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I know this is simple, but I just cannot get my head around. Can anyone explain the following?

if $\mathcal{C} \subset\mathcal{B}$ and $\sigma(\mathcal{C})=\mathcal{B}$, then $h^{-1}:\mathcal{C}\rightarrow\Sigma$, => h is $\Sigma$ measurable.

The proof in the book:

Let $\mathcal{E}$ be the class of elements B in $\mathcal{B}$ such that $h^{-1}(B)\in \Sigma$. Then $\mathcal{E}$ is a Sigma algebra. Then $\mathcal{C} \subset \mathcal{E}$.

So how does this prove it?

user1559897
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1 Answers1

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Theorem: Let $(X,\mathcal{M})$ and $(Y,\mathcal{N})$ be given. Assume $\mathcal{N}$ is generated by $\mathcal{E}$ (i.e. $\mathcal{N}=\sigma(\mathcal{E})$). If $f\colon X\rightarrow Y$ satisfies $$ E\text{ is in }\mathcal{E}\implies f^{-1}(E)\text{ is in }\mathcal{M}, $$ $f$ is measurable.

Proof: Define the pullback $$ f_{*}(\mathcal{M})=\left\{ E\subset Y\colon f^{-1}(E)\text{ is in }\mathcal{M}\right\} . $$ The above is a $\sigma$-algebra (convince yourself of this). By assumption, if $E$ is in $\mathcal{E}$, then $E$ is in $f_{*}(\mathcal{M})$. Therefore, $\mathcal{E}\subset f_{*}(\mathcal{M})$. Since $\mathcal{E}$ generates $\mathcal{N}$, $\mathcal{E}\subset\mathcal{N}\subset f_{*}(\mathcal{M})$. This implies that all sets in $\mathcal{N}$ are measurable.

parsiad
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