Equivalent definitions of Lebesgue measurability I'm having difficulty understanding why these two definitions of Lebesgue measurability of a set $E \subseteq \mathbb{R}^{d}$ are equivalent:
1) For every $\epsilon >0$, there exist an open set $O$ such that $m_{*}(O\setminus E) \le \epsilon$.
2) For every set $A \subseteq \mathbb{R}^d$, the following holds:  $m_{*}(A) = m_{*}(A \cap E) + m_{*}(A \cap E^{c}).$ 
 A: The second definition is the more common one. You should also note that $m_*$ is an outer measure here; indeed, $m(A)$ is not even defined for all sets $A \subset \mathbb{R}^n$. One is first required to show that open sets are measurable with this definition (I suggest you try) -- how you prove this depends on your construction of general measures.
Note that, in 2), by subadditivity of outer measure, you always have
$$m_*(A) = m_*((A \cap E) \cup (A \cap E^c)) \leq m_*(A\cap E) + m_*(A\cap E^c),$$
so you only have to show one side of the inequality.
(1 $\Rightarrow$ 2.) Suppose that $E$ is a measurable set. For any $\epsilon > 0$, there is an open set $O$ such that $E \subset O$ and $m(O\setminus E)=m(O \cap E^c) < \epsilon$. Since open sets are measurable, we have
$$m_*(A) = m_*(A \cap O) + m_*(A \cap O^c).$$
Hence, 
\begin{align*}
m(A) + \epsilon &> m(A) + m(O\cap E^c) \\
&= m(A \cap O) + m(A \cap O^c) + m(O\cap E^c) \\
&\geq m(A \cap E) + m(A \cap O^c \cap E^c) + m(O\cap E^c \cap A)  \\
&= m(A \cap E) + m(A \cap E^c) \tag{by measurability of O}
\end{align*}
where the third line follows because intersecting makes sets (and hence their measure) smaller (for the second two terms), and $E \subset O$ by assumption (for the first term). Since $\epsilon$ is arbitrary, we have the desired inequality. 
(2 $\Rightarrow$ 1)
This is a general property of all ``Borel - regular'' measures, which the Lebesgue measure is. In fact, you can say that, for measurable and bounded $E$, 
$m(E) = \inf\{m(O) : E \subset O, O\;open\} = \sup\{m(K) : K \subset O, K\;compact\}.$
Let me know if you want some help / sources where to find the rest. 
A: To hopefully complement snar's excellent answer, here's something.

Prerequisites to the proof
The knowledge that every open set satisfies Caratheodory's criterion (which is indeed a bit of work to prove) can be forgone, provided that the following is instead known:


*

*For any set $A$, there exists a sequence of open sets $(G_n)$ such that $A \subseteq \bigcap_n G_n$ and, moreover, $m_*(A) = m_*\big( \bigcap_n G_n \big)$.

*If $(G_n)$ is a sequence of open sets, then $\bigcap_n G_n$ is measurable in the sense of definition 1.

*If sets $A$ and $B$ are each measurable in the sense of definition 1, then so are the sets $A \cap B$ and $A \setminus B$.

*For any disjoint sets $A$ and $B$ which are each measurable in the sense of definition 1, we have $m(A \cup B) = m(A) + m(B).$

*The outer measure is monotone.


The first and last of these facts is quite easy to prove just from the definition of the outer measure, but 2-4 can be more difficult to prove.

Proof
Give me a measurable set $E$. I have to show that $m_*(A) = m_*(A \cap E) + m_*(A \setminus E)$ for any set $A$. So, give me $A$. By fact 1., we have
$$ m_*(A) = m(G) $$
for some set $G$ containing $A$, and by fact 2., this set $G$ is measurable (in the sense of definition 1).
Therefore, because $E$ is also (by assumption) measurable in the sense of definition 1, we have that so are $E \cap G$ and $E \setminus G$ (by fact 3.). Hence, according to fact 4.,
$$ m(G) = m(G \cap E) + m(G \setminus E). $$
But, by fact 5.,
$$ m(G \cap E) + m(G \setminus E) \geq m_*(A \cap E) + m_*(A \setminus E). $$
In conjunction,
$$ m_*(A) \geq m_*(A \cap E) + m_*(A \setminus E), $$
as required.
A: This is not an answer but a comment which is too long. 
I just want to provide another (equivalent) notion of measurability  not very common in standard courses on integration but which has the advantage that it can be generalized easily to setting of Integration on vector valued measures and to Stochastic Calculus. The idea goes back to Daniell, Littlewood, Stone, etc.
I will just give a brief outline. If you are interested in this things I will give some references at the end.

Several steps need to happen first: (1) Construction the integral first through Daniell's procedure; (2)  making use of Littlewood's observations:

Suppose your ambient space is $\Omega$, and that the Daniell integral is constructed base on an elementary integral $(\mathcal{E},I)$. ($\mathcal{E}$ is vector lattice of bounded functions closed under the operation $\min(f,1)$; $I$ is a linear functional on $\mathcal{E}$ that has some continuity properties, for instance $f_n\searrow0$, and $f_n\in\mathcal{E}$ implies that $I(f_n)\rightarrow0$.)
Example:
$\mathcal{E}$ the continuous bounded functions in $[0,1]$ and $I$ the Riemman integral acting on $\mathcal{E}$.
(a) Construction of the Daniell mean involves a sup and inf type operations, first for functions $\mathcal{E}^\uparrow$ that are monotone limits of functions in $\mathcal{E}$, $I(h)=\sup\{I(f): f\in\mathcal{E},\,f\leq h\}$. Then, for any function $f$, $I(f)=\inf\{I(h):h\in\mathcal{E}^\uparrow,\, f\leq h\}$. For any function $f$ define
$$
\|f\|^*=I(|f|)
$$
It turns out that $\|\;\|$ defines a nice pseudonorm when restricted to the space $\mathcal{F}$ of functions such that $\|f\|^*<\infty$. Now, the space of integrable function $\mathcal{L}_1(I)$ is defined as the closure of $\mathcal{E}$ in $(\mathcal{F},\|\;\|^*)$. A set $A$ is integrable if $\mathbb{1}_A\in\mathcal{L}_1$. There is a natural procedure then to define $I(f)$ for any $f\in \mathcal{L}_1$ via Cauchy sequences in $\mathcal{E}$. 
(b) Littlewood's observations can be summarize as follows: (I) $f$ is integrable if for any $\varepsilon>0$ there is a set $U\in\mathcal{E}^\uparrow$ such that $\|U\|<\varepsilon$, and on $\Omega\setminus U$, $f$ is the uniform limit of a sequence in $\mathcal{E}$; (II) for any $\varepsilon>0$, if $f_n$ is a sequence of integrable functions that converge a.s (a set $B$ is a.s. empty if $\|B\|^*=0$) to $f$ in an integrable $A$, then there is another integrable set $A_0$ contained in $A$ such that $\|A\setminus A_0\|^*<\varepsilon$ and on $A_0$, $f_n$ converges uniformly to $f$.
Definition: A function $f ∈ \mathbb{R}^\Omega$ is measurable on an integrable set $A$ if for any $\varepsilon>0$, there is another integrable set $A_0$ contained in $A$ and a function $g$ in the uniform closure  of $\mathcal{E}$ (the closure of $\mathcal{E}$ as a subspace of the space of bounded functions with the sup norm) such that $\|A\setminus A_0\|^*<\varepsilon$ and $f=g$ on $A_0$.
A function $f$ is $\|\;\|^*$-measurable if it is measurable on any integrable set.
A set $B\subset\Omega$ is measurable whenever $\mathbb{1}_B$ is measurbale.

It can be shown that Daniell measurability is equivalent to the Lebesgue-Caratheodory notion that is very common to discuss in integration courses.

A good place to look at the Daniel integral is Bichteler's Integration theory, a functional approach. Another good reference, although it is more dense is Dunford & Schwatrz Linear Operators Volume I. The latter construct integration for vector--valued measures using an approach reminiscent to Daniell's/
