Let $(\Omega, \mathcal{A}, P)$ be a probability space, $(\hat{\Omega}, \hat{\mathcal{A}})$ be a measurable space and $X : \Omega \rightarrow \hat{\Omega}$ be a random variable.
We say that an event $A \in \mathcal{A}$ is a possible property of $X$ if and only if, for all $w \in \Omega$, knowing the value of $X(w)$ is enough information to unambigously decide whether $w \in A$ or $w \in A^{c}$ ( = whether $A$ happened or did not happen). In other words, $A \in \mathcal{A}$ is a possible property of $X$ if and only if there is a set $E \subseteq \hat{\Omega}$ such that $A = X^{-1}(E)$. If $A$ is a possible property of $X$, one can always choose $E \subseteq X(\Omega)$.
We denote the set of all possible properties of $X$ as $\mathcal{B}(X)$. I am trying to characterize $\mathcal{B}(X)$ and understand its relation to the $\sigma$-algebra generated by $X$.
The $\sigma$-algebra generated by $X$ is the smallest $\sigma$-algebra such that $X$ is measurable and is given by $$ \sigma(X) := X^{-1}(\hat{\mathcal{A}}) =\{X^{-1}(\hat{A}) \ \vert \ \hat{A} \in \hat{\mathcal{A}} \}. $$
Clearly, all events in $\sigma(X)$ are possible properties of $X$ and it holds that $\sigma(X) \subseteq \mathcal{B}(X) \subseteq \mathcal{A} $. Note, that $\sigma(X)$ depends on $X$ and $\hat{\mathcal{A}}$. I wonder what necessary and sufficient conditions on $X$ and $\hat{\mathcal{A}}$ are such that $$ \sigma(X) = \mathcal{B}(X). $$
It is not hard to see that $$ \mathcal{B}(X) = X^{-1}(\mathcal{P}(\hat{\Omega})) \cap \mathcal{A}, $$ where $\mathcal{P}$ denotes the power set. In particular, this implies that $\mathcal{B}(X)$ is always a $\sigma$-algebra. If we write $$ \sigma(X) = X^{-1}(\hat{\mathcal{A}}) \cap \mathcal{A} ,$$ it becomes clear that $\hat{\mathcal{A}} = \mathcal{P}(\hat{\Omega})$ is a sufficient condition for $ \sigma(X) = \mathcal{B}(X)$. However, I do not think this condition is necessary and it is rarely fulfilled in practical applications.
What are conditions on $X$ and $\hat{\mathcal{A}}$ for $ \sigma(X) = \mathcal{B}(X)$ that are both necessary and sufficient?
I'm happy to hear as many thoughts as possible on this!
Kind regards, Joker
I came up with a necessary and sufficient condition. After it's proof, there are some results and examples that came to my mind, which led me to the answer I give here.
[NOTATION: I changed $\mathcal B(X)$ to $\mathfrak B(X)$ in order to avoid confusion with 'borelian sets of X'—a notion that makes sense whenever a topology is explicitly defined or assumed by default for $X$; although $X$ is a function, it is itself a set—like everything in ZFC—and then it always admits the notion of a topology.)
Proof: To see that it is a necessary condition, let's assume (on the contrary), that there is a set $A_0\in \mathcal A$ such that $$X^{-1}\big(X(A_0)\big)\in \mathcal A \quad \wedge \quad X(A_0) \notin \hat{\mathcal A}.$$
Then, while $X^{-1}\big(X(A_0)\big)$ is a set in $\mathcal A$ of the form $X^{-1}(\hat A)$ with $\hat A \subset \hat\Omega$, and so $$X^{-1}\big(X(A_0)\big) \in \mathfrak B(X),$$ this is not a set in $\sigma(X)$, since is not of the form $X^{-1}(\hat A)$ and with $\hat A\in \hat{\mathcal A}$. This comes from the fact that $$X^{-1}(\hat A)=X^{-1}\big(X(A_0)\big)$$ implies $$\hat A=X\big(X^{-1}(\hat A)\big)=X\Big(X^{-1}\big(X(A_0)\big)\Big)=X(A_0)$$ and so, by hypothesis $\hat A=X(A_0)\notin \hat{\mathcal A}$. Since $X^{-1}\big(X(A_0)\big)$ cannot belong to $\sigma(X)$, it turns out that $\sigma(X)\subsetneq\mathfrak B(X)$. This shows the condition is necessary.
Now, lets assume the condition holds.
Let $A_0 \in \mathfrak B(X)$, which means $$A_0 \in \mathcal A \quad \wedge \quad \exists \hat A_0 \subseteq\hat \Omega \colon \; A_0=X^{-1}(\hat A_0)$$ and the last equality implies that $X(A_0)=\hat A_0$.
We want to see that there is some $\hat A_1 \in \hat{\mathcal A}$ such that $A_0=X^{-1}(\hat A_1)$. But $$A_0=X^{-1}(\hat A_0)=X^{-1}\big(X(A_0)\big)\in \mathcal A$$ (because $A_0 \in \mathcal A$). But the condition (assumed by hypothesis), implies that $X(A_0)\in \hat{\mathcal A}$.
So that $A_0=X^{-1}(\hat A)$ with $\hat A \in \hat{\mathcal A}$, which shows that $A_0 \in \sigma(X)$. In conclusion $\mathfrak B(X)\subseteq \sigma(X)$, and since the opposite inclusion always holds this means that $$\mathfrak B(X) = \sigma(X),$$ and so the condition is sufficient.
COMMENTS on examples and partial results I explored
A few examples I considered helped me to successively weaken the sufficient condition $\hat{\mathcal A}=\mathcal P(\hat \Omega)$.
1) It is easy to see that this it not necessary, since if you have any probability space $(\Omega,\mathcal A, P)$ with $\mathcal A \subsetneq \mathcal P(\Omega)$ and a measurable space $(\hat\Omega,\hat{\mathcal A})$ with $\hat\Omega=\Omega$ and $\hat{\mathcal{A}}=\mathcal A$, then defining $X\colon\Omega\to\hat\Omega$ as the identity (well defined, since domain and codomain actually are the same—or we could just call it the 'inclusion map'—) gives $\sigma(X)=\mathcal A$ and since any event $A\in \mathcal A$ is also one of the elements of $\hat{\mathcal A}$ and also $X^{-1}(A)=A$, $\sigma(X)=\mathfrak B(X)$ trivially.
I considered then an example that suggested a weaker sufficient condition.
2) Let's take $\Omega=\hat\Omega=\{0,1\}$, $\hat{\mathcal A}=\mathcal P(\Omega)$, $P$ can be any probability (the general case would be $P(\{1\})=p$, $P(\{0\})=q$, with $p,q\in [0,1]$ and $p+q=1$—and of course $P(\emptyset)=0$ and $P(\Omega)=1$—), and take $\hat{\mathcal A}=\{\emptyset,\{0,1\}\}$, which is trivially a $\sigma$-algebra over $\hat\Omega$. It is immediate to check that $\sigma(X)=\{\emptyset,\{0,1\}\}\subsetneq\Omega$ but $\mathfrak B(X)=\Omega$.
One might think that the problem here is that, again, $\hat{A}$ was taken strictly smaller than $\mathcal P (\Omega)$, but the simple structure of the example let us see that the problem is that it was too small relative to $\mathcal A$. The same would have happened for a larger $\Omega$ maybe infinite, even uncountable, say $\mathbb R$, and $\mathcal A$ none of the trivial or the discrete $\sigma$-algebras, say $\mathcal B(\mathbb R)$. If we take $\hat {\mathcal A}=\{\emptyset,\mathbb R\}$, or any $\sigma$-algebra smaller than the set of borelians, then the same situation will arise.
3) In general, the 'problem' in the previous example seemed to be the existence of sets $A\in \mathcal A$ such that $X(A)$ is not a member of $\hat {\mathcal A}$.
This gives a (weaker) sufficient condition:
This can be proved by taking $A_0\in\mathcal B(X)$, that is, $A_0 \in \mathcal A$ and there exists $\hat A_0 \in \hat{\Omega}$ such that $A_0=X^{-1}(\hat A_0)$.
But this implies $X(A_0)=X\big(X^{-1}(\hat A_0)\big)=\hat A_0$, and since by hypothesis $X(A_0)\in \hat{\mathcal A}$, in fact $\hat A_0\in \hat{\mathcal A}$. Then, $A_0=X^{-1}(\hat A_0) \in \sigma(X)$, as we wanted to prove.
4) At first it seemed intuitive to me that this would also be a necessary condition, but I got stucked while looking for a proof (which of course did not exist).
I quickly reasoned that if there is $A_0\in \mathcal A$ such that $X(A_0)\notin \hat{\mathcal A}$, then $A_1=X^{-1}\big(X(A_0)\big) \in B(X)$ but since $A_1=X^{-1}(\hat A_1)$ implies $\hat A_1= X(A_0) \notin \hat{\mathcal A}$, then $A_1$ **is in $\mathfrak B(X)$ but not in $\sigma(X)$.
That would be great... if it wasn't completely wrong. Of course $X(A_0)\subset \hat\Omega$, but too conclude $A_1\in\mathfrak(X)$ we should also check that $A_1 \in \mathcal A$. And guess what: no, I couldn't. (Just in case: $X^{-1}\big(X(A)) \supset A$, but equality need not hold—yet it could hold in particular cases, as for injective $X$).
Not being able to prove that important detail, I started looking for an example where this exact situation arose.
5) This put me to consider more interesting examples involving more complex $\sigma$-algebras. I avoided borelian sets for there's no simple general expression to work with them (of course, you can use a characterization involving ordinal numbers and think of sets of class $G_{\delta\sigma\delta}$ or $F_{\sigma\delta\sigma\delta\ldots}$ or worst... but I said 'simple'). So I relied on the $\sigma$-algebras of countable sets and their complements for a less finitistic example.
Consider now the probability space $\big([0,1],\hat{\mathcal A},P\big)$, with $$\hat{\mathcal A}=\{A\subset [0,1]\colon A \text{ is countable or } [0,1]\setminus A \text{ is countable}\}$$ and any possible $P$ (Borel measure, for instance), and have the measure space $\big([0,1],\hat{\mathcal A}\big)$ where $$\hat{\mathcal A}= \big\{\emptyset,\{0\},(0,1],[0,1]\big\}.$$
I considered several r.v. until some thoughts about uncountability of the preimages of elements in $\hat{\mathcal A}$ and also of their complements led me to consider $X\colon [0,1]\to [0,1]$ as the Cantor function.
Before going further, some additional words on the Cantor function. For the sake of brevity let's just remember that if $X$ is the Cantor function, then:
-- $X\big([0,1]\big)=[0,1]$ ($X$ is surjective), -- $X^{-1}\big(\{0\}\big)=\{0\}\in\hat{\mathcal A}$ and $X^{-1}\big((0,1]\big)=(0,1]$, and so $X$ is measurable;
While constructing and analyzing this example the main goal of doing so was not well defined and actually changed frequently, but in a sense I wanted a more complex case without the property mentioned in where I tried to see if $\sigma(X)=\mathfrak B(X)$ too, or not. My intuition was that $\hat{\mathcal A}$ is too small, since there are plenty of sets $A\in \mathcal A$ for which $X(A)\notin \hat{\mathcal A}$. Basically, all the uncountably many events $A$ but $\{0\}$, $(0,1]$ and of course $\emptyset$ and $[0,1]$ have their image outside $\hat{\mathcal A}$. But counterexamples were not obvious...
Suppose $A_0=X^{-1}(\hat A_0)$, where $\hat A_0$ is a subset of $\hat \Omega\setminus \hat{\mathcal A}$. We want to find such an $A_0$ which belongs to $\mathcal A$, since this would be the element of $\mathfrak B(X)$ which won't be in $\sigma(X)$.
I worked out some special cases ($\mathcal C$ stands for the Cantor set):
All this made me considere the schema $$A_0 \in \mathcal A \colon \quad A_1=X^{-1}\big(X(A_0)\big)$$
In order to find counterexamples we had to discard those cases in which:
What this said is that the sufficient condition $X(A)\in \hat{\mathcal A}$ only becomes relevant for those $A\in \mathcal A$ such that $X^{-1}\big(X(A)\big)\in \mathcal A$ too. Or, in other words, that whenever we have $A\in \mathcal A$ and $X(A) \notin \hat{\mathcal A}$ we need $X^{-1}\big(X(A)\big)\notin \mathcal A$ in order to avoid a counterexample to $\mathfrak B(X) = \sigma(X)$.
This conclusion actually allowed me to arrange a counterexample: - If $A_0$ is a countable nonempty subset of $\mathcal C_u$ (and is not exactly $\{0\}$, the injectivity of $X$ over $\mathcal C_u$ implies that while $A_0 \in \mathcal A$, again $X(A_0)\notin \hat{\mathcal A}$; but this time this becomes relevant because $X^{-1}\big(X(A_0)\big) \in \mathcal A$, too (in fact this is the same as $A_0$).
A general argument in the same spirit gave the proof of necessity that appears in the first part.
I've also seen that considering sets $A$ which were not in $\mathcal A$ did not strengthen this necessary condition since every time we find $A_0\notin \mathcal A$ for which $X(A_0)\notin \hat{\mathcal A}$ but $X^{-1}\big(X(A_0)\big)\in \mathcal A$, we would be able to also take $A_1=X^{-1}\big(X(A_0)\big)\in \mathcal A$, and this would give a counterexample: indeed, $$X(A_1)=X\Big(X^{-1}\big(X(A_0)\big)\Big)=X(A_0) \notin \hat{\mathcal A}$$ and $$X^{-1}\big(X(A_1)\big)=X^{-1}\bigg(X\Big(X^{-1}\big(X(A_0)\big)\Big)\bigg)=X^{-1}\big(X(A_0)\big) \in \hat{\mathcal A},$$ Then, $A_1$ is also a counterexample, but with $A_1$ being an event (unlike $A_0$).
Of course, all this took many steps forward and backwards. I finally crossed my fingers for this new condition to be also sufficient, and luck did its part.