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In Kharazishvili's "Nonmeasurable Sets and Functions" there is the following theorem:

There exists a subset $X$ of $\mathbb{R}$ which is a Vitali set and a Bernstein set.

The proof is as follows:

Let $\alpha$ denote the first ordinal of cardinality the continuum. Let {$x_{\xi} : \xi < \alpha$} be an injective family of all points of $\mathbb{R}$ and let {$F_{\xi} : \xi < \alpha$} denote an injctive family of all uncountable closed subsets of $\mathbb{R}$. Similarly to the classical Bernstein construction, we define, by applying the method of transfinite recursion, an injective family {$x_{\xi} : \xi < \alpha$} of points in $\mathbb{R}$. Suppose that, for an ordinal $\beta < \alpha$, the partial family of points {$x_{\xi} : \xi < \beta$} has already been constructed. Consider the set

$Z_{\beta} = \cup${$x_{\xi} + \mathbb{Q} : \xi < \beta$}.

Obviously, $card(Z_{\beta}) \leq card(\beta)\centerdot \omega < c$.

Since $card(F_{\beta}) = c$, we have $F_{\beta}\backslash Z_{\beta} \neq 0$.

Take any element $z \in F_{\beta}\backslash Z_{\beta}$ and put $x_{\beta} = z$. In this way, the required family of points
{$x_{\xi} : \xi < \alpha$} will be constructed. Now we define:

$X' =$ {$x_{\xi} : \xi < \alpha$}.

Let us remark that in view of our construction any equivalence class of the Vitali partition {$x + \mathbb{Q} : x \in \mathbb{R}$} contains at most one point from $X'$. Moreover, our construction implies at once that the set $\mathbb{R}\backslash X'$ is totally imperfect in $\mathbb{R}$. In other words, $X'$ turns out to be a partial selector of the Vitali partition whose complement is totally imperfect. Evidently, we can extend $X'$ to a selector of the same partition. We denote by $X$ the selector obtained in this manner.

Kharazishvili goes on to prove that $X$ is a Bernstein set.

However, my question is, is $X \backslash X'$ a null set?

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1 Answer 1

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This might depend on exactly what choices you make during the construction. It is easy to choose the $x_\beta$'s in such a way that $X\setminus X'$ is not a null set but in fact is another Bernstein set and thus is not disjoint from any set of positive measure. To see this, imagine doing Kharazishvili's construction with some extra steps interleaved as follows. Immediately after choosing $x_\beta$, choose another point $y_\beta\in F_\beta$, from a different coset modulo $\mathbb Q$; also make sure that $x_\beta$ and $y_\beta$ are not in the cosets of any previous $x_\xi$'s or $y_\xi$'s. This guarantees that the set $Y=\{y_\beta:\beta<\alpha\}$ is a second Bernstein-Vitali set just like $X'$, but meeting only cosets that $X'$ didn't meet. So you can choose $X$ to include $X'\cup Y$. (Note that, if you do this construction but show me only your $x_\beta$'s, keeping your $y_\beta$'s secret, it will look to me as if you're exactly following Kharazishvili's construction.)

The remaining question is whether Kharazishvili's construction can be done in such a way that $X\setminus X'$ is a null set; that will require more thought.

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