_*(A) := \sup \{ m^*(B): B \subset A, B \text{ measurable}\}$, and let $m^*(A)$ denote the outer measure of $A$. 
One approach here is to show that $m^*(A) = m_*(A)$ since this implies that $A$ is Lebesgue measurable. (This is sometimes taken as the definition of measurability from which Caratheodory's definition follows.) 
We are given that $A \subset I = [b,c]$.  Using the fact that for any $n \in \mathbb{N}$ there exists an open set $G_n \supset I\setminus A$ such that $m^*(G) < m^*(I \setminus A)+1/n$ and we can produce a measurable set $C = \bigcap_{n=1}^\infty G_n$ such that $C \supset I \setminus A$ and $m^*(I \setminus A) = m^*(C)$.
Since $C$ and $I \setminus C$ are measurable we have 
$$m^*(I) = m^*(I \setminus C) + m^*(C) =m^*(I \setminus C) + m^*(I \setminus A)$$
Also, since $I \setminus C \subset A$ we have $m^*(I \setminus C) \leqslant m_*(A)$ by the definition of inner measure.
Thus,
$$\underbrace{m^*(A) + m^*(I \setminus A)}_{|A| + |[b,c]\setminus A| } = c-b = m^*(I) \leqslant m_*(A)+ m^*(I \setminus A)$$
This implies that $m^*(A) \leqslant m_*(A)$.  Since it always holds that $m_*(A) \leqslant m^*(A)$ it follows that $m_*(A) = m^*(A)$ and $A$ is Lebesgue measurable.
