The question is quite "simple". Let $\lambda^\*$ denote the usual Lebesgue outer measure on $\mathbb R.$ Let $E\subseteq [0,1]$ be a non-measurable subset. Do we **always** have
$$
\lambda^\*(E) +\lambda^\* ([0,1]\backslash E) >1?
$$
Are there examples of non-measurable sets such that equality $\lambda^\*(E) +\lambda^\* ([0,1]\backslash E) =1$ holds?


Although this is very easy to state, it is very hard to think of what examples do we have. After all, the construction of a non-measurable set alone is not that easy. The Vitali set does not provide the example asked for above.



Summability issues of measure when we decompose a measurable set into two non-measurable parts
Suppose $E\subseteq[0,1]$, $\ F=[0,1]\setminus E$, $\ \lambda^\*(E)=a$, $\ \lambda^\*(F)=b$, $\ a+b=1$.


There are Lebesgue measurable sets ($G\_\delta$ sets) $A,B\subseteq[0,1]$ such that $E\subseteq A$, $\ \lambda(A)=\lambda^\*(E)=a$, $\ F\subseteq B$, $\ \lambda(B)=\lambda^\*(F)=b$.


Now $\lambda(A\cap B)=\lambda(A)+\lambda(B)-\lambda(A\cup B)=a+b-1=0$, and $A\setminus E\subseteq A\cap B$, so $A\setminus E$ is Lebesgue measurable, and so is $E=A\setminus(A\setminus E)$.


Therefore, if a subset $E$ of $[0,1]$ is nonmeasurable, then $\lambda^\*(E)+\lambda^\*([0,1]\setminus E)\gt1$.


