Given $\Omega=\{1,2\}$, and $\mathscr E_1=\emptyset$,$\mathscr E_2=\{\emptyset\}$ Then i want to generate the smallest sigma field for each given case. The answer is:

$$ \mathscr F_1 = \{\emptyset,\emptyset^c=\Omega\} $$ $$ \mathscr F_2 = \{\emptyset,\Omega\} $$ That is, $\mathscr F_1 = \mathscr F_2$. However, I find this to be counterintuitive, as $\emptyset \ne \{\emptyset\}$, so shouldn't it be:

$$ \mathscr F_2^s = \{\emptyset,\Omega,\{\emptyset\},\{\Omega\}\} $$

Although I know this doesn't make sense either as $\mathscr F_2^s$ is not a subset of the power set of $\Omega$....

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    $\begingroup$ All $\sigma$-algebras contain $X$ and $\emptyset$. $\endgroup$ – copper.hat Mar 8 '15 at 6:38
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    $\begingroup$ You can have the same sigma field be generated by different subsets. $\endgroup$ – Alan Mar 8 '15 at 6:45
  • $\begingroup$ @copper.hat I don't see why your comment is useful here, I know that the sigma algebra contains the null set.. $\endgroup$ – dimebucker Mar 8 '15 at 8:06
  • $\begingroup$ Please define the sigma fields, at least as a pair. Conventional naming choices aren't definitions. Don't start a conversation about $n$ and $a, b$ expecting people to assume the first is a natural and the others are reals. You also didn't say what it means to "generate the smallest sigma field for each given case." Case of what? What is it to generate for something? The actual sets of subsets corresponding to each generating set are introduced without stating any properties that tie them to the rest of the question, not even stating that they're generated by other things. $\endgroup$ – Loki Clock Mar 8 '15 at 8:07
  • $\begingroup$ @LokiClock I think it's pretty clear that for example $\mathscr F_1 = \sigma (\mathscr E_1)$.. $\endgroup$ – dimebucker Mar 8 '15 at 8:09

The OP, as far as I can understand, insists that we ought to differentiate between the two $\sigma$-algebras. At the same time the OP admits that he couldn'n describe the right solution.

Here is my answer:

(1) By definition, a $\sigma$-algebra is a nonempty class. That is $$\sigma(\emptyset)=\sigma(\mathscr E_1)\not=\emptyset.$$ (2) By definition, the $\sigma$-algebra generated by a class is the intersection of all $\sigma$-algebras containing that class. As a result $$\sigma(\emptyset)=\sigma(\mathscr E_1)=\{\emptyset,\Omega\}.$$ and for the same reason$$\sigma(\{\emptyset\})=\sigma(\mathscr E_2)=\{\emptyset,\Omega\}.$$

This is so, because $$\emptyset \subset \text{of any set},$$ at the same time $$\{\emptyset\} \subset \text{of any }\sigma\text{ algebra}.$$


Here's a more intuitive version of the same problem: What is the smallest sigma-field $S$ over $\Omega$ such that $\{\Omega\} \subseteq S$?

  • $\begingroup$ Sorry for my previous answer. I answered by memorizing answers instead of thinking about what I was saying. $\endgroup$ – Loki Clock Mar 8 '15 at 8:57
  • $\begingroup$ So that would be $S=\{\Omega,\emptyset\}$ $\endgroup$ – dimebucker Mar 8 '15 at 9:50
  • $\begingroup$ @dimebucker91 Right. Does that ease your suspicions? $\endgroup$ – Loki Clock Mar 8 '15 at 9:51
  • $\begingroup$ So the smallest sigma algebra containing $\emptyset$ is the trivial sigma algebra $S=\{\emptyset,\Omega\}$, My question is why is the smallest sigma algebra that contains the set containing the empty set also $S=\{\emptyset,\Omega\}$, as $\emptyset \ne \{\emptyset\}$ $\endgroup$ – dimebucker Mar 8 '15 at 10:10
  • $\begingroup$ @dimebucker91 $\{\Omega\} \neq \{\emptyset\}$ also holds. The reason for $\sigma(\emptyset)=\sigma(\{\emptyset\})$ consists of any proof that $\sigma(\emptyset)=\{\emptyset, \Omega\}$ together with any proof that $\sigma(\{\emptyset\}) = \{\emptyset, \Omega\}$. $\endgroup$ – Loki Clock Mar 8 '15 at 22:37

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