If $E_1 \cup E_2$ is measurable with $λ^\ast(E_1 \cup E_2) = λ^\ast(E_1)+ λ^\ast(E_2)$, then $E_1$ and $E_2$ are measurable. Denote by $\lambda^\ast$the Lebesgue outer measure on $\mathbb{R}^m$. Assume
that $E_1, E_2 \subset \mathbb{R}^m$ satisfy $E_1 \cap E_2 = \phi$, $E_1 \cup E_2$ is Lebesgue measurable with $λ^\ast(E_1 \cup E_2) <\infty$ , and $λ^\ast(E_1 \cup E_2) = λ^\ast(E_1)+ λ^\ast(E_2)$. Prove that $E_1, E_2$ are Lebesgue measurable.
$\textbf{Question}$:
Does $λ^\ast(E_1 \cup E_2) = λ^\ast(E_1)+ λ^\ast(E_2)$ implies that for any set $A$, we can write  $λ^\ast\big((A\cap E_1) \cup (A\cap E_2)\big) = λ^\ast((A\cap E_1))+ λ^\ast((A\cap E_2))$ ? (note that I don't have measurability of $E_1$ and $E_2$).
$\textbf{My attempt,  which relies on above doubt}$:  by def we have that for any set $A$ we have
$$λ^\ast(A) = λ^\ast\big((A\cap (E_1 \cup  E_2)\big) +λ^\ast\big((A\cap (E_1 \cup  E_2)^c\big)$$
using the following identities;
$ A\cap (E_1 \cup  E_2) = (A\cap E_1) \cup (A\cap E_1^c \cap  E_2)$, where 
$(A\cap E_1) \cap (A\cap E_1^c \cap  E_2)= \phi$
if the above question is correct, I can write using, $(A\cap E_1^c \cap  E_2) \cap (A\cap E_1^c \cap  E_2^c)= \phi$
\begin{align}
λ^\ast(A)
& = λ^\ast((A\cap E_1))+ λ^\ast(A\cap E_1^c \cap  E_2) +λ^\ast\big((A\cap (E_1 \cup  E_2)^c\big)\\
& = λ^\ast((A\cap E_1))+ λ^\ast\big[(A\cap E_1^c \cap  E_2) \cup (A\cap E_1^c \cap  E_2^c)\big]\\
& = λ^\ast((A\cap E_1))+ λ^\ast\big[(A\cap E_1^c \cup  (E_2 \cap E_2^c)\big]\\
& \geq λ^\ast((A\cap E_1))+ λ^\ast (A\cap E_1^c) \\
\end{align}
Then $E_1$ is measurable. similarly for $E_2$.
 A: Choose a $G_\delta$ set $G$ such that $E_{1}\subseteq G$ and $\lambda^{\ast}(E_{1})=\lambda(G)$. In particular, $G$ is measurable.
Now let $H=(E_{1}\cup E_{2})\cap G$ so $H$ is measurable. We have $E_{1}\subseteq H\subseteq G$ and hence $\lambda^{\ast}(E_{1})\leq\lambda(H)\leq\lambda(G)=\lambda^{\ast}(E_{1})$, so $\lambda^{\ast}(E_{1})=\lambda(H)$.
Now 
\begin{align*}
\lambda^{\ast}(E_{1})+\lambda^{\ast}(E_{2})&=\lambda^{\ast}(E_{1}\cup E_{2})\\
&=\lambda^{\ast}((E_{1}\cup E_{2})\cap H)+\lambda^{\ast}((E_{1}\cup E_{2})-H)\\
&=\lambda^{\ast}(H)+\lambda^{\ast}((E_{1}\cup E_{2})-H)\\
&=\lambda^{\ast}(E_{1})+\lambda^{\ast}((E_{1}\cup E_{2})-H).
\end{align*}
So
\begin{align*}
\lambda^{\ast}(E_{2})=\lambda((E_{1}\cup E_{2})-H).
\end{align*}
On the other hand, we have $(E_{1}\cup E_{2})-H\subseteq E_{2}$.
Now 
\begin{align*}
\lambda^{\ast}(E_{2})&=\lambda^{\ast}(E_{2}\cap((E_{1}\cup E_{2})-H))+\lambda^{\ast}(E_{2}-((E_{1}\cup E_{2})-H))\\
&=\lambda^{\ast}((E_{1}\cup E_{2})-H)+\lambda^{\ast}(E_{2}-((E_{1}\cup E_{2})-H))\\
&=\lambda^{\ast}(E_{2})+\lambda^{\ast}(E_{2}-((E_{1}\cup E_{2})-H)),
\end{align*}
so $\lambda^{\ast}(E_{2}-((E_{1}\cup E_{2}-H)))=0$ and hence $E_{2}-((E_{1}\cup E_{2})-H)$ is measurable and hence $E_{2}=((E_{1}\cup E_{2})-H)\cup(E_{2}-((E_{1}\cup E_{2})-H))$ is also measurable.
