Measurable set, inner and outer measure I am trying to show the following
"If $E \subset \mathbb R^n$ is Lebesgue measurable then for every $A \subset E$, $$m(E)=m_i(A)+m_e(E \setminus A)$$
I got stuck trying to show this, I know that the following inequalities are satisfied $$m(E)=m_e(E) \leq m_e(A)+m_e(E \setminus A),$$$$m(E)=m_i(E) \geq m_i(A)+m_i(E\setminus A)$$
Any suggestions would be appreciated
 A: So, as the above poster pointed out, the idea of why this should be true is that the exterior measure of a set is that it looks at open sets containing it, while the interior measure of a set looks at closed sets contained within it.  If we take complements, then it seems like we can get from a closed set inside $A$ to an open set containing $E - A$.  Somehow we should be able to use this correspondence to get both bounds needed to solve this problem.  A proof would look something like this:
Assume that $|A|_i < +\infty$ and $|E-A|_e < +\infty$ (if not, then the result is trivial because $|E|$ would also be infinity).  Let $\varepsilon > 0$.  Then we have that there exists a closed set $F$ such that $|F| > |A|_i - \varepsilon$.  Now consider the set $E-F$.  As both sets are measurable, we have that $|E-F| = |E|-|F| < |E| - |A|_i + \varepsilon$.  Furthermore, $E-A \subset E-F$, so $|E-A|_e < |E| - |A|_i + \varepsilon$.  As the choice of $\varepsilon$ was arbitrary, we have $|E-A|_e \leq |E| - |A|_i \Rightarrow |E-A|_e + |A|_i \leq |E|$.
Now note that there exists an open set $G$ such that $G \supset E-A$ and $|G| < |E-A|_e + \varepsilon$.  Consider then the set $E-G$.  We have $A \supset E-G$, so $|A|_i \geq |E-G| = |E| - |G| > |E| - |E-A|_e - \varepsilon$.  As the choice of $\varepsilon$ was arbitrary, we have that $|A|_i \geq |E| - |E-A|_e \Rightarrow |E-A|_e + |A|_i \geq |E|$, so the two are equal, as desired.
A: Hint: Let $E \subset U$, with $U$ open and $m(U \setminus E) < \epsilon/2$.
Find a closed set $G \subset A$ with $m(G) > m_i(A) - \epsilon/2$.  Note that $U \setminus G$ is an open set containing $E \setminus A$.
