# Is a $\sigma$-algebra measurable? [closed]

As class of measurable sets is a $\sigma$-algebra and in the proof we use the definition of measurable set, so can we also make this statement as "$\sigma$-algebra is a measurable set"? Does both statements mean the same?

I want to elaborate this concept in a more rational way.

## closed as unclear what you're asking by Did, TheGeekGreek, Leucippus, Namaste, ShaileshMay 28 '17 at 2:12

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• "As class of measurable sets is a $\sigma$-algebra"? Do you mean the class of all measurable sets or any class of measurable sets? Big difference. – Nap D. Lover May 27 '17 at 19:00
• I suppose that you want to say "formal" instead of "rational" – Masacroso May 27 '17 at 19:01
• Firstly let me know what is the difference? – Fernando May 27 '17 at 19:02
• @Masacroso I want to know the logic behind this stuff, so used rational – Fernando May 27 '17 at 19:05
• @Fernando the logic behind mathematics is represented through it formalization, then the most used word is "formal" instead of "rational". The concept of "rational" is imprecise out of mathematics, and in mathematics "rational" means different things but not directly related to what you want to say. – Masacroso May 27 '17 at 19:17

A measurable space is a pair $(X,\mathfrak F)$. where $\mathfrak F$ is a subset of $\mathcal P(X)=2^X$ satisfying some properties.
The elements of $\mathfrak F$ are called measurable sets. So in order to speak about measurability of a set $A$ we already need to have the measure space $(X,\mathfrak F)$ and we need to know that $A\subseteq X$.
So in general, it makes no sense to talk about whether $\mathfrak F$ is measurable or not, because $\mathfrak F$ is not a subset of $X$.