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What would be an example of a $\sigma$-algebra on some sample space $\Omega$ that is not the smallest $\sigma$-algebra on that space?

This is an extension of the question at Understanding Borel sets. I would've posted in the comments but I don't have the rep to do so.

Community
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Learning...
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    Sorry, I guess my understanding is flawed. I will edit the question. – Learning... Dec 06 '16 at 04:53
  • By smallest, do you mean cardinality? If so, wouldn't the trivial $\sigma$-algebra be the smallest? Then the Borel $\sigma$-algebra on $\bb{R}$ is an example of what you seek. – user332239 Dec 06 '16 at 04:57
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    Do you mean to ask for an example of a $\sigma$-algebra containing some specified set (e.g., the set of all open sets) but which is not the smallest $\sigma$-algebra containing it? – manthanomen Dec 06 '16 at 05:02
  • yes, that is what I meant to ask... – Learning... Dec 06 '16 at 05:44

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For any space $X$, $\{\emptyset, X\}$ will be the smallest sigma algeba on $X$. So choose any other non-trivial sigma algebra, and it's not the smallest. For instance, the Borel sigma algebra on $\mathbb{R}$ is not the smallest sigma algebra on $\mathbb{R}$.

Your question seems to be asking what is meant by the smallest sigma algebra containing some specified collection of sets, though. Say we want a sigma algebra $\mathcal{F}$ on $X$ that contains all the sets in some collection $\mathcal{A}$ of subsets of $X$. Given any two sigma algebras $\mathcal{F}_1$ and $\mathcal{F}_2$ on $X$ such that $\mathcal{A}\subseteq \mathcal{F}_1$ and $\mathcal{A}\subseteq \mathcal{F}_2$, $\mathcal{F}_1\cap \mathcal{F}_2$ is also a sigma algebra on $X$ that contains $\mathcal{A}$ (check this!). So $$\mathcal{F}:=\bigcap_{\text{sigma algebras } \mathcal{G}\supseteq\mathcal{A}} \mathcal{G}$$ is a sigma algebra on $X$ containing $\mathcal{A}$, and it is the smallest such. Why the smallest? Well if $\mathcal{F}'$ is any other such sigma algebra, it is one of the sigma algebras in the intersection defining $\mathcal{F}$, so $\mathcal{F}\subseteq \mathcal{F}'$.

As an exercise, you can convince yourself that the smallest sigma algebra on $X=[0,1]$ containing $[0,\frac{1}{2}]$ is $\mathcal{F}=\{\emptyset,[0,1],[0,\frac{1}{2}],(\frac{1}{2},1]\}$.

kccu
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  • Thanks. If I add $[\frac{1}{5}, \frac{1}{4}]$ to $\mathcal{F}$ and call it $\mathcal{F_1}$, this would still be a $\sigma$-algebra? But $[\frac{1}{5}, \frac{1}{4}]$ wouldn't be a Borel set? – Learning... Dec 06 '16 at 05:08
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    In general if you add a set to a sigma algebra, it will no longer be a sigma algebra. For instance, you will need to add the set's complement, and all intersections of that set with other sets in the sigma algebra, and more. And $[\frac{1}{5},\frac{1}{4}]$ is a Borel set. It is the complement of $[0,\frac{1}{5})\cup ( \frac{1}{4},1]$ (an open set in $[0,1]$). – kccu Dec 06 '16 at 05:27
  • If $T$ is a topology on $\Omega,$ the set $B_T$ of Borel sets (with respect to $T$) is the smallest $\sigma$-algebra $S$ on $\Omega$ such that $S\supset T.$ So any subset of $\Omega$ that is open or closed with respect to $T$ belongs to $B_T.$ ... When $T$ is the standard topology on $\mathbb R,$ the Axiom of Choice implies that $B_T$ is a proper subset of the $\sigma$-algebra of Lebesgue-measurable sets, which is a proper subset of $P(\mathbb R)$ (the set of all real subsets), which is the largest $\sigma$-algebra on $\mathbb R.$ – DanielWainfleet Dec 06 '16 at 08:30
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Start with a set $X \subset \mathbb{R}$ that is not a Borel set, and consider the $\sigma$-algebra it generates. The result is (by definition) a $\sigma$-algebra, but because it contains a set that is not Borel it cannot be the smallest.

Reese Johnston
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  • I don't see why you need to choose $X$ not Borel. Any $X \neq \emptyset, \mathbb{R}$ will generate a $\sigma$-algebra that is nontrivial and thus not the smallest. – kccu Dec 06 '16 at 05:04
  • @kccu From context, I assumed OP was considering only algebras containing all open sets - it looks like in your answer you addressed that more explicitly. Kudos! – Reese Johnston Dec 06 '16 at 05:14