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Simple question: Let $(E,\mathcal E)$ be a measurable space and $\Delta\not\in E$. Now let $E_\Delta:=E\cup\{\Delta\}$ and $\mathcal E_\Delta$ denote the smallest $\sigma$-algebra on $E_\Delta$ containing $\mathcal E$.

Question: Can we given an explicit formula for $\mathcal E_\Delta$?
We've clearly got $\{\Delta\}\in\mathcal E_\Delta$, since $E,E_\Delta\in\mathcal E_\Delta$.

BTW: Is there a better, established notation for the $\sigma$-algebra generated by $\mathcal E$ on $E_\Delta$? Writing $\sigma(\mathcal E)$ wouldn't make clear whether we mean the $\sigma$-algebra generated on $E$ or a larger space like $E_\Delta$.

amWhy
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0xbadf00d
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    What about $\mathcal E_{\Delta} = {B \cup {\Delta}:: B \in \mathcal E},$ or am I overlooking something obvious? Regarding $``$ Writing $\sigma(\mathcal E)$ wouldn't make clear whether we mean the $\sigma$-algebra generated on $E$ or a larger space like $E_{\Delta}",$ maybe ${\sigma}{E}(\mathcal E)$ and ${\sigma}{E_{\Delta}}(\mathcal E),$ or $\sigma(\mathcal E, , E)$ and $\sigma(\mathcal E,, E_{\Delta})$ (although if the latter, I'd probably put $\mathcal E$ as the second coordinate, since the ambient space is more primary than the collection of subsets under consideration). – Dave L. Renfro Mar 19 '22 at 18:52
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    Thank you for your answer. Your formula for $\mathcal E_\Delta$ must be wrong, since it only contains sets which are supersets of ${\Delta}$. – 0xbadf00d Mar 20 '22 at 05:33
  • You're correct, I lost track of the distinction between sets and collections of sets, and a correct formulation is given by @Paul Klass. Note that this is similar to ${\mathcal P}(X \cup{a});=;{\mathcal P}(X);\cup;{B\cup{a}:;B\in{\mathcal P}(X)},$ where $\mathcal P$ is the power set operation and $X$ is a set such that $a \notin X,$ which can be used to show that if a finite set $X$ has $n$ many subsets, then $X\cup{a}$ has $2n$ many subsets (e.g. see here and here). – Dave L. Renfro Mar 20 '22 at 18:32

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A notation that makes clear,that what you are lookin at is a $\sigma$-Algebra over $E_\Delta$ is: $\sigma(\mathcal{E} \cup\{ \{\Delta \} \})$. Because $\mathcal{E} \cup\{ \{\Delta \} \}$ is a family of subsets of $E_\Delta$ but not $E$. It's also discribes the problem you are lookin at. An explicit formula could be:

$\sigma(\mathcal{E} \cup \{\{\Delta \} \})= \mathcal{E} \cup \{B \cup \{\Delta \}: B \in \mathcal{E}\}$

Tho it remains to be shown that the right side is a sigma algebra and subset of the left side.

Paul Klass
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  • Thank you for your answer. I've got a subsequent question: https://math.stackexchange.com/q/4408385/47771. Would be great if you could take a look! – 0xbadf00d Mar 20 '22 at 09:20