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I'm having some trouble trying to understand the validity of the following lemma:

Lemma: Let $(S,\mathcal{A})$ and $(T,\mathcal{B})$ be two measurable spaces and let $f:S\to T$. Suppose that $\mathcal{F}\subseteq \mathcal{B}$ is such that $\sigma(\mathcal{F})=\mathcal{B}$. If $f^{-1}(F)\in\mathcal{A}$ for all $F\in\mathcal{F}$, then $f$ is measurable.

The proof revolves around proving that $\tilde{\mathcal{B}} = \mathcal{B}$, with $\tilde{\mathcal{B}} = \{B\in\mathcal{B}:f^{-1}(B)\in\mathcal{A}\}.$ I just don't understand why $f^{-1}(F)\in\mathcal{A}$ for all $F\in\mathcal{F}$, means that $f$ is measurable (considering that $\mathcal{B} = \tilde{\mathcal{B}}$). I guess that it somehow shows that we have $f^{-1}(B)\in\mathcal{A}$ for all $B\in\mathcal{B}$, but I don't see it.

Question: Why can this lemma be proved by showing that $\mathcal{B} = \{B\in\mathcal{B}:f^{-1}(B)\in\mathcal{A}\}$?

Thanks!

titusAdam
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2 Answers2

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If $f^{-1}(F)\in\mathcal A$ for all $F\in\mathcal F$ then we can denote that with: $$f^{-1}(\mathcal F)\subseteq\mathcal A\tag0$$

Combining this with the fact that $\mathcal A$ is a $\sigma$-algebra it can be concluded that also: $$\sigma(f^{-1}(\mathcal F))\subseteq\mathcal A\tag1$$

Fortunately it can be proved that: $$\sigma(f^{-1}(\mathcal F))=f^{-1}(\sigma(\mathcal F))\tag2$$

For a proof of that see this answer.

Then combining $(1)$ and $(2)$ with $\mathcal B=\sigma(\mathcal F)$ we find:$$f^{-1}(\mathcal B)\subseteq\mathcal A$$which is exactly the statement that $f$ is measurable.

drhab
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By definition of measurability, $f$ is measurable if $f^{-1}(B) \in \mathcal{A}$ for all $B \in \cal{B}$. On the other hand, by definition of $\widetilde{B}$, any $B \in \widetilde{\mathcal{B}}$ satisfies already $f^{-1}(B) \in \cal{A}$. Thus, if $\widetilde{\mathcal{B}}= \mathcal{B}$, then $f^{-1}(B) \in \cal{A}$ for all $\mathcal{B}$.

p4sch
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