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