Here are my thoughts. We want to show that $f^{-1}$ maps Borel sets to Borel sets.
Let $f:\mathbb R\to\mathbb R$ be a continuous function and $\mathcal B(\mathbb R)$ a Borel $\sigma$-algebra. Let $B\in\mathcal B(\mathbb R)$ and $B$ open. Then $f^{-1}(B)$ is open as well, that is, there exists neighborhoods around every point of $f^{-1}(B)$ contained within. Since these neighborhoods are members of $\mathcal B(\mathbb R)$, their union are also members of $\mathcal B(\mathbb R)$. Hence $f^{-1}(B)$ is a member of $\mathcal B(\mathbb R)$.
Here's where I'm stuck. I don't know how to show countably many such members can have their unions taken to give me $f^{-1}(B)$. I also don't know what to do with the closed sets in $\mathcal B(\mathbb R)$, especially singletons and those only generatable via complements.
Hints?
Best Answer
You need to establish the following lemma:
Proof: The only if part is trivial. For the other direction, observe that the $\sigma$-algebra generated by the set $\{S \subseteq Y\mid f^{-1}[S] \in \Sigma\}$ contains $\mathcal{E}$, so it also contains $\mathrm{T}$.
By substituting $\Sigma,\mathrm{T}$ with the Borel sets, and $\mathcal{E}$ with the open sets (a.k.a. topology) on $\mathbb{R}$, the proof is complete.