I was wondering if the following proof works to prove the following statement:
Prove the following are equivalent for $E \subseteq \mathbb{R}$, where $m$ denotes the Lebesgue outer measure.
(i) $E$ is measurable (definition taken to be satisfying the Caratheodory criterion).
(ii) $m(B) = m(B \cap E) + m(B \cap E^c)$ for all Borel sets $B$.
(iii) $m(U) = m(U \cap E) + m(U \cap E^c)$ for all open sets $U$.
Proof: (i) $\Rightarrow$ (ii), (i) $\Rightarrow$ (iii), (ii) $\Rightarrow$ (iii) are all trivial. (iii) $\Rightarrow$ (ii) follows from the fact that the $\sigma$-algebra generated by the open sets is equal to the Borel algebra. To see (iii) $\Rightarrow$ (i), cover $A$ by open intervals $I_k$ such that $m(A) + \epsilon \geq \sum_k I_k $, then we have: $$ m(A) + \epsilon \geq m(\cup_k I_k) = m(\cup_k I_k \cap E) + m(\cup_k I_k \cap E^c) $$ by our hypothesis, but then since $A \subseteq \cup_k I_k$ we have that $m(\cup_k I_k \cap E) \geq m(A \cap E)$ and likewise for $E^c$ by monotonicity, thus giving: $$ m(A) + \epsilon \geq m(A \cap E) + m(A \cap E^c) $$ Letting $\epsilon \rightarrow 0$ thus gives the inequality and hence the result (reverse inequality trivially follows from countable sub-additivity). This completes the proof.
I'm fairly new to measure theory so any feedback, corrections or help would be greatly appreciated!
You have to delete the sentence 'iii) implies ii) follows from the fact that the sigma algebra generated by the open sets is equal to the Borel sigma algebra'. This is not clear and it is not required. [For this to work you have to prove that the class of all sets for which the equation in ii) holds is a sigma algebra]. Once you prove that iii) implies i) you also have iii) implies ii) because i) implies ii). Otherwise, your proof is fine.