Equivalence of two definitions of Measurable Sets (1) A subset $A$ of $\mathbb{R}^d$ is measurable if given $\epsilon>0$, there exists an open set $\mathcal{O}$, such that $A\subseteq \mathcal{O}$ and $m^*(\mathcal{O}-A)<\epsilon$.
(2) A subset $A$ of $\mathbb{R}^d$ is measurable if for every subset $X$ of $\mathbb{R}^d$, $$m^*(X)=m^*(A\cap X)+m^*(A^c\cap X).$$
Question: How can we prove that these two definitions are equivalent.
(one may give a hint also).
I am not too much familiar with the techniques of measure theory, I didn't get any direction.
 A: One important key is that $m^*(\mathcal O - A) = m^*(\mathcal O \cap A^c)$. 
Two the Lebesgue Measure is defined via volume of open boxes, and open sets, like $\mathcal O$, are made up of open boxes. 
Three in (2) $m^*(A)\leq m^*(A \cap X) + m^*(A \cap X^c)$ is always true. So the only relevant proof is one for $m^*(A) \geq m^*(A \cap X) + m^*(A \cap X^c)$.
Using these three it should be simple to get an idea, the rest is technical.
Note: a box is always open in the context below.
Start by proving (2) for every $A$ is a box, and $X$ only range every open box of $\mathbb R^d$. It is simple as then $A \cap X$ is another box, and $A \cap X^c$ is made up of finite decomposition of boxes $I_n$, which should give you a hint about what the Lebesgue measures of those two parts are. So you now have a finite sequence of boxes $I_n$ and a single box $I$ such that $m^*(A \cap X) = m^*(I)$ and $m^*(A \cap X^c) = \Sigma m^*(I_n)$, These follows from definition.
So $m^*(A \cap X)+ m^*(A \cap X^c) = \Sigma m^*(I_n)+ m^*(I)$.
But $I_n \cup I = A$  you use also covers $A$ (remember $A$ is a box) perfectly, meaning $m^*(A) =  \Sigma m^*(I_n)+ m^*(I)$.
Then we are going to extend it to every $X \subseteq \mathbb R^d$, instead of just boxes. $A$ is still just a box, however. But that is trivial, because we can cover $X$ with sequences of open boxes, which will give an upper bound of the measure $m^*(A \cap X)$ and $m^*(A \cap X^c$. By the result we just proved, and sigma subadditivity, we can obtain
$$m^*(A \cap X)+ m^*(A \cap X^c) \leq m^*(A).$$
Any open set is a countable union of open intervals. It then extends that (2) holds for ANY open sets. (Measurable sets form a $\sigma$-algebra for any measure. The proof is application of subadditivity).
Now (1) $\implies$ (2) is trivial. (Note "boxes" are open in our context) As we know (2) holds for $\mathcal O$. $\forall \varepsilon > 0$, we can find $\mathcal O_\varepsilon$
$$ m^*(O_\varepsilon) - m^*(A)$$
$$ = m^*((A) \cup (\mathcal O_\varepsilon-A)) - m^*(A)$$
$$ \leq m^*(A)+m^*(\mathcal O_\varepsilon -A) - m^*(A)$$
$$ = m^*(O_\varepsilon-A) <\varepsilon$$.
Then with $A \cap X \subseteq \mathcal O \cap X$, we get $m^*(A \cap X) \leq m^*(\mathcal O_\varepsilon \cap X)$, same for $X^c$, it follows that
$$m^*(A \cap X)+ m^*(A \cap X^c)$$
$$ \leq m^*(\mathcal O_\varepsilon \cap X)+ m^*(\mathcal O_\varepsilon \cap X^c) $$
$$ = m^*(\mathcal O_\varepsilon) = m^*(A) + \varepsilon \text{   ((2) for all open sets)}$$
$\varepsilon$ is arbitrary, meaning $m^*(A \cap X)+ m^*(A \cap X^c) \leq m^*(A)$, completing (1) $\implies$ (2).
For the other direction, you will need to create $\mathcal O$ from open boxes, specifically those used in the definition of $m^*(A)$. Since $m^*(A) = \inf\{$ total measure of sequence of boxes$\}$, meaning for every $\varepsilon$, we can find a sequence of boxes $\mathcal O = \bigcup I_n$ which measure $m^*(\mathcal O) - m^*(A) < \varepsilon$.
Sorry for a super long answer, but the three key above, and subaddivity of Lebesgue Measure. Hope it helps.
