Can this informal argument (for the fact that almost all reals in the unit interval are irrational) be saved? In the textbook from which I am teaching a Discrete Math course, the authors propose randomly generating an infinite sequence of decimal digits $d_1, d_2, \dots$.  We are to think of this as the decimal expansion of a real number in the unit interval.  
They propose that "it is overwhelmingly likely" that the resulting sequence is non-periodic, and thus that the number in question is irrational.  This is intended to motivate the idea that there are more irrational numbers in [0,1] than rationals in [0,1].  
This argument is appealing until you realize that it's nonsense, since we are somehow taking a uniform probability distribution on an uncountable set.  
My question is whether there is any way to interpret this so that it is not nonsense.  
Let me make a couple of quick points, in the hopes that this question will not be closed out of hand.  I know the irrationals are uncountable, and I know the rationals are a set of measure zero.  What I am asking is whether the above argument can be put on a legitimate footing.  I don't insist that it be possible to put it on a legitimate footing which I could explain to my Discrete Math students --that seems clearly out of reach-- but whether or not there is some legitimate footing will alter how I want to present this argument.  
The textbook in question, if anyone is interested, is A Discrete Transition to Advanced Mathematics, by Richmond and Richmond.  I like it, and will probably use it again.  
 A: You can make sense of the uniform probability distribution on lots of infinite sets, notably any compact topological group $G$, where "uniform probability distribution" should mean "normalized Haar measure." Here the group is an infinite direct product of copies of $\mathbb{Z}/10\mathbb{Z}$, which is compact by Tychonoff's theorem. You've probably also considered the uniform probability distribution on $S^1$ at some point in your life. 
The failure of groups like $\mathbb{Z}$ and $\mathbb{R}$ to have a uniform probability distribution isn't just due to infinitude, it's also due to lack of compactness. 
A: Short answer: The countable product of probability measures is a well-defined object so the countability of periodic sequences is enough to conclude that the probability of drawing a periodic sequence is zero. 
Slightly longer answer: Let $\mu$ be the uniform measure on $\{0,1,2,3,4,5,6,7,8,9\}$ since we want to talk about ordinary decimal numbers. In this situation we want to treat the sequences $09999...$ and $100000....$ as distinct since they are different outcomes in the experiment. This won't worry us because there are again only countably many such pairs so we will only be disregarding a probability zero event. The crux of the matter is the product $\prod_{i=1}^{\infty} \mu_i$ where all the $\mu_i = \mu$. But this is the conclusion of the Kolmogorov Extension Theorem which has two main conditions which are satisfied by the assumption that $\mu_i=\mu$ and that we are  taking the limit of the product measures, $\prod_{i=1}^{n} \mu_i$.
