# Is every $\sigma$-algebra generated by some measurable function? [closed]

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?

• 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 at 16:20
• 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 at 16:24
• 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 at 16:28
• And if I consider $|\mathcal{A}|\leq|\mathcal{B}|$? Nov 14 at 16:29
• 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 at 16:43