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We define measurable function from a measure space to a topological space as space which pulls back open sets to measurable sets. How can we prove measurable functions pulls back Borel set also to measurable sets. where Borel sets are elements in sigma algebra generated by topology

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2 Answers 2

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Define

$$ \Sigma_f := \{M \in \mathcal{B} \mid f^{-1}(M) \in \mathcal{A} \}, $$

where $\mathcal{A}$ is the chosen $\sigma$-algebra on your measurable space.

Show that $\Sigma_f$ is a $\sigma$-algebra. Why does that help you?

EDIT: The same proof shows that it suffices to show $f^{-1}(M) \in \mathcal{A}$ for all $M \in \mathcal{M}$ for any family $\mathcal{M}$ of subsets to conclude that $f$ is measurable w.r.t $\mathcal{A}$ and the generated $\sigma$-algebra $\sigma(\mathcal{M})$.

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  • $\begingroup$ Ok if we are able to show this is sigma algebra. Then because open set belongs to it and Borel algebra is smallest algebra containing open set. So we are done. Am I right? $\endgroup$
    – Sushil
    Sep 11, 2014 at 14:49
  • $\begingroup$ Exactly. You just missed one "$\sigma$" in front of "algebra" :) $\endgroup$
    – PhoemueX
    Sep 11, 2014 at 15:05
  • $\begingroup$ Oh yes while typing I just forgot. $\endgroup$
    – Sushil
    Sep 11, 2014 at 15:14
  • $\begingroup$ Ok I got the complete answer. This is sigma algebra because of fact inverse images have good property under complement and union.(Sorry can't write it here) $\endgroup$
    – Sushil
    Sep 11, 2014 at 16:16
  • $\begingroup$ Yes, exactly. Normally, this is formulated as "taking inverse images commutes with all usual operations of set-theory", i.e. $f^{-1}(\bigcup_i M_i) = \bigcup_i f^{-1}(M_i)$ and $f^{-1}(M^c) = (f^{-1}(M))^c$. The same also holds for intersection. $\endgroup$
    – PhoemueX
    Sep 11, 2014 at 16:27
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Let $f:X\rightarrow Y$ be a function and let $\mathcal{O}\subseteq\wp\left(Y\right)$. In your case $\mathcal O$ is the topology.

It comes to proving that $\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)=f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$ where in general $\sigma\left(\mathcal{V}\right)$ denotes the $\sigma$-algebra generated by collection $\mathcal{V}$.

  • If $\mathcal{A}$ is a $\sigma$-algebra on $X$ then $\left\{ A\in\wp\left(Y\right)\mid f^{-1}\left(A\right)\in\mathcal{A}\right\} $ is a $\sigma$-algebra on $Y$.

Filling in $\mathcal{A}=\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)$ we find this collection to be a $\sigma$-algebra containing $\mathcal{O}$. Then it also contains $\sigma\left(\mathcal{O}\right)$. This is exactly the statement that $f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)\subseteq\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)$.

  • If $\mathcal{B}$ is a $\sigma$-algebra on $Y$ then $f^{-1}\left(\mathcal{B}\right)$ is a $\sigma$-algebra on $X$.

Filling in $\mathcal{B}=\sigma\left(\mathcal{O}\right)$ we find $f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$ to be a $\sigma$-algebra with $f^{-1}\left(\mathcal{O}\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$ and consequently $\sigma\left(f^{-1}\left(\mathcal{O}\right)\right)\subseteq f^{-1}\left(\sigma\left(\mathcal{O}\right)\right)$.

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