While reading the probability space in Wikipedia, I'd found the usual formulation is a triplet, which is ${\displaystyle (\Omega ,{\mathcal {F}},P)}$.
Upon my understanding, the middle ${\mathcal {F}}$ is a power set of $\Omega$ which will be allocated with real-valued probabiilty by $P$.
If every set in this nature has power set, there might be no necessity of introduction of ${\mathcal {F}}$ I guess however, I've never thought of a set which doesn't have its power set.
Is there any set that doesn't have power set? or if not, which means every set has its power set, is there any plausible reason that ${\mathcal {F}}$ is introduced in probability formulation?
Best Answer
In standard mathematics, every set has a power set. This is encoded in the Axiom of Power Set. However, your confusion lies with the definition of a probability space, not with set theory.
The set $\mathcal F$ in a probability space $(\Omega, \mathcal F, P)$ is not necessarily the power set of $\Omega$. The set $\mathcal F$ is a subset of the power set $\mathcal P(\Omega)$. This $\mathcal F$ is required to be a so-called sigma algebra, which tells you that it shares some properties in common with the full power set, but it need not be the full power set at all.
In particular, for any $\Omega$, you can take $\mathcal F = \{\emptyset, \Omega\}$, and this will be a sigma algebra on $\Omega$. Unless $|\Omega| \leq 1$, it will not be the power set.