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The author writes:

One conceivable and seemingly natural way of constructing $\sigma(\mathcal{A})$ from $\mathcal{A}$ is as follows. For a class $\mathcal{E}$ of subsets of $\Omega$, denote by $\hat{\mathcal{E}}$ the class of subsets of $\Omega$ consisting of the sets contained in $\mathcal{E}$, their complements, and finite or countable unions of the sets in $\mathcal{E}$. Define $\mathcal{A}_0 = \mathcal{A}, \mathcal{A}_1 = \hat{\mathcal{A}_0}, \mathcal{A}_2 = \hat{\mathcal{A}_1}, \text{etc.}$ Clearly, for each $n$ the system $\mathcal{A}_n$ is contained in $\sigma(\mathcal{A})$, and one might expect that $\mathcal{A}_n = \sigma(\mathcal{A})$ for some $n$ or, at least, $\bigcup_{n} \mathcal{A}_n = \sigma(\mathcal{A})$.

For context here $\mathcal{A}$ is an algebra. What I do not understand is: Shouldn't $\mathcal{A_2} = \hat{\mathcal{A_1}} = \mathcal{A_1}$ where the last equality holds because $\mathcal{A_1}$ is already a sigma-algebra? What am I missing?

Thank you a lot for your help

Samael Manasseh
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1 Answers1

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The construction s/he used is related to the Borel hierarchy. Say $\mathcal A_0$ are all the open or closed sets, then $\mathcal A_1$ are all the $G_\delta$ or $F_\sigma$ sets. In their notation $\mathcal A_\alpha=\Sigma_\alpha\cup\Delta_\alpha$ for every ordinal $\alpha$.

After all finite operations we reach $\mathcal A_\omega=\bigcup_{n=1}^\infty\mathcal A_n$, however it may not (and very likely not!) closed under countable union nor countable intersection. You can pick some special sets $E_n\in\mathcal A_n$ for $n=1,2,3,\dots$ such that $\bigcup_n E_n\notin\mathcal A_\omega$.

For space like Borel $\sigma$-algebra of $\mathbb R$ this won't be finished in any countable operations. See this post.

Liding Yao
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