Chapter 1.2; Ross: 2nd Course in Probability I would like to clarify on Ross' example of a non-measurable event. 

Consider a circle having radius equal to one. We say that two points
  on the edge of the circle are in the same family if you can go from
  one point to the other point by taking steps of length one unit around
  the edge of the circle. By this we mean each step you take moves you
  an angle of exactly one radian degree around the circle, and you are
  allowed to keep looping around the circle in either direction.
Suppose each family elects one of its members to be the head of the
  family. Here is the question: what is the probability a point $X$ is
  selected uniformly at random along the edge of the circle is the head
  of its family? It turns out this question has no answer.

I can understand this example well. Ross then further clarified why the probability is undefined using the following equation:

$$ \begin{align} 1 = P(A) + \sum_{i=1}^\infty(P(A_i) + P(B_i))
\end{align} $$
  Thus if $x = P(A)$, we get $1 = x + \sum_{i=1}^\infty 2x$, which has
  no solution where $0 \leq x \leq 1$.

Which I also can understand. 
My confusion here is that Ross' example seems to be overly and unnecessarily complicated. The idea here is that for disjoint events $A_1,A_2,\dots$, we must have $P(\bigcup_{i=1}^\infty A_i) = \sum_{i=1}^\infty P(A_i) = 1$. I believe that the following example is much simpler: Consider the "uniform distribution over $\mathbb{Q}$", i.e. (roughly speaking) each rational number has an equal probability of being chosen. Then the probability of choosing a specific rational number must be undefined as $\mathbb{Q}$ is countable.
Am I missing something here?
 A: You are missing the most important thing, where you speak roughly : no such uniform distribution exists over the rationals, or even over a superset of the rationals, simply for the reason that the rationals are countable, and a countable sum $\sum_{n=1}^\infty  x$ is either $0$ or $\infty$ for $0 \leq x \leq 1$. So there is no measure space on which we can conduct the operations specified by you. This is why Sheldon Ross can't give this particular thing as an example, but he can do something similar to the example he gives and which you attempt to give, which is done in other books.
Namely, take $[0,1]$ with the usual measure, and now say two points are in the same family if their absolute difference is a rational number. Then pick a head of each family, put the heads together in one set, and the probability of that set of heads $A \subset [0,1]$ can't be defined i.e. it is a non-measurable event.
The reason remains roughly the same : the sets of the form $x+A$ , for $x \in \mathbb Q \cap [0,1]$, are disjoint, have the same measure as $A$ and have union $[0,1]$. Therefore, if $P(A) = y$ then $\sum_{x \in \mathbb Q \cap [0,1]} y = 1$, which can't be satisfied as the sum is countable.
Such a set $A$ is called a Vitali set (Note : A rational translate of a Vitali set is also a Vitali set, so no uniqueness holds). Note that we used the axiom of choice to "pick" a member of each family to put in the Vitali set. Without the axiom of choice, it's nice to know that Solovay has constructed a set model in which all sets are Lebesgue measurable, but the model doesn't include the axiom of choice.
For another example of a non-measurable set , see the Sierpinski construction of a non-measurable set.
