I have some curiosity that for a Vitali set $V$, whether the set $V+V$ is measurable or not. Heuristically, I think there is no reason to be measurable.
I tried to express each element as the sum of representive and rational number, but this method fails since it depends on choice function.. Is there any idea to get a clever answer? Or is this undecidable in standard ZFC axioms?
It is possible to have a Vitali set $V\subset[0,1]$ such that $V+V$ contains all of $(0,2)$ and thus is measurable. To do this, enumerate $(0,2)$ as $(x_\alpha)_{\alpha<\mathfrak{c}}$. We can then recursively pick $a_\alpha,b_\alpha\in[0,1]$ such that $a_\alpha+b_\alpha=x_\alpha$ and no two of the $a_\alpha$ and $b_\alpha$ have rational difference (so that the set $\{a_\alpha,b_\alpha:\alpha<\mathfrak{c}\}$ can then be completed to a Vitali set). To do so, just note that there are $\mathfrak{c}$ different pairs $(a,b)\in[0,1]^2$ such that $a+b=x_\alpha$, and fewer than $\mathfrak{c}$ have $a$ or $b$ such that $a-b$ is rational or $a$ or $b$ differs from any $a_\beta$ or $b_\beta$ for $\beta<\alpha$ by a rational number. So, it is always possible to pick $(a,b)$ to be $a_\alpha$ and $b_\alpha$ to not form any rational differences.
On the other hand, as Wojowu pointed out in a comment, you can also have a Vitali set such that $V+V$ is not measurable. For instance, you could have $1\in V$ but every other element of $V$ is in $[0,1/2]$. Then $(V+V)\cap[1,2]=V+1$ is not measurable so $V+V$ is not measurable.