If $m(E)=m_*(E_1)+m_*(E_2)$ , then $E_1$ and $E_2$ are measurable
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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 $m_*(O-E_1)lt epsilon$ ?
real-analysis lebesgue-measure
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asked Jan 8 at 16:43
J.Guo J.Guo
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