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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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    @Thomas: Robert Israel's construction works just fine and is essentially the same as the one given in ccc's answer (I believe RI's construction is a bit easier in the details).2012-06-14

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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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    @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.2012-06-14