Measure Definition Equivalency Given the definition of measurability of a set $F$:
$(1)$ A set $F \subset \mathbb{R}^d$ is said to be measurable if for every $\epsilon > 0$, $\exists$ an open set $O$, containing $F$, such that $\mu^*(O - F) \leq \epsilon$.
Measurability is also defined as:
$(2)$ A set $F \subset \mathbb{R}^d$ is said to be measurable if for every $A \subset \mathbb{R}^d$, with $\mu^*(A)<\infty $, the following is true
$$\mu^*(A)=\mu^*(A \cap F)+\mu^*(A - F)$$
How does $(1) \implies (2)$?
$\mu^*$ denotes the outer-measure here
LIMITATIONS:
(1) The outer measure is defined as the infimum of the volume of cubes covering the set.
(2) Only an open set is the disjoint union of cubes (which provides for additivity).

You can show using the definition of the outer measure that
$$\mu^*(A) = \inf \{ \mu^*(O) \mid A \subset O, \text{ $O$ open}\}.$$ According to your comment you are able to show that if $O$ is open and $F$ is measurable, then $$\mu^*(O \cap F) + \mu^*(O \setminus F) = \mu^*(O).$$ If $F$ is measurable and $A$ is arbitrary, you can let $O$ be an open set containing $A$ to find
$$\mu^*(A \cap F) + \mu^*(A \setminus F) \le \mu^*(O \cap F) + \mu^*(O \setminus F) = \mu^*(O).$$ Take the infimum over all such $O$ to obtain
$$\mu^*(A \cap F) + \mu^*(A \setminus F) \le \mu^*(A).$$
