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Can someone check my proof? I'm trying to prove this statement:

Let $f : E \rightarrow F$, define $\sigma: \mathcal P(F) \to \mathcal P(F)$ through: $$\sigma(\varepsilon) = \bigcap_{B \supset \varepsilon} B \tag*{(where the Bs are $\sigma$-algebras)}$$ That is, $\sigma(\varepsilon)$ is the smallest $\sigma$-subalgebra of $\mathcal P(F)$ containing $\varepsilon$. Then: $$f^{-1}(\sigma(\varepsilon)) = \sigma(f^{-1}(\varepsilon))$$ That is $\sigma$ commutes with $f^{-1}$.

My attempt: $$f^{-1}(\sigma(\boldsymbol{\varepsilon})) = f^{-1}\left(\bigcap_{B \supset\boldsymbol{\varepsilon} } B \right)= \bigcap_{B \supset \boldsymbol{\varepsilon}} f^{-1}(B)=\sigma(f^{-1}(\boldsymbol{\varepsilon}))$$ Because $f^{-1}(B)$ are $\sigma$-algebras containing $f^{-1}(\boldsymbol{\varepsilon})$. Is this proof valid?

TC159
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HellBoy
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  • https://math.stackexchange.com/questions/7881/preimage-of-generated-sigma-algebra – Math1000 Aug 13 '23 at 15:42
  • @Math1000 i asked if what i wrote is enough or nonsense? – HellBoy Aug 13 '23 at 15:46
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    Does this answer your question? Prove the commutation of smallest sigma-algebra and inverse – Snoop Aug 13 '23 at 15:47
  • @Snoop im asking to proof check, not give me an alternative proof – HellBoy Aug 13 '23 at 15:52
  • @HellBoy you can look at the posted references for valid proofs – Snoop Aug 13 '23 at 15:59
  • @Snoop i saw them i dont want those, i want someone to check the validity of mine – HellBoy Aug 13 '23 at 16:00
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    It looks fishy. You are saying that the $\sigma$-algebra generated by $f^{-1}(\epsilon)$ (a subset of $E$) is found by interesecting $f^{-1}(B_i)$ when $B_i$ runs over $\sigma$-algebras in $F,.$ Never seen such a proof. Case closed. – Kurt G. Aug 13 '23 at 16:30
  • @KurtG. yes because $f^{-1}(B_i)$ are $\sigma$-algebras containing $f^{-1}(\boldsymbol{\epsilon})$ you can recheck the definitions of $\sigma(\boldsymbol{\varepsilon})$ – HellBoy Aug 13 '23 at 16:33
  • But they are not all such $\sigma$-algebras :) – Kurt G. Aug 13 '23 at 16:35
  • @KurtG. wym not all sigma algebras?, each $B_i$ is a sigma algebra containing $\boldsymbol{\varepsilon}$ and an earlier lemma said that for any sigma algebra B over F, then $f^{-1}(B)$ is a sigma algebra over E – HellBoy Aug 13 '23 at 16:36
  • To find $\sigma(f^{-1}(\epsilon))$ we have to intersect all $\sigma$-algebras in $E$ that contain $f^{-1}(\epsilon),.$ Over to you. If you show that any $\sigma$-algebra we need to intersect here is of the form $f^{-1}(B_i)$ I will give in. – Kurt G. Aug 13 '23 at 16:38
  • @KurtG. i literally just did, why do you fail to see that $f^{-1}(\boldsymbol{\varepsilon}) \subset f^{-1}(B_i)$ – HellBoy Aug 13 '23 at 16:40
  • I don't fail to see that. My point is that in $\bigcap_{\epsilon\subset B_i}$ the collection of $\sigma$-algebras over which you take that intersection is not what one expects. – Kurt G. Aug 13 '23 at 16:43
  • i fail to see why that doesnt give us whats expected @KurtG. – HellBoy Aug 13 '23 at 16:44
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    If you say so. As I said: case closed. – Kurt G. Aug 13 '23 at 16:45

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