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By definition, the sets in a $\sigma$-algebra are called measurable; so, consider the power set of $[0,1]$. This is a $\sigma$-algebra, so the sets in this collection should be measurable. But a Vitali set is a subset of $[0,1]$, so it should then be called measurable as well.

Why then do we say that a Vitali set is non-measurable, even though it is a member of the power set of $[0,1]$?

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closed as unclear what you're asking by Namaste, Did, José Carlos Santos, Rebellos, Shailesh Dec 2 '18 at 0:07

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  • $\begingroup$ I'm not specialist of the subject, but I recall it has to do with breaking the invariance by translation of the measure. $\endgroup$ – zwim Dec 1 '18 at 13:13
  • $\begingroup$ For a $\sigma$-algebra $\Sigma,$ a set being $\Sigma$-measurable just means it is a member of $\Sigma.$ However when we just say a set of real numbers is measurable, we mean with respect to a specific $\sigma$-algebra, namely the Lebesgue $\sigma$-algebra. The Vitali set, like every subset of $[0,1]$, is $\mathcal P([0,1])$-measurable, however it is not Lebesgue measurable. We don’t care much about $\mathcal P([0,1])$ as a $\sigma$-algebra because there are not many interesting measures on it. On the other hand the Lebesgue measure on the Lebesgue algebra is usually what we’re interested in. $\endgroup$ – spaceisdarkgreen Dec 20 '18 at 2:06
  • $\begingroup$ The construction of the Vitali set shows it cannot be a member of any sigma algebra that has a nontrivial translation-invariant measure. If our aim to extend the idea of interval length to more obscure subsets of the reals, this tells we cannot extend it to all subsets of the reals. $\endgroup$ – spaceisdarkgreen Dec 20 '18 at 2:14
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A measurable space is a pair $(X,\mathcal A)$ where $X$ is a set and $\mathcal A\subseteq\wp(X)$ is a $\sigma$-algebra. In that context a subset of $X$ is measurable iff it is an element of $\mathcal A$.

Looking at the Vitali set $V$ as a subset of $[0,1]$ it is a measurable subset wrt measurable space $([0,1],\mathcal A)$ iff $V\in\mathcal A$ .

So actually not measures "decide" whether $V$ is a measurable set or is not.

You can start with $([0,1],\mathcal A)$ where $V\in\mathcal A$ (making $V$ measurable in advance) and then go looking for measures on that measurable space.

Examples are:

  • $\mathcal A=\wp([0,1])$ and $\mu$ is the counting measure.
  • $\mathcal A=\wp([0,1])$ and $\mu$ is the zero measure (your suggestion).
  • $\mathcal A=\{\varnothing,V,V^{\complement},[0,1]\}$ and $\mu$ is the measure determined by e.g. $\mu(V)=4$ and $\mu(V^{\complement})=\pi$.

For completeness let me mention that a triple $(X,\mathcal A,\mu)$ where $(X,\mathcal A)$ is a measurable space and $\mu$ is a measure on it (i.e. a function $\mathcal A\mapsto[0,\infty]$ that has certain properties) is a measure space.


On the other hand you can start with a set $X$ and some function on its subsets that "has the looks" of a measure and then go for finding a $\sigma$-algebra such that the function restricted to it is indeed a measure.

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