I came across this problem.
If we throw two dice the sample space is $\Omega = \{ (i, j), 1 \leq i, j \leq 6 \}$. The $\sigma$-algebra is generated by the events $A_k = \{ (i,j) : \max(i,j) = k \}$
Show that $x_1=max(i,j)$ is a random variable and $X_2=i+j$ isn't a random variable
There's a short explanation which explains the relation within the borel sets of the random variable and their preimages.
"Actually, a function $X$ from some measurable space $(\Omega, \Sigma)$ to $\mathbb{R}$ equipped with the Borel $\sigma$-algebra is defined as a random variable if it's measurable; that is, for every Borel set B, the preimage $X^{-1}(B) = \{\omega : \omega \in \Omega, X(\omega) ∈ B\}$ is measurable."
"Notice the focus on $X^{-1}(B)$ (preimages of Borel sets) rather than $X^{-1}(x)$(preimages of individual real numbers)"
1) Whenever you have a random variable, are the events of the random variable always borel sets?
I mean if we have that the probability space of the random variable is $(\Omega_X,F_X,P_X)$, then for all the random variables, the family of events $F_X$ is always the $\sigma$-borel algebra?
2) If the preimages $X_1^{-1}(x)$ are $\left \{ 1,2,3,4,5,6 \right \}$, what are the preimages $X_2^{-1}(x)$? And what would be the preimages of $X_1^{-1}(B)$ and $X_2^{-1}(B)$?
How would you show that $X_1$ is a random variable and $X_2$ isn't?
Generators of the $\sigma$-algebra, first case:
$A_1=\{(1,1)\}$
$A_2=\{(1,2),(2,2),(2,1)\}$
$A_3=\{(1,3),(2,3),(3,3),(3,2),(3,1)\}$
$A_4=\{(1,4),(2,4),(3,4),(4,4),(4,3),(4,2),(4,1)\}$
$A_5=\{(1,5),(2,5),(3,5),(4,5),(5,5),(5,4),(5,3),(5,2),(5,1)\}$
$A_6=\{(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2),(6,1)\}$
Here $A_k:=\{(i,j)\in\Omega\mid\max(i,j)=k\}$
Generators of the $\sigma$-algebra, second case
$B_2=\{(1,1)\}$
$B_3=\{(1,2),(2,1)\}$
$B_4=\{(1,3),(2,2),(3,1)\}$
$B_5=\{(1,4),(2,3),(4,1),(3,2)\}$
$B_6=\{(1,5),(2,4),(3,3),(5,1),(4,2)\}$
$B_7=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}$
$B_8=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$
$B_9=\{(3,6),(4,5),(6,3),(5,4)\}$
$B_{10}=\{(4,6),(5,5),(6,4)\}$
$B_{11}=\{(5,5),(6,5)\}$
$B_{12}=\{(6,6)\}$
Here $B_k:=\{(i,j)\in\Omega\mid i+j=k\}$
Best Answer
1)
If by $\Omega_X$ you mean $\mathbb R$ and by $P_X$ you mean the probability measure prescribed by $B\mapsto P(X\in B)$ then: "yes, we usually go for $\mathcal F_X=\mathcal B(\mathbb R)$".
$X$ is usually by definition a random variable if it is a function that takes real values and is measurable wrt the Borel $\sigma$-algebra.
2)
Looking at $X_1$ the preimage of $B\in\mathcal B(\mathbb R)$ wrt $X_1$ is the set:$$X_1^{-1}(B)=\{(i,j)\in\{1,2,3,4,5,6\}^2\mid max(i,j)\in B\}$$
If you take singleton $B=\{x\}$ then we get:$$X_1^{-1}(B)=\{(i,j)\in\{1,2,3,4,5,6\}^2\mid max(i,j)=x\}$$
For $X_2$ you will get similar expressions where $\max(i,j)$ is replaced by $i+j$.
You cannot speak of "the preimages of $X_1^{-1}(B)$".
The correct wording is that "$X_1^{-1}(B)$ is the preimage of $B$ under (or with respect to) $X_1$".
The $\sigma$-algebra generated by the events $A_k$ is formally the smallest $\sigma$-algebra on $\Omega$ that contains these sets
Now observe that every preimage $X_1^{-1}(B)$ can be written as a union of these sets. This tells us that $X_1$ is measurable wrt to this $\sigma$-algebra. That means that $X_1$ can be classified as a random variable if $\Omega$ is equippes with the $\sigma$-algebra. Actually the $\sigma$-algebra generated by the $A_k$ can be shown to be the collection $$X_1^{-1}(\mathcal B(\mathbb R)):=\{X_1^{-1}(B)\mid B\in\mathcal B(\mathbb R)\}=$$$$\{\{(i,j)\in\Omega\mid \max(i,j)\in B\}\mid B\in\mathcal B(\mathbb R)\}\tag1$$
However it is not possible to write e.g. $X_2^{-1}(\{4\})=\{(i,j)\mid i+j=4\}$ as an element of this $\sigma$-algebra.
That's why $X_2$ cannot be classified as a random variable.
edit:
Working $(1)$ out we find that for every $B\in\mathcal B(\mathbb R)$ we can find a set $I=I_B\subseteq\{1,2,3,4,5,6\}$ such that: $$X_1^{-1}(B)=\bigcup_{i\in I}A_i$$
So actually: $$X_1^{-1}(\mathcal B(\mathbb R))=\left\{\bigcup_{i\in I}A_i\mid I\subseteq\{1,2,3,4,5,6\}\right\}$$
For e.g. preimage $X_2^{-1}(\{4\})=\{(1,3),(2,2),(3,1)\}$ we cannot find such a set $I$.