$A$ Lebesgue measurable iff $\forall\varepsilon>0\ \exists G=\bigcup_{n=1}^{N}I_n$ bounded open intervals: $|A\setminus G|+|G\setminus A|<\varepsilon$ "Suppose $A\subset\mathbb{R}$ and $|A|<\infty$. Prove that $A$ is Lebesgue measurable if and only if for every $\varepsilon>0$ there exists a set $G$ that is the union of finitely many disjoint bounded open intervals such that $|A\setminus G|+ |G\setminus A|<\varepsilon$"
$\Rightarrow$ Since $|A|<\infty$ there must be a sequence of open bounded intervals containing $A$ because if all of the sequences of open intervals containing $A$ all contained at least one unbounded interval then by definition of outer measure we would have $|A|=\infty$, contradiction.
Let now $\varepsilon>0$: then, among these sequences of open bounded intervals there must be one $I_1,I_2,\dots$ such that $\sum_{n=1}^{\infty}\ell(I_k)\leq |A|+\frac{\varepsilon}{2}$ (if there existed $\bar{\varepsilon}>0$ such that for every bounded sequence of open intervals containing $A$ we had $\sum_{n=1}^{\infty}\ell(I_k)>|A|+\bar{\varepsilon}$ then by taking the $\inf$ of both sides we would have $|A|\geq |A|+\bar{\varepsilon}$ so $\bar{\varepsilon}\leq 0$, a contradiction).
Since $\bigcup_{n=1}^{\infty}I_n$, being the union of open sets, is an open set, we know that we can write it as the disjoint union of other open intervals $J_n$, which must thus also be bounded so
$\bigcup_{n=1}^{\infty}I_k=\bigcup_{n=1}^{\infty} J_k$, $\sum_{n=1}^{\infty}\ell(J_k)=|\bigcup_{n=1}^{\infty} J_k|= |\bigcup_{n=1}^{\infty}I_k|<|A|+\frac{\varepsilon}{2}<\infty$ and since $\sum_{n=1}^{\infty}\ell(J_k)$ is convergent there must be $N\geq 1$ such that $\sum_{n=N+1}^{\infty}\ell(J_k)<\frac{\varepsilon}{2}$ thus if we take $G=\bigcup_{n=1}^{N}J_n$ we have $$|A\setminus G|+|G\setminus A|=|A\setminus\bigcup_{n=1}^{N}J_n|+|(\bigcup_{n=1}^{N}J_n)\setminus A|\leq |\bigcup_{n=1}^{\infty}J_n\setminus\bigcup_{n=1}^{N}J_n|+|\bigcup_{n=1}^{\infty}J_n\setminus A|\overset{A\in\mathcal{L},|A|<\infty}{=}|\bigcup_{n=N+1}^{\infty}J_n|+|\bigcup_{n=1}^{\infty}J_n|-|A|<|\bigcup_{n=N+1}^{\infty}J_n|+\frac{\varepsilon}{2}\leq\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$$ as desired.

$\Leftarrow$ By hypothesis we know that for every $n\geq 1$ there exist $j_n$ bounded disjoint open intervals $G_{1_n}, G_{2_n},\dots, G_{j_n}$ such that $G_{n}=\bigcup_{k=1}^{j_n}G_{k_n}$ and $|A\setminus G_n|+|G_n\setminus A|<\frac{1}{n}$. Consider now $H=\bigcup_{n=1}^{\infty}G_n$: being the union of open sets it is open hence Borel and we have that $A\setminus H\subset A\setminus G_n$ so $|A\setminus H|\leq |A\setminus G_n|\leq\frac{1}{n}-|G_n\setminus A|\leq\frac{1}{n}$ for every $n\geq 1$ so $|A\setminus H|=0$ \
EDIT 15 September now if I could show that $H\subset A$ I could conclude that $A$ is Lebesgue measurable. One way to do this is to try by contradiction so suppose that $H\setminus A\neq\emptyset$: now I believe that if $H\setminus A$ contained a countable number of elements we could take them out of $H$ and obtain a Borel set $H'\subset A$ for which $|A\setminus H'|=0$ but what if the number of elements in $H\setminus A$ is uncountable? Would this method still work?

DEF. (Lebesgue measurable set)
A set $A\subset\mathbb{R}$ is called Lebesgue measurable if there exists a Borel set $B\subset A$ such that $|A\setminus B|=0$.
 A: In your edit your  method does not work because  $H$ is an open set so maybe not be a subset of $A$
Take for instance $A$ to be a set that contains only irrational numbers and has positive outer measure.
A proof of the leftwards implication goes like this.
Let $\epsilon >0$.
Then there exists an open set $G \supset A$ such that $m^*(G)<m^*(A)+\epsilon$
Let $J=\bigcup_{k=1}^mI_m$ the union of interval in hypothesis and $U=G \cap J$
By subadditivity $m^*(G\Delta A) \leq m^*(G \Delta U)+m^*(U\Delta A)(*)$
Since $U \subseteq J$ we have that $U \setminus A \subseteq J \setminus A$ and since $A \subseteq G$ we have that $A \setminus U=A \setminus J$
So $U\Delta A\subseteq  J\Delta A$ and thus $m^*(U \Delta A)<\epsilon$
But $A \subseteq U \cup (U\Delta A)$ so $m^*(A)<m^*(U)+\epsilon$
So by $(*)$ $m^*(G \Delta U)=m^*(G \setminus U)=m^*(G)-m^*(U) \leq m^*(A)-m^*(U)+\epsilon<2\epsilon$
Then again by $(*)$ we have $m^*(G\setminus A)=m^*(G \Delta A)<3\epsilon$
So $A$ is measurable.
A: A version of Marios' proof of left implication distilled by avoiding the use of symmetric difference:
Let $\varepsilon > 0$ and $J=\bigcup_{k=1}^mI_k$ such that $m^*(J\setminus A) + m^*(A\setminus J) < \varepsilon$.
By definition of outer measure, there exists an open set $G \subset A$ such that $m^*(G)<m^*(A)+\varepsilon$
By subadditivity $$m^*(G\setminus A) \leq m^*(G\setminus J) + m^*(J \setminus A) < m^*(G \setminus J) + \varepsilon.$$
Moreover, since $G$ and $J$ are Lebesgue measurable, $$m^*(G\setminus J) = m^*(G\setminus (G\cap J)) = m^*(G) - m^*(G\cap J) < m^*(A) - m^*(G \cap J) + \varepsilon \leq m^*(A \setminus (G \cap J)) + \varepsilon = m^*(A \setminus J) + \varepsilon \leq 2\varepsilon.$$ Where the last equality is motivated by $A \subset G.$
Thus, $m^*(G \setminus A) < m^*(G \setminus J) + \varepsilon < 3\varepsilon$, proving that $A$ is Lebesgue measurable.
