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.

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

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

  • $\begingroup$ Could you suggest a reference for the result used in the second paragraph? (Every set has a Lebesgue measurable subset with the same outer measure). Thank you. $\endgroup$
    – Ma Joad
    Jan 5, 2021 at 9:03
  • $\begingroup$ Oh I make a mistake. Thank you for the answer. $\endgroup$
    – Ma Joad
    Jan 5, 2021 at 9:15

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