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 $m_*(O-E_1)\lt \epsilon$ ?


1 Answer 1


Let $\epsilon>0$ be arbitrarily given and $U_i$, $i=1,2$ be open sets such that $E_i\subset U_i$ and $$ m_*E_i\le mU_i<m_*E_i +\epsilon $$ 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 $i=1,2$.


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