1

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?

freebird
  • 11
  • 1
    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 $\mu^(X)=\mu^(X\cap A)+\mu^(X\setminus A)$. – Berci Sep 23 '15 at 22:19
  • @Berci- What is the outer measure for the Vitali set then? – freebird Sep 23 '15 at 22:20
  • 1
    http://math.stackexchange.com/questions/182870/what-is-the-outer-measure-of-vitali-set – parsiad Sep 23 '15 at 22:21

1 Answers1

2

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.

parsiad
  • 25,154