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Let $f:X\to Y$ be a function, $\mathcal C$ be a family of subsets of $Y$. I am convinced that $f^{-1}(\sigma(\mathcal C))=\sigma(f^{-1}(\mathcal C))$, where $\sigma(\mathcal A)$ is the $\sigma$-algebra generated by $\mathcal A$.

The part $\sigma(f^{-1}(\mathcal C))\subseteq f^{-1}(\sigma(\mathcal C))$ is quite straight forward since $f^{-1}$ behave nicely with set-theoretic operations but I cannot see how to prove its converse. I tried to characterize $\sigma(\mathcal A)$ by seeing how it can be constructively define but my effort was not very fruitful. (From this post by Mr.Asaf, it seems that I must use higher cardinal.)

So how should I prove $f^{-1}(\sigma(\mathcal C))\subseteq\sigma(f^{-1}(\mathcal C))$ ? It is really hard or perhaps I missed something silly? Any hint would be appreciate. Thank you in advance.

Community
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BigbearZzz
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1 Answers1

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Define $\mathcal A=\{E\subseteq Y:\; f^{-1}(E)\in \sigma(f^{-1}(\mathcal C)\}.$ Then it is not hard to see that $\mathcal A$ is a $\sigma$-algebra. Try to do it yourself. Since $\mathcal A$ contains $\mathcal C$, it contains the smallest $\sigma$-algebra that contains $\mathcal C$. Therefore $\sigma(\mathcal C)\subseteq \mathcal E$ so that $f^{-1}(\sigma(\mathcal F))\subseteq \sigma(f^{-1}(\mathcal C)).$

Marko
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