I need to prove that A is Lebesgue-measurable I'm struggling with this measure theory problem. It says:

Let $B$ be a Lebesgue-measurable set, with $m(B)<\infty$ and
$A\subset B$. Prove that $A$ is Lebesgue-measurable if and only if
$m(B)=m$* $(A)+m$* $(B\backslash A)$. ($m$ is the measure application
and $m$*$ $ the exterior measure application in $\Bbb R$).

The implication $A$ is Lebesgue-measurable $\longrightarrow$ $m(B)=m$* $(A)+m$* $(B\backslash A)$ is easy, because if $A$ is Lebesgue-measurable, then $m$* $(F)=m$* $(F\cap A)+m$* $(F\backslash A), \forall F\subseteq \Bbb R$.
I have problems with the reverse implication. The problem gives a hint: "you can use that, if $A\subseteq \Bbb R, m$* $(A)<\infty$, then $\exists B$ Lebesgue-measurable with $A\subseteq B$ that verifies $m(B)=m$* $(A)$".
 A: As the hint says, there exists $E$ measurable with $A\subset E$ and $m^*(A)=m(E)$. From this we get that we can replace $B$ by $B\cap E$ in
$$\tag1
m(B)=m^*(A)+m^*(B\cap A^c). 
$$
Indeed, 
$$\tag2
m(E)=m^*(A)\leq m(B\cap E)\leq m(E), 
$$
so $m(E)=m(B\cap E)$. Then
\begin{align}
m^*(A)+m^*(B\cap E\cap A^c)
&=m^*(A)+m^*(B\cap A^c\cap E)+m^*(B\cap A^c\cap E^c)\\[0.25cm]&\ \ \ \ \ \ \ \ \ \ \ \ \ \ -m^*(B\cap A^c\cap E^c)\\[0.3cm]
&=m^*(A)+m^*(B\cap A^c)-m^*(B\cap A^c\cap E^c)\\[0.3cm]
&=m^*(B)-m^*(B\cap A^c\cap E^c)\\[0.3cm]
&=m^*(B)-m^*(B\cap  E^c)\\[0.3cm]
&=m(B\cap (B\cap E^c)^c)\\[0.3cm]
&=m(B\cap E)=m^*(A). 
\end{align}
So we have $m^*(B\cap E\cap A^c)=0$.  
Now,  for any $F$, \begin{align}
m^*(F\cap A)+m^*(F\cap A^c)
&\leq m^*(F\cap  B\cap E)+m^*(F\cap ((B\cap E)^c\cup (B\cap E)\setminus A) )\\[0.3cm]
&\leq m^*(F\cap  B\cap E)+m^*(F\cap (B\cap E)^c)+m^*(F\cap B\cap E\cap A^c)\\[0.3cm]
&\leq m^*(F\cap  B\cap E)+m^*(F\cap (B\cap E)^c)+m^*( B\cap E\cap A^c)\\[0.3cm]
&=m^*(F\cap  B\cap E)+m^*(F\cap (B\cap E)^c)\\[0.3cm]
&=m^*(F).
\end{align}
