The mapping $P$ is a measure on, what? If $(\Omega,\mathcal{F},P)$ is a probability space, do we say that $P$ is a measure on $\Omega$ or $\mathcal{F}$?
I know that it is a measure on $(\Omega,\mathcal{F})$, but I want to be sure which one of them it is to be precise.
 A: Rigorously speaking, $P$ is a measure on $(\Omega, \mathcal{F})$. It can't just be a measure on $\Omega$ because it's not necessairly defined on every subset.
Colloquially, on the other hand, people sometimes speak of a measure $P$ on a set $\Omega$. Either it is understood that the $\sigma$-algebra is something standard (like the entire power set $\mathcal{P}(\Omega)$, or the Borel $\sigma$-algebra), or else it is understood that the measurable sets are implicitly included in the measure $P$. In this latter case, you can think of $P$ as being a partial function.
A: If $(\Omega,\mathscr F,\mathbb P)$ is a probability space, then the probability measure $\mathbb P:\mathscr F\to[0,1]$ is a function whose domain is $\mathscr F$, satisfying $\mathbb P(\varnothing)=0$, $\mathbb P(\Omega)=1$, and $\mathbb P(\bigcup_{n=1}^{\infty} A_n)=\sum_{n=1}^{\infty}\mathbb P(A_n)$ whenever $A_1,A_2,\ldots\in\mathscr F$ are pairwise disjoint.
On the other hand, even if you say that “$\mathbb P$ is a measure on $\Omega$,” which one might say is incorrect from a purist’s point of view, everybody will know what you’re talking about as long as the underlying $\sigma$-algebra $\mathscr F$ is clearly understood.
