Probability of $x$ being rational Let the sequence $(x_1, x_2,...)$ be generated by sampling (uniformly and independently) the integers $\{0,1,2,...,9\}$.
From the sequence we construct the real number $x =  0.\overline{x_1x_2x_3..}$
What is the probability that $x$ is rational?
I just came up with this question, and realized that I do not even know how to start thinking about it. Intuitively, since the rationals are countable and the irrationals are uncountable, one might guess that the probability of $x$ being rational is 0.
Does the question even make sense from a probability theory's perspective?
 A: The set of all rational numbers in $[0,1]$ is countably infinite.  So by Axiom 3 of probability theory, we can just add up the probabilities one by one.  Let $X$ be your random variable.  Then
$$ P[X \in [0,1]\cap \mathbb{Q}] = \sum_{x \in [0,1]\cap \mathbb{Q}} \underbrace{P[X=x]}_{0} = 0 $$
where we use the fact that $P[X=x]=0$ for all $x \in [0,1]$ (indeed, a comment by Ivan mentions that $X$ is uniformly distributed over $[0,1]$). The same is true for any random variable $X$ with a continuous CDF:  The probability that $X$ is irrational is 1.
For your situation, you can see that $P[X=x]=0$ for all $x \in [0,1]$ by noting that for your i.i.d. random variables $\{X_i\}_{i=1}^{\infty}$ we have $X= \sum_{i=1}^{\infty} X_i 10^{-i}$.
Now a prior comment notes that every $x \in [0,1]$ has at most two representations $0.a_1a_2a_3...$ (there can be two if one has an infinite tail of 9s).  But we observe that for any particular sequence $\{a_1, a_2, a_3, ...\}$ such that $a_i \in \{0, 1, 2, ..., 9\}$ for all $i$ we get
\begin{align}
P[X_1=a_1, X_2=a_2, X_3=a_3, ...] &= P[\cap_{i=1}^{\infty} \{X_i=a_i\}] \\
&\overset{(a)}{=} \prod_{i=1}^{\infty} \underbrace{P[X_i=a_i]}_{1/10} \\
&= 0
\end{align}
where step (a) holds by mutual independence of $\{X_i\}_{i=1}^{\infty}$.
A: Sketch for a solution: this is just an sketched answer as to fill all the details will be extremely lenghty. You can find an exposition with all the details for this in many books of probability theory.
Let $\Omega :=\{0,\ldots ,9\}$ equiped with the uniform probability measure in the powerset, that is, we have that $P(k)=\frac1{10}$ for each $k\in \Omega $. Now, you want to build from $(\Omega ,2^{\Omega },P)$ a $\sigma $-algebra and a probability measure $Q$ in $J:=\Omega ^{\mathbb{N}}$, the space of sequences.
This construction is done as follow: first we equip $J$ with the product topology and from here with the Borel $\sigma $-algebra. Now, from all Borel sets, we construct $Q$ first in the cylinders, that is, sets $G\subset J$ of the form $G=\prod_{k\geqslant 1}G_k $ such that $G_k \subset \Omega $ for each $k$ and such that there is a $N\in \mathbb{N}$ such that $G_k=\Omega $ for each $k>N$. Then in the cylinders we set $Q(G):=\prod_{k=1}^N P(G_k)$.
From there it can be shown that $Q$ extends to a probability measure in $J$, known as the product measure. With this construction we have that, by example, the probability that a sequence starts by $4$ is $Q(\{4\}\times \Omega ^{\mathbb{N}})=P(4)=\frac1{10}$, the probability that a sequence starts with $1,2,3$ is $Q(\{1\}\times \{2\}\times \{3\}\times \Omega ^{\mathbb{N}})=P(1)\cdot P(2)\cdot P(3)=\frac1{10}\cdot \frac1{10}\cdot \frac1{10}=\frac1{1000}$. The probability that a sequence start by $1$ or $2$ and that in the third position appear an even number is $Q(\{1,2\}\times \Omega \times \{0,2,4,6,8\}\times \Omega ^{\mathbb{N}})=P(\{1,2\})\cdot P(\Omega )\cdot P(\{0,2,4,6,8\})=\frac{2}{10}\cdot 1\cdot \frac{5}{10}=\frac1{10}$, etc...
Then, from here, its easy to see that the probability that a sequence becomes periodic at some point is zero, and so, after you find some measurable surjection $f:J\to [0,1]$ you can conclude that the measure of all sequences that represent the rationals in $[0,1]$ have zero measure.
