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Question Source: Extension of Example 1b in Section 6 from "Measure Theory" by Halmos

For a fixed subset $A\subset X$, $E$ is the class of all sets of which $A$ is a subset, i.e. $E=\{F:A\subset F\}$. What is the $\sigma$-algebra generated by the class $E$ of sets here described?

My Question: I was wondering if there's a neat characterization or what Halmos intended for us to do/see here.

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    Neither of the "extremes" makes sense. Your ${\Omega,\varnothing}$ does not contain $E$ as a subset, so it cannot possibly be the "$\sigma$-algebra generated by $E$." The collection $\mathcal{P}(X)$ is too large, and it makes no sense to simply declare to be "the $\sigma$-algebra generated by $E$". Do you understand what "the $\sigma$-algebra generated by a collection $E$" is? – Arturo Magidin May 24 '21 at 19:47
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    Perhaps this answer will help clarify what is going on, without telling you what the answer is. – Arturo Magidin May 24 '21 at 19:48
  • I meant ${X,\emptyset}$ which would be the $\sigma$-algebra generated by $E={X}$. For the latter, I was just wrong. –  May 24 '21 at 20:05
  • Note that "$A\in X$" should be $A\subset X$. It's a subset, not an element. – Arturo Magidin May 24 '21 at 20:12
  • Actually, for the second case, I am still not sure why it doesn't make sense. If $A=\emptyset$, then isn't $E=\mathcal{P}(X)$ and so the intersection of all such algebras containing the power set is the power set itself? –  May 24 '21 at 20:12
  • It was a bit discombobulated; perhaps my comment was not accurate, as I took it to mean "the extremes of what the $\sigma$-algebra generated could be." – Arturo Magidin May 24 '21 at 20:13

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Let $\mathcal G$ be the $\sigma$-algebra generated by $E$. It should necessarily contains the complements of the sets of $E$, that is, the sets contained in $X\setminus A$. If $S\subset X\setminus A$, then $S\in\mathcal G$ and $S\cup A\in\mathcal G$. Therefore, we can try $$ \mathcal G=\left\{S\subset X, S\subset X\setminus A\right\}\cup \left\{A\cup S, S\subset X\setminus A\right\} $$ and check that such a $\mathcal G$ is a $\sigma$-algebra containing $E$, and it is the smallest for the inclusion doing this job.

Davide Giraudo
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