# Uniform probability measure on integers and arithmetic progressions

Does there exist a probability measure on the integers such that, the probability of any two arithmetic progressions with the same difference part, is the same?

We assume the probability measure is defined over the power set of the integers. Hence, a probability measure corresponds to a sequence of positive (non-negative) numbers $$\{p_z\}_{z \in \mathbb{Z}}$$ which sum up to one.

• How do you define the probability of an arithmetic progression ? sum of the probabilities of all the number in the progression ? Commented Sep 21, 2021 at 21:17

If I understood well, denoting $$p$$ this probability measure defined on $$\mathcal P(\mathbb Z)$$, you want that for all $$a,b,c\in\mathbb Z$$, $$p(a+c\mathbb Z)=p(b+c\mathbb Z)$$?
Indeed, let $$x\in\mathbb Z$$. For all $$n\in\mathbb N^*$$, you have $$p(n\mathbb Z)=p(1+n\mathbb Z)=\cdots=p(n-1+n\mathbb Z),$$ and $$p(n\mathbb Z)+p(1+n\mathbb Z)+\cdots+p(n-1+n\mathbb Z)=p(\mathbb Z)=1,$$ hence $$p(x+n\mathbb Z)=\frac1n\cdot$$
We deduce that $$p(\{x\})\le p(x+n\mathbb Z)\le\frac1n\underset{n\to+\infty}{\longrightarrow}0,$$ and therefore $$p(\{x\})=0$$ for all $$x\in\mathbb Z$$, which implies $$p(\mathbb Z)=0$$.