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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.

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    $\begingroup$ A measure on $\mathbb{Q}$ is not a function $\mathbb{Q}\to\mathbb{R}$. What the answer in the link tells is that $\mu$ is defined by$$\mu(E)=\sum_{k=1}^{\infty}\frac{1}{2^k}\delta_{q_k}(E)$$for each subset $E$ of $\mathbb{Q}$. Also, $\delta_a$ is the point mass (a.k.a. Dirac delta) concentrated at $a$, i.e., $$\delta_a(E)=\begin{cases}1,&a\in E\\0,&a\notin E\end{cases}$$ So, we can equally define $\mu$ as $$\mu(E)=\sum_{k:q_k\in E}\frac{1}{2^k},$$ where the sum is taken for all $k$ for which $q_k\in E$. $\endgroup$ Commented Nov 2, 2020 at 21:52
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    $\begingroup$ I think the given formula involves the dirac measure $\endgroup$ Commented Nov 2, 2020 at 21:53
  • $\begingroup$ I am gonna edit the question to correct the mistakes. $\endgroup$
    – EoDmnFOr3q
    Commented Nov 2, 2020 at 21:55

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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 $$ P(\{q\}) = \frac{1}{2^{b(q)}}. $$

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  • $\begingroup$ Thank you very much for your useful answer. Could you please explain how to extend your answer to both $\mathbb{Q}$? Also, could you explain how to extend it to $\mathbb{Q}^N$, in the event that it's possible? $\endgroup$
    – EoDmnFOr3q
    Commented Nov 2, 2020 at 22:15
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    $\begingroup$ @Héctor $\mathbb{Q}$ and $\mathbb{Q}^n$ are countably infinite - that is, there's a bijection to $N$. So you can transfer this measure on $N$ to either. You can't extend the measure from $\mathbb{Q}$ for several reasons. In particular, that isn't a subset of $\mathbb{Q}^n$ for $n > 1$. $\endgroup$ Commented Nov 2, 2020 at 22:30
  • $\begingroup$ Thank you for your comment. I'm stuck with the the idea that "I can transfer this measure on $\mathbb{N}$ to $\mathbb{Q}^N$ because there exists a bijection between $\mathbb{N}$ and $\mathbb{Q}^N$". For $\mathbb{Q}$, I should assign any singleton $n\in\mathbb{Q}$ a probability $1/2^n$, correct? But what should this probability be for an element in $\mathbb{Q}^N$ instead? I don't see it. Should I instead just order the elements in either $\mathbb{Q}$ or $\mathbb{Q}^N$ as I please and then use their index according to this order to substitute the value $n$ in $1/2^n$? Thank you very much again. $\endgroup$
    – EoDmnFOr3q
    Commented Nov 2, 2020 at 22:39
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    $\begingroup$ @Héctor Your "instead" is right. See my edit. $\endgroup$ Commented Nov 3, 2020 at 0:50

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