Let $\mu^*$ be the outer Lebesgue measure on $\mathbb{R}$, then $A\subseteq\mathbb{R}$ is measurable iff $\mu^*(I)=\mu^*(I\cap{}A)+\mu^*(A\backslash{}I)$ for every open interval $I$.
The first direction is trivial. The other one is troubling.
My attempt is the following: Suppose $A$ satisfies the above equality and let $B$ be any subset of $\mathbb{R}$. For every $\epsilon>0$ there exists an open set $U$ such that $B\subseteq{}U$ and $\mu^*(B)\leq\mu^*(U)<\mu^*(B)+\epsilon$. Then $B\cap{}A\subseteq{}U\cap{}A$ and thus $\mu^*(U\cap{}A)<\mu^*(B\cap{}A)+\epsilon$. Doing the same for $A^{c}$ we get $\mu^*(U\cap{}A^c)<\mu^*(B\cap{}A^c)+\epsilon$ and thus $\mu^*(B)\leq{}\mu^*(U)=\mu^*(U\cap{}A)+\mu^*(U\cap{}A^c)<\mu^*(B\cap{}A)+\mu^*(B\cap{}A^c)+2\epsilon$. But from this I only get that $\mu^*(B)\leq{}\mu^*(B\cap{}A)+\mu^*(B\backslash{}A)$
How can I continue?
If you just write your inequalities in the reverse direction, you will get your result.
As $B\subset U \implies B \cap A \subset U \cap A$ $\&$ $B \cap A^c \subset U \cap A^c $. Hence $\mu^*(U \cap A)\geq \mu^*(B \cap A) $ and similarly $\mu^*(U \cap A^c)\geq \mu^*(B \cap A^c) $.
So what you get is that $\mu^*(B) \geq \mu^*(U) - \epsilon = \mu^*(U \cap A) + \mu^*(U \cap A^c) -\epsilon \geq \mu^*(B \cap A) + \mu^*(B \cap A^c) -\epsilon $. As $\epsilon$ is arbitrary, your result is hence proved.
You are also subtly using one more result, which I think should be mentioned in the answer, that the equality $\mu^*(U)= \mu^*(U \cap A) + \mu^*(U \cap A^c)$ is true for any open set $U$ as $U$ can be written as a union of open intervals.