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

Question

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 $[0,1]$?

Answer 1

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.

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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.