Your observation is reasonable, but your suggested cure for the problem, making $(\Omega,\mathcal F,P)$ the domain of $X$, won't work because the domain of a function needs to be a set (or a type or something like that). My impression is that, when people refer to a function $X:\Omega\to\mathbb R$ as a random variable, they always do so in the context of a probability-space structure ($\mathcal F$ and $P$) on $\Omega$. If no such structure is given, then I wouldn't call $X$ a random variable. And if there is uncertainty about which structure is intended, then, as you said, notions like "distribution" of $X$ will not be well-defined.
If I had to formalize the notion of random variable, in Bourbaki style, I would probably say that a random variable is a pair consisting of a probability space $(\Omega,\mathcal F,P)$ together with a function $X:\Omega\to\mathbb R$. As with many mathematical concepts, one often omits mentioning part of an entity (in this case the probability space) when it is understood from the context.
I am not entirely convinced with the line the sample space is also called the support of a random variable
That looks quite wrong to me.
What is even more confusing is, when we talk about support, do we mean that of $X$ or that of the distribution function $Pr$?
In rather informal terms, the "support" of a random variable $X$ is defined as the support (in the function sense) of the density function $f_X(x)$.
I say, in rather informal terms, because the density function is a quite intuitive and practical concept for dealing with probabilities, but no so much when speaking of probability in general and formal terms. For one thing, it's not a proper function for "discrete distributions" (again, a practical but loose concept).
In more formal/strict terms, the comment of Stefan fits the bill.
Do we interpret the support to be
- the set of outcomes in Ω which have a non-zero probability,
- the set of values that X can take with non-zero probability?
Neither, actually. Consider a random variable that has a uniform density in $[0,1]$, with $\Omega = \mathbb{R}$.
Then the support is the full interval $[0,1]$ - which is a subset of $\Omega$. But, then, of course, say $x=1/2$ belongs to the support. But the probability that $X$ takes this value is zero.
Best Answer
This was originally posted as a comment:
To which Leo answered this: