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The product $\sigma$-algebra, say on $\{0,1\}^\mathbb{N}$, is the smallest $\sigma$-algebra that contains all of the cylinder sets.

What I'd like, for the purposes of learning and being able to spot counterexamples, is an example of a subset of $\{0,1\}^\mathbb{N}$ that is not measurable according to this $\sigma$-algebra.

I'm aware that there's an answer to this question along the lines of "start with a Vitali subset of $[0,1]$ and take the binary expansion of each element". But this is very indirect, since it uses the fact that the Cantor space and the unit interval are isomorphic as measurable spaces, and the indirectness means I find it hard to get much insight from it.

So I'm hoping for a more elementary way to construct counterexamples by applying the axiom of choice to sequences of digits directly somehow, instead of going via the reals. (And a proof that it's unmeasurable that doesn't go via the reals either.) Is there a straightforward way to do that?

N. Virgo
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  • Note that despite the similar title, https://math.stackexchange.com/questions/3027642/example-of-a-set-which-is-not-in-the-product-sigma-algebra is a different question, since that one asks about a finite product of copies of $\mathbb{R}$ while I'm asking about an infinite product of copies of ${0,1}$. – N. Virgo Aug 21 '24 at 03:34
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    Not an answer, but related: https://math.stackexchange.com/questions/1925288 – Anne Bauval Aug 21 '24 at 04:34
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    @AnneBauval thanks! The link in the second answer there looks like it might contain an answer to my question. I'll go through it. – N. Virgo Aug 21 '24 at 04:39
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    It seems like the link I mentioned (a blog post by Jyotirmoy Bhattacharya) proposes a way to answer my question. The proof is not completely satisfying to me, because it works by assuming on ${0,1}^\mathbb{N}$, whereas I'd ideally like to see directly that the set isn't a member of the product $\sigma$-algebra, without needing to consider measures. Maybe this isn't possible though, I don't know. I'll post a self-answer if I find a way. – N. Virgo Aug 21 '24 at 04:49
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    ("assuming a measure on ${0,1}^\mathbb{N}$", that should have said) – N. Virgo Aug 21 '24 at 05:38
  • This sounds like a job for a free ultrafilter. – Eric Towers Aug 21 '24 at 08:07

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As Eric Towers suggests, from the paper "A non-measurable tail set" by D. Blackwell and P. Diaconis: For an element $a\in\{0,1\}^{\Bbb N}$ define $N_a:=\{i\in\Bbb N: a_i=1\}$. Let $\mathscr U$ be a free ultrafilter on $\Bbb N$. Then $E:=\{a\in\{0,1\}^{\Bbb N}: N_a\in\mathscr U\}$ is non-measurable. They reach this conclusion by showing that both $E$ and its complement have inner measure $0$ with respect to the fair-coin-tossing measure on $\{0,1\}^{\Bbb N}$.

John Dawkins
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