# Showing $B$ Borel set, then $f^{-1}(B)$ is a Borel set with $f$ continuous function.

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?...

• 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. – 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.