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?