Independence of Random Variables in measure theoretic sense In preparation for my graduate school, I am studying probability theory myself but am confused about the independence of random variables and it has been weeks since I am stuck on this. Could you please all share your thoughts?
Definition of independence of events

So this says that two random variables are independent if the events  {X $ \in A$} and {Y$\in B$} are independent. But here this assumes that there is same probability measure "P" defined on domain space $\Omega$. Therefore, using P on these sets makes sense. How would this definition make sense when we want to show whether two Normal Random variables in R are independent? What should be the underlying ($\Omega$,$\wp$, P) in that case?
Moreover, doesn't each Normal RV generate a different probability measure on R with Borel Sigma-Algebra, then does this make sense to speak about independence when we have different measures?
 A: Imagine throwing a fair, six-sided die. We model the possible outcomes and their probabilities using a probability space, where  $\Omega=\{1,2,3,4,5,6\}$, $\mathcal p$ is the power set of $\Omega$, and $P(A):=\frac{\#A}{6}$, where $\#A$ is the number of elements of $A$ (so it's a uniform probability distribution). This is our underlying probability space.
Once we have this probability space (and not a second earlier!) we can define random variables on it. This underlying space is crucial, since random variables are functions with this probability space as their domain. For instance, we could define the random variable $X$ which measures wether the roll yields an even number, so $X(\omega)=1$ if $\omega\in\Omega$ is even, and $X(\omega)=0$ otherwise. Similarly, we could define a completely different random variable $Y$ on the same probability space, measuring wether the roll yields a prime. These two random variables are not independent, since the chance that the roll yields an even prime is $1/6$ (since 2 is the only even prime), but if they were independent it would be $P(X=1,Y=1)=P(X=1)P(Y=1)=\frac12 \cdot\frac12=\frac14$.
Usually when working on modeling any problem in probability, you will be working with one single underlying probability space on which all necessary random variables are defined. Often this space won't be mentioned explicitly. It's just a given that it's there. It's also not important how exactly it looks. The only thing that's important is how the random variables we define on it are distributed (that's the probability distribution you mentioned, which does depend on the random variable) and how they interact. But we can take two completely different probability spaces and define sets of random variables on both such that their distributions and interactions are exactly the same, so it really doesn't matter how the probability space looks. The only thing it has to do is be there. But we rarely have to actually work with it, since it's not our object if interest - the random variables are.
