Skip to main content
10 events
when toggle format what by license comment
Nov 15, 2023 at 12:45 comment added Joel David Hamkins That $\sigma$-algebra is not countably generated (a fun exercise), and so by Gerald's observation this is not generated by any measurable function via the Borel sets, which are countably generated.
Nov 14, 2023 at 18:55 history closed Jochen Wengenroth
Gerald Edgar
Christian Remling
Max Horn
Tobias Fritz
Not suitable for this site
Nov 14, 2023 at 16:49 comment added MatEZ I am asking because I read, that $\mathcal{C}:=\{ C\subset \mathbb{R}: C\text{ Is countable or } C^c \text{ is countable}\}$ Is not generated by any function $f:\mathbb{R}\rightarrow(\mathbb{R},\mathcal{B}(\mathbb{R}))$ And I didn't believe it. Thanks for clarification
Nov 14, 2023 at 16:48 review Close votes
Nov 14, 2023 at 18:55
Nov 14, 2023 at 16:43 comment added Gerald Edgar This is the wrong forum for the question..... Answer: YES, in the case: real-valued measurable function $f$ and countably-generated $\sigma$-algebra $\mathcal A$. But otherwise, perhaps not. So, if $\mathcal B$ is countably generated, then so is $\sigma(f)$.
Nov 14, 2023 at 16:29 comment added MatEZ And if I consider $|\mathcal{A}|\leq|\mathcal{B}|$?
Nov 14, 2023 at 16:28 comment added Jochen Wengenroth Of course not. For $X=\{0,1\}$ and $\mathcal B$ the power set, every $\sigma$-algebra of $\sigma(f)$ has at most four elements.
Nov 14, 2023 at 16:24 comment added MatEZ Given measurable spaces $(X,\mathcal{B})$ and $(\Omega,\mathcal{A})$, does there exists a function $f:\Omega\rightarrow(X,\mathcal{B})$ such that $f$ generates $\mathcal{A}$?
Nov 14, 2023 at 16:20 comment added Joel David Hamkins Can you clarify the quantifiers? You are asking whether there is such a measurable space $(X,\mathcal{B})$, which will generate every $(\Omega,\mathcal{A})$ via some measurable $f$? (Also, what is a "general measurable space"?)
Nov 14, 2023 at 16:13 history asked MatEZ CC BY-SA 4.0