TITLE: Abuse of notation when we denote $\mathcal{F}_n $-measurability of $X_n$ by "$X_n \in \mathcal{F}_n$"?
QUESTION [6 upvotes]: I am confused about a notation that I am seeing in many probability texts when authors write $X_n \in \mathcal{F}_n$ to express that $X_n$ is  $\mathcal{F}_n$-measurable. The elements in $\mathcal{F}_n$ are subsets of $\Omega$, our probability space, so it does not make any sense for $X_n$ to be an element of $\mathcal{F}_n$... If anything, what we should write is $X^{-1}_n(\omega) \in \mathcal{F}_n$ for every $\omega \in \Omega$ to denote $\mathcal{F}_n$-measurability. So my question is, do authors just write $X_n \in \mathcal{F}_n$ as some sort of shorthand convenience/abuse of notation, or am I missing something deeper here? Thanks in advance!

REPLY [4 votes]: Yes, this is a common abuse of notation in probability theory. I do not know the history of this notation, but I can give some arguments which suggest why it is sensible.

With this notation, we have the implication$$
X\in \mathcal F\text{ and }\mathcal F\subseteq \mathcal G\implies X\in \mathcal G,
$$
which is what we would hope to be true for a binary relation notated with $\in$.

To say that $X$ is $\mathcal F$-measurable is to say that for all $B\in \mathcal B(\mathbb R)$, that $X^{-1}(B)\in \mathcal F$. That is,
$$
X\in \mathcal F \iff X^{-1}(B)\in \mathcal F\text{ for all }B\in \mathcal B(\mathbb R)
$$
In other words, when we write $X\in \mathcal F$, we are not saying that $X$ itself is an element of $\mathcal F$, but that the inverse image of every Borel set under $X$ is an element of $\mathcal F$. This is similar to how when we write $3\le X\le 5$, we are not saying that $X$ itself is an element of the closed interval $[3,5]$ (which is nonsense, since $X$ is a function, not a number), but rather that the image of every $\omega$ under $X$ is in this interval.

A good intuition for $\sigma$-algebras is that they represent knowledge; knowing $\mathcal F$ means knowing, for each $A\in \mathcal F$, whether or not $A$ has occurred (that is, whether or not $\omega\in A$). If $X$ is $\mathcal F$-measurable, then knowing $\mathcal F$ gives you enough information to deduce $X(\omega)$. Therefore, we could roughly say that
$$
\text{The information needed to deduce $X$ is contained in $\mathcal F$,}
$$
but this is wordy, so we abbreviate it as $X\in \mathcal F$.


Whether you find these arguments convincing is a purely subjective matter.