# Why is the function $\mu(E)=\sum_{k=1}^{\infty}(1/2^k)\delta_{q_k}(E)$ a probability measure over $\mathbb{Q}$?

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.

• 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 $q_k\in E$. Commented Nov 2, 2020 at 21:52
• I think the given formula involves the dirac measure Commented Nov 2, 2020 at 21:53
• I am gonna edit the question to correct the mistakes. Commented Nov 2, 2020 at 21:55

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

• 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? Commented Nov 2, 2020 at 22:15
• @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 $n > 1$. Commented Nov 2, 2020 at 22:30
• 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. Commented Nov 2, 2020 at 22:39
• @Héctor Your "instead" is right. See my edit. Commented Nov 3, 2020 at 0:50