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I've read several math SE posts about the construction of the $\sigma$-field generated by a class $\mathcal A$ of sets in $\Omega$, i.e. $\sigma(\mathcal A)$, including this one

https://math.stackexchange.com/a/54179/235690

But I am still confused. In particular I'm not sure and want to confirm if the answer simply means that

"$\sigma(\mathcal A)$ can be constructed by starting with sets in $\mathcal A$ and performing any countable number of set operations (i.e. union & complement) on them."

Can someone confirm if the above statement is correct? Or if it's not, why, and how should it be corrected?

By definition, $\sigma(\mathcal A)$ is the smallest $\sigma$-field containing $\mathcal A$, so any subset of $\Omega$ that can be obtained by a countable number of set operations on sets of $\mathcal A$ should be in $\sigma(\mathcal A)$. What I am not sure about is if $\sigma(\mathcal A)$ includes anything more than these? My understanding is that Borel $\sigma$-field ($\mathcal B$) on the real line does not. That is $\mathcal B$ is the smallest $\sigma$-field containing any open intervals, $(a,b)$, and is generated by any countable set operations on open intervals. Is my understanding correct? Is this true for any $\sigma(\mathcal A)$?

I'd greatly appreciate a confirmation/refutation/correction. Thanks a lot!

syeh_106
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2 Answers2

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That is exactly correct, and it is true for any generating collection of sets $\mathcal{A}$. The Borel $\sigma$-field is generated in the way you describe by countable unions and complements of all basic open sets (or closed sets if you prefer).

The Borel $\sigma$-field is the smallest $\sigma$-field containing all the open intervals in $\mathbb{R}$, but the completion of the Borel $\sigma$-field leads to the Lebesgue $\sigma$-field, and the definition of the Lebesgue integral on $\mathbb{R}$.

smb3
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By definition, a sigma-algebra (over a set $X$) is a collection of subsets (including the empty set) which are closed under complementation (relative to $X$) and countable unions (and intersections).

"$σ(A)$ can be constructed by starting with sets in $A$ and performing any countable number of set operations (i.e. union & complement) on them."

That describes the process to find the closure of $A$ under countable set operations.   The result will definitely be a sigma-algebra.

By definition, $σ(A)$ is the smallest $σ$-field containing $A$, so any subset of $Ω$ that can be obtained by a countable number of set operations on sets of $A$ should be in $σ(A)$. What I am not sure about is if $σ(A)$ includes anything more than these?

If the collection contains anything other than what it must contain, then it will not be the smallest possible.   The only things it must contain are the sets in $A$ and (in order to be a sigma-algebra) their completion under countable set operations.   That is all.

Graham Kemp
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