Set of all subset is a sigma algebra? In the book Measure, Integration and Real Analysis (Axler, 2020), the author claims that:

Suppose $X$ is a set. Then the set of all subsets of $X$ is a $\sigma-$ algebra on $X$

Then he defines the measurable set:

A measurable space is an ordered pair $(X,S)$ where $X$ is a set and $S$ is a $\sigma -$ algebra on $X$.
An element of $S$ is called an $S$ measurable set (or just a measurable set if $S$ is clear from the context.

I have a question concerning the measurablity of every subset of $\mathbb{R}$.
As we know, not every subset of $\mathbb{R}$ is measurable ($i.e$ it does not belong to any $\sigma -$ algebra. So the set of all subset of $\mathbb{R}$ can't be a $\sigma -$ algebra on $\mathbb{R}$ which contradicts the claim of the author.
Could you please explain me what does the author mean ? Or is there any misunderstanding in my mind when reading the book ?
Thank you very much for your help!
 A: The assertion “not every subset of $\Bbb R$ is measurable” is valid for the Lebesgue measure. Anyway, here Axler is talking about $\sigma$-algebras, not about measures. And, yes, $\mathcal P(\Bbb R)$ is a $\sigma$-algebra. And, on this $\sigma$-algebra you can define the measure $m(X)=\#X$; with respect to this measure, $\Bbb R$ is measurable.
A: The set   $\mathcal{P}(X)$ of all subsets of $X$ is a sigma algebra because it satisfies the axioms for that kind of structure. That is independent of any discussion of measures.
The definition of a measure space is a pair $(X,S)$ where $S$ is some subset of $\mathcal{P}(X)$ that happens to be a sigma algebra. It may or may not be all of $\mathcal{P}(X)$.
When you set out to define Lebesgue measure on $\mathbb{R}$ you soon discover that you can't get all the properties you want if you try to assign a "length" to every subset of $\mathbb{R}$. So you restrict the definition of "length" to a subalgebra $S$ of $\mathcal{P}(\mathbb{R})$ and call those the measurable sets.
