Every set in a sigma-algebra not measurable? I'm studying Axler's measure theory book and there a set is said to be measurable if it belongs to a $\sigma$-algebra. A power set is always a $\sigma$-algebra, and combining these, it would seem that every set in $\mathcal{P}(\mathbb R)$ is measurable, which is not true. What am i not understanding, why don't these definitions contradict each other?
 A: A measure space must specify three things:

*

*An underlying set;

*A $\sigma$-algebra of subsets which are measurable in this measure space;

*A measure (a function on this $\sigma$-algebra).

When we say "a measurable subset of the real numbers" we typically mean "a subset of the real numbers measurable with respect to the Lebesgue measure", which is one way to make these choices. Rather than say "measurable subset", we could have said "Lebesgue-measurable subset" to avoid confusion.
You can instead define a different measure on the real numbers. You could choose to use the $\sigma$-algebra $\mathcal P(\mathbb R)$, and for example decide that the measure of a set is equal to the number of integers it contains. When you're working with this measure, every subset of the real numbers is measurable, but now you're not talking about the same thing as everyone else.
A: The family of measurable sets form a $\sigma$-algebra but not all sets belonging in any $\sigma$-algebra are measurable. So to fully define a measure space we need a set $X$, a $\sigma$-algebra $\mathcal{A}$ on $X$ and a function $\mu: \mathcal{A} \to [0, \infty]$ which satisfies some properties. A set $A \subseteq X$ is called $\mu$-measurable (or simply measurable if this doesn't cause confusion) if it belongs to the $\sigma$-algebra $\mathcal{A}$. You can't define a measure space without clearly stating what each of these three things is.
