A while ago I asked this question on mathstackexchange:

Let $\mathscr{C}\subset \mathscr{P}(\Omega)$ be a class of subsets of a nonempty set $\Omega$ containing $\Omega$ and $\varnothing$. Define $\mathscr{C}_0=\mathscr{C}$ and for each $n\geq 1$ define $$ \mathscr{C}_{n+1}=\mathscr{C}_n\cup \{A^c \mathrel: A\in \mathscr{C}_n\}\cup \{\bigcup_{i=1}^\infty A_{i} \mathrel: \{A_{i}\}_{i=1}^\infty\subset \mathscr{C}_n\}. $$

Question: I am looking for an example of $\Omega$ such that $\mathscr{A}(\mathscr{C})=\bigcup_{i=1}^\infty \mathscr{C}_n$ does not equal $\sigma(\mathscr{C})$, the sigma algebra generated by $\mathscr{C}$.

I first tried to pick a $\Omega=\mathbb{R}$ and consider $\mathscr{C}$ the class given by the finite and cofinite sets. It turns out that by applying the above procedure we get the the class of the contable and co-countable subsets, which is a sigma álgebra ….

I know that must exists examples that $\mathscr{A}(\mathscr{C})=\bigcup_{i=1}^\infty \mathscr{C}_n$ does not equal $\sigma(\mathscr{C})$ because in general we need an ordinal argument in the construction (Borel sigma algebra) ….

At that time I had no answer but some colleagues suggested to me that the answer to this question relies on something called Borel hierarchy ….

Now I repeat the question here: is there a "simple" argument to show that $\bigcup_{\alpha < \beta} \Sigma_\alpha^0 \varsubsetneq \mathcal{B}(\mathbb{R})$ for every countable ordinal $\beta$? How can I convince myself that the countable levels all bring something new?

Here "simple" means: an undergraduate student who knows transfinite induction arguments can understand it.

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    $\begingroup$ There's a summary at mathoverflow.net/questions/376292/…. I'm not sure there is a really elementary argument, the idea of a universal set seems to be fairly essential. $\endgroup$ Apr 6, 2021 at 3:01
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    $\begingroup$ For the very few places I know of that specifically address the fact that the union of the finite Borel level classes is not equal to collection of all Borel sets, see my answer to Constructing sigma algebras in countably many steps. I do not know whether a simpler proof of either nonequality at the first transfinite level or nonequality at all countable transfinite levels can be given for some other (possibly designed specifically for this purpose) algebra and $\sigma$-algebra of sets. $\endgroup$ Apr 6, 2021 at 6:09
  • $\begingroup$ @DaveLRenfro I see your post, Billingsley's example is exactly what I was looking for. Thank you very much $\endgroup$
    – Eduardo
    Apr 6, 2021 at 13:45


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