My understanding is that a random variable is actually a function $X: \Omega \to T$, where $\Omega$ is the sample space of some random experiment and $T$ is the set from which the possible values of the random variable are taken.
Regarding the set of values that the random variable can actually take, it is the image of the function $X$. If the image is finite, then $X$ must be a discrete random variable. However, if it is an infinite set, then $X$ may or may not be a continuous random variable.
Whether it is depends on whether the image is countable or not. If it is countable, then $X$ is a discrete random variable; whereas if it is not, then $X$ is continuous.
Assuming that my understanding is correct, why does the fact that the image is uncountable imply that $Pr(X = x) = 0$.
I would have thought that the fact that the image is infinite, regardless of whether it is countable or not, would already imply that $Pr(X = x) = 0$ since if it is infinite, then the domain $\Omega$ must also be infinite, and therefore
$$Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}} = \frac{\text{# outcomes of the experiment where X = x}}{|\Omega|} = \frac{\text{# outcomes of the experiment where X = x}}{\infty} = 0$$
What is wrong with my argument?
Why does the probability that a continuous random variable takes on a specific value actually equal zero?
Best Answer
The problem begins with your use of the formula
$$ Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}}\;. $$
This is the principle of indifference. It is often a good way to obtain probabilities in concrete situations, but it is not an axiom of probability, and probability distributions can take many other forms. A probability distribution that satisfies the principle of indifference is a uniform distribution; any outcome is equally likely. You are right that there is no uniform distribution over a countably infinite set. There are, however, non-uniform distributions over countably infinite sets, for instance the distribution $p(n)=6/(n\pi)^2$ over $\mathbb N$.
For uncountable sets, on the other hand, there cannot be any distribution, uniform or not, that assigns non-zero probability to uncountably many elements. This can be shown as follows:
Consider all elements whose probability lies in $(1/(n+1),1/n]$ for $n\in\mathbb N$. The union of all these intervals is $(0,1]$. If there were finitely many such elements for each $n\in\mathbb N$, then we could enumerate all the elements by first enumerating the ones for $n=1$, then for $n=2$ and so on. Thus, since we can't enumerate the uncountably many elements, there must be an infinite (in fact uncountably infinite) number of elements in at least one of these classes. But then by countable additivity their probabilities would sum up to more than $1$, which is impossible. Thus there cannot be such a probability distribution.