Proving the *Caratheodory Criterion* for *Lebesgue Measurability* I'm following Terry Tao's An introduction to measure theory. Here he has defined lebesgue measurability as:

Definition 1 (Lebesgue measurability): A set $E$ is said to be lebesgue measurable if for every $\epsilon > 0$, $\exists$ an open set $U$, containing $E$, such that $m^*(U \setminus E) \leq \epsilon$.

where $m^*$ denotes the outer lebesgue measure.
Then he has stated the Caratheodory Criterion:

Definition 2 (Lebesgue measurability): A set $E$ is said to be lebesgue measurable if for every elementary set $A$, we have 
  $$m(A)=m^*(A \cap E)+m^*(A \setminus E)$$

where an elementary set is a finite union of boxes.

The problem is to show the equivalence of the two definitions. I have shown that Def $(1)$ $\implies$ Def $(2)$, but having difficulty in showing the reverse, i.e. Def $(2)$ $\implies$ Def $(1)$. Any help (full/brief solution or even some hints) would be greatly appreciated! Thanks in advance.


I know that this is an old post but the existing answers don't seem to use the 'restricted' version of Caratheodory.
Pick an open set such that $m^*(O)\leq m^*(E) + \epsilon$ for some $\epsilon > 0$.
$$m^*(O)=\color{red}{ \Sigma m^*(B_n) \geq \Sigma(m^*(B_n \cap E) + m^*(B_n-E))}\geq m^*(\cup B_n\cap E)+m^*(\cup B_n-E)\\=m^*(E)+m^*(O-E)$$
This leads to $$m^*(E)+\epsilon \geq m^*(O)=m^*(E)+m^*(O-E)$$
Or
$$\epsilon \geq m^*(O-E)$$
We have not even used the Caratheodory criterion for elementary sets. We have only used them for boxes. We have used the fact that open sets are just disjoint, countable boxes and the countable subadditivity of outer measure. The place where Caratheodory criterion is used is highlighted in red.
