In this question, the OP asks how to construct a probability measure $\mu(k):[\mathbb{Q},\mathcal{F},P]\to\mathbb{R}$ over the set $\mathbb{Q}$ of Rational numbers, where $[\mathbb{Q},\mathcal{F},P]$ is a probability space. One of the answers suggests to let $\{q_n\}_{n=1}^\infty$ be an enumeration over $\mathbb{Q}$ and then defines the measure
\begin{gather}
\mu(E)=\sum_{k=1}^{\infty}\frac{1}{2^k}\delta_{q_k}(E),
\end{gather}
for each subset $E\subset\mathbb{Q}$, where $\delta_{q_k}(E)$ is the point mass (i.e., Dirac delta). The answer does not provide much context, and it does not explain why such a measure satisfies countable additivity nor why it returns results in the unit interval $[0, 1]\cap\mathbb{R}$, returning $0$ for the empty set and $1$ for the entire space. Could anybody please explain why such a measure is a probability measure over $\mathbb{Q}$?
BONUS QUESTION: Since the Cartesian product of any two countable spaces is also countable, could such a measure be extended to the $n$-dimensional set $\mathbb{Q}^N$?
Thank you all for your help.
Best Answer
I think the easiest way to think about this question is to start by constructing a nice measure on the set of positive integers. Assign to the singleton $n$ the probability $1/2^n$. Then define the probability $P$ of any subset to be the sum of the probabilities of the elements it contains.
Then $$ P(\mathbb{N}) = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = 1. $$
The countable additivity of $P$ is straightforward - you're just summing subsets of the terms in that absolutely convergent geometric series.
To define a probability measure on a countably infinite set (like the rationals) just use a bijection to the positive integers to transfer that one. That settles the bonus question.
Edit in response to comment.
To "transfer" this measure from $\mathbb{N}$ to $\mathbb{Q}$, start with your favorite bijection $$ b: \mathbb{Q} \to \mathbb{N}. $$ Then for each $q \in \mathbb{Q}$ let $$ P(\{q\}) = \frac{1}{2^{b(q)}}. $$