# Sigma-algebra generated by integer multiples

I came across this question and I'd like to check if my answer is correct. The problem is the following:

Consider the collection $$\mathcal{A}$$ of subsets $$A_1,A_2,...$$ of $$\mathbb{Z}$$ such that $$A_i = \{ni \,|\,n \text{ is an integer}\}.$$ Determine the sigma-algebra generated by $$\mathcal{A}$$, denoted by $$\sigma(\mathcal{A})$$.

My answer: Since the sets $$A_i$$ are symmetric around $$\{0\}$$, I first noticed that, for any $$k\in\mathbb{N}$$, I can write $$\{-k,k\}=A_k\backslash\bigcup_{i=1}^\infty A_{i+k}\in\sigma(\mathcal{A}).$$ Note also that, when $$k=0$$, $$A_0$$ is not defined, but we can write $$\{0\}=\bigcap_{i=1}^\infty A_i\in\sigma(\mathcal{A}).$$ Thus any complement, countable union or intersection of sets in $$\mathcal{A}$$ can be written as a union of sets of the form $$\{-k,k\}$$ and $$\{0\}$$, which are all in $$\sigma(\mathcal{A})$$. Hence, $$\sigma(\mathcal{A})$$ consists of all subsets $$B$$ in $$\mathbb{Z}$$ such that for each interger $$k\in B$$, we have $$-k\in B$$.

Is this enough?

• You showed that all those sets are in $\sigma(\mathcal{A})$. You only have left to show that the sets you describe actually form a $\sigma$-algebra! – csprun Jan 24 at 1:13

To make your argument precise consider $$\{B \in \sigma (\mathcal A): -B=B\}$$. ($$-B$$ stands for $$\{-b: b\in B\}$$). Verify that this is a sigma algebra and that it contains $$\mathcal A$$. Conclusion: $$-B=B$$ holds for every set $$B$$ in $$\sigma (\mathcal A)$$. Conversely, $$-B=B$$ implies $$B=\cup_{k \in A} \{-k,k\} \cup \{0\}$$ where $$A=\{n \in B: n>0\}$$. Thus, $$B$$ is a countable union of sets in $$\sigma (\mathcal A)$$ proving that $$B \in \sigma (\mathcal A)$$.
• Thank you. What about the argument regarding the 'smallest $\sigma$-algebra', what can be said? – sam wolfe Jan 24 at 20:30