# If $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable

Suppose $$E$$ is measurable with $$m(E) \lt \infty$$ , and $$E=E_1 \cup E_2 , E_1 \cap E_2 \text{is empty}$$ Show that if $$m(E)=m_*(E_1)+m_*(E_2)$$ , then $$E_1$$ and $$E_2$$ are measurable.

My attempt:
Since $$E_2=E-E_1$$ , it suffice to prove $$E_1$$ is measurable . And for $$E_1$$ , since $$m_*(E_1)\le m(E) \lt \infty$$ , for any $$\epsilon$$ , we can find an open set $$O$$ such taht $$E_1 \subset O$$ and $$m(O)-m_*(E_1) \lt \epsilon$$ , but how to show that $$m_*(O-E_1)\lt \epsilon$$ ?

Let $$\epsilon>0$$ be arbitrarily given and $$U_i$$, $$i=1,2$$ be open sets such that $$E_i\subset U_i$$ and $$m_*E_i\le mU_i for $$i=1,2$$. Since $$E\subset U_1\cup U_2,$$ we have $$mE \leq m(U_1\cup U_2)=mU_1+mU_2-m(U_1\cap U_2).$$ This gives $$m(U_1\cap U_2)\le mU_1+mU_2-mE<2\epsilon$$ by the assumption that $$mE=m_*E_1+m_* E_2$$. Now, observe that $$U_i\setminus E_i \subset \left(U_1\cap U_2\right)\cup \left(\left(U_1\cup U_2\right)\setminus E\right).$$ This implies $$\begin{eqnarray} m_*(U_i\setminus E_i)&\le& m_*\left( \left(U_1\cap U_2\right)\cup \left(\left(U_1\cup U_2\right)\setminus E\right)\right)\\ &\le& m(U_1\cap U_2)+m(\left(U_1\cup U_2\right)\setminus E)\\ &\le& 2\epsilon +m(U_1\cup U_2)-mE\\ &\le& 2\epsilon + mU_1+mU_2-mE<4\epsilon. \end{eqnarray}$$ Since $$\epsilon>0$$ was arbitrary, it says that $$E_i$$ are measurable for $$i=1,2$$.