In standard mathematical notation, a probability space follows the same structure of a measure space - i.e. a measur-able space, $(S,\mathcal A)$, plus a measure, $\mu$. It is formulated as a triple within parentheses. In the case of a probability space it looks like $(\Omega, \mathcal{F},\mathbb P)$, corresponding to the sample space ($\Omega$), a sigma algebra ($\mathcal F$) of measurable events, and finally the probability measure with a total mass of $1$.
On this probability space a random variable, $X$, is defined as a function or mapping from the subsets of $\Omega$ in the sigma algebra $\mathcal F$ to the real line, $\mathbb R$, expressed as $X: \Omega \rightarrow \mathbb R$. Intuition would have it that the ending of the prior sentence would be $\mathbb P$, which is the last entry in the triple $(\Omega, \mathcal{F},\mathbb P)$. But it, instead an $\mathbb R$. But this is not the actual question...
In the simple case of tossing two die, and counting the notation would make reference to:
- Sample space composed of the set of sets $\{\{1,2\},\{1,3\},\{1,4\},\cdots\{6,6\}\}$.
- Contrarily to the continuous case where the sample space is $\mathbb R$ and we can't assign probabilities to the immense powerset, we can now seamlessly consider the power set of $\Omega$, i.e. $2^{\Omega}$ as the events included in the sigma algebra in a discrete problem such as the one at hand. So the events would be the same as the sets in the sample space.
- The random variable would map each one of these sets to the real line ($\mathbb R$) in the obvious way of $X:\{2,3\}\mapsto 5$.
- Finally the cumulative density function (CDF) would be the measure $\mathbb P$ assigned to $X=5$, which I think happens to be $1/9$.
So the questions are:
- Why is the random variable $X$ not included in the $(\Omega, \mathcal{F},\mathbb P)$ - for example, $(\Omega, \mathcal{F},\color{blue}{X},\mathbb P) $ - but rather left implicit?
- Why is it stated that: A function $X:\Omega \rightarrow \mathbb R$ is measurable with respect to $\mathcal F$ if for any $x \in \mathbb R$ , the set $\{X \leq x\}=\{\omega\in\Omega:X(\omega)\leq x\}$ is measurable, or belongs to $\mathcal F$ - in other words, it is an event? That it needs to be an event is clear from the extrapolation to continuous random variables, but why does the nomenclature recurs back to $\omega$ and $\Omega$ in this expression... Hadn't the sample space been "tamed" into the more malleable events in the sigma algebra? Why not say, $\{X \leq x\}=\{\color{blue}{f}\in \color{blue}{\mathcal F}:X(\color{blue}{f})\leq x\}$?
Quick recap after Matthew's answer: The random variable maps outcomes (in the sample space) to the real line. The probability space maps events to [0,1]. So if we draw from an exponential distribution modeling waiting times with a $\lambda = 30 $ sec., the probability of an outcome of exactly $30$ sec. is $0$, because $30$ is not an event. Yet, evidently we have timed whatever process (random experiment) and mapped it to $\mathbb R$ as a random variable.
