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Let $X$ be a set, let $\mathscr{F} \subset \mathscr{P}(X)$, and let $\Omega$ be the collection of all $\sigma$-algebras on $X$ which contain $\mathscr{F}$. Then $$\sigma(\mathscr{F}) = \bigcap_{\mathfrak{M} \in \Omega} \mathfrak{M}$$ is the smallest $\sigma$-algebra on $X$ which contains $\mathscr{F}$.

Question: When proving the statement above, Rudin points out that $\Omega$ is nonempty. Why is this relevant?

Randy Randerson
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    If $\Omega$ is nonempty, then what is the meaning of the intersection? –  Nov 13 '15 at 19:09
  • $\mathscr{P}(X)$ http://math.stackexchange.com/questions/370188/empty-intersection-and-empty-union – Randy Randerson Nov 13 '15 at 19:09
  • I don't have Rudin at hand and I've forgotten: how does he define the $\sigma$-algebra generated by $\mathscr{F}\subseteq \mathscr{P}(X)$? – BrianO Nov 13 '15 at 19:13
  • As the smallest $\sigma$-algebra on $X$ containing $\mathscr{F}$, constructed as above. – Randy Randerson Nov 13 '15 at 19:13
  • It's worth mentioning only because the intersection of the empty collection of subsets of $X$ is, as you note, $\mathscr{P}(X)$. However, the intersection of the empty collection without regard to any containing superset is the whole universe of set theory, which isn't a set. Maybe/probably that's what he's guarding against. In any case, $\Omega$ is not empty, though, because $\mathscr{P}(X)$ is a member of it. It doesn't get you much, e.g. if $\mathscr{F} = \mathscr{P}(X)$ then it's the only member. – BrianO Nov 13 '15 at 19:21
  • @BrianO So the proof remains correct without mentioning it? – Randy Randerson Nov 13 '15 at 19:25
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    It depends on this subtle & almost trivial point about how $\bigcap\emptyset$ is to be construed. The $\bigcap$ operator isn't "decorated" with $X$, so I believe he points out that $\Omega$ is nonempty because otherwise the intersection is everything in the universe. Check his definition of $\bigcap$ in the early pages. He's very careful and economical, so he would only mention it for a reason. – BrianO Nov 13 '15 at 19:29
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    @BrianO Thanks, that clarifies it. – Randy Randerson Nov 13 '15 at 19:31
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    Last word: the issue is discussed here. I infer that Rudin does not adhere to the convention which that wikipedia article section settles on. – BrianO Nov 13 '15 at 19:45

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