For any subset $E\subseteq I=[0,1]$ define the inner Lebesgue measure by $m_{*}(E):=1-m^*(I\setminus E)$, where $m^*$ is the outer Lebesgue measure. Show that $$ E\subseteq I \mbox{ is }m^*- \mbox{measurable }\Leftrightarrow m_*(E)=m^*(E). $$


Let $E\subseteq I$ be measurable relating to $m^*$, that means $$ m^*(Q)=m^*(Q\cap E)+m^*(Q\cap E^C)~\forall Q\subseteq I. $$ Choose $Q=I$. Then $$ m_*(E)=1-m^*(I\setminus E)\\=1-m^*(I\cap E^C)\\1-(m^*(I)-m^*(I\cap E))\\=1-m^*(I)+m^*(E). $$ When I look athe the definition of $m^*$ then $$ m^*(I)=\inf\left\{\sum\limits_{n\geq 1}\lambda(A_n): A_n\subseteq\mathcal{I},n\in\mathbb{N},I\subset\bigcup\limits_{n\geq 1}A_n\right\}, $$ where $\lambda$ is the Lebesgue content on the half ring $\mathcal{I}=\left\{(a,b],a,b\in\mathbb{R},a\leq b\right\}$ and the infimum is taken over the covers of $I$.

And this infimum is 1 - the left boundary going from the left side to 0.

So it follows $$ m_*(E)=1-m^*(I)+m^*(E)=1-1+m^*(E)=m^*(E). $$

My problem is to show the direction "$\Leftarrow$".

Can you give me a help how to show that direction please?


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