Why is $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ called a measurable space when actually is not? I get confused when I put the following three notes together:


*

*Power set of any set is a $\sigma$-algebra.

*If $X$ is a set and $\Sigma$ is a $\sigma$-algebra over $X$, then the pair $(X, \Sigma)$ is a measurable space.

*Vitali set is known as a counterexample that there is no measure on all the subsets of $\mathbb{R}$.


By (1) and (2), one may think that $(\mathbb{R}, \mathcal{P}(\mathbb{R}))$ is a measurable space, which intuitively concludes that there must be a measure on all the subsets of $\mathbb{R}$. However, (3) says the opposite. Can anyone help me understand what is going on?
 A: It is an unfortunate choice of terminology.


*

*The term measureable space, taken as a single unit, is defined to mean a set $X$ with a chosen sigma-algebra $\Sigma$. There is no specific choice of measure involved, and saying that $(X,\Sigma)$ is a measurable space is not intended to claim that it is possible to define a measure with that set and that sigma-algebra (even though, as it turns out, it is always possible: link).Therefore it is absolutely correct to say that $(\mathbb{R},\mathcal{P}(\mathbb{R}))$ is a "measurable space", but you should not take that as saying anything about a specific measure, or even whether it is possible to define a measure, despite the typical semantics of the word "measurable".

*When we say that the Vitali set $V$ is "non-measurable", that is (implicitly) taken to mean with respect to the Lebesgue measure on the set $\mathbb{R}$ with the Lebesgue sigma-algebra $\mathscr{L}$. In other words, it is a claim that $V\notin \mathscr{L}$. The Vitali set is not a counterexample to what you said, because it is possible to define a measure on the set $\mathbb{R}$ with the sigma-algebra $\mathcal{P}(\mathbb{R})$, as I pointed out earlier (link).  The correct statement is that the Vitali set shows that one cannot define a measure $\mu$ on $(\mathbb{R},\mathcal{P}(\mathbb{R}))$ that also satsifies certain desired properties (such as $\mu((a,b))=b-a$, among others.) It is absolutely possible to define a measure on $(\mathbb{R},\mathcal{P}(\mathbb{R}))$ that does not have those nice properties.
A: the reason for your confusion could be, that you forgot some conditions for 3.
Vitali sets are a counterexample that there is no measure on all the subsets of $\mathbb R$ with the following conditions:


*

*the measure of an open interval (a,b) should be b-a

*the measure should be invariant w.r.t to translation

*the measure is countable additive

