Let $R$ be a ring and $\sigma(R)$ the ring generated by $R$. Is there an easy counter example to show that a proper subset of $B\in\sigma(R)$ is not in the $\sigma$-ring?
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1What kind of rings are you talking about? – Rob Arthan Jul 08 '22 at 20:49
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Any ring such that set difference and countable unions are close. Preferably the Borel sets in $\mathbb{R}$ which belong to a $\sigma$-ring. This was mentioned in a comment on a previous question here https://math.stackexchange.com/questions/4488659/mu-measurable-set-in-the-sigma-ring/4488678?noredirect=1#comment9415450_4488678 – user2820579 Jul 09 '22 at 09:13
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Indeed, for the above example, is there a proper subset $A$ of the interval $(a,b)$ such that $A\notin \sigma$? – user2820579 Jul 09 '22 at 09:16
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1As I said in the comments, this should answer your question Lebesgue measurable set that is not a Borel measurable set – Mr.Gandalf Sauron Jul 10 '22 at 11:07