Summability issues of measure when we decompose a measurable set into two non-measurable parts

Question

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.

Answer 1

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 $\lambda^*(E)+\lambda^*([0,1]\setminus E)\gt1$.