Suppose $E\subset\mathbb{R}$ is a Lebesgue measurable set, $m(E)<\infty$. $F\subset E$, prove that $m(E)=m^*(F)+m^*(E\setminus F)$ $\iff$ $F$ is measurable.
Here is the same question asked before: Prove: A subset of $\mathbb{R}$ is measurable if and only if $m(E) = m^\ast (F) + m^\ast (E-F)$
However, I have no idea how to prove "$\Rightarrow$". I consider the interior measure, but I'm not quite familiar with it.
The definition of Lebesgue measurable set $E$: for any set $A$, $m^*(A)=m^*(A\cap E)+m^*(A\cap E^\mathrm{c})$
If $F$ is measurable then $m(F) = m_*(F) = m^*(F)$ by definition of lebesgue measurable sets.
Since $F$ is measurable, we have, $$m(E) = m(F) + m(E/F)$$ $$m(E) = m^*(F) + m(E/F)$$
Since $E \setminus F$ is also measurable, we have that $m(E\setminus F) = m_*(E \setminus F) = m^*(E \setminus F)$ $$m(E) = m^*(F) + m^*(E/F)$$
Now follow the link in the question for the proof in the other way around. I am repeating it here with more details.
Choose, $A_n \subseteq F$ and hence $\cup_n A_n \subseteq F$ such that $m_*(F) = \lim_k m(\cup_{i=1}^k A_i)$ by definition of inner measure.
Choose, $B_n \supseteq E \setminus F$ and hence $\cap_n B_n \supseteq E \setminus F$ such that $m^*(E \setminus F) = \lim_k m(\cap_{i=1}^k B_i)$ by definition of outer measure.
Now choose $C_n = (E \setminus (B_n \cap E)) \cup A_n$. Now $F \supseteq C_n \supseteq A_n$ and $E \setminus F \subseteq E \setminus C_n \subseteq B_n$.
Hence $$\lim_k m(\cup_{i=1}^k C_i) = m_*(F)$$ and $$\lim_k m(E \setminus \cup_{i=1}^k C_i) = m^*(E \setminus F).$$
Hence we have (since $\cup_{i=1}^k C_i$ is measurable.) $$m(E) = m(\cup_{i=1}^k C_i) + m(E \setminus \cup_{i=1}^k C_i)$$ $$m(E) = \lim_k m(\cup_{i=1}^k C_i) + \lim_k m(E \setminus \cup_{i=1}^k C_i)$$ $$m(E) = m_*(F) + m^*(E \setminus F).$$
Since we also have (by hypothesis in the question:) $$m(E) = m^*(F) + m^*(E \setminus F).$$ we have that $$m_*(F) = m^*(F)$$
Hence $F$ is measurable.