Several textbooks don't include in the definition of a $\sigma$-algebra $\mathcal{F}$ on a set $\Omega$, that $\Omega$ must be non-empty. I wonder if that's a requirement. Because if $\Omega=\emptyset$, $\mathcal{F}$ can only be $\{\emptyset\}$. I know that $\{\emptyset, \Omega\}$ is the trivial $\sigma$-algebra on any non-empty $\Omega$, but if $\Omega$ itself is $\emptyset$, then the set containing the empty set itself would be a $\sigma$-algebra. Have never heard that (have you?), but it doesn't seem to contradict the definition.

Solution: As Willie pointed out in 2), I was missing the point that $\{\emptyset\}=\{\emptyset,\emptyset\}=\{\emptyset,\Omega\}$ so the $\sigma$-algebra generated is the trivial one again.

  • $\begingroup$ (1) $\emptyset \subseteq \emptyset$, so $\mathcal{F}$ can contain $\emptyset$ (this is not the same as saying that $\mathcal{F}$ is the empty set). (2) $\{a,a\} = \{a\}$. $\endgroup$ – Willie Wong Sep 12 '16 at 19:48
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    $\begingroup$ Hello and welcome. This is a well-formatted question. As to your question: If $\Omega = \emptyset$, then there is no integration theory for this set and the textbook would be much shorter. Maybe that' s the reason? $\endgroup$ – Hans Engler Sep 12 '16 at 19:53
  • $\begingroup$ Thanks, Willie. 1) okay, sure. 2) ah... so $\mathcal{F}=\{\emptyset\}=\{\emptyset, \Omega\}$ for $\Omega = \emptyset$ and thus $\mathcal{F}$ is actually the trivial $\sigma$-algebra again. So one does indeed not require $\Omega$ to be non-empty $\endgroup$ – Marius Hofert Sep 12 '16 at 19:53
  • $\begingroup$ Just a quick question. So, in that case, we have $\emptyset^c=\emptyset$? $\endgroup$ – sam wolfe Jan 24 at 20:58
  • $\begingroup$ I'd say so, yes, the complement of the empty set within the empty set (= $\Omega$) is again the empty set. $\endgroup$ – Marius Hofert Jan 24 at 22:32

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