I have been looking for a direct answer for this question a little while. Ichecked a lot of topics and none seemed to answer that in really direct way.

Why does the measurability definition exclude Vitali sets? I`d like to see a proof(It could be a reference) that Vitali sets are not measurable(in all that desirable way...) using the definition(*) of a measurable set.

Also, I`d like some intuition on it. Deeply, I want to understand why the definition of measurable sets works well in the task of saying wether a set behaves nicely or not in the sense of a desirable measure.

What I`ve thought so far: it all lies on the fact that Vitali sets are based on picking representatives which are arbitrary. Am I in the right direction on the interpreatation?

Edit:(*) by definition of measurable set I meant:

A set E is said to be measurable if for each set A we have $$m^∗A=m^∗(A∩E)+m^∗(A∩Ec)$$

where $m^*$ denotes the exterior measure.

  • $\begingroup$ Which definition of measurable sets is "the" measurability definition? $\endgroup$ – bof Mar 26 '17 at 20:03
  • $\begingroup$ I was refering to: A set E is said to be measurable if for each set A we have $$m*A=m*(A \cap E)+m*(A \cap E^{c})$$ where m* denotes the exterior measure. $\endgroup$ – Marcelo Broinizi Mar 26 '17 at 20:07
  • 1
    $\begingroup$ You may find the discussion in the question and my answer at math.stackexchange.com/questions/1420440/… helpful. $\endgroup$ – Eric Wofsey Mar 26 '17 at 21:18

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