The Uncountable and Probability Suppose we draw a random uniformly number from $[0,1]$, if we do this countable many times, how many times will we get $1$, I suspect $0$?
If we do it uncountable many times, how often will we get $1$? $0$, finite, countable or uncountable?
Suppose we define $f(x)=0$ if $x$ irrational and $7$ otherwise.
Also $S=0$.
Now, for every real number in $[0,1]$ we draw a random number $r$ in $[0,1]$ and add $f(r)$ to $S$, once this process is finished, what is the probability distribution for $S$?
 A: These questions make little sense from the classical (Kolmogorov's axiomatic) probability theory point of view. 
For example, let $\xi$ be $U[0,1]$, and $\eta = \xi$ if $\xi$ is irrational and $1/\pi$ otherwise. Then these variables have the same distribution, yet $\eta$ never takes rational values.
So the answers to both questions would be $0$. 
However, it's not that simple. Suppose we are trying to formulate this in the classical way. We set $\Omega = [0,1]^{[0,1]}$ (as we draw a random number $\omega_r$ from $[0,1]$ for each $r\in[0,1]$), $\mathcal F$ is the $\sigma$-algebra generated by cylindrical subsets of $2^\Omega$, and $\mathsf P$ is  the Caratheodory extension of the measure defined on cylindrical sets: for each distinct $r_1,\dots,r_n\in[0,1]$ and each $B\in \mathcal B([0,1]^n)$ $$\mathsf P(\{\omega: (\omega_{r_1},\dots,\omega_{r_n})\in B\}) = \lambda_n(B);$$
$\lambda_n$ is the Lebesgue measure on $\mathbb R^n$.
There is a full description of random events in this model. Namely, $A\in \mathcal F$ iff there exists a countable collection of points $r_1,r_2,\dots \in[0,1]$ and a set $C\in \mathcal B(\mathbb R^\infty)$ (I won't define what's this, since it is not very relevant for our question) such that $$A = \{\omega:  (\omega_{r_1},\omega_{r_2},\dots)\in C\}.$$
This essentially means that in this model you are only allowed to ask what happens to a countable subcollection of these random variables. 
Thus, if you want to answer these questions, you first need to define what do they mean precisely.
