I am trying to prove the first part of exercise 33, ch. 1 in Stein and Shakarchi (Real Analysis). I am running into some difficulties following the hint though. Here is the problem (note, $N$ is a Vitali set constructed in $[0,1]$):

Show that the set $[0,1]-N$ has outer measure $m_*(N^c)=1$. [Hint: argue by contradiction, and pick a measurable set such that $N^c \subset U \subset [0,1]$ and $m_*(U) \le 1-\epsilon$.

I know that both $N$ and its complement are not measurable, so neither are countable. I know that measurable subsets of non-measurable sets have measure 0. I am not sure how to proceed given the proof though. A point to note: the book does not work with the inner measure at all, and even though I have used the inner measure in a previous course, I do not think I am allowed to for this proof.

  • 2
    $\begingroup$ You wrote something false. Take a Vitali set in $[0,\frac12]$, union the interval $[\frac12,1]$. This set is certainly non-measurable, but it has a measurable subset of measure $\frac12$. It is true, however, that the Vitali set has inner measure zero. $\endgroup$ – Asaf Karagila Sep 30 '12 at 22:30
  • $\begingroup$ One way to see that Vitali set have inner measure zero is Lemma 2 in here $\endgroup$ – leo Oct 1 '12 at 22:56

Hint: By Theorem 3.2, $m_*(I)=m_*(U)+m_*(U^c).$


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