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I thought the Vitali set does not have an outer measure (as such an outer measure cannot be defined. However, I was told that it does indeed have an outer measure. However, it is not measurable.

Could someone please clarify this for me?

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    $\begingroup$ Yes, as I remember, the formula for outer measure is meaningful for all sets. And a set $A$ is measurable, if it cuts every set $X$ nicely w.r.t. the outer measure $\mu^*$, i.e. if $\mu^*(X)=\mu^*(X\cap A)+\mu^*(X\setminus A)$. $\endgroup$ – Berci Sep 23 '15 at 22:19
  • $\begingroup$ @Berci- What is the outer measure for the Vitali set then? $\endgroup$ – freebird Sep 23 '15 at 22:20
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    $\begingroup$ math.stackexchange.com/questions/182870/… $\endgroup$ – parsiad Sep 23 '15 at 22:21
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See, for example, the definition of Lebesgue measure on Wikipedia.

Note that $\lambda^\star$, the outer measure, is well-defined for any input. However, an additional condition is required for a set to be measurable.

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