Equivalence of measurability definitions The definition of Lebesgue measurable set in my real analysis textbook is:
$E \subseteq \mathbb{R}^d$ is called lebesgue measurable if for all $\varepsilon > 0$ there is an open set $E \subseteq O$ of $\mathbb{R}^d$ such that $m_{*} (O / E) \leq \varepsilon$ (and from there it's kinda easy to prove that the family of all the measurable sets is a sigma algebra containing all the borel sets).
BUT my abstract measure theory book uses the Carathéodory criterion: $E \subseteq \mathbb{R}^d$ is lebesgue measurable if for all $A \subseteq \mathbb{R}^d$ we have $m_{*} (A) = m_{*} (A \cap E) + m_{*} (A \cap E^c)$. How can i prove that these two definitions are equivalent?
 A: Let $O_n$ be the sequence such that $E \subset O_n$ and $m^*(O_n-E) \to 0$. (I am going to use $m^*$ for the outer measure).
By subadditivity, we have that
$$m^*(A) \le m^*(A \cap E) + m^*(A \cap E^c).$$
For the other inequality, since open sets are Lebesgue measurable (by either definition - you should check this), we have that for any $A \subset \mathbb{R}^d$, we have that
$$m^*(A) = m^*(A \cap O_n) + m^*(A \cap O_n^c).$$
We note that $E \subset O_n$, so $m^*(A \cap O_n) \ge m^*(A \cap E)$ and
$$m^*(E^c \cap A) \le m^*(O_n^c \cap A) + m^*((O_n - E) \cap A) \Rightarrow m^*(O_n^c \cap A) \ge m^*(E^c \cap A) -  m^*((O_n - E) \cap A).$$
Thus, we get that
$$m^*(A) \ge m^*(A \cap E) + m^*(A \cap E^c).$$

For the reverse direction, assume that $m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$. By the definition of the outer measure, we have that
$$m^*(E) = \inf \left\{ \sum m^*(R_i) : E \subset \bigcup R_i, R_i \text{ is a rectangle}\right\}.$$
Let $(\{R^n_i\})_n$ be a sequences of covers of $E$ such that
$$m^*\left(\bigcup_i R_i^n\right) \le \sum_i m^*(R^n_i) \le m^*(E) + \frac{1}{n}.$$
Using this define $O_n = \bigcup_i R_i^n$ and apply the Caratheodory condition with $O_n$ to get
$$m^*(E) + \frac{1}{n} \ge m^*(O_n) = m^*(O_n \cap E) + m^*(O_n \cap E^c) = m^*(E) + m^*(O_n \cap E^c). $$
This implies that
$$ \frac{1}{n} \ge m^*(O_n \cap E^c) = m^*(O_n - E). $$
Thus, we have the needed sequence of sets. Thus, we are done.
A: You wan't to prove that the sigma-algebras coming from these two different definitions of measurability are the same. It seems that you already know that the sigma algebra for the first definition is the completion of the Borel sigma algebra with respect to Lebesgue measure $$\overline{B_{\mathbb{R}^d}} = \{E \cup N : E \in B_{\mathbb{R}^d}, N \subset S \in B_{\mathbb{R}^d}, m(S) = 0\}.$$
It remains to show that the Caratheodory sigma algebra, which i'll denote $M$, is $\overline{B_{\mathbb{R}^d}}$. This is true for any measure generated by the Caratheodory construction from a sigma-finite premeasure. See here for a proof: The completion of the Borel $\sigma$-algebra the same as the completion of the Lebesgue outer measure?
