I think this statement should be intuitively true, but I can't prove it myself or find the proof elsewhere. Could you help me, please?

Consider a general measurable space $(X,\mathcal{B})$ and any other measurable space $(\Omega,\mathcal{A})$. Does there exist (for every $\sigma$-algebra $\mathcal{A}$) a function (measurable) $f:\Omega\rightarrow(X,\mathcal{B})$ such that $\mathcal{A}=\sigma(f)$? In other words, is every $\sigma$-algebra generated by some measurable function?

  • $\begingroup$ 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"?) $\endgroup$ Nov 14 at 16:20
  • $\begingroup$ 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 $\mathcal{A}$? $\endgroup$
    – MatEZ
    Nov 14 at 16:24
  • 4
    $\begingroup$ Of course not. For $X=\{0,1\}$ and $\mathcal B$ the power set, every $\sigma$-algebra of $\sigma(f)$ has at most four elements. $\endgroup$ Nov 14 at 16:28
  • $\begingroup$ And if I consider $|\mathcal{A}|\leq|\mathcal{B}|$? $\endgroup$
    – MatEZ
    Nov 14 at 16:29
  • 1
    $\begingroup$ 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 $\sigma(f)$. $\endgroup$ Nov 14 at 16:43


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