$A\subset (b,c)\subset \mathbb{R}$ is Lebesgue measurable if and only if $|A|+ |(b, c)\setminus A|= c −b$ I am trying to prove the following statement and I am having difficulties proving the leftward implication so I would appreciate an hint about how to do it:
"Suppose $b < c$ and $A\subset (b, c)$. Prove that $A$ is Lebesgue measurable if and
only if $|A|+ |(b, c)\setminus A|= c −b$."
What I have done:
$\fbox{$\Rightarrow$}$: $(b, c)\setminus A$ and $A$ are disjoint subsets of $\mathbb{R}$ of which $A$ is Lebesgue measurable: since we know that if $A$ and $B$ are disjoint subsets of $\mathbb{R}$ and $B$ is Lebesgue measurable then $|A\cup B|=|A|+|B|$ it follows that $c-b=|(b,c)|=|((b,c)\setminus A)\cup A|=|(b,c)\setminus A|+|A|$, as desired.
$\fbox{$\Leftarrow$}$: For the leftward implication I have tried to argue by contradiction by using the fact that if $A$ is not Lebesgue measurable then there exists some $\bar{\varepsilon}>0$ such that $|G\setminus A|\geq\bar{\varepsilon}$ for every $G\supset A$ open so in particular $|(b,c)\setminus A|\geq\bar{\varepsilon}$ but I haven't been able to obtain a contradiction and I am now stuck so an hint about how to better understand this problem is welcome.
EDIT: I guess that another way to see the leftward implication is to note that $|A|+ |(b, c)\setminus A|=c−b=|(b,c)|=|A\cup (b,c)\setminus A|$ so outer measure is additive and it wouldn't be if $A$ were not Lebesgue measurable (cfr Vitali set). Now, while compelling, this is not a rigorous explanation: how to make it so? Thanks

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 $|A\setminus B|= 0$.
$|\cdot|$ denotes outer measure
 A: Your proof for $\fbox{$\Rightarrow$}$ is correct.
Now let us look at the $\fbox{$\Leftarrow$}$ part.
First, note that this part is true only if $c- b < +\infty$, this means, if $c$ and $b$ are finite. In fact, if $b= -\infty$ or $c = +\infty$, the equation $|A|+ |(b, c)\setminus A|= c -b$ becomes trivially true for all subsets $A \subseteq (b,c)$.
Now, assume  $c- b < +\infty$, $A \subseteq (b,c)$ and $|A|+ |A^c|= c - b$ (where $A^c$ means  $(b,c) \setminus A$). Let us prove $A$ is Lebesgue measurable.
Note that, since $|(b,c)| < +\infty$, we have that $|A| <+\infty$ and $|A^c| <+\infty$.
Given any set $S \subseteq (b,c)$, such that $|S| <+\infty$, then there is a Borel set $T \subseteq (b,c)$, such that $S \subseteq T$ and $|S| =|T|$. This is a direct consequence of the definition of outer measure and its is true for all $S$ such that $|S| +\infty$, no matter $S$ is Lebesgue measurable or not. (Attention, in general, $|S| =|T|$ does not imply $|T \setminus S| = 0$).
The simplest way to prove the $\fbox{$\Leftarrow$}$ part is apply the above property to $A$ and $A^c$ and use the fact that $|A|+ |A^c|= c − b$.
Here is a detailed proof:
Since  $|A| <+\infty$, let $E \subseteq (b,c)$ be a Borel set, such that $A \subseteq E$ and $|A| =|E|$. Since $E$ is a Borel set, we have
$c -b = |(b,c)| = |E| + |E^c| $.
Since   $|A|+ |A^c|= c - b$, we have
$$|E^c| = (c-b) -|E| = (|A|+ |A^c|) - |A| = |A^c|$$
In a similar way, since  $|A^c| <+\infty$, let $F \subseteq (b,c)$ be a Borel set, such that $A^c \subseteq F$ and $|A^c| =|F|$. Since $F$ is a Borel set, we have
$c -b = |(b,c)| = |F| + |F^c| $.
Since   $|A|+ |A^c|= c - b$, we have
$$|F^c| = (c-b) -|F| = (|A|+ |A^c|) - |A^c| = |A|$$
Now, note that $F^c$ and $E$ are Borel sets and that $F^c \subseteq A \subseteq E$. So
$$|E| = |F^c| + |E\setminus F^c| $$
But $|E|=|A|=|F^c|$. So
$$|E\setminus F^c| =0$$
Since $A \subseteq E$, we have that  $(A \setminus F^c)  \subseteq (E\setminus F^c)$, and so $|A \setminus F^c| =0$.
So we have that $F^c$ is a Borel set, $F^c \subseteq A$ and $|A \setminus F^c| =0$. So $A$ is Lebesgue measurable.
