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Consider the basic setup for a real valued random varibel i.e $(\Omega,\mathcal{F},P)$ and some r.v $X$ defined on $\Omega$. We then define $P_{X}$ on $(\mathbb{R},\mathcal{B})$ by $P(\omega ; X(\omega) \in B)$ as the probabilty dist. of $X$. Why do we even have the space $(\Omega,\mathcal{F},P)$ why dont just consider the events in $(\mathbb{R},\mathcal{B})$ with probablities w.r.t $P_{X}$ directly instead?

Operations on r.v's should work fine in this spaces aswell, im stuck in some kind of "chicken and the egg"-loop.

Why do we make this move and define all these objects, which to me contain very similar info. Ultimalty one wants to control $X\in B$, I suppose.

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While just using the law for $X$ is a perfectly good probability space for a universe consisting of a single random variable, what about situations with two, three, four, countably infinite, uncountably infinite numbers of random variables?

In your situation with a single random variable, you can just take $(\Omega,\mathcal{F}, P) = (\mathbb{R},\mathcal{B},P_X)$ as your probability space. $P_X$ is just a specified probability measure on $(\mathbb{R},\mathcal{B}),$ so there's no circularity here (you can define it in terms of a distribution function if you wish). Similarly, if you have a universe with two random variables, you can take $(\Omega,\mathcal{F}, P) = (\mathbb{R^2},\mathcal{B(\mathbb{R}}^2),P_{X_1,X_2})$ for the appropriate probability measure $P_{X_1,X_2}.$ The pattern goes on through situations where there's an infinite number of RVs or even uncountably (like continuous time stochastic processes) though the functional analysis gets challenging. What doesn't change as the complication increases is that it can be described a probability space ... in the case of a continuous time like Brownian motion the natural one would look like $(\Omega,\mathcal{F}, P)=$ (all functions B(t), some sigma algebra, some measure on the space of functions). Since there are many different possible random systems you want to be able to describe, with probability spaces of varying complexity, it's easier to keep the notation abstract. However if you have a concrete situation with a natural set of random variables, you almost always think in terms of the natural probability spaces described above, perhaps in terms of distribution functions instead of measures (but those two are equivalent).

So why bother with the $P_X(B) = P(\omega | X(\omega)\in B)$ definition? Well, $X$ might not be the only thing in the universe and you might want to lift it out to consider it in isolation. (Of course, they are equivalent if $(\Omega,\mathcal{F}, P) = (\mathbb{R},\mathcal{B},P_X)$.) Or you might have chosen a different representation of the probability space than the natural one. For instance, maybe you take $$(\Omega,\mathcal{F}, P) = ([0,1], \mathcal{B}([0,1]), \mathrm{Lebesgue}).$$ This space can be used to model lots of stuff. You can model a random sequence of zeros and ones (whose "natural" outcome space is not $[0,1]$ but rather $\{0,1\}^{\mathbb{N}}$) by mapping $\omega \in [0,1]$ to its decimal expansion. Or you can model a Gaussian RV by mapping $\omega \in [0,1]$ to $X = \Phi^{-1}(\omega) \in \mathbb{R}$ with $\Phi$ the normal CDF (like in inverse transform sampling). You can even use some trickery to model a sequence of random variables with a given distribution. As you alluded to in your question, the two representations have the same information and are equivalent. Since there are many possible spaces to represent a given random model and it doesn't matter which you choose, it's natural to be abstract and write $(\Omega,\mathcal{F}, P).$

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  • $\begingroup$ these can be regarded as "single" objects aswell? with thier own $P$'s on some Borel algebra. And I guess independece and conditionals could be formulated aswell. $\endgroup$ – user1 Dec 30 '16 at 8:12
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    $\begingroup$ I guess there are two different questions here: 1. Why not make your probability space the joint law of a set of random variables? (answer: you are totally free to if that describes your set up. It's just that people like to be more abstract than that and not commit to a concrete representation of this sort). 2. What are the benefits of the measure theory foundation in the first place? (partially answered last comment) $\endgroup$ – spaceisdarkgreen Dec 30 '16 at 9:09
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    $\begingroup$ @user32423 But it's not immediately clear what random variables one can introduce. You need to actually pick a probability space and see what kind of RVs can live there. As I mentioned in my answer, the probability space $([0,1], \mathcal B, Leb)$ is (perhaps somewhat surprisingly) capable of supporting a great many sets of random variables and joint distributions for them, but seeing this requires constructing some elaborate encodings. However in practice we do normally just assume our space is "big enough" to have all the random variables we are interested in. $\endgroup$ – spaceisdarkgreen Feb 13 '18 at 16:40
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    $\begingroup$ @user32423 in fact, we usually effectively work (or imagine we work) in the probability space defined by some tuple of random variables and their distribution. But we don't need to be committed to that precise representation: any space and definition of the RVs on it is fine provided the push forward measure of the RVs is what we want it to be. $\endgroup$ – spaceisdarkgreen Feb 13 '18 at 16:58
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    $\begingroup$ @user32423 Yes. $\endgroup$ – spaceisdarkgreen Feb 15 '18 at 0:51

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