Prove or disprove the following statement.
Let $E\subset [0,1]$, if $ m^*(E)+m^*([0,1]\setminus E)=1$ ,then $E$ is a measurable set, where $m^*$ is Lebesgue outer measure.
At first glance, I thought it is triavially true.
That is, the set $E$ is measurable provided for any set $A$, $$m^*(A \cap E)+m^*(A\cap E^c)=m^*(A)..............(*),$$
and thus
$$m^*(E)+m^*([0,1]\setminus E) = m^*([0,1]\cap E)+m^*([0,1]\setminus E)=m([0,1])=1.$$
Hence $E$ is measurable. But I realized that using the whole set $[0,1]$ as $A$, i.e., $A=[0,1]$ is just a particular case and that I need to show that $(*)$ is true for any set $ A$.
Can someone give me an idea how to go by this?
For any set $A$; $m^*(A)+m_*([0,1]\setminus A)=1$
Also it is given that $m^*(A)+m^*([0,1]\setminus A)=1$
Hence we have $m^*([0,1]\setminus A)=m_*([0,1]\setminus A)\implies [0,1]\setminus A $ is measurable $\implies A$ is measurable