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.
Best Answer
$\newcommand{\c}{\mathfrak{c}}$ 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.