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suppose $X:\Omega\rightarrow\Omega'$ where $(\Omega,\beta)$ and $(\Omega',\beta')$ are two measurable space. suppose $C'$ generate $\beta'$;how can show $X$ is measurable iff $X^{-1}(C')\subset\beta$

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    Only if should be obvious, for the "if" part, let $\gamma = \{A \in \beta'\mid X^{-1}[A] \in \beta\}$ and show that $\gamma$ is a $\sigma$-algebra containing $C'$ (and your result follows).2012-08-20

2 Answers 2

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Let $C'$ be a system of subsets of $\Omega'$ such that $\sigma(C')=\beta'$. Suppose that $X$ is $(\beta,\beta')$-measureable, i.e. $X^{-1}(A)\in\beta$ for all $A\in\beta'$. Then we obviously have that $X^{-1}(A)\in\beta$ for all $A\in C'$ since $C'\subseteq \beta'$.

Now suppose that $X^{-1}(A)\in \beta$ for all $A\in C'$. If we define $$ \Lambda=\{A\subseteq \Omega' \mid X^{-1}(A)\in\beta\}, $$ then obviously $C'\subseteq \Lambda$. Now show that $\Lambda$ is a $\sigma$-field (by using the properties of pre-images) because then $\beta'=\sigma(C')\subseteq \Lambda$ and hence $X$ is $(\beta,\beta')$-measureable.

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One implication is easy.

The other follows from the following fact :

If $f : X \to Y$ is a map, $\mathcal{E}$ is a subset of $\mathcal{P}(Y)$, then $f^{-1}(\sigma(\mathcal{E})) = \sigma(f^{-1}(\mathcal{E}))$.

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    Is it trivial to show that function inverse commutes with the $\sigma$ operation? One inclusion is obvious...2012-08-20
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    It's not completely trivial, it uses the same trick given in others answers below. If you set $\mathcal{B} = \{B \subset Y \, / \, f^{-1}(B) \in \sigma(f^{-1}(\mathcal{E})) \}$, then you can prove that it's a $\sigma$-algebra containing $\mathcal{E}$. It follows that $\sigma(\mathcal{E}) \subset \mathcal{B}$, and the definition of $\mathcal{B}$ gives the desired inclusion.2012-08-20
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    Thanks, I was afraid I would have to lose myself in ordinals!2012-08-20