I'm just starting to learn measure theory and we have defined a random variable $X$ as a measurable function from some set $\Omega$ to some other set $E$ which is usually $\mathbb{R}$. We defined $\mathbb{P}(X \in S)$ for $S$ measurable in $E$ as being $\mu(\{\omega: X(\omega) \in S\})$ where $\mu$ is the probability measure on $\Omega$.
I'm struggling to understand how this relates to the "usual" pre-measure theory definition of a random variable. For example if we had a standard Gaussian R.V what would $\Omega$ be and what would the map $X$ be?
My intuitions tell me we should take $\Omega$ to be $(0,1)$ with the standard measure and then $X$ sends $\omega \in \Omega$ to $\inf\limits_{x \in \mathbb{R}} \{x: \omega \leq F(x)\}$ where $F$ is the C.D.F of the Gaussian. However I found a very similar question on this site which said we take $\Omega$ to be $\mathbb{R}$ and $X$ as the identity map which didn't really make sense to me.
Any clarifications are much appreciated.
Best Answer
There is not generally a 'right answer' for what probability space to take in order to support some particular random variable. I think the probability space $(\mathbb R, \mathcal B(\mathbb R), P)$ where $$P(A) = \int_A \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx,$$ where we take $X$ to be the identity function on $\mathbb R,$ is a fine choice.
Your choice works just fine as well. The only important thing is that for any $x\in \mathbb R,$ $$\mu(\{\omega\in \Omega :X(\omega)\le x\}) = \int_{-\infty}^x \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx.$$
The first example I gave (which is presumably the same one in the other post) is a special probability space called the law of $X.$ It is given abstractly in terms of any underlying probability space $(\Omega,\Sigma,\mu)$ with $X:\Omega\to \mathbb R$ as$$ P(A) = \mu(X^{-1}(A))$$ for any Borel-measureable $A\subseteq R.$ In other words, it is the push-forward measure of $\mu$ by the $\Sigma-\mathcal{B}(\mathbb{R})$-measurable function $X.$