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This is a different question from my previous question Difference between a partial selector and a selector, however I am going to repeat the preamble...

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 begins 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.

We are now going to depart from the proof a little...

Let $A \subseteq \mathbb{R}$ be uncountable. Consider the set

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

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

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

My question is can $X$ be non-measurable?

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Since you allow $A$ to be any uncountable subset of the reals, how do you ensure that $F_\beta\setminus Z_\beta$ is nonempty so that you can choose $z$? – Andreas Blass Jun 17 '11 at 16:13
Sorry yes, I had thought of that, but not followed it through properly... in the case $F_{\beta}\backslash Z_{\beta}$ is empty we put $x_{\beta} = 0$... it's a bit ugly, but I don't think it makes any difference to the question (other than to ensure X is defined)... – George Lazou Jun 17 '11 at 17:39

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