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I know that for any $\varepsilon\in (0,1]$ we can find a non-measurable subset (w.r.t Lebesgue measure) of $[0,1]$ so that its outer-measure equals exactly $\varepsilon$. It is done basicly with the traditional Vitali construction inside the interval $[0,\varepsilon]$ and noticing that such a set carries zero inner-mass, and thus its complement in $[0,\varepsilon]$ (being non-measurable as well) must carry the full outer-mass of $[0,\varepsilon]$.

However, this resulting non-measurable set is a complement of the traditional Vitali constructed set. My question asks if the Vitali construction itself can yield a non-measurable set with outer-measure of exactly $1$ (or any before-hand decided number from $(0,1]$). Some modifications can be done inside the construction of course, but in particular I would like to stay away from taking complements. Maybe someone knows how this could be done?

Any references and input is appreciated. Thanks in advance.

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    $\begingroup$ The person who down-voted is welcome to leave a comment and explain why so. $\endgroup$
    – T. Eskin
    Jun 12, 2012 at 21:44
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    $\begingroup$ The thread math.stackexchange.com/q/14591 contains a few constructions among which the ones ruled out in the question. However, Jonas Meyer's answer links to a sci.math post by Robert Israel giving the description of a Vitali set (a set of coset representatives of $\mathbb R /\mathbb Q$) of full outer measure in $[0,1]$: groups.google.com/group/sci.math/browse_frm/thread/… $\endgroup$
    – user33592
    Jun 12, 2012 at 22:55
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    $\begingroup$ While not an actual answer, I recall that free ultrafilters on $\mathbb N$, when seen as subsets of $2^{\mathbb N}$ are non-measurable with outer measure $1$ and inner measure $0$. (At least some ultrafilters should have this property, I think that outer measure $1$ is provable but inner measure may change.) $\endgroup$
    – Asaf Karagila ♦
    Jun 12, 2012 at 23:00
  • $\begingroup$ @GiuseppeVitali: Do you know if anyone has gone through the details of this construction and checked whether it works and gives the desired set? $\endgroup$
    – T. Eskin
    Jun 13, 2012 at 6:07
  • $\begingroup$ @AsafKaragila: Thanks. Do you have some references for this where I could possibly look it up? $\endgroup$
    – T. Eskin
    Jun 13, 2012 at 6:07

1 Answer 1

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$\newcommand{\c}{\mathfrak{c}}$ Let $\c$ denote the cardinal of the continuum and wellorder the Borel subsets of $[0,1]$ as $(B_\alpha)_{\alpha < \c}$. We build by transfinite recursion a sequence $(x_\alpha)_{\alpha < \c}$ of elements of $[0,1]$ such that:

(a) $x_\alpha$ is Vitali inequivalent to $x_\beta$ for all $\beta < \alpha$, and

(b) $x_\alpha \in [0,1] \setminus B_\alpha$ if $[0,1] \setminus B_\alpha$ is uncountable.

Note that this process can't get stuck, since if the complement of $B_\alpha$ is uncountable then it has cardinality $\c$, and thus it meets an unused Vitali equivalence class (since at most $|\alpha| < \c$ have been used so far). Then by setting $X = \{x_\alpha : \alpha < c\}$ we obtain a set such that whenever $B$ is a Borel set with $X \subseteq B$, then $B$ has countable complement (and in particular has measure $1$). So $X$ has outer measure $1$ as desired.

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    $\begingroup$ Of course by restricting your attention to the subspace $[0,a] \subseteq [0,1]$, this construction gives a Vitali set with outer measure $a$. $\endgroup$
    – user83827
    Jun 13, 2012 at 8:37
  • $\begingroup$ Nice. This is similar to what Robert Israel does in the sci.math thread mentioned by "Giuseppe Vitali" (ha-ha...). R.I. replaces the Borel sets in your construction by the open sets of measure less than one. the technical details seem a bit simpler: Outer measure $1$ is also guaranteed by the same diagonalization argument, the perfect set property is easier to prove for closed sets, and counting the open sets is also easier than counting the Borel sets. A further minor point: do you guarantee that you hit all Vitali equivalence classes? I think you have to add the missing classes at the end. $\endgroup$
    – t.b.
    Jun 14, 2012 at 1:09
  • $\begingroup$ @t.b., I see, you're right that it's essentially the same as Robert Israel's simpler argument. You're also right that this does not in general meet every Vitali equivalence class, but it meets Thomas E.'s request for a set with inner measure $0$ and outer measure $1$. Of course you can just extend to a "traditional" Vitali set at the end, or you can wellorder the classes themselves and ensure that you always choose the least available option as you go. $\endgroup$
    – user83827
    Jun 14, 2012 at 7:41

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