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

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.

• If you mean outer measure: en.wikipedia.org/wiki/Carath%C3%A9odory%27s_criterion Jan 4, 2021 at 23:27
• @OtisChodosh It's different. Carathéodory's criterion requires the condition to be true for all sets $A$. I am asking about one particular set $[0,1].$ Jan 4, 2021 at 23:36
• @bof Oh yes - I actually mean Lebesgue outer measure. I have now edited my question. Jan 5, 2021 at 0:28

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