How can certain choices of sets $\mathbb {S}$ produce paradoxes? Preface: I am not a mathematician, and I don't have any knowledge of measure theory. If this question requires too much knowledge to answer directly I'd be happy if you would point me in the right direction.
I am currently reading "Deep Learning" by Ian Goodfellow, and Yoshua Bengio and Aaron Courville. Section 3.12 describe technical details of contiuous variables. Here they say that the probability of a continuous vector-valued $x$ lying in some set $\mathbb{S}$ is given by the integral of $ p(x)$. Here they mention that some choices of this set S can produce paradoxes.
How can the sets $\mathbb{S}_1$ and $\mathbb{S}_2$ produce paradoxes if they are defined as follows?
$p(x\in\mathbb{S}_1) + (x\in\mathbb{S}_2)  > 1$ but $\mathbb{S}_2 \cap \mathbb{S}_2 = \emptyset$
How does the infinite precision of real numbers come into play here? How can fractal-shaped sets or sets that are defined  by transforming the set of rational numbers cause this paradox?
Does this mean that fractals hold no place in probability theory?
 A: Let's get concrete.
Consider the interval $[0, 1]$. We consider two points $x, y$ on the interval to be "equivalent" if $x - y$ is a rational number.
The "equivalence class" of $x$ is the set of all $y$ s.t. $y$ is equivalent to $x$. That is, $Equiv(x) = \{y \in [0, 1] : x - y$ is rational$\}$. Note that $Equiv(x) = Equiv(y)$ iff $x = y$.
Now consider the collection of all equivalence classes $S = \{Equiv(x) : x \in [0, 1]\}$. Each equivalence class has an element (in particular, $x \in Equiv(x)$). By the axiom of choice, we may take a function $f : S \to [0, 1]$ s.t. for every $x$, $f(Equiv(x)) \in Equiv(x)$.
Now consider the set $J = \{f(Equiv(x)) : x \in [0, 1]\}$. And consider the uniform distribution over the set $[-1, 2]$. What is $P(x \in J)$?
I claim that $P(x \in J)$ cannot be zero. For consider sets of the form $J_q = \{j + q : j \in J\}$, where $q$ is a rational number s.t. $-1 \leq 1$. It can be demonstrated that $[0, 1] \in \bigcup\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} J_q$ and that $J_q, J_{q'}$ are disjoint whenever $q \neq q'$. Therefore, $P(x \in [0, 1]) = 1/3 \leq P(x \in \bigcup\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} J_q) = \sum\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} P(x \in J_q)$. Since $J_q$ is just a translation of $J$ and we're dealing with a uniform distribution, it must be the case that $P(x \in J_q) = P(x \in J)$. Then we have $1/3 \leq \sum\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} P(x \in J_q) = \sum\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} P(x \in J)$.
But if $P(x \in J) = 0$, then we would have $1/3 \leq 0$; this is a contradiction. Then $P(x \in J) > 0$.
But in that case, we would have $P(x \in \bigcup\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} J_q) = \sum\limits_{q \in \mathbb{Q}, -1 \leq q \leq 1} P(x \in J) = \infty$, since we're adding up infinitely many of the same positive value. This, too, is a contradiction.
Thus, the only sensible thing to do is say that $P(x \in J)$ is undefined. This illustrates the principle that we must carefully consider for what $J$ the statement $P(x \in J)$ is defined. Careful analysis of these considerations lead mathematicians to the notion of a $\sigma$ algebra.
This construction is known as the "Vitali Set construction".
