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Math Help - Vitali sets

  1. #1
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    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?
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  2. #2
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    There are restrictions on the choice function C.

    Quote Originally Posted by nomadreid View Post
    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.
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  3. #3
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    Thanks, emakarov. That clears everything up.
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