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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?

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    $\begingroup$ 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! $\endgroup$ – csprun Jan 24 at 1:13
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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 $B \in \sigma (\mathcal A)$.

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  • $\begingroup$ Thank you. What about the argument regarding the 'smallest $\sigma$-algebra', what can be said? $\endgroup$ – sam wolfe Jan 24 at 20:30

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