Let $f:\mathbb{R}\to [-\infty,\infty]$ continuous function.

(a) Let $\Omega=\left\{E:f^{-1}(E) \text{ is a Borel set } \right\}$. Show that $\Omega$ is $\sigma$-algebra. I already proved this. :)

(b) Let $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set.

For (b) I have this. $f$ continuous then $f$ is measurable, because $B$ is a Borel set, then $f^{-1}(B)$ is a Borel set. It is correct?...

  • $\begingroup$ No, your logic is assailable. Since $f$ is continuous, $\Omega$ contains the open sets, and since it is a $\sigma$-field, it contains all the Borel sets. $\endgroup$ – copper.hat Dec 4 '18 at 0:00

It appears that you are assuming what you are supposed to prove. The purpose of this exercise is to prove that a continuous function is Borel measurable. The correct argument is to use a): the sigma algebra in a) contains all open sets because $f^{-1}(U)$ is open for any open set $U$; if a sigma algebra contains all open sets it contains all Borel sets because Borel sigma algebra is the smallest sigma algebra containing all open sets.


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