1. ## Vitali sets

This is measure theory, so I am not sure if it should go here, under Set Theory, or Analysis. I will opt for Topology.

I am a little dubious on a point on the proof that a Vitali set is non-measurable. I look at the definition of it being a subset of [0,1] defined as

for all r in R, let $v_r$ = r + $q_r$ for some rational $q_r$. (This requires a choice function C).
Then $V_C$ = {v_r : r in R} $\cap$ [0,1].

Then there are proofs based on a combination of translation invariance of length and additivity of a measure to show that each V is non-measurable. But can we say that this is true for all C? For example, although it is highly unlikely, nonetheless the function that chooses the same value of $q_r$ for all r, although it would be a boring choice function, would give us an interval that clearly had measure one. For example, if C gave us $q_r$ = 0 for all r, then $V_C$ = [0,1]. So, shouldn't we put on some other conditions in the definition to ensure non-measurability?

2. There are restrictions on the choice function C.

for all r in R, let $v_r$ = r + $q_r$ for some rational $q_r$. (This requires a choice function C).
First, given r, we don't choose some $q_r$ and then take the intersection $\{r + q_r\mid r\in\mathbb{R}\}\cap[0,1]$. We must choose $q_r$ so that $r+q_r\in[0,1]$.
More importantly, after we have chosen $q_r$ for some $r$, the choice for all $r'$ such that $r-r'\in\mathbb{Q}$ is fixed: we must have $r'+q_{r'}=r+q_r$. This is because, in Wikipedia description, the choice function is from equivalence classes $r+\mathbb{Q}$ to [0,1]: all reals that differ by a rational number produce a single element in the resulting Vitali set V.