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The definition of a $\sigma$-algebra is easy enough to state in a way that doesn't require a whole lot of prerequisite knowledge to understand (calculus, plus set theory at the level of Baby Rudin Chap. 2, should suffice). But I was unable to turn up any good concrete example of a $\sigma$-algebra that's on the one hand not trivial (trivial like "the power set of a finite set") but on the other hand doesn't require significantly more background to understand (more than the aforementioned background of undergrad calculus and basic set theory).

In trying to come up with such a "medium-sophistication" concrete example on my own, I landed on the following:

Define a sub-interval of $[0, 1]$ to be any interval $[a, b]$ or $(a, b]$ or $[a, b)$ or $(a, b)$ such that $0 \leq a \leq 1$ and $0 \leq b \leq 1$, and $a \leq b$ (so, allowing an "interval" to potentially be a single point or even the empty set). And any countable union of such sub-intervals I'll call a "compound sub-interval" of $[0, 1]$ (my own made-up nomenclature, sorry). Then it seems like the set of all such "compound sub-intervals" of $[0, 1]$ satisfies the definition of a $\sigma$-algebra (I don't have a fully rigorous proof, but have convinced myself of its plausibility).

Is this right? (is this in fact a $\sigma$-algebra?). Also does anyone know of a book or something that contains other such "medium-sophistication" examples (preferrably including some actual proof or explanation showing that the given example is in fact a $\sigma$-algebra)? Thanks.

Update:

It's clear now that my example was wrong -- the Cantor set is a great counter-example (thanks to the commenters). For anyone else who might be interested, I did eventually come across a different non-trivial but easily comprehensible example:

If $\mathbb{X}$ is an uncountable set, then one possible $\sigma$-algebra is the set of all $A \subset \mathbb{X}$ such that $A$ is countable or $A^c$ is countable.

Asaf Karagila
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NikS
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    No, you are wrong. Any $\sigma-$ algebra which contains intervals of any one type automatically contains all Borel sets and it is not possible describe all sets in this $\sigma-$ algebra. To put it briefly abstraction cannot be avoided in dealing with (non-trivial) $\sigma-$ algebras. – Kavi Rama Murthy Sep 08 '22 at 05:23
  • The $\sigma-$ algebra generated by a partition ${A_1,A_2,...}$ of the space can be described explicitly (as the collection of all possible unions of $A_i$'s). – Kavi Rama Murthy Sep 08 '22 at 05:39
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    There's no point in studying a $\sigma$-algebra simpler than the Borel $\sigma$-algebra of $\mathbb{R}$, say; those are the kinds of $\sigma$-algebras you need to deal with to understand their application to measure theory. $\sigma$-algebras are not a structure people are interested in for their own sake; the application to measure theory is it. – Qiaochu Yuan Sep 08 '22 at 05:42
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    Note the set of irrational numbers in $[0,1]$ isn't a "compound subinterval" – Brian Moehring Sep 08 '22 at 06:00
  • The Cantor set is a countable intersection of complements of open intervals. It isnt’t a “compound subinterval“, but it’s in your example $\sigma$-algebra, which is just the Borel sets of the unit interval. – BrianO Sep 08 '22 at 06:10
  • This answer I gave earlier was specific to stochastic processes and probability theory, but might be useful for an example of an easy to understand $\sigma$-algebra besides a power set. You might consider it a trivial example, though, because it was more to illustrate filtrations than $\sigma$-algebras: https://math.stackexchange.com/questions/3703811/stochastic-processes-why-do-we-need-filtration/3705947#3705947 – user6247850 Sep 08 '22 at 16:47

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