I want to prove that a continuous function maps Borel sets to Borel sets. Royden's Real Analysis divided it into three questions:
- Show that a strictly increasing function that is defined on an interval has a continuous inverse.
- Let $f$ be a continuous function and $B$ be a Borel set. Show that $f^{-1}(B)$ is a Borel set. (Show that the collection of sets $E$ for which $f^{-1}(E)$ is Borel is a $\sigma$-algebra containing the open sets.)
- Show that a continuous strictly increasing function that is defined on an interval maps Borel sets to Borel sets
But I wonder whether $G=\{E|f(E)\text{ is Borel}\}$ itself form a $\sigma$-algebra? I have tried like this. Suppose $E_1,E_2,\cdots\in G$. It can be verified that $f(\bigcup_{i=1}^{\infty}E_i)=\bigcup_{i=1}^{\infty}f(E_i)$ is Borel. So $\bigcup_{i=1}^{\infty}E_i\in G$. Also, $\bigcap_{i=1}^{\infty}E_i\in G$ and $E^c\in G$. If $\mathcal{O}$ is open, then $f(\mathcal{O})$ is open.So $G$ is a $\sigma$-algebra containing all open sets?
If it is correct, things seem to be more simple. But I haven't find any wrong with it. If something goes wrong, is there a counterexample? Thanks!